InternBootcamp/examples/data/InternBootcamp_eval/cipher_merged.jsonl

{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string without punctuation.\n- Preparation:\n    - Multitap Code Table\n        | Letter | Multitap Code |\n        | --- | --- |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Multitap Code Table:\n            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.\n\n \noriginal information: pyritic encrypt to encoded text: 7^19^37^34^38^14^32^3\noriginal information: nervii encrypt to encoded text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 1, "source_filename": "icl_with_rule_encode_Kor_rule3_MultiTapPhoneCode_en.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "NERVII", "extra_args": {}, "output": "原始文本: nervii\n处理后的文本(仅大写字母): NERVII\n开始逐字符加密:\n字符 N 对应的多击编码是: 6^2\n字符 E 对应的多击编码是: 3^2\n字符 R 对应的多击编码是: 7^3\n字符 V 对应的多击编码是: 8^3\n字符 I 对应的多击编码是: 4^3\n字符 I 对应的多击编码是: 4^3\n最终加密结果: 6^23^27^38^34^34^3\n", "ground_truth": "6^23^27^38^34^34^3"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string without punctuation.\n- Preparation:\n    - Multitap Code Table\n        | Letter | Multitap Code |\n        | --- | --- |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Multitap Code Table:\n            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.\n\n \nplain text: pyritic encrypt into encrypted text: 7^19^37^34^38^14^32^3\nplain text: nonculpability encrypt into encrypted text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 2, "source_filename": "icl_with_rule_encode_Kor_rule3_MultiTapPhoneCode_en.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "NONCULPABILITY", "extra_args": {}, "output": "原始文本: nonculpability\n处理后的文本(仅大写字母): NONCULPABILITY\n开始逐字符加密:\n字符 N 对应的多击编码是: 6^2\n字符 O 对应的多击编码是: 6^3\n字符 N 对应的多击编码是: 6^2\n字符 C 对应的多击编码是: 2^3\n字符 U 对应的多击编码是: 8^2\n字符 L 对应的多击编码是: 5^3\n字符 P 对应的多击编码是: 7^1\n字符 A 对应的多击编码是: 2^1\n字符 B 对应的多击编码是: 2^2\n字符 I 对应的多击编码是: 4^3\n字符 L 对应的多击编码是: 5^3\n字符 I 对应的多击编码是: 4^3\n字符 T 对应的多击编码是: 8^1\n字符 Y 对应的多击编码是: 9^3\n最终加密结果: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3\n", "ground_truth": "6^26^36^22^38^25^37^12^12^24^35^34^38^19^3"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string without punctuation.\n- Preparation:\n    - Multitap Code Table\n        | Letter | Multitap Code |\n        | --- | --- |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Multitap Code Table:\n            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.\n\n \nplain text: nervii encode to encrypted text: 6^23^27^38^34^34^3\nplain text: nervii encode to encrypted text: 6^23^27^38^34^34^3\nplain text: nonculpability encode to encrypted text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3\nplain text: pyritic encode to encrypted text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 3, "source_filename": "icl_with_rule_encode_Kor_rule3_MultiTapPhoneCode_en.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "PYRITIC", "extra_args": {}, "output": "原始文本: pyritic\n处理后的文本(仅大写字母): PYRITIC\n开始逐字符加密:\n字符 P 对应的多击编码是: 7^1\n字符 Y 对应的多击编码是: 9^3\n字符 R 对应的多击编码是: 7^3\n字符 I 对应的多击编码是: 4^3\n字符 T 对应的多击编码是: 8^1\n字符 I 对应的多击编码是: 4^3\n字符 C 对应的多击编码是: 2^3\n最终加密结果: 7^19^37^34^38^14^32^3\n", "ground_truth": "7^19^37^34^38^14^32^3"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Encryption Steps:\n    - Fill in the first plaintext in the first position of the first line.\n    - Two ways to fill:\n        - Downward Fill: Fill the first row with a \"#\" (removing the first column, since the first position is already filled with a plaintext letter), then fill the plaintext downwards from the second row to the last row (fifth row).\n        - Upwards: the plaintext is filled upwards from the last line (fifth line) to the second line, and then the first line is filled with a \"#\".\n    - For each letter in the plaintext (except the first letter which has been filled in the first position), perform downward padding first, fill one column, then turn to upward padding, then turn to downward padding, and so on alternately, until all the letters have been filled.\n        - For example, HELLOWORLD.The content of the five lines after filling is\n        H L #\n        E R D\n        L O\n        L W\n        O #\n    - Filled out, read line by line, first read the contents of the first line, after reading the contents of each line are added a * sign, marking the end of the line; and then read the contents of the second line, and so on, read all the lines, the formation of the final password text.\n        H L #\n        E R D\n        L O\n        L W\n        O #\n        - Read the final cipher text line by line as \"HL#*ERD*LO*LW*O#*\" (i.e., there must be five *'s in the final cipher text to indicate that there are five lines)\n\n \nplain text: pyritic encrypt to cipher text: P*Y*RC*II*T#*\nplain text: nonculpability encrypt to cipher text: NB#*OAI*NPL*CLIY*U#T#*\nplain text: pyritic encrypt to cipher text: P*Y*RC*II*T#*\nplain text: nervii encrypt to cipher text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 4, "source_filename": "icl_with_rule_encode_Kor_rule16_RedefenceFigureCipher_en.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "NERVII", "extra_args": {}, "output": "加密步骤开始:\n1. 处理输入文本为大写字母: NERVII\n2. 准备5行矩阵用于填充\n3. 在第一行第一个位置填入第一个字母: N\n4. 开始交替向下和向上填充字母:\n  向下填充:\n    在第2行填入: E\n    在第3行填入: R\n    在第4行填入: V\n    在第5行填入: I\n  向上填充:\n    在第5行填入#\n    在第4行填入最后一个字母: I\n\n5. 最终矩阵:\n  N _ _ _ _ _\n  E _ _ _ _ _\n  R _ _ _ _ _\n  V I _ _ _ _\n  I # _ _ _ _\n\n6. 按行读取并添加*号分隔符\n最终密文: N*E*R*VI*I#*\n", "ground_truth": "N*E*R*VI*I#*"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Encryption Steps:\n    - Fill in the first plaintext in the first position of the first line.\n    - Two ways to fill:\n        - Downward Fill: Fill the first row with a \"#\" (removing the first column, since the first position is already filled with a plaintext letter), then fill the plaintext downwards from the second row to the last row (fifth row).\n        - Upwards: the plaintext is filled upwards from the last line (fifth line) to the second line, and then the first line is filled with a \"#\".\n    - For each letter in the plaintext (except the first letter which has been filled in the first position), perform downward padding first, fill one column, then turn to upward padding, then turn to downward padding, and so on alternately, until all the letters have been filled.\n        - For example, HELLOWORLD.The content of the five lines after filling is\n        H L #\n        E R D\n        L O\n        L W\n        O #\n    - Filled out, read line by line, first read the contents of the first line, after reading the contents of each line are added a * sign, marking the end of the line; and then read the contents of the second line, and so on, read all the lines, the formation of the final password text.\n        H L #\n        E R D\n        L O\n        L W\n        O #\n        - Read the final cipher text line by line as \"HL#*ERD*LO*LW*O#*\" (i.e., there must be five *'s in the final cipher text to indicate that there are five lines)\n\n \nclear text: nervii encrypt into encrypted text: N*E*R*VI*I#*\nclear text: pyritic encrypt into encrypted text: P*Y*RC*II*T#*\nclear text: nonculpability encrypt into encrypted text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 5, "source_filename": "icl_with_rule_encode_Kor_rule16_RedefenceFigureCipher_en.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "加密步骤开始:\n1. 处理输入文本为大写字母: NONCULPABILITY\n2. 准备5行矩阵用于填充\n3. 在第一行第一个位置填入第一个字母: N\n4. 开始交替向下和向上填充字母:\n  向下填充:\n    在第2行填入: O\n    在第3行填入: N\n    在第4行填入: C\n    在第5行填入: U\n  向上填充:\n    在第5行填入#\n    在第4行填入: L\n    在第3行填入: P\n    在第2行填入: A\n    在第1行填入: B\n  向下填充:\n    在第1行填入#\n    在第2行填入: I\n    在第3行填入: L\n    在第4行填入: I\n    在第5行填入: T\n  向上填充:\n    在第5行填入#\n    在第4行填入最后一个字母: Y\n\n5. 最终矩阵:\n  N B # _ _ _ _ _ _ _ _ _ _ _\n  O A I _ _ _ _ _ _ _ _ _ _ _\n  N P L _ _ _ _ _ _ _ _ _ _ _\n  C L I Y _ _ _ _ _ _ _ _ _ _\n  U # T # _ _ _ _ _ _ _ _ _ _\n\n6. 按行读取并添加*号分隔符\n最终密文: NB#*OAI*NPL*CLIY*U#T#*\n", "ground_truth": "NB#*OAI*NPL*CLIY*U#T#*"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Encryption Steps:\n    - Fill in the first plaintext in the first position of the first line.\n    - Two ways to fill:\n        - Downward Fill: Fill the first row with a \"#\" (removing the first column, since the first position is already filled with a plaintext letter), then fill the plaintext downwards from the second row to the last row (fifth row).\n        - Upwards: the plaintext is filled upwards from the last line (fifth line) to the second line, and then the first line is filled with a \"#\".\n    - For each letter in the plaintext (except the first letter which has been filled in the first position), perform downward padding first, fill one column, then turn to upward padding, then turn to downward padding, and so on alternately, until all the letters have been filled.\n        - For example, HELLOWORLD.The content of the five lines after filling is\n        H L #\n        E R D\n        L O\n        L W\n        O #\n    - Filled out, read line by line, first read the contents of the first line, after reading the contents of each line are added a * sign, marking the end of the line; and then read the contents of the second line, and so on, read all the lines, the formation of the final password text.\n        H L #\n        E R D\n        L O\n        L W\n        O #\n        - Read the final cipher text line by line as \"HL#*ERD*LO*LW*O#*\" (i.e., there must be five *'s in the final cipher text to indicate that there are five lines)\n\n \nclear text: nervii encode into encrypted text: N*E*R*VI*I#*\nclear text: pyritic encode into encrypted text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 6, "source_filename": "icl_with_rule_encode_Kor_rule16_RedefenceFigureCipher_en.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "PYRITIC", "extra_args": {}, "output": "加密步骤开始:\n1. 处理输入文本为大写字母: PYRITIC\n2. 准备5行矩阵用于填充\n3. 在第一行第一个位置填入第一个字母: P\n4. 开始交替向下和向上填充字母:\n  向下填充:\n    在第2行填入: Y\n    在第3行填入: R\n    在第4行填入: I\n    在第5行填入: T\n  向上填充:\n    在第5行填入#\n    在第4行填入: I\n    在第3行填入最后一个字母: C\n\n5. 最终矩阵:\n  P _ _ _ _ _ _\n  Y _ _ _ _ _ _\n  R C _ _ _ _ _\n  I I _ _ _ _ _\n  T # _ _ _ _ _\n\n6. 按行读取并添加*号分隔符\n最终密文: P*Y*RC*II*T#*\n", "ground_truth": "P*Y*RC*II*T#*"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(所有行和列号从0开始计数):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - 位于所有行第一个字母的MKGWB是行首字母\n        - 位于所有列最后一个字母的PQUCL是列尾字母\n- 加密步骤:\n    - 移除非加密信息中的空格、标点和字母J，并将所有字母转换为大写\n    - 对非加密信息中的每个字母p:\n        - 在网格中找到字母p的位置，然后找到相应的行首和列尾字符\n        - 将行首和列尾字符连接成二元组作为该字母p的加密消息\n        - 例如，如果字母p是H，它在第4行，行首字符是B；它在第3列，列尾字符是H，所以加密消息是BH\n    \n    连接所有加密消息作为最终加密文本输出\n解密算法:\n解密规则:\n- 输入:\n    - 加密文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 每次从加密文本中取两个字母c1,c2\n        - 字母c1标识解密字母p在网格中的行位置，找到c1作为行首的行\n        - 字母c2标识解密字母p在网格中的列位置，找到c2在列底的列\n        - 在网格中找到这个行列位置的字母，即为c1,c2的解密消息p\n        - 例如，如果c1,c2=BH，B是行首的行是第4行，H是行底的行是第2列，(4,2)处的字母是H，所以BH解密为H\n    - 两个两个地解密加密文本中的字母，并将解密消息连接起来作为最终非加密信息输出\n加密文本: KTGEKTWIGIKHMIGTBBBIKHBIBTWE 解码为非加密信息: nonculpability\n加密文本: MIWEWTBIBTBIWI 解码为非加密信息: pyritic\n加密文本: MIWEWTBIBTBIWI 解码为非加密信息: pyritic\n加密文本: KTBEWTWHBIBI 解码为非加密信息是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 7, "source_filename": "icl_with_rule_decode_Kor_rule15_CollonCipher_cn.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "KTBEWTWHBIBI", "extra_args": {}, "output": "开始解密过程...\n加密文本: KTBEWTWHBIBI\n\n每次解密两个字符:\n\n处理字符对: KT\n行首字符: K, 列尾字符: T\n-> 在第2行第2列找到原文字符: N\n\n处理字符对: BE\n行首字符: B, 列尾字符: E\n-> 在第5行第3列找到原文字符: E\n\n处理字符对: WT\n行首字符: W, 列尾字符: T\n-> 在第4行第2列找到原文字符: R\n\n处理字符对: WH\n行首字符: W, 列尾字符: H\n-> 在第4行第4列找到原文字符: V\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 隐文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(所有行和列号从0开始计数):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - 位于所有行第一个字母的MKGWB是行首字母\n        - 位于所有列最后一个字母的PQUCL是列尾字母\n- 加密步骤:\n    - 移除非加密信息中的空格、标点和字母J，并将所有字母转换为大写\n    - 对非加密信息中的每个字母p:\n        - 在网格中找到字母p的位置，然后找到相应的行首和列尾字符\n        - 将行首和列尾字符连接成二元组作为该字母p的加密消息\n        - 例如，如果字母p是H，它在第4行，行首字符是B；它在第3列，列尾字符是H，所以加密消息是BH\n    \n    连接所有加密消息作为最终隐文输出\n解密算法:\n解密规则:\n- 输入:\n    - 隐文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 每次从隐文中取两个字母c1,c2\n        - 字母c1标识解密字母p在网格中的行位置，找到c1作为行首的行\n        - 字母c2标识解密字母p在网格中的列位置，找到c2在列底的列\n        - 在网格中找到这个行列位置的字母，即为c1,c2的解密消息p\n        - 例如，如果c1,c2=BH，B是行首的行是第4行，H是行底的行是第2列，(4,2)处的字母是H，所以BH解密为H\n    - 两个两个地解密隐文中的字母，并将解密消息连接起来作为最终非加密信息输出\n隐文: KTBEWTWHBIBI 解码为非加密信息: nervii\n隐文: KTGEKTWIGIKHMIGTBBBIKHBIBTWE 解码为非加密信息是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 8, "source_filename": "icl_with_rule_decode_Kor_rule15_CollonCipher_cn.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "KTGEKTWIGIKHMIGTBBBIKHBIBTWE", "extra_args": {}, "output": "开始解密过程...\n加密文本: KTGEKTWIGIKHMIGTBBBIKHBIBTWE\n\n每次解密两个字符:\n\n处理字符对: KT\n行首字符: K, 列尾字符: T\n-> 在第2行第2列找到原文字符: N\n\n处理字符对: GE\n行首字符: G, 列尾字符: E\n-> 在第3行第3列找到原文字符: O\n\n处理字符对: KT\n行首字符: K, 列尾字符: T\n-> 在第2行第2列找到原文字符: N\n\n处理字符对: WI\n行首字符: W, 列尾字符: I\n-> 在第4行第5列找到原文字符: C\n\n处理字符对: GI\n行首字符: G, 列尾字符: I\n-> 在第3行第5列找到原文字符: U\n\n处理字符对: KH\n行首字符: K, 列尾字符: H\n-> 在第2行第4列找到原文字符: L\n\n处理字符对: MI\n行首字符: M, 列尾字符: I\n-> 在第1行第5列找到原文字符: P\n\n处理字符对: GT\n行首字符: G, 列尾字符: T\n-> 在第3行第2列找到原文字符: A\n\n处理字符对: BB\n行首字符: B, 列尾字符: B\n-> 在第5行第1列找到原文字符: B\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: KH\n行首字符: K, 列尾字符: H\n-> 在第2行第4列找到原文字符: L\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: BT\n行首字符: B, 列尾字符: T\n-> 在第5行第2列找到原文字符: T\n\n处理字符对: WE\n行首字符: W, 列尾字符: E\n-> 在第4行第3列找到原文字符: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(所有行和列号从0开始计数):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - 位于所有行第一个字母的MKGWB是行首字母\n        - 位于所有列最后一个字母的PQUCL是列尾字母\n- 加密步骤:\n    - 移除非加密信息中的空格、标点和字母J，并将所有字母转换为大写\n    - 对非加密信息中的每个字母p:\n        - 在网格中找到字母p的位置，然后找到相应的行首和列尾字符\n        - 将行首和列尾字符连接成二元组作为该字母p的加密消息\n        - 例如，如果字母p是H，它在第4行，行首字符是B；它在第3列，列尾字符是H，所以加密消息是BH\n    \n    连接所有加密消息作为最终暗码输出\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 输出:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 每次从暗码中取两个字母c1,c2\n        - 字母c1标识解密字母p在网格中的行位置，找到c1作为行首的行\n        - 字母c2标识解密字母p在网格中的列位置，找到c2在列底的列\n        - 在网格中找到这个行列位置的字母，即为c1,c2的解密消息p\n        - 例如，如果c1,c2=BH，B是行首的行是第4行，H是行底的行是第2列，(4,2)处的字母是H，所以BH解密为H\n    - 两个两个地解密暗码中的字母，并将解密消息连接起来作为最终非加密信息输出\n暗码: KTBEWTWHBIBI 解密为非加密信息: nervii\n暗码: KTGEKTWIGIKHMIGTBBBIKHBIBTWE 解密为非加密信息: nonculpability\n暗码: MIWEWTBIBTBIWI 解密为非加密信息是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 9, "source_filename": "icl_with_rule_decode_Kor_rule15_CollonCipher_cn.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "MIWEWTBIBTBIWI", "extra_args": {}, "output": "开始解密过程...\n加密文本: MIWEWTBIBTBIWI\n\n每次解密两个字符:\n\n处理字符对: MI\n行首字符: M, 列尾字符: I\n-> 在第1行第5列找到原文字符: P\n\n处理字符对: WE\n行首字符: W, 列尾字符: E\n-> 在第4行第3列找到原文字符: Y\n\n处理字符对: WT\n行首字符: W, 列尾字符: T\n-> 在第4行第2列找到原文字符: R\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: BT\n行首字符: B, 列尾字符: T\n-> 在第5行第2列找到原文字符: T\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: WI\n行首字符: W, 列尾字符: I\n-> 在第4行第5列找到原文字符: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗码: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - e: 263\n    - n: 299\n- 加密步骤:\n    - 对初始文本中的每个字母p:\n        - 获取字母p对应的ASCII码的十进制数x\n        - 计算x^e mod n作为该字母的暗码数字y，这里^表示乘法运算\n    - 最后，将所有y用逗号连接形成最终暗码\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 暗码: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 输出:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格\n- 准备:\n    - e: 263\n    - n: 299\n- 解密步骤:\n    - 对暗码中的每个数字c:\n        - 计算z = c^e mod n，这里^表示乘法运算\n        - 根据z的十进制值，使用ASCII码找到对应的字母作为初始文本字母p\n    - 最后，将所有p连接得到最终初始文本\n\n暗码: 156,23,62,226,213,213 解密成为初始文本是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 10, "source_filename": "icl_with_rule_decode_Kor_rule23_RSACipher_cn.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "156,23,62,226,213,213", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 156,23,62,226,213,213\n使用参数: e=263, n=299\n分割后的数字序列: ['156', '23', '62', '226', '213', '213']\n\n逐数字解密:\n计算 156^263 mod 299 = 78\n对应的字符为: N\n计算 23^263 mod 299 = 69\n对应的字符为: E\n计算 62^263 mod 299 = 82\n对应的字符为: R\n计算 226^263 mod 299 = 86\n对应的字符为: V\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 213^263 mod 299 = 73\n对应的字符为: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 隐文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - e: 263\n    - n: 299\n- 加密步骤:\n    - 对明文中的每个字母p:\n        - 获取字母p对应的ASCII码的十进制数x\n        - 计算x^e mod n作为该字母的隐文数字y，这里^表示乘法运算\n    - 最后，将所有y用逗号连接形成最终隐文\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 隐文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 输出:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 准备:\n    - e: 263\n    - n: 299\n- 解密步骤:\n    - 对隐文中的每个数字c:\n        - 计算z = c^e mod n，这里^表示乘法运算\n        - 根据z的十进制值，使用ASCII码找到对应的字母作为明文字母p\n    - 最后，将所有p连接得到最终明文\n\n隐文: 228,84,62,213,89,213,241 解密为明文: pyritic\n隐文: 228,84,62,213,89,213,241 解密为明文: pyritic\n隐文: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 解密为明文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 11, "source_filename": "icl_with_rule_decode_Kor_rule23_RSACipher_cn.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "156,53,156,241,197,240,228,247,222,213,240,213,89,84", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 156,53,156,241,197,240,228,247,222,213,240,213,89,84\n使用参数: e=263, n=299\n分割后的数字序列: ['156', '53', '156', '241', '197', '240', '228', '247', '222', '213', '240', '213', '89', '84']\n\n逐数字解密:\n计算 156^263 mod 299 = 78\n对应的字符为: N\n计算 53^263 mod 299 = 79\n对应的字符为: O\n计算 156^263 mod 299 = 78\n对应的字符为: N\n计算 241^263 mod 299 = 67\n对应的字符为: C\n计算 197^263 mod 299 = 85\n对应的字符为: U\n计算 240^263 mod 299 = 76\n对应的字符为: L\n计算 228^263 mod 299 = 80\n对应的字符为: P\n计算 247^263 mod 299 = 65\n对应的字符为: A\n计算 222^263 mod 299 = 66\n对应的字符为: B\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 240^263 mod 299 = 76\n对应的字符为: L\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 89^263 mod 299 = 84\n对应的字符为: T\n计算 84^263 mod 299 = 89\n对应的字符为: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - e: 263\n    - n: 299\n- 加密步骤:\n    - 对明文中的每个字母p:\n        - 获取字母p对应的ASCII码的十进制数x\n        - 计算x^e mod n作为该字母的暗文数字y，这里^表示乘法运算\n    - 最后，将所有y用逗号连接形成最终暗文\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 暗文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 输出:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 准备:\n    - e: 263\n    - n: 299\n- 解密步骤:\n    - 对暗文中的每个数字c:\n        - 计算z = c^e mod n，这里^表示乘法运算\n        - 根据z的十进制值，使用ASCII码找到对应的字母作为明文字母p\n    - 最后，将所有p连接得到最终明文\n\n暗文: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 解码为明文: nonculpability\n暗文: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 解码为明文: nonculpability\n暗文: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 解码为明文: nonculpability\n暗文: 228,84,62,213,89,213,241 解码为明文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 12, "source_filename": "icl_with_rule_decode_Kor_rule23_RSACipher_cn.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "228,84,62,213,89,213,241", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 228,84,62,213,89,213,241\n使用参数: e=263, n=299\n分割后的数字序列: ['228', '84', '62', '213', '89', '213', '241']\n\n逐数字解密:\n计算 228^263 mod 299 = 80\n对应的字符为: P\n计算 84^263 mod 299 = 89\n对应的字符为: Y\n计算 62^263 mod 299 = 82\n对应的字符为: R\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 89^263 mod 299 = 84\n对应的字符为: T\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 241^263 mod 299 = 67\n对应的字符为: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 原始信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - Polybius方阵:\n    1  2  3  4  5\n1   R  T  X  F  S\n2   W  C  M  V  H\n3   Z  J  A  P  B\n4   L  Q  Y  G  K\n5   N  E  U  D  I\n\n- 加密步骤:\n    - 对于每个原始信息字符p:\n        - 如果p是存在于Polybius方阵中的大写字母:\n            - 用字符在方阵中的行号和列号(都从1开始计数)替换该字符\n        - 特别地，字母O不存在于方阵中，用66替换\n    \n解密算法:\n\n解密规则:\n- 输入:\n    - 暗码: 字符串\n- 输出:\n    - 原始信息: 大写字母字符串\n- 准备:\n    - Polybius方阵(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 对于暗码中的每对数字CrCc:\n        - 根据CrCc表示的行列号在Polybius方阵中找到对应字母\n        - 如果CrCc=66，替换为\"O\"\n    \n暗码: 34431155125522 解密成原始信息: pyritic\n暗码: 34431155125522 解密成原始信息: pyritic\n暗码: 5166512253413433355541551243 解密成原始信息: nonculpability\n暗码: 515211245555 解密成原始信息是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 13, "source_filename": "icl_with_rule_decode_Kor_rule4_PolybiusSquareCipher_cn.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "515211245555", "extra_args": {}, "output": "开始解密文本: 515211245555\n开始逐对数字解密:\n遇到数字对: 51，对应方阵位置: 第5行第1列，解密为: N\n遇到数字对: 52，对应方阵位置: 第5行第2列，解密为: E\n遇到数字对: 11，对应方阵位置: 第1行第1列，解密为: R\n遇到数字对: 24，对应方阵位置: 第2行第4列，解密为: V\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 隐文: 大写字母字符串\n- 准备:\n    - Polybius方阵:\n    1  2  3  4  5\n1   R  T  X  F  S\n2   W  C  M  V  H\n3   Z  J  A  P  B\n4   L  Q  Y  G  K\n5   N  E  U  D  I\n\n- 加密步骤:\n    - 对于每个初始文本字符p:\n        - 如果p是存在于Polybius方阵中的大写字母:\n            - 用字符在方阵中的行号和列号(都从1开始计数)替换该字符\n        - 特别地，字母O不存在于方阵中，用66替换\n    \n解密算法:\n\n解密规则:\n- 输入:\n    - 隐文: 字符串\n- 输出:\n    - 初始文本: 大写字母字符串\n- 准备:\n    - Polybius方阵(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 对于隐文中的每对数字CrCc:\n        - 根据CrCc表示的行列号在Polybius方阵中找到对应字母\n        - 如果CrCc=66，替换为\"O\"\n    \n隐文: 34431155125522 解密为初始文本: pyritic\n隐文: 5166512253413433355541551243 解密为初始文本: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 14, "source_filename": "icl_with_rule_decode_Kor_rule4_PolybiusSquareCipher_cn.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "5166512253413433355541551243", "extra_args": {}, "output": "开始解密文本: 5166512253413433355541551243\n开始逐对数字解密:\n遇到数字对: 51，对应方阵位置: 第5行第1列，解密为: N\n遇到数字对: 66，解密为: O\n遇到数字对: 51，对应方阵位置: 第5行第1列，解密为: N\n遇到数字对: 22，对应方阵位置: 第2行第2列，解密为: C\n遇到数字对: 53，对应方阵位置: 第5行第3列，解密为: U\n遇到数字对: 41，对应方阵位置: 第4行第1列，解密为: L\n遇到数字对: 34，对应方阵位置: 第3行第4列，解密为: P\n遇到数字对: 33，对应方阵位置: 第3行第3列，解密为: A\n遇到数字对: 35，对应方阵位置: 第3行第5列，解密为: B\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 41，对应方阵位置: 第4行第1列，解密为: L\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 12，对应方阵位置: 第1行第2列，解密为: T\n遇到数字对: 43，对应方阵位置: 第4行第3列，解密为: Y\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - Polybius方阵:\n    1  2  3  4  5\n1   R  T  X  F  S\n2   W  C  M  V  H\n3   Z  J  A  P  B\n4   L  Q  Y  G  K\n5   N  E  U  D  I\n\n- 加密步骤:\n    - 对于每个初始文本字符p:\n        - 如果p是存在于Polybius方阵中的大写字母:\n            - 用字符在方阵中的行号和列号(都从1开始计数)替换该字符\n        - 特别地，字母O不存在于方阵中，用66替换\n    \n解密算法:\n\n解密规则:\n- 输入:\n    - 暗码: 字符串\n- 输出:\n    - 初始文本: 大写字母字符串\n- 准备:\n    - Polybius方阵(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 对于暗码中的每对数字CrCc:\n        - 根据CrCc表示的行列号在Polybius方阵中找到对应字母\n        - 如果CrCc=66，替换为\"O\"\n    \n暗码: 5166512253413433355541551243 解密成为初始文本: nonculpability\n暗码: 34431155125522 解密成为初始文本: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 15, "source_filename": "icl_with_rule_decode_Kor_rule4_PolybiusSquareCipher_cn.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "34431155125522", "extra_args": {}, "output": "开始解密文本: 34431155125522\n开始逐对数字解密:\n遇到数字对: 34，对应方阵位置: 第3行第4列，解密为: P\n遇到数字对: 43，对应方阵位置: 第4行第3列，解密为: Y\n遇到数字对: 11，对应方阵位置: 第1行第1列，解密为: R\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 12，对应方阵位置: 第1行第2列，解密为: T\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 22，对应方阵位置: 第2行第2列，解密为: C\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 非加密信息: 大写字母字符串,不含标点和空格\n    - 输出:\n        - 加密文本: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列:\n            - 54张卡牌的列表,包括52张花色牌和2张可区分的王牌(A王牌和B王牌)。花色牌按四种花色顺序编号1-52,王牌值为53和54。\n            - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n        - 密钥流算法:\n            该算法通过移动卡牌生成密钥流值。算法是确定性的,意味着密钥流值仅取决于卡牌的初始顺序。卡牌组被视为循环数组,允许需要移到底部的卡牌绕到顶部。\n            \n            执行以下步骤生成密钥流的一个字符:\n            1. 找到A王牌并下移一位。如果是最后一张牌,则成为第二张牌。不能成为第一张牌。\n            2. 找到B王牌并下移两位。如果是倒数第二张牌,则绕到第二位。如果是最后一张牌,则成为第三张牌。不能成为第一张牌。\n            3. 进行\"三切\":用王牌作为边界将牌组分成三部分,然后交换顶部和底部。王牌本身及其之间的牌保持不变。\n            4. 进行\"计数切\":检查牌组底牌。如果是王牌(53/54),其值固定为53。从牌组顶部取出该数量的牌,插入到最后一张牌的上方。\n            5. 查看顶牌的值。同样,任何王牌计为53。计算该牌下方的位置数,使用该位置的牌值作为下一个密钥流值。如果计算出的牌是王牌,忽略它并重复密钥流算法。\n            6. 返回生成的密钥流值\n    - 加密步骤:\n        - cards=初始卡牌序列\n        - 对每个非加密信息字符p:\n            - 使用字母表将p转换为对应的位置值x(从0开始)\n            - 使用初始卡牌序列为p生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 当密钥流值y加到位置值x时,应用模26运算得到z:\n                - z=(y+x) % 26\n            - 使用LETTERS列表返回对应位置z的字母\n            - 将其附加到加密文本\n解密算法:\n解密规则:\n    - 输入:\n        - 加密文本: 大写字母字符串\n    - 输出:\n        - 非加密信息: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列(与加密相同)\n        - 密钥流算法(与加密相同)\n    - 解密步骤(与加密步骤完全相反):\n        - cards=初始卡牌序列\n        - 对每个加密文本字符c:\n            - 使用LETTERS将c转换为对应的位置值z(从0开始)\n            - 为c生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 从加密文本字符c计算原始位置值x:\n                - x=(z-y) mod 26\n            - 使用LETTERS列表返回对应位置x的字母\n            - 将其附加到解密非加密信息\n加密文本: FMMLWKFPPPDEOF 解密成为非加密信息: nonculpability\n加密文本: FMMLWKFPPPDEOF 解密成为非加密信息: nonculpability\n加密文本: AVPUUPJ 解密成为非加密信息: pyritic\n加密文本: FBPMFP 解密成为非加密信息: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 16, "source_filename": "icl_with_rule_decode_Kor_rule6_SolitaireCipher_cn.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "FBPMFP", "extra_args": {}, "output": "开始解密密文: FBPMFP\n\n处理第1个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (11 - 14) % 26 = 23\n解密后的字符是: N\n\n处理第2个字符 'B':\n密文字符'B'在字母表中的位置是: 20\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n解密计算: (20 - 7) % 26 = 13\n解密后的字符是: E\n\n处理第3个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n解密计算: (18 - 13) % 26 = 5\n解密后的字符是: R\n\n处理第4个字符 'M':\n密文字符'M'在字母表中的位置是: 10\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n解密计算: (10 - 17) % 26 = 19\n解密后的字符是: V\n\n处理第5个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (11 - 38) % 26 = 25\n解密后的字符是: I\n\n处理第6个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (18 - 45) % 26 = 25\n解密后的字符是: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 非加密信息: 大写字母字符串,不含标点和空格\n    - 输出:\n        - 隐文: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列:\n            - 54张卡牌的列表,包括52张花色牌和2张可区分的王牌(A王牌和B王牌)。花色牌按四种花色顺序编号1-52,王牌值为53和54。\n            - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n        - 密钥流算法:\n            该算法通过移动卡牌生成密钥流值。算法是确定性的,意味着密钥流值仅取决于卡牌的初始顺序。卡牌组被视为循环数组,允许需要移到底部的卡牌绕到顶部。\n            \n            执行以下步骤生成密钥流的一个字符:\n            1. 找到A王牌并下移一位。如果是最后一张牌,则成为第二张牌。不能成为第一张牌。\n            2. 找到B王牌并下移两位。如果是倒数第二张牌,则绕到第二位。如果是最后一张牌,则成为第三张牌。不能成为第一张牌。\n            3. 进行\"三切\":用王牌作为边界将牌组分成三部分,然后交换顶部和底部。王牌本身及其之间的牌保持不变。\n            4. 进行\"计数切\":检查牌组底牌。如果是王牌(53/54),其值固定为53。从牌组顶部取出该数量的牌,插入到最后一张牌的上方。\n            5. 查看顶牌的值。同样,任何王牌计为53。计算该牌下方的位置数,使用该位置的牌值作为下一个密钥流值。如果计算出的牌是王牌,忽略它并重复密钥流算法。\n            6. 返回生成的密钥流值\n    - 加密步骤:\n        - cards=初始卡牌序列\n        - 对每个非加密信息字符p:\n            - 使用字母表将p转换为对应的位置值x(从0开始)\n            - 使用初始卡牌序列为p生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 当密钥流值y加到位置值x时,应用模26运算得到z:\n                - z=(y+x) % 26\n            - 使用LETTERS列表返回对应位置z的字母\n            - 将其附加到隐文\n解密算法:\n解密规则:\n    - 输入:\n        - 隐文: 大写字母字符串\n    - 输出:\n        - 非加密信息: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列(与加密相同)\n        - 密钥流算法(与加密相同)\n    - 解密步骤(与加密步骤完全相反):\n        - cards=初始卡牌序列\n        - 对每个隐文字符c:\n            - 使用LETTERS将c转换为对应的位置值z(从0开始)\n            - 为c生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 从隐文字符c计算原始位置值x:\n                - x=(z-y) mod 26\n            - 使用LETTERS列表返回对应位置x的字母\n            - 将其附加到解密非加密信息\n隐文: FBPMFP 解码为非加密信息: nervii\n隐文: FMMLWKFPPPDEOF 解码为非加密信息是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 17, "source_filename": "icl_with_rule_decode_Kor_rule6_SolitaireCipher_cn.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "FMMLWKFPPPDEOF", "extra_args": {}, "output": "开始解密密文: FMMLWKFPPPDEOF\n\n处理第1个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (11 - 14) % 26 = 23\n解密后的字符是: N\n\n处理第2个字符 'M':\n密文字符'M'在字母表中的位置是: 10\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n解密计算: (10 - 7) % 26 = 3\n解密后的字符是: O\n\n处理第3个字符 'M':\n密文字符'M'在字母表中的位置是: 10\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n解密计算: (10 - 13) % 26 = 23\n解密后的字符是: N\n\n处理第4个字符 'L':\n密文字符'L'在字母表中的位置是: 24\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n解密计算: (24 - 17) % 26 = 7\n解密后的字符是: C\n\n处理第5个字符 'W':\n密文字符'W'在字母表中的位置是: 2\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (2 - 38) % 26 = 16\n解密后的字符是: U\n\n处理第6个字符 'K':\n密文字符'K'在字母表中的位置是: 17\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (17 - 45) % 26 = 24\n解密后的字符是: L\n\n处理第7个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (11 - 45) % 26 = 18\n解密后的字符是: P\n\n处理第8个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (18 - 38) % 26 = 6\n解密后的字符是: A\n\n处理第9个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 50\n解密计算: (18 - 50) % 26 = 20\n解密后的字符是: B\n\n处理第10个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (18 - 45) % 26 = 25\n解密后的字符是: I\n\n处理第11个字符 'D':\n密文字符'D'在字母表中的位置是: 1\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 29\n解密计算: (1 - 29) % 26 = 24\n解密后的字符是: L\n\n处理第12个字符 'E':\n密文字符'E'在字母表中的位置是: 13\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (13 - 14) % 26 = 25\n解密后的字符是: I\n\n处理第13个字符 'O':\n密文字符'O'在字母表中的位置是: 3\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n解密计算: (3 - 25) % 26 = 4\n解密后的字符是: T\n\n处理第14个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n解密计算: (11 - 25) % 26 = 12\n解密后的字符是: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 原文: 大写字母字符串,不含标点和空格\n    - 输出:\n        - 暗码: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列:\n            - 54张卡牌的列表,包括52张花色牌和2张可区分的王牌(A王牌和B王牌)。花色牌按四种花色顺序编号1-52,王牌值为53和54。\n            - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n        - 密钥流算法:\n            该算法通过移动卡牌生成密钥流值。算法是确定性的,意味着密钥流值仅取决于卡牌的初始顺序。卡牌组被视为循环数组,允许需要移到底部的卡牌绕到顶部。\n            \n            执行以下步骤生成密钥流的一个字符:\n            1. 找到A王牌并下移一位。如果是最后一张牌,则成为第二张牌。不能成为第一张牌。\n            2. 找到B王牌并下移两位。如果是倒数第二张牌,则绕到第二位。如果是最后一张牌,则成为第三张牌。不能成为第一张牌。\n            3. 进行\"三切\":用王牌作为边界将牌组分成三部分,然后交换顶部和底部。王牌本身及其之间的牌保持不变。\n            4. 进行\"计数切\":检查牌组底牌。如果是王牌(53/54),其值固定为53。从牌组顶部取出该数量的牌,插入到最后一张牌的上方。\n            5. 查看顶牌的值。同样,任何王牌计为53。计算该牌下方的位置数,使用该位置的牌值作为下一个密钥流值。如果计算出的牌是王牌,忽略它并重复密钥流算法。\n            6. 返回生成的密钥流值\n    - 加密步骤:\n        - cards=初始卡牌序列\n        - 对每个原文字符p:\n            - 使用字母表将p转换为对应的位置值x(从0开始)\n            - 使用初始卡牌序列为p生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 当密钥流值y加到位置值x时,应用模26运算得到z:\n                - z=(y+x) % 26\n            - 使用LETTERS列表返回对应位置z的字母\n            - 将其附加到暗码\n解密算法:\n解密规则:\n    - 输入:\n        - 暗码: 大写字母字符串\n    - 输出:\n        - 原文: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列(与加密相同)\n        - 密钥流算法(与加密相同)\n    - 解密步骤(与加密步骤完全相反):\n        - cards=初始卡牌序列\n        - 对每个暗码字符c:\n            - 使用LETTERS将c转换为对应的位置值z(从0开始)\n            - 为c生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 从暗码字符c计算原始位置值x:\n                - x=(z-y) mod 26\n            - 使用LETTERS列表返回对应位置x的字母\n            - 将其附加到解密原文\n暗码: FBPMFP 解密为原文: nervii\n暗码: FBPMFP 解密为原文: nervii\n暗码: FMMLWKFPPPDEOF 解密为原文: nonculpability\n暗码: AVPUUPJ 解密为原文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 18, "source_filename": "icl_with_rule_decode_Kor_rule6_SolitaireCipher_cn.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "AVPUUPJ", "extra_args": {}, "output": "开始解密密文: AVPUUPJ\n\n处理第1个字符 'A':\n密文字符'A'在字母表中的位置是: 6\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (6 - 14) % 26 = 18\n解密后的字符是: P\n\n处理第2个字符 'V':\n密文字符'V'在字母表中的位置是: 19\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n解密计算: (19 - 7) % 26 = 12\n解密后的字符是: Y\n\n处理第3个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n解密计算: (18 - 13) % 26 = 5\n解密后的字符是: R\n\n处理第4个字符 'U':\n密文字符'U'在字母表中的位置是: 16\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n解密计算: (16 - 17) % 26 = 25\n解密后的字符是: I\n\n处理第5个字符 'U':\n密文字符'U'在字母表中的位置是: 16\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (16 - 38) % 26 = 4\n解密后的字符是: T\n\n处理第6个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (18 - 45) % 26 = 25\n解密后的字符是: I\n\n处理第7个字符 'J':\n密文字符'J'在字母表中的位置是: 0\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (0 - 45) % 26 = 7\n解密后的字符是: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds\n        - Gird1\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n        - Gird2\n            - E  C  H  O  A\n            B  D  F  G  I\n            J  K  L  M  N\n            P  R  S  T  U\n            V  W  X  Y  Z\n        - Gird3\n            - V  O  R  T  E\n            X  A  B  C  D\n            F  G  H  I  J\n            K  L  M  N  P\n            S  U  W  Y  Z\n        - Gird4\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n- Encryption Steps:\n    - Cleans up plaintext, removing spaces and non-alphabetic characters, removing the letter Q, and converting all letters to uppercase.\n    - If the plaintext is of odd length, add a letter 'X' to make it of even length.\n    - Split the cleaned plaintext into two-letter groups.\n    - For each double-letter group p1,p2 of the plaintext:\n        - In Gird1 and Gird4, find the position of the first letter and the second letter.\n        - In Gird2 and Gird3, find the letters corresponding to these two positions, use these two letters as the result of encrypting the double letter combination.\n    - Concatenate all the encrypted double letter groups to form the final ciphertext.\n\n \nplain text: pyritic encrypt into encrypted text: BIDYGYSH\nplain text: nervii encrypt into encrypted text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 19, "source_filename": "icl_with_rule_encode_Kor_rule11_FourSquareCipher_en.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n原始文本: nervii\n清理后的文本(移除非字母字符并转大写，移除Q): NERVII\n\n开始逐对字母加密:\n\n处理字母对: NE\n第一个字母 N 在原始表格中的位置: (0, 3)\n第二个字母 E 在原始表格中的位置: (3, 4)\n加密后的字母对: OP\n\n处理字母对: RV\n第一个字母 R 在原始表格中的位置: (1, 1)\n第二个字母 V 在原始表格中的位置: (2, 0)\n加密后的字母对: DF\n\n处理字母对: II\n第一个字母 I 在原始表格中的位置: (4, 3)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: YY\n\n最终加密结果: OPDFYY\n", "ground_truth": "OPDFYY"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds\n        - Gird1\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n        - Gird2\n            - E  C  H  O  A\n            B  D  F  G  I\n            J  K  L  M  N\n            P  R  S  T  U\n            V  W  X  Y  Z\n        - Gird3\n            - V  O  R  T  E\n            X  A  B  C  D\n            F  G  H  I  J\n            K  L  M  N  P\n            S  U  W  Y  Z\n        - Gird4\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n- Encryption Steps:\n    - Cleans up plaintext, removing spaces and non-alphabetic characters, removing the letter Q, and converting all letters to uppercase.\n    - If the plaintext is of odd length, add a letter 'X' to make it of even length.\n    - Split the cleaned plaintext into two-letter groups.\n    - For each double-letter group p1,p2 of the plaintext:\n        - In Gird1 and Gird4, find the position of the first letter and the second letter.\n        - In Gird2 and Gird3, find the letters corresponding to these two positions, use these two letters as the result of encrypting the double letter combination.\n    - Concatenate all the encrypted double letter groups to form the final ciphertext.\n\n \nclear text: nervii encode to encoded text: OPDFYY\nclear text: pyritic encode to encoded text: BIDYGYSH\nclear text: nervii encode to encoded text: OPDFYY\nclear text: nonculpability encode to encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 20, "source_filename": "icl_with_rule_encode_Kor_rule11_FourSquareCipher_en.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n原始文本: nonculpability\n清理后的文本(移除非字母字符并转大写，移除Q): NONCULPABILITY\n\n开始逐对字母加密:\n\n处理字母对: NO\n第一个字母 N 在原始表格中的位置: (0, 3)\n第二个字母 O 在原始表格中的位置: (0, 4)\n加密后的字母对: OE\n\n处理字母对: NC\n第一个字母 N 在原始表格中的位置: (0, 3)\n第二个字母 C 在原始表格中的位置: (3, 2)\n加密后的字母对: OM\n\n处理字母对: UL\n第一个字母 U 在原始表格中的位置: (1, 4)\n第二个字母 L 在原始表格中的位置: (0, 1)\n加密后的字母对: IO\n\n处理字母对: PA\n第一个字母 P 在原始表格中的位置: (1, 0)\n第二个字母 A 在原始表格中的位置: (3, 0)\n加密后的字母对: BK\n\n处理字母对: BI\n第一个字母 B 在原始表格中的位置: (3, 1)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: RY\n\n处理字母对: LI\n第一个字母 L 在原始表格中的位置: (0, 1)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: CY\n\n处理字母对: TY\n第一个字母 T 在原始表格中的位置: (1, 3)\n第二个字母 Y 在原始表格中的位置: (2, 3)\n加密后的字母对: GI\n\n最终加密结果: OEOMIOBKRYCYGI\n", "ground_truth": "OEOMIOBKRYCYGI"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds\n        - Gird1\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n        - Gird2\n            - E  C  H  O  A\n            B  D  F  G  I\n            J  K  L  M  N\n            P  R  S  T  U\n            V  W  X  Y  Z\n        - Gird3\n            - V  O  R  T  E\n            X  A  B  C  D\n            F  G  H  I  J\n            K  L  M  N  P\n            S  U  W  Y  Z\n        - Gird4\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n- Encryption Steps:\n    - Cleans up plaintext, removing spaces and non-alphabetic characters, removing the letter Q, and converting all letters to uppercase.\n    - If the plaintext is of odd length, add a letter 'X' to make it of even length.\n    - Split the cleaned plaintext into two-letter groups.\n    - For each double-letter group p1,p2 of the plaintext:\n        - In Gird1 and Gird4, find the position of the first letter and the second letter.\n        - In Gird2 and Gird3, find the letters corresponding to these two positions, use these two letters as the result of encrypting the double letter combination.\n    - Concatenate all the encrypted double letter groups to form the final ciphertext.\n\n \nplain text: nervii encode into encoded text: OPDFYY\nplain text: pyritic encode into encoded text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 21, "source_filename": "icl_with_rule_encode_Kor_rule11_FourSquareCipher_en.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "PYRITICX", "extra_args": {}, "output": "开始加密过程...\n原始文本: pyritic\n清理后的文本(移除非字母字符并转大写，移除Q): PYRITIC\n文本长度为奇数，添加X: PYRITICX\n\n开始逐对字母加密:\n\n处理字母对: PY\n第一个字母 P 在原始表格中的位置: (1, 0)\n第二个字母 Y 在原始表格中的位置: (2, 3)\n加密后的字母对: BI\n\n处理字母对: RI\n第一个字母 R 在原始表格中的位置: (1, 1)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: DY\n\n处理字母对: TI\n第一个字母 T 在原始表格中的位置: (1, 3)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: GY\n\n处理字母对: CX\n第一个字母 C 在原始表格中的位置: (3, 2)\n第二个字母 X 在原始表格中的位置: (2, 2)\n加密后的字母对: SH\n\n最终加密结果: BIDYGYSH\n", "ground_truth": "BIDYGYSH"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串,不含标点和空格\n- 输出:\n    - 加密信息: 大写字母字符串\n- 准备:\n    - 5x5网格(不含字母J,行列坐标在0到4之间):\n        - U B I L A\n        N T C D E\n        F G H K M\n        O P Q R S\n        V W X Y Z\n- 加密步骤:\n    - 去除明文中的标点、空格和字母J\n    - 对明文中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)\n    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142\n    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成加密信息。例如对于0142,加密信息对应(0,1)是B,(4,2)是X。最终加密信息是BX\n明文: nervii 编码为加密信息: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 22, "source_filename": "icl_with_rule_encode_Kor_rule13_BifidCipher_cn.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "NERVII", "extra_args": {}, "output": "加密步骤:\n1. 去除标点和空格,将J替换为I后的文本: NERVII\n2. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n3. 获取每个字母的行列坐标: [1, 1, 3, 4, 0, 0, 0, 4, 3, 0, 2, 2]\n4. 最终加密结果: TSUAOH\n", "ground_truth": "TSUAOH"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串,不含标点和空格\n- 输出:\n    - 隐文: 大写字母字符串\n- 准备:\n    - 5x5网格(不含字母J,行列坐标在0到4之间):\n        - U B I L A\n        N T C D E\n        F G H K M\n        O P Q R S\n        V W X Y Z\n- 加密步骤:\n    - 去除初始文本中的标点、空格和字母J\n    - 对初始文本中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)\n    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142\n    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成隐文。例如对于0142,隐文对应(0,1)是B,(4,2)是X。最终隐文是BX\n初始文本: nervii 加密成隐文: TSUAOH\n初始文本: pyritic 加密成隐文: SONTRGH\n初始文本: pyritic 加密成隐文: SONTRGH\n初始文本: nonculpability 加密成隐文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 23, "source_filename": "icl_with_rule_encode_Kor_rule13_BifidCipher_cn.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "加密步骤:\n1. 去除标点和空格,将J替换为I后的文本: NONCULPABILITY\n2. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n3. 获取每个字母的行列坐标: [1, 3, 1, 1, 0, 0, 3, 0, 0, 0, 0, 0, 1, 4, 0, 0, 0, 2, 0, 3, 1, 4, 1, 2, 3, 2, 1, 3]\n4. 最终加密结果: DTUOUUEUILECQD\n", "ground_truth": "DTUOUUEUILECQD"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 大写字母字符串,不含标点和空格\n- 输出:\n    - 加密信息: 大写字母字符串\n- 准备:\n    - 5x5网格(不含字母J,行列坐标在0到4之间):\n        - U B I L A\n        N T C D E\n        F G H K M\n        O P Q R S\n        V W X Y Z\n- 加密步骤:\n    - 去除原始信息中的标点、空格和字母J\n    - 对原始信息中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)\n    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142\n    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成加密信息。例如对于0142,加密信息对应(0,1)是B,(4,2)是X。最终加密信息是BX\n原始信息: nervii 加密成为加密信息: TSUAOH\n原始信息: nervii 加密成为加密信息: TSUAOH\n原始信息: pyritic 加密成为加密信息: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 24, "source_filename": "icl_with_rule_encode_Kor_rule13_BifidCipher_cn.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "PYRITIC", "extra_args": {}, "output": "加密步骤:\n1. 去除标点和空格,将J替换为I后的文本: PYRITIC\n2. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n3. 获取每个字母的行列坐标: [3, 4, 3, 0, 1, 0, 1, 1, 3, 3, 2, 1, 2, 2]\n4. 最终加密结果: SONTRGH\n", "ground_truth": "SONTRGH"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对明文进行加密\n加密规则:\n    - 输入:\n        - 明文: 不含标点和空格的大写字母字符串\n    - 输出:\n        - 隐文: 不含标点和空格的字符串\n    - 准备:\n        - 转置序列表:\n            - [1, 4, 0, 6, 5, 2, 3]\n            - 转置序列表用于按顺序逐行写入明文，然后根据转置序列表调整列的顺序，使每行中的字符按给定顺序排列。\n            - 列从0开始计数。\n    - 加密步骤:\n        - [1, 4, 0, 6, 5, 2, 3]转置序列表共7位，表示一行应写入7个字母。\n        - 按顺序逐行写入明文，每行7个。当不足7个时，最后一行用$填充。可以得到一个写入网格。\n        - 根据转置序列表调整列的顺序，即现在列的顺序为[原列1，原列4，原列0，原列6，原列5，原列2，原列3]，可以得到调整列顺序后的网格。\n        - 逐行读取网格并连接起来得到最终隐文。(注意需要保留$)\n明文: nervii 编码为隐文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 25, "source_filename": "icl_with_rule_encode_Kor_rule20_TranspositionCipher_cn.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n原始网格:\nNERVII$\n转置后的网格:\nEIN$IRV\n最终加密结果: EIN$IRV\n", "ground_truth": "EIN$IRV"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对原始信息进行加密\n加密规则:\n    - 输入:\n        - 原始信息: 不含标点和空格的大写字母字符串\n    - 输出:\n        - 加密信息: 不含标点和空格的字符串\n    - 准备:\n        - 转置序列表:\n            - [1, 4, 0, 6, 5, 2, 3]\n            - 转置序列表用于按顺序逐行写入原始信息，然后根据转置序列表调整列的顺序，使每行中的字符按给定顺序排列。\n            - 列从0开始计数。\n    - 加密步骤:\n        - [1, 4, 0, 6, 5, 2, 3]转置序列表共7位，表示一行应写入7个字母。\n        - 按顺序逐行写入原始信息，每行7个。当不足7个时，最后一行用$填充。可以得到一个写入网格。\n        - 根据转置序列表调整列的顺序，即现在列的顺序为[原列1，原列4，原列0，原列6，原列5，原列2，原列3]，可以得到调整列顺序后的网格。\n        - 逐行读取网格并连接起来得到最终加密信息。(注意需要保留$)\n原始信息: nervii 加密成为加密信息: EIN$IRV\n原始信息: nonculpability 加密成为加密信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 26, "source_filename": "icl_with_rule_encode_Kor_rule20_TranspositionCipher_cn.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 2 行来存放文本\n原始网格:\nNONCULP\nABILITY\n转置后的网格:\nOUNPLNC\nBIAYTIL\n最终加密结果: OUNPLNCBIAYTIL\n", "ground_truth": "OUNPLNCBIAYTIL"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原文进行加密\n加密规则:\n    - 输入:\n        - 原文: 不含标点和空格的大写字母字符串\n    - 输出:\n        - 暗码: 不含标点和空格的字符串\n    - 准备:\n        - 转置序列表:\n            - [1, 4, 0, 6, 5, 2, 3]\n            - 转置序列表用于按顺序逐行写入原文，然后根据转置序列表调整列的顺序，使每行中的字符按给定顺序排列。\n            - 列从0开始计数。\n    - 加密步骤:\n        - [1, 4, 0, 6, 5, 2, 3]转置序列表共7位，表示一行应写入7个字母。\n        - 按顺序逐行写入原文，每行7个。当不足7个时，最后一行用$填充。可以得到一个写入网格。\n        - 根据转置序列表调整列的顺序，即现在列的顺序为[原列1，原列4，原列0，原列6，原列5，原列2，原列3]，可以得到调整列顺序后的网格。\n        - 逐行读取网格并连接起来得到最终暗码。(注意需要保留$)\n原文: nonculpability 编码为暗码: OUNPLNCBIAYTIL\n原文: pyritic 编码为暗码: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 27, "source_filename": "icl_with_rule_encode_Kor_rule20_TranspositionCipher_cn.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n原始网格:\nPYRITIC\n转置后的网格:\nYTPCIRI\n最终加密结果: YTPCIRI\n", "ground_truth": "YTPCIRI"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 初始文本: 仅包含大写字母的字符串，不含标点符号和空格\n    - 输出:\n        - 暗文: 十六进制字符串（包含小写字母a-e）\n    - 准备:\n        - 密钥(SHA哈希值)\n            - 对\"SECRET_KEY\"执行SHA-256运算，得到'73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'\n    - 加密步骤:\n        - 将初始文本字符串转换为字节序列（ASCII编码）\n        - 使用密钥对每个字节进行异或（XOR）运算加密。重复使用密钥使其长度与初始文本字节数相同\n        - 将加密后的字节序列转换为十六进制字符串作为暗文输出\n解密算法:\n解密规则:\n    - 输入:\n        - 暗文: 十六进制字符串（包含小写字母a-e）\n    - 输出:\n        - 初始文本: 仅包含大写字母的字符串，不含标点符号和空格\n    - 准备:\n        - 密钥（与加密相同，是通过SHA-256获得的哈希值）\n    - 解密步骤:\n        - 将暗文字符串转换为字节序列\n        - 使用密钥对每个字节进行异或（XOR）运算解密（解密过程与加密过程相同）\n        - 将解密后的字节序列转换为初始文本字符串\n暗文: 3da0640d88362ffe45b41326fbc0 解密成为初始文本: nonculpability\n暗文: 23b6780789333c 解密成为初始文本: pyritic\n暗文: 3daa78189433 解密成为初始文本是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 28, "source_filename": "icl_with_rule_decode_Kor_rule25_SHACipher_cn.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "3daa78189433", "extra_args": {}, "output": "开始解密过程...\n加密的十六进制文本: 3daa78189433\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n十六进制转换为字节序列: b'=\\xaax\\x18\\x943'\n开始XOR解密...\n解密后的字节序列: b'NERVII'\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 明文: 仅包含大写字母的字符串，不含标点符号和空格\n    - 输出:\n        - 加密信息: 十六进制字符串（包含小写字母a-e）\n    - 准备:\n        - 密钥(SHA哈希值)\n            - 对\"SECRET_KEY\"执行SHA-256运算，得到'73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'\n    - 加密步骤:\n        - 将明文字符串转换为字节序列（ASCII编码）\n        - 使用密钥对每个字节进行异或（XOR）运算加密。重复使用密钥使其长度与明文字节数相同\n        - 将加密后的字节序列转换为十六进制字符串作为加密信息输出\n解密算法:\n解密规则:\n    - 输入:\n        - 加密信息: 十六进制字符串（包含小写字母a-e）\n    - 输出:\n        - 明文: 仅包含大写字母的字符串，不含标点符号和空格\n    - 准备:\n        - 密钥（与加密相同，是通过SHA-256获得的哈希值）\n    - 解密步骤:\n        - 将加密信息字符串转换为字节序列\n        - 使用密钥对每个字节进行异或（XOR）运算解密（解密过程与加密过程相同）\n        - 将解密后的字节序列转换为明文字符串\n加密信息: 23b6780789333c 解密为明文: pyritic\n加密信息: 3da0640d88362ffe45b41326fbc0 解密为明文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 29, "source_filename": "icl_with_rule_decode_Kor_rule25_SHACipher_cn.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "3da0640d88362ffe45b41326fbc0", "extra_args": {}, "output": "开始解密过程...\n加密的十六进制文本: 3da0640d88362ffe45b41326fbc0\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n十六进制转换为字节序列: b'=\\xa0d\\r\\x886/\\xfeE\\xb4\\x13&\\xfb\\xc0'\n开始XOR解密...\n解密后的字节序列: b'NONCULPABILITY'\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 原文: 仅包含大写字母的字符串，不含标点符号和空格\n    - 输出:\n        - 密文: 十六进制字符串（包含小写字母a-e）\n    - 准备:\n        - 密钥(SHA哈希值)\n            - 对\"SECRET_KEY\"执行SHA-256运算，得到'73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'\n    - 加密步骤:\n        - 将原文字符串转换为字节序列（ASCII编码）\n        - 使用密钥对每个字节进行异或（XOR）运算加密。重复使用密钥使其长度与原文字节数相同\n        - 将加密后的字节序列转换为十六进制字符串作为密文输出\n解密算法:\n解密规则:\n    - 输入:\n        - 密文: 十六进制字符串（包含小写字母a-e）\n    - 输出:\n        - 原文: 仅包含大写字母的字符串，不含标点符号和空格\n    - 准备:\n        - 密钥（与加密相同，是通过SHA-256获得的哈希值）\n    - 解密步骤:\n        - 将密文字符串转换为字节序列\n        - 使用密钥对每个字节进行异或（XOR）运算解密（解密过程与加密过程相同）\n        - 将解密后的字节序列转换为原文字符串\n密文: 3da0640d88362ffe45b41326fbc0 解密为原文: nonculpability\n密文: 3da0640d88362ffe45b41326fbc0 解密为原文: nonculpability\n密文: 3da0640d88362ffe45b41326fbc0 解密为原文: nonculpability\n密文: 23b6780789333c 解密为原文: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 30, "source_filename": "icl_with_rule_decode_Kor_rule25_SHACipher_cn.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "23b6780789333c", "extra_args": {}, "output": "开始解密过程...\n加密的十六进制文本: 23b6780789333c\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n十六进制转换为字节序列: b'#\\xb6x\\x07\\x893<'\n开始XOR解密...\n解密后的字节序列: b'PYRITIC'\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\nPlease decrypt the password based on the encryption and decryption algorithms.\n\nEncryption Algorithm:\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Convert the plaintext to uppercase, retain only the letter characters, and split them into pairs of two letters.\n3. Encrypt each pair of letters:\n   - Find the position of each letter in the default matrix, swap their y-coordinates,\n   - Retrieve the encrypted letter pair from the M1 and M2 matrices based on the new coordinates.\n4. Output the encrypted text.\n\nDecryption Algorithm:\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Group the ciphertext into pairs of two letters.\n3. Decrypt each pair of encrypted letters:\n   - Find the positions of the encrypted letters in M1 and M2, swap their column coordinates,\n   - Retrieve the corresponding plaintext letters from the default matrix.\n4. Output the decrypted text.\nencrypted text: JK FI LL KS CE GE NY extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to plain text: nonculpability\nencrypted text: JK FI LL KS CE GE NY extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to plain text: nonculpability\nencrypted text: RV RB RD WX extra parameter: {'str1': 'sawbill', 'str2': 'houri'} decode to plain text: pyritic\nencrypted text: JV KT CA extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} decode to plain text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 31, "source_filename": "icl_with_rule_decode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "JV KT CA", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将密文分成两个字母一组: ['JV', 'KT', 'CA']\n\nStep 4.1: 解密字母对 JV\n  - J 在 M1 中的位置: (2, 4)\n  - V 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 4) 的字母: E\n  - 解密结果为: NE\n\nStep 4.2: 解密字母对 KT\n  - K 在 M1 中的位置: (3, 0)\n  - T 在 M2 中的位置: (4, 1)\n  - 交换列索引后新位置: (3, 1) 和 (4, 0)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (4, 0) 的字母: V\n  - 解密结果为: RV\n\nStep 4.3: 解密字母对 CA\n  - C 在 M1 中的位置: (1, 3)\n  - A 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: II\n\n解密完成！最终解密结果为: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\nPlease decrypt the password based on the encryption and decryption algorithms.\n\nEncryption Algorithm:\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Convert the plaintext to uppercase, retain only the letter characters, and split them into pairs of two letters.\n3. Encrypt each pair of letters:\n   - Find the position of each letter in the default matrix, swap their y-coordinates,\n   - Retrieve the encrypted letter pair from the M1 and M2 matrices based on the new coordinates.\n4. Output the encrypted text.\n\nDecryption Algorithm:\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Group the ciphertext into pairs of two letters.\n3. Decrypt each pair of encrypted letters:\n   - Find the positions of the encrypted letters in M1 and M2, swap their column coordinates,\n   - Retrieve the corresponding plaintext letters from the default matrix.\n4. Output the decrypted text.\nencrypted text: JV KT CA secret key: {'str1': 'parametral', 'str2': 'quiverful'} decrypt to original information: nervii\nencrypted text: JV KT CA secret key: {'str1': 'parametral', 'str2': 'quiverful'} decrypt to original information: nervii\nencrypted text: JV KT CA secret key: {'str1': 'parametral', 'str2': 'quiverful'} decrypt to original information: nervii\nencrypted text: JK FI LL KS CE GE NY secret key: {'str1': 'trichophytia', 'str2': 'sphinges'} decrypt to original information is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 32, "source_filename": "icl_with_rule_decode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "JK FI LL KS CE GE NY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['JK', 'FI', 'LL', 'KS', 'CE', 'GE', 'NY']\n\nStep 4.1: 解密字母对 JK\n  - J 在 M1 中的位置: (2, 4)\n  - K 在 M2 中的位置: (2, 3)\n  - 交换列索引后新位置: (2, 3) 和 (2, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (2, 4) 的字母: O\n  - 解密结果为: NO\n\nStep 4.2: 解密字母对 FI\n  - F 在 M1 中的位置: (2, 2)\n  - I 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 2)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 解密结果为: NC\n\nStep 4.3: 解密字母对 LL\n  - L 在 M1 中的位置: (3, 1)\n  - L 在 M2 中的位置: (2, 4)\n  - 交换列索引后新位置: (3, 4) 和 (2, 1)\n  - 在 M_T 中查找位置 (3, 4) 的字母: U\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 解密结果为: UL\n\nStep 4.4: 解密字母对 KS\n  - K 在 M1 中的位置: (3, 0)\n  - S 在 M2 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (0, 0) 的字母: A\n  - 解密结果为: PA\n\nStep 4.5: 解密字母对 CE\n  - C 在 M1 中的位置: (0, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (0, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (0, 1) 的字母: B\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: BI\n\nStep 4.6: 解密字母对 GE\n  - G 在 M1 中的位置: (2, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (2, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: LI\n\nStep 4.7: 解密字母对 NY\n  - N 在 M1 中的位置: (3, 3)\n  - Y 在 M2 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: TY\n\n解密完成！最终解密结果为: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\nPlease decrypt the password based on the encryption and decryption algorithms.\n\nEncryption Algorithm:\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Convert the plaintext to uppercase, retain only the letter characters, and split them into pairs of two letters.\n3. Encrypt each pair of letters:\n   - Find the position of each letter in the default matrix, swap their y-coordinates,\n   - Retrieve the encrypted letter pair from the M1 and M2 matrices based on the new coordinates.\n4. Output the encrypted text.\n\nDecryption Algorithm:\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Group the ciphertext into pairs of two letters.\n3. Decrypt each pair of encrypted letters:\n   - Find the positions of the encrypted letters in M1 and M2, swap their column coordinates,\n   - Retrieve the corresponding plaintext letters from the default matrix.\n4. Output the decrypted text.\nencoded text: JK FI LL KS CE GE NY extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to original information: nonculpability\nencoded text: JK FI LL KS CE GE NY extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to original information: nonculpability\nencoded text: JK FI LL KS CE GE NY extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to original information: nonculpability\nencoded text: RV RB RD WX extra parameter: {'str1': 'sawbill', 'str2': 'houri'} decode to original information is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 33, "source_filename": "icl_with_rule_decode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "RV RB RD WX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['RV', 'RB', 'RD', 'WX']\n\nStep 4.1: 解密字母对 RV\n  - R 在 M1 中的位置: (3, 3)\n  - V 在 M2 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: PY\n\nStep 4.2: 解密字母对 RB\n  - R 在 M1 中的位置: (3, 3)\n  - B 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (3, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: RI\n\nStep 4.3: 解密字母对 RD\n  - R 在 M1 中的位置: (3, 3)\n  - D 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: TI\n\nStep 4.4: 解密字母对 WX\n  - W 在 M1 中的位置: (0, 2)\n  - X 在 M2 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 在 M_T 中查找位置 (4, 2) 的字母: X\n  - 解密结果为: CX\n\n解密完成！最终解密结果为: PYRITICX\n", "ground_truth": "PYRITICX"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - Grid and Template:\n        - Prepare a blank grid and a template with holes (fence).\n        - The template used is:\n            \n            ▮  ▮  ▮  ▮\n            ▮  ▮  ▯  ▯\n            ▮  ▮  ▮  ▯\n            ▯  ▮  ▮  ▮\n            \n            Where the white ones are holes, place the template on a blank grid, only through the white holes, and fill in the corresponding positions of the grid with characters.\n            \n- Encryption Steps:\n    - Divide the plaintext into blocks of 16 letters one by one.(If the plaintext is less than 16 long it is one  block)\n        - For example, the plaintext message \"LOVEHELLOWORLD\" is less than 16 and is one block.\n    - For each block, perform the following encryption operation:\n        - Place the template with holes on the blank grid.\n        - Fill the grid with the letters of the plaintext message in order through the holes in the template.\n            - For example, the plaintext is \"LOVEHELLOWORLD\", through the four holes, after filling in the LOVE grid is\n                \n                [['', '', '', ''],\n                ['', '', 'L', 'O'],\n                ['', '', '', 'V'],\n                ['E', '', '', '']]\n                \n        - The template has a total of four holes, so after filling out the four letters, it can not be filled out, then you need to rotate the template counterclockwise 90 degrees.\n            - For example, after filling in LOVE after rotating the template after the template for:\n                \n                ▮  ▯  ▯  ▮\n                ▮  ▯  ▮  ▮\n                ▮  ▮  ▮  ▮\n                ▮  ▮  ▮  ▯\n                \n        - Repeat filling the visible holes in the grid with the next letter of the plaintext message and rotating the template until the entire grid is completely filled. This will perform 4 fills + rotations and eventually the template will turn to the original template. If the message is not enough to fill the entire grid, it is made up with filler characters (e.g. '#').\n            - For example, if the plain text is \"LOVEHELLOWORLD\", after several fills and rotations, the final grid will be\n                \n                [['L', 'H', 'E', 'O'],\n                ['W', 'L', 'L', 'O'],\n                ['O', 'R', 'D', 'V'],\n                ['E', '#', '#', 'L']]\n                \n        - Once the grid is filled, read the contents of the grid line by line as the encrypted message for that block.\n        - When moving on to the next block, empty the contents of the grid and redo the entire fill and rotate operation.\n        \n        Finally, the encrypted messages of all the blocks are connected together to be the final ciphertext.\n        \n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters  string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Grid and Template (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into blocks of 16 letters one by one.\n    - For each block, perform the following operations:\n        - Write the 16 letters row by row to fill the grid.\n        - Place the template with holes on the filled grid.\n        - Read the letters visible through the holes to get a portion of the plaintext message.\n        - Since there are only four holes, at this point you need to rotate the template 90 degrees counterclockwise to read the next set of letters.\n        - Repeat the reading step four times to get this fast decrypted message.\n    - Connecting the decrypted messages of all the blocks gives the final plaintext.\nencrypted text: TULBIPNOLIYNC##A decode to clear text: nonculpability\nencrypted text: #TI##CPY###RI### decode to clear text: pyritic\nencrypted text: #TI##CPY###RI### decode to clear text: pyritic\nencrypted text: #II###NE###RV### decode to clear text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 34, "source_filename": "icl_with_rule_decode_Kor_rule18_RotatingGridCipher_en.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "#II###NE###RV###", "extra_args": {}, "output": "使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始解密过程:\n\n处理第1个块:\n当前块的密文: #II###NE###RV###\n\n重构的网格:\n[['#' 'I' 'I' '#']\n ['#' '#' 'N' 'E']\n ['#' '#' '#' 'R']\n ['V' '#' '#' '#']]\n\n第1次旋转后读取:\n从位置(1,2)读取字符: N\n从位置(1,3)读取字符: E\n从位置(2,3)读取字符: R\n从位置(3,0)读取字符: V\n\n第2次旋转后读取:\n从位置(0,1)读取字符: I\n从位置(0,2)读取字符: I\n从位置(1,1)读取字符: #\n从位置(3,3)读取字符: #\n\n第3次旋转后读取:\n从位置(0,3)读取字符: #\n从位置(1,0)读取字符: #\n从位置(2,0)读取字符: #\n从位置(2,1)读取字符: #\n\n第4次旋转后读取:\n从位置(0,0)读取字符: #\n从位置(2,2)读取字符: #\n从位置(3,1)读取字符: #\n从位置(3,2)读取字符: #\n当前块解密结果: NERVII##########\n\n最终明文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - Grid and Template:\n        - Prepare a blank grid and a template with holes (fence).\n        - The template used is:\n            \n            ▮  ▮  ▮  ▮\n            ▮  ▮  ▯  ▯\n            ▮  ▮  ▮  ▯\n            ▯  ▮  ▮  ▮\n            \n            Where the white ones are holes, place the template on a blank grid, only through the white holes, and fill in the corresponding positions of the grid with characters.\n            \n- Encryption Steps:\n    - Divide the plaintext into blocks of 16 letters one by one.(If the plaintext is less than 16 long it is one  block)\n        - For example, the plaintext message \"LOVEHELLOWORLD\" is less than 16 and is one block.\n    - For each block, perform the following encryption operation:\n        - Place the template with holes on the blank grid.\n        - Fill the grid with the letters of the plaintext message in order through the holes in the template.\n            - For example, the plaintext is \"LOVEHELLOWORLD\", through the four holes, after filling in the LOVE grid is\n                \n                [['', '', '', ''],\n                ['', '', 'L', 'O'],\n                ['', '', '', 'V'],\n                ['E', '', '', '']]\n                \n        - The template has a total of four holes, so after filling out the four letters, it can not be filled out, then you need to rotate the template counterclockwise 90 degrees.\n            - For example, after filling in LOVE after rotating the template after the template for:\n                \n                ▮  ▯  ▯  ▮\n                ▮  ▯  ▮  ▮\n                ▮  ▮  ▮  ▮\n                ▮  ▮  ▮  ▯\n                \n        - Repeat filling the visible holes in the grid with the next letter of the plaintext message and rotating the template until the entire grid is completely filled. This will perform 4 fills + rotations and eventually the template will turn to the original template. If the message is not enough to fill the entire grid, it is made up with filler characters (e.g. '#').\n            - For example, if the plain text is \"LOVEHELLOWORLD\", after several fills and rotations, the final grid will be\n                \n                [['L', 'H', 'E', 'O'],\n                ['W', 'L', 'L', 'O'],\n                ['O', 'R', 'D', 'V'],\n                ['E', '#', '#', 'L']]\n                \n        - Once the grid is filled, read the contents of the grid line by line as the encrypted message for that block.\n        - When moving on to the next block, empty the contents of the grid and redo the entire fill and rotate operation.\n        \n        Finally, the encrypted messages of all the blocks are connected together to be the final ciphertext.\n        \n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters  string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Grid and Template (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into blocks of 16 letters one by one.\n    - For each block, perform the following operations:\n        - Write the 16 letters row by row to fill the grid.\n        - Place the template with holes on the filled grid.\n        - Read the letters visible through the holes to get a portion of the plaintext message.\n        - Since there are only four holes, at this point you need to rotate the template 90 degrees counterclockwise to read the next set of letters.\n        - Repeat the reading step four times to get this fast decrypted message.\n    - Connecting the decrypted messages of all the blocks gives the final plaintext.\ncipher text: #TI##CPY###RI### decrypt to clear text: pyritic\ncipher text: TULBIPNOLIYNC##A decrypt to clear text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 35, "source_filename": "icl_with_rule_decode_Kor_rule18_RotatingGridCipher_en.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "TULBIPNOLIYNC##A", "extra_args": {}, "output": "使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始解密过程:\n\n处理第1个块:\n当前块的密文: TULBIPNOLIYNC##A\n\n重构的网格:\n[['T' 'U' 'L' 'B']\n ['I' 'P' 'N' 'O']\n ['L' 'I' 'Y' 'N']\n ['C' '#' '#' 'A']]\n\n第1次旋转后读取:\n从位置(1,2)读取字符: N\n从位置(1,3)读取字符: O\n从位置(2,3)读取字符: N\n从位置(3,0)读取字符: C\n\n第2次旋转后读取:\n从位置(0,1)读取字符: U\n从位置(0,2)读取字符: L\n从位置(1,1)读取字符: P\n从位置(3,3)读取字符: A\n\n第3次旋转后读取:\n从位置(0,3)读取字符: B\n从位置(1,0)读取字符: I\n从位置(2,0)读取字符: L\n从位置(2,1)读取字符: I\n\n第4次旋转后读取:\n从位置(0,0)读取字符: T\n从位置(2,2)读取字符: Y\n从位置(3,1)读取字符: #\n从位置(3,2)读取字符: #\n当前块解密结果: NONCULPABILITY##\n\n最终明文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - Grid and Template:\n        - Prepare a blank grid and a template with holes (fence).\n        - The template used is:\n            \n            ▮  ▮  ▮  ▮\n            ▮  ▮  ▯  ▯\n            ▮  ▮  ▮  ▯\n            ▯  ▮  ▮  ▮\n            \n            Where the white ones are holes, place the template on a blank grid, only through the white holes, and fill in the corresponding positions of the grid with characters.\n            \n- Encryption Steps:\n    - Divide the plaintext into blocks of 16 letters one by one.(If the plaintext is less than 16 long it is one  block)\n        - For example, the plaintext message \"LOVEHELLOWORLD\" is less than 16 and is one block.\n    - For each block, perform the following encryption operation:\n        - Place the template with holes on the blank grid.\n        - Fill the grid with the letters of the plaintext message in order through the holes in the template.\n            - For example, the plaintext is \"LOVEHELLOWORLD\", through the four holes, after filling in the LOVE grid is\n                \n                [['', '', '', ''],\n                ['', '', 'L', 'O'],\n                ['', '', '', 'V'],\n                ['E', '', '', '']]\n                \n        - The template has a total of four holes, so after filling out the four letters, it can not be filled out, then you need to rotate the template counterclockwise 90 degrees.\n            - For example, after filling in LOVE after rotating the template after the template for:\n                \n                ▮  ▯  ▯  ▮\n                ▮  ▯  ▮  ▮\n                ▮  ▮  ▮  ▮\n                ▮  ▮  ▮  ▯\n                \n        - Repeat filling the visible holes in the grid with the next letter of the plaintext message and rotating the template until the entire grid is completely filled. This will perform 4 fills + rotations and eventually the template will turn to the original template. If the message is not enough to fill the entire grid, it is made up with filler characters (e.g. '#').\n            - For example, if the plain text is \"LOVEHELLOWORLD\", after several fills and rotations, the final grid will be\n                \n                [['L', 'H', 'E', 'O'],\n                ['W', 'L', 'L', 'O'],\n                ['O', 'R', 'D', 'V'],\n                ['E', '#', '#', 'L']]\n                \n        - Once the grid is filled, read the contents of the grid line by line as the encrypted message for that block.\n        - When moving on to the next block, empty the contents of the grid and redo the entire fill and rotate operation.\n        \n        Finally, the encrypted messages of all the blocks are connected together to be the final ciphertext.\n        \n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters  string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Grid and Template (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into blocks of 16 letters one by one.\n    - For each block, perform the following operations:\n        - Write the 16 letters row by row to fill the grid.\n        - Place the template with holes on the filled grid.\n        - Read the letters visible through the holes to get a portion of the plaintext message.\n        - Since there are only four holes, at this point you need to rotate the template 90 degrees counterclockwise to read the next set of letters.\n        - Repeat the reading step four times to get this fast decrypted message.\n    - Connecting the decrypted messages of all the blocks gives the final plaintext.\ncipher text: TULBIPNOLIYNC##A decrypt to original information: nonculpability\ncipher text: #II###NE###RV### decrypt to original information: nervii\ncipher text: #TI##CPY###RI### decrypt to original information: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 36, "source_filename": "icl_with_rule_decode_Kor_rule18_RotatingGridCipher_en.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "#TI##CPY###RI###", "extra_args": {}, "output": "使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始解密过程:\n\n处理第1个块:\n当前块的密文: #TI##CPY###RI###\n\n重构的网格:\n[['#' 'T' 'I' '#']\n ['#' 'C' 'P' 'Y']\n ['#' '#' '#' 'R']\n ['I' '#' '#' '#']]\n\n第1次旋转后读取:\n从位置(1,2)读取字符: P\n从位置(1,3)读取字符: Y\n从位置(2,3)读取字符: R\n从位置(3,0)读取字符: I\n\n第2次旋转后读取:\n从位置(0,1)读取字符: T\n从位置(0,2)读取字符: I\n从位置(1,1)读取字符: C\n从位置(3,3)读取字符: #\n\n第3次旋转后读取:\n从位置(0,3)读取字符: #\n从位置(1,0)读取字符: #\n从位置(2,0)读取字符: #\n从位置(2,1)读取字符: #\n\n第4次旋转后读取:\n从位置(0,0)读取字符: #\n从位置(2,2)读取字符: #\n从位置(3,1)读取字符: #\n从位置(3,2)读取字符: #\n当前块解密结果: PYRITIC#########\n\n最终明文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (矩阵中的行和列从0开始计数)\n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'], \n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'], \n        ['6', 'J', 'V', 'O', 'E', 'M']]\n    - 加密字符集: \"ADFGVX\"\n- 加密步骤:\n    - 对明文中的每个字符:\n        - 在6X6矩阵中找到该字符的行数和列数。例如，A的行数为0，列数为2。\n        - 加密字符集中的字符位置标记为0-6。使用加密字符集中对应位置的字符替换行数和列数得到加密二元组。\n            - 例如，A的行数为0对应加密字符集中的A，列数为2对应加密字符集中的F，所以A的加密二元组为(A,F)。\n        - 读取所有加密二元组的行，然后读取所有加密二元组的列得到最终加密文本。\n            - 例如，加密二元组为(A,F)(X,V)，最后读取为AXFV，所以最终加密文本为AXFV。\n明文: nervii 编码为加密文本: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 37, "source_filename": "icl_with_rule_encode_Kor_rule19_ADFGVX_cn.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n\n加密过程:\n1. 获取每个字符的行列位置并转换为加密字符:\n字符 N 在矩阵中的位置是 (4,5)，转换为加密二元组 (V,X)\n字符 E 在矩阵中的位置是 (5,4)，转换为加密二元组 (X,V)\n字符 R 在矩阵中的位置是 (0,0)，转换为加密二元组 (A,A)\n字符 V 在矩阵中的位置是 (5,2)，转换为加密二元组 (X,F)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n\n2. 合并所有加密二元组的行和列:\n先读取所有行: VXAXVV\n再读取所有列: XVAFFF\n最终密文: VXAXVVXVAFFF\n", "ground_truth": "VXAXVVXVAFFF"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 隐文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (矩阵中的行和列从0开始计数)\n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'], \n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'], \n        ['6', 'J', 'V', 'O', 'E', 'M']]\n    - 加密字符集: \"ADFGVX\"\n- 加密步骤:\n    - 对明文中的每个字符:\n        - 在6X6矩阵中找到该字符的行数和列数。例如，A的行数为0，列数为2。\n        - 加密字符集中的字符位置标记为0-6。使用加密字符集中对应位置的字符替换行数和列数得到加密二元组。\n            - 例如，A的行数为0对应加密字符集中的A，列数为2对应加密字符集中的F，所以A的加密二元组为(A,F)。\n        - 读取所有加密二元组的行，然后读取所有加密二元组的列得到最终隐文。\n            - 例如，加密二元组为(A,F)(X,V)，最后读取为AXFV，所以最终隐文为AXFV。\n明文: nervii 加密成为隐文: VXAXVVXVAFFF\n明文: nonculpability 加密成为隐文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 38, "source_filename": "icl_with_rule_encode_Kor_rule19_ADFGVX_cn.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n\n加密过程:\n1. 获取每个字符的行列位置并转换为加密字符:\n字符 N 在矩阵中的位置是 (4,5)，转换为加密二元组 (V,X)\n字符 O 在矩阵中的位置是 (5,3)，转换为加密二元组 (X,G)\n字符 N 在矩阵中的位置是 (4,5)，转换为加密二元组 (V,X)\n字符 C 在矩阵中的位置是 (3,1)，转换为加密二元组 (G,D)\n字符 U 在矩阵中的位置是 (0,1)，转换为加密二元组 (A,D)\n字符 L 在矩阵中的位置是 (3,5)，转换为加密二元组 (G,X)\n字符 P 在矩阵中的位置是 (4,4)，转换为加密二元组 (V,V)\n字符 A 在矩阵中的位置是 (0,2)，转换为加密二元组 (A,F)\n字符 B 在矩阵中的位置是 (0,5)，转换为加密二元组 (A,X)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 L 在矩阵中的位置是 (3,5)，转换为加密二元组 (G,X)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 T 在矩阵中的位置是 (2,4)，转换为加密二元组 (F,V)\n字符 Y 在矩阵中的位置是 (3,0)，转换为加密二元组 (G,A)\n\n2. 合并所有加密二元组的行和列:\n先读取所有行: VXVGAGVAAVGVFG\n再读取所有列: XGXDDXVFXFXFVA\n最终密文: VXVGAGVAAVGVFGXGXDDXVFXFXFVA\n", "ground_truth": "VXVGAGVAAVGVFGXGXDDXVFXFXFVA"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原文进行加密\n加密规则:\n- 输入:\n    - 原文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (矩阵中的行和列从0开始计数)\n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'], \n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'], \n        ['6', 'J', 'V', 'O', 'E', 'M']]\n    - 加密字符集: \"ADFGVX\"\n- 加密步骤:\n    - 对原文中的每个字符:\n        - 在6X6矩阵中找到该字符的行数和列数。例如，A的行数为0，列数为2。\n        - 加密字符集中的字符位置标记为0-6。使用加密字符集中对应位置的字符替换行数和列数得到加密二元组。\n            - 例如，A的行数为0对应加密字符集中的A，列数为2对应加密字符集中的F，所以A的加密二元组为(A,F)。\n        - 读取所有加密二元组的行，然后读取所有加密二元组的列得到最终暗码。\n            - 例如，加密二元组为(A,F)(X,V)，最后读取为AXFV，所以最终暗码为AXFV。\n原文: nonculpability 加密成暗码: VXVGAGVAAVGVFGXGXDDXVFXFXFVA\n原文: nervii 加密成暗码: VXAXVVXVAFFF\n原文: pyritic 加密成暗码是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 39, "source_filename": "icl_with_rule_encode_Kor_rule19_ADFGVX_cn.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n\n加密过程:\n1. 获取每个字符的行列位置并转换为加密字符:\n字符 P 在矩阵中的位置是 (4,4)，转换为加密二元组 (V,V)\n字符 Y 在矩阵中的位置是 (3,0)，转换为加密二元组 (G,A)\n字符 R 在矩阵中的位置是 (0,0)，转换为加密二元组 (A,A)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 T 在矩阵中的位置是 (2,4)，转换为加密二元组 (F,V)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 C 在矩阵中的位置是 (3,1)，转换为加密二元组 (G,D)\n\n2. 合并所有加密二元组的行和列:\n先读取所有行: VGAVFVG\n再读取所有列: VAAFVFD\n最终密文: VGAVFVGVAAFVFD\n", "ground_truth": "VGAVFVGVAAFVFD"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids:\n        \n        Grid0:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid1:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid2:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n        Grid3:\n        \n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        \n        Grid4:\n        \n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        \n        Grid5:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid6:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid7:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n- Encryption Steps:\n    - The plaintext is grouped into blocks of 5 characters, numbered from 0.\n    - For blocks of 5 characters:\n        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.\n        - For each character in the current block:\n            - If the character is \"J\", it is not encrypted and is appended directly to the encrypted block.\n            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.\n            - Add the encrypted letter to the encryption block.\n    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into 5-character blocks:\n        - For example, if the cipher text is \"KHOORTQHQTHHSAUQ\", the 0th block is \"KHOOR\", the 1st block is \"TQHQH\", and so on. (Numbering starts at 0)\n    - Determine the grid used for the current block:\n    Calculate grid_index = (i // 5) % 8 to select the appropriate grid from the grid list. i is the block number.\n    - For each character in the block:\n        - If the character is \"J\": add \"J\" directly to the decrypted block without decryption.\n        - Otherwise find the position of the character in the grid and obtain it by moving one grid to the upper left (or continue on the opposite side of the corresponding boundary if it crosses the boundary). the letter at the position after the move is used as the decrypted letter.\n        - Append the decrypted letter to the decryption block.\n    - After processing all characters in the block, append the decrypted block to the decrypted message list. Form the final decrypted message.\nencrypted text: OWOMVEBGKDEDZS decode into original information: nonculpability\nencrypted text: OWOMVEBGKDEDZS decode into original information: nonculpability\nencrypted text: OWOMVEBGKDEDZS decode into original information: nonculpability\nencrypted text: OFYHDD decode into original information: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 40, "source_filename": "icl_with_rule_decode_Kor_rule7_PhillipsFigureCipher_en.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "OFYHDD", "extra_args": {}, "output": "需要解密的文本: OFYHDD\n\n处理第0个块: OFYHD\n使用第0号网格\n字母O在位置(3,0)，向左上移动一格得到N\n字母F在位置(2,0)，向左上移动一格得到E\n字母Y在位置(4,3)，向左上移动一格得到R\n字母H在位置(0,1)，向左上移动一格得到V\n字母D在位置(1,3)，向左上移动一格得到I\n解密后的块: NERVI\n\n处理第1个块: D\n使用第1号网格\n字母D在位置(2,3)，向左上移动一格得到I\n解密后的块: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids:\n        \n        Grid0:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid1:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid2:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n        Grid3:\n        \n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        \n        Grid4:\n        \n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        \n        Grid5:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid6:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid7:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n- Encryption Steps:\n    - The plaintext is grouped into blocks of 5 characters, numbered from 0.\n    - For blocks of 5 characters:\n        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.\n        - For each character in the current block:\n            - If the character is \"J\", it is not encrypted and is appended directly to the encrypted block.\n            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.\n            - Add the encrypted letter to the encryption block.\n    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into 5-character blocks:\n        - For example, if the cipher text is \"KHOORTQHQTHHSAUQ\", the 0th block is \"KHOOR\", the 1st block is \"TQHQH\", and so on. (Numbering starts at 0)\n    - Determine the grid used for the current block:\n    Calculate grid_index = (i // 5) % 8 to select the appropriate grid from the grid list. i is the block number.\n    - For each character in the block:\n        - If the character is \"J\": add \"J\" directly to the decrypted block without decryption.\n        - Otherwise find the position of the character in the grid and obtain it by moving one grid to the upper left (or continue on the opposite side of the corresponding boundary if it crosses the boundary). the letter at the position after the move is used as the decrypted letter.\n        - Append the decrypted letter to the decryption block.\n    - After processing all characters in the block, append the decrypted block to the decrypted message list. Form the final decrypted message.\nencoded text: OFYHDD decrypt to clear text: nervii\nencoded text: OFYHDD decrypt to clear text: nervii\nencoded text: OWOMVEBGKDEDZS decrypt to clear text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 41, "source_filename": "icl_with_rule_decode_Kor_rule7_PhillipsFigureCipher_en.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "OWOMVEBGKDEDZS", "extra_args": {}, "output": "需要解密的文本: OWOMVEBGKDEDZS\n\n处理第0个块: OWOMV\n使用第0号网格\n字母O在位置(3,0)，向左上移动一格得到N\n字母W在位置(4,1)，向左上移动一格得到O\n字母O在位置(3,0)，向左上移动一格得到N\n字母M在位置(2,3)，向左上移动一格得到C\n字母V在位置(4,0)，向左上移动一格得到U\n解密后的块: NONCU\n\n处理第1个块: EBGKD\n使用第1号网格\n字母E在位置(2,4)，向左上移动一格得到L\n字母B在位置(2,1)，向左上移动一格得到P\n字母G在位置(3,1)，向左上移动一格得到A\n字母K在位置(3,2)，向左上移动一格得到B\n字母D在位置(2,3)，向左上移动一格得到I\n解密后的块: LPABI\n\n处理第2个块: EDZS\n使用第2号网格\n字母E在位置(3,4)，向左上移动一格得到L\n字母D在位置(3,3)，向左上移动一格得到I\n字母Z在位置(1,4)，向左上移动一格得到T\n字母S在位置(2,4)，向左上移动一格得到Y\n解密后的块: LITY\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids:\n        \n        Grid0:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid1:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid2:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n        Grid3:\n        \n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        \n        Grid4:\n        \n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        \n        Grid5:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid6:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid7:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n- Encryption Steps:\n    - The plaintext is grouped into blocks of 5 characters, numbered from 0.\n    - For blocks of 5 characters:\n        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.\n        - For each character in the current block:\n            - If the character is \"J\", it is not encrypted and is appended directly to the encrypted block.\n            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.\n            - Add the encrypted letter to the encryption block.\n    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into 5-character blocks:\n        - For example, if the cipher text is \"KHOORTQHQTHHSAUQ\", the 0th block is \"KHOOR\", the 1st block is \"TQHQH\", and so on. (Numbering starts at 0)\n    - Determine the grid used for the current block:\n    Calculate grid_index = (i // 5) % 8 to select the appropriate grid from the grid list. i is the block number.\n    - For each character in the block:\n        - If the character is \"J\": add \"J\" directly to the decrypted block without decryption.\n        - Otherwise find the position of the character in the grid and obtain it by moving one grid to the upper left (or continue on the opposite side of the corresponding boundary if it crosses the boundary). the letter at the position after the move is used as the decrypted letter.\n        - Append the decrypted letter to the decryption block.\n    - After processing all characters in the block, append the decrypted block to the decrypted message list. Form the final decrypted message.\nencrypted text: OFYHDD decrypt into plain text: nervii\nencrypted text: BSYDZDM decrypt into plain text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 42, "source_filename": "icl_with_rule_decode_Kor_rule7_PhillipsFigureCipher_en.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "BSYDZDM", "extra_args": {}, "output": "需要解密的文本: BSYDZDM\n\n处理第0个块: BSYDZ\n使用第0号网格\n字母B在位置(1,1)，向左上移动一格得到P\n字母S在位置(0,4)，向左上移动一格得到Y\n字母Y在位置(4,3)，向左上移动一格得到R\n字母D在位置(1,3)，向左上移动一格得到I\n字母Z在位置(4,4)，向左上移动一格得到T\n解密后的块: PYRIT\n\n处理第1个块: DM\n使用第1号网格\n字母D在位置(2,3)，向左上移动一格得到I\n字母M在位置(3,3)，向左上移动一格得到C\n解密后的块: IC\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - e: 263\n    - n: 299\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x of the value corresponding to the ascii code of p.\n        - Compute x^e mod n as the ciphertext number y for that letter p . Here ^ denotes the multiplication operation.\n    - Finally, concatenate all the y's and separate them with commas for the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - e: 263\n    - n: 299\n- Decryption Steps:\n    - For each number c in the ciphertext:\n        - Calculate z = c^e mod n. Here ^ means multiplication.\n        Based on the decimal number represented by z, use the ascii code to find the corresponding letter as the plaintext letter p.\n    - Finally, concatenate all p's to get the final plaintext.\nencoded text: 156,23,62,226,213,213 decrypt into original information is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 43, "source_filename": "icl_with_rule_decode_Kor_rule23_RSACipher_en.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "156,23,62,226,213,213", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 156,23,62,226,213,213\n使用参数: e=263, n=299\n分割后的数字序列: ['156', '23', '62', '226', '213', '213']\n\n逐数字解密:\n计算 156^263 mod 299 = 78\n对应的字符为: N\n计算 23^263 mod 299 = 69\n对应的字符为: E\n计算 62^263 mod 299 = 82\n对应的字符为: R\n计算 226^263 mod 299 = 86\n对应的字符为: V\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 213^263 mod 299 = 73\n对应的字符为: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - e: 263\n    - n: 299\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x of the value corresponding to the ascii code of p.\n        - Compute x^e mod n as the ciphertext number y for that letter p . Here ^ denotes the multiplication operation.\n    - Finally, concatenate all the y's and separate them with commas for the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - e: 263\n    - n: 299\n- Decryption Steps:\n    - For each number c in the ciphertext:\n        - Calculate z = c^e mod n. Here ^ means multiplication.\n        Based on the decimal number represented by z, use the ascii code to find the corresponding letter as the plaintext letter p.\n    - Finally, concatenate all p's to get the final plaintext.\nencoded text: 228,84,62,213,89,213,241 decode into original information: pyritic\nencoded text: 228,84,62,213,89,213,241 decode into original information: pyritic\nencoded text: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 decode into original information: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 44, "source_filename": "icl_with_rule_decode_Kor_rule23_RSACipher_en.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "156,53,156,241,197,240,228,247,222,213,240,213,89,84", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 156,53,156,241,197,240,228,247,222,213,240,213,89,84\n使用参数: e=263, n=299\n分割后的数字序列: ['156', '53', '156', '241', '197', '240', '228', '247', '222', '213', '240', '213', '89', '84']\n\n逐数字解密:\n计算 156^263 mod 299 = 78\n对应的字符为: N\n计算 53^263 mod 299 = 79\n对应的字符为: O\n计算 156^263 mod 299 = 78\n对应的字符为: N\n计算 241^263 mod 299 = 67\n对应的字符为: C\n计算 197^263 mod 299 = 85\n对应的字符为: U\n计算 240^263 mod 299 = 76\n对应的字符为: L\n计算 228^263 mod 299 = 80\n对应的字符为: P\n计算 247^263 mod 299 = 65\n对应的字符为: A\n计算 222^263 mod 299 = 66\n对应的字符为: B\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 240^263 mod 299 = 76\n对应的字符为: L\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 89^263 mod 299 = 84\n对应的字符为: T\n计算 84^263 mod 299 = 89\n对应的字符为: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - e: 263\n    - n: 299\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x of the value corresponding to the ascii code of p.\n        - Compute x^e mod n as the ciphertext number y for that letter p . Here ^ denotes the multiplication operation.\n    - Finally, concatenate all the y's and separate them with commas for the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - e: 263\n    - n: 299\n- Decryption Steps:\n    - For each number c in the ciphertext:\n        - Calculate z = c^e mod n. Here ^ means multiplication.\n        Based on the decimal number represented by z, use the ascii code to find the corresponding letter as the plaintext letter p.\n    - Finally, concatenate all p's to get the final plaintext.\nencrypted text: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 decrypt into original information: nonculpability\nencrypted text: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 decrypt into original information: nonculpability\nencrypted text: 156,53,156,241,197,240,228,247,222,213,240,213,89,84 decrypt into original information: nonculpability\nencrypted text: 228,84,62,213,89,213,241 decrypt into original information is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 45, "source_filename": "icl_with_rule_decode_Kor_rule23_RSACipher_en.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "228,84,62,213,89,213,241", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 228,84,62,213,89,213,241\n使用参数: e=263, n=299\n分割后的数字序列: ['228', '84', '62', '213', '89', '213', '241']\n\n逐数字解密:\n计算 228^263 mod 299 = 80\n对应的字符为: P\n计算 84^263 mod 299 = 89\n对应的字符为: Y\n计算 62^263 mod 299 = 82\n对应的字符为: R\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 89^263 mod 299 = 84\n对应的字符为: T\n计算 213^263 mod 299 = 73\n对应的字符为: I\n计算 241^263 mod 299 = 67\n对应的字符为: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - period: Defines how often the inner disc rotates. Periodicity indicates that after every number of characters processed in the encryption process, the inner disc will rotate once according to the incremental value.\n    - increment: Defines the number of characters the inner disc rotates each time. At the end of each cycle, the inner disc will rotate to the right by the corresponding number of characters based on the increment value. For example, if the increment is 4, the inner disc will rotate 4 characters to the right for each cycle that passes (e.g., 5 characters are processed).\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Encryption Steps:\n    - For each character p in the plaintext:\n        - Find the character in the outer_disk.\n        - Replace it with the character at the corresponding position on the inner_disk.\n        - After encrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\n\n \nclear text: pyritic encode into cipher text: MCZBXSU\nclear text: pyritic encode into cipher text: MCZBXSU\nclear text: nonculpability encode into cipher text: YNYEVZFGOSBHDF\nclear text: nervii encode into cipher text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 46, "source_filename": "icl_with_rule_encode_Kor_rule9_AlertiCipher_en.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n加密过程:\n字符 N 在外圈位置 22 对应内圈字符 Y\n字符 E 在外圈位置 2 对应内圈字符 L\n字符 R 在外圈位置 3 对应内圈字符 Z\n字符 V 在外圈位置 20 对应内圈字符 R\n字符 I 在外圈位置 7 对应内圈字符 B\n已处理5个字符，内圈向右旋转4个位置\n字符 I 在外圈位置 7 对应内圈字符 S\n\n最终加密结果: YLZRBS\n", "ground_truth": "YLZRBS"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - period: Defines how often the inner disc rotates. Periodicity indicates that after every number of characters processed in the encryption process, the inner disc will rotate once according to the incremental value.\n    - increment: Defines the number of characters the inner disc rotates each time. At the end of each cycle, the inner disc will rotate to the right by the corresponding number of characters based on the increment value. For example, if the increment is 4, the inner disc will rotate 4 characters to the right for each cycle that passes (e.g., 5 characters are processed).\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Encryption Steps:\n    - For each character p in the plaintext:\n        - Find the character in the outer_disk.\n        - Replace it with the character at the corresponding position on the inner_disk.\n        - After encrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\n\n \nplain text: nervii encode into encoded text: YLZRBS\nplain text: nonculpability encode into encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 47, "source_filename": "icl_with_rule_encode_Kor_rule9_AlertiCipher_en.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n加密过程:\n字符 N 在外圈位置 22 对应内圈字符 Y\n字符 O 在外圈位置 8 对应内圈字符 N\n字符 N 在外圈位置 22 对应内圈字符 Y\n字符 C 在外圈位置 19 对应内圈字符 E\n字符 U 在外圈位置 6 对应内圈字符 V\n已处理5个字符，内圈向右旋转4个位置\n字符 L 在外圈位置 25 对应内圈字符 Z\n字符 P 在外圈位置 9 对应内圈字符 F\n字符 A 在外圈位置 10 对应内圈字符 G\n字符 B 在外圈位置 21 对应内圈字符 O\n字符 I 在外圈位置 7 对应内圈字符 S\n已处理5个字符，内圈向右旋转4个位置\n字符 L 在外圈位置 25 对应内圈字符 B\n字符 I 在外圈位置 7 对应内圈字符 H\n字符 T 在外圈位置 4 对应内圈字符 D\n字符 Y 在外圈位置 5 对应内圈字符 F\n\n最终加密结果: YNYEVZFGOSBHDF\n", "ground_truth": "YNYEVZFGOSBHDF"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - period: Defines how often the inner disc rotates. Periodicity indicates that after every number of characters processed in the encryption process, the inner disc will rotate once according to the incremental value.\n    - increment: Defines the number of characters the inner disc rotates each time. At the end of each cycle, the inner disc will rotate to the right by the corresponding number of characters based on the increment value. For example, if the increment is 4, the inner disc will rotate 4 characters to the right for each cycle that passes (e.g., 5 characters are processed).\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Encryption Steps:\n    - For each character p in the plaintext:\n        - Find the character in the outer_disk.\n        - Replace it with the character at the corresponding position on the inner_disk.\n        - After encrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\n\n \noriginal information: pyritic encrypt into encrypted text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 48, "source_filename": "icl_with_rule_encode_Kor_rule9_AlertiCipher_en.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n加密过程:\n字符 P 在外圈位置 9 对应内圈字符 M\n字符 Y 在外圈位置 5 对应内圈字符 C\n字符 R 在外圈位置 3 对应内圈字符 Z\n字符 I 在外圈位置 7 对应内圈字符 B\n字符 T 在外圈位置 4 对应内圈字符 X\n已处理5个字符，内圈向右旋转4个位置\n字符 I 在外圈位置 7 对应内圈字符 S\n字符 C 在外圈位置 19 对应内圈字符 U\n\n最终加密结果: MCZBXSU\n", "ground_truth": "MCZBXSU"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n原始信息: pyritic 加密成暗文: OCAX..C\n原始信息: nervii 加密成暗文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 49, "source_filename": "icl_encode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "NERVII", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NERVII\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  E -> 层: 1, 列: 2, 行: 2\n  R -> 层: 2, 列: 3, 行: 3\n  V -> 层: 3, 列: 1, 行: 2\n  I -> 层: 1, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\nStep 3: 合并坐标:\n  层坐标: 212311\n  列坐标: 223133\n  行坐标: 223233\nStep 4: 重组坐标: ['212', '311', '223', '133', '223', '233']\nStep 5: 根据新坐标获取密文:\n  212 -> M\n  311 -> S\n  223 -> Q\n  133 -> I\n  223 -> Q\n  233 -> R\n", "ground_truth": "MSQIQR"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n初始文本: nervii 加密为暗文: MSQIQR\n初始文本: pyritic 加密为暗文: OCAX..C\n初始文本: nonculpability 加密为暗文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 50, "source_filename": "icl_encode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NONCULPABILITY\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  O -> 层: 2, 列: 3, 行: 2\n  N -> 层: 2, 列: 2, 行: 2\n  C -> 层: 1, 列: 3, 行: 1\n  U -> 层: 3, 列: 3, 行: 1\n  L -> 层: 2, 列: 3, 行: 1\n  P -> 层: 2, 列: 1, 行: 3\n  A -> 层: 1, 列: 1, 行: 1\n  B -> 层: 1, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  L -> 层: 2, 列: 3, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  Y -> 层: 3, 列: 1, 行: 3\nStep 3: 合并坐标:\n  层坐标: 22213221112133\n  列坐标: 23233311233321\n  行坐标: 22211131131313\nStep 4: 重组坐标: ['222', '132', '211', '121', '332', '323', '331', '123', '332', '122', '211', '131', '131', '313']\nStep 5: 根据新坐标获取密文:\n  222 -> N\n  132 -> F\n  211 -> J\n  121 -> B\n  332 -> X\n  323 -> Z\n  331 -> U\n  123 -> H\n  332 -> X\n  122 -> E\n  211 -> J\n  131 -> C\n  131 -> C\n  313 -> Y\n", "ground_truth": "NFJBXZUHXEJCCY"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n非加密信息: nonculpability 加密为密文: NFJBXZUHXEJCCY\n非加密信息: nonculpability 加密为密文: NFJBXZUHXEJCCY\n非加密信息: nonculpability 加密为密文: NFJBXZUHXEJCCY\n非加密信息: pyritic 加密为密文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 51, "source_filename": "icl_encode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "PYRITIC", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: PYRITIC\nStep 2: 获取每个字符的坐标:\n  P -> 层: 2, 列: 1, 行: 3\n  Y -> 层: 3, 列: 1, 行: 3\n  R -> 层: 2, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  C -> 层: 1, 列: 3, 行: 1\nStep 3: 合并坐标:\n  层坐标: 2321311\n  列坐标: 1133233\n  行坐标: 3333131\nStep 4: 重组坐标: ['232', '131', '111', '332', '333', '333', '131']\nStep 5: 根据新坐标获取密文:\n  232 -> O\n  131 -> C\n  111 -> A\n  332 -> X\n  333 -> .\n  333 -> .\n  131 -> C\n", "ground_truth": "OCAX..C"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nclear text: nonculpability secret key or extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} encode to encoded text: JK FI LL KS CE GE NY\nclear text: pyritic secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode to encoded text: RV RB RD WX\nclear text: pyritic secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode to encoded text: RV RB RD WX\nclear text: pyritic secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode to encoded text: RV RB RD WX\nclear text: pyritic secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode to encoded text: RV RB RD WX\nclear text: nervii secret key or extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} encode to encoded text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 52, "source_filename": "icl_encode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "NERVII", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将明文分成两个字母一组: ['NE', 'RV', 'II']\n\nStep 4.1: 加密字母对 NE\n  - N 在 M_T 中的位置: (2, 3)\n  - E 在 M_T 中的位置: (0, 4)\n  - 交换列索引后新位置: (2, 4) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (0, 3) 的字母: V\n  - 加密结果为: JV\n\nStep 4.2: 加密字母对 RV\n  - R 在 M_T 中的位置: (3, 1)\n  - V 在 M_T 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 1)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (4, 1) 的字母: T\n  - 加密结果为: KT\n\nStep 4.3: 加密字母对 II\n  - I 在 M_T 中的位置: (1, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (1, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 3) 的字母: A\n  - 加密结果为: CA\n\n加密完成！最终加密结果为: JV KT CA\n", "ground_truth": "JV KT CA"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\noriginal information: pyritic extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode into cipher text: RV RB RD WX\noriginal information: nervii extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} encode into cipher text: JV KT CA\noriginal information: pyritic extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode into cipher text: RV RB RD WX\noriginal information: pyritic extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode into cipher text: RV RB RD WX\noriginal information: nonculpability extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} encode into cipher text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 53, "source_filename": "icl_encode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "NONCULPABILITY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将明文分成两个字母一组: ['NO', 'NC', 'UL', 'PA', 'BI', 'LI', 'TY']\n\nStep 4.1: 加密字母对 NO\n  - N 在 M_T 中的位置: (2, 3)\n  - O 在 M_T 中的位置: (2, 4)\n  - 交换列索引后新位置: (2, 4) 和 (2, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (2, 3) 的字母: K\n  - 加密结果为: JK\n\nStep 4.2: 加密字母对 NC\n  - N 在 M_T 中的位置: (2, 3)\n  - C 在 M_T 中的位置: (0, 2)\n  - 交换列索引后新位置: (2, 2) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 2) 的字母: F\n  - 在 M2 中查找位置 (0, 3) 的字母: I\n  - 加密结果为: FI\n\nStep 4.3: 加密字母对 UL\n  - U 在 M_T 中的位置: (3, 4)\n  - L 在 M_T 中的位置: (2, 1)\n  - 交换列索引后新位置: (3, 1) 和 (2, 4)\n  - 在 M1 中查找位置 (3, 1) 的字母: L\n  - 在 M2 中查找位置 (2, 4) 的字母: L\n  - 加密结果为: LL\n\nStep 4.4: 加密字母对 PA\n  - P 在 M_T 中的位置: (3, 0)\n  - A 在 M_T 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (0, 0) 的字母: S\n  - 加密结果为: KS\n\nStep 4.5: 加密字母对 BI\n  - B 在 M_T 中的位置: (0, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (0, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (0, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: CE\n\nStep 4.6: 加密字母对 LI\n  - L 在 M_T 中的位置: (2, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (2, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (2, 3) 的字母: G\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: GE\n\nStep 4.7: 加密字母对 TY\n  - T 在 M_T 中的位置: (3, 3)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: N\n  - 在 M2 中查找位置 (4, 3) 的字母: Y\n  - 加密结果为: NY\n\n加密完成！最终加密结果为: JK FI LL KS CE GE NY\n", "ground_truth": "JK FI LL KS CE GE NY"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\noriginal information: nervii secret key or extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} encode to encrypted text: JV KT CA\noriginal information: nonculpability secret key or extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} encode to encrypted text: JK FI LL KS CE GE NY\noriginal information: nervii secret key or extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} encode to encrypted text: JV KT CA\noriginal information: pyritic secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encode to encrypted text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 54, "source_filename": "icl_encode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "PYRITICX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\n明文字符个数为奇数，补一个字符 X\nStep 3: 将明文分成两个字母一组: ['PY', 'RI', 'TI', 'CX']\n\nStep 4.1: 加密字母对 PY\n  - P 在 M_T 中的位置: (3, 0)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 0)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (4, 0) 的字母: V\n  - 加密结果为: RV\n\nStep 4.2: 加密字母对 RI\n  - R 在 M_T 中的位置: (3, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 1) 的字母: B\n  - 加密结果为: RB\n\nStep 4.3: 加密字母对 TI\n  - T 在 M_T 中的位置: (3, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 3) 的字母: D\n  - 加密结果为: RD\n\nStep 4.4: 加密字母对 CX\n  - C 在 M_T 中的位置: (0, 2)\n  - X 在 M_T 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M1 中查找位置 (0, 2) 的字母: W\n  - 在 M2 中查找位置 (4, 2) 的字母: X\n  - 加密结果为: WX\n\n加密完成！最终加密结果为: RV RB RD WX\n", "ground_truth": "RV RB RD WX"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: A string of uppercase letters and spaces.\n- Output:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - S_BOX\n        - S_BOX = {\n        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n        }\n- Encryption Steps:\n    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \\x00 (null character) to make its length a multiple of 8 bytes.\n    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.\n    3. Block Encryption:\n        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.\n        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.\n        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.\n    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.\n    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.\n\n\nclear text: nervii encode into encrypted text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 55, "source_filename": "icl_with_rule_encode_Kor_rule22_SBOXCipher_en.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n1. 规范化输入文本: NERVII\n2. 开始分块处理...\n\n处理第1个块: NERVII\n  - 填充后的块: NERVII\u0000\u0000\n  - 与密钥XOR后: 7F7761627C7F3738\n  - S盒替换后: 0C00DAD7010C505E\n  - 左移置换后: 1800B5AF0218A0BC\n  - 最终与密钥XOR后: 2932869B372E9784\n\n最终加密结果: 2932869B372E9784\n", "ground_truth": "2932869B372E9784"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: A string of uppercase letters and spaces.\n- Output:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - S_BOX\n        - S_BOX = {\n        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n        }\n- Encryption Steps:\n    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \\x00 (null character) to make its length a multiple of 8 bytes.\n    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.\n    3. Block Encryption:\n        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.\n        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.\n        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.\n    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.\n    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.\n\n\noriginal information: nervii encode into encoded text: 2932869B372E9784\noriginal information: nervii encode into encoded text: 2932869B372E9784\noriginal information: nonculpability encode into encoded text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 56, "source_filename": "icl_with_rule_encode_Kor_rule22_SBOXCipher_en.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n1. 规范化输入文本: NONCULPABILITY\n2. 开始分块处理...\n\n处理第1个块: NONCULPA\n  - 填充后的块: NONCULPA\n  - 与密钥XOR后: 7F7D7D77607A6779\n  - S盒替换后: 0C020200DF04D008\n  - 左移置换后: 18040400BF08A110\n  - 最终与密钥XOR后: 293637348A3E9628\n\n处理第2个块: BILITY\n  - 填充后的块: BILITY\u0000\u0000\n  - 与密钥XOR后: 737B7F7D616F3738\n  - S盒替换后: 05060C02DADC505E\n  - 左移置换后: 0A0C1804B5B9A0BC\n  - 最终与密钥XOR后: 3B3E2B30808F9784\n\n最终加密结果: 293637348A3E96283B3E2B30808F9784\n", "ground_truth": "293637348A3E96283B3E2B30808F9784"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: A string of uppercase letters and spaces.\n- Output:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - S_BOX\n        - S_BOX = {\n        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n        }\n- Encryption Steps:\n    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \\x00 (null character) to make its length a multiple of 8 bytes.\n    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.\n    3. Block Encryption:\n        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.\n        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.\n        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.\n    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.\n    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.\n\n\nclear text: pyritic encrypt to encoded text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 57, "source_filename": "icl_with_rule_encode_Kor_rule22_SBOXCipher_en.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n1. 规范化输入文本: PYRITIC\n2. 开始分块处理...\n\n处理第1个块: PYRITIC\n  - 填充后的块: PYRITIC\u0000\n  - 与密钥XOR后: 616B617D617F7438\n  - S盒替换后: DAD6DA02DA0C095E\n  - 左移置换后: B5ADB504B51812BC\n  - 最终与密钥XOR后: 849F8630802E2584\n\n最终加密结果: 849F8630802E2584\n", "ground_truth": "849F8630802E2584"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - standard_alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Encryption Steps:\n    - For each letter p in the given Plaintext:\n    - (1) Use reversed_alphabet for reverse mapping. Find its position in the standard_alphabet and replace it with the letter in the corresponding position in reversed_alphabet. For example, A is mapped to Z and B is mapped to Y.\n    - (2) Move the letter obtained in (1) forward 4 places in the standard_alphabet order. For example, if p=A, after (1) is mapped to Z, then Z is shifted forward 4 positions in the standard_alphabet to get D.\n    - (3) Replace the letter obtained from (2) by finding its position in standard_alphabet and using the corresponding letter in substitution_alphabet, resulting in the final ciphertext letter. For example, if the letter obtained by going through (2) is D, it is mapped as J.\n\n**Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Decryption Steps (exact opposite of encryption steps):\n    - (1) For each letter c in Ciphertext, find its position in substitution_alphabet and replace it with the corresponding letter in standard_alphabet.\n    - (2) Shift the letter obtained from (1) backward by 4 positions according to the standard_alphabet order.\n    - (3) Replace the letter obtained from (2) by finding its position in reversed_alphabet, then replace it with the corresponding letter in standard_alphabet. For example, Z maps to A, Y maps to B.\ncipher text: ONOFEQMJDVQVGH decrypt to clear text: nonculpability\ncipher text: OZKCVV decrypt to clear text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 58, "source_filename": "icl_with_rule_decode_Kor_rule1_InverseShiftSubstitutionCipher_en.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "OZKCVV", "extra_args": {}, "output": "标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n解密字符 O:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q\n2. 将得到的字母向后移动4位: Q -> M\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N\n\n解密字符 Z:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Z -> Z\n2. 将得到的字母向后移动4位: Z -> V\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: V -> E\n\n解密字符 K:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: K -> M\n2. 将得到的字母向后移动4位: M -> I\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: I -> R\n\n解密字符 C:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: C -> I\n2. 将得到的字母向后移动4位: I -> E\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: E -> V\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - standard_alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Encryption Steps:\n    - For each letter p in the given Plaintext:\n    - (1) Use reversed_alphabet for reverse mapping. Find its position in the standard_alphabet and replace it with the letter in the corresponding position in reversed_alphabet. For example, A is mapped to Z and B is mapped to Y.\n    - (2) Move the letter obtained in (1) forward 4 places in the standard_alphabet order. For example, if p=A, after (1) is mapped to Z, then Z is shifted forward 4 positions in the standard_alphabet to get D.\n    - (3) Replace the letter obtained from (2) by finding its position in standard_alphabet and using the corresponding letter in substitution_alphabet, resulting in the final ciphertext letter. For example, if the letter obtained by going through (2) is D, it is mapped as J.\n\n**Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Decryption Steps (exact opposite of encryption steps):\n    - (1) For each letter c in Ciphertext, find its position in substitution_alphabet and replace it with the corresponding letter in standard_alphabet.\n    - (2) Shift the letter obtained from (1) backward by 4 positions according to the standard_alphabet order.\n    - (3) Replace the letter obtained from (2) by finding its position in reversed_alphabet, then replace it with the corresponding letter in standard_alphabet. For example, Z maps to A, Y maps to B.\nencrypted text: MHKVGVF decode to clear text: pyritic\nencrypted text: ONOFEQMJDVQVGH decode to clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 59, "source_filename": "icl_with_rule_decode_Kor_rule1_InverseShiftSubstitutionCipher_en.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "ONOFEQMJDVQVGH", "extra_args": {}, "output": "标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n解密字符 O:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q\n2. 将得到的字母向后移动4位: Q -> M\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N\n\n解密字符 N:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: N -> P\n2. 将得到的字母向后移动4位: P -> L\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: L -> O\n\n解密字符 O:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q\n2. 将得到的字母向后移动4位: Q -> M\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N\n\n解密字符 F:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: F -> B\n2. 将得到的字母向后移动4位: B -> X\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: X -> C\n\n解密字符 E:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: E -> J\n2. 将得到的字母向后移动4位: J -> F\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: F -> U\n\n解密字符 Q:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Q -> S\n2. 将得到的字母向后移动4位: S -> O\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: O -> L\n\n解密字符 M:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: M -> O\n2. 将得到的字母向后移动4位: O -> K\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: K -> P\n\n解密字符 J:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: J -> D\n2. 将得到的字母向后移动4位: D -> Z\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: Z -> A\n\n解密字符 D:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: D -> C\n2. 将得到的字母向后移动4位: C -> Y\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: Y -> B\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 Q:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Q -> S\n2. 将得到的字母向后移动4位: S -> O\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: O -> L\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 G:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: G -> K\n2. 将得到的字母向后移动4位: K -> G\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: G -> T\n\n解密字符 H:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: H -> F\n2. 将得到的字母向后移动4位: F -> B\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: B -> Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - standard_alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Encryption Steps:\n    - For each letter p in the given Plaintext:\n    - (1) Use reversed_alphabet for reverse mapping. Find its position in the standard_alphabet and replace it with the letter in the corresponding position in reversed_alphabet. For example, A is mapped to Z and B is mapped to Y.\n    - (2) Move the letter obtained in (1) forward 4 places in the standard_alphabet order. For example, if p=A, after (1) is mapped to Z, then Z is shifted forward 4 positions in the standard_alphabet to get D.\n    - (3) Replace the letter obtained from (2) by finding its position in standard_alphabet and using the corresponding letter in substitution_alphabet, resulting in the final ciphertext letter. For example, if the letter obtained by going through (2) is D, it is mapped as J.\n\n**Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Decryption Steps (exact opposite of encryption steps):\n    - (1) For each letter c in Ciphertext, find its position in substitution_alphabet and replace it with the corresponding letter in standard_alphabet.\n    - (2) Shift the letter obtained from (1) backward by 4 positions according to the standard_alphabet order.\n    - (3) Replace the letter obtained from (2) by finding its position in reversed_alphabet, then replace it with the corresponding letter in standard_alphabet. For example, Z maps to A, Y maps to B.\nencoded text: OZKCVV decrypt into clear text: nervii\nencoded text: MHKVGVF decrypt into clear text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 60, "source_filename": "icl_with_rule_decode_Kor_rule1_InverseShiftSubstitutionCipher_en.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "MHKVGVF", "extra_args": {}, "output": "标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n解密字符 M:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: M -> O\n2. 将得到的字母向后移动4位: O -> K\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: K -> P\n\n解密字符 H:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: H -> F\n2. 将得到的字母向后移动4位: F -> B\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: B -> Y\n\n解密字符 K:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: K -> M\n2. 将得到的字母向后移动4位: M -> I\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: I -> R\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 G:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: G -> K\n2. 将得到的字母向后移动4位: K -> G\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: G -> T\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 F:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: F -> B\n2. 将得到的字母向后移动4位: B -> X\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: X -> C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n        加密规则:\n        - 输入:\n            - 原文: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 输出:\n            - 暗文: 仅包含0和1的二进制字符串\n        - 准备:\n            - 固定密钥: 8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 加密步骤:\n            1. 将每个字符转换为二进制:\n                - 将每个字符转换为ASCII值\n                - 将ASCII值转换为8位二进制字符串\n            2. XOR运算:\n                - 对字符的8位二进制表示与固定密钥进行XOR运算\n                - 如果需要，重复密钥以匹配二进制表示的长度\n            3. 置换:\n                - 对XOR结果应用置换表得到每个字符的最终加密二进制字符串\n            4. 合并二进制字符串:\n                - 将所有字符的二进制字符串连接形成最终暗文\n        \n解密算法:\n\n        解密规则:\n        - 输入:\n            - 暗文: 仅包含0和1的二进制字符串\n        - 输出:\n            - 原文: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 准备:\n            - 固定密钥: 与加密使用的相同8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 解密步骤:\n            1. 将暗文分块:\n                - 将二进制暗文分成8位一组，每组代表一个加密字符\n            2. 逆置换:\n                - 对每个8位块应用逆置换表以还原加密时的置换\n            3. XOR运算:\n                - 对置换后的二进制块与固定密钥进行XOR运算\n            4. 二进制转字符:\n                - 将得到的二进制字符串转换为十进制值\n                - 将十进制值转换为对应的ASCII字符\n            5. 合并字符:\n                - 将每个二进制块得到的字符连接形成最终原文\n        \n暗文: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101 解密成为原文: nonculpability\n暗文: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101 解密成为原文: nonculpability\n暗文: 110100101101111111111000111110101101010111010101 解密成为原文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 61, "source_filename": "icl_with_rule_decode_Kor_rule21_XORCipher_cn.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "110100101101111111111000111110101101010111010101", "extra_args": {}, "output": "1. 接收到的加密文本: 110100101101111111111000111110101101010111010101\n\n处理8位二进制块: 11010010\n2. 应用逆置换: 11100100\n3. 与密钥进行XOR运算: 01001110\n4. 转换为字符: N\n\n处理8位二进制块: 11011111\n2. 应用逆置换: 11101111\n3. 与密钥进行XOR运算: 01000101\n4. 转换为字符: E\n\n处理8位二进制块: 11111000\n2. 应用逆置换: 11111000\n3. 与密钥进行XOR运算: 01010010\n4. 转换为字符: R\n\n处理8位二进制块: 11111010\n2. 应用逆置换: 11111100\n3. 与密钥进行XOR运算: 01010110\n4. 转换为字符: V\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n        加密规则:\n        - 输入:\n            - 非加密信息: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 输出:\n            - 加密文本: 仅包含0和1的二进制字符串\n        - 准备:\n            - 固定密钥: 8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 加密步骤:\n            1. 将每个字符转换为二进制:\n                - 将每个字符转换为ASCII值\n                - 将ASCII值转换为8位二进制字符串\n            2. XOR运算:\n                - 对字符的8位二进制表示与固定密钥进行XOR运算\n                - 如果需要，重复密钥以匹配二进制表示的长度\n            3. 置换:\n                - 对XOR结果应用置换表得到每个字符的最终加密二进制字符串\n            4. 合并二进制字符串:\n                - 将所有字符的二进制字符串连接形成最终加密文本\n        \n解密算法:\n\n        解密规则:\n        - 输入:\n            - 加密文本: 仅包含0和1的二进制字符串\n        - 输出:\n            - 非加密信息: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 准备:\n            - 固定密钥: 与加密使用的相同8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 解密步骤:\n            1. 将加密文本分块:\n                - 将二进制加密文本分成8位一组，每组代表一个加密字符\n            2. 逆置换:\n                - 对每个8位块应用逆置换表以还原加密时的置换\n            3. XOR运算:\n                - 对置换后的二进制块与固定密钥进行XOR运算\n            4. 二进制转字符:\n                - 将得到的二进制字符串转换为十进制值\n                - 将十进制值转换为对应的ASCII字符\n            5. 合并字符:\n                - 将每个二进制块得到的字符连接形成最终非加密信息\n        \n加密文本: 11111100111101011111100011010101111111101101010111011001 解码为非加密信息: pyritic\n加密文本: 11111100111101011111100011010101111111101101010111011001 解码为非加密信息: pyritic\n加密文本: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101 解码为非加密信息是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 62, "source_filename": "icl_with_rule_decode_Kor_rule21_XORCipher_cn.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101", "extra_args": {}, "output": "1. 接收到的加密文本: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n\n处理8位二进制块: 11010010\n2. 应用逆置换: 11100100\n3. 与密钥进行XOR运算: 01001110\n4. 转换为字符: N\n\n处理8位二进制块: 11010011\n2. 应用逆置换: 11100101\n3. 与密钥进行XOR运算: 01001111\n4. 转换为字符: O\n\n处理8位二进制块: 11010010\n2. 应用逆置换: 11100100\n3. 与密钥进行XOR运算: 01001110\n4. 转换为字符: N\n\n处理8位二进制块: 11011001\n2. 应用逆置换: 11101001\n3. 与密钥进行XOR运算: 01000011\n4. 转换为字符: C\n\n处理8位二进制块: 11111111\n2. 应用逆置换: 11111111\n3. 与密钥进行XOR运算: 01010101\n4. 转换为字符: U\n\n处理8位二进制块: 11010110\n2. 应用逆置换: 11100110\n3. 与密钥进行XOR运算: 01001100\n4. 转换为字符: L\n\n处理8位二进制块: 11111100\n2. 应用逆置换: 11111010\n3. 与密钥进行XOR运算: 01010000\n4. 转换为字符: P\n\n处理8位二进制块: 11011101\n2. 应用逆置换: 11101011\n3. 与密钥进行XOR运算: 01000001\n4. 转换为字符: A\n\n处理8位二进制块: 11011000\n2. 应用逆置换: 11101000\n3. 与密钥进行XOR运算: 01000010\n4. 转换为字符: B\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11010110\n2. 应用逆置换: 11100110\n3. 与密钥进行XOR运算: 01001100\n4. 转换为字符: L\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11111110\n2. 应用逆置换: 11111110\n3. 与密钥进行XOR运算: 01010100\n4. 转换为字符: T\n\n处理8位二进制块: 11110101\n2. 应用逆置换: 11110011\n3. 与密钥进行XOR运算: 01011001\n4. 转换为字符: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n        加密规则:\n        - 输入:\n            - 非加密信息: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 输出:\n            - 暗码: 仅包含0和1的二进制字符串\n        - 准备:\n            - 固定密钥: 8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 加密步骤:\n            1. 将每个字符转换为二进制:\n                - 将每个字符转换为ASCII值\n                - 将ASCII值转换为8位二进制字符串\n            2. XOR运算:\n                - 对字符的8位二进制表示与固定密钥进行XOR运算\n                - 如果需要，重复密钥以匹配二进制表示的长度\n            3. 置换:\n                - 对XOR结果应用置换表得到每个字符的最终加密二进制字符串\n            4. 合并二进制字符串:\n                - 将所有字符的二进制字符串连接形成最终暗码\n        \n解密算法:\n\n        解密规则:\n        - 输入:\n            - 暗码: 仅包含0和1的二进制字符串\n        - 输出:\n            - 非加密信息: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 准备:\n            - 固定密钥: 与加密使用的相同8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 解密步骤:\n            1. 将暗码分块:\n                - 将二进制暗码分成8位一组，每组代表一个加密字符\n            2. 逆置换:\n                - 对每个8位块应用逆置换表以还原加密时的置换\n            3. XOR运算:\n                - 对置换后的二进制块与固定密钥进行XOR运算\n            4. 二进制转字符:\n                - 将得到的二进制字符串转换为十进制值\n                - 将十进制值转换为对应的ASCII字符\n            5. 合并字符:\n                - 将每个二进制块得到的字符连接形成最终非加密信息\n        \n暗码: 11111100111101011111100011010101111111101101010111011001 解密成非加密信息: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 63, "source_filename": "icl_with_rule_decode_Kor_rule21_XORCipher_cn.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "11111100111101011111100011010101111111101101010111011001", "extra_args": {}, "output": "1. 接收到的加密文本: 11111100111101011111100011010101111111101101010111011001\n\n处理8位二进制块: 11111100\n2. 应用逆置换: 11111010\n3. 与密钥进行XOR运算: 01010000\n4. 转换为字符: P\n\n处理8位二进制块: 11110101\n2. 应用逆置换: 11110011\n3. 与密钥进行XOR运算: 01011001\n4. 转换为字符: Y\n\n处理8位二进制块: 11111000\n2. 应用逆置换: 11111000\n3. 与密钥进行XOR运算: 01010010\n4. 转换为字符: R\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11111110\n2. 应用逆置换: 11111110\n3. 与密钥进行XOR运算: 01010100\n4. 转换为字符: T\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11011001\n2. 应用逆置换: 11101001\n3. 与密钥进行XOR运算: 01000011\n4. 转换为字符: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nEncrypt the plaintext using the following encryption method:\nThe Atbash cipher works by substituting each letter with its opposite counterpart from the alphabet. For instance, if you see 'A', write 'Z'; if you see 'B', write 'Y'; and so forth. Any characters that aren't letters should remain unchanged.\nplain text: pyritic encode to encoded text: kbirgrx\nplain text: nonculpability encode to encoded text: mlmxfokzyrorgb\nplain text: nervii encode to encoded text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 64, "source_filename": "icl_with_rule_encode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "nervii", "extra_args": {}, "output": "编码过程:\n转换单词: nervii\n  'n' -> 'm'\n  'e' -> 'v'\n  'r' -> 'i'\n  'v' -> 'e'\n  'i' -> 'r'\n  'i' -> 'r'\n编码后的文本: mvierr\n", "ground_truth": "mvierr"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nEncrypt the plaintext using the following cipher:\nThe Atbash cipher works by substituting each letter with its opposite counterpart from the alphabet. For instance, 'A' is replaced with 'Z', 'B' becomes 'Y', and so on in this pattern. Any character that isn't a letter remains unchanged in the output.\nplain text: nervii encode to encoded text: mvierr\nplain text: nervii encode to encoded text: mvierr\nplain text: nonculpability encode to encoded text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 65, "source_filename": "icl_with_rule_encode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "nonculpability", "extra_args": {}, "output": "编码过程:\n转换单词: nonculpability\n  'n' -> 'm'\n  'o' -> 'l'\n  'n' -> 'm'\n  'c' -> 'x'\n  'u' -> 'f'\n  'l' -> 'o'\n  'p' -> 'k'\n  'a' -> 'z'\n  'b' -> 'y'\n  'i' -> 'r'\n  'l' -> 'o'\n  'i' -> 'r'\n  't' -> 'g'\n  'y' -> 'b'\n编码后的文本: mlmxfokzyrorgb\n", "ground_truth": "mlmxfokzyrorgb"}, "data_source": "Cipher"}
{"prompt": "Encrypt the plaintext using the Atbash cipher method. In this encryption scheme, each letter is replaced with its corresponding letter from the opposite end of the alphabet. For instance, 'A' is encoded as 'Z', 'B' becomes 'Y', and so forth. Any characters that are not letters remain unchanged in the output.\nclear text: pyritic encrypt to encoded text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 66, "source_filename": "icl_with_rule_encode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "pyritic", "extra_args": {}, "output": "编码过程:\n转换单词: pyritic\n  'p' -> 'k'\n  'y' -> 'b'\n  'r' -> 'i'\n  'i' -> 'r'\n  't' -> 'g'\n  'i' -> 'r'\n  'c' -> 'x'\n编码后的文本: kbirgrx\n", "ground_truth": "kbirgrx"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string.\n- Preparation:\n    - Numerical-Index Mapping Table\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - morse_code Table\n        - morse_code = {\n                'A': '.-',     'B': '-...',   'C': '-.-.',   'D': '-..',\n                'E': '.',      'F': '..-.',   'G': '--.',    'H': '....',\n                'I': '..',     'J': '.---',   'K': '-.-',    'L': '.-..',\n                'M': '--',     'N': '-.',     'O': '---',    'P': '.--.',\n                'Q': '--.-',   'R': '.-.',    'S': '...',    'T': '-',\n                'U': '..-',    'V': '...-',   'W': '.--',    'X': '-..-',\n                'Y': '-.--',   'Z': '--..',\n            }\n- Encryption Steps:\n    - Convert each character in the plaintext into Morse code according to the morse_code Table, with each character separated by a /, e.g., the Morse code corresponding to AB is '.-/-...'.\n    - Divide Morse code into two pairs of characters. If the length of the Morse code is odd, the last character, without the following mapping, is finally added to the end of the ciphertext.\n    - Each pair of characters is converted into a corresponding numeric string according to the numeric index mapping table.\n    - The encrypted message is represented by a string.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A numeric string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Numerical-Index Mapping Table (Same as encryption)\n    - morse_code Table (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Each digit in the ciphertext is converted to the corresponding character pair according to the digit-index mapping table. If there is a non-numeric character at the end of the ciphertext, it is not processed. By this point, the complete Morse code is obtained.\n    - Obtain the Morse code for each character by splitting the Morse code by /.\n    - Convert the Morse code of each character to the corresponding plaintext character according to the morse_code table.\n    - The final plaintext character is an uppercase string.\nencoded text: 485834472743247641732173645 decrypt to clear text: nonculpability\nencoded text: 485834472743247641732173645 decrypt to clear text: nonculpability\nencoded text: 2482623187344 decrypt to clear text: pyritic\nencoded text: 477471617. decrypt to clear text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 67, "source_filename": "icl_with_rule_decode_Kor_rule12_MorbitCipher_en.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "477471617.", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的加密文本: 477471617.\n2. 将数字转换回摩斯密码: -././.-./...-/../..\n3. 将摩斯密码转换为原文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string.\n- Preparation:\n    - Numerical-Index Mapping Table\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - morse_code Table\n        - morse_code = {\n                'A': '.-',     'B': '-...',   'C': '-.-.',   'D': '-..',\n                'E': '.',      'F': '..-.',   'G': '--.',    'H': '....',\n                'I': '..',     'J': '.---',   'K': '-.-',    'L': '.-..',\n                'M': '--',     'N': '-.',     'O': '---',    'P': '.--.',\n                'Q': '--.-',   'R': '.-.',    'S': '...',    'T': '-',\n                'U': '..-',    'V': '...-',   'W': '.--',    'X': '-..-',\n                'Y': '-.--',   'Z': '--..',\n            }\n- Encryption Steps:\n    - Convert each character in the plaintext into Morse code according to the morse_code Table, with each character separated by a /, e.g., the Morse code corresponding to AB is '.-/-...'.\n    - Divide Morse code into two pairs of characters. If the length of the Morse code is odd, the last character, without the following mapping, is finally added to the end of the ciphertext.\n    - Each pair of characters is converted into a corresponding numeric string according to the numeric index mapping table.\n    - The encrypted message is represented by a string.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A numeric string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Numerical-Index Mapping Table (Same as encryption)\n    - morse_code Table (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Each digit in the ciphertext is converted to the corresponding character pair according to the digit-index mapping table. If there is a non-numeric character at the end of the ciphertext, it is not processed. By this point, the complete Morse code is obtained.\n    - Obtain the Morse code for each character by splitting the Morse code by /.\n    - Convert the Morse code of each character to the corresponding plaintext character according to the morse_code table.\n    - The final plaintext character is an uppercase string.\ncipher text: 2482623187344 decrypt into original information: pyritic\ncipher text: 2482623187344 decrypt into original information: pyritic\ncipher text: 485834472743247641732173645 decrypt into original information is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 68, "source_filename": "icl_with_rule_decode_Kor_rule12_MorbitCipher_en.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "485834472743247641732173645", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的加密文本: 485834472743247641732173645\n2. 将数字转换回摩斯密码: -./---/-./-.-./..-/.-../.--./.-/-.../../.-../../-/-.--\n3. 将摩斯密码转换为原文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string.\n- Preparation:\n    - Numerical-Index Mapping Table\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - morse_code Table\n        - morse_code = {\n                'A': '.-',     'B': '-...',   'C': '-.-.',   'D': '-..',\n                'E': '.',      'F': '..-.',   'G': '--.',    'H': '....',\n                'I': '..',     'J': '.---',   'K': '-.-',    'L': '.-..',\n                'M': '--',     'N': '-.',     'O': '---',    'P': '.--.',\n                'Q': '--.-',   'R': '.-.',    'S': '...',    'T': '-',\n                'U': '..-',    'V': '...-',   'W': '.--',    'X': '-..-',\n                'Y': '-.--',   'Z': '--..',\n            }\n- Encryption Steps:\n    - Convert each character in the plaintext into Morse code according to the morse_code Table, with each character separated by a /, e.g., the Morse code corresponding to AB is '.-/-...'.\n    - Divide Morse code into two pairs of characters. If the length of the Morse code is odd, the last character, without the following mapping, is finally added to the end of the ciphertext.\n    - Each pair of characters is converted into a corresponding numeric string according to the numeric index mapping table.\n    - The encrypted message is represented by a string.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A numeric string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Numerical-Index Mapping Table (Same as encryption)\n    - morse_code Table (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Each digit in the ciphertext is converted to the corresponding character pair according to the digit-index mapping table. If there is a non-numeric character at the end of the ciphertext, it is not processed. By this point, the complete Morse code is obtained.\n    - Obtain the Morse code for each character by splitting the Morse code by /.\n    - Convert the Morse code of each character to the corresponding plaintext character according to the morse_code table.\n    - The final plaintext character is an uppercase string.\nencrypted text: 485834472743247641732173645 decode into plain text: nonculpability\nencrypted text: 2482623187344 decode into plain text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 69, "source_filename": "icl_with_rule_decode_Kor_rule12_MorbitCipher_en.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "2482623187344", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的加密文本: 2482623187344\n2. 将数字转换回摩斯密码: .--./-.--/.-./../-/../-.-.\n3. 将摩斯密码转换为原文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list:\n        - [1, 4, 0, 6, 5, 2, 3]\n        - The transposed sequence  list is used to write the plaintext in order one line at a time, and then the columns are adjusted according to the transposed sequence table so that the characters in each line are arranged in the given order.\n        - Columns count from 0.\n- Encryption Steps:\n    - The [1, 4, 0, 6, 5, 2, 3] transposed sequence table has a total of 7 bits, indicating that 7 letters should be written on a line.\n    - Write the plaintext in order, line by line, 7 per line. When there are less than 7, the last line is filled with $. A written grid can be obtained.\n    - According to the transposed sequence table, adjust the order of the columns, that is, the order of the columns is now [the original column 1, the original column 4, the original column 0, the original column 6, the original column 5, the original column 2, the original column 3], you can get after the adjustment of the columns after the order of the grid.\n    - Read the grid row by row and connect them to get the final ciphertext.(Note that $ needs to be retained.)\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list (Same as encryption)\n- Decryption Steps:\n    - Write the ciphertext sequentially, line by line, 7 letters per line.\n    - Read line by line, but when reading each line, read a character in column 2 corresponding to 0, then a character in column 0 corresponding to 1, then a character in column 6 corresponding to 2, etc., etc., etc., etc., etc., etc.\n    - Eventually read the message line by line, remove the $ at the end, you can get the decrypted plaintext.\nencrypted text: EIN$IRV decrypt to plain text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 70, "source_filename": "icl_with_rule_decode_Kor_rule20_TranspositionCipher_en.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "EIN$IRV", "extra_args": {}, "output": "需要解密的文本: EIN$IRV\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n加密的网格:\nEIN$IRV\n还原后的网格:\nNERVII$\n解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list:\n        - [1, 4, 0, 6, 5, 2, 3]\n        - The transposed sequence  list is used to write the plaintext in order one line at a time, and then the columns are adjusted according to the transposed sequence table so that the characters in each line are arranged in the given order.\n        - Columns count from 0.\n- Encryption Steps:\n    - The [1, 4, 0, 6, 5, 2, 3] transposed sequence table has a total of 7 bits, indicating that 7 letters should be written on a line.\n    - Write the plaintext in order, line by line, 7 per line. When there are less than 7, the last line is filled with $. A written grid can be obtained.\n    - According to the transposed sequence table, adjust the order of the columns, that is, the order of the columns is now [the original column 1, the original column 4, the original column 0, the original column 6, the original column 5, the original column 2, the original column 3], you can get after the adjustment of the columns after the order of the grid.\n    - Read the grid row by row and connect them to get the final ciphertext.(Note that $ needs to be retained.)\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list (Same as encryption)\n- Decryption Steps:\n    - Write the ciphertext sequentially, line by line, 7 letters per line.\n    - Read line by line, but when reading each line, read a character in column 2 corresponding to 0, then a character in column 0 corresponding to 1, then a character in column 6 corresponding to 2, etc., etc., etc., etc., etc., etc.\n    - Eventually read the message line by line, remove the $ at the end, you can get the decrypted plaintext.\nencoded text: YTPCIRI decrypt into plain text: pyritic\nencoded text: OUNPLNCBIAYTIL decrypt into plain text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 71, "source_filename": "icl_with_rule_decode_Kor_rule20_TranspositionCipher_en.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "OUNPLNCBIAYTIL", "extra_args": {}, "output": "需要解密的文本: OUNPLNCBIAYTIL\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 2 行来存放文本\n加密的网格:\nOUNPLNC\nBIAYTIL\n还原后的网格:\nNONCULP\nABILITY\n解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list:\n        - [1, 4, 0, 6, 5, 2, 3]\n        - The transposed sequence  list is used to write the plaintext in order one line at a time, and then the columns are adjusted according to the transposed sequence table so that the characters in each line are arranged in the given order.\n        - Columns count from 0.\n- Encryption Steps:\n    - The [1, 4, 0, 6, 5, 2, 3] transposed sequence table has a total of 7 bits, indicating that 7 letters should be written on a line.\n    - Write the plaintext in order, line by line, 7 per line. When there are less than 7, the last line is filled with $. A written grid can be obtained.\n    - According to the transposed sequence table, adjust the order of the columns, that is, the order of the columns is now [the original column 1, the original column 4, the original column 0, the original column 6, the original column 5, the original column 2, the original column 3], you can get after the adjustment of the columns after the order of the grid.\n    - Read the grid row by row and connect them to get the final ciphertext.(Note that $ needs to be retained.)\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list (Same as encryption)\n- Decryption Steps:\n    - Write the ciphertext sequentially, line by line, 7 letters per line.\n    - Read line by line, but when reading each line, read a character in column 2 corresponding to 0, then a character in column 0 corresponding to 1, then a character in column 6 corresponding to 2, etc., etc., etc., etc., etc., etc.\n    - Eventually read the message line by line, remove the $ at the end, you can get the decrypted plaintext.\nencrypted text: EIN$IRV decode to clear text: nervii\nencrypted text: OUNPLNCBIAYTIL decode to clear text: nonculpability\nencrypted text: EIN$IRV decode to clear text: nervii\nencrypted text: YTPCIRI decode to clear text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 72, "source_filename": "icl_with_rule_decode_Kor_rule20_TranspositionCipher_en.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "YTPCIRI", "extra_args": {}, "output": "需要解密的文本: YTPCIRI\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n加密的网格:\nYTPCIRI\n还原后的网格:\nPYRITIC\n解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原文进行加密\n加密规则:\n    - 输入:\n        - 原文: 大写字母字符串,不含标点和空格\n    - 输出:\n        - 暗码: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列:\n            - 54张卡牌的列表,包括52张花色牌和2张可区分的王牌(A王牌和B王牌)。花色牌按四种花色顺序编号1-52,王牌值为53和54。\n            - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n        - 密钥流算法:\n            该算法通过移动卡牌生成密钥流值。算法是确定性的,意味着密钥流值仅取决于卡牌的初始顺序。卡牌组被视为循环数组,允许需要移到底部的卡牌绕到顶部。\n            \n            执行以下步骤生成密钥流的一个字符:\n            1. 找到A王牌并下移一位。如果是最后一张牌,则成为第二张牌。不能成为第一张牌。\n            2. 找到B王牌并下移两位。如果是倒数第二张牌,则绕到第二位。如果是最后一张牌,则成为第三张牌。不能成为第一张牌。\n            3. 进行\"三切\":用王牌作为边界将牌组分成三部分,然后交换顶部和底部。王牌本身及其之间的牌保持不变。\n            4. 进行\"计数切\":检查牌组底牌。如果是王牌(53/54),其值固定为53。从牌组顶部取出该数量的牌,插入到最后一张牌的上方。\n            5. 查看顶牌的值。同样,任何王牌计为53。计算该牌下方的位置数,使用该位置的牌值作为下一个密钥流值。如果计算出的牌是王牌,忽略它并重复密钥流算法。\n            6. 返回生成的密钥流值\n    - 加密步骤:\n        - cards=初始卡牌序列\n        - 对每个原文字符p:\n            - 使用字母表将p转换为对应的位置值x(从0开始)\n            - 使用初始卡牌序列为p生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 当密钥流值y加到位置值x时,应用模26运算得到z:\n                - z=(y+x) % 26\n            - 使用LETTERS列表返回对应位置z的字母\n            - 将其附加到暗码\n原文: nonculpability 加密为暗码: FMMLWKFPPPDEOF\n原文: nonculpability 加密为暗码: FMMLWKFPPPDEOF\n原文: nervii 加密为暗码: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 73, "source_filename": "icl_with_rule_encode_Kor_rule6_SolitaireCipher_cn.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n开始加密过程:\n\n处理第1个字符 'N':\n字符'N'在字母表中的位置是: 23\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (23 + 14) % 26 = 11\n加密后的字符是: F\n\n处理第2个字符 'E':\n字符'E'在字母表中的位置是: 13\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n加密计算: (13 + 7) % 26 = 20\n加密后的字符是: B\n\n处理第3个字符 'R':\n字符'R'在字母表中的位置是: 5\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n加密计算: (5 + 13) % 26 = 18\n加密后的字符是: P\n\n处理第4个字符 'V':\n字符'V'在字母表中的位置是: 19\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n加密计算: (19 + 17) % 26 = 10\n加密后的字符是: M\n\n处理第5个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (25 + 38) % 26 = 11\n加密后的字符是: F\n\n处理第6个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (25 + 45) % 26 = 18\n加密后的字符是: P\n\n最终加密结果: FBPMFP\n", "ground_truth": "FBPMFP"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对明文进行加密\n加密规则:\n    - 输入:\n        - 明文: 大写字母字符串,不含标点和空格\n    - 输出:\n        - 暗码: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列:\n            - 54张卡牌的列表,包括52张花色牌和2张可区分的王牌(A王牌和B王牌)。花色牌按四种花色顺序编号1-52,王牌值为53和54。\n            - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n        - 密钥流算法:\n            该算法通过移动卡牌生成密钥流值。算法是确定性的,意味着密钥流值仅取决于卡牌的初始顺序。卡牌组被视为循环数组,允许需要移到底部的卡牌绕到顶部。\n            \n            执行以下步骤生成密钥流的一个字符:\n            1. 找到A王牌并下移一位。如果是最后一张牌,则成为第二张牌。不能成为第一张牌。\n            2. 找到B王牌并下移两位。如果是倒数第二张牌,则绕到第二位。如果是最后一张牌,则成为第三张牌。不能成为第一张牌。\n            3. 进行\"三切\":用王牌作为边界将牌组分成三部分,然后交换顶部和底部。王牌本身及其之间的牌保持不变。\n            4. 进行\"计数切\":检查牌组底牌。如果是王牌(53/54),其值固定为53。从牌组顶部取出该数量的牌,插入到最后一张牌的上方。\n            5. 查看顶牌的值。同样,任何王牌计为53。计算该牌下方的位置数,使用该位置的牌值作为下一个密钥流值。如果计算出的牌是王牌,忽略它并重复密钥流算法。\n            6. 返回生成的密钥流值\n    - 加密步骤:\n        - cards=初始卡牌序列\n        - 对每个明文字符p:\n            - 使用字母表将p转换为对应的位置值x(从0开始)\n            - 使用初始卡牌序列为p生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 当密钥流值y加到位置值x时,应用模26运算得到z:\n                - z=(y+x) % 26\n            - 使用LETTERS列表返回对应位置z的字母\n            - 将其附加到暗码\n明文: nervii 加密成暗码: FBPMFP\n明文: nonculpability 加密成暗码: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 74, "source_filename": "icl_with_rule_encode_Kor_rule6_SolitaireCipher_cn.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n开始加密过程:\n\n处理第1个字符 'N':\n字符'N'在字母表中的位置是: 23\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (23 + 14) % 26 = 11\n加密后的字符是: F\n\n处理第2个字符 'O':\n字符'O'在字母表中的位置是: 3\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n加密计算: (3 + 7) % 26 = 10\n加密后的字符是: M\n\n处理第3个字符 'N':\n字符'N'在字母表中的位置是: 23\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n加密计算: (23 + 13) % 26 = 10\n加密后的字符是: M\n\n处理第4个字符 'C':\n字符'C'在字母表中的位置是: 7\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n加密计算: (7 + 17) % 26 = 24\n加密后的字符是: L\n\n处理第5个字符 'U':\n字符'U'在字母表中的位置是: 16\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (16 + 38) % 26 = 2\n加密后的字符是: W\n\n处理第6个字符 'L':\n字符'L'在字母表中的位置是: 24\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (24 + 45) % 26 = 17\n加密后的字符是: K\n\n处理第7个字符 'P':\n字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (18 + 45) % 26 = 11\n加密后的字符是: F\n\n处理第8个字符 'A':\n字符'A'在字母表中的位置是: 6\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (6 + 38) % 26 = 18\n加密后的字符是: P\n\n处理第9个字符 'B':\n字符'B'在字母表中的位置是: 20\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 50\n加密计算: (20 + 50) % 26 = 18\n加密后的字符是: P\n\n处理第10个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (25 + 45) % 26 = 18\n加密后的字符是: P\n\n处理第11个字符 'L':\n字符'L'在字母表中的位置是: 24\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 29\n加密计算: (24 + 29) % 26 = 1\n加密后的字符是: D\n\n处理第12个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (25 + 14) % 26 = 13\n加密后的字符是: E\n\n处理第13个字符 'T':\n字符'T'在字母表中的位置是: 4\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n加密计算: (4 + 25) % 26 = 3\n加密后的字符是: O\n\n处理第14个字符 'Y':\n字符'Y'在字母表中的位置是: 12\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n加密计算: (12 + 25) % 26 = 11\n加密后的字符是: F\n\n最终加密结果: FMMLWKFPPPDEOF\n", "ground_truth": "FMMLWKFPPPDEOF"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对非加密信息进行加密\n加密规则:\n    - 输入:\n        - 非加密信息: 大写字母字符串,不含标点和空格\n    - 输出:\n        - 暗文: 大写字母字符串\n    - 准备:\n        - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y', 'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n        - 将每个字母与其在LETTERS中的位置关联(从0开始):\n            J->0, D->1, W->2, O->3, T->4, R->5, A->6, C->7, X->8, Q->9, M->10, F->11, Y->12, E->13, Z->14, G->15, U->16, K->17, P->18, V->19, B->20, S->21, H->22, N->23, L->24, I->25\n        - 初始卡牌序列:\n            - 54张卡牌的列表,包括52张花色牌和2张可区分的王牌(A王牌和B王牌)。花色牌按四种花色顺序编号1-52,王牌值为53和54。\n            - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n        - 密钥流算法:\n            该算法通过移动卡牌生成密钥流值。算法是确定性的,意味着密钥流值仅取决于卡牌的初始顺序。卡牌组被视为循环数组,允许需要移到底部的卡牌绕到顶部。\n            \n            执行以下步骤生成密钥流的一个字符:\n            1. 找到A王牌并下移一位。如果是最后一张牌,则成为第二张牌。不能成为第一张牌。\n            2. 找到B王牌并下移两位。如果是倒数第二张牌,则绕到第二位。如果是最后一张牌,则成为第三张牌。不能成为第一张牌。\n            3. 进行\"三切\":用王牌作为边界将牌组分成三部分,然后交换顶部和底部。王牌本身及其之间的牌保持不变。\n            4. 进行\"计数切\":检查牌组底牌。如果是王牌(53/54),其值固定为53。从牌组顶部取出该数量的牌,插入到最后一张牌的上方。\n            5. 查看顶牌的值。同样,任何王牌计为53。计算该牌下方的位置数,使用该位置的牌值作为下一个密钥流值。如果计算出的牌是王牌,忽略它并重复密钥流算法。\n            6. 返回生成的密钥流值\n    - 加密步骤:\n        - cards=初始卡牌序列\n        - 对每个非加密信息字符p:\n            - 使用字母表将p转换为对应的位置值x(从0开始)\n            - 使用初始卡牌序列为p生成密钥流值y:\n                - y, cards = 密钥流算法(cards)\n                - 该算法修改卡牌顺序,下次执行使用新顺序\n            - 当密钥流值y加到位置值x时,应用模26运算得到z:\n                - z=(y+x) % 26\n            - 使用LETTERS列表返回对应位置z的字母\n            - 将其附加到暗文\n非加密信息: nervii 加密成为暗文: FBPMFP\n非加密信息: nervii 加密成为暗文: FBPMFP\n非加密信息: nonculpability 加密成为暗文: FMMLWKFPPPDEOF\n非加密信息: pyritic 加密成为暗文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 75, "source_filename": "icl_with_rule_encode_Kor_rule6_SolitaireCipher_cn.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n开始加密过程:\n\n处理第1个字符 'P':\n字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (18 + 14) % 26 = 6\n加密后的字符是: A\n\n处理第2个字符 'Y':\n字符'Y'在字母表中的位置是: 12\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n加密计算: (12 + 7) % 26 = 19\n加密后的字符是: V\n\n处理第3个字符 'R':\n字符'R'在字母表中的位置是: 5\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n加密计算: (5 + 13) % 26 = 18\n加密后的字符是: P\n\n处理第4个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n加密计算: (25 + 17) % 26 = 16\n加密后的字符是: U\n\n处理第5个字符 'T':\n字符'T'在字母表中的位置是: 4\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (4 + 38) % 26 = 16\n加密后的字符是: U\n\n处理第6个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (25 + 45) % 26 = 18\n加密后的字符是: P\n\n处理第7个字符 'C':\n字符'C'在字母表中的位置是: 7\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (7 + 45) % 26 = 0\n加密后的字符是: J\n\n最终加密结果: AVPUUPJ\n", "ground_truth": "AVPUUPJ"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - k_q_x : 12\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x corresponding to the ascii code of p.\n        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.\n    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - k_q_x : 12\n- Decryption Steps:\n    - For each number c in the ciphertext:\n        - Compute z = c // k_q_x. where // is used in integer division operations to return the integer part of the quotient (i.e., the result of removing the fractional part).\n        - Based on the decimal number represented by z, use ascii to find the corresponding letter as the plaintext letter p.\n    - Finally, concatenate all p's to get the final plaintext.\nencoded text: 960,1068,984,876,1008,876,804 decode to original information: pyritic\nencoded text: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 decode to original information: nonculpability\nencoded text: 960,1068,984,876,1008,876,804 decode to original information: pyritic\nencoded text: 936,828,984,1032,876,876 decode to original information is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 76, "source_filename": "icl_with_rule_decode_Kor_rule24_ECCCipher_en.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "936,828,984,1032,876,876", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 936,828,984,1032,876,876\n使用预设的k_q_x值: 12\n\n处理数字: 936\n除以k_q_x后的值: 78\n对应的字符: N\n\n处理数字: 828\n除以k_q_x后的值: 69\n对应的字符: E\n\n处理数字: 984\n除以k_q_x后的值: 82\n对应的字符: R\n\n处理数字: 1032\n除以k_q_x后的值: 86\n对应的字符: V\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - k_q_x : 12\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x corresponding to the ascii code of p.\n        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.\n    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - k_q_x : 12\n- Decryption Steps:\n    - For each number c in the ciphertext:\n        - Compute z = c // k_q_x. where // is used in integer division operations to return the integer part of the quotient (i.e., the result of removing the fractional part).\n        - Based on the decimal number represented by z, use ascii to find the corresponding letter as the plaintext letter p.\n    - Finally, concatenate all p's to get the final plaintext.\nencoded text: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 decrypt into original information is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 77, "source_filename": "icl_with_rule_decode_Kor_rule24_ECCCipher_en.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068\n使用预设的k_q_x值: 12\n\n处理数字: 936\n除以k_q_x后的值: 78\n对应的字符: N\n\n处理数字: 948\n除以k_q_x后的值: 79\n对应的字符: O\n\n处理数字: 936\n除以k_q_x后的值: 78\n对应的字符: N\n\n处理数字: 804\n除以k_q_x后的值: 67\n对应的字符: C\n\n处理数字: 1020\n除以k_q_x后的值: 85\n对应的字符: U\n\n处理数字: 912\n除以k_q_x后的值: 76\n对应的字符: L\n\n处理数字: 960\n除以k_q_x后的值: 80\n对应的字符: P\n\n处理数字: 780\n除以k_q_x后的值: 65\n对应的字符: A\n\n处理数字: 792\n除以k_q_x后的值: 66\n对应的字符: B\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 912\n除以k_q_x后的值: 76\n对应的字符: L\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 1008\n除以k_q_x后的值: 84\n对应的字符: T\n\n处理数字: 1068\n除以k_q_x后的值: 89\n对应的字符: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - k_q_x : 12\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x corresponding to the ascii code of p.\n        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.\n    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - k_q_x : 12\n- Decryption Steps:\n    - For each number c in the ciphertext:\n        - Compute z = c // k_q_x. where // is used in integer division operations to return the integer part of the quotient (i.e., the result of removing the fractional part).\n        - Based on the decimal number represented by z, use ascii to find the corresponding letter as the plaintext letter p.\n    - Finally, concatenate all p's to get the final plaintext.\nencrypted text: 936,828,984,1032,876,876 decrypt into original information: nervii\nencrypted text: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 decrypt into original information: nonculpability\nencrypted text: 960,1068,984,876,1008,876,804 decrypt into original information: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 78, "source_filename": "icl_with_rule_decode_Kor_rule24_ECCCipher_en.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "960,1068,984,876,1008,876,804", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 960,1068,984,876,1008,876,804\n使用预设的k_q_x值: 12\n\n处理数字: 960\n除以k_q_x后的值: 80\n对应的字符: P\n\n处理数字: 1068\n除以k_q_x后的值: 89\n对应的字符: Y\n\n处理数字: 984\n除以k_q_x后的值: 82\n对应的字符: R\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 1008\n除以k_q_x后的值: 84\n对应的字符: T\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 804\n除以k_q_x后的值: 67\n对应的字符: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原文进行加密\n加密方案概述：将字符串转换为ASCII编码。\n原文: nonculpability 加密成加密信息: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n原文: nervii 加密成加密信息是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 79, "source_filename": "icl_with_rule_encode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": "nervii", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   e, ASCII Code: 101\nCharacter:   r, ASCII Code: 114\nCharacter:   v, ASCII Code: 118\nCharacter:   i, ASCII Code: 105\nCharacter:   i, ASCII Code: 105\nASCII List: [110, 101, 114, 118, 105, 105]\n", "ground_truth": [110, 101, 114, 118, 105, 105]}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对初始文本进行加密\n加密方案概述：将字符串转换为ASCII编码。\n初始文本: nervii 编码为加密文本: [110, 101, 114, 118, 105, 105]\n初始文本: nonculpability 编码为加密文本是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 80, "source_filename": "icl_with_rule_encode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": "nonculpability", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   o, ASCII Code: 111\nCharacter:   n, ASCII Code: 110\nCharacter:   c, ASCII Code: 99\nCharacter:   u, ASCII Code: 117\nCharacter:   l, ASCII Code: 108\nCharacter:   p, ASCII Code: 112\nCharacter:   a, ASCII Code: 97\nCharacter:   b, ASCII Code: 98\nCharacter:   i, ASCII Code: 105\nCharacter:   l, ASCII Code: 108\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   y, ASCII Code: 121\nASCII List: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n", "ground_truth": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对原始信息进行加密\n加密方案概述：将字符串转换为ASCII编码。\n原始信息: nervii 编码为隐文: [110, 101, 114, 118, 105, 105]\n原始信息: pyritic 编码为隐文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 81, "source_filename": "icl_with_rule_encode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": "pyritic", "extra_args": {}, "output": "Character:   p, ASCII Code: 112\nCharacter:   y, ASCII Code: 121\nCharacter:   r, ASCII Code: 114\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   i, ASCII Code: 105\nCharacter:   c, ASCII Code: 99\nASCII List: [112, 121, 114, 105, 116, 105, 99]\n", "ground_truth": [112, 121, 114, 105, 116, 105, 99]}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules**\n\n- Input:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Output:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Preparation:\n    - Fixed Key: A binary string of length 8 (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Encryption Steps:\n    1. Convert Each Character to Binary:\n        - For each character in the plaintext, convert it to its ASCII value.\n        - Convert this ASCII value to an 8-bit binary string.\n    2. XOR Operation:\n        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.\n        - Ensure the key is repeated to match the length of the binary representation if needed.\n    3. Permutation:\n        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.\n    4. Combine Binary Strings:\n        - Concatenate the binary strings of all characters to form the final ciphertext.\n\n**Decryption Rules**\n\n- Input:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Output:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Preparation:\n    - Fixed Key: Same binary string of length 8 used for encryption (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Decryption Steps:\n    1. Split Ciphertext into Chunks:\n        - Split the binary ciphertext into 8-bit chunks, each representing an encrypted character.\n    2. Inverse Permutation:\n        - Apply the inverse permutation table to each 8-bit chunk to reverse the permutation applied during encryption.\n    3. XOR Operation:\n        - Perform an XOR operation between the permuted binary chunk and the fixed key.\n    4. Convert Binary to Character:\n        - Convert the resulting binary string back to its decimal equivalent.\n        - Convert this decimal value to its corresponding ASCII character.\n    5. Combine Characters:\n        - Concatenate the characters obtained from each binary chunk to form the final plaintext.\ncipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101 decode into plain text: nonculpability\ncipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101 decode into plain text: nonculpability\ncipher text: 110100101101111111111000111110101101010111010101 decode into plain text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 82, "source_filename": "icl_with_rule_decode_Kor_rule21_XORCipher_en.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "110100101101111111111000111110101101010111010101", "extra_args": {}, "output": "1. 接收到的加密文本: 110100101101111111111000111110101101010111010101\n\n处理8位二进制块: 11010010\n2. 应用逆置换: 11100100\n3. 与密钥进行XOR运算: 01001110\n4. 转换为字符: N\n\n处理8位二进制块: 11011111\n2. 应用逆置换: 11101111\n3. 与密钥进行XOR运算: 01000101\n4. 转换为字符: E\n\n处理8位二进制块: 11111000\n2. 应用逆置换: 11111000\n3. 与密钥进行XOR运算: 01010010\n4. 转换为字符: R\n\n处理8位二进制块: 11111010\n2. 应用逆置换: 11111100\n3. 与密钥进行XOR运算: 01010110\n4. 转换为字符: V\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\n**Encryption Rules**\n\n- Input:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Output:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Preparation:\n    - Fixed Key: A binary string of length 8 (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Encryption Steps:\n    1. Convert Each Character to Binary:\n        - For each character in the plaintext, convert it to its ASCII value.\n        - Convert this ASCII value to an 8-bit binary string.\n    2. XOR Operation:\n        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.\n        - Ensure the key is repeated to match the length of the binary representation if needed.\n    3. Permutation:\n        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.\n    4. Combine Binary Strings:\n        - Concatenate the binary strings of all characters to form the final ciphertext.\n\n**Decryption Rules**\n\n- Input:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Output:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Preparation:\n    - Fixed Key: Same binary string of length 8 used for encryption (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Decryption Steps:\n    1. Split Ciphertext into Chunks:\n        - Split the binary ciphertext into 8-bit chunks, each representing an encrypted character.\n    2. Inverse Permutation:\n        - Apply the inverse permutation table to each 8-bit chunk to reverse the permutation applied during encryption.\n    3. XOR Operation:\n        - Perform an XOR operation between the permuted binary chunk and the fixed key.\n    4. Convert Binary to Character:\n        - Convert the resulting binary string back to its decimal equivalent.\n        - Convert this decimal value to its corresponding ASCII character.\n    5. Combine Characters:\n        - Concatenate the characters obtained from each binary chunk to form the final plaintext.\nencoded text: 11111100111101011111100011010101111111101101010111011001 decode to plain text: pyritic\nencoded text: 11111100111101011111100011010101111111101101010111011001 decode to plain text: pyritic\nencoded text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101 decode to plain text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 83, "source_filename": "icl_with_rule_decode_Kor_rule21_XORCipher_en.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101", "extra_args": {}, "output": "1. 接收到的加密文本: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n\n处理8位二进制块: 11010010\n2. 应用逆置换: 11100100\n3. 与密钥进行XOR运算: 01001110\n4. 转换为字符: N\n\n处理8位二进制块: 11010011\n2. 应用逆置换: 11100101\n3. 与密钥进行XOR运算: 01001111\n4. 转换为字符: O\n\n处理8位二进制块: 11010010\n2. 应用逆置换: 11100100\n3. 与密钥进行XOR运算: 01001110\n4. 转换为字符: N\n\n处理8位二进制块: 11011001\n2. 应用逆置换: 11101001\n3. 与密钥进行XOR运算: 01000011\n4. 转换为字符: C\n\n处理8位二进制块: 11111111\n2. 应用逆置换: 11111111\n3. 与密钥进行XOR运算: 01010101\n4. 转换为字符: U\n\n处理8位二进制块: 11010110\n2. 应用逆置换: 11100110\n3. 与密钥进行XOR运算: 01001100\n4. 转换为字符: L\n\n处理8位二进制块: 11111100\n2. 应用逆置换: 11111010\n3. 与密钥进行XOR运算: 01010000\n4. 转换为字符: P\n\n处理8位二进制块: 11011101\n2. 应用逆置换: 11101011\n3. 与密钥进行XOR运算: 01000001\n4. 转换为字符: A\n\n处理8位二进制块: 11011000\n2. 应用逆置换: 11101000\n3. 与密钥进行XOR运算: 01000010\n4. 转换为字符: B\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11010110\n2. 应用逆置换: 11100110\n3. 与密钥进行XOR运算: 01001100\n4. 转换为字符: L\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11111110\n2. 应用逆置换: 11111110\n3. 与密钥进行XOR运算: 01010100\n4. 转换为字符: T\n\n处理8位二进制块: 11110101\n2. 应用逆置换: 11110011\n3. 与密钥进行XOR运算: 01011001\n4. 转换为字符: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules**\n\n- Input:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Output:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Preparation:\n    - Fixed Key: A binary string of length 8 (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Encryption Steps:\n    1. Convert Each Character to Binary:\n        - For each character in the plaintext, convert it to its ASCII value.\n        - Convert this ASCII value to an 8-bit binary string.\n    2. XOR Operation:\n        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.\n        - Ensure the key is repeated to match the length of the binary representation if needed.\n    3. Permutation:\n        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.\n    4. Combine Binary Strings:\n        - Concatenate the binary strings of all characters to form the final ciphertext.\n\n**Decryption Rules**\n\n- Input:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Output:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Preparation:\n    - Fixed Key: Same binary string of length 8 used for encryption (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Decryption Steps:\n    1. Split Ciphertext into Chunks:\n        - Split the binary ciphertext into 8-bit chunks, each representing an encrypted character.\n    2. Inverse Permutation:\n        - Apply the inverse permutation table to each 8-bit chunk to reverse the permutation applied during encryption.\n    3. XOR Operation:\n        - Perform an XOR operation between the permuted binary chunk and the fixed key.\n    4. Convert Binary to Character:\n        - Convert the resulting binary string back to its decimal equivalent.\n        - Convert this decimal value to its corresponding ASCII character.\n    5. Combine Characters:\n        - Concatenate the characters obtained from each binary chunk to form the final plaintext.\ncipher text: 11111100111101011111100011010101111111101101010111011001 decrypt to clear text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 84, "source_filename": "icl_with_rule_decode_Kor_rule21_XORCipher_en.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "11111100111101011111100011010101111111101101010111011001", "extra_args": {}, "output": "1. 接收到的加密文本: 11111100111101011111100011010101111111101101010111011001\n\n处理8位二进制块: 11111100\n2. 应用逆置换: 11111010\n3. 与密钥进行XOR运算: 01010000\n4. 转换为字符: P\n\n处理8位二进制块: 11110101\n2. 应用逆置换: 11110011\n3. 与密钥进行XOR运算: 01011001\n4. 转换为字符: Y\n\n处理8位二进制块: 11111000\n2. 应用逆置换: 11111000\n3. 与密钥进行XOR运算: 01010010\n4. 转换为字符: R\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11111110\n2. 应用逆置换: 11111110\n3. 与密钥进行XOR运算: 01010100\n4. 转换为字符: T\n\n处理8位二进制块: 11010101\n2. 应用逆置换: 11100011\n3. 与密钥进行XOR运算: 01001001\n4. 转换为字符: I\n\n处理8位二进制块: 11011001\n2. 应用逆置换: 11101001\n3. 与密钥进行XOR运算: 01000011\n4. 转换为字符: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对原文进行加密\n\n加密规则:\n\n输入:\n- 原文: 大写字母字符串，不含标点和空格\n- period: 定义内圈多久旋转一次。周期性表示在加密过程中每处理指定数量的字符后，内圈将根据增量值旋转一次\n- increment: 定义内圈每次旋转的字符数。在每个周期结束时，内圈将根据增量值向右旋转相应数量的字符\n\n输出:\n- 暗码: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n加密步骤:\n- 对原文中的每个字符p:\n    - 在外圈找到该字符\n    - 用内圈对应位置的字符替换它\n    - 每加密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n原文: pyritic 编码为暗码: MCZBXSU\n原文: pyritic 编码为暗码: MCZBXSU\n原文: nonculpability 编码为暗码: YNYEVZFGOSBHDF\n原文: nervii 编码为暗码是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 85, "source_filename": "icl_with_rule_encode_Kor_rule9_AlertiCipher_cn.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n加密过程:\n字符 N 在外圈位置 22 对应内圈字符 Y\n字符 E 在外圈位置 2 对应内圈字符 L\n字符 R 在外圈位置 3 对应内圈字符 Z\n字符 V 在外圈位置 20 对应内圈字符 R\n字符 I 在外圈位置 7 对应内圈字符 B\n已处理5个字符，内圈向右旋转4个位置\n字符 I 在外圈位置 7 对应内圈字符 S\n\n最终加密结果: YLZRBS\n", "ground_truth": "YLZRBS"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对非加密信息进行加密\n\n加密规则:\n\n输入:\n- 非加密信息: 大写字母字符串，不含标点和空格\n- period: 定义内圈多久旋转一次。周期性表示在加密过程中每处理指定数量的字符后，内圈将根据增量值旋转一次\n- increment: 定义内圈每次旋转的字符数。在每个周期结束时，内圈将根据增量值向右旋转相应数量的字符\n\n输出:\n- 暗码: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n加密步骤:\n- 对非加密信息中的每个字符p:\n    - 在外圈找到该字符\n    - 用内圈对应位置的字符替换它\n    - 每加密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n非加密信息: nervii 编码为暗码: YLZRBS\n非加密信息: nonculpability 编码为暗码: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 86, "source_filename": "icl_with_rule_encode_Kor_rule9_AlertiCipher_cn.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n加密过程:\n字符 N 在外圈位置 22 对应内圈字符 Y\n字符 O 在外圈位置 8 对应内圈字符 N\n字符 N 在外圈位置 22 对应内圈字符 Y\n字符 C 在外圈位置 19 对应内圈字符 E\n字符 U 在外圈位置 6 对应内圈字符 V\n已处理5个字符，内圈向右旋转4个位置\n字符 L 在外圈位置 25 对应内圈字符 Z\n字符 P 在外圈位置 9 对应内圈字符 F\n字符 A 在外圈位置 10 对应内圈字符 G\n字符 B 在外圈位置 21 对应内圈字符 O\n字符 I 在外圈位置 7 对应内圈字符 S\n已处理5个字符，内圈向右旋转4个位置\n字符 L 在外圈位置 25 对应内圈字符 B\n字符 I 在外圈位置 7 对应内圈字符 H\n字符 T 在外圈位置 4 对应内圈字符 D\n字符 Y 在外圈位置 5 对应内圈字符 F\n\n最终加密结果: YNYEVZFGOSBHDF\n", "ground_truth": "YNYEVZFGOSBHDF"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n\n加密规则:\n\n输入:\n- 初始文本: 大写字母字符串，不含标点和空格\n- period: 定义内圈多久旋转一次。周期性表示在加密过程中每处理指定数量的字符后，内圈将根据增量值旋转一次\n- increment: 定义内圈每次旋转的字符数。在每个周期结束时，内圈将根据增量值向右旋转相应数量的字符\n\n输出:\n- 暗码: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n加密步骤:\n- 对初始文本中的每个字符p:\n    - 在外圈找到该字符\n    - 用内圈对应位置的字符替换它\n    - 每加密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n初始文本: pyritic 编码为暗码是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 87, "source_filename": "icl_with_rule_encode_Kor_rule9_AlertiCipher_cn.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n加密过程:\n字符 P 在外圈位置 9 对应内圈字符 M\n字符 Y 在外圈位置 5 对应内圈字符 C\n字符 R 在外圈位置 3 对应内圈字符 Z\n字符 I 在外圈位置 7 对应内圈字符 B\n字符 T 在外圈位置 4 对应内圈字符 X\n已处理5个字符，内圈向右旋转4个位置\n字符 I 在外圈位置 7 对应内圈字符 S\n字符 C 在外圈位置 19 对应内圈字符 U\n\n最终加密结果: MCZBXSU\n", "ground_truth": "MCZBXSU"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets\n        \n        Use the following 13 cipher alphabets, each associated with two letters:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n        \n    - Standard Alphabet\n        - ABCDEFGHIJKLMNOPQRSTUVWXYZ\n- Encryption Steps:\n    - Pair each letter in the key with each letter in the plaintext. If the key is shorter than the plaintext, repeat the key.\n    - For each given plaintext character p:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark the choice AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of p in the standard alphabet and replace it with the letter at the same position in the cipher alphabet.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets (Same as encryption)\n    - Standard Alphabet(Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Pair each letter in the key with each letter in the ciphertext. If the key is shorter than the plaintext, repeat the key.\n    - For each given ciphertext character c:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark AB Option: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of c in the cipher alphabet and reduce it to the plaintext letter using the same position in the standard alphabet.\nencoded text: LILTBUP secret key or extra parameter: {'key': 'supercanonization'} decode into clear text: pyritic\nencoded text: LILTBUP secret key or extra parameter: {'key': 'supercanonization'} decode into clear text: pyritic\nencoded text: LILTBUP secret key or extra parameter: {'key': 'supercanonization'} decode into clear text: pyritic\nencoded text: HRMISY secret key or extra parameter: {'key': 'paraguayans'} decode into clear text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 88, "source_filename": "icl_with_rule_decode_Kor_rule8_PortaCipher_en.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "HRMISY", "extra_args": {"key": "paraguayans"}, "output": "开始解密文本: HRMISY\n使用密钥: paraguayans\n解密步骤:\n  第1个字符 H 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将密文字符 H 还原为明文字符 N\n  第2个字符 R 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将密文字符 R 还原为明文字符 E\n  第3个字符 M 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将密文字符 M 还原为明文字符 R\n  第4个字符 I 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将密文字符 I 还原为明文字符 V\n  第5个字符 S 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 S 还原为明文字符 I\n  第6个字符 Y 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 Y 还原为明文字符 I\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets\n        \n        Use the following 13 cipher alphabets, each associated with two letters:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n        \n    - Standard Alphabet\n        - ABCDEFGHIJKLMNOPQRSTUVWXYZ\n- Encryption Steps:\n    - Pair each letter in the key with each letter in the plaintext. If the key is shorter than the plaintext, repeat the key.\n    - For each given plaintext character p:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark the choice AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of p in the standard alphabet and replace it with the letter at the same position in the cipher alphabet.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets (Same as encryption)\n    - Standard Alphabet(Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Pair each letter in the key with each letter in the ciphertext. If the key is shorter than the plaintext, repeat the key.\n    - For each given ciphertext character c:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark AB Option: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of c in the cipher alphabet and reduce it to the plaintext letter using the same position in the standard alphabet.\ncipher text: LILTBUP extra parameter: {'key': 'supercanonization'} decode to clear text: pyritic\ncipher text: LILTBUP extra parameter: {'key': 'supercanonization'} decode to clear text: pyritic\ncipher text: HRMISY extra parameter: {'key': 'paraguayans'} decode to clear text: nervii\ncipher text: ALFZLSFVPVOQJC extra parameter: {'key': 'bulgingly'} decode to clear text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 89, "source_filename": "icl_with_rule_decode_Kor_rule8_PortaCipher_en.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "ALFZLSFVPVOQJC", "extra_args": {"key": "bulgingly"}, "output": "开始解密文本: ALFZLSFVPVOQJC\n使用密钥: bulgingly\n解密步骤:\n  第1个字符 A 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将密文字符 A 还原为明文字符 N\n  第2个字符 L 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 L 还原为明文字符 O\n  第3个字符 F 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将密文字符 F 还原为明文字符 N\n  第4个字符 Z 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 Z 还原为明文字符 C\n  第5个字符 L 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将密文字符 L 还原为明文字符 U\n  第6个字符 S 使用密钥字符 N:\n    - 查找密钥字符 N 对应的替换表\n    - 将密文字符 S 还原为明文字符 L\n  第7个字符 F 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 F 还原为明文字符 P\n  第8个字符 V 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将密文字符 V 还原为明文字符 A\n  第9个字符 P 使用密钥字符 Y:\n    - 查找密钥字符 Y 对应的替换表\n    - 将密文字符 P 还原为明文字符 B\n  第10个字符 V 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将密文字符 V 还原为明文字符 I\n  第11个字符 O 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 O 还原为明文字符 L\n  第12个字符 Q 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将密文字符 Q 还原为明文字符 I\n  第13个字符 J 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 J 还原为明文字符 T\n  第14个字符 C 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将密文字符 C 还原为明文字符 Y\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets\n        \n        Use the following 13 cipher alphabets, each associated with two letters:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n        \n    - Standard Alphabet\n        - ABCDEFGHIJKLMNOPQRSTUVWXYZ\n- Encryption Steps:\n    - Pair each letter in the key with each letter in the plaintext. If the key is shorter than the plaintext, repeat the key.\n    - For each given plaintext character p:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark the choice AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of p in the standard alphabet and replace it with the letter at the same position in the cipher alphabet.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets (Same as encryption)\n    - Standard Alphabet(Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Pair each letter in the key with each letter in the ciphertext. If the key is shorter than the plaintext, repeat the key.\n    - For each given ciphertext character c:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark AB Option: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of c in the cipher alphabet and reduce it to the plaintext letter using the same position in the standard alphabet.\ncipher text: LILTBUP extra parameter: {'key': 'supercanonization'} decrypt into original information is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 90, "source_filename": "icl_with_rule_decode_Kor_rule8_PortaCipher_en.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "LILTBUP", "extra_args": {"key": "supercanonization"}, "output": "开始解密文本: LILTBUP\n使用密钥: supercanonization\n解密步骤:\n  第1个字符 L 使用密钥字符 S:\n    - 查找密钥字符 S 对应的替换表\n    - 将密文字符 L 还原为明文字符 P\n  第2个字符 I 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 I 还原为明文字符 Y\n  第3个字符 L 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将密文字符 L 还原为明文字符 R\n  第4个字符 T 使用密钥字符 E:\n    - 查找密钥字符 E 对应的替换表\n    - 将密文字符 T 还原为明文字符 I\n  第5个字符 B 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将密文字符 B 还原为明文字符 T\n  第6个字符 U 使用密钥字符 C:\n    - 查找密钥字符 C 对应的替换表\n    - 将密文字符 U 还原为明文字符 I\n  第7个字符 P 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将密文字符 P 还原为明文字符 C\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\noriginal information: nonculpability encrypt to encrypted text: mlmxfokzyrorgb\noriginal information: nonculpability encrypt to encrypted text: mlmxfokzyrorgb\noriginal information: nervii encrypt to encrypted text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 91, "source_filename": "icl_encode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "nervii", "extra_args": {}, "output": "编码过程:\n转换单词: nervii\n  'n' -> 'm'\n  'e' -> 'v'\n  'r' -> 'i'\n  'v' -> 'e'\n  'i' -> 'r'\n  'i' -> 'r'\n编码后的文本: mvierr\n", "ground_truth": "mvierr"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\nclear text: pyritic encrypt to encoded text: kbirgrx\nclear text: nervii encrypt to encoded text: mvierr\nclear text: pyritic encrypt to encoded text: kbirgrx\nclear text: nervii encrypt to encoded text: mvierr\nclear text: pyritic encrypt to encoded text: kbirgrx\nclear text: nonculpability encrypt to encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 92, "source_filename": "icl_encode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "nonculpability", "extra_args": {}, "output": "编码过程:\n转换单词: nonculpability\n  'n' -> 'm'\n  'o' -> 'l'\n  'n' -> 'm'\n  'c' -> 'x'\n  'u' -> 'f'\n  'l' -> 'o'\n  'p' -> 'k'\n  'a' -> 'z'\n  'b' -> 'y'\n  'i' -> 'r'\n  'l' -> 'o'\n  'i' -> 'r'\n  't' -> 'g'\n  'y' -> 'b'\n编码后的文本: mlmxfokzyrorgb\n", "ground_truth": "mlmxfokzyrorgb"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\nclear text: nonculpability encrypt into encoded text: mlmxfokzyrorgb\nclear text: nervii encrypt into encoded text: mvierr\nclear text: pyritic encrypt into encoded text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 93, "source_filename": "icl_encode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "pyritic", "extra_args": {}, "output": "编码过程:\n转换单词: pyritic\n  'p' -> 'k'\n  'y' -> 'b'\n  'r' -> 'i'\n  'i' -> 'r'\n  't' -> 'g'\n  'i' -> 'r'\n  'c' -> 'x'\n编码后的文本: kbirgrx\n", "ground_truth": "kbirgrx"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对原文进行加密\n加密规则:\n- 输入:\n    - 原文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(Grid0-Grid7)\n- 加密步骤:\n    - 原文按5个字符分组，从0开始编号\n    - 对于5个字符的块:\n        - 使用的网格由grid_index = (i // 5) % 8确定，其中i是块号。整数除法运算符//将左边的数除以右边的数，向下取整结果\n        - 对于当前块中的每个字符:\n            - 如果字符是\"J\"，不加密直接添加到加密块\n            - 否则，在当前网格中找到字符的位置。然后向右下方移动一个网格位置(row+1,col+1)(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为加密字母\n            - 将加密字母添加到加密块\n    - 处理完所有块后，连接加密块形成最终加密消息\n原文: nonculpability 加密为密文: OWOMVEBGKDEDZS\n原文: nonculpability 加密为密文: OWOMVEBGKDEDZS\n原文: nervii 加密为密文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 94, "source_filename": "icl_with_rule_encode_Kor_rule7_PhillipsFigureCipher_cn.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n\n处理第0个块: NERVI\n使用第0号网格\n字母N在位置(2,4)，向右下移动一格得到O\n字母E在位置(1,4)，向右下移动一格得到F\n字母R在位置(3,2)，向右下移动一格得到Y\n字母V在位置(4,0)，向右下移动一格得到H\n字母I在位置(0,2)，向右下移动一格得到D\n加密后的块: OFYHD\n\n处理第1个块: I\n使用第1号网格\n字母I在位置(1,2)，向右下移动一格得到D\n加密后的块: D\n\n最终加密结果: OFYHDD\n", "ground_truth": "OFYHDD"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(Grid0-Grid7)\n- 加密步骤:\n    - 初始文本按5个字符分组，从0开始编号\n    - 对于5个字符的块:\n        - 使用的网格由grid_index = (i // 5) % 8确定，其中i是块号。整数除法运算符//将左边的数除以右边的数，向下取整结果\n        - 对于当前块中的每个字符:\n            - 如果字符是\"J\"，不加密直接添加到加密块\n            - 否则，在当前网格中找到字符的位置。然后向右下方移动一个网格位置(row+1,col+1)(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为加密字母\n            - 将加密字母添加到加密块\n    - 处理完所有块后，连接加密块形成最终加密消息\n初始文本: pyritic 加密成暗文: BSYDZDM\n初始文本: pyritic 加密成暗文: BSYDZDM\n初始文本: pyritic 加密成暗文: BSYDZDM\n初始文本: nonculpability 加密成暗文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 95, "source_filename": "icl_with_rule_encode_Kor_rule7_PhillipsFigureCipher_cn.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n\n处理第0个块: NONCU\n使用第0号网格\n字母N在位置(2,4)，向右下移动一格得到O\n字母O在位置(3,0)，向右下移动一格得到W\n字母N在位置(2,4)，向右下移动一格得到O\n字母C在位置(1,2)，向右下移动一格得到M\n字母U在位置(3,4)，向右下移动一格得到V\n加密后的块: OWOMV\n\n处理第1个块: LPABI\n使用第1号网格\n字母L在位置(1,3)，向右下移动一格得到E\n字母P在位置(1,0)，向右下移动一格得到B\n字母A在位置(2,0)，向右下移动一格得到G\n字母B在位置(2,1)，向右下移动一格得到K\n字母I在位置(1,2)，向右下移动一格得到D\n加密后的块: EBGKD\n\n处理第2个块: LITY\n使用第2号网格\n字母L在位置(2,3)，向右下移动一格得到E\n字母I在位置(2,2)，向右下移动一格得到D\n字母T在位置(0,3)，向右下移动一格得到Z\n字母Y在位置(1,3)，向右下移动一格得到S\n加密后的块: EDZS\n\n最终加密结果: OWOMVEBGKDEDZS\n", "ground_truth": "OWOMVEBGKDEDZS"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(Grid0-Grid7)\n- 加密步骤:\n    - 初始文本按5个字符分组，从0开始编号\n    - 对于5个字符的块:\n        - 使用的网格由grid_index = (i // 5) % 8确定，其中i是块号。整数除法运算符//将左边的数除以右边的数，向下取整结果\n        - 对于当前块中的每个字符:\n            - 如果字符是\"J\"，不加密直接添加到加密块\n            - 否则，在当前网格中找到字符的位置。然后向右下方移动一个网格位置(row+1,col+1)(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为加密字母\n            - 将加密字母添加到加密块\n    - 处理完所有块后，连接加密块形成最终加密消息\n初始文本: nonculpability 编码为暗文: OWOMVEBGKDEDZS\n初始文本: nervii 编码为暗文: OFYHDD\n初始文本: nervii 编码为暗文: OFYHDD\n初始文本: pyritic 编码为暗文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 96, "source_filename": "icl_with_rule_encode_Kor_rule7_PhillipsFigureCipher_cn.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n\n处理第0个块: PYRIT\n使用第0号网格\n字母P在位置(0,0)，向右下移动一格得到B\n字母Y在位置(4,3)，向右下移动一格得到S\n字母R在位置(3,2)，向右下移动一格得到Y\n字母I在位置(0,2)，向右下移动一格得到D\n字母T在位置(3,3)，向右下移动一格得到Z\n加密后的块: BSYDZ\n\n处理第1个块: IC\n使用第1号网格\n字母I在位置(1,2)，向右下移动一格得到D\n字母C在位置(2,2)，向右下移动一格得到M\n加密后的块: DM\n\n最终加密结果: BSYDZDM\n", "ground_truth": "BSYDZDM"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对初始文本进行加密\n加密规则:\n    - 输入:\n        - 初始文本: 仅包含大写字母的字符串，不含标点符号和空格\n    - 输出:\n        - 暗文: 十六进制字符串（包含小写字母a-e）\n    - 准备:\n        - 密钥(SHA哈希值)\n            - 对\"SECRET_KEY\"执行SHA-256运算，得到'73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'\n    - 加密步骤:\n        - 将初始文本字符串转换为字节序列（ASCII编码）\n        - 使用密钥对每个字节进行异或（XOR）运算加密。重复使用密钥使其长度与初始文本字节数相同\n        - 将加密后的字节序列转换为十六进制字符串作为暗文输出\n初始文本: nonculpability 加密为暗文: 3da0640d88362ffe45b41326fbc0\n初始文本: nonculpability 加密为暗文: 3da0640d88362ffe45b41326fbc0\n初始文本: pyritic 加密为暗文: 23b6780789333c\n初始文本: nervii 加密为暗文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 97, "source_filename": "icl_with_rule_encode_Kor_rule25_SHACipher_cn.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n原始输入文本: nervii\n处理后的输入文本: NERVII\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n文本转换为字节序列: b'NERVII'\n开始XOR加密...\n加密后的字节序列: b'=\\xaax\\x18\\x943'\n最终加密结果(十六进制): 3daa78189433\n", "ground_truth": "3daa78189433"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对明文进行加密\n加密规则:\n    - 输入:\n        - 明文: 仅包含大写字母的字符串，不含标点符号和空格\n    - 输出:\n        - 密文: 十六进制字符串（包含小写字母a-e）\n    - 准备:\n        - 密钥(SHA哈希值)\n            - 对\"SECRET_KEY\"执行SHA-256运算，得到'73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'\n    - 加密步骤:\n        - 将明文字符串转换为字节序列（ASCII编码）\n        - 使用密钥对每个字节进行异或（XOR）运算加密。重复使用密钥使其长度与明文字节数相同\n        - 将加密后的字节序列转换为十六进制字符串作为密文输出\n明文: nonculpability 加密成密文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 98, "source_filename": "icl_with_rule_encode_Kor_rule25_SHACipher_cn.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n原始输入文本: nonculpability\n处理后的输入文本: NONCULPABILITY\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n文本转换为字节序列: b'NONCULPABILITY'\n开始XOR加密...\n加密后的字节序列: b'=\\xa0d\\r\\x886/\\xfeE\\xb4\\x13&\\xfb\\xc0'\n最终加密结果(十六进制): 3da0640d88362ffe45b41326fbc0\n", "ground_truth": "3da0640d88362ffe45b41326fbc0"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n加密规则:\n    - 输入:\n        - 初始文本: 仅包含大写字母的字符串，不含标点符号和空格\n    - 输出:\n        - 密文: 十六进制字符串（包含小写字母a-e）\n    - 准备:\n        - 密钥(SHA哈希值)\n            - 对\"SECRET_KEY\"执行SHA-256运算，得到'73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'\n    - 加密步骤:\n        - 将初始文本字符串转换为字节序列（ASCII编码）\n        - 使用密钥对每个字节进行异或（XOR）运算加密。重复使用密钥使其长度与初始文本字节数相同\n        - 将加密后的字节序列转换为十六进制字符串作为密文输出\n初始文本: nonculpability 加密为密文: 3da0640d88362ffe45b41326fbc0\n初始文本: nonculpability 加密为密文: 3da0640d88362ffe45b41326fbc0\n初始文本: nonculpability 加密为密文: 3da0640d88362ffe45b41326fbc0\n初始文本: pyritic 加密为密文: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 99, "source_filename": "icl_with_rule_encode_Kor_rule25_SHACipher_cn.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n原始输入文本: pyritic\n处理后的输入文本: PYRITIC\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n文本转换为字节序列: b'PYRITIC'\n开始XOR加密...\n加密后的字节序列: b'#\\xb6x\\x07\\x893<'\n最终加密结果(十六进制): 23b6780789333c\n", "ground_truth": "23b6780789333c"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Maximum number of characters per line: 5\n- Encryption Steps:\n    - Lines are counted from 1.\n    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.\n        - For example, for the plaintext \"LIDAHELLOWORLD\", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows\n        LIDAH\n        WOLLE\n        ORLD\n    - Then read by column, each column is read from the top down, each read a column plus a \"#\", read out the contents of the final ciphertext.\n        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Decryption Steps (exact opposite of encryption steps):\n    - For each character in the ciphertext, it is written from top to bottom; if it encounters #, it switches to the next column and continues writing until the entire ciphertext has been written.\n        - For example, for the ciphertext LWO#IOR#DLL#ALD#HE#,it is written as\n        LIDAH\n        WOLLE\n        ORLD\n    - Then according to the odd-numbered rows from left to right read, the even number of rows from right to left read the order of sequential reading, the final result is the decrypted plaintext.\n        - For example, for the above written content, the first line, read from left to right as LIDAH, the second line, read from right to left as ELLOW, the third line, read from left to right as ORLD, and ultimately the entire content connected, decrypted plaintext as LIDAHELLOWORLD.\nencrypted text: N#E#R#V#II# decrypt to original information: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 100, "source_filename": "icl_with_rule_decode_Kor_rule17_PathCipher_en.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "N#E#R#V#II#", "extra_args": {}, "output": "1. 收到密文: N#E#R#V#II#\n2. 将密文分割成5列: ['N', 'E', 'R', 'V', 'II']\n3. 按列填充网格:\n   第1列: N \n   第2列: E \n   第3列: R \n   第4列: V \n   第5列: I I \n4. 按特定顺序读取:\n   第1行从左到右读取: N E R V I \n   第2行从右到左读取: I \n5. 最终明文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Maximum number of characters per line: 5\n- Encryption Steps:\n    - Lines are counted from 1.\n    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.\n        - For example, for the plaintext \"LIDAHELLOWORLD\", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows\n        LIDAH\n        WOLLE\n        ORLD\n    - Then read by column, each column is read from the top down, each read a column plus a \"#\", read out the contents of the final ciphertext.\n        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Decryption Steps (exact opposite of encryption steps):\n    - For each character in the ciphertext, it is written from top to bottom; if it encounters #, it switches to the next column and continues writing until the entire ciphertext has been written.\n        - For example, for the ciphertext LWO#IOR#DLL#ALD#HE#,it is written as\n        LIDAH\n        WOLLE\n        ORLD\n    - Then according to the odd-numbered rows from left to right read, the even number of rows from right to left read the order of sequential reading, the final result is the decrypted plaintext.\n        - For example, for the above written content, the first line, read from left to right as LIDAH, the second line, read from right to left as ELLOW, the third line, read from left to right as ORLD, and ultimately the entire content connected, decrypted plaintext as LIDAHELLOWORLD.\nencoded text: NIL#OBI#NAT#CPY#UL# decrypt to plain text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 101, "source_filename": "icl_with_rule_decode_Kor_rule17_PathCipher_en.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "NIL#OBI#NAT#CPY#UL#", "extra_args": {}, "output": "1. 收到密文: NIL#OBI#NAT#CPY#UL#\n2. 将密文分割成5列: ['NIL', 'OBI', 'NAT', 'CPY', 'UL']\n3. 按列填充网格:\n   第1列: N I L \n   第2列: O B I \n   第3列: N A T \n   第4列: C P Y \n   第5列: U L \n4. 按特定顺序读取:\n   第1行从左到右读取: N O N C U \n   第2行从右到左读取: L P A B I \n   第3行从左到右读取: L I T Y \n5. 最终明文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Maximum number of characters per line: 5\n- Encryption Steps:\n    - Lines are counted from 1.\n    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.\n        - For example, for the plaintext \"LIDAHELLOWORLD\", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows\n        LIDAH\n        WOLLE\n        ORLD\n    - Then read by column, each column is read from the top down, each read a column plus a \"#\", read out the contents of the final ciphertext.\n        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Decryption Steps (exact opposite of encryption steps):\n    - For each character in the ciphertext, it is written from top to bottom; if it encounters #, it switches to the next column and continues writing until the entire ciphertext has been written.\n        - For example, for the ciphertext LWO#IOR#DLL#ALD#HE#,it is written as\n        LIDAH\n        WOLLE\n        ORLD\n    - Then according to the odd-numbered rows from left to right read, the even number of rows from right to left read the order of sequential reading, the final result is the decrypted plaintext.\n        - For example, for the above written content, the first line, read from left to right as LIDAH, the second line, read from right to left as ELLOW, the third line, read from left to right as ORLD, and ultimately the entire content connected, decrypted plaintext as LIDAHELLOWORLD.\nencoded text: N#E#R#V#II# decode to plain text: nervii\nencoded text: NIL#OBI#NAT#CPY#UL# decode to plain text: nonculpability\nencoded text: P#Y#R#IC#TI# decode to plain text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 102, "source_filename": "icl_with_rule_decode_Kor_rule17_PathCipher_en.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "P#Y#R#IC#TI#", "extra_args": {}, "output": "1. 收到密文: P#Y#R#IC#TI#\n2. 将密文分割成5列: ['P', 'Y', 'R', 'IC', 'TI']\n3. 按列填充网格:\n   第1列: P \n   第2列: Y \n   第3列: R \n   第4列: I C \n   第5列: T I \n4. 按特定顺序读取:\n   第1行从左到右读取: P Y R I T \n   第2行从右到左读取: I C \n5. 最终明文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 隐文: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 加密步骤:\n    - 对初始文本中的每个字母p:\n    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。\n    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。\n    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的隐文字母。例如，如果经过步骤(2)得到字母D，则映射为J。\n初始文本: pyritic 加密成隐文: MHKVGVF\n初始文本: nonculpability 加密成隐文: ONOFEQMJDVQVGH\n初始文本: nervii 加密成隐文是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 103, "source_filename": "icl_with_rule_encode_Kor_rule1_InverseShiftSubstitutionCipher_cn.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n加密字符 N:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: N -> M\n2. 将得到的字母向前移动4位: M -> Q\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Q -> O\n\n加密字符 E:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: E -> V\n2. 将得到的字母向前移动4位: V -> Z\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Z -> Z\n\n加密字符 R:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: R -> I\n2. 将得到的字母向前移动4位: I -> M\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: M -> K\n\n加密字符 V:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: V -> E\n2. 将得到的字母向前移动4位: E -> I\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: I -> C\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n最终加密结果: OZKCVV\n", "ground_truth": "OZKCVV"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对原文进行加密\n加密规则:\n- 输入:\n    - 原文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 加密步骤:\n    - 对原文中的每个字母p:\n    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。\n    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。\n    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的暗码字母。例如，如果经过步骤(2)得到字母D，则映射为J。\n原文: nonculpability 加密成暗码是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 104, "source_filename": "icl_with_rule_encode_Kor_rule1_InverseShiftSubstitutionCipher_cn.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n加密字符 N:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: N -> M\n2. 将得到的字母向前移动4位: M -> Q\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Q -> O\n\n加密字符 O:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: O -> L\n2. 将得到的字母向前移动4位: L -> P\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: P -> N\n\n加密字符 N:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: N -> M\n2. 将得到的字母向前移动4位: M -> Q\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Q -> O\n\n加密字符 C:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: C -> X\n2. 将得到的字母向前移动4位: X -> B\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: B -> F\n\n加密字符 U:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: U -> F\n2. 将得到的字母向前移动4位: F -> J\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: J -> E\n\n加密字符 L:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: L -> O\n2. 将得到的字母向前移动4位: O -> S\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: S -> Q\n\n加密字符 P:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: P -> K\n2. 将得到的字母向前移动4位: K -> O\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: O -> M\n\n加密字符 A:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: A -> Z\n2. 将得到的字母向前移动4位: Z -> D\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: D -> J\n\n加密字符 B:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: B -> Y\n2. 将得到的字母向前移动4位: Y -> C\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: C -> D\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 L:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: L -> O\n2. 将得到的字母向前移动4位: O -> S\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: S -> Q\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 T:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: T -> G\n2. 将得到的字母向前移动4位: G -> K\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: K -> G\n\n加密字符 Y:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: Y -> B\n2. 将得到的字母向前移动4位: B -> F\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: F -> H\n\n最终加密结果: ONOFEQMJDVQVGH\n", "ground_truth": "ONOFEQMJDVQVGH"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 加密步骤:\n    - 对原始信息中的每个字母p:\n    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。\n    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。\n    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的加密文本字母。例如，如果经过步骤(2)得到字母D，则映射为J。\n原始信息: nervii 加密成为加密文本: OZKCVV\n原始信息: nervii 加密成为加密文本: OZKCVV\n原始信息: pyritic 加密成为加密文本: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 105, "source_filename": "icl_with_rule_encode_Kor_rule1_InverseShiftSubstitutionCipher_cn.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n加密字符 P:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: P -> K\n2. 将得到的字母向前移动4位: K -> O\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: O -> M\n\n加密字符 Y:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: Y -> B\n2. 将得到的字母向前移动4位: B -> F\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: F -> H\n\n加密字符 R:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: R -> I\n2. 将得到的字母向前移动4位: I -> M\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: M -> K\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 T:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: T -> G\n2. 将得到的字母向前移动4位: G -> K\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: K -> G\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 C:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: C -> X\n2. 将得到的字母向前移动4位: X -> B\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: B -> F\n\n最终加密结果: MHKVGVF\n", "ground_truth": "MHKVGVF"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\nencoded text: JK FI LL KS CE GE NY secret key or extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decrypt to clear text: nonculpability\nencoded text: JK FI LL KS CE GE NY secret key or extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decrypt to clear text: nonculpability\nencoded text: RV RB RD WX secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} decrypt to clear text: pyritic\nencoded text: JV KT CA secret key or extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} decrypt to clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 106, "source_filename": "icl_decode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "JV KT CA", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将密文分成两个字母一组: ['JV', 'KT', 'CA']\n\nStep 4.1: 解密字母对 JV\n  - J 在 M1 中的位置: (2, 4)\n  - V 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 4) 的字母: E\n  - 解密结果为: NE\n\nStep 4.2: 解密字母对 KT\n  - K 在 M1 中的位置: (3, 0)\n  - T 在 M2 中的位置: (4, 1)\n  - 交换列索引后新位置: (3, 1) 和 (4, 0)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (4, 0) 的字母: V\n  - 解密结果为: RV\n\nStep 4.3: 解密字母对 CA\n  - C 在 M1 中的位置: (1, 3)\n  - A 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: II\n\n解密完成！最终解密结果为: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\nencrypted text: RV RB RD WX secret key or extra parameter: {'str1': 'sawbill', 'str2': 'houri'} decrypt into original information: pyritic\nencrypted text: JV KT CA secret key or extra parameter: {'str1': 'parametral', 'str2': 'quiverful'} decrypt into original information: nervii\nencrypted text: JK FI LL KS CE GE NY secret key or extra parameter: {'str1': 'trichophytia', 'str2': 'sphinges'} decrypt into original information: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 107, "source_filename": "icl_decode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "JK FI LL KS CE GE NY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['JK', 'FI', 'LL', 'KS', 'CE', 'GE', 'NY']\n\nStep 4.1: 解密字母对 JK\n  - J 在 M1 中的位置: (2, 4)\n  - K 在 M2 中的位置: (2, 3)\n  - 交换列索引后新位置: (2, 3) 和 (2, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (2, 4) 的字母: O\n  - 解密结果为: NO\n\nStep 4.2: 解密字母对 FI\n  - F 在 M1 中的位置: (2, 2)\n  - I 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 2)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 解密结果为: NC\n\nStep 4.3: 解密字母对 LL\n  - L 在 M1 中的位置: (3, 1)\n  - L 在 M2 中的位置: (2, 4)\n  - 交换列索引后新位置: (3, 4) 和 (2, 1)\n  - 在 M_T 中查找位置 (3, 4) 的字母: U\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 解密结果为: UL\n\nStep 4.4: 解密字母对 KS\n  - K 在 M1 中的位置: (3, 0)\n  - S 在 M2 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (0, 0) 的字母: A\n  - 解密结果为: PA\n\nStep 4.5: 解密字母对 CE\n  - C 在 M1 中的位置: (0, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (0, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (0, 1) 的字母: B\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: BI\n\nStep 4.6: 解密字母对 GE\n  - G 在 M1 中的位置: (2, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (2, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: LI\n\nStep 4.7: 解密字母对 NY\n  - N 在 M1 中的位置: (3, 3)\n  - Y 在 M2 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: TY\n\n解密完成！最终解密结果为: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\nencrypted text: JV KT CA secret key: {'str1': 'parametral', 'str2': 'quiverful'} decode to original information: nervii\nencrypted text: JK FI LL KS CE GE NY secret key: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to original information: nonculpability\nencrypted text: JK FI LL KS CE GE NY secret key: {'str1': 'trichophytia', 'str2': 'sphinges'} decode to original information: nonculpability\nencrypted text: RV RB RD WX secret key: {'str1': 'sawbill', 'str2': 'houri'} decode to original information: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 108, "source_filename": "icl_decode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "RV RB RD WX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['RV', 'RB', 'RD', 'WX']\n\nStep 4.1: 解密字母对 RV\n  - R 在 M1 中的位置: (3, 3)\n  - V 在 M2 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: PY\n\nStep 4.2: 解密字母对 RB\n  - R 在 M1 中的位置: (3, 3)\n  - B 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (3, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: RI\n\nStep 4.3: 解密字母对 RD\n  - R 在 M1 中的位置: (3, 3)\n  - D 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: TI\n\nStep 4.4: 解密字母对 WX\n  - W 在 M1 中的位置: (0, 2)\n  - X 在 M2 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 在 M_T 中查找位置 (4, 2) 的字母: X\n  - 解密结果为: CX\n\n解密完成！最终解密结果为: PYRITICX\n", "ground_truth": "PYRITICX"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对原文进行加密\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将原文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n原文: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成暗文: RV RB RD WX\n原文: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成暗文: RV RB RD WX\n原文: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成暗文: RV RB RD WX\n原文: nervii 密钥: {'str1': 'parametral', 'str2': 'quiverful'} 加密成暗文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 109, "source_filename": "icl_with_rule_encode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "NERVII", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将明文分成两个字母一组: ['NE', 'RV', 'II']\n\nStep 4.1: 加密字母对 NE\n  - N 在 M_T 中的位置: (2, 3)\n  - E 在 M_T 中的位置: (0, 4)\n  - 交换列索引后新位置: (2, 4) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (0, 3) 的字母: V\n  - 加密结果为: JV\n\nStep 4.2: 加密字母对 RV\n  - R 在 M_T 中的位置: (3, 1)\n  - V 在 M_T 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 1)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (4, 1) 的字母: T\n  - 加密结果为: KT\n\nStep 4.3: 加密字母对 II\n  - I 在 M_T 中的位置: (1, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (1, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 3) 的字母: A\n  - 加密结果为: CA\n\n加密完成！最终加密结果为: JV KT CA\n", "ground_truth": "JV KT CA"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对初始文本进行加密\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将初始文本文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n初始文本: pyritic 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 编码为隐文: RV RB RD WX\n初始文本: nervii 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 编码为隐文: JV KT CA\n初始文本: pyritic 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 编码为隐文: RV RB RD WX\n初始文本: nonculpability 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 编码为隐文: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 110, "source_filename": "icl_with_rule_encode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "NONCULPABILITY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将明文分成两个字母一组: ['NO', 'NC', 'UL', 'PA', 'BI', 'LI', 'TY']\n\nStep 4.1: 加密字母对 NO\n  - N 在 M_T 中的位置: (2, 3)\n  - O 在 M_T 中的位置: (2, 4)\n  - 交换列索引后新位置: (2, 4) 和 (2, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (2, 3) 的字母: K\n  - 加密结果为: JK\n\nStep 4.2: 加密字母对 NC\n  - N 在 M_T 中的位置: (2, 3)\n  - C 在 M_T 中的位置: (0, 2)\n  - 交换列索引后新位置: (2, 2) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 2) 的字母: F\n  - 在 M2 中查找位置 (0, 3) 的字母: I\n  - 加密结果为: FI\n\nStep 4.3: 加密字母对 UL\n  - U 在 M_T 中的位置: (3, 4)\n  - L 在 M_T 中的位置: (2, 1)\n  - 交换列索引后新位置: (3, 1) 和 (2, 4)\n  - 在 M1 中查找位置 (3, 1) 的字母: L\n  - 在 M2 中查找位置 (2, 4) 的字母: L\n  - 加密结果为: LL\n\nStep 4.4: 加密字母对 PA\n  - P 在 M_T 中的位置: (3, 0)\n  - A 在 M_T 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (0, 0) 的字母: S\n  - 加密结果为: KS\n\nStep 4.5: 加密字母对 BI\n  - B 在 M_T 中的位置: (0, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (0, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (0, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: CE\n\nStep 4.6: 加密字母对 LI\n  - L 在 M_T 中的位置: (2, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (2, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (2, 3) 的字母: G\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: GE\n\nStep 4.7: 加密字母对 TY\n  - T 在 M_T 中的位置: (3, 3)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: N\n  - 在 M2 中查找位置 (4, 3) 的字母: Y\n  - 加密结果为: NY\n\n加密完成！最终加密结果为: JK FI LL KS CE GE NY\n", "ground_truth": "JK FI LL KS CE GE NY"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对明文进行加密\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n明文: pyritic 额外参数: {'str1': 'sawbill', 'str2': 'houri'} 加密为暗文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 111, "source_filename": "icl_with_rule_encode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "PYRITICX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\n明文字符个数为奇数，补一个字符 X\nStep 3: 将明文分成两个字母一组: ['PY', 'RI', 'TI', 'CX']\n\nStep 4.1: 加密字母对 PY\n  - P 在 M_T 中的位置: (3, 0)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 0)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (4, 0) 的字母: V\n  - 加密结果为: RV\n\nStep 4.2: 加密字母对 RI\n  - R 在 M_T 中的位置: (3, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 1) 的字母: B\n  - 加密结果为: RB\n\nStep 4.3: 加密字母对 TI\n  - T 在 M_T 中的位置: (3, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 3) 的字母: D\n  - 加密结果为: RD\n\nStep 4.4: 加密字母对 CX\n  - C 在 M_T 中的位置: (0, 2)\n  - X 在 M_T 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M1 中查找位置 (0, 2) 的字母: W\n  - 在 M2 中查找位置 (4, 2) 的字母: X\n  - 加密结果为: WX\n\n加密完成！最终加密结果为: RV RB RD WX\n", "ground_truth": "RV RB RD WX"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对原文进行加密\n\n加密规则:\n- 输入:\n    - 原文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 密文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - e: 263\n    - n: 299\n- 加密步骤:\n    - 对原文中的每个字母p:\n        - 获取字母p对应的ASCII码的十进制数x\n        - 计算x^e mod n作为该字母的密文数字y，这里^表示乘法运算\n    - 最后，将所有y用逗号连接形成最终密文\n\n原文: pyritic 编码为密文: 228,84,62,213,89,213,241\n原文: nervii 编码为密文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 112, "source_filename": "icl_with_rule_encode_Kor_rule23_RSACipher_cn.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NERVII\n使用参数: e=263, n=299\n\n逐字符加密:\n字符 N 的ASCII码为 78\n计算 78^263 mod 299 = 156\n字符 E 的ASCII码为 69\n计算 69^263 mod 299 = 23\n字符 R 的ASCII码为 82\n计算 82^263 mod 299 = 62\n字符 V 的ASCII码为 86\n计算 86^263 mod 299 = 226\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n\n最终加密结果: 156,23,62,226,213,213\n", "ground_truth": "156,23,62,226,213,213"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对非加密信息进行加密\n\n加密规则:\n- 输入:\n    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 隐文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - e: 263\n    - n: 299\n- 加密步骤:\n    - 对非加密信息中的每个字母p:\n        - 获取字母p对应的ASCII码的十进制数x\n        - 计算x^e mod n作为该字母的隐文数字y，这里^表示乘法运算\n    - 最后，将所有y用逗号连接形成最终隐文\n\n非加密信息: pyritic 加密为隐文: 228,84,62,213,89,213,241\n非加密信息: nervii 加密为隐文: 156,23,62,226,213,213\n非加密信息: pyritic 加密为隐文: 228,84,62,213,89,213,241\n非加密信息: nonculpability 加密为隐文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 113, "source_filename": "icl_with_rule_encode_Kor_rule23_RSACipher_cn.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NONCULPABILITY\n使用参数: e=263, n=299\n\n逐字符加密:\n字符 N 的ASCII码为 78\n计算 78^263 mod 299 = 156\n字符 O 的ASCII码为 79\n计算 79^263 mod 299 = 53\n字符 N 的ASCII码为 78\n计算 78^263 mod 299 = 156\n字符 C 的ASCII码为 67\n计算 67^263 mod 299 = 241\n字符 U 的ASCII码为 85\n计算 85^263 mod 299 = 197\n字符 L 的ASCII码为 76\n计算 76^263 mod 299 = 240\n字符 P 的ASCII码为 80\n计算 80^263 mod 299 = 228\n字符 A 的ASCII码为 65\n计算 65^263 mod 299 = 247\n字符 B 的ASCII码为 66\n计算 66^263 mod 299 = 222\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 L 的ASCII码为 76\n计算 76^263 mod 299 = 240\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 T 的ASCII码为 84\n计算 84^263 mod 299 = 89\n字符 Y 的ASCII码为 89\n计算 89^263 mod 299 = 84\n\n最终加密结果: 156,53,156,241,197,240,228,247,222,213,240,213,89,84\n", "ground_truth": "156,53,156,241,197,240,228,247,222,213,240,213,89,84"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对初始文本进行加密\n\n加密规则:\n- 输入:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密信息: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - e: 263\n    - n: 299\n- 加密步骤:\n    - 对初始文本中的每个字母p:\n        - 获取字母p对应的ASCII码的十进制数x\n        - 计算x^e mod n作为该字母的加密信息数字y，这里^表示乘法运算\n    - 最后，将所有y用逗号连接形成最终加密信息\n\n初始文本: nervii 加密成为加密信息: 156,23,62,226,213,213\n初始文本: nonculpability 加密成为加密信息: 156,53,156,241,197,240,228,247,222,213,240,213,89,84\n初始文本: pyritic 加密成为加密信息: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 114, "source_filename": "icl_with_rule_encode_Kor_rule23_RSACipher_cn.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: PYRITIC\n使用参数: e=263, n=299\n\n逐字符加密:\n字符 P 的ASCII码为 80\n计算 80^263 mod 299 = 228\n字符 Y 的ASCII码为 89\n计算 89^263 mod 299 = 84\n字符 R 的ASCII码为 82\n计算 82^263 mod 299 = 62\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 T 的ASCII码为 84\n计算 84^263 mod 299 = 89\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 C 的ASCII码为 67\n计算 67^263 mod 299 = 241\n\n最终加密结果: 228,84,62,213,89,213,241\n", "ground_truth": "228,84,62,213,89,213,241"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 明文: 大写字母和空格组成的字符串\n    - 输出:\n        - 密文: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - S盒\n            - S_BOX = {\n            0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n            0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n            0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n            0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n            }\n    - 加密步骤:\n        1. 填充: 如果明文长度不是8字节的倍数，用\\x00(空字符)填充使其长度成为8字节的倍数\n        2. 分块: 将填充后的明文分成8字节的块\n        3. 块加密:\n            - 转换为字节: 使用ASCII编码将每个块转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 替换: 使用S盒替换每个字节的高4位和低4位并拼接\n            - 置换: 通过将每个字节左移1位进行简单置换\n            - 与密钥XOR: 置换后的字节块再次与固定密钥进行XOR运算\n        4. 十六进制编码: 将加密后的字节块转换为十六进制字符串\n        5. 拼接: 将所有加密块的十六进制字符串拼接形成最终密文\n解密算法:\n解密规则:\n    - 输入:\n        - 密文: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 输出:\n        - 明文: 大写字母和空格组成的字符串\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - 逆S盒\n            - INV_S_BOX = {\n            0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,\n            0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,\n            0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,\n            0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F\n            }\n    - 解密步骤:\n        1. 分块: 将密文分成16字符(8字节)的块\n        2. 块解密:\n            - 转换为字节: 将每个块从十六进制字符串转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 逆置换: 通过将每个字节右移1位进行逆置换\n            - 替换: 使用逆S盒替换块中字节的高4位和低4位并拼接\n            - 与密钥XOR: 字节块再次与固定密钥进行XOR运算\n        3. 转换为文本: 使用ASCII解码将解密后的字节块转换回文本\n        4. 移除填充: 从解密后的明文末尾移除填充字符(\\x00)\n        5. 拼接: 将所有解密块拼接形成最终明文\n密文: 293637348A3E96283B3E2B30808F9784 解码为明文: nonculpability\n密文: 293637348A3E96283B3E2B30808F9784 解码为明文: nonculpability\n密文: 2932869B372E9784 解码为明文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 115, "source_filename": "icl_with_rule_decode_Kor_rule22_SBOXCipher_cn.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "2932869B372E9784", "extra_args": {}, "output": "开始解密过程...\n1. 接收到的密文: 2932869B372E9784\n2. 开始分块处理...\n\n处理第1个块: 2932869B372E9784\n  - 转换为字节: 2932869B372E9784\n  - 与密钥XOR后: 1800B5AF0218A0BC\n  - 右移置换后: 0C00DAD7010C505E\n  - 逆S盒替换后: 7F7761627C7F3738\n  - 最终与密钥XOR后: 4E45525649490000\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 原始信息: 大写字母和空格组成的字符串\n    - 输出:\n        - 加密信息: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - S盒\n            - S_BOX = {\n            0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n            0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n            0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n            0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n            }\n    - 加密步骤:\n        1. 填充: 如果原始信息长度不是8字节的倍数，用\\x00(空字符)填充使其长度成为8字节的倍数\n        2. 分块: 将填充后的原始信息分成8字节的块\n        3. 块加密:\n            - 转换为字节: 使用ASCII编码将每个块转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 替换: 使用S盒替换每个字节的高4位和低4位并拼接\n            - 置换: 通过将每个字节左移1位进行简单置换\n            - 与密钥XOR: 置换后的字节块再次与固定密钥进行XOR运算\n        4. 十六进制编码: 将加密后的字节块转换为十六进制字符串\n        5. 拼接: 将所有加密块的十六进制字符串拼接形成最终加密信息\n解密算法:\n解密规则:\n    - 输入:\n        - 加密信息: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 输出:\n        - 原始信息: 大写字母和空格组成的字符串\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - 逆S盒\n            - INV_S_BOX = {\n            0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,\n            0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,\n            0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,\n            0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F\n            }\n    - 解密步骤:\n        1. 分块: 将加密信息分成16字符(8字节)的块\n        2. 块解密:\n            - 转换为字节: 将每个块从十六进制字符串转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 逆置换: 通过将每个字节右移1位进行逆置换\n            - 替换: 使用逆S盒替换块中字节的高4位和低4位并拼接\n            - 与密钥XOR: 字节块再次与固定密钥进行XOR运算\n        3. 转换为文本: 使用ASCII解码将解密后的字节块转换回文本\n        4. 移除填充: 从解密后的原始信息末尾移除填充字符(\\x00)\n        5. 拼接: 将所有解密块拼接形成最终原始信息\n加密信息: 2932869B372E9784 解密成为原始信息: nervii\n加密信息: 2932869B372E9784 解密成为原始信息: nervii\n加密信息: 849F8630802E2584 解密成为原始信息: pyritic\n加密信息: 293637348A3E96283B3E2B30808F9784 解密成为原始信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 116, "source_filename": "icl_with_rule_decode_Kor_rule22_SBOXCipher_cn.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "293637348A3E96283B3E2B30808F9784", "extra_args": {}, "output": "开始解密过程...\n1. 接收到的密文: 293637348A3E96283B3E2B30808F9784\n2. 开始分块处理...\n\n处理第1个块: 293637348A3E9628\n  - 转换为字节: 293637348A3E9628\n  - 与密钥XOR后: 18040400BF08A110\n  - 右移置换后: 0C020200DF04D008\n  - 逆S盒替换后: 7F7D7D77607A6779\n  - 最终与密钥XOR后: 4E4F4E43554C5041\n\n处理第2个块: 3B3E2B30808F9784\n  - 转换为字节: 3B3E2B30808F9784\n  - 与密钥XOR后: 0A0C1804B5B9A0BC\n  - 右移置换后: 05060C02DADC505E\n  - 逆S盒替换后: 737B7F7D616F3738\n  - 最终与密钥XOR后: 42494C4954590000\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 初始文本: 大写字母和空格组成的字符串\n    - 输出:\n        - 加密文本: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - S盒\n            - S_BOX = {\n            0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n            0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n            0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n            0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n            }\n    - 加密步骤:\n        1. 填充: 如果初始文本长度不是8字节的倍数，用\\x00(空字符)填充使其长度成为8字节的倍数\n        2. 分块: 将填充后的初始文本分成8字节的块\n        3. 块加密:\n            - 转换为字节: 使用ASCII编码将每个块转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 替换: 使用S盒替换每个字节的高4位和低4位并拼接\n            - 置换: 通过将每个字节左移1位进行简单置换\n            - 与密钥XOR: 置换后的字节块再次与固定密钥进行XOR运算\n        4. 十六进制编码: 将加密后的字节块转换为十六进制字符串\n        5. 拼接: 将所有加密块的十六进制字符串拼接形成最终加密文本\n解密算法:\n解密规则:\n    - 输入:\n        - 加密文本: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 输出:\n        - 初始文本: 大写字母和空格组成的字符串\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - 逆S盒\n            - INV_S_BOX = {\n            0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,\n            0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,\n            0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,\n            0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F\n            }\n    - 解密步骤:\n        1. 分块: 将加密文本分成16字符(8字节)的块\n        2. 块解密:\n            - 转换为字节: 将每个块从十六进制字符串转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 逆置换: 通过将每个字节右移1位进行逆置换\n            - 替换: 使用逆S盒替换块中字节的高4位和低4位并拼接\n            - 与密钥XOR: 字节块再次与固定密钥进行XOR运算\n        3. 转换为文本: 使用ASCII解码将解密后的字节块转换回文本\n        4. 移除填充: 从解密后的初始文本末尾移除填充字符(\\x00)\n        5. 拼接: 将所有解密块拼接形成最终初始文本\n加密文本: 2932869B372E9784 解密成初始文本: nervii\n加密文本: 293637348A3E96283B3E2B30808F9784 解密成初始文本: nonculpability\n加密文本: 849F8630802E2584 解密成初始文本: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 117, "source_filename": "icl_with_rule_decode_Kor_rule22_SBOXCipher_cn.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "849F8630802E2584", "extra_args": {}, "output": "开始解密过程...\n1. 接收到的密文: 849F8630802E2584\n2. 开始分块处理...\n\n处理第1个块: 849F8630802E2584\n  - 转换为字节: 849F8630802E2584\n  - 与密钥XOR后: B5ADB504B51812BC\n  - 右移置换后: DAD6DA02DA0C095E\n  - 逆S盒替换后: 616B617D617F7438\n  - 最终与密钥XOR后: 5059524954494300\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n暗码: mlmxfokzyrorgb 解密为原始信息: nonculpability\n暗码: mlmxfokzyrorgb 解密为原始信息: nonculpability\n暗码: kbirgrx 解密为原始信息: pyritic\n暗码: mlmxfokzyrorgb 解密为原始信息: nonculpability\n暗码: mvierr 解密为原始信息是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 118, "source_filename": "icl_decode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "mvierr", "extra_args": {}, "output": "解码过程:\n转换单词: mvierr\n  'm' -> 'n'\n  'v' -> 'e'\n  'i' -> 'r'\n  'e' -> 'v'\n  'r' -> 'i'\n  'r' -> 'i'\n解码后的文本: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n加密文本: mvierr 解码为原始信息: nervii\n加密文本: mvierr 解码为原始信息: nervii\n加密文本: mvierr 解码为原始信息: nervii\n加密文本: mlmxfokzyrorgb 解码为原始信息是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 119, "source_filename": "icl_decode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "mlmxfokzyrorgb", "extra_args": {}, "output": "解码过程:\n转换单词: mlmxfokzyrorgb\n  'm' -> 'n'\n  'l' -> 'o'\n  'm' -> 'n'\n  'x' -> 'c'\n  'f' -> 'u'\n  'o' -> 'l'\n  'k' -> 'p'\n  'z' -> 'a'\n  'y' -> 'b'\n  'r' -> 'i'\n  'o' -> 'l'\n  'r' -> 'i'\n  'g' -> 't'\n  'b' -> 'y'\n解码后的文本: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "加密信息: mvierr 解密为初始文本: nervii\n加密信息: kbirgrx 解密为初始文本是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 120, "source_filename": "icl_decode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "kbirgrx", "extra_args": {}, "output": "解码过程:\n转换单词: kbirgrx\n  'k' -> 'p'\n  'b' -> 'y'\n  'i' -> 'r'\n  'r' -> 'i'\n  'g' -> 't'\n  'r' -> 'i'\n  'x' -> 'c'\n解码后的文本: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - Grid and Template:\n        - Prepare a blank grid and a template with holes (fence).\n        - The template used is:\n            \n            ▮  ▮  ▮  ▮\n            ▮  ▮  ▯  ▯\n            ▮  ▮  ▮  ▯\n            ▯  ▮  ▮  ▮\n            \n            Where the white ones are holes, place the template on a blank grid, only through the white holes, and fill in the corresponding positions of the grid with characters.\n            \n- Encryption Steps:\n    - Divide the plaintext into blocks of 16 letters one by one.(If the plaintext is less than 16 long it is one  block)\n        - For example, the plaintext message \"LOVEHELLOWORLD\" is less than 16 and is one block.\n    - For each block, perform the following encryption operation:\n        - Place the template with holes on the blank grid.\n        - Fill the grid with the letters of the plaintext message in order through the holes in the template.\n            - For example, the plaintext is \"LOVEHELLOWORLD\", through the four holes, after filling in the LOVE grid is\n                \n                [['', '', '', ''],\n                ['', '', 'L', 'O'],\n                ['', '', '', 'V'],\n                ['E', '', '', '']]\n                \n        - The template has a total of four holes, so after filling out the four letters, it can not be filled out, then you need to rotate the template counterclockwise 90 degrees.\n            - For example, after filling in LOVE after rotating the template after the template for:\n                \n                ▮  ▯  ▯  ▮\n                ▮  ▯  ▮  ▮\n                ▮  ▮  ▮  ▮\n                ▮  ▮  ▮  ▯\n                \n        - Repeat filling the visible holes in the grid with the next letter of the plaintext message and rotating the template until the entire grid is completely filled. This will perform 4 fills + rotations and eventually the template will turn to the original template. If the message is not enough to fill the entire grid, it is made up with filler characters (e.g. '#').\n            - For example, if the plain text is \"LOVEHELLOWORLD\", after several fills and rotations, the final grid will be\n                \n                [['L', 'H', 'E', 'O'],\n                ['W', 'L', 'L', 'O'],\n                ['O', 'R', 'D', 'V'],\n                ['E', '#', '#', 'L']]\n                \n        - Once the grid is filled, read the contents of the grid line by line as the encrypted message for that block.\n        - When moving on to the next block, empty the contents of the grid and redo the entire fill and rotate operation.\n        \n        Finally, the encrypted messages of all the blocks are connected together to be the final ciphertext.\n        \n\n \noriginal information: nervii encrypt into cipher text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 121, "source_filename": "icl_with_rule_encode_Kor_rule18_RotatingGridCipher_en.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始加密过程:\n\n处理第1个块:\n当前块的文本: NERVII\n文本长度不足16，用#补充\n补充后的文本: NERVII##########\n\n填充后的网格:\n[['#' 'I' 'I' '#']\n ['#' '#' 'N' 'E']\n ['#' '#' '#' 'R']\n ['V' '#' '#' '#']]\n当前块的密文: #II###NE###RV###\n\n最终密文: #II###NE###RV###\n", "ground_truth": "#II###NE###RV###"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - Grid and Template:\n        - Prepare a blank grid and a template with holes (fence).\n        - The template used is:\n            \n            ▮  ▮  ▮  ▮\n            ▮  ▮  ▯  ▯\n            ▮  ▮  ▮  ▯\n            ▯  ▮  ▮  ▮\n            \n            Where the white ones are holes, place the template on a blank grid, only through the white holes, and fill in the corresponding positions of the grid with characters.\n            \n- Encryption Steps:\n    - Divide the plaintext into blocks of 16 letters one by one.(If the plaintext is less than 16 long it is one  block)\n        - For example, the plaintext message \"LOVEHELLOWORLD\" is less than 16 and is one block.\n    - For each block, perform the following encryption operation:\n        - Place the template with holes on the blank grid.\n        - Fill the grid with the letters of the plaintext message in order through the holes in the template.\n            - For example, the plaintext is \"LOVEHELLOWORLD\", through the four holes, after filling in the LOVE grid is\n                \n                [['', '', '', ''],\n                ['', '', 'L', 'O'],\n                ['', '', '', 'V'],\n                ['E', '', '', '']]\n                \n        - The template has a total of four holes, so after filling out the four letters, it can not be filled out, then you need to rotate the template counterclockwise 90 degrees.\n            - For example, after filling in LOVE after rotating the template after the template for:\n                \n                ▮  ▯  ▯  ▮\n                ▮  ▯  ▮  ▮\n                ▮  ▮  ▮  ▮\n                ▮  ▮  ▮  ▯\n                \n        - Repeat filling the visible holes in the grid with the next letter of the plaintext message and rotating the template until the entire grid is completely filled. This will perform 4 fills + rotations and eventually the template will turn to the original template. If the message is not enough to fill the entire grid, it is made up with filler characters (e.g. '#').\n            - For example, if the plain text is \"LOVEHELLOWORLD\", after several fills and rotations, the final grid will be\n                \n                [['L', 'H', 'E', 'O'],\n                ['W', 'L', 'L', 'O'],\n                ['O', 'R', 'D', 'V'],\n                ['E', '#', '#', 'L']]\n                \n        - Once the grid is filled, read the contents of the grid line by line as the encrypted message for that block.\n        - When moving on to the next block, empty the contents of the grid and redo the entire fill and rotate operation.\n        \n        Finally, the encrypted messages of all the blocks are connected together to be the final ciphertext.\n        \n\n \noriginal information: nonculpability encrypt to encoded text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 122, "source_filename": "icl_with_rule_encode_Kor_rule18_RotatingGridCipher_en.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始加密过程:\n\n处理第1个块:\n当前块的文本: NONCULPABILITY\n文本长度不足16，用#补充\n补充后的文本: NONCULPABILITY##\n\n填充后的网格:\n[['T' 'U' 'L' 'B']\n ['I' 'P' 'N' 'O']\n ['L' 'I' 'Y' 'N']\n ['C' '#' '#' 'A']]\n当前块的密文: TULBIPNOLIYNC##A\n\n最终密文: TULBIPNOLIYNC##A\n", "ground_truth": "TULBIPNOLIYNC##A"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - Grid and Template:\n        - Prepare a blank grid and a template with holes (fence).\n        - The template used is:\n            \n            ▮  ▮  ▮  ▮\n            ▮  ▮  ▯  ▯\n            ▮  ▮  ▮  ▯\n            ▯  ▮  ▮  ▮\n            \n            Where the white ones are holes, place the template on a blank grid, only through the white holes, and fill in the corresponding positions of the grid with characters.\n            \n- Encryption Steps:\n    - Divide the plaintext into blocks of 16 letters one by one.(If the plaintext is less than 16 long it is one  block)\n        - For example, the plaintext message \"LOVEHELLOWORLD\" is less than 16 and is one block.\n    - For each block, perform the following encryption operation:\n        - Place the template with holes on the blank grid.\n        - Fill the grid with the letters of the plaintext message in order through the holes in the template.\n            - For example, the plaintext is \"LOVEHELLOWORLD\", through the four holes, after filling in the LOVE grid is\n                \n                [['', '', '', ''],\n                ['', '', 'L', 'O'],\n                ['', '', '', 'V'],\n                ['E', '', '', '']]\n                \n        - The template has a total of four holes, so after filling out the four letters, it can not be filled out, then you need to rotate the template counterclockwise 90 degrees.\n            - For example, after filling in LOVE after rotating the template after the template for:\n                \n                ▮  ▯  ▯  ▮\n                ▮  ▯  ▮  ▮\n                ▮  ▮  ▮  ▮\n                ▮  ▮  ▮  ▯\n                \n        - Repeat filling the visible holes in the grid with the next letter of the plaintext message and rotating the template until the entire grid is completely filled. This will perform 4 fills + rotations and eventually the template will turn to the original template. If the message is not enough to fill the entire grid, it is made up with filler characters (e.g. '#').\n            - For example, if the plain text is \"LOVEHELLOWORLD\", after several fills and rotations, the final grid will be\n                \n                [['L', 'H', 'E', 'O'],\n                ['W', 'L', 'L', 'O'],\n                ['O', 'R', 'D', 'V'],\n                ['E', '#', '#', 'L']]\n                \n        - Once the grid is filled, read the contents of the grid line by line as the encrypted message for that block.\n        - When moving on to the next block, empty the contents of the grid and redo the entire fill and rotate operation.\n        \n        Finally, the encrypted messages of all the blocks are connected together to be the final ciphertext.\n        \n\n \noriginal information: nervii encode to encrypted text: #II###NE###RV###\noriginal information: nonculpability encode to encrypted text: TULBIPNOLIYNC##A\noriginal information: nonculpability encode to encrypted text: TULBIPNOLIYNC##A\noriginal information: pyritic encode to encrypted text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 123, "source_filename": "icl_with_rule_encode_Kor_rule18_RotatingGridCipher_en.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始加密过程:\n\n处理第1个块:\n当前块的文本: PYRITIC\n文本长度不足16，用#补充\n补充后的文本: PYRITIC#########\n\n填充后的网格:\n[['#' 'T' 'I' '#']\n ['#' 'C' 'P' 'Y']\n ['#' '#' '#' 'R']\n ['I' '#' '#' '#']]\n当前块的密文: #TI##CPY###RI###\n\n最终密文: #TI##CPY###RI###\n", "ground_truth": "#TI##CPY###RI###"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n原始信息: nonculpability 密钥: {'str1': 'trichophytia', 'str2': 'sphinges'} 加密成加密文本: JK FI LL KS CE GE NY\n原始信息: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成加密文本: RV RB RD WX\n原始信息: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成加密文本: RV RB RD WX\n原始信息: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成加密文本: RV RB RD WX\n原始信息: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密成加密文本: RV RB RD WX\n原始信息: nervii 密钥: {'str1': 'parametral', 'str2': 'quiverful'} 加密成加密文本是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 124, "source_filename": "icl_encode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "NERVII", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将明文分成两个字母一组: ['NE', 'RV', 'II']\n\nStep 4.1: 加密字母对 NE\n  - N 在 M_T 中的位置: (2, 3)\n  - E 在 M_T 中的位置: (0, 4)\n  - 交换列索引后新位置: (2, 4) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (0, 3) 的字母: V\n  - 加密结果为: JV\n\nStep 4.2: 加密字母对 RV\n  - R 在 M_T 中的位置: (3, 1)\n  - V 在 M_T 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 1)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (4, 1) 的字母: T\n  - 加密结果为: KT\n\nStep 4.3: 加密字母对 II\n  - I 在 M_T 中的位置: (1, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (1, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 3) 的字母: A\n  - 加密结果为: CA\n\n加密完成！最终加密结果为: JV KT CA\n", "ground_truth": "JV KT CA"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n初始文本: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密为密文: RV RB RD WX\n初始文本: nervii 密钥: {'str1': 'parametral', 'str2': 'quiverful'} 加密为密文: JV KT CA\n初始文本: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密为密文: RV RB RD WX\n初始文本: pyritic 密钥: {'str1': 'sawbill', 'str2': 'houri'} 加密为密文: RV RB RD WX\n初始文本: nonculpability 密钥: {'str1': 'trichophytia', 'str2': 'sphinges'} 加密为密文: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 125, "source_filename": "icl_encode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "NONCULPABILITY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将明文分成两个字母一组: ['NO', 'NC', 'UL', 'PA', 'BI', 'LI', 'TY']\n\nStep 4.1: 加密字母对 NO\n  - N 在 M_T 中的位置: (2, 3)\n  - O 在 M_T 中的位置: (2, 4)\n  - 交换列索引后新位置: (2, 4) 和 (2, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (2, 3) 的字母: K\n  - 加密结果为: JK\n\nStep 4.2: 加密字母对 NC\n  - N 在 M_T 中的位置: (2, 3)\n  - C 在 M_T 中的位置: (0, 2)\n  - 交换列索引后新位置: (2, 2) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 2) 的字母: F\n  - 在 M2 中查找位置 (0, 3) 的字母: I\n  - 加密结果为: FI\n\nStep 4.3: 加密字母对 UL\n  - U 在 M_T 中的位置: (3, 4)\n  - L 在 M_T 中的位置: (2, 1)\n  - 交换列索引后新位置: (3, 1) 和 (2, 4)\n  - 在 M1 中查找位置 (3, 1) 的字母: L\n  - 在 M2 中查找位置 (2, 4) 的字母: L\n  - 加密结果为: LL\n\nStep 4.4: 加密字母对 PA\n  - P 在 M_T 中的位置: (3, 0)\n  - A 在 M_T 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (0, 0) 的字母: S\n  - 加密结果为: KS\n\nStep 4.5: 加密字母对 BI\n  - B 在 M_T 中的位置: (0, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (0, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (0, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: CE\n\nStep 4.6: 加密字母对 LI\n  - L 在 M_T 中的位置: (2, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (2, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (2, 3) 的字母: G\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: GE\n\nStep 4.7: 加密字母对 TY\n  - T 在 M_T 中的位置: (3, 3)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: N\n  - 在 M2 中查找位置 (4, 3) 的字母: Y\n  - 加密结果为: NY\n\n加密完成！最终加密结果为: JK FI LL KS CE GE NY\n", "ground_truth": "JK FI LL KS CE GE NY"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n非加密信息: nervii 额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 编码为暗文: JV KT CA\n非加密信息: nonculpability 额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 编码为暗文: JK FI LL KS CE GE NY\n非加密信息: nervii 额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 编码为暗文: JV KT CA\n非加密信息: pyritic 额外参数: {'str1': 'sawbill', 'str2': 'houri'} 编码为暗文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 126, "source_filename": "icl_encode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "PYRITICX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\n明文字符个数为奇数，补一个字符 X\nStep 3: 将明文分成两个字母一组: ['PY', 'RI', 'TI', 'CX']\n\nStep 4.1: 加密字母对 PY\n  - P 在 M_T 中的位置: (3, 0)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 0)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (4, 0) 的字母: V\n  - 加密结果为: RV\n\nStep 4.2: 加密字母对 RI\n  - R 在 M_T 中的位置: (3, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 1) 的字母: B\n  - 加密结果为: RB\n\nStep 4.3: 加密字母对 TI\n  - T 在 M_T 中的位置: (3, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 3) 的字母: D\n  - 加密结果为: RD\n\nStep 4.4: 加密字母对 CX\n  - C 在 M_T 中的位置: (0, 2)\n  - X 在 M_T 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M1 中查找位置 (0, 2) 的字母: W\n  - 在 M2 中查找位置 (4, 2) 的字母: X\n  - 加密结果为: WX\n\n加密完成！最终加密结果为: RV RB RD WX\n", "ground_truth": "RV RB RD WX"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n非加密信息: nonculpability 加密为密文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n非加密信息: nonculpability 加密为密文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n非加密信息: pyritic 加密为密文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n非加密信息: pyritic 加密为密文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n非加密信息: nonculpability 加密为密文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n非加密信息: nervii 加密为密文是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 127, "source_filename": "icl_encode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "NERVII", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'E'，它位于第 1 行第 5 列，编码为 '. .....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'V'，它位于第 5 行第 1 列，编码为 '..... .'\n步骤 5：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n最终步骤：加密完成，加密后的消息是：... ...  . .....  .... ..  ..... .  .. ....  .. ....\n", "ground_truth": "... ...  . .....  .... ..  ..... .  .. ....  .. ...."}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n原文: pyritic 加密成加密信息: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n原文: pyritic 加密成加密信息: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n原文: nonculpability 加密成加密信息是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 128, "source_filename": "icl_encode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "NONCULPABILITY", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'O'，它位于第 3 行第 4 列，编码为 '... ....'\n步骤 3：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 4：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n步骤 5：正在加密字母 'U'，它位于第 4 行第 5 列，编码为 '.... .....'\n步骤 6：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 7：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 8：正在加密字母 'A'，它位于第 1 行第 1 列，编码为 '. .'\n步骤 9：正在加密字母 'B'，它位于第 1 行第 2 列，编码为 '. ..'\n步骤 10：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 11：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 12：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 13：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 14：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n最终步骤：加密完成，加密后的消息是：... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n", "ground_truth": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ...."}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n明文: nonculpability 加密成为加密文本: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n明文: nervii 加密成为加密文本: ... ...  . .....  .... ..  ..... .  .. ....  .. ....\n明文: pyritic 加密成为加密文本是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 129, "source_filename": "icl_encode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "PYRITIC", "extra_args": {}, "output": "步骤 1：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 2：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 5：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 7：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n最终步骤：加密完成，加密后的消息是：... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n", "ground_truth": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ..."}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对非加密信息进行加密\n加密规则：\n1. 将输入文本中的每个字母向后移动指定的位数（shift）。\n2. 输出加密后的文本。\n非加密信息: nervii 密钥: {'shift': 7} 加密为隐文是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 130, "source_filename": "icl_with_rule_encode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "nervii", "extra_args": {"shift": 7}, "output": "开始编码过程，文本为: nervii，位移数为: 7\n处理字符: n\n字符 n 向后移动 7 位，变为 u\n处理字符: e\n字符 e 向后移动 7 位，变为 l\n处理字符: r\n字符 r 向后移动 7 位，变为 y\n处理字符: v\n字符 v 向后移动 7 位，变为 c\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n编码完成，结果为: ulycpp\n", "ground_truth": "ulycpp"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对非加密信息进行加密\n加密规则：\n1. 将输入文本中的每个字母向后移动指定的位数（shift）。\n2. 输出加密后的文本。\n非加密信息: nervii 额外参数: {'shift': 7} 加密成暗文: ulycpp\n非加密信息: nervii 额外参数: {'shift': 7} 加密成暗文: ulycpp\n非加密信息: nonculpability 额外参数: {'shift': 15} 加密成暗文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 131, "source_filename": "icl_with_rule_encode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "nonculpability", "extra_args": {"shift": 15}, "output": "开始编码过程，文本为: nonculpability，位移数为: 15\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: o\n字符 o 向后移动 15 位，变为 d\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: c\n字符 c 向后移动 15 位，变为 r\n处理字符: u\n字符 u 向后移动 15 位，变为 j\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: p\n字符 p 向后移动 15 位，变为 e\n处理字符: a\n字符 a 向后移动 15 位，变为 p\n处理字符: b\n字符 b 向后移动 15 位，变为 q\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: t\n字符 t 向后移动 15 位，变为 i\n处理字符: y\n字符 y 向后移动 15 位，变为 n\n编码完成，结果为: cdcrjaepqxaxin\n", "ground_truth": "cdcrjaepqxaxin"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对明文进行加密\n加密规则：\n1. 将输入文本中的每个字母向后移动指定的位数（shift）。\n2. 输出加密后的文本。\n明文: nonculpability 密钥: {'shift': 15} 加密为隐文: cdcrjaepqxaxin\n明文: nonculpability 密钥: {'shift': 15} 加密为隐文: cdcrjaepqxaxin\n明文: pyritic 密钥: {'shift': 1} 加密为隐文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 132, "source_filename": "icl_with_rule_encode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "pyritic", "extra_args": {"shift": 1}, "output": "开始编码过程，文本为: pyritic，位移数为: 1\n处理字符: p\n字符 p 向后移动 1 位，变为 q\n处理字符: y\n字符 y 向后移动 1 位，变为 z\n处理字符: r\n字符 r 向后移动 1 位，变为 s\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: t\n字符 t 向后移动 1 位，变为 u\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: c\n字符 c 向后移动 1 位，变为 d\n编码完成，结果为: qzsjujd\n", "ground_truth": "qzsjujd"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n    - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n        X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n        P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n        A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n    - A: 3\n    - B: 5\n    - A_inv: 9\n- 加密步骤:\n    - 对于每个明文字符p:\n        - 设x为其在仿射字母表中的位置\n        - 应用仿射编码函数计算y:\n            - y = (Ax + B) mod 26\n        - 在仿射字母表中找到位置y对应的字母，形成加密消息\n解密算法:\n解密规则:\n    - 输入:\n        - 暗文: 大写字母字符串\n    - 输出:\n        - 明文: 大写字母字符串\n    - 准备:\n        - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n        - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n            X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n            P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n            A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n        - A: 3\n        - B: 5\n        - A_inv: 9\n    - 解密步骤:\n        - 对于每个暗文字符c:\n            - 设y为其在仿射字母表中的位置\n            - 计算x:\n                - x = A_inv * (y - B) % n\n            - 用仿射字母表中位置x处的字母替换c，形成解密消息\n暗文: XEXAKJQMSLJLVH 解密成明文: nonculpability\n暗文: QHOLVLA 解密成明文: pyritic\n暗文: XEXAKJQMSLJLVH 解密成明文: nonculpability\n暗文: XYOFLL 解密成明文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 133, "source_filename": "icl_with_rule_decode_Kor_rule5_AffineCipher_cn.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "XYOFLL", "extra_args": {}, "output": "加密文本: XYOFLL\n开始解密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5, A的逆元=9\n字符 X -> 位置 0 -> 计算9*(0-5)%26=7 -> 解密为 N\n字符 Y -> 位置 22 -> 计算9*(22-5)%26=23 -> 解密为 E\n字符 O -> 位置 6 -> 计算9*(6-5)%26=9 -> 解密为 R\n字符 F -> 位置 15 -> 计算9*(15-5)%26=12 -> 解密为 V\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n解密完成，结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n        X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n        P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n        A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n    - A: 3\n    - B: 5\n    - A_inv: 9\n- 加密步骤:\n    - 对于每个非加密信息字符p:\n        - 设x为其在仿射字母表中的位置\n        - 应用仿射编码函数计算y:\n            - y = (Ax + B) mod 26\n        - 在仿射字母表中找到位置y对应的字母，形成加密消息\n解密算法:\n解密规则:\n    - 输入:\n        - 加密文本: 大写字母字符串\n    - 输出:\n        - 非加密信息: 大写字母字符串\n    - 准备:\n        - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n        - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n            X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n            P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n            A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n        - A: 3\n        - B: 5\n        - A_inv: 9\n    - 解密步骤:\n        - 对于每个加密文本字符c:\n            - 设y为其在仿射字母表中的位置\n            - 计算x:\n                - x = A_inv * (y - B) % n\n            - 用仿射字母表中位置x处的字母替换c，形成解密消息\n加密文本: XYOFLL 解密为非加密信息: nervii\n加密文本: QHOLVLA 解密为非加密信息: pyritic\n加密文本: XYOFLL 解密为非加密信息: nervii\n加密文本: XEXAKJQMSLJLVH 解密为非加密信息: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 134, "source_filename": "icl_with_rule_decode_Kor_rule5_AffineCipher_cn.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "XEXAKJQMSLJLVH", "extra_args": {}, "output": "加密文本: XEXAKJQMSLJLVH\n开始解密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5, A的逆元=9\n字符 X -> 位置 0 -> 计算9*(0-5)%26=7 -> 解密为 N\n字符 E -> 位置 23 -> 计算9*(23-5)%26=6 -> 解密为 O\n字符 X -> 位置 0 -> 计算9*(0-5)%26=7 -> 解密为 N\n字符 A -> 位置 16 -> 计算9*(16-5)%26=21 -> 解密为 C\n字符 K -> 位置 17 -> 计算9*(17-5)%26=4 -> 解密为 U\n字符 J -> 位置 2 -> 计算9*(2-5)%26=25 -> 解密为 L\n字符 Q -> 位置 3 -> 计算9*(3-5)%26=8 -> 解密为 P\n字符 M -> 位置 1 -> 计算9*(1-5)%26=16 -> 解密为 A\n字符 S -> 位置 18 -> 计算9*(18-5)%26=13 -> 解密为 B\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 J -> 位置 2 -> 计算9*(2-5)%26=25 -> 解密为 L\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 V -> 位置 12 -> 计算9*(12-5)%26=11 -> 解密为 T\n字符 H -> 位置 19 -> 计算9*(19-5)%26=22 -> 解密为 Y\n解密完成，结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n        X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n        P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n        A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n    - A: 3\n    - B: 5\n    - A_inv: 9\n- 加密步骤:\n    - 对于每个原文字符p:\n        - 设x为其在仿射字母表中的位置\n        - 应用仿射编码函数计算y:\n            - y = (Ax + B) mod 26\n        - 在仿射字母表中找到位置y对应的字母，形成加密消息\n解密算法:\n解密规则:\n    - 输入:\n        - 暗码: 大写字母字符串\n    - 输出:\n        - 原文: 大写字母字符串\n    - 准备:\n        - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n        - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n            X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n            P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n            A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n        - A: 3\n        - B: 5\n        - A_inv: 9\n    - 解密步骤:\n        - 对于每个暗码字符c:\n            - 设y为其在仿射字母表中的位置\n            - 计算x:\n                - x = A_inv * (y - B) % n\n            - 用仿射字母表中位置x处的字母替换c，形成解密消息\n暗码: XYOFLL 解密成原文: nervii\n暗码: XYOFLL 解密成原文: nervii\n暗码: XYOFLL 解密成原文: nervii\n暗码: QHOLVLA 解密成原文: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 135, "source_filename": "icl_with_rule_decode_Kor_rule5_AffineCipher_cn.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "QHOLVLA", "extra_args": {}, "output": "加密文本: QHOLVLA\n开始解密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5, A的逆元=9\n字符 Q -> 位置 3 -> 计算9*(3-5)%26=8 -> 解密为 P\n字符 H -> 位置 19 -> 计算9*(19-5)%26=22 -> 解密为 Y\n字符 O -> 位置 6 -> 计算9*(6-5)%26=9 -> 解密为 R\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 V -> 位置 12 -> 计算9*(12-5)%26=11 -> 解密为 T\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 A -> 位置 16 -> 计算9*(16-5)%26=21 -> 解密为 C\n解密完成，结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 密文: 数字字符串，不含标点和空格\n- 准备:\n    - 3个网格(所有行列号从0开始计数):\n        - 网格1 (3x9):\n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n        - 网格2 (9x3):\n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n        - 网格3 (3x3):\n            1 2 3\n            4 5 6\n            7 8 9\n- 加密步骤:\n    - 移除所有空格和标点，将文本转换为大写字母\n    - 将明文切分为6个字符一组，如果最后一组不足6个字符，用#填充\n    - 将每组6个字符分成3个二元组\n    - 对每个二元组(L1, L2)执行以下操作:\n        - 确定L1在网格1中的行列号(l1_row, l1_col)\n        - 确定L2在网格2中的行列号(l2_row, l2_col)\n        - 在网格3中用l1_row和l2_col找到对应数字num3\n        - 输出三元组(l1_col, num3, l2_row)\n    - 将所有三元组连接成一个数字串作为加密信息\n明文: pyritic 编码为密文: 061322422398898898\n明文: nervii 编码为密文: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 136, "source_filename": "icl_with_rule_encode_Kor_rule14_DigrafidCipher_cn.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "NERVII", "extra_args": {}, "output": "\n开始加密过程:\n原始文本: nervii\n预处理后的文本: NERVII\n补充#后的文本: NERVII\n分割成二元组: ['NE', 'RV', 'II']\n处理字符对 NE:\n- 在grid1中找到N的位置: 行=2, 列=6\n- 在grid2中找到E的位置: 行=0, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (6, 9, 0)\n处理字符对 RV:\n- 在grid1中找到R的位置: 行=0, 列=3\n- 在grid2中找到V的位置: 行=7, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (3, 2, 7)\n处理字符对 II:\n- 在grid1中找到I的位置: 行=0, 列=7\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (7, 2, 2)\n\n最终加密结果: 690327722\n", "ground_truth": "690327722"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密文本: 数字字符串，不含标点和空格\n- 准备:\n    - 3个网格(所有行列号从0开始计数):\n        - 网格1 (3x9):\n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n        - 网格2 (9x3):\n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n        - 网格3 (3x3):\n            1 2 3\n            4 5 6\n            7 8 9\n- 加密步骤:\n    - 移除所有空格和标点，将文本转换为大写字母\n    - 将明文切分为6个字符一组，如果最后一组不足6个字符，用#填充\n    - 将每组6个字符分成3个二元组\n    - 对每个二元组(L1, L2)执行以下操作:\n        - 确定L1在网格1中的行列号(l1_row, l1_col)\n        - 确定L2在网格2中的行列号(l2_row, l2_col)\n        - 在网格3中用l1_row和l2_col找到对应数字num3\n        - 输出三元组(l1_col, num3, l2_row)\n    - 将所有三元组连接成一个数字串作为加密信息\n明文: nervii 编码为加密文本: 690327722\n明文: pyritic 编码为加密文本: 061322422398898898\n明文: nonculpability 编码为加密文本是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 137, "source_filename": "icl_with_rule_encode_Kor_rule14_DigrafidCipher_cn.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "\n开始加密过程:\n原始文本: nonculpability\n预处理后的文本: NONCULPABILITY\n补充#后的文本: NONCULPABILITY####\n分割成二元组: ['NO', 'NC', 'UL', 'PA', 'BI', 'LI', 'TY', '##', '##']\n处理字符对 NO:\n- 在grid1中找到N的位置: 行=2, 列=6\n- 在grid2中找到O的位置: 行=2, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (6, 9, 2)\n处理字符对 NC:\n- 在grid1中找到N的位置: 行=2, 列=6\n- 在grid2中找到C的位置: 行=7, 列=0\n- 在grid3中找到对应数字: 7 (使用grid1的行2和grid2的列0)\n- 生成三元组: (6, 7, 7)\n处理字符对 UL:\n- 在grid1中找到U的位置: 行=0, 列=6\n- 在grid2中找到L的位置: 行=6, 列=0\n- 在grid3中找到对应数字: 1 (使用grid1的行0和grid2的列0)\n- 生成三元组: (6, 1, 6)\n处理字符对 PA:\n- 在grid1中找到P的位置: 行=1, 列=0\n- 在grid2中找到A的位置: 行=3, 列=1\n- 在grid3中找到对应数字: 5 (使用grid1的行1和grid2的列1)\n- 生成三元组: (0, 5, 3)\n处理字符对 BI:\n- 在grid1中找到B的位置: 行=2, 列=5\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 8 (使用grid1的行2和grid2的列1)\n- 生成三元组: (5, 8, 2)\n处理字符对 LI:\n- 在grid1中找到L的位置: 行=2, 列=0\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 8 (使用grid1的行2和grid2的列1)\n- 生成三元组: (0, 8, 2)\n处理字符对 TY:\n- 在grid1中找到T的位置: 行=0, 列=4\n- 在grid2中找到Y的位置: 行=1, 列=2\n- 在grid3中找到对应数字: 3 (使用grid1的行0和grid2的列2)\n- 生成三元组: (4, 3, 1)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n\n最终加密结果: 692677616053582082431898898\n", "ground_truth": "692677616053582082431898898"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗码: 数字字符串，不含标点和空格\n- 准备:\n    - 3个网格(所有行列号从0开始计数):\n        - 网格1 (3x9):\n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n        - 网格2 (9x3):\n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n        - 网格3 (3x3):\n            1 2 3\n            4 5 6\n            7 8 9\n- 加密步骤:\n    - 移除所有空格和标点，将文本转换为大写字母\n    - 将明文切分为6个字符一组，如果最后一组不足6个字符，用#填充\n    - 将每组6个字符分成3个二元组\n    - 对每个二元组(L1, L2)执行以下操作:\n        - 确定L1在网格1中的行列号(l1_row, l1_col)\n        - 确定L2在网格2中的行列号(l2_row, l2_col)\n        - 在网格3中用l1_row和l2_col找到对应数字num3\n        - 输出三元组(l1_col, num3, l2_row)\n    - 将所有三元组连接成一个数字串作为加密信息\n明文: nonculpability 加密为暗码: 692677616053582082431898898\n明文: nonculpability 加密为暗码: 692677616053582082431898898\n明文: nervii 加密为暗码: 690327722\n明文: pyritic 加密为暗码: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 138, "source_filename": "icl_with_rule_encode_Kor_rule14_DigrafidCipher_cn.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "PYRITIC", "extra_args": {}, "output": "\n开始加密过程:\n原始文本: pyritic\n预处理后的文本: PYRITIC\n补充#后的文本: PYRITIC#####\n分割成二元组: ['PY', 'RI', 'TI', 'C#', '##', '##']\n处理字符对 PY:\n- 在grid1中找到P的位置: 行=1, 列=0\n- 在grid2中找到Y的位置: 行=1, 列=2\n- 在grid3中找到对应数字: 6 (使用grid1的行1和grid2的列2)\n- 生成三元组: (0, 6, 1)\n处理字符对 RI:\n- 在grid1中找到R的位置: 行=0, 列=3\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (3, 2, 2)\n处理字符对 TI:\n- 在grid1中找到T的位置: 行=0, 列=4\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (4, 2, 2)\n处理字符对 C#:\n- 在grid1中找到C的位置: 行=2, 列=3\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (3, 9, 8)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n\n最终加密结果: 061322422398898898\n", "ground_truth": "061322422398898898"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 隐文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - k_q_x值: 12\n- 加密步骤:\n    - 对原始信息中的每个字母p:\n        - 获取字母p对应的ASCII码值x\n        - 计算x * k_q_x作为该字母对应的隐文数字y\n    - 最后将所有y值用逗号连接得到最终隐文\n原始信息: pyritic 编码为隐文: 960,1068,984,876,1008,876,804\n原始信息: pyritic 编码为隐文: 960,1068,984,876,1008,876,804\n原始信息: nonculpability 编码为隐文: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068\n原始信息: nervii 编码为隐文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 139, "source_filename": "icl_with_rule_encode_Kor_rule24_ECCCipher_cn.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NERVII\n使用预设的k_q_x值: 12\n\n处理字符: N\n对应的ASCII值: 78\n加密后的值: 936\n\n处理字符: E\n对应的ASCII值: 69\n加密后的值: 828\n\n处理字符: R\n对应的ASCII值: 82\n加密后的值: 984\n\n处理字符: V\n对应的ASCII值: 86\n加密后的值: 1032\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n最终加密结果: 936,828,984,1032,876,876\n", "ground_truth": "936,828,984,1032,876,876"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密信息: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - k_q_x值: 12\n- 加密步骤:\n    - 对明文中的每个字母p:\n        - 获取字母p对应的ASCII码值x\n        - 计算x * k_q_x作为该字母对应的加密信息数字y\n    - 最后将所有y值用逗号连接得到最终加密信息\n明文: pyritic 编码为加密信息: 960,1068,984,876,1008,876,804\n明文: pyritic 编码为加密信息: 960,1068,984,876,1008,876,804\n明文: pyritic 编码为加密信息: 960,1068,984,876,1008,876,804\n明文: nonculpability 编码为加密信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 140, "source_filename": "icl_with_rule_encode_Kor_rule24_ECCCipher_cn.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NONCULPABILITY\n使用预设的k_q_x值: 12\n\n处理字符: N\n对应的ASCII值: 78\n加密后的值: 936\n\n处理字符: O\n对应的ASCII值: 79\n加密后的值: 948\n\n处理字符: N\n对应的ASCII值: 78\n加密后的值: 936\n\n处理字符: C\n对应的ASCII值: 67\n加密后的值: 804\n\n处理字符: U\n对应的ASCII值: 85\n加密后的值: 1020\n\n处理字符: L\n对应的ASCII值: 76\n加密后的值: 912\n\n处理字符: P\n对应的ASCII值: 80\n加密后的值: 960\n\n处理字符: A\n对应的ASCII值: 65\n加密后的值: 780\n\n处理字符: B\n对应的ASCII值: 66\n加密后的值: 792\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: L\n对应的ASCII值: 76\n加密后的值: 912\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: T\n对应的ASCII值: 84\n加密后的值: 1008\n\n处理字符: Y\n对应的ASCII值: 89\n加密后的值: 1068\n\n最终加密结果: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068\n", "ground_truth": "936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - k_q_x值: 12\n- 加密步骤:\n    - 对原始信息中的每个字母p:\n        - 获取字母p对应的ASCII码值x\n        - 计算x * k_q_x作为该字母对应的密文数字y\n    - 最后将所有y值用逗号连接得到最终密文\n原始信息: nervii 加密成密文: 936,828,984,1032,876,876\n原始信息: pyritic 加密成密文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 141, "source_filename": "icl_with_rule_encode_Kor_rule24_ECCCipher_cn.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: PYRITIC\n使用预设的k_q_x值: 12\n\n处理字符: P\n对应的ASCII值: 80\n加密后的值: 960\n\n处理字符: Y\n对应的ASCII值: 89\n加密后的值: 1068\n\n处理字符: R\n对应的ASCII值: 82\n加密后的值: 984\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: T\n对应的ASCII值: 84\n加密后的值: 1008\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: C\n对应的ASCII值: 67\n加密后的值: 804\n\n最终加密结果: 960,1068,984,876,1008,876,804\n", "ground_truth": "960,1068,984,876,1008,876,804"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\ncipher text: mlmxfokzyrorgb decode to original information: nonculpability\ncipher text: mlmxfokzyrorgb decode to original information: nonculpability\ncipher text: kbirgrx decode to original information: pyritic\ncipher text: mlmxfokzyrorgb decode to original information: nonculpability\ncipher text: mvierr decode to original information is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 142, "source_filename": "icl_decode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "mvierr", "extra_args": {}, "output": "解码过程:\n转换单词: mvierr\n  'm' -> 'n'\n  'v' -> 'e'\n  'i' -> 'r'\n  'e' -> 'v'\n  'r' -> 'i'\n  'r' -> 'i'\n解码后的文本: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\nencrypted text: mvierr decode into plain text: nervii\nencrypted text: mvierr decode into plain text: nervii\nencrypted text: mvierr decode into plain text: nervii\nencrypted text: mlmxfokzyrorgb decode into plain text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 143, "source_filename": "icl_decode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "mlmxfokzyrorgb", "extra_args": {}, "output": "解码过程:\n转换单词: mlmxfokzyrorgb\n  'm' -> 'n'\n  'l' -> 'o'\n  'm' -> 'n'\n  'x' -> 'c'\n  'f' -> 'u'\n  'o' -> 'l'\n  'k' -> 'p'\n  'z' -> 'a'\n  'y' -> 'b'\n  'r' -> 'i'\n  'o' -> 'l'\n  'r' -> 'i'\n  'g' -> 't'\n  'b' -> 'y'\n解码后的文本: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\nencrypted text: mvierr decode into clear text: nervii\nencrypted text: kbirgrx decode into clear text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 144, "source_filename": "icl_decode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "kbirgrx", "extra_args": {}, "output": "解码过程:\n转换单词: kbirgrx\n  'k' -> 'p'\n  'b' -> 'y'\n  'i' -> 'r'\n  'r' -> 'i'\n  'g' -> 't'\n  'r' -> 'i'\n  'x' -> 'c'\n解码后的文本: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "Please encrypt the plaintext using the encryption algorithm.\nTap Code is a simple cipher technique where each letter is represented by a dot pattern. The pattern is encoded by counting the dots in the rows and columns. For example, the letter 'A' is represented by a single dot in the first row and first column.\nplain text: nonculpability encode to encoded text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\nplain text: nonculpability encode to encoded text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\nplain text: nervii encode to encoded text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 145, "source_filename": "icl_with_rule_encode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "NERVII", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'E'，它位于第 1 行第 5 列，编码为 '. .....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'V'，它位于第 5 行第 1 列，编码为 '..... .'\n步骤 5：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n最终步骤：加密完成，加密后的消息是：... ...  . .....  .... ..  ..... .  .. ....  .. ....\n", "ground_truth": "... ...  . .....  .... ..  ..... .  .. ....  .. ...."}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\nPlease encrypt the plaintext using the encryption algorithm.\nTap Code is a simple cipher technique where each letter is represented by a dot pattern. The pattern is encoded by counting the dots in the rows and columns. For example, the letter 'A' is represented by a single dot in the first row and first column.\noriginal information: pyritic encrypt into cipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\noriginal information: nervii encrypt into cipher text: ... ...  . .....  .... ..  ..... .  .. ....  .. ....\noriginal information: nervii encrypt into cipher text: ... ...  . .....  .... ..  ..... .  .. ....  .. ....\noriginal information: nonculpability encrypt into cipher text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 146, "source_filename": "icl_with_rule_encode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "NONCULPABILITY", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'O'，它位于第 3 行第 4 列，编码为 '... ....'\n步骤 3：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 4：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n步骤 5：正在加密字母 'U'，它位于第 4 行第 5 列，编码为 '.... .....'\n步骤 6：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 7：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 8：正在加密字母 'A'，它位于第 1 行第 1 列，编码为 '. .'\n步骤 9：正在加密字母 'B'，它位于第 1 行第 2 列，编码为 '. ..'\n步骤 10：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 11：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 12：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 13：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 14：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n最终步骤：加密完成，加密后的消息是：... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n", "ground_truth": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ...."}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nPlease encrypt the plaintext using the encryption algorithm.\nTap Code is a simple cipher technique where each letter is represented by a dot pattern. The pattern is encoded by counting the dots in the rows and columns. For example, the letter 'A' is represented by a single dot in the first row and first column.\noriginal information: pyritic encrypt to encrypted text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 147, "source_filename": "icl_with_rule_encode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "PYRITIC", "extra_args": {}, "output": "步骤 1：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 2：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 5：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 7：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n最终步骤：加密完成，加密后的消息是：... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n", "ground_truth": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ..."}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\nPlease decrypt the password based on the encryption and decryption algorithms.\nEncryption Algorithm:\nOverview of the encryption scheme: Convert the string to ASCII encoding.\nDecryption Algorithm:\nOverview of the decryption scheme: Traverse the ASCII encoded string and return the converted string.\nencrypted text: [112, 121, 114, 105, 116, 105, 99] decode to plain text: pyritic\nencrypted text: [112, 121, 114, 105, 116, 105, 99] decode to plain text: pyritic\nencrypted text: [110, 101, 114, 118, 105, 105] decode to plain text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 148, "source_filename": "icl_with_rule_decode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": [110, 101, 114, 118, 105, 105], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 101, Character: e\nASCII   Code: 114, Character: r\nASCII   Code: 118, Character: v\nASCII   Code: 105, Character: i\nASCII   Code: 105, Character: i\nResult: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\nPlease decrypt the password based on the encryption and decryption algorithms.\nEncryption Algorithm:\nOverview of the encryption scheme: Convert the string to ASCII encoding.\nDecryption Algorithm:\nOverview of the decryption scheme: Traverse the ASCII encoded string and return the converted string.\ncipher text: [110, 101, 114, 118, 105, 105] decrypt to clear text: nervii\ncipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] decrypt to clear text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 149, "source_filename": "icl_with_rule_decode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 111, Character: o\nASCII   Code: 110, Character: n\nASCII   Code: 99, Character: c\nASCII   Code: 117, Character: u\nASCII   Code: 108, Character: l\nASCII   Code: 112, Character: p\nASCII   Code: 97, Character: a\nASCII   Code: 98, Character: b\nASCII   Code: 105, Character: i\nASCII   Code: 108, Character: l\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 121, Character: y\nResult: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\nPlease decrypt the password based on the encryption and decryption algorithms.\nEncryption Algorithm:\nOverview of the encryption scheme: Convert the string to ASCII encoding.\nDecryption Algorithm:\nOverview of the decryption scheme: Traverse the ASCII encoded string and return the converted string.\nencoded text: [110, 101, 114, 118, 105, 105] decrypt into clear text: nervii\nencoded text: [110, 101, 114, 118, 105, 105] decrypt into clear text: nervii\nencoded text: [112, 121, 114, 105, 116, 105, 99] decrypt into clear text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 150, "source_filename": "icl_with_rule_decode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": [112, 121, 114, 105, 116, 105, 99], "extra_args": {}, "output": "ASCII   Code: 112, Character: p\nASCII   Code: 121, Character: y\nASCII   Code: 114, Character: r\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 105, Character: i\nASCII   Code: 99, Character: c\nResult: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 隐文: 不含标点和空格的字符串\n- 准备:\n    - 每行最大字符数: 5\n- 加密步骤:\n    - 行数从1开始计数\n    - 初始文本按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母\n        - 例如，对于初始文本\"LIDAHELLOWORLD\"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按列读取，每列从上到下读取，每读完一列加一个\"#\"，读出的内容即为最终隐文\n        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#\n初始文本: nonculpability 加密成为隐文: NIL#OBI#NAT#CPY#UL#\n初始文本: nonculpability 加密成为隐文: NIL#OBI#NAT#CPY#UL#\n初始文本: nervii 加密成为隐文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 151, "source_filename": "icl_with_rule_encode_Kor_rule17_PathCipher_cn.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "NERVII", "extra_args": {}, "output": "1. 输入文本转换为大写并去除非字母字符: NERVII\n2. 确定网格大小: 2行 x 5列\n3. 开始填充网格:\n   第1行从左到右填充: N E R V I \n   第2行从右到左填充: I \n4. 按列读取并添加#号:\n   第1列: N##\n   第2列: E##\n   第3列: R##\n   第4列: V##\n   第5列: II#\n5. 最终密文: N#E#R#V#II#\n", "ground_truth": "N#E#R#V#II#"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对非加密信息进行加密\n加密规则:\n- 输入:\n    - 非加密信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密文本: 不含标点和空格的字符串\n- 准备:\n    - 每行最大字符数: 5\n- 加密步骤:\n    - 行数从1开始计数\n    - 非加密信息按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母\n        - 例如，对于非加密信息\"LIDAHELLOWORLD\"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按列读取，每列从上到下读取，每读完一列加一个\"#\"，读出的内容即为最终加密文本\n        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#\n非加密信息: nonculpability 加密为加密文本是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 152, "source_filename": "icl_with_rule_encode_Kor_rule17_PathCipher_cn.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "1. 输入文本转换为大写并去除非字母字符: NONCULPABILITY\n2. 确定网格大小: 3行 x 5列\n3. 开始填充网格:\n   第1行从左到右填充: N O N C U \n   第2行从右到左填充: L P A B I \n   第3行从左到右填充: L I T Y \n4. 按列读取并添加#号:\n   第1列: NIL#\n   第2列: OBI#\n   第3列: NAT#\n   第4列: CPY#\n   第5列: UL##\n5. 最终密文: NIL#OBI#NAT#CPY#UL#\n", "ground_truth": "NIL#OBI#NAT#CPY#UL#"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 隐文: 不含标点和空格的字符串\n- 准备:\n    - 每行最大字符数: 5\n- 加密步骤:\n    - 行数从1开始计数\n    - 初始文本按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母\n        - 例如，对于初始文本\"LIDAHELLOWORLD\"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按列读取，每列从上到下读取，每读完一列加一个\"#\"，读出的内容即为最终隐文\n        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#\n初始文本: nervii 加密成为隐文: N#E#R#V#II#\n初始文本: pyritic 加密成为隐文: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 153, "source_filename": "icl_with_rule_encode_Kor_rule17_PathCipher_cn.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "PYRITIC", "extra_args": {}, "output": "1. 输入文本转换为大写并去除非字母字符: PYRITIC\n2. 确定网格大小: 2行 x 5列\n3. 开始填充网格:\n   第1行从左到右填充: P Y R I T \n   第2行从右到左填充: I C \n4. 按列读取并添加#号:\n   第1列: P##\n   第2列: Y##\n   第3列: R##\n   第4列: IC#\n   第5列: TI#\n5. 最终密文: P#Y#R#IC#TI#\n", "ground_truth": "P#Y#R#IC#TI#"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - standard_alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Encryption Steps:\n    - For each letter p in the given Plaintext:\n    - (1) Use reversed_alphabet for reverse mapping. Find its position in the standard_alphabet and replace it with the letter in the corresponding position in reversed_alphabet. For example, A is mapped to Z and B is mapped to Y.\n    - (2) Move the letter obtained in (1) forward 4 places in the standard_alphabet order. For example, if p=A, after (1) is mapped to Z, then Z is shifted forward 4 positions in the standard_alphabet to get D.\n    - (3) Replace the letter obtained from (2) by finding its position in standard_alphabet and using the corresponding letter in substitution_alphabet, resulting in the final ciphertext letter. For example, if the letter obtained by going through (2) is D, it is mapped as J.\n\n\nplain text: pyritic encode to cipher text: MHKVGVF\nplain text: nonculpability encode to cipher text: ONOFEQMJDVQVGH\nplain text: nervii encode to cipher text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 154, "source_filename": "icl_with_rule_encode_Kor_rule1_InverseShiftSubstitutionCipher_en.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n加密字符 N:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: N -> M\n2. 将得到的字母向前移动4位: M -> Q\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Q -> O\n\n加密字符 E:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: E -> V\n2. 将得到的字母向前移动4位: V -> Z\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Z -> Z\n\n加密字符 R:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: R -> I\n2. 将得到的字母向前移动4位: I -> M\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: M -> K\n\n加密字符 V:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: V -> E\n2. 将得到的字母向前移动4位: E -> I\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: I -> C\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n最终加密结果: OZKCVV\n", "ground_truth": "OZKCVV"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - standard_alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Encryption Steps:\n    - For each letter p in the given Plaintext:\n    - (1) Use reversed_alphabet for reverse mapping. Find its position in the standard_alphabet and replace it with the letter in the corresponding position in reversed_alphabet. For example, A is mapped to Z and B is mapped to Y.\n    - (2) Move the letter obtained in (1) forward 4 places in the standard_alphabet order. For example, if p=A, after (1) is mapped to Z, then Z is shifted forward 4 positions in the standard_alphabet to get D.\n    - (3) Replace the letter obtained from (2) by finding its position in standard_alphabet and using the corresponding letter in substitution_alphabet, resulting in the final ciphertext letter. For example, if the letter obtained by going through (2) is D, it is mapped as J.\n\n\noriginal information: nonculpability encrypt to encrypted text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 155, "source_filename": "icl_with_rule_encode_Kor_rule1_InverseShiftSubstitutionCipher_en.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n加密字符 N:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: N -> M\n2. 将得到的字母向前移动4位: M -> Q\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Q -> O\n\n加密字符 O:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: O -> L\n2. 将得到的字母向前移动4位: L -> P\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: P -> N\n\n加密字符 N:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: N -> M\n2. 将得到的字母向前移动4位: M -> Q\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: Q -> O\n\n加密字符 C:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: C -> X\n2. 将得到的字母向前移动4位: X -> B\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: B -> F\n\n加密字符 U:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: U -> F\n2. 将得到的字母向前移动4位: F -> J\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: J -> E\n\n加密字符 L:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: L -> O\n2. 将得到的字母向前移动4位: O -> S\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: S -> Q\n\n加密字符 P:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: P -> K\n2. 将得到的字母向前移动4位: K -> O\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: O -> M\n\n加密字符 A:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: A -> Z\n2. 将得到的字母向前移动4位: Z -> D\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: D -> J\n\n加密字符 B:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: B -> Y\n2. 将得到的字母向前移动4位: Y -> C\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: C -> D\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 L:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: L -> O\n2. 将得到的字母向前移动4位: O -> S\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: S -> Q\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 T:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: T -> G\n2. 将得到的字母向前移动4位: G -> K\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: K -> G\n\n加密字符 Y:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: Y -> B\n2. 将得到的字母向前移动4位: B -> F\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: F -> H\n\n最终加密结果: ONOFEQMJDVQVGH\n", "ground_truth": "ONOFEQMJDVQVGH"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - standard_alphabet: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - reversed_alphabet: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - substitution_alphabet: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- Encryption Steps:\n    - For each letter p in the given Plaintext:\n    - (1) Use reversed_alphabet for reverse mapping. Find its position in the standard_alphabet and replace it with the letter in the corresponding position in reversed_alphabet. For example, A is mapped to Z and B is mapped to Y.\n    - (2) Move the letter obtained in (1) forward 4 places in the standard_alphabet order. For example, if p=A, after (1) is mapped to Z, then Z is shifted forward 4 positions in the standard_alphabet to get D.\n    - (3) Replace the letter obtained from (2) by finding its position in standard_alphabet and using the corresponding letter in substitution_alphabet, resulting in the final ciphertext letter. For example, if the letter obtained by going through (2) is D, it is mapped as J.\n\n\nclear text: nervii encrypt into encoded text: OZKCVV\nclear text: nervii encrypt into encoded text: OZKCVV\nclear text: pyritic encrypt into encoded text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 156, "source_filename": "icl_with_rule_encode_Kor_rule1_InverseShiftSubstitutionCipher_en.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n加密字符 P:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: P -> K\n2. 将得到的字母向前移动4位: K -> O\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: O -> M\n\n加密字符 Y:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: Y -> B\n2. 将得到的字母向前移动4位: B -> F\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: F -> H\n\n加密字符 R:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: R -> I\n2. 将得到的字母向前移动4位: I -> M\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: M -> K\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 T:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: T -> G\n2. 将得到的字母向前移动4位: G -> K\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: K -> G\n\n加密字符 I:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: I -> R\n2. 将得到的字母向前移动4位: R -> V\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: V -> V\n\n加密字符 C:\n1. 在标准字母表中找到位置并用反转字母表对应位置字母替换: C -> X\n2. 将得到的字母向前移动4位: X -> B\n3. 在标准字母表中找到位置并用替换字母表对应位置字母替换: B -> F\n\n最终加密结果: MHKVGVF\n", "ground_truth": "MHKVGVF"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n原文: nonculpability 加密为暗码: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n原文: nonculpability 加密为暗码: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n原文: pyritic 加密为暗码: [112, 121, 114, 105, 116, 105, 99]\n原文: nervii 加密为暗码是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 157, "source_filename": "icl_encode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": "nervii", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   e, ASCII Code: 101\nCharacter:   r, ASCII Code: 114\nCharacter:   v, ASCII Code: 118\nCharacter:   i, ASCII Code: 105\nCharacter:   i, ASCII Code: 105\nASCII List: [110, 101, 114, 118, 105, 105]\n", "ground_truth": [110, 101, 114, 118, 105, 105]}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n原文: nervii 加密为暗文: [110, 101, 114, 118, 105, 105]\n原文: nervii 加密为暗文: [110, 101, 114, 118, 105, 105]\n原文: nonculpability 加密为暗文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 158, "source_filename": "icl_encode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": "nonculpability", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   o, ASCII Code: 111\nCharacter:   n, ASCII Code: 110\nCharacter:   c, ASCII Code: 99\nCharacter:   u, ASCII Code: 117\nCharacter:   l, ASCII Code: 108\nCharacter:   p, ASCII Code: 112\nCharacter:   a, ASCII Code: 97\nCharacter:   b, ASCII Code: 98\nCharacter:   i, ASCII Code: 105\nCharacter:   l, ASCII Code: 108\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   y, ASCII Code: 121\nASCII List: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n", "ground_truth": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n原始信息: nervii 加密为暗码: [110, 101, 114, 118, 105, 105]\n原始信息: nervii 加密为暗码: [110, 101, 114, 118, 105, 105]\n原始信息: nonculpability 加密为暗码: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n原始信息: nervii 加密为暗码: [110, 101, 114, 118, 105, 105]\n原始信息: pyritic 加密为暗码: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 159, "source_filename": "icl_encode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": "pyritic", "extra_args": {}, "output": "Character:   p, ASCII Code: 112\nCharacter:   y, ASCII Code: 121\nCharacter:   r, ASCII Code: 114\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   i, ASCII Code: 105\nCharacter:   c, ASCII Code: 99\nASCII List: [112, 121, 114, 105, 116, 105, 99]\n", "ground_truth": [112, 121, 114, 105, 116, 105, 99]}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list:\n        - [1, 4, 0, 6, 5, 2, 3]\n        - The transposed sequence  list is used to write the plaintext in order one line at a time, and then the columns are adjusted according to the transposed sequence table so that the characters in each line are arranged in the given order.\n        - Columns count from 0.\n- Encryption Steps:\n    - The [1, 4, 0, 6, 5, 2, 3] transposed sequence table has a total of 7 bits, indicating that 7 letters should be written on a line.\n    - Write the plaintext in order, line by line, 7 per line. When there are less than 7, the last line is filled with $. A written grid can be obtained.\n    - According to the transposed sequence table, adjust the order of the columns, that is, the order of the columns is now [the original column 1, the original column 4, the original column 0, the original column 6, the original column 5, the original column 2, the original column 3], you can get after the adjustment of the columns after the order of the grid.\n    - Read the grid row by row and connect them to get the final ciphertext.(Note that $ needs to be retained.)\n\n \nclear text: nervii encode into encrypted text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 160, "source_filename": "icl_with_rule_encode_Kor_rule20_TranspositionCipher_en.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n原始网格:\nNERVII$\n转置后的网格:\nEIN$IRV\n最终加密结果: EIN$IRV\n", "ground_truth": "EIN$IRV"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list:\n        - [1, 4, 0, 6, 5, 2, 3]\n        - The transposed sequence  list is used to write the plaintext in order one line at a time, and then the columns are adjusted according to the transposed sequence table so that the characters in each line are arranged in the given order.\n        - Columns count from 0.\n- Encryption Steps:\n    - The [1, 4, 0, 6, 5, 2, 3] transposed sequence table has a total of 7 bits, indicating that 7 letters should be written on a line.\n    - Write the plaintext in order, line by line, 7 per line. When there are less than 7, the last line is filled with $. A written grid can be obtained.\n    - According to the transposed sequence table, adjust the order of the columns, that is, the order of the columns is now [the original column 1, the original column 4, the original column 0, the original column 6, the original column 5, the original column 2, the original column 3], you can get after the adjustment of the columns after the order of the grid.\n    - Read the grid row by row and connect them to get the final ciphertext.(Note that $ needs to be retained.)\n\n \noriginal information: nervii encode into encoded text: EIN$IRV\noriginal information: nonculpability encode into encoded text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 161, "source_filename": "icl_with_rule_encode_Kor_rule20_TranspositionCipher_en.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 2 行来存放文本\n原始网格:\nNONCULP\nABILITY\n转置后的网格:\nOUNPLNC\nBIAYTIL\n最终加密结果: OUNPLNCBIAYTIL\n", "ground_truth": "OUNPLNCBIAYTIL"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Transposed sequence list:\n        - [1, 4, 0, 6, 5, 2, 3]\n        - The transposed sequence  list is used to write the plaintext in order one line at a time, and then the columns are adjusted according to the transposed sequence table so that the characters in each line are arranged in the given order.\n        - Columns count from 0.\n- Encryption Steps:\n    - The [1, 4, 0, 6, 5, 2, 3] transposed sequence table has a total of 7 bits, indicating that 7 letters should be written on a line.\n    - Write the plaintext in order, line by line, 7 per line. When there are less than 7, the last line is filled with $. A written grid can be obtained.\n    - According to the transposed sequence table, adjust the order of the columns, that is, the order of the columns is now [the original column 1, the original column 4, the original column 0, the original column 6, the original column 5, the original column 2, the original column 3], you can get after the adjustment of the columns after the order of the grid.\n    - Read the grid row by row and connect them to get the final ciphertext.(Note that $ needs to be retained.)\n\n \nclear text: nonculpability encode into encoded text: OUNPLNCBIAYTIL\nclear text: pyritic encode into encoded text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 162, "source_filename": "icl_with_rule_encode_Kor_rule20_TranspositionCipher_en.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n原始网格:\nPYRITIC\n转置后的网格:\nYTPCIRI\n最终加密结果: YTPCIRI\n", "ground_truth": "YTPCIRI"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n输入:\n    - 原文: 大写字母字符串，不含标点和空格\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 密文: 大写字母字符串\n准备工作:\n    - 密码替换表:\n        使用以下13个密码替换表，每个表对应两个字母:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n    - 标准字母表:\n        ABCDEFGHIJKLMNOPQRSTUVWXYZ\n加密步骤:\n    - 将密钥中的每个字母与原文中的每个字母配对。如果密钥比原文短，重复使用密钥\n    - 对于每个原文字符p:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在标准字母表中找到p的位置，用密码替换表中相同位置的字母替换它\n解密算法:\n解密规则:\n输入:\n    - 密文: 大写字母字符串\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 原文: 大写字母字符串\n准备工作:\n    - 密码替换表: (与加密相同)\n    - 标准字母表: (与加密相同)\n解密步骤:\n    - 将密钥中的每个字母与密文中的每个字母配对。如果密钥比密文短，重复使用密钥\n    - 对于每个密文字符c:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在密码替换表中找到c的位置，用标准字母表中相同位置的字母还原它\n密文: LILTBUP 密钥: {'key': 'supercanonization'} 解密成原文: pyritic\n密文: LILTBUP 密钥: {'key': 'supercanonization'} 解密成原文: pyritic\n密文: LILTBUP 密钥: {'key': 'supercanonization'} 解密成原文: pyritic\n密文: HRMISY 密钥: {'key': 'paraguayans'} 解密成原文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 163, "source_filename": "icl_with_rule_decode_Kor_rule8_PortaCipher_cn.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "HRMISY", "extra_args": {"key": "paraguayans"}, "output": "开始解密文本: HRMISY\n使用密钥: paraguayans\n解密步骤:\n  第1个字符 H 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将密文字符 H 还原为明文字符 N\n  第2个字符 R 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将密文字符 R 还原为明文字符 E\n  第3个字符 M 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将密文字符 M 还原为明文字符 R\n  第4个字符 I 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将密文字符 I 还原为明文字符 V\n  第5个字符 S 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 S 还原为明文字符 I\n  第6个字符 Y 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 Y 还原为明文字符 I\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n输入:\n    - 原始信息: 大写字母字符串，不含标点和空格\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 加密信息: 大写字母字符串\n准备工作:\n    - 密码替换表:\n        使用以下13个密码替换表，每个表对应两个字母:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n    - 标准字母表:\n        ABCDEFGHIJKLMNOPQRSTUVWXYZ\n加密步骤:\n    - 将密钥中的每个字母与原始信息中的每个字母配对。如果密钥比原始信息短，重复使用密钥\n    - 对于每个原始信息字符p:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在标准字母表中找到p的位置，用密码替换表中相同位置的字母替换它\n解密算法:\n解密规则:\n输入:\n    - 加密信息: 大写字母字符串\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 原始信息: 大写字母字符串\n准备工作:\n    - 密码替换表: (与加密相同)\n    - 标准字母表: (与加密相同)\n解密步骤:\n    - 将密钥中的每个字母与加密信息中的每个字母配对。如果密钥比加密信息短，重复使用密钥\n    - 对于每个加密信息字符c:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在密码替换表中找到c的位置，用标准字母表中相同位置的字母还原它\n加密信息: LILTBUP 密钥: {'key': 'supercanonization'} 解密成原始信息: pyritic\n加密信息: LILTBUP 密钥: {'key': 'supercanonization'} 解密成原始信息: pyritic\n加密信息: HRMISY 密钥: {'key': 'paraguayans'} 解密成原始信息: nervii\n加密信息: ALFZLSFVPVOQJC 密钥: {'key': 'bulgingly'} 解密成原始信息是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 164, "source_filename": "icl_with_rule_decode_Kor_rule8_PortaCipher_cn.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "ALFZLSFVPVOQJC", "extra_args": {"key": "bulgingly"}, "output": "开始解密文本: ALFZLSFVPVOQJC\n使用密钥: bulgingly\n解密步骤:\n  第1个字符 A 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将密文字符 A 还原为明文字符 N\n  第2个字符 L 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 L 还原为明文字符 O\n  第3个字符 F 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将密文字符 F 还原为明文字符 N\n  第4个字符 Z 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 Z 还原为明文字符 C\n  第5个字符 L 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将密文字符 L 还原为明文字符 U\n  第6个字符 S 使用密钥字符 N:\n    - 查找密钥字符 N 对应的替换表\n    - 将密文字符 S 还原为明文字符 L\n  第7个字符 F 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 F 还原为明文字符 P\n  第8个字符 V 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将密文字符 V 还原为明文字符 A\n  第9个字符 P 使用密钥字符 Y:\n    - 查找密钥字符 Y 对应的替换表\n    - 将密文字符 P 还原为明文字符 B\n  第10个字符 V 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将密文字符 V 还原为明文字符 I\n  第11个字符 O 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 O 还原为明文字符 L\n  第12个字符 Q 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将密文字符 Q 还原为明文字符 I\n  第13个字符 J 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将密文字符 J 还原为明文字符 T\n  第14个字符 C 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将密文字符 C 还原为明文字符 Y\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n输入:\n    - 原文: 大写字母字符串，不含标点和空格\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 暗文: 大写字母字符串\n准备工作:\n    - 密码替换表:\n        使用以下13个密码替换表，每个表对应两个字母:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n    - 标准字母表:\n        ABCDEFGHIJKLMNOPQRSTUVWXYZ\n加密步骤:\n    - 将密钥中的每个字母与原文中的每个字母配对。如果密钥比原文短，重复使用密钥\n    - 对于每个原文字符p:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在标准字母表中找到p的位置，用密码替换表中相同位置的字母替换它\n解密算法:\n解密规则:\n输入:\n    - 暗文: 大写字母字符串\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 原文: 大写字母字符串\n准备工作:\n    - 密码替换表: (与加密相同)\n    - 标准字母表: (与加密相同)\n解密步骤:\n    - 将密钥中的每个字母与暗文中的每个字母配对。如果密钥比暗文短，重复使用密钥\n    - 对于每个暗文字符c:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在密码替换表中找到c的位置，用标准字母表中相同位置的字母还原它\n暗文: LILTBUP 额外参数: {'key': 'supercanonization'} 解密为原文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 165, "source_filename": "icl_with_rule_decode_Kor_rule8_PortaCipher_cn.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "LILTBUP", "extra_args": {"key": "supercanonization"}, "output": "开始解密文本: LILTBUP\n使用密钥: supercanonization\n解密步骤:\n  第1个字符 L 使用密钥字符 S:\n    - 查找密钥字符 S 对应的替换表\n    - 将密文字符 L 还原为明文字符 P\n  第2个字符 I 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将密文字符 I 还原为明文字符 Y\n  第3个字符 L 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将密文字符 L 还原为明文字符 R\n  第4个字符 T 使用密钥字符 E:\n    - 查找密钥字符 E 对应的替换表\n    - 将密文字符 T 还原为明文字符 I\n  第5个字符 B 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将密文字符 B 还原为明文字符 T\n  第6个字符 U 使用密钥字符 C:\n    - 查找密钥字符 C 对应的替换表\n    - 将密文字符 U 还原为明文字符 I\n  第7个字符 P 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将密文字符 P 还原为明文字符 C\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "明文: nonculpability 编码为加密文本: mlmxfokzyrorgb\n明文: nonculpability 编码为加密文本: mlmxfokzyrorgb\n明文: nervii 编码为加密文本是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 166, "source_filename": "icl_encode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "nervii", "extra_args": {}, "output": "编码过程:\n转换单词: nervii\n  'n' -> 'm'\n  'e' -> 'v'\n  'r' -> 'i'\n  'v' -> 'e'\n  'i' -> 'r'\n  'i' -> 'r'\n编码后的文本: mvierr\n", "ground_truth": "mvierr"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n初始文本: pyritic 加密成暗码: kbirgrx\n初始文本: nervii 加密成暗码: mvierr\n初始文本: pyritic 加密成暗码: kbirgrx\n初始文本: nervii 加密成暗码: mvierr\n初始文本: pyritic 加密成暗码: kbirgrx\n初始文本: nonculpability 加密成暗码: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 167, "source_filename": "icl_encode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "nonculpability", "extra_args": {}, "output": "编码过程:\n转换单词: nonculpability\n  'n' -> 'm'\n  'o' -> 'l'\n  'n' -> 'm'\n  'c' -> 'x'\n  'u' -> 'f'\n  'l' -> 'o'\n  'p' -> 'k'\n  'a' -> 'z'\n  'b' -> 'y'\n  'i' -> 'r'\n  'l' -> 'o'\n  'i' -> 'r'\n  't' -> 'g'\n  'y' -> 'b'\n编码后的文本: mlmxfokzyrorgb\n", "ground_truth": "mlmxfokzyrorgb"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n原始信息: nonculpability 加密成为隐文: mlmxfokzyrorgb\n原始信息: nervii 加密成为隐文: mvierr\n原始信息: pyritic 加密成为隐文: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 168, "source_filename": "icl_encode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "pyritic", "extra_args": {}, "output": "编码过程:\n转换单词: pyritic\n  'p' -> 'k'\n  'y' -> 'b'\n  'r' -> 'i'\n  'i' -> 'r'\n  't' -> 'g'\n  'i' -> 'r'\n  'c' -> 'x'\n编码后的文本: kbirgrx\n", "ground_truth": "kbirgrx"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Polybius_square:\n```\n+----+----+----+----+----+\n|    | 1  | 2  | 3  | 4  | 5  |\n+----+----+----+----+----+\n| 1  | R  | T  | X  | F  | S  |\n| 2  | W  | C  | M  | V  | H  |\n| 3  | Z  | J  | A  | P  | B  |\n| 4  | L  | Q  | Y  | G  | K  |\n| 5  | N  | E  | U  | D  | I  |\n+----+----+----+----+----+\n```\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Polybius square:\n            - Replace `p` with the number of rows and columns (both counted from 1) in which the character appears.\n        - In particular, O, which does not exist in the Polybius square, is replaced by 66.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation: Polybius_square (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each of the two numbers CrCc in the given ciphertext.\n        - Based on the rows and columns denoted by CrCc, find the corresponding letters in the Polybius square grid.\n        - If CrCc=66, replace with \"O\".\ncipher text: 34431155125522 decrypt to plain text: pyritic\ncipher text: 34431155125522 decrypt to plain text: pyritic\ncipher text: 5166512253413433355541551243 decrypt to plain text: nonculpability\ncipher text: 515211245555 decrypt to plain text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 169, "source_filename": "icl_with_rule_decode_Kor_rule4_PolybiusSquareCipher_en.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "515211245555", "extra_args": {}, "output": "开始解密文本: 515211245555\n开始逐对数字解密:\n遇到数字对: 51，对应方阵位置: 第5行第1列，解密为: N\n遇到数字对: 52，对应方阵位置: 第5行第2列，解密为: E\n遇到数字对: 11，对应方阵位置: 第1行第1列，解密为: R\n遇到数字对: 24，对应方阵位置: 第2行第4列，解密为: V\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Polybius_square:\n```\n+----+----+----+----+----+\n|    | 1  | 2  | 3  | 4  | 5  |\n+----+----+----+----+----+\n| 1  | R  | T  | X  | F  | S  |\n| 2  | W  | C  | M  | V  | H  |\n| 3  | Z  | J  | A  | P  | B  |\n| 4  | L  | Q  | Y  | G  | K  |\n| 5  | N  | E  | U  | D  | I  |\n+----+----+----+----+----+\n```\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Polybius square:\n            - Replace `p` with the number of rows and columns (both counted from 1) in which the character appears.\n        - In particular, O, which does not exist in the Polybius square, is replaced by 66.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation: Polybius_square (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each of the two numbers CrCc in the given ciphertext.\n        - Based on the rows and columns denoted by CrCc, find the corresponding letters in the Polybius square grid.\n        - If CrCc=66, replace with \"O\".\ncipher text: 34431155125522 decrypt to clear text: pyritic\ncipher text: 5166512253413433355541551243 decrypt to clear text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 170, "source_filename": "icl_with_rule_decode_Kor_rule4_PolybiusSquareCipher_en.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "5166512253413433355541551243", "extra_args": {}, "output": "开始解密文本: 5166512253413433355541551243\n开始逐对数字解密:\n遇到数字对: 51，对应方阵位置: 第5行第1列，解密为: N\n遇到数字对: 66，解密为: O\n遇到数字对: 51，对应方阵位置: 第5行第1列，解密为: N\n遇到数字对: 22，对应方阵位置: 第2行第2列，解密为: C\n遇到数字对: 53，对应方阵位置: 第5行第3列，解密为: U\n遇到数字对: 41，对应方阵位置: 第4行第1列，解密为: L\n遇到数字对: 34，对应方阵位置: 第3行第4列，解密为: P\n遇到数字对: 33，对应方阵位置: 第3行第3列，解密为: A\n遇到数字对: 35，对应方阵位置: 第3行第5列，解密为: B\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 41，对应方阵位置: 第4行第1列，解密为: L\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 12，对应方阵位置: 第1行第2列，解密为: T\n遇到数字对: 43，对应方阵位置: 第4行第3列，解密为: Y\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Polybius_square:\n```\n+----+----+----+----+----+\n|    | 1  | 2  | 3  | 4  | 5  |\n+----+----+----+----+----+\n| 1  | R  | T  | X  | F  | S  |\n| 2  | W  | C  | M  | V  | H  |\n| 3  | Z  | J  | A  | P  | B  |\n| 4  | L  | Q  | Y  | G  | K  |\n| 5  | N  | E  | U  | D  | I  |\n+----+----+----+----+----+\n```\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Polybius square:\n            - Replace `p` with the number of rows and columns (both counted from 1) in which the character appears.\n        - In particular, O, which does not exist in the Polybius square, is replaced by 66.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation: Polybius_square (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each of the two numbers CrCc in the given ciphertext.\n        - Based on the rows and columns denoted by CrCc, find the corresponding letters in the Polybius square grid.\n        - If CrCc=66, replace with \"O\".\ncipher text: 5166512253413433355541551243 decrypt into plain text: nonculpability\ncipher text: 34431155125522 decrypt into plain text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 171, "source_filename": "icl_with_rule_decode_Kor_rule4_PolybiusSquareCipher_en.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "34431155125522", "extra_args": {}, "output": "开始解密文本: 34431155125522\n开始逐对数字解密:\n遇到数字对: 34，对应方阵位置: 第3行第4列，解密为: P\n遇到数字对: 43，对应方阵位置: 第4行第3列，解密为: Y\n遇到数字对: 11，对应方阵位置: 第1行第1列，解密为: R\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 12，对应方阵位置: 第1行第2列，解密为: T\n遇到数字对: 55，对应方阵位置: 第5行第5列，解密为: I\n遇到数字对: 22，对应方阵位置: 第2行第2列，解密为: C\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string.\n- Preparation:\n    - Numerical-Index Mapping Table\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - morse_code Table\n        - morse_code = {\n                'A': '.-',     'B': '-...',   'C': '-.-.',   'D': '-..',\n                'E': '.',      'F': '..-.',   'G': '--.',    'H': '....',\n                'I': '..',     'J': '.---',   'K': '-.-',    'L': '.-..',\n                'M': '--',     'N': '-.',     'O': '---',    'P': '.--.',\n                'Q': '--.-',   'R': '.-.',    'S': '...',    'T': '-',\n                'U': '..-',    'V': '...-',   'W': '.--',    'X': '-..-',\n                'Y': '-.--',   'Z': '--..',\n            }\n- Encryption Steps:\n    - Convert each character in the plaintext into Morse code according to the morse_code Table, with each character separated by a /, e.g., the Morse code corresponding to AB is '.-/-...'.\n    - Divide Morse code into two pairs of characters. If the length of the Morse code is odd, the last character, without the following mapping, is finally added to the end of the ciphertext.\n    - Each pair of characters is converted into a corresponding numeric string according to the numeric index mapping table.\n    - The encrypted message is represented by a string.\n\n \nplain text: nonculpability encrypt to cipher text: 485834472743247641732173645\nplain text: pyritic encrypt to cipher text: 2482623187344\nplain text: nervii encrypt to cipher text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 172, "source_filename": "icl_with_rule_encode_Kor_rule12_MorbitCipher_en.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "NERVII", "extra_args": {}, "output": "加密步骤开始:\n1. 将输入文本标准化为大写字母: NERVII\n2. 将文本转换为摩斯密码: -././.-./...-/../..\n3. 根据数字索引映射表将摩斯密码对转换为数字: 477471617.\n", "ground_truth": "477471617."}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string.\n- Preparation:\n    - Numerical-Index Mapping Table\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - morse_code Table\n        - morse_code = {\n                'A': '.-',     'B': '-...',   'C': '-.-.',   'D': '-..',\n                'E': '.',      'F': '..-.',   'G': '--.',    'H': '....',\n                'I': '..',     'J': '.---',   'K': '-.-',    'L': '.-..',\n                'M': '--',     'N': '-.',     'O': '---',    'P': '.--.',\n                'Q': '--.-',   'R': '.-.',    'S': '...',    'T': '-',\n                'U': '..-',    'V': '...-',   'W': '.--',    'X': '-..-',\n                'Y': '-.--',   'Z': '--..',\n            }\n- Encryption Steps:\n    - Convert each character in the plaintext into Morse code according to the morse_code Table, with each character separated by a /, e.g., the Morse code corresponding to AB is '.-/-...'.\n    - Divide Morse code into two pairs of characters. If the length of the Morse code is odd, the last character, without the following mapping, is finally added to the end of the ciphertext.\n    - Each pair of characters is converted into a corresponding numeric string according to the numeric index mapping table.\n    - The encrypted message is represented by a string.\n\n \nclear text: nonculpability encrypt into cipher text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 173, "source_filename": "icl_with_rule_encode_Kor_rule12_MorbitCipher_en.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "加密步骤开始:\n1. 将输入文本标准化为大写字母: NONCULPABILITY\n2. 将文本转换为摩斯密码: -./---/-./-.-./..-/.-../.--./.-/-.../../.-../../-/-.--\n3. 根据数字索引映射表将摩斯密码对转换为数字: 485834472743247641732173645\n", "ground_truth": "485834472743247641732173645"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string.\n- Preparation:\n    - Numerical-Index Mapping Table\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - morse_code Table\n        - morse_code = {\n                'A': '.-',     'B': '-...',   'C': '-.-.',   'D': '-..',\n                'E': '.',      'F': '..-.',   'G': '--.',    'H': '....',\n                'I': '..',     'J': '.---',   'K': '-.-',    'L': '.-..',\n                'M': '--',     'N': '-.',     'O': '---',    'P': '.--.',\n                'Q': '--.-',   'R': '.-.',    'S': '...',    'T': '-',\n                'U': '..-',    'V': '...-',   'W': '.--',    'X': '-..-',\n                'Y': '-.--',   'Z': '--..',\n            }\n- Encryption Steps:\n    - Convert each character in the plaintext into Morse code according to the morse_code Table, with each character separated by a /, e.g., the Morse code corresponding to AB is '.-/-...'.\n    - Divide Morse code into two pairs of characters. If the length of the Morse code is odd, the last character, without the following mapping, is finally added to the end of the ciphertext.\n    - Each pair of characters is converted into a corresponding numeric string according to the numeric index mapping table.\n    - The encrypted message is represented by a string.\n\n \nplain text: pyritic encode into encoded text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 174, "source_filename": "icl_with_rule_encode_Kor_rule12_MorbitCipher_en.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "PYRITIC", "extra_args": {}, "output": "加密步骤开始:\n1. 将输入文本标准化为大写字母: PYRITIC\n2. 将文本转换为摩斯密码: .--./-.--/.-./../-/../-.-.\n3. 根据数字索引映射表将摩斯密码对转换为数字: 2482623187344\n", "ground_truth": "2482623187344"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n加密规则:\n输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 暗码: 大写字母字符串\n准备工作:\n    - 密码替换表:\n        使用以下13个密码替换表，每个表对应两个字母:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n    - 标准字母表:\n        ABCDEFGHIJKLMNOPQRSTUVWXYZ\n加密步骤:\n    - 将密钥中的每个字母与初始文本中的每个字母配对。如果密钥比初始文本短，重复使用密钥\n    - 对于每个初始文本字符p:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在标准字母表中找到p的位置，用密码替换表中相同位置的字母替换它\n初始文本: nonculpability 密钥或额外参数: {'key': 'bulgingly'} 编码为暗码: ALFZLSFVPVOQJC\n初始文本: nonculpability 密钥或额外参数: {'key': 'bulgingly'} 编码为暗码: ALFZLSFVPVOQJC\n初始文本: pyritic 密钥或额外参数: {'key': 'supercanonization'} 编码为暗码: LILTBUP\n初始文本: nervii 密钥或额外参数: {'key': 'paraguayans'} 编码为暗码: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 175, "source_filename": "icl_with_rule_encode_Kor_rule8_PortaCipher_cn.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "NERVII", "extra_args": {"key": "paraguayans"}, "output": "开始加密文本: nervii\n使用密钥: paraguayans\n预处理后的文本: NERVII\n加密步骤:\n  第1个字符 N 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将明文字符 N 替换为密文字符 H\n  第2个字符 E 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将明文字符 E 替换为密文字符 R\n  第3个字符 R 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将明文字符 R 替换为密文字符 M\n  第4个字符 V 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将明文字符 V 替换为密文字符 I\n  第5个字符 I 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 I 替换为密文字符 S\n  第6个字符 I 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 I 替换为密文字符 Y\n最终加密结果: HRMISY\n", "ground_truth": "HRMISY"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对非加密信息进行加密\n加密规则:\n输入:\n    - 非加密信息: 大写字母字符串，不含标点和空格\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 隐文: 大写字母字符串\n准备工作:\n    - 密码替换表:\n        使用以下13个密码替换表，每个表对应两个字母:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n    - 标准字母表:\n        ABCDEFGHIJKLMNOPQRSTUVWXYZ\n加密步骤:\n    - 将密钥中的每个字母与非加密信息中的每个字母配对。如果密钥比非加密信息短，重复使用密钥\n    - 对于每个非加密信息字符p:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在标准字母表中找到p的位置，用密码替换表中相同位置的字母替换它\n非加密信息: nonculpability 密钥或额外参数: {'key': 'bulgingly'} 加密成隐文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 176, "source_filename": "icl_with_rule_encode_Kor_rule8_PortaCipher_cn.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "NONCULPABILITY", "extra_args": {"key": "bulgingly"}, "output": "开始加密文本: nonculpability\n使用密钥: bulgingly\n预处理后的文本: NONCULPABILITY\n加密步骤:\n  第1个字符 N 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将明文字符 N 替换为密文字符 A\n  第2个字符 O 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 O 替换为密文字符 L\n  第3个字符 N 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将明文字符 N 替换为密文字符 F\n  第4个字符 C 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 C 替换为密文字符 Z\n  第5个字符 U 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将明文字符 U 替换为密文字符 L\n  第6个字符 L 使用密钥字符 N:\n    - 查找密钥字符 N 对应的替换表\n    - 将明文字符 L 替换为密文字符 S\n  第7个字符 P 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 P 替换为密文字符 F\n  第8个字符 A 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将明文字符 A 替换为密文字符 V\n  第9个字符 B 使用密钥字符 Y:\n    - 查找密钥字符 Y 对应的替换表\n    - 将明文字符 B 替换为密文字符 P\n  第10个字符 I 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将明文字符 I 替换为密文字符 V\n  第11个字符 L 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 L 替换为密文字符 O\n  第12个字符 I 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将明文字符 I 替换为密文字符 Q\n  第13个字符 T 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 T 替换为密文字符 J\n  第14个字符 Y 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将明文字符 Y 替换为密文字符 C\n最终加密结果: ALFZLSFVPVOQJC\n", "ground_truth": "ALFZLSFVPVOQJC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n加密规则:\n输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n    - 密钥: 用于选择替换表的字符串\n输出:\n    - 加密文本: 大写字母字符串\n准备工作:\n    - 密码替换表:\n        使用以下13个密码替换表，每个表对应两个字母:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n    - 标准字母表:\n        ABCDEFGHIJKLMNOPQRSTUVWXYZ\n加密步骤:\n    - 将密钥中的每个字母与初始文本中的每个字母配对。如果密钥比初始文本短，重复使用密钥\n    - 对于每个初始文本字符p:\n        - 根据与之配对的密钥字母找到对应的密码替换表\n        - 在标准字母表中找到p的位置，用密码替换表中相同位置的字母替换它\n初始文本: nonculpability 密钥或额外参数: {'key': 'bulgingly'} 加密成为加密文本: ALFZLSFVPVOQJC\n初始文本: pyritic 密钥或额外参数: {'key': 'supercanonization'} 加密成为加密文本是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 177, "source_filename": "icl_with_rule_encode_Kor_rule8_PortaCipher_cn.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "PYRITIC", "extra_args": {"key": "supercanonization"}, "output": "开始加密文本: pyritic\n使用密钥: supercanonization\n预处理后的文本: PYRITIC\n加密步骤:\n  第1个字符 P 使用密钥字符 S:\n    - 查找密钥字符 S 对应的替换表\n    - 将明文字符 P 替换为密文字符 L\n  第2个字符 Y 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 Y 替换为密文字符 I\n  第3个字符 R 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将明文字符 R 替换为密文字符 L\n  第4个字符 I 使用密钥字符 E:\n    - 查找密钥字符 E 对应的替换表\n    - 将明文字符 I 替换为密文字符 T\n  第5个字符 T 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将明文字符 T 替换为密文字符 B\n  第6个字符 I 使用密钥字符 C:\n    - 查找密钥字符 C 对应的替换表\n    - 将明文字符 I 替换为密文字符 U\n  第7个字符 C 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将明文字符 C 替换为密文字符 P\n最终加密结果: LILTBUP\n", "ground_truth": "LILTBUP"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对非加密信息进行加密\n\n        加密规则:\n        - 输入:\n            - 非加密信息: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 输出:\n            - 暗码: 仅包含0和1的二进制字符串\n        - 准备:\n            - 固定密钥: 8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 加密步骤:\n            1. 将每个字符转换为二进制:\n                - 将每个字符转换为ASCII值\n                - 将ASCII值转换为8位二进制字符串\n            2. XOR运算:\n                - 对字符的8位二进制表示与固定密钥进行XOR运算\n                - 如果需要，重复密钥以匹配二进制表示的长度\n            3. 置换:\n                - 对XOR结果应用置换表得到每个字符的最终加密二进制字符串\n            4. 合并二进制字符串:\n                - 将所有字符的二进制字符串连接形成最终暗码\n        \n非加密信息: pyritic 加密为暗码: 11111100111101011111100011010101111111101101010111011001\n非加密信息: nonculpability 加密为暗码: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n非加密信息: nervii 加密为暗码是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 178, "source_filename": "icl_with_rule_encode_Kor_rule21_XORCipher_cn.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "NERVII", "extra_args": {}, "output": "1. 输入文本处理为: NERVII\n\n处理字符: N\n2. 转换为8位二进制: 01001110\n3. 与密钥进行XOR运算: 11100100\n4. 应用置换表: 11010010\n\n处理字符: E\n2. 转换为8位二进制: 01000101\n3. 与密钥进行XOR运算: 11101111\n4. 应用置换表: 11011111\n\n处理字符: R\n2. 转换为8位二进制: 01010010\n3. 与密钥进行XOR运算: 11111000\n4. 应用置换表: 11111000\n\n处理字符: V\n2. 转换为8位二进制: 01010110\n3. 与密钥进行XOR运算: 11111100\n4. 应用置换表: 11111010\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n最终加密结果: 110100101101111111111000111110101101010111010101\n", "ground_truth": "110100101101111111111000111110101101010111010101"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对初始文本进行加密\n\n        加密规则:\n        - 输入:\n            - 初始文本: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 输出:\n            - 暗码: 仅包含0和1的二进制字符串\n        - 准备:\n            - 固定密钥: 8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 加密步骤:\n            1. 将每个字符转换为二进制:\n                - 将每个字符转换为ASCII值\n                - 将ASCII值转换为8位二进制字符串\n            2. XOR运算:\n                - 对字符的8位二进制表示与固定密钥进行XOR运算\n                - 如果需要，重复密钥以匹配二进制表示的长度\n            3. 置换:\n                - 对XOR结果应用置换表得到每个字符的最终加密二进制字符串\n            4. 合并二进制字符串:\n                - 将所有字符的二进制字符串连接形成最终暗码\n        \n初始文本: nervii 编码为暗码: 110100101101111111111000111110101101010111010101\n初始文本: nervii 编码为暗码: 110100101101111111111000111110101101010111010101\n初始文本: nervii 编码为暗码: 110100101101111111111000111110101101010111010101\n初始文本: nonculpability 编码为暗码是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 179, "source_filename": "icl_with_rule_encode_Kor_rule21_XORCipher_cn.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "1. 输入文本处理为: NONCULPABILITY\n\n处理字符: N\n2. 转换为8位二进制: 01001110\n3. 与密钥进行XOR运算: 11100100\n4. 应用置换表: 11010010\n\n处理字符: O\n2. 转换为8位二进制: 01001111\n3. 与密钥进行XOR运算: 11100101\n4. 应用置换表: 11010011\n\n处理字符: N\n2. 转换为8位二进制: 01001110\n3. 与密钥进行XOR运算: 11100100\n4. 应用置换表: 11010010\n\n处理字符: C\n2. 转换为8位二进制: 01000011\n3. 与密钥进行XOR运算: 11101001\n4. 应用置换表: 11011001\n\n处理字符: U\n2. 转换为8位二进制: 01010101\n3. 与密钥进行XOR运算: 11111111\n4. 应用置换表: 11111111\n\n处理字符: L\n2. 转换为8位二进制: 01001100\n3. 与密钥进行XOR运算: 11100110\n4. 应用置换表: 11010110\n\n处理字符: P\n2. 转换为8位二进制: 01010000\n3. 与密钥进行XOR运算: 11111010\n4. 应用置换表: 11111100\n\n处理字符: A\n2. 转换为8位二进制: 01000001\n3. 与密钥进行XOR运算: 11101011\n4. 应用置换表: 11011101\n\n处理字符: B\n2. 转换为8位二进制: 01000010\n3. 与密钥进行XOR运算: 11101000\n4. 应用置换表: 11011000\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: L\n2. 转换为8位二进制: 01001100\n3. 与密钥进行XOR运算: 11100110\n4. 应用置换表: 11010110\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: T\n2. 转换为8位二进制: 01010100\n3. 与密钥进行XOR运算: 11111110\n4. 应用置换表: 11111110\n\n处理字符: Y\n2. 转换为8位二进制: 01011001\n3. 与密钥进行XOR运算: 11110011\n4. 应用置换表: 11110101\n\n最终加密结果: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n", "ground_truth": "1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对原文进行加密\n\n        加密规则:\n        - 输入:\n            - 原文: 仅包含大写字母(A-Z)的字符串，不含标点和空格\n        - 输出:\n            - 暗文: 仅包含0和1的二进制字符串\n        - 准备:\n            - 固定密钥: 8位二进制字符串(例如'10101010')\n            - 置换表:\n                - 置换表: (2, 0, 3, 1, 4, 6, 5, 7)\n                - 逆置换表: (1, 3, 0, 2, 4, 6, 5, 7)\n        - 加密步骤:\n            1. 将每个字符转换为二进制:\n                - 将每个字符转换为ASCII值\n                - 将ASCII值转换为8位二进制字符串\n            2. XOR运算:\n                - 对字符的8位二进制表示与固定密钥进行XOR运算\n                - 如果需要，重复密钥以匹配二进制表示的长度\n            3. 置换:\n                - 对XOR结果应用置换表得到每个字符的最终加密二进制字符串\n            4. 合并二进制字符串:\n                - 将所有字符的二进制字符串连接形成最终暗文\n        \n原文: nonculpability 加密成暗文: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n原文: nonculpability 加密成暗文: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n原文: nonculpability 加密成暗文: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n原文: pyritic 加密成暗文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 180, "source_filename": "icl_with_rule_encode_Kor_rule21_XORCipher_cn.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "PYRITIC", "extra_args": {}, "output": "1. 输入文本处理为: PYRITIC\n\n处理字符: P\n2. 转换为8位二进制: 01010000\n3. 与密钥进行XOR运算: 11111010\n4. 应用置换表: 11111100\n\n处理字符: Y\n2. 转换为8位二进制: 01011001\n3. 与密钥进行XOR运算: 11110011\n4. 应用置换表: 11110101\n\n处理字符: R\n2. 转换为8位二进制: 01010010\n3. 与密钥进行XOR运算: 11111000\n4. 应用置换表: 11111000\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: T\n2. 转换为8位二进制: 01010100\n3. 与密钥进行XOR运算: 11111110\n4. 应用置换表: 11111110\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: C\n2. 转换为8位二进制: 01000011\n3. 与密钥进行XOR运算: 11101001\n4. 应用置换表: 11011001\n\n最终加密结果: 11111100111101011111100011010101111111101101010111011001\n", "ground_truth": "11111100111101011111100011010101111111101101010111011001"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Polybius_square:\n```\n+----+----+----+----+----+\n|    | 1  | 2  | 3  | 4  | 5  |\n+----+----+----+----+----+\n| 1  | R  | T  | X  | F  | S  |\n| 2  | W  | C  | M  | V  | H  |\n| 3  | Z  | J  | A  | P  | B  |\n| 4  | L  | Q  | Y  | G  | K  |\n| 5  | N  | E  | U  | D  | I  |\n+----+----+----+----+----+\n```\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Polybius square:\n            - Replace `p` with the number of rows and columns (both counted from 1) in which the character appears.\n        - In particular, O, which does not exist in the Polybius square, is replaced by 66.\n\n \nclear text: pyritic encode into encrypted text: 34431155125522\nclear text: nervii encode into encrypted text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 181, "source_filename": "icl_with_rule_encode_Kor_rule4_PolybiusSquareCipher_en.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密文本: nervii\n预处理后的文本: NERVII\n开始逐字符加密:\n处理字符: N\n字符 N 在方阵中的位置是: 第5行第1列，替换为: 51\n处理字符: E\n字符 E 在方阵中的位置是: 第5行第2列，替换为: 52\n处理字符: R\n字符 R 在方阵中的位置是: 第1行第1列，替换为: 11\n处理字符: V\n字符 V 在方阵中的位置是: 第2行第4列，替换为: 24\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n最终加密结果: 515211245555\n", "ground_truth": "515211245555"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Polybius_square:\n```\n+----+----+----+----+----+\n|    | 1  | 2  | 3  | 4  | 5  |\n+----+----+----+----+----+\n| 1  | R  | T  | X  | F  | S  |\n| 2  | W  | C  | M  | V  | H  |\n| 3  | Z  | J  | A  | P  | B  |\n| 4  | L  | Q  | Y  | G  | K  |\n| 5  | N  | E  | U  | D  | I  |\n+----+----+----+----+----+\n```\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Polybius square:\n            - Replace `p` with the number of rows and columns (both counted from 1) in which the character appears.\n        - In particular, O, which does not exist in the Polybius square, is replaced by 66.\n\n \nclear text: pyritic encrypt into encoded text: 34431155125522\nclear text: nonculpability encrypt into encoded text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 182, "source_filename": "icl_with_rule_encode_Kor_rule4_PolybiusSquareCipher_en.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密文本: nonculpability\n预处理后的文本: NONCULPABILITY\n开始逐字符加密:\n处理字符: N\n字符 N 在方阵中的位置是: 第5行第1列，替换为: 51\n处理字符: O\n字符 O 不在Polybius方阵中，替换为: 66\n处理字符: N\n字符 N 在方阵中的位置是: 第5行第1列，替换为: 51\n处理字符: C\n字符 C 在方阵中的位置是: 第2行第2列，替换为: 22\n处理字符: U\n字符 U 在方阵中的位置是: 第5行第3列，替换为: 53\n处理字符: L\n字符 L 在方阵中的位置是: 第4行第1列，替换为: 41\n处理字符: P\n字符 P 在方阵中的位置是: 第3行第4列，替换为: 34\n处理字符: A\n字符 A 在方阵中的位置是: 第3行第3列，替换为: 33\n处理字符: B\n字符 B 在方阵中的位置是: 第3行第5列，替换为: 35\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: L\n字符 L 在方阵中的位置是: 第4行第1列，替换为: 41\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: T\n字符 T 在方阵中的位置是: 第1行第2列，替换为: 12\n处理字符: Y\n字符 Y 在方阵中的位置是: 第4行第3列，替换为: 43\n最终加密结果: 5166512253413433355541551243\n", "ground_truth": "5166512253413433355541551243"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Polybius_square:\n```\n+----+----+----+----+----+\n|    | 1  | 2  | 3  | 4  | 5  |\n+----+----+----+----+----+\n| 1  | R  | T  | X  | F  | S  |\n| 2  | W  | C  | M  | V  | H  |\n| 3  | Z  | J  | A  | P  | B  |\n| 4  | L  | Q  | Y  | G  | K  |\n| 5  | N  | E  | U  | D  | I  |\n+----+----+----+----+----+\n```\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Polybius square:\n            - Replace `p` with the number of rows and columns (both counted from 1) in which the character appears.\n        - In particular, O, which does not exist in the Polybius square, is replaced by 66.\n\n \noriginal information: nervii encrypt to encrypted text: 515211245555\noriginal information: nervii encrypt to encrypted text: 515211245555\noriginal information: nervii encrypt to encrypted text: 515211245555\noriginal information: pyritic encrypt to encrypted text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 183, "source_filename": "icl_with_rule_encode_Kor_rule4_PolybiusSquareCipher_en.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密文本: pyritic\n预处理后的文本: PYRITIC\n开始逐字符加密:\n处理字符: P\n字符 P 在方阵中的位置是: 第3行第4列，替换为: 34\n处理字符: Y\n字符 Y 在方阵中的位置是: 第4行第3列，替换为: 43\n处理字符: R\n字符 R 在方阵中的位置是: 第1行第1列，替换为: 11\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: T\n字符 T 在方阵中的位置是: 第1行第2列，替换为: 12\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: C\n字符 C 在方阵中的位置是: 第2行第2列，替换为: 22\n最终加密结果: 34431155125522\n", "ground_truth": "34431155125522"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\nAtbash 密码的编码规则是将每个字母替换为其在字母表中的反向对应字母。例如，'A' 变成 'Z'，'B' 变成 'Y'，依此类推。非字母字符保持不变。\n解密算法:\nAtbash 密码的解码规则是将每个字母替换为其在字母表中的反向对应字母。由于 Atbash 密码是对称的，解码过程与编码过程相同。例如，'Z' 变成 'A'，'Y' 变成 'B'，依此类推。非字母字符保持不变。\n暗文: mlmxfokzyrorgb 解码为非加密信息: nonculpability\n暗文: kbirgrx 解码为非加密信息: pyritic\n暗文: kbirgrx 解码为非加密信息: pyritic\n暗文: mvierr 解码为非加密信息: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 184, "source_filename": "icl_with_rule_decode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "mvierr", "extra_args": {}, "output": "解码过程:\n转换单词: mvierr\n  'm' -> 'n'\n  'v' -> 'e'\n  'i' -> 'r'\n  'e' -> 'v'\n  'r' -> 'i'\n  'r' -> 'i'\n解码后的文本: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\nAtbash 密码的编码规则是将每个字母替换为其在字母表中的反向对应字母。例如，'A' 变成 'Z'，'B' 变成 'Y'，依此类推。非字母字符保持不变。\n解密算法:\nAtbash 密码的解码规则是将每个字母替换为其在字母表中的反向对应字母。由于 Atbash 密码是对称的，解码过程与编码过程相同。例如，'Z' 变成 'A'，'Y' 变成 'B'，依此类推。非字母字符保持不变。\n隐文: mvierr 解密成为原始信息: nervii\n隐文: kbirgrx 解密成为原始信息: pyritic\n隐文: mlmxfokzyrorgb 解密成为原始信息是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 185, "source_filename": "icl_with_rule_decode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "mlmxfokzyrorgb", "extra_args": {}, "output": "解码过程:\n转换单词: mlmxfokzyrorgb\n  'm' -> 'n'\n  'l' -> 'o'\n  'm' -> 'n'\n  'x' -> 'c'\n  'f' -> 'u'\n  'o' -> 'l'\n  'k' -> 'p'\n  'z' -> 'a'\n  'y' -> 'b'\n  'r' -> 'i'\n  'o' -> 'l'\n  'r' -> 'i'\n  'g' -> 't'\n  'b' -> 'y'\n解码后的文本: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\nAtbash 密码的编码规则是将每个字母替换为其在字母表中的反向对应字母。例如，'A' 变成 'Z'，'B' 变成 'Y'，依此类推。非字母字符保持不变。\n解密算法:\nAtbash 密码的解码规则是将每个字母替换为其在字母表中的反向对应字母。由于 Atbash 密码是对称的，解码过程与编码过程相同。例如，'Z' 变成 'A'，'Y' 变成 'B'，依此类推。非字母字符保持不变。\n密文: kbirgrx 解密成为原始信息是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 186, "source_filename": "icl_with_rule_decode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "kbirgrx", "extra_args": {}, "output": "解码过程:\n转换单词: kbirgrx\n  'k' -> 'p'\n  'b' -> 'y'\n  'i' -> 'r'\n  'r' -> 'i'\n  'g' -> 't'\n  'r' -> 'i'\n  'x' -> 'c'\n解码后的文本: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对明文进行加密\n\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 不含标点和空格的字符串\n- 准备:\n    - 行数: 5\n- 加密步骤:\n    - 在第一行第一个位置填入第一个明文字母\n    - 两种填充方式:\n        - 向下填充: 在第一行填入\"#\"(除去第一列，因为第一个位置已经填入明文字母)，然后从第二行到最后一行(第五行)向下填充明文\n        - 向上填充: 从最后一行(第五行)到第二行向上填充明文，然后在第一行填入\"#\"\n    - 对于明文中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入\n    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终暗码\n\n明文: pyritic 加密为暗码: P*Y*RC*II*T#*\n明文: nonculpability 加密为暗码: NB#*OAI*NPL*CLIY*U#T#*\n明文: pyritic 加密为暗码: P*Y*RC*II*T#*\n明文: nervii 加密为暗码: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 187, "source_filename": "icl_with_rule_encode_Kor_rule16_RedefenceFigureCipher_cn.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "NERVII", "extra_args": {}, "output": "加密步骤开始:\n1. 处理输入文本为大写字母: NERVII\n2. 准备5行矩阵用于填充\n3. 在第一行第一个位置填入第一个字母: N\n4. 开始交替向下和向上填充字母:\n  向下填充:\n    在第2行填入: E\n    在第3行填入: R\n    在第4行填入: V\n    在第5行填入: I\n  向上填充:\n    在第5行填入#\n    在第4行填入最后一个字母: I\n\n5. 最终矩阵:\n  N _ _ _ _ _\n  E _ _ _ _ _\n  R _ _ _ _ _\n  V I _ _ _ _\n  I # _ _ _ _\n\n6. 按行读取并添加*号分隔符\n最终密文: N*E*R*VI*I#*\n", "ground_truth": "N*E*R*VI*I#*"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对明文进行加密\n\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 不含标点和空格的字符串\n- 准备:\n    - 行数: 5\n- 加密步骤:\n    - 在第一行第一个位置填入第一个明文字母\n    - 两种填充方式:\n        - 向下填充: 在第一行填入\"#\"(除去第一列，因为第一个位置已经填入明文字母)，然后从第二行到最后一行(第五行)向下填充明文\n        - 向上填充: 从最后一行(第五行)到第二行向上填充明文，然后在第一行填入\"#\"\n    - 对于明文中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入\n    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终暗文\n\n明文: nervii 加密成暗文: N*E*R*VI*I#*\n明文: pyritic 加密成暗文: P*Y*RC*II*T#*\n明文: nonculpability 加密成暗文: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 188, "source_filename": "icl_with_rule_encode_Kor_rule16_RedefenceFigureCipher_cn.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "加密步骤开始:\n1. 处理输入文本为大写字母: NONCULPABILITY\n2. 准备5行矩阵用于填充\n3. 在第一行第一个位置填入第一个字母: N\n4. 开始交替向下和向上填充字母:\n  向下填充:\n    在第2行填入: O\n    在第3行填入: N\n    在第4行填入: C\n    在第5行填入: U\n  向上填充:\n    在第5行填入#\n    在第4行填入: L\n    在第3行填入: P\n    在第2行填入: A\n    在第1行填入: B\n  向下填充:\n    在第1行填入#\n    在第2行填入: I\n    在第3行填入: L\n    在第4行填入: I\n    在第5行填入: T\n  向上填充:\n    在第5行填入#\n    在第4行填入最后一个字母: Y\n\n5. 最终矩阵:\n  N B # _ _ _ _ _ _ _ _ _ _ _\n  O A I _ _ _ _ _ _ _ _ _ _ _\n  N P L _ _ _ _ _ _ _ _ _ _ _\n  C L I Y _ _ _ _ _ _ _ _ _ _\n  U # T # _ _ _ _ _ _ _ _ _ _\n\n6. 按行读取并添加*号分隔符\n最终密文: NB#*OAI*NPL*CLIY*U#T#*\n", "ground_truth": "NB#*OAI*NPL*CLIY*U#T#*"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对原始信息进行加密\n\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 不含标点和空格的字符串\n- 准备:\n    - 行数: 5\n- 加密步骤:\n    - 在第一行第一个位置填入第一个原始信息字母\n    - 两种填充方式:\n        - 向下填充: 在第一行填入\"#\"(除去第一列，因为第一个位置已经填入原始信息字母)，然后从第二行到最后一行(第五行)向下填充原始信息\n        - 向上填充: 从最后一行(第五行)到第二行向上填充原始信息，然后在第一行填入\"#\"\n    - 对于原始信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入\n    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终暗文\n\n原始信息: nervii 编码为暗文: N*E*R*VI*I#*\n原始信息: pyritic 编码为暗文是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 189, "source_filename": "icl_with_rule_encode_Kor_rule16_RedefenceFigureCipher_cn.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "PYRITIC", "extra_args": {}, "output": "加密步骤开始:\n1. 处理输入文本为大写字母: PYRITIC\n2. 准备5行矩阵用于填充\n3. 在第一行第一个位置填入第一个字母: P\n4. 开始交替向下和向上填充字母:\n  向下填充:\n    在第2行填入: Y\n    在第3行填入: R\n    在第4行填入: I\n    在第5行填入: T\n  向上填充:\n    在第5行填入#\n    在第4行填入: I\n    在第3行填入最后一个字母: C\n\n5. 最终矩阵:\n  P _ _ _ _ _ _\n  Y _ _ _ _ _ _\n  R C _ _ _ _ _\n  I I _ _ _ _ _\n  T # _ _ _ _ _\n\n6. 按行读取并添加*号分隔符\n最终密文: P*Y*RC*II*T#*\n", "ground_truth": "P*Y*RC*II*T#*"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\nPlease encrypt the plaintext according to the encryption algorithm.\n\nOverview of the encryption scheme:\n1. Filter out non-alphabetic characters and convert the remaining characters to uppercase.\n2. Convert each letter to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the ciphertext.\n5. Output the ciphertext according to the encryption logic.\nclear text: pyritic encode into cipher text: OCAX..C\nclear text: nonculpability encode into cipher text: NFJBXZUHXEJCCY\nclear text: nervii encode into cipher text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 190, "source_filename": "icl_with_rule_encode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "NERVII", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NERVII\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  E -> 层: 1, 列: 2, 行: 2\n  R -> 层: 2, 列: 3, 行: 3\n  V -> 层: 3, 列: 1, 行: 2\n  I -> 层: 1, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\nStep 3: 合并坐标:\n  层坐标: 212311\n  列坐标: 223133\n  行坐标: 223233\nStep 4: 重组坐标: ['212', '311', '223', '133', '223', '233']\nStep 5: 根据新坐标获取密文:\n  212 -> M\n  311 -> S\n  223 -> Q\n  133 -> I\n  223 -> Q\n  233 -> R\n", "ground_truth": "MSQIQR"}, "data_source": "Cipher"}
{"prompt": "Please encrypt the plaintext according to the encryption algorithm.\n\nOverview of the encryption scheme:\n1. Filter out non-alphabetic characters and convert the remaining characters to uppercase.\n2. Convert each letter to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the ciphertext.\n5. Output the ciphertext according to the encryption logic.\noriginal information: pyritic encrypt into cipher text: OCAX..C\noriginal information: nonculpability encrypt into cipher text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 191, "source_filename": "icl_with_rule_encode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NONCULPABILITY\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  O -> 层: 2, 列: 3, 行: 2\n  N -> 层: 2, 列: 2, 行: 2\n  C -> 层: 1, 列: 3, 行: 1\n  U -> 层: 3, 列: 3, 行: 1\n  L -> 层: 2, 列: 3, 行: 1\n  P -> 层: 2, 列: 1, 行: 3\n  A -> 层: 1, 列: 1, 行: 1\n  B -> 层: 1, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  L -> 层: 2, 列: 3, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  Y -> 层: 3, 列: 1, 行: 3\nStep 3: 合并坐标:\n  层坐标: 22213221112133\n  列坐标: 23233311233321\n  行坐标: 22211131131313\nStep 4: 重组坐标: ['222', '132', '211', '121', '332', '323', '331', '123', '332', '122', '211', '131', '131', '313']\nStep 5: 根据新坐标获取密文:\n  222 -> N\n  132 -> F\n  211 -> J\n  121 -> B\n  332 -> X\n  323 -> Z\n  331 -> U\n  123 -> H\n  332 -> X\n  122 -> E\n  211 -> J\n  131 -> C\n  131 -> C\n  313 -> Y\n", "ground_truth": "NFJBXZUHXEJCCY"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\nPlease encrypt the plaintext according to the encryption algorithm.\n\nOverview of the encryption scheme:\n1. Filter out non-alphabetic characters and convert the remaining characters to uppercase.\n2. Convert each letter to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the ciphertext.\n5. Output the ciphertext according to the encryption logic.\nplain text: nonculpability encode to cipher text: NFJBXZUHXEJCCY\nplain text: nonculpability encode to cipher text: NFJBXZUHXEJCCY\nplain text: nonculpability encode to cipher text: NFJBXZUHXEJCCY\nplain text: pyritic encode to cipher text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 192, "source_filename": "icl_with_rule_encode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "PYRITIC", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: PYRITIC\nStep 2: 获取每个字符的坐标:\n  P -> 层: 2, 列: 1, 行: 3\n  Y -> 层: 3, 列: 1, 行: 3\n  R -> 层: 2, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  C -> 层: 1, 列: 3, 行: 1\nStep 3: 合并坐标:\n  层坐标: 2321311\n  列坐标: 1133233\n  行坐标: 3333131\nStep 4: 重组坐标: ['232', '131', '111', '332', '333', '333', '131']\nStep 5: 根据新坐标获取密文:\n  232 -> O\n  131 -> C\n  111 -> A\n  332 -> X\n  333 -> .\n  333 -> .\n  131 -> C\n", "ground_truth": "OCAX..C"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Reels\n        \n        Reel1:   \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n        Reel2:   \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n        Reel3:   \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n        Reel4:   \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n        Reel5:   \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n        Reel6:   \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n        Reel7:    \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n        Reel8:    \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n        Reel9:    \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n        Reel10:   \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n        Reel11:    \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n        Reel12:    \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n        Reel13:    \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n        Reel14:    \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n        Reel15:    \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n        Reel16:    \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n        Reel17:    \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n        Reel18:    \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n        Reel19:    \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n        Reel20:    \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n        Reel21:     \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n        Reel22:    \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n        Reel23:    \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n        Reel24:    \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n        Reel25:    \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        \n- Encryption Steps:\n    - Initially select the Reel1.\n    - For each character p in the plaintext:\n        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.\n        - If the current character is at the end of the Reel, go around to the beginning of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Reels (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Initially select the Reel1.\n    - For each character c in the ciphertext:\n        - Find character c on the Reel and replace it with the previous character  on the Reel to get the plaintext character.\n        - If the current character is at the beginning of the Reel, go around to the end of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the decryption process.\ncipher text: WGTFJXF decrypt to plain text: pyritic\ncipher text: WGTFJXF decrypt to plain text: pyritic\ncipher text: AHTPTX decrypt to plain text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 193, "source_filename": "icl_with_rule_decode_Kor_rule10_JeffersonCipher_en.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "AHTPTX", "extra_args": {}, "output": "初始化Jefferson密码轮...\n开始解密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 A 替换为前一个字符 N\n  * 在第2个密码轮上，将字符 H 替换为前一个字符 E\n  * 在第3个密码轮上，将字符 T 替换为前一个字符 R\n  * 在第4个密码轮上，将字符 P 替换为前一个字符 V\n  * 在第5个密码轮上，将字符 T 替换为前一个字符 I\n  * 在第6个密码轮上，将字符 X 替换为前一个字符 I\n解密完成，结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Reels\n        \n        Reel1:   \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n        Reel2:   \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n        Reel3:   \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n        Reel4:   \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n        Reel5:   \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n        Reel6:   \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n        Reel7:    \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n        Reel8:    \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n        Reel9:    \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n        Reel10:   \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n        Reel11:    \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n        Reel12:    \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n        Reel13:    \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n        Reel14:    \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n        Reel15:    \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n        Reel16:    \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n        Reel17:    \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n        Reel18:    \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n        Reel19:    \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n        Reel20:    \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n        Reel21:     \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n        Reel22:    \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n        Reel23:    \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n        Reel24:    \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n        Reel25:    \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        \n- Encryption Steps:\n    - Initially select the Reel1.\n    - For each character p in the plaintext:\n        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.\n        - If the current character is at the end of the Reel, go around to the beginning of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Reels (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Initially select the Reel1.\n    - For each character c in the ciphertext:\n        - Find character c on the Reel and replace it with the previous character  on the Reel to get the plaintext character.\n        - If the current character is at the beginning of the Reel, go around to the end of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the decryption process.\nencrypted text: AUXBKUVIXCTSSR decode into original information: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 194, "source_filename": "icl_with_rule_decode_Kor_rule10_JeffersonCipher_en.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "AUXBKUVIXCTSSR", "extra_args": {}, "output": "初始化Jefferson密码轮...\n开始解密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 A 替换为前一个字符 N\n  * 在第2个密码轮上，将字符 U 替换为前一个字符 O\n  * 在第3个密码轮上，将字符 X 替换为前一个字符 N\n  * 在第4个密码轮上，将字符 B 替换为前一个字符 C\n  * 在第5个密码轮上，将字符 K 替换为前一个字符 U\n  * 在第6个密码轮上，将字符 U 替换为前一个字符 L\n  * 在第7个密码轮上，将字符 V 替换为前一个字符 P\n  * 在第8个密码轮上，将字符 I 替换为前一个字符 A\n  * 在第9个密码轮上，将字符 X 替换为前一个字符 B\n  * 在第10个密码轮上，将字符 C 替换为前一个字符 I\n  * 在第11个密码轮上，将字符 T 替换为前一个字符 L\n  * 在第12个密码轮上，将字符 S 替换为前一个字符 I\n  * 在第13个密码轮上，将字符 S 替换为前一个字符 T\n  * 在第14个密码轮上，将字符 R 替换为前一个字符 Y\n解密完成，结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Reels\n        \n        Reel1:   \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n        Reel2:   \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n        Reel3:   \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n        Reel4:   \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n        Reel5:   \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n        Reel6:   \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n        Reel7:    \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n        Reel8:    \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n        Reel9:    \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n        Reel10:   \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n        Reel11:    \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n        Reel12:    \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n        Reel13:    \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n        Reel14:    \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n        Reel15:    \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n        Reel16:    \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n        Reel17:    \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n        Reel18:    \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n        Reel19:    \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n        Reel20:    \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n        Reel21:     \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n        Reel22:    \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n        Reel23:    \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n        Reel24:    \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n        Reel25:    \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        \n- Encryption Steps:\n    - Initially select the Reel1.\n    - For each character p in the plaintext:\n        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.\n        - If the current character is at the end of the Reel, go around to the beginning of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Reels (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Initially select the Reel1.\n    - For each character c in the ciphertext:\n        - Find character c on the Reel and replace it with the previous character  on the Reel to get the plaintext character.\n        - If the current character is at the beginning of the Reel, go around to the end of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the decryption process.\ncipher text: AHTPTX decrypt into clear text: nervii\ncipher text: AHTPTX decrypt into clear text: nervii\ncipher text: AUXBKUVIXCTSSR decrypt into clear text: nonculpability\ncipher text: WGTFJXF decrypt into clear text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 195, "source_filename": "icl_with_rule_decode_Kor_rule10_JeffersonCipher_en.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "WGTFJXF", "extra_args": {}, "output": "初始化Jefferson密码轮...\n开始解密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 W 替换为前一个字符 P\n  * 在第2个密码轮上，将字符 G 替换为前一个字符 Y\n  * 在第3个密码轮上，将字符 T 替换为前一个字符 R\n  * 在第4个密码轮上，将字符 F 替换为前一个字符 I\n  * 在第5个密码轮上，将字符 J 替换为前一个字符 T\n  * 在第6个密码轮上，将字符 X 替换为前一个字符 I\n  * 在第7个密码轮上，将字符 F 替换为前一个字符 C\n解密完成，结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n    - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n        X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n        P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n        A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n    - A: 3\n    - B: 5\n    - A_inv: 9\n- 加密步骤:\n    - 对于每个明文字符p:\n        - 设x为其在仿射字母表中的位置\n        - 应用仿射编码函数计算y:\n            - y = (Ax + B) mod 26\n        - 在仿射字母表中找到位置y对应的字母，形成加密消息\n明文: nonculpability 编码为暗文: XEXAKJQMSLJLVH\n明文: nonculpability 编码为暗文: XEXAKJQMSLJLVH\n明文: nervii 编码为暗文: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 196, "source_filename": "icl_with_rule_encode_Kor_rule5_AffineCipher_cn.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "NERVII", "extra_args": {}, "output": "原始文本: NERVII\n开始加密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5\n字符 N -> 位置 7 -> 计算(3*7+5)%26=0 -> 加密为 X\n字符 E -> 位置 23 -> 计算(3*23+5)%26=22 -> 加密为 Y\n字符 R -> 位置 9 -> 计算(3*9+5)%26=6 -> 加密为 O\n字符 V -> 位置 12 -> 计算(3*12+5)%26=15 -> 加密为 F\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n加密完成，结果: XYOFLL\n", "ground_truth": "XYOFLL"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对非加密信息进行加密\n加密规则:\n- 输入:\n    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n        X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n        P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n        A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n    - A: 3\n    - B: 5\n    - A_inv: 9\n- 加密步骤:\n    - 对于每个非加密信息字符p:\n        - 设x为其在仿射字母表中的位置\n        - 应用仿射编码函数计算y:\n            - y = (Ax + B) mod 26\n        - 在仿射字母表中找到位置y对应的字母，形成加密消息\n非加密信息: pyritic 加密成为加密文本: QHOLVLA\n非加密信息: nonculpability 加密成为加密文本: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 197, "source_filename": "icl_with_rule_encode_Kor_rule5_AffineCipher_cn.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "原始文本: NONCULPABILITY\n开始加密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5\n字符 N -> 位置 7 -> 计算(3*7+5)%26=0 -> 加密为 X\n字符 O -> 位置 6 -> 计算(3*6+5)%26=23 -> 加密为 E\n字符 N -> 位置 7 -> 计算(3*7+5)%26=0 -> 加密为 X\n字符 C -> 位置 21 -> 计算(3*21+5)%26=16 -> 加密为 A\n字符 U -> 位置 4 -> 计算(3*4+5)%26=17 -> 加密为 K\n字符 L -> 位置 25 -> 计算(3*25+5)%26=2 -> 加密为 J\n字符 P -> 位置 8 -> 计算(3*8+5)%26=3 -> 加密为 Q\n字符 A -> 位置 16 -> 计算(3*16+5)%26=1 -> 加密为 M\n字符 B -> 位置 13 -> 计算(3*13+5)%26=18 -> 加密为 S\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 L -> 位置 25 -> 计算(3*25+5)%26=2 -> 加密为 J\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 T -> 位置 11 -> 计算(3*11+5)%26=12 -> 加密为 V\n字符 Y -> 位置 22 -> 计算(3*22+5)%26=19 -> 加密为 H\n加密完成，结果: XEXAKJQMSLJLVH\n", "ground_truth": "XEXAKJQMSLJLVH"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n    - 仿射字母表 = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - 将每个字母与其在仿射字母表中的位置关联（从0开始）:\n        X->0, M->1, J->2, Q->3, U->4, D->5, O->6, N->7,\n        P->8, R->9, G->10, T->11, V->12, B->13, W->14, F->15,\n        A->16, K->17, S->18, H->19, Z->20, C->21, Y->22, E->23, I->24, L->25\n    - A: 3\n    - B: 5\n    - A_inv: 9\n- 加密步骤:\n    - 对于每个原始信息字符p:\n        - 设x为其在仿射字母表中的位置\n        - 应用仿射编码函数计算y:\n            - y = (Ax + B) mod 26\n        - 在仿射字母表中找到位置y对应的字母，形成加密消息\n原始信息: pyritic 编码为密文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 198, "source_filename": "icl_with_rule_encode_Kor_rule5_AffineCipher_cn.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "PYRITIC", "extra_args": {}, "output": "原始文本: PYRITIC\n开始加密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5\n字符 P -> 位置 8 -> 计算(3*8+5)%26=3 -> 加密为 Q\n字符 Y -> 位置 22 -> 计算(3*22+5)%26=19 -> 加密为 H\n字符 R -> 位置 9 -> 计算(3*9+5)%26=6 -> 加密为 O\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 T -> 位置 11 -> 计算(3*11+5)%26=12 -> 加密为 V\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 C -> 位置 21 -> 计算(3*21+5)%26=16 -> 加密为 A\n加密完成，结果: QHOLVLA\n", "ground_truth": "QHOLVLA"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):\n        - U  B  I  L  A\n        N  T  C  D  E\n        F  G  H  K  M\n        O  P  Q  R  S\n        V  W  X  Y  Z\n- Encryption Steps:\n    - Remove punctuation, spaces, and the letter J from plaintext.\n    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).\n    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.\n    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 grid (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each letter c in the ciphertext, find its corresponding row and column coordinates (between 0 and 4) in the grid to obtain a sequence of coordinates.\n    - Divide the coordinate sequence into two halves, the first half is all the original row coordinates, the second half is all the original column coordinates, read one from the original row coordinates, read one from the original column coordinates, and find the corresponding letter in the grid according to the original row and column coordinates to form the plaintext.\n        - For example [0,1,4,2], the first half as row coordinates: [0, 1], the second half as column coordinates: [4,2].\n        - Read one from the row coordinates, one from the column coordinates, and so on alternately. Obtain (0,4),(1,2), respectively, and find the corresponding letters in the grid as AC.\ncipher text: DTUOUUEUILECQD decode into plain text: nonculpability\ncipher text: TSUAOH decode into plain text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 199, "source_filename": "icl_with_rule_decode_Kor_rule13_BifidCipher_en.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "TSUAOH", "extra_args": {}, "output": "解密步骤:\n1. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n2. 获取密文字母的坐标: [1, 3, 0, 0, 3, 2, 1, 4, 0, 4, 0, 2]\n3. 还原原始坐标: ['1', '0', '1', '4', '3', '3', '4', '0', '0', '2', '0', '2']\n4. 最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):\n        - U  B  I  L  A\n        N  T  C  D  E\n        F  G  H  K  M\n        O  P  Q  R  S\n        V  W  X  Y  Z\n- Encryption Steps:\n    - Remove punctuation, spaces, and the letter J from plaintext.\n    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).\n    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.\n    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 grid (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each letter c in the ciphertext, find its corresponding row and column coordinates (between 0 and 4) in the grid to obtain a sequence of coordinates.\n    - Divide the coordinate sequence into two halves, the first half is all the original row coordinates, the second half is all the original column coordinates, read one from the original row coordinates, read one from the original column coordinates, and find the corresponding letter in the grid according to the original row and column coordinates to form the plaintext.\n        - For example [0,1,4,2], the first half as row coordinates: [0, 1], the second half as column coordinates: [4,2].\n        - Read one from the row coordinates, one from the column coordinates, and so on alternately. Obtain (0,4),(1,2), respectively, and find the corresponding letters in the grid as AC.\nencoded text: SONTRGH decrypt to plain text: pyritic\nencoded text: DTUOUUEUILECQD decrypt to plain text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 200, "source_filename": "icl_with_rule_decode_Kor_rule13_BifidCipher_en.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "DTUOUUEUILECQD", "extra_args": {}, "output": "解密步骤:\n1. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n2. 获取密文字母的坐标: [1, 1, 0, 3, 0, 0, 1, 0, 0, 0, 1, 1, 3, 1, 3, 1, 0, 0, 0, 0, 4, 0, 2, 3, 4, 2, 2, 3]\n3. 还原原始坐标: ['1', '0', '3', '0', '1', '0', '1', '2', '0', '0', '0', '3', '3', '1', '0', '4', '0', '1', '0', '2', '0', '3', '0', '2', '1', '1', '4', '3']\n4. 最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):\n        - U  B  I  L  A\n        N  T  C  D  E\n        F  G  H  K  M\n        O  P  Q  R  S\n        V  W  X  Y  Z\n- Encryption Steps:\n    - Remove punctuation, spaces, and the letter J from plaintext.\n    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).\n    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.\n    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 grid (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each letter c in the ciphertext, find its corresponding row and column coordinates (between 0 and 4) in the grid to obtain a sequence of coordinates.\n    - Divide the coordinate sequence into two halves, the first half is all the original row coordinates, the second half is all the original column coordinates, read one from the original row coordinates, read one from the original column coordinates, and find the corresponding letter in the grid according to the original row and column coordinates to form the plaintext.\n        - For example [0,1,4,2], the first half as row coordinates: [0, 1], the second half as column coordinates: [4,2].\n        - Read one from the row coordinates, one from the column coordinates, and so on alternately. Obtain (0,4),(1,2), respectively, and find the corresponding letters in the grid as AC.\nencoded text: DTUOUUEUILECQD decode to original information: nonculpability\nencoded text: SONTRGH decode to original information: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 201, "source_filename": "icl_with_rule_decode_Kor_rule13_BifidCipher_en.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "SONTRGH", "extra_args": {}, "output": "解密步骤:\n1. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n2. 获取密文字母的坐标: [3, 3, 1, 1, 3, 2, 2, 4, 0, 0, 1, 3, 1, 2]\n3. 还原原始坐标: ['3', '1', '4', '3', '3', '3', '0', '2', '1', '1', '0', '2', '1', '2']\n4. 最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\ncipher text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... decode to clear text: nonculpability\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decode to clear text: pyritic\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decode to clear text: pyritic\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decode to clear text: pyritic\ncipher text: ... ...  . .....  .... ..  ..... .  .. ....  .. .... decode to clear text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 202, "source_filename": "icl_decode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "... ...  . .....  .... ..  ..... .  .. ....  .. ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '. .....'，它表示第 1 行第 5 列，对应的字母是 'E'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '..... .'，它表示第 5 行第 1 列，对应的字母是 'V'\n步骤 5：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n最终步骤：解码完成，解码后的消息是：NERVII \n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decrypt to original information: pyritic\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decrypt to original information: pyritic\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decrypt to original information: pyritic\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decrypt to original information: pyritic\ncipher text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... decrypt to original information: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 203, "source_filename": "icl_decode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '... ....'，它表示第 3 行第 4 列，对应的字母是 'O'\n步骤 3：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 4：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n步骤 5：正在解码 '.... .....'，它表示第 4 行第 5 列，对应的字母是 'U'\n步骤 6：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 7：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 8：正在解码 '. .'，它表示第 1 行第 1 列，对应的字母是 'A'\n步骤 9：正在解码 '. ..'，它表示第 1 行第 2 列，对应的字母是 'B'\n步骤 10：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 11：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 12：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 13：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 14：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n最终步骤：解码完成，解码后的消息是：NONCULPABILITY \n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\nencoded text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... decode into original information: nonculpability\nencoded text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... decode into original information: nonculpability\nencoded text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... decode into original information: nonculpability\nencoded text: ... ...  . .....  .... ..  ..... .  .. ....  .. .... decode into original information: nervii\nencoded text: ... ...  . .....  .... ..  ..... .  .. ....  .. .... decode into original information: nervii\nencoded text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decode into original information is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 204, "source_filename": "icl_decode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...", "extra_args": {}, "output": "步骤 1：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 2：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 5：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 7：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n最终步骤：解码完成，解码后的消息是：PYRITIC \n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n密文: OCAX..C 解密成非加密信息: pyritic\n密文: NFJBXZUHXEJCCY 解密成非加密信息: nonculpability\n密文: NFJBXZUHXEJCCY 解密成非加密信息: nonculpability\n密文: MSQIQR 解密成非加密信息: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 205, "source_filename": "icl_decode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "MSQIQR", "extra_args": {}, "output": "步骤1: 将密文转换为大写: MSQIQR\n步骤2: 将每个字母转换为坐标: 212311223133223233\n步骤3: 计算列数: 6\n步骤4: 将坐标组织成3xN的矩阵:\n212311\n223133\n223233\n步骤5: 转置矩阵:\n222\n122\n233\n312\n133\n133\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (1, 2, 2), (2, 3, 3), (3, 1, 2), (1, 3, 3), (1, 3, 3)]\n步骤7: 将坐标转换回字母: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n加密信息: MSQIQR 解码为初始文本: nervii\n加密信息: MSQIQR 解码为初始文本: nervii\n加密信息: MSQIQR 解码为初始文本: nervii\n加密信息: OCAX..C 解码为初始文本: pyritic\n加密信息: NFJBXZUHXEJCCY 解码为初始文本是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 206, "source_filename": "icl_decode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "NFJBXZUHXEJCCY", "extra_args": {}, "output": "步骤1: 将密文转换为大写: NFJBXZUHXEJCCY\n步骤2: 将每个字母转换为坐标: 222132211121332323331123332122211131131313\n步骤3: 计算列数: 14\n步骤4: 将坐标组织成3xN的矩阵:\n22213221112133\n23233311233321\n22211131131313\n步骤5: 转置矩阵:\n222\n232\n222\n131\n331\n231\n213\n111\n121\n133\n231\n133\n321\n313\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (2, 3, 2), (2, 2, 2), (1, 3, 1), (3, 3, 1), (2, 3, 1), (2, 1, 3), (1, 1, 1), (1, 2, 1), (1, 3, 3), (2, 3, 1), (1, 3, 3), (3, 2, 1), (3, 1, 3)]\n步骤7: 将坐标转换回字母: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n隐文: MSQIQR 解密成明文: nervii\n隐文: MSQIQR 解密成明文: nervii\n隐文: MSQIQR 解密成明文: nervii\n隐文: MSQIQR 解密成明文: nervii\n隐文: MSQIQR 解密成明文: nervii\n隐文: OCAX..C 解密成明文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 207, "source_filename": "icl_decode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "OCAX..C", "extra_args": {}, "output": "步骤1: 将密文转换为大写: OCAX..C\n步骤2: 将每个字母转换为坐标: 232131111332333333131\n步骤3: 计算列数: 7\n步骤4: 将坐标组织成3xN的矩阵:\n2321311\n1133233\n3333131\n步骤5: 转置矩阵:\n213\n313\n233\n133\n321\n133\n131\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 1, 3), (3, 1, 3), (2, 3, 3), (1, 3, 3), (3, 2, 1), (1, 3, 3), (1, 3, 1)]\n步骤7: 将坐标转换回字母: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - e: 263\n    - n: 299\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x of the value corresponding to the ascii code of p.\n        - Compute x^e mod n as the ciphertext number y for that letter p . Here ^ denotes the multiplication operation.\n    - Finally, concatenate all the y's and separate them with commas for the final ciphertext.\n\n \nclear text: pyritic encrypt into encrypted text: 228,84,62,213,89,213,241\nclear text: nervii encrypt into encrypted text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 208, "source_filename": "icl_with_rule_encode_Kor_rule23_RSACipher_en.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NERVII\n使用参数: e=263, n=299\n\n逐字符加密:\n字符 N 的ASCII码为 78\n计算 78^263 mod 299 = 156\n字符 E 的ASCII码为 69\n计算 69^263 mod 299 = 23\n字符 R 的ASCII码为 82\n计算 82^263 mod 299 = 62\n字符 V 的ASCII码为 86\n计算 86^263 mod 299 = 226\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n\n最终加密结果: 156,23,62,226,213,213\n", "ground_truth": "156,23,62,226,213,213"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - e: 263\n    - n: 299\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x of the value corresponding to the ascii code of p.\n        - Compute x^e mod n as the ciphertext number y for that letter p . Here ^ denotes the multiplication operation.\n    - Finally, concatenate all the y's and separate them with commas for the final ciphertext.\n\n \noriginal information: pyritic encode to cipher text: 228,84,62,213,89,213,241\noriginal information: nervii encode to cipher text: 156,23,62,226,213,213\noriginal information: pyritic encode to cipher text: 228,84,62,213,89,213,241\noriginal information: nonculpability encode to cipher text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 209, "source_filename": "icl_with_rule_encode_Kor_rule23_RSACipher_en.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NONCULPABILITY\n使用参数: e=263, n=299\n\n逐字符加密:\n字符 N 的ASCII码为 78\n计算 78^263 mod 299 = 156\n字符 O 的ASCII码为 79\n计算 79^263 mod 299 = 53\n字符 N 的ASCII码为 78\n计算 78^263 mod 299 = 156\n字符 C 的ASCII码为 67\n计算 67^263 mod 299 = 241\n字符 U 的ASCII码为 85\n计算 85^263 mod 299 = 197\n字符 L 的ASCII码为 76\n计算 76^263 mod 299 = 240\n字符 P 的ASCII码为 80\n计算 80^263 mod 299 = 228\n字符 A 的ASCII码为 65\n计算 65^263 mod 299 = 247\n字符 B 的ASCII码为 66\n计算 66^263 mod 299 = 222\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 L 的ASCII码为 76\n计算 76^263 mod 299 = 240\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 T 的ASCII码为 84\n计算 84^263 mod 299 = 89\n字符 Y 的ASCII码为 89\n计算 89^263 mod 299 = 84\n\n最终加密结果: 156,53,156,241,197,240,228,247,222,213,240,213,89,84\n", "ground_truth": "156,53,156,241,197,240,228,247,222,213,240,213,89,84"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - e: 263\n    - n: 299\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x of the value corresponding to the ascii code of p.\n        - Compute x^e mod n as the ciphertext number y for that letter p . Here ^ denotes the multiplication operation.\n    - Finally, concatenate all the y's and separate them with commas for the final ciphertext.\n\n \nplain text: nervii encrypt into encoded text: 156,23,62,226,213,213\nplain text: nonculpability encrypt into encoded text: 156,53,156,241,197,240,228,247,222,213,240,213,89,84\nplain text: pyritic encrypt into encoded text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 210, "source_filename": "icl_with_rule_encode_Kor_rule23_RSACipher_en.jsonl", "cipher_name": "Kor_rule23_RSACipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: PYRITIC\n使用参数: e=263, n=299\n\n逐字符加密:\n字符 P 的ASCII码为 80\n计算 80^263 mod 299 = 228\n字符 Y 的ASCII码为 89\n计算 89^263 mod 299 = 84\n字符 R 的ASCII码为 82\n计算 82^263 mod 299 = 62\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 T 的ASCII码为 84\n计算 84^263 mod 299 = 89\n字符 I 的ASCII码为 73\n计算 73^263 mod 299 = 213\n字符 C 的ASCII码为 67\n计算 67^263 mod 299 = 241\n\n最终加密结果: 228,84,62,213,89,213,241\n", "ground_truth": "228,84,62,213,89,213,241"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 原文: 不含标点和空格的大写字母字符串\n    - 输出:\n        - 暗码: 不含标点和空格的字符串\n    - 准备:\n        - 转置序列表:\n            - [1, 4, 0, 6, 5, 2, 3]\n            - 转置序列表用于按顺序逐行写入原文，然后根据转置序列表调整列的顺序，使每行中的字符按给定顺序排列。\n            - 列从0开始计数。\n    - 加密步骤:\n        - [1, 4, 0, 6, 5, 2, 3]转置序列表共7位，表示一行应写入7个字母。\n        - 按顺序逐行写入原文，每行7个。当不足7个时，最后一行用$填充。可以得到一个写入网格。\n        - 根据转置序列表调整列的顺序，即现在列的顺序为[原列1，原列4，原列0，原列6，原列5，原列2，原列3]，可以得到调整列顺序后的网格。\n        - 逐行读取网格并连接起来得到最终暗码。(注意需要保留$)\n解密算法:\n解密规则:\n    - 输入:\n        - 暗码: 不含标点和空格的字符串\n    - 输出:\n        - 原文: 不含标点和空格的大写字母字符串\n    - 准备:\n        - 转置序列表(与加密相同)\n    - 解密步骤:\n        - 按顺序逐行写入暗码，每行7个字母。\n        - 逐行读取，但读取每行时，先读取对应0的第2列的字符，然后读取对应1的第0列的字符，然后读取对应2的第6列的字符，依此类推。\n        - 最终逐行读取信息，去掉末尾的$，即可得到解密后的原文。\n暗码: EIN$IRV 解密为原文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 211, "source_filename": "icl_with_rule_decode_Kor_rule20_TranspositionCipher_cn.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "EIN$IRV", "extra_args": {}, "output": "需要解密的文本: EIN$IRV\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n加密的网格:\nEIN$IRV\n还原后的网格:\nNERVII$\n解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 原始信息: 不含标点和空格的大写字母字符串\n    - 输出:\n        - 暗码: 不含标点和空格的字符串\n    - 准备:\n        - 转置序列表:\n            - [1, 4, 0, 6, 5, 2, 3]\n            - 转置序列表用于按顺序逐行写入原始信息，然后根据转置序列表调整列的顺序，使每行中的字符按给定顺序排列。\n            - 列从0开始计数。\n    - 加密步骤:\n        - [1, 4, 0, 6, 5, 2, 3]转置序列表共7位，表示一行应写入7个字母。\n        - 按顺序逐行写入原始信息，每行7个。当不足7个时，最后一行用$填充。可以得到一个写入网格。\n        - 根据转置序列表调整列的顺序，即现在列的顺序为[原列1，原列4，原列0，原列6，原列5，原列2，原列3]，可以得到调整列顺序后的网格。\n        - 逐行读取网格并连接起来得到最终暗码。(注意需要保留$)\n解密算法:\n解密规则:\n    - 输入:\n        - 暗码: 不含标点和空格的字符串\n    - 输出:\n        - 原始信息: 不含标点和空格的大写字母字符串\n    - 准备:\n        - 转置序列表(与加密相同)\n    - 解密步骤:\n        - 按顺序逐行写入暗码，每行7个字母。\n        - 逐行读取，但读取每行时，先读取对应0的第2列的字符，然后读取对应1的第0列的字符，然后读取对应2的第6列的字符，依此类推。\n        - 最终逐行读取信息，去掉末尾的$，即可得到解密后的原始信息。\n暗码: YTPCIRI 解密成原始信息: pyritic\n暗码: OUNPLNCBIAYTIL 解密成原始信息是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 212, "source_filename": "icl_with_rule_decode_Kor_rule20_TranspositionCipher_cn.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "OUNPLNCBIAYTIL", "extra_args": {}, "output": "需要解密的文本: OUNPLNCBIAYTIL\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 2 行来存放文本\n加密的网格:\nOUNPLNC\nBIAYTIL\n还原后的网格:\nNONCULP\nABILITY\n解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n    - 输入:\n        - 明文: 不含标点和空格的大写字母字符串\n    - 输出:\n        - 暗文: 不含标点和空格的字符串\n    - 准备:\n        - 转置序列表:\n            - [1, 4, 0, 6, 5, 2, 3]\n            - 转置序列表用于按顺序逐行写入明文，然后根据转置序列表调整列的顺序，使每行中的字符按给定顺序排列。\n            - 列从0开始计数。\n    - 加密步骤:\n        - [1, 4, 0, 6, 5, 2, 3]转置序列表共7位，表示一行应写入7个字母。\n        - 按顺序逐行写入明文，每行7个。当不足7个时，最后一行用$填充。可以得到一个写入网格。\n        - 根据转置序列表调整列的顺序，即现在列的顺序为[原列1，原列4，原列0，原列6，原列5，原列2，原列3]，可以得到调整列顺序后的网格。\n        - 逐行读取网格并连接起来得到最终暗文。(注意需要保留$)\n解密算法:\n解密规则:\n    - 输入:\n        - 暗文: 不含标点和空格的字符串\n    - 输出:\n        - 明文: 不含标点和空格的大写字母字符串\n    - 准备:\n        - 转置序列表(与加密相同)\n    - 解密步骤:\n        - 按顺序逐行写入暗文，每行7个字母。\n        - 逐行读取，但读取每行时，先读取对应0的第2列的字符，然后读取对应1的第0列的字符，然后读取对应2的第6列的字符，依此类推。\n        - 最终逐行读取信息，去掉末尾的$，即可得到解密后的明文。\n暗文: EIN$IRV 解密为明文: nervii\n暗文: OUNPLNCBIAYTIL 解密为明文: nonculpability\n暗文: EIN$IRV 解密为明文: nervii\n暗文: YTPCIRI 解密为明文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 213, "source_filename": "icl_with_rule_decode_Kor_rule20_TranspositionCipher_cn.jsonl", "cipher_name": "Kor_rule20_TranspositionCipher", "input": "YTPCIRI", "extra_args": {}, "output": "需要解密的文本: YTPCIRI\n使用转置序列: [1, 4, 0, 6, 5, 2, 3]\n需要 1 行来存放文本\n加密的网格:\nYTPCIRI\n还原后的网格:\nPYRITIC\n解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n加密信息: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解码为明文: nonculpability\n加密信息: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解码为明文: nonculpability\n加密信息: RV RB RD WX 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解码为明文: pyritic\n加密信息: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解码为明文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 214, "source_filename": "icl_decode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "JV KT CA", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将密文分成两个字母一组: ['JV', 'KT', 'CA']\n\nStep 4.1: 解密字母对 JV\n  - J 在 M1 中的位置: (2, 4)\n  - V 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 4) 的字母: E\n  - 解密结果为: NE\n\nStep 4.2: 解密字母对 KT\n  - K 在 M1 中的位置: (3, 0)\n  - T 在 M2 中的位置: (4, 1)\n  - 交换列索引后新位置: (3, 1) 和 (4, 0)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (4, 0) 的字母: V\n  - 解密结果为: RV\n\nStep 4.3: 解密字母对 CA\n  - C 在 M1 中的位置: (1, 3)\n  - A 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: II\n\n解密完成！最终解密结果为: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n密文: RV RB RD WX 额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解密成非加密信息: pyritic\n密文: JV KT CA 额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密成非加密信息: nervii\n密文: JK FI LL KS CE GE NY 额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成非加密信息是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 215, "source_filename": "icl_decode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "JK FI LL KS CE GE NY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['JK', 'FI', 'LL', 'KS', 'CE', 'GE', 'NY']\n\nStep 4.1: 解密字母对 JK\n  - J 在 M1 中的位置: (2, 4)\n  - K 在 M2 中的位置: (2, 3)\n  - 交换列索引后新位置: (2, 3) 和 (2, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (2, 4) 的字母: O\n  - 解密结果为: NO\n\nStep 4.2: 解密字母对 FI\n  - F 在 M1 中的位置: (2, 2)\n  - I 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 2)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 解密结果为: NC\n\nStep 4.3: 解密字母对 LL\n  - L 在 M1 中的位置: (3, 1)\n  - L 在 M2 中的位置: (2, 4)\n  - 交换列索引后新位置: (3, 4) 和 (2, 1)\n  - 在 M_T 中查找位置 (3, 4) 的字母: U\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 解密结果为: UL\n\nStep 4.4: 解密字母对 KS\n  - K 在 M1 中的位置: (3, 0)\n  - S 在 M2 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (0, 0) 的字母: A\n  - 解密结果为: PA\n\nStep 4.5: 解密字母对 CE\n  - C 在 M1 中的位置: (0, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (0, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (0, 1) 的字母: B\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: BI\n\nStep 4.6: 解密字母对 GE\n  - G 在 M1 中的位置: (2, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (2, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: LI\n\nStep 4.7: 解密字母对 NY\n  - N 在 M1 中的位置: (3, 3)\n  - Y 在 M2 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: TY\n\n解密完成！最终解密结果为: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n隐文: JV KT CA 额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密成原始信息: nervii\n隐文: JK FI LL KS CE GE NY 额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成原始信息: nonculpability\n隐文: JK FI LL KS CE GE NY 额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成原始信息: nonculpability\n隐文: RV RB RD WX 额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解密成原始信息: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 216, "source_filename": "icl_decode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "RV RB RD WX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['RV', 'RB', 'RD', 'WX']\n\nStep 4.1: 解密字母对 RV\n  - R 在 M1 中的位置: (3, 3)\n  - V 在 M2 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: PY\n\nStep 4.2: 解密字母对 RB\n  - R 在 M1 中的位置: (3, 3)\n  - B 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (3, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: RI\n\nStep 4.3: 解密字母对 RD\n  - R 在 M1 中的位置: (3, 3)\n  - D 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: TI\n\nStep 4.4: 解密字母对 WX\n  - W 在 M1 中的位置: (0, 2)\n  - X 在 M2 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 在 M_T 中查找位置 (4, 2) 的字母: X\n  - 解密结果为: CX\n\n解密完成！最终解密结果为: PYRITICX\n", "ground_truth": "PYRITICX"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密方案概述：\n        1. 过滤掉非字母字符并转换为大写。\n        2. 将每个字母转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成隐文。\n        5. 按照加密逻辑，将隐文输出。\n        \n解密算法:\n解密方案概述：\n        1. 将隐文转换为大写字母。\n        2. 将隐文转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成原文。\n        5. 按照解密逻辑，将原文输出。\n        \n隐文: NFJBXZUHXEJCCY 解密成为原文: nonculpability\n隐文: OCAX..C 解密成为原文: pyritic\n隐文: NFJBXZUHXEJCCY 解密成为原文: nonculpability\n隐文: MSQIQR 解密成为原文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 217, "source_filename": "icl_with_rule_decode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "MSQIQR", "extra_args": {}, "output": "步骤1: 将密文转换为大写: MSQIQR\n步骤2: 将每个字母转换为坐标: 212311223133223233\n步骤3: 计算列数: 6\n步骤4: 将坐标组织成3xN的矩阵:\n212311\n223133\n223233\n步骤5: 转置矩阵:\n222\n122\n233\n312\n133\n133\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (1, 2, 2), (2, 3, 3), (3, 1, 2), (1, 3, 3), (1, 3, 3)]\n步骤7: 将坐标转换回字母: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密方案概述：\n        1. 过滤掉非字母字符并转换为大写。\n        2. 将每个字母转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成密文。\n        5. 按照加密逻辑，将密文输出。\n        \n解密算法:\n解密方案概述：\n        1. 将密文转换为大写字母。\n        2. 将密文转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成非加密信息。\n        5. 按照解密逻辑，将非加密信息输出。\n        \n密文: OCAX..C 解码为非加密信息: pyritic\n密文: MSQIQR 解码为非加密信息: nervii\n密文: MSQIQR 解码为非加密信息: nervii\n密文: NFJBXZUHXEJCCY 解码为非加密信息: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 218, "source_filename": "icl_with_rule_decode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "NFJBXZUHXEJCCY", "extra_args": {}, "output": "步骤1: 将密文转换为大写: NFJBXZUHXEJCCY\n步骤2: 将每个字母转换为坐标: 222132211121332323331123332122211131131313\n步骤3: 计算列数: 14\n步骤4: 将坐标组织成3xN的矩阵:\n22213221112133\n23233311233321\n22211131131313\n步骤5: 转置矩阵:\n222\n232\n222\n131\n331\n231\n213\n111\n121\n133\n231\n133\n321\n313\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (2, 3, 2), (2, 2, 2), (1, 3, 1), (3, 3, 1), (2, 3, 1), (2, 1, 3), (1, 1, 1), (1, 2, 1), (1, 3, 3), (2, 3, 1), (1, 3, 3), (3, 2, 1), (3, 1, 3)]\n步骤7: 将坐标转换回字母: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密方案概述：\n        1. 过滤掉非字母字符并转换为大写。\n        2. 将每个字母转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成隐文。\n        5. 按照加密逻辑，将隐文输出。\n        \n解密算法:\n解密方案概述：\n        1. 将隐文转换为大写字母。\n        2. 将隐文转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成初始文本。\n        5. 按照解密逻辑，将初始文本输出。\n        \n隐文: MSQIQR 解码为初始文本: nervii\n隐文: NFJBXZUHXEJCCY 解码为初始文本: nonculpability\n隐文: NFJBXZUHXEJCCY 解码为初始文本: nonculpability\n隐文: OCAX..C 解码为初始文本是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 219, "source_filename": "icl_with_rule_decode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "OCAX..C", "extra_args": {}, "output": "步骤1: 将密文转换为大写: OCAX..C\n步骤2: 将每个字母转换为坐标: 232131111332333333131\n步骤3: 计算列数: 7\n步骤4: 将坐标组织成3xN的矩阵:\n2321311\n1133233\n3333131\n步骤5: 转置矩阵:\n213\n313\n233\n133\n321\n133\n131\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 1, 3), (3, 1, 3), (2, 3, 3), (1, 3, 3), (3, 2, 1), (1, 3, 3), (1, 3, 1)]\n步骤7: 将坐标转换回字母: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\nPlease decrypt the password based on the encryption and decryption algorithms.\n\nEncryption Algorithm:\nOverview of the encryption scheme:\n1. Filter out non-alphabetic characters and convert to uppercase.\n2. Convert each letter to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the ciphertext.\n5. Output the ciphertext according to the encryption logic.\n\nDecryption Algorithm:\nOverview of the decryption scheme:\n1. Convert the ciphertext to uppercase letters.\n2. Convert the ciphertext to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the plaintext.\n5. Output the plaintext according to the decryption logic.\nencoded text: NFJBXZUHXEJCCY decode to plain text: nonculpability\nencoded text: OCAX..C decode to plain text: pyritic\nencoded text: NFJBXZUHXEJCCY decode to plain text: nonculpability\nencoded text: MSQIQR decode to plain text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 220, "source_filename": "icl_with_rule_decode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "MSQIQR", "extra_args": {}, "output": "步骤1: 将密文转换为大写: MSQIQR\n步骤2: 将每个字母转换为坐标: 212311223133223233\n步骤3: 计算列数: 6\n步骤4: 将坐标组织成3xN的矩阵:\n212311\n223133\n223233\n步骤5: 转置矩阵:\n222\n122\n233\n312\n133\n133\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (1, 2, 2), (2, 3, 3), (3, 1, 2), (1, 3, 3), (1, 3, 3)]\n步骤7: 将坐标转换回字母: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\nPlease decrypt the password based on the encryption and decryption algorithms.\n\nEncryption Algorithm:\nOverview of the encryption scheme:\n1. Filter out non-alphabetic characters and convert to uppercase.\n2. Convert each letter to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the ciphertext.\n5. Output the ciphertext according to the encryption logic.\n\nDecryption Algorithm:\nOverview of the decryption scheme:\n1. Convert the ciphertext to uppercase letters.\n2. Convert the ciphertext to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the plaintext.\n5. Output the plaintext according to the decryption logic.\ncipher text: OCAX..C decrypt into clear text: pyritic\ncipher text: MSQIQR decrypt into clear text: nervii\ncipher text: MSQIQR decrypt into clear text: nervii\ncipher text: NFJBXZUHXEJCCY decrypt into clear text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 221, "source_filename": "icl_with_rule_decode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "NFJBXZUHXEJCCY", "extra_args": {}, "output": "步骤1: 将密文转换为大写: NFJBXZUHXEJCCY\n步骤2: 将每个字母转换为坐标: 222132211121332323331123332122211131131313\n步骤3: 计算列数: 14\n步骤4: 将坐标组织成3xN的矩阵:\n22213221112133\n23233311233321\n22211131131313\n步骤5: 转置矩阵:\n222\n232\n222\n131\n331\n231\n213\n111\n121\n133\n231\n133\n321\n313\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (2, 3, 2), (2, 2, 2), (1, 3, 1), (3, 3, 1), (2, 3, 1), (2, 1, 3), (1, 1, 1), (1, 2, 1), (1, 3, 3), (2, 3, 1), (1, 3, 3), (3, 2, 1), (3, 1, 3)]\n步骤7: 将坐标转换回字母: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\nPlease decrypt the password based on the encryption and decryption algorithms.\n\nEncryption Algorithm:\nOverview of the encryption scheme:\n1. Filter out non-alphabetic characters and convert to uppercase.\n2. Convert each letter to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the ciphertext.\n5. Output the ciphertext according to the encryption logic.\n\nDecryption Algorithm:\nOverview of the decryption scheme:\n1. Convert the ciphertext to uppercase letters.\n2. Convert the ciphertext to its corresponding 3D coordinates.\n3. Reorganize the 3D coordinates, grouping every three characters together.\n4. Find the corresponding letters in the 3D cube based on the new coordinate groups to generate the plaintext.\n5. Output the plaintext according to the decryption logic.\nencrypted text: MSQIQR decrypt into plain text: nervii\nencrypted text: NFJBXZUHXEJCCY decrypt into plain text: nonculpability\nencrypted text: NFJBXZUHXEJCCY decrypt into plain text: nonculpability\nencrypted text: OCAX..C decrypt into plain text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 222, "source_filename": "icl_with_rule_decode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "OCAX..C", "extra_args": {}, "output": "步骤1: 将密文转换为大写: OCAX..C\n步骤2: 将每个字母转换为坐标: 232131111332333333131\n步骤3: 计算列数: 7\n步骤4: 将坐标组织成3xN的矩阵:\n2321311\n1133233\n3333131\n步骤5: 转置矩阵:\n213\n313\n233\n133\n321\n133\n131\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 1, 3), (3, 1, 3), (2, 3, 3), (1, 3, 3), (3, 2, 1), (1, 3, 3), (1, 3, 1)]\n步骤7: 将坐标转换回字母: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 隐文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(Grid0-Grid7)\n- 加密步骤:\n    - 明文按5个字符分组，从0开始编号\n    - 对于5个字符的块:\n        - 使用的网格由grid_index = (i // 5) % 8确定，其中i是块号。整数除法运算符//将左边的数除以右边的数，向下取整结果\n        - 对于当前块中的每个字符:\n            - 如果字符是\"J\"，不加密直接添加到加密块\n            - 否则，在当前网格中找到字符的位置。然后向右下方移动一个网格位置(row+1,col+1)(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为加密字母\n            - 将加密字母添加到加密块\n    - 处理完所有块后，连接加密块形成最终加密消息\n解密算法:\n解密规则:\n- 输入:\n    - 隐文: 大写字母字符串\n- 输出:\n    - 明文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(与加密相同)\n- 解密步骤:\n    - 将隐文分成5个字符的块:\n        - 例如，如果隐文是\"KHOORTQHQTHHSAUQ\"，第0块是\"KHOOR\"，第1块是\"TQHQH\"，以此类推(从0开始编号)\n    - 确定用于当前块的网格:\n        计算grid_index = (i // 5) % 8从网格列表中选择适当的网格。i是块号\n    - 对于块中的每个字符:\n        - 如果字符是\"J\": 直接将\"J\"添加到解密块，不进行解密\n        - 否则在网格中找到字符的位置，通过向左上方移动一格获得(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为解密字母\n        - 将解密字母添加到解密块\n    - 处理完块中所有字符后，将解密块添加到解密消息列表。形成最终解密消息\n隐文: OWOMVEBGKDEDZS 解密成为明文: nonculpability\n隐文: OWOMVEBGKDEDZS 解密成为明文: nonculpability\n隐文: OWOMVEBGKDEDZS 解密成为明文: nonculpability\n隐文: OFYHDD 解密成为明文: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 223, "source_filename": "icl_with_rule_decode_Kor_rule7_PhillipsFigureCipher_cn.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "OFYHDD", "extra_args": {}, "output": "需要解密的文本: OFYHDD\n\n处理第0个块: OFYHD\n使用第0号网格\n字母O在位置(3,0)，向左上移动一格得到N\n字母F在位置(2,0)，向左上移动一格得到E\n字母Y在位置(4,3)，向左上移动一格得到R\n字母H在位置(0,1)，向左上移动一格得到V\n字母D在位置(1,3)，向左上移动一格得到I\n解密后的块: NERVI\n\n处理第1个块: D\n使用第1号网格\n字母D在位置(2,3)，向左上移动一格得到I\n解密后的块: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(Grid0-Grid7)\n- 加密步骤:\n    - 明文按5个字符分组，从0开始编号\n    - 对于5个字符的块:\n        - 使用的网格由grid_index = (i // 5) % 8确定，其中i是块号。整数除法运算符//将左边的数除以右边的数，向下取整结果\n        - 对于当前块中的每个字符:\n            - 如果字符是\"J\"，不加密直接添加到加密块\n            - 否则，在当前网格中找到字符的位置。然后向右下方移动一个网格位置(row+1,col+1)(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为加密字母\n            - 将加密字母添加到加密块\n    - 处理完所有块后，连接加密块形成最终加密消息\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 大写字母字符串\n- 输出:\n    - 明文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(与加密相同)\n- 解密步骤:\n    - 将密文分成5个字符的块:\n        - 例如，如果密文是\"KHOORTQHQTHHSAUQ\"，第0块是\"KHOOR\"，第1块是\"TQHQH\"，以此类推(从0开始编号)\n    - 确定用于当前块的网格:\n        计算grid_index = (i // 5) % 8从网格列表中选择适当的网格。i是块号\n    - 对于块中的每个字符:\n        - 如果字符是\"J\": 直接将\"J\"添加到解密块，不进行解密\n        - 否则在网格中找到字符的位置，通过向左上方移动一格获得(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为解密字母\n        - 将解密字母添加到解密块\n    - 处理完块中所有字符后，将解密块添加到解密消息列表。形成最终解密消息\n密文: OFYHDD 解密为明文: nervii\n密文: OFYHDD 解密为明文: nervii\n密文: OWOMVEBGKDEDZS 解密为明文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 224, "source_filename": "icl_with_rule_decode_Kor_rule7_PhillipsFigureCipher_cn.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "OWOMVEBGKDEDZS", "extra_args": {}, "output": "需要解密的文本: OWOMVEBGKDEDZS\n\n处理第0个块: OWOMV\n使用第0号网格\n字母O在位置(3,0)，向左上移动一格得到N\n字母W在位置(4,1)，向左上移动一格得到O\n字母O在位置(3,0)，向左上移动一格得到N\n字母M在位置(2,3)，向左上移动一格得到C\n字母V在位置(4,0)，向左上移动一格得到U\n解密后的块: NONCU\n\n处理第1个块: EBGKD\n使用第1号网格\n字母E在位置(2,4)，向左上移动一格得到L\n字母B在位置(2,1)，向左上移动一格得到P\n字母G在位置(3,1)，向左上移动一格得到A\n字母K在位置(3,2)，向左上移动一格得到B\n字母D在位置(2,3)，向左上移动一格得到I\n解密后的块: LPABI\n\n处理第2个块: EDZS\n使用第2号网格\n字母E在位置(3,4)，向左上移动一格得到L\n字母D在位置(3,3)，向左上移动一格得到I\n字母Z在位置(1,4)，向左上移动一格得到T\n字母S在位置(2,4)，向左上移动一格得到Y\n解密后的块: LITY\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(Grid0-Grid7)\n- 加密步骤:\n    - 明文按5个字符分组，从0开始编号\n    - 对于5个字符的块:\n        - 使用的网格由grid_index = (i // 5) % 8确定，其中i是块号。整数除法运算符//将左边的数除以右边的数，向下取整结果\n        - 对于当前块中的每个字符:\n            - 如果字符是\"J\"，不加密直接添加到加密块\n            - 否则，在当前网格中找到字符的位置。然后向右下方移动一个网格位置(row+1,col+1)(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为加密字母\n            - 将加密字母添加到加密块\n    - 处理完所有块后，连接加密块形成最终加密消息\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 大写字母字符串\n- 输出:\n    - 明文: 大写字母字符串\n- 准备:\n    - 字母表 = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(不包含字母J)\n    - 8个网格(与加密相同)\n- 解密步骤:\n    - 将暗码分成5个字符的块:\n        - 例如，如果暗码是\"KHOORTQHQTHHSAUQ\"，第0块是\"KHOOR\"，第1块是\"TQHQH\"，以此类推(从0开始编号)\n    - 确定用于当前块的网格:\n        计算grid_index = (i // 5) % 8从网格列表中选择适当的网格。i是块号\n    - 对于块中的每个字符:\n        - 如果字符是\"J\": 直接将\"J\"添加到解密块，不进行解密\n        - 否则在网格中找到字符的位置，通过向左上方移动一格获得(如果越界则在对应边界的另一侧继续)，移动后位置的字母作为解密字母\n        - 将解密字母添加到解密块\n    - 处理完块中所有字符后，将解密块添加到解密消息列表。形成最终解密消息\n暗码: OFYHDD 解密为明文: nervii\n暗码: BSYDZDM 解密为明文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 225, "source_filename": "icl_with_rule_decode_Kor_rule7_PhillipsFigureCipher_cn.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "BSYDZDM", "extra_args": {}, "output": "需要解密的文本: BSYDZDM\n\n处理第0个块: BSYDZ\n使用第0号网格\n字母B在位置(1,1)，向左上移动一格得到P\n字母S在位置(0,4)，向左上移动一格得到Y\n字母Y在位置(4,3)，向左上移动一格得到R\n字母D在位置(1,3)，向左上移动一格得到I\n字母Z在位置(4,4)，向左上移动一格得到T\n解密后的块: PYRIT\n\n处理第1个块: DM\n使用第1号网格\n字母D在位置(2,3)，向左上移动一格得到I\n字母M在位置(3,3)，向左上移动一格得到C\n解密后的块: IC\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets\n        \n        Use the following 13 cipher alphabets, each associated with two letters:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n        \n    - Standard Alphabet\n        - ABCDEFGHIJKLMNOPQRSTUVWXYZ\n- Encryption Steps:\n    - Pair each letter in the key with each letter in the plaintext. If the key is shorter than the plaintext, repeat the key.\n    - For each given plaintext character p:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark the choice AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of p in the standard alphabet and replace it with the letter at the same position in the cipher alphabet.\n\n \nplain text: nonculpability extra parameter: {'key': 'bulgingly'} encode into cipher text: ALFZLSFVPVOQJC\nplain text: nonculpability extra parameter: {'key': 'bulgingly'} encode into cipher text: ALFZLSFVPVOQJC\nplain text: pyritic extra parameter: {'key': 'supercanonization'} encode into cipher text: LILTBUP\nplain text: nervii extra parameter: {'key': 'paraguayans'} encode into cipher text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 226, "source_filename": "icl_with_rule_encode_Kor_rule8_PortaCipher_en.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "NERVII", "extra_args": {"key": "paraguayans"}, "output": "开始加密文本: nervii\n使用密钥: paraguayans\n预处理后的文本: NERVII\n加密步骤:\n  第1个字符 N 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将明文字符 N 替换为密文字符 H\n  第2个字符 E 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将明文字符 E 替换为密文字符 R\n  第3个字符 R 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将明文字符 R 替换为密文字符 M\n  第4个字符 V 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将明文字符 V 替换为密文字符 I\n  第5个字符 I 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 I 替换为密文字符 S\n  第6个字符 I 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 I 替换为密文字符 Y\n最终加密结果: HRMISY\n", "ground_truth": "HRMISY"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets\n        \n        Use the following 13 cipher alphabets, each associated with two letters:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n        \n    - Standard Alphabet\n        - ABCDEFGHIJKLMNOPQRSTUVWXYZ\n- Encryption Steps:\n    - Pair each letter in the key with each letter in the plaintext. If the key is shorter than the plaintext, repeat the key.\n    - For each given plaintext character p:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark the choice AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of p in the standard alphabet and replace it with the letter at the same position in the cipher alphabet.\n\n \nplain text: nonculpability secret key: {'key': 'bulgingly'} encode into cipher text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 227, "source_filename": "icl_with_rule_encode_Kor_rule8_PortaCipher_en.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "NONCULPABILITY", "extra_args": {"key": "bulgingly"}, "output": "开始加密文本: nonculpability\n使用密钥: bulgingly\n预处理后的文本: NONCULPABILITY\n加密步骤:\n  第1个字符 N 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将明文字符 N 替换为密文字符 A\n  第2个字符 O 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 O 替换为密文字符 L\n  第3个字符 N 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将明文字符 N 替换为密文字符 F\n  第4个字符 C 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 C 替换为密文字符 Z\n  第5个字符 U 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将明文字符 U 替换为密文字符 L\n  第6个字符 L 使用密钥字符 N:\n    - 查找密钥字符 N 对应的替换表\n    - 将明文字符 L 替换为密文字符 S\n  第7个字符 P 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 P 替换为密文字符 F\n  第8个字符 A 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将明文字符 A 替换为密文字符 V\n  第9个字符 B 使用密钥字符 Y:\n    - 查找密钥字符 Y 对应的替换表\n    - 将明文字符 B 替换为密文字符 P\n  第10个字符 I 使用密钥字符 B:\n    - 查找密钥字符 B 对应的替换表\n    - 将明文字符 I 替换为密文字符 V\n  第11个字符 L 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 L 替换为密文字符 O\n  第12个字符 I 使用密钥字符 L:\n    - 查找密钥字符 L 对应的替换表\n    - 将明文字符 I 替换为密文字符 Q\n  第13个字符 T 使用密钥字符 G:\n    - 查找密钥字符 G 对应的替换表\n    - 将明文字符 T 替换为密文字符 J\n  第14个字符 Y 使用密钥字符 I:\n    - 查找密钥字符 I 对应的替换表\n    - 将明文字符 Y 替换为密文字符 C\n最终加密结果: ALFZLSFVPVOQJC\n", "ground_truth": "ALFZLSFVPVOQJC"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - Key: The key is a string used to select the alphabets. Each letter in the key will be used to select the corresponding alphabet for encryption or decryption.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Cipher Alphabets\n        \n        Use the following 13 cipher alphabets, each associated with two letters:\n        AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        CD: ZNOPQRSTUVWXYBCDEFGHIJKLMA\n        EF: YZNOPQRSTUVWXCDEFGHIJKLMAB\n        GH: XYZNOPQRSTUVWDEFGHIJKLMABC\n        IJ: WXYZNOPQRSTUVEFGHIJKLMABCD\n        KL: VWXYZNOPQRSTUFGHIJKLMABCDE\n        MN: UVWXYZNOPQRSTGHIJKLMABCDEF\n        OP: TUVWXYZNOPQRSHIJKLMABCDEFG\n        QR: STUVWXYZNOPQRIJKLMABCDEFGH\n        ST: RSTUVWXYZNOPQJKLMABCDEFGHI\n        UV: QRSTUVWXYZNOPKLMABCDEFGHIJ\n        WX: PQRSTUVWXYZNOLMABCDEFGHIJK\n        YZ: OPQRSTUVWXYZNMABCDEFGHIJKL\n        \n    - Standard Alphabet\n        - ABCDEFGHIJKLMNOPQRSTUVWXYZ\n- Encryption Steps:\n    - Pair each letter in the key with each letter in the plaintext. If the key is shorter than the plaintext, repeat the key.\n    - For each given plaintext character p:\n        - Find the corresponding cipher alphabet based on the key letter paired with it. If the key letter is A, mark the choice AB: NOPQRSTUVWXYZABCDEFGHIJKLM\n        - Find the position of p in the standard alphabet and replace it with the letter at the same position in the cipher alphabet.\n\n \noriginal information: nonculpability secret key: {'key': 'bulgingly'} encrypt into cipher text: ALFZLSFVPVOQJC\noriginal information: pyritic secret key: {'key': 'supercanonization'} encrypt into cipher text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 228, "source_filename": "icl_with_rule_encode_Kor_rule8_PortaCipher_en.jsonl", "cipher_name": "Kor_rule8_PortaCipher", "input": "PYRITIC", "extra_args": {"key": "supercanonization"}, "output": "开始加密文本: pyritic\n使用密钥: supercanonization\n预处理后的文本: PYRITIC\n加密步骤:\n  第1个字符 P 使用密钥字符 S:\n    - 查找密钥字符 S 对应的替换表\n    - 将明文字符 P 替换为密文字符 L\n  第2个字符 Y 使用密钥字符 U:\n    - 查找密钥字符 U 对应的替换表\n    - 将明文字符 Y 替换为密文字符 I\n  第3个字符 R 使用密钥字符 P:\n    - 查找密钥字符 P 对应的替换表\n    - 将明文字符 R 替换为密文字符 L\n  第4个字符 I 使用密钥字符 E:\n    - 查找密钥字符 E 对应的替换表\n    - 将明文字符 I 替换为密文字符 T\n  第5个字符 T 使用密钥字符 R:\n    - 查找密钥字符 R 对应的替换表\n    - 将明文字符 T 替换为密文字符 B\n  第6个字符 I 使用密钥字符 C:\n    - 查找密钥字符 C 对应的替换表\n    - 将明文字符 I 替换为密文字符 U\n  第7个字符 C 使用密钥字符 A:\n    - 查找密钥字符 A 对应的替换表\n    - 将明文字符 C 替换为密文字符 P\n最终加密结果: LILTBUP\n", "ground_truth": "LILTBUP"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Preparation:\n    - 3 Girds(All row and column numbers are counted from 0.):\n        - Grid 1 (3x9):\n            \n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n            \n        - Grid 2 (9x3):\n            \n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n            \n        - Grid 3 (3x3):\n            \n            1 2 3\n            4 5 6\n            7 8 9\n            \n- Encryption Steps:\n    - Remove all spaces and punctuation from the plain text and convert them to capital letters.\n    - Cut the plaintext into groups of 6 characters, and if a group is less than 6 characters long, fill it with `#`.\n    - Split each group of 6 characters into 3 binary groups.\n    - For each binary group `(L1, L2)` do the following:\n        - Determine the row and column numbers  of `L1` in `grid1` as `(l1_row, l1_col)`.\n        - Determine the row and column numbers of `L2` in `grid2` as `(l2_row, l2_col)`.\n        - Find the corresponding number `num3` in `grid3` from `l1_row` and `l2_col`.\n        - Output the triple `(l1_col, num3, l2_row)` of the three numbers.\n            - For example, for the binary (T,H), the row and column numbers of T in `grid1` are (0,4), the row and column numbers of H in `grid2` are (5,0).\n            - Based on `l1_row` (0) and `l2_col`(0) , the corresponding number `num3` found in `grid3` is 1, and the final ternary is (4, 1, 5).\n    - After performing the previous step multiple times and converting all the binary groups to triples, read them in turn to form a string of numbers that can be used as encrypted information. For example (4, 1, 5), (2, 3, 3), (8, 1, 8), the final encrypted message is 415233818.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 3 Girds (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into triples of three numbers.\n    - For each ternary `(x, y, z)` do the following:\n        - Find the row number of y in Gird3 where the value of L1_row is equal to the row number;\n        - Find the column number of y in Gird3, with the value of L2_col equal to that column number;\n        - The value of L1_col is equal to x, the value of L2_row is equal to z.\n        - Find the corresponding letter p1 in Gird1 according to the determined (L1_row,L1_col).\n        - Based on the determined (L2_row,L2_col), find the corresponding letter p2 in Gird2.\n        - p1p2 is the decrypted message of the triad.\n        - For example, for the ternary (4,1,5), in Gird3, the row number corresponding to 1 is 0, so L1_row=0, in Gird3, the column number corresponding to 1 is 0, so L2_col=0. L1_col=4, L2_row=5. According to (0,4) in Gird1, we find the corresponding letter T, and according to (5,0) in Gird2, we decrypt all the ternaries to get the letter p2 p1p2 as the message after decrypting the ternary. We find the corresponding letter H in Gird2 according to (5,0), so the final decryption is TH.\n    - String all the decrypted messages together, remove the `#` at the end (these characters are padding characters added to make the message length a multiple of 6), to form the decrypted plaintext.\ncipher text: 690327722 decrypt to plain text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 229, "source_filename": "icl_with_rule_decode_Kor_rule14_DigrafidCipher_en.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "690327722", "extra_args": {}, "output": "\n开始解密过程:\n加密文本: 690327722\n分割成三元组: ['690', '327', '722']\n\n解密三元组 (6, 9, 0):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: N (使用行=2, 列=6)\n- 在grid2中找到字符: E (使用行=0, 列=2)\n- 解密结果: NE\n\n解密三元组 (3, 2, 7):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: R (使用行=0, 列=3)\n- 在grid2中找到字符: V (使用行=7, 列=1)\n- 解密结果: RV\n\n解密三元组 (7, 2, 2):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: I (使用行=0, 列=7)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: II\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Preparation:\n    - 3 Girds(All row and column numbers are counted from 0.):\n        - Grid 1 (3x9):\n            \n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n            \n        - Grid 2 (9x3):\n            \n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n            \n        - Grid 3 (3x3):\n            \n            1 2 3\n            4 5 6\n            7 8 9\n            \n- Encryption Steps:\n    - Remove all spaces and punctuation from the plain text and convert them to capital letters.\n    - Cut the plaintext into groups of 6 characters, and if a group is less than 6 characters long, fill it with `#`.\n    - Split each group of 6 characters into 3 binary groups.\n    - For each binary group `(L1, L2)` do the following:\n        - Determine the row and column numbers  of `L1` in `grid1` as `(l1_row, l1_col)`.\n        - Determine the row and column numbers of `L2` in `grid2` as `(l2_row, l2_col)`.\n        - Find the corresponding number `num3` in `grid3` from `l1_row` and `l2_col`.\n        - Output the triple `(l1_col, num3, l2_row)` of the three numbers.\n            - For example, for the binary (T,H), the row and column numbers of T in `grid1` are (0,4), the row and column numbers of H in `grid2` are (5,0).\n            - Based on `l1_row` (0) and `l2_col`(0) , the corresponding number `num3` found in `grid3` is 1, and the final ternary is (4, 1, 5).\n    - After performing the previous step multiple times and converting all the binary groups to triples, read them in turn to form a string of numbers that can be used as encrypted information. For example (4, 1, 5), (2, 3, 3), (8, 1, 8), the final encrypted message is 415233818.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 3 Girds (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into triples of three numbers.\n    - For each ternary `(x, y, z)` do the following:\n        - Find the row number of y in Gird3 where the value of L1_row is equal to the row number;\n        - Find the column number of y in Gird3, with the value of L2_col equal to that column number;\n        - The value of L1_col is equal to x, the value of L2_row is equal to z.\n        - Find the corresponding letter p1 in Gird1 according to the determined (L1_row,L1_col).\n        - Based on the determined (L2_row,L2_col), find the corresponding letter p2 in Gird2.\n        - p1p2 is the decrypted message of the triad.\n        - For example, for the ternary (4,1,5), in Gird3, the row number corresponding to 1 is 0, so L1_row=0, in Gird3, the column number corresponding to 1 is 0, so L2_col=0. L1_col=4, L2_row=5. According to (0,4) in Gird1, we find the corresponding letter T, and according to (5,0) in Gird2, we decrypt all the ternaries to get the letter p2 p1p2 as the message after decrypting the ternary. We find the corresponding letter H in Gird2 according to (5,0), so the final decryption is TH.\n    - String all the decrypted messages together, remove the `#` at the end (these characters are padding characters added to make the message length a multiple of 6), to form the decrypted plaintext.\ncipher text: 061322422398898898 decrypt into clear text: pyritic\ncipher text: 061322422398898898 decrypt into clear text: pyritic\ncipher text: 690327722 decrypt into clear text: nervii\ncipher text: 692677616053582082431898898 decrypt into clear text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 230, "source_filename": "icl_with_rule_decode_Kor_rule14_DigrafidCipher_en.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "692677616053582082431898898", "extra_args": {}, "output": "\n开始解密过程:\n加密文本: 692677616053582082431898898\n分割成三元组: ['692', '677', '616', '053', '582', '082', '431', '898', '898']\n\n解密三元组 (6, 9, 2):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: N (使用行=2, 列=6)\n- 在grid2中找到字符: O (使用行=2, 列=2)\n- 解密结果: NO\n\n解密三元组 (6, 7, 7):\n- 在grid3中找到7的位置: 行=2, 列=0\n- 在grid1中找到字符: N (使用行=2, 列=6)\n- 在grid2中找到字符: C (使用行=7, 列=0)\n- 解密结果: NC\n\n解密三元组 (6, 1, 6):\n- 在grid3中找到1的位置: 行=0, 列=0\n- 在grid1中找到字符: U (使用行=0, 列=6)\n- 在grid2中找到字符: L (使用行=6, 列=0)\n- 解密结果: UL\n\n解密三元组 (0, 5, 3):\n- 在grid3中找到5的位置: 行=1, 列=1\n- 在grid1中找到字符: P (使用行=1, 列=0)\n- 在grid2中找到字符: A (使用行=3, 列=1)\n- 解密结果: PA\n\n解密三元组 (5, 8, 2):\n- 在grid3中找到8的位置: 行=2, 列=1\n- 在grid1中找到字符: B (使用行=2, 列=5)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: BI\n\n解密三元组 (0, 8, 2):\n- 在grid3中找到8的位置: 行=2, 列=1\n- 在grid1中找到字符: L (使用行=2, 列=0)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: LI\n\n解密三元组 (4, 3, 1):\n- 在grid3中找到3的位置: 行=0, 列=2\n- 在grid1中找到字符: T (使用行=0, 列=4)\n- 在grid2中找到字符: Y (使用行=1, 列=2)\n- 解密结果: TY\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Preparation:\n    - 3 Girds(All row and column numbers are counted from 0.):\n        - Grid 1 (3x9):\n            \n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n            \n        - Grid 2 (9x3):\n            \n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n            \n        - Grid 3 (3x3):\n            \n            1 2 3\n            4 5 6\n            7 8 9\n            \n- Encryption Steps:\n    - Remove all spaces and punctuation from the plain text and convert them to capital letters.\n    - Cut the plaintext into groups of 6 characters, and if a group is less than 6 characters long, fill it with `#`.\n    - Split each group of 6 characters into 3 binary groups.\n    - For each binary group `(L1, L2)` do the following:\n        - Determine the row and column numbers  of `L1` in `grid1` as `(l1_row, l1_col)`.\n        - Determine the row and column numbers of `L2` in `grid2` as `(l2_row, l2_col)`.\n        - Find the corresponding number `num3` in `grid3` from `l1_row` and `l2_col`.\n        - Output the triple `(l1_col, num3, l2_row)` of the three numbers.\n            - For example, for the binary (T,H), the row and column numbers of T in `grid1` are (0,4), the row and column numbers of H in `grid2` are (5,0).\n            - Based on `l1_row` (0) and `l2_col`(0) , the corresponding number `num3` found in `grid3` is 1, and the final ternary is (4, 1, 5).\n    - After performing the previous step multiple times and converting all the binary groups to triples, read them in turn to form a string of numbers that can be used as encrypted information. For example (4, 1, 5), (2, 3, 3), (8, 1, 8), the final encrypted message is 415233818.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 3 Girds (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Divide the ciphertext into triples of three numbers.\n    - For each ternary `(x, y, z)` do the following:\n        - Find the row number of y in Gird3 where the value of L1_row is equal to the row number;\n        - Find the column number of y in Gird3, with the value of L2_col equal to that column number;\n        - The value of L1_col is equal to x, the value of L2_row is equal to z.\n        - Find the corresponding letter p1 in Gird1 according to the determined (L1_row,L1_col).\n        - Based on the determined (L2_row,L2_col), find the corresponding letter p2 in Gird2.\n        - p1p2 is the decrypted message of the triad.\n        - For example, for the ternary (4,1,5), in Gird3, the row number corresponding to 1 is 0, so L1_row=0, in Gird3, the column number corresponding to 1 is 0, so L2_col=0. L1_col=4, L2_row=5. According to (0,4) in Gird1, we find the corresponding letter T, and according to (5,0) in Gird2, we decrypt all the ternaries to get the letter p2 p1p2 as the message after decrypting the ternary. We find the corresponding letter H in Gird2 according to (5,0), so the final decryption is TH.\n    - String all the decrypted messages together, remove the `#` at the end (these characters are padding characters added to make the message length a multiple of 6), to form the decrypted plaintext.\ncipher text: 692677616053582082431898898 decrypt to clear text: nonculpability\ncipher text: 692677616053582082431898898 decrypt to clear text: nonculpability\ncipher text: 690327722 decrypt to clear text: nervii\ncipher text: 061322422398898898 decrypt to clear text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 231, "source_filename": "icl_with_rule_decode_Kor_rule14_DigrafidCipher_en.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "061322422398898898", "extra_args": {}, "output": "\n开始解密过程:\n加密文本: 061322422398898898\n分割成三元组: ['061', '322', '422', '398', '898', '898']\n\n解密三元组 (0, 6, 1):\n- 在grid3中找到6的位置: 行=1, 列=2\n- 在grid1中找到字符: P (使用行=1, 列=0)\n- 在grid2中找到字符: Y (使用行=1, 列=2)\n- 解密结果: PY\n\n解密三元组 (3, 2, 2):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: R (使用行=0, 列=3)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: RI\n\n解密三元组 (4, 2, 2):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: T (使用行=0, 列=4)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: TI\n\n解密三元组 (3, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: C (使用行=2, 列=3)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: C#\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\ncipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] decrypt to clear text: nonculpability\ncipher text: [112, 121, 114, 105, 116, 105, 99] decrypt to clear text: pyritic\ncipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] decrypt to clear text: nonculpability\ncipher text: [112, 121, 114, 105, 116, 105, 99] decrypt to clear text: pyritic\ncipher text: [112, 121, 114, 105, 116, 105, 99] decrypt to clear text: pyritic\ncipher text: [110, 101, 114, 118, 105, 105] decrypt to clear text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 232, "source_filename": "icl_decode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": [110, 101, 114, 118, 105, 105], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 101, Character: e\nASCII   Code: 114, Character: r\nASCII   Code: 118, Character: v\nASCII   Code: 105, Character: i\nASCII   Code: 105, Character: i\nResult: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\nencoded text: [110, 101, 114, 118, 105, 105] decode into original information: nervii\nencoded text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] decode into original information: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 233, "source_filename": "icl_decode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 111, Character: o\nASCII   Code: 110, Character: n\nASCII   Code: 99, Character: c\nASCII   Code: 117, Character: u\nASCII   Code: 108, Character: l\nASCII   Code: 112, Character: p\nASCII   Code: 97, Character: a\nASCII   Code: 98, Character: b\nASCII   Code: 105, Character: i\nASCII   Code: 108, Character: l\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 121, Character: y\nResult: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\nencrypted text: [110, 101, 114, 118, 105, 105] decode into plain text: nervii\nencrypted text: [110, 101, 114, 118, 105, 105] decode into plain text: nervii\nencrypted text: [110, 101, 114, 118, 105, 105] decode into plain text: nervii\nencrypted text: [112, 121, 114, 105, 116, 105, 99] decode into plain text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 234, "source_filename": "icl_decode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": [112, 121, 114, 105, 116, 105, 99], "extra_args": {}, "output": "ASCII   Code: 112, Character: p\nASCII   Code: 121, Character: y\nASCII   Code: 114, Character: r\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 105, Character: i\nASCII   Code: 99, Character: c\nResult: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗码: 数字字符串，不含标点和空格\n- 准备:\n    - 3个网格(所有行列号从0开始计数):\n        - 网格1 (3x9):\n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n        - 网格2 (9x3):\n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n        - 网格3 (3x3):\n            1 2 3\n            4 5 6\n            7 8 9\n- 加密步骤:\n    - 移除所有空格和标点，将文本转换为大写字母\n    - 将非加密信息切分为6个字符一组，如果最后一组不足6个字符，用#填充\n    - 将每组6个字符分成3个二元组\n    - 对每个二元组(L1, L2)执行以下操作:\n        - 确定L1在网格1中的行列号(l1_row, l1_col)\n        - 确定L2在网格2中的行列号(l2_row, l2_col)\n        - 在网格3中用l1_row和l2_col找到对应数字num3\n        - 输出三元组(l1_col, num3, l2_row)\n    - 将所有三元组连接成一个数字串作为加密信息\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 数字字符串，不含标点和空格\n- 输出:\n    - 非加密信息: 大写字母字符串，不含标点和空格\n- 准备:\n    - 3个网格(与加密相同)\n- 解密步骤:\n    - 将暗码分成三个数字一组\n    - 对每个三元组(x, y, z)执行以下操作:\n        - 在网格3中找到y的行号作为L1_row\n        - 在网格3中找到y的列号作为L2_col\n        - L1_col等于x，L2_row等于z\n        - 根据确定的(L1_row,L1_col)在网格1中找到对应字母p1\n        - 根据确定的(L2_row,L2_col)在网格2中找到对应字母p2\n        - p1p2为该三元组解密后的消息\n    - 将所有解密后的消息连接起来，移除末尾的#(这些字符是为使消息长度为6的倍数而添加的填充字符)，形成解密后的非加密信息\n暗码: 690327722 解密为非加密信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 235, "source_filename": "icl_with_rule_decode_Kor_rule14_DigrafidCipher_cn.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "690327722", "extra_args": {}, "output": "\n开始解密过程:\n加密文本: 690327722\n分割成三元组: ['690', '327', '722']\n\n解密三元组 (6, 9, 0):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: N (使用行=2, 列=6)\n- 在grid2中找到字符: E (使用行=0, 列=2)\n- 解密结果: NE\n\n解密三元组 (3, 2, 7):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: R (使用行=0, 列=3)\n- 在grid2中找到字符: V (使用行=7, 列=1)\n- 解密结果: RV\n\n解密三元组 (7, 2, 2):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: I (使用行=0, 列=7)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: II\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 密文: 数字字符串，不含标点和空格\n- 准备:\n    - 3个网格(所有行列号从0开始计数):\n        - 网格1 (3x9):\n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n        - 网格2 (9x3):\n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n        - 网格3 (3x3):\n            1 2 3\n            4 5 6\n            7 8 9\n- 加密步骤:\n    - 移除所有空格和标点，将文本转换为大写字母\n    - 将初始文本切分为6个字符一组，如果最后一组不足6个字符，用#填充\n    - 将每组6个字符分成3个二元组\n    - 对每个二元组(L1, L2)执行以下操作:\n        - 确定L1在网格1中的行列号(l1_row, l1_col)\n        - 确定L2在网格2中的行列号(l2_row, l2_col)\n        - 在网格3中用l1_row和l2_col找到对应数字num3\n        - 输出三元组(l1_col, num3, l2_row)\n    - 将所有三元组连接成一个数字串作为加密信息\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 数字字符串，不含标点和空格\n- 输出:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 准备:\n    - 3个网格(与加密相同)\n- 解密步骤:\n    - 将密文分成三个数字一组\n    - 对每个三元组(x, y, z)执行以下操作:\n        - 在网格3中找到y的行号作为L1_row\n        - 在网格3中找到y的列号作为L2_col\n        - L1_col等于x，L2_row等于z\n        - 根据确定的(L1_row,L1_col)在网格1中找到对应字母p1\n        - 根据确定的(L2_row,L2_col)在网格2中找到对应字母p2\n        - p1p2为该三元组解密后的消息\n    - 将所有解密后的消息连接起来，移除末尾的#(这些字符是为使消息长度为6的倍数而添加的填充字符)，形成解密后的初始文本\n密文: 061322422398898898 解密为初始文本: pyritic\n密文: 061322422398898898 解密为初始文本: pyritic\n密文: 690327722 解密为初始文本: nervii\n密文: 692677616053582082431898898 解密为初始文本是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 236, "source_filename": "icl_with_rule_decode_Kor_rule14_DigrafidCipher_cn.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "692677616053582082431898898", "extra_args": {}, "output": "\n开始解密过程:\n加密文本: 692677616053582082431898898\n分割成三元组: ['692', '677', '616', '053', '582', '082', '431', '898', '898']\n\n解密三元组 (6, 9, 2):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: N (使用行=2, 列=6)\n- 在grid2中找到字符: O (使用行=2, 列=2)\n- 解密结果: NO\n\n解密三元组 (6, 7, 7):\n- 在grid3中找到7的位置: 行=2, 列=0\n- 在grid1中找到字符: N (使用行=2, 列=6)\n- 在grid2中找到字符: C (使用行=7, 列=0)\n- 解密结果: NC\n\n解密三元组 (6, 1, 6):\n- 在grid3中找到1的位置: 行=0, 列=0\n- 在grid1中找到字符: U (使用行=0, 列=6)\n- 在grid2中找到字符: L (使用行=6, 列=0)\n- 解密结果: UL\n\n解密三元组 (0, 5, 3):\n- 在grid3中找到5的位置: 行=1, 列=1\n- 在grid1中找到字符: P (使用行=1, 列=0)\n- 在grid2中找到字符: A (使用行=3, 列=1)\n- 解密结果: PA\n\n解密三元组 (5, 8, 2):\n- 在grid3中找到8的位置: 行=2, 列=1\n- 在grid1中找到字符: B (使用行=2, 列=5)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: BI\n\n解密三元组 (0, 8, 2):\n- 在grid3中找到8的位置: 行=2, 列=1\n- 在grid1中找到字符: L (使用行=2, 列=0)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: LI\n\n解密三元组 (4, 3, 1):\n- 在grid3中找到3的位置: 行=0, 列=2\n- 在grid1中找到字符: T (使用行=0, 列=4)\n- 在grid2中找到字符: Y (使用行=1, 列=2)\n- 解密结果: TY\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密信息: 数字字符串，不含标点和空格\n- 准备:\n    - 3个网格(所有行列号从0开始计数):\n        - 网格1 (3x9):\n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n        - 网格2 (9x3):\n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n        - 网格3 (3x3):\n            1 2 3\n            4 5 6\n            7 8 9\n- 加密步骤:\n    - 移除所有空格和标点，将文本转换为大写字母\n    - 将初始文本切分为6个字符一组，如果最后一组不足6个字符，用#填充\n    - 将每组6个字符分成3个二元组\n    - 对每个二元组(L1, L2)执行以下操作:\n        - 确定L1在网格1中的行列号(l1_row, l1_col)\n        - 确定L2在网格2中的行列号(l2_row, l2_col)\n        - 在网格3中用l1_row和l2_col找到对应数字num3\n        - 输出三元组(l1_col, num3, l2_row)\n    - 将所有三元组连接成一个数字串作为加密信息\n解密算法:\n解密规则:\n- 输入:\n    - 加密信息: 数字字符串，不含标点和空格\n- 输出:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 准备:\n    - 3个网格(与加密相同)\n- 解密步骤:\n    - 将加密信息分成三个数字一组\n    - 对每个三元组(x, y, z)执行以下操作:\n        - 在网格3中找到y的行号作为L1_row\n        - 在网格3中找到y的列号作为L2_col\n        - L1_col等于x，L2_row等于z\n        - 根据确定的(L1_row,L1_col)在网格1中找到对应字母p1\n        - 根据确定的(L2_row,L2_col)在网格2中找到对应字母p2\n        - p1p2为该三元组解密后的消息\n    - 将所有解密后的消息连接起来，移除末尾的#(这些字符是为使消息长度为6的倍数而添加的填充字符)，形成解密后的初始文本\n加密信息: 692677616053582082431898898 解密成为初始文本: nonculpability\n加密信息: 692677616053582082431898898 解密成为初始文本: nonculpability\n加密信息: 690327722 解密成为初始文本: nervii\n加密信息: 061322422398898898 解密成为初始文本: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 237, "source_filename": "icl_with_rule_decode_Kor_rule14_DigrafidCipher_cn.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "061322422398898898", "extra_args": {}, "output": "\n开始解密过程:\n加密文本: 061322422398898898\n分割成三元组: ['061', '322', '422', '398', '898', '898']\n\n解密三元组 (0, 6, 1):\n- 在grid3中找到6的位置: 行=1, 列=2\n- 在grid1中找到字符: P (使用行=1, 列=0)\n- 在grid2中找到字符: Y (使用行=1, 列=2)\n- 解密结果: PY\n\n解密三元组 (3, 2, 2):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: R (使用行=0, 列=3)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: RI\n\n解密三元组 (4, 2, 2):\n- 在grid3中找到2的位置: 行=0, 列=1\n- 在grid1中找到字符: T (使用行=0, 列=4)\n- 在grid2中找到字符: I (使用行=2, 列=1)\n- 解密结果: TI\n\n解密三元组 (3, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: C (使用行=2, 列=3)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: C#\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n解密三元组 (8, 9, 8):\n- 在grid3中找到9的位置: 行=2, 列=2\n- 在grid1中找到字符: # (使用行=2, 列=8)\n- 在grid2中找到字符: # (使用行=8, 列=2)\n- 解密结果: ##\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n暗文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密成为初始文本: nonculpability\n暗文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密成为初始文本: pyritic\n暗文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密成为初始文本: pyritic\n暗文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密成为初始文本: pyritic\n暗文: ... ...  . .....  .... ..  ..... .  .. ....  .. .... 解密成为初始文本: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 238, "source_filename": "icl_decode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "... ...  . .....  .... ..  ..... .  .. ....  .. ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '. .....'，它表示第 1 行第 5 列，对应的字母是 'E'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '..... .'，它表示第 5 行第 1 列，对应的字母是 'V'\n步骤 5：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n最终步骤：解码完成，解码后的消息是：NERVII \n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n密文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密为明文: pyritic\n密文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密为明文: pyritic\n密文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密为明文: pyritic\n密文: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密为明文: pyritic\n密文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密为明文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 239, "source_filename": "icl_decode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '... ....'，它表示第 3 行第 4 列，对应的字母是 'O'\n步骤 3：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 4：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n步骤 5：正在解码 '.... .....'，它表示第 4 行第 5 列，对应的字母是 'U'\n步骤 6：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 7：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 8：正在解码 '. .'，它表示第 1 行第 1 列，对应的字母是 'A'\n步骤 9：正在解码 '. ..'，它表示第 1 行第 2 列，对应的字母是 'B'\n步骤 10：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 11：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 12：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 13：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 14：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n最终步骤：解码完成，解码后的消息是：NONCULPABILITY \n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n暗码: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密成为原文: nonculpability\n暗码: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密成为原文: nonculpability\n暗码: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密成为原文: nonculpability\n暗码: ... ...  . .....  .... ..  ..... .  .. ....  .. .... 解密成为原文: nervii\n暗码: ... ...  . .....  .... ..  ..... .  .. ....  .. .... 解密成为原文: nervii\n暗码: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密成为原文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 240, "source_filename": "icl_decode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...", "extra_args": {}, "output": "步骤 1：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 2：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 5：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 7：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n最终步骤：解码完成，解码后的消息是：PYRITIC \n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nPlease encrypt the plaintext using the encryption algorithm.\n\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Convert the plaintext to uppercase, retain only the letter characters, and split them into pairs of two letters.\n3. Encrypt each pair of letters:\n   - Find the position of each letter in the default matrix, swap their y-coordinates,\n   - Retrieve the encrypted letter pair from the M1 and M2 matrices based on the new coordinates.\n4. Output the encrypted text.\noriginal information: pyritic secret key: {'str1': 'sawbill', 'str2': 'houri'} encrypt into cipher text: RV RB RD WX\noriginal information: pyritic secret key: {'str1': 'sawbill', 'str2': 'houri'} encrypt into cipher text: RV RB RD WX\noriginal information: pyritic secret key: {'str1': 'sawbill', 'str2': 'houri'} encrypt into cipher text: RV RB RD WX\noriginal information: nervii secret key: {'str1': 'parametral', 'str2': 'quiverful'} encrypt into cipher text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 241, "source_filename": "icl_with_rule_encode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "NERVII", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将明文分成两个字母一组: ['NE', 'RV', 'II']\n\nStep 4.1: 加密字母对 NE\n  - N 在 M_T 中的位置: (2, 3)\n  - E 在 M_T 中的位置: (0, 4)\n  - 交换列索引后新位置: (2, 4) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (0, 3) 的字母: V\n  - 加密结果为: JV\n\nStep 4.2: 加密字母对 RV\n  - R 在 M_T 中的位置: (3, 1)\n  - V 在 M_T 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 1)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (4, 1) 的字母: T\n  - 加密结果为: KT\n\nStep 4.3: 加密字母对 II\n  - I 在 M_T 中的位置: (1, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (1, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 3) 的字母: A\n  - 加密结果为: CA\n\n加密完成！最终加密结果为: JV KT CA\n", "ground_truth": "JV KT CA"}, "data_source": "Cipher"}
{"prompt": "Please encrypt the plaintext using the encryption algorithm.\n\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Convert the plaintext to uppercase, retain only the letter characters, and split them into pairs of two letters.\n3. Encrypt each pair of letters:\n   - Find the position of each letter in the default matrix, swap their y-coordinates,\n   - Retrieve the encrypted letter pair from the M1 and M2 matrices based on the new coordinates.\n4. Output the encrypted text.\nplain text: pyritic secret key: {'str1': 'sawbill', 'str2': 'houri'} encode to cipher text: RV RB RD WX\nplain text: nervii secret key: {'str1': 'parametral', 'str2': 'quiverful'} encode to cipher text: JV KT CA\nplain text: pyritic secret key: {'str1': 'sawbill', 'str2': 'houri'} encode to cipher text: RV RB RD WX\nplain text: nonculpability secret key: {'str1': 'trichophytia', 'str2': 'sphinges'} encode to cipher text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 242, "source_filename": "icl_with_rule_encode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "NONCULPABILITY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将明文分成两个字母一组: ['NO', 'NC', 'UL', 'PA', 'BI', 'LI', 'TY']\n\nStep 4.1: 加密字母对 NO\n  - N 在 M_T 中的位置: (2, 3)\n  - O 在 M_T 中的位置: (2, 4)\n  - 交换列索引后新位置: (2, 4) 和 (2, 3)\n  - 在 M1 中查找位置 (2, 4) 的字母: J\n  - 在 M2 中查找位置 (2, 3) 的字母: K\n  - 加密结果为: JK\n\nStep 4.2: 加密字母对 NC\n  - N 在 M_T 中的位置: (2, 3)\n  - C 在 M_T 中的位置: (0, 2)\n  - 交换列索引后新位置: (2, 2) 和 (0, 3)\n  - 在 M1 中查找位置 (2, 2) 的字母: F\n  - 在 M2 中查找位置 (0, 3) 的字母: I\n  - 加密结果为: FI\n\nStep 4.3: 加密字母对 UL\n  - U 在 M_T 中的位置: (3, 4)\n  - L 在 M_T 中的位置: (2, 1)\n  - 交换列索引后新位置: (3, 1) 和 (2, 4)\n  - 在 M1 中查找位置 (3, 1) 的字母: L\n  - 在 M2 中查找位置 (2, 4) 的字母: L\n  - 加密结果为: LL\n\nStep 4.4: 加密字母对 PA\n  - P 在 M_T 中的位置: (3, 0)\n  - A 在 M_T 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M1 中查找位置 (3, 0) 的字母: K\n  - 在 M2 中查找位置 (0, 0) 的字母: S\n  - 加密结果为: KS\n\nStep 4.5: 加密字母对 BI\n  - B 在 M_T 中的位置: (0, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (0, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (0, 3) 的字母: C\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: CE\n\nStep 4.6: 加密字母对 LI\n  - L 在 M_T 中的位置: (2, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (2, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (2, 3) 的字母: G\n  - 在 M2 中查找位置 (1, 1) 的字母: E\n  - 加密结果为: GE\n\nStep 4.7: 加密字母对 TY\n  - T 在 M_T 中的位置: (3, 3)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: N\n  - 在 M2 中查找位置 (4, 3) 的字母: Y\n  - 加密结果为: NY\n\n加密完成！最终加密结果为: JK FI LL KS CE GE NY\n", "ground_truth": "JK FI LL KS CE GE NY"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\nPlease encrypt the plaintext using the encryption algorithm.\n\nSolution:\n1. Construct three matrices: the default alphabet matrix M_T, and the encryption matrices M1 and M2 (generated from str1 and str2).\n2. Convert the plaintext to uppercase, retain only the letter characters, and split them into pairs of two letters.\n3. Encrypt each pair of letters:\n   - Find the position of each letter in the default matrix, swap their y-coordinates,\n   - Retrieve the encrypted letter pair from the M1 and M2 matrices based on the new coordinates.\n4. Output the encrypted text.\nclear text: pyritic extra parameter: {'str1': 'sawbill', 'str2': 'houri'} encrypt to encrypted text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 243, "source_filename": "icl_with_rule_encode_FourSquareCipher_en.jsonl", "cipher_name": "FourSquareCipher", "input": "PYRITICX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将明文文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\n明文字符个数为奇数，补一个字符 X\nStep 3: 将明文分成两个字母一组: ['PY', 'RI', 'TI', 'CX']\n\nStep 4.1: 加密字母对 PY\n  - P 在 M_T 中的位置: (3, 0)\n  - Y 在 M_T 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 0)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (4, 0) 的字母: V\n  - 加密结果为: RV\n\nStep 4.2: 加密字母对 RI\n  - R 在 M_T 中的位置: (3, 1)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 1)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 1) 的字母: B\n  - 加密结果为: RB\n\nStep 4.3: 加密字母对 TI\n  - T 在 M_T 中的位置: (3, 3)\n  - I 在 M_T 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M1 中查找位置 (3, 3) 的字母: R\n  - 在 M2 中查找位置 (1, 3) 的字母: D\n  - 加密结果为: RD\n\nStep 4.4: 加密字母对 CX\n  - C 在 M_T 中的位置: (0, 2)\n  - X 在 M_T 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M1 中查找位置 (0, 2) 的字母: W\n  - 在 M2 中查找位置 (4, 2) 的字母: X\n  - 加密结果为: WX\n\n加密完成！最终加密结果为: RV RB RD WX\n", "ground_truth": "RV RB RD WX"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对非加密信息进行加密\n\n加密规则:\n- 输入:\n    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格。\n- 输出:\n    - 加密文本: 不含标点的字符串。\n- 准备:\n    - 多击编码表\n        | 字母 | 多击编码 |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- 加密步骤:\n    - 对于每个非加密信息字符p:\n        - 如果p是大写字母且存在于多击编码表中:\n            - 用多击编码表中对应的多击编码替换p。\n\n非加密信息: pyritic 加密成为加密文本: 7^19^37^34^38^14^32^3\n非加密信息: nervii 加密成为加密文本是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 244, "source_filename": "icl_with_rule_encode_Kor_rule3_MultiTapPhoneCode_cn.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "NERVII", "extra_args": {}, "output": "原始文本: nervii\n处理后的文本(仅大写字母): NERVII\n开始逐字符加密:\n字符 N 对应的多击编码是: 6^2\n字符 E 对应的多击编码是: 3^2\n字符 R 对应的多击编码是: 7^3\n字符 V 对应的多击编码是: 8^3\n字符 I 对应的多击编码是: 4^3\n字符 I 对应的多击编码是: 4^3\n最终加密结果: 6^23^27^38^34^34^3\n", "ground_truth": "6^23^27^38^34^34^3"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对初始文本进行加密\n\n加密规则:\n- 输入:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格。\n- 输出:\n    - 隐文: 不含标点的字符串。\n- 准备:\n    - 多击编码表\n        | 字母 | 多击编码 |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- 加密步骤:\n    - 对于每个初始文本字符p:\n        - 如果p是大写字母且存在于多击编码表中:\n            - 用多击编码表中对应的多击编码替换p。\n\n初始文本: pyritic 加密成隐文: 7^19^37^34^38^14^32^3\n初始文本: nonculpability 加密成隐文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 245, "source_filename": "icl_with_rule_encode_Kor_rule3_MultiTapPhoneCode_cn.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "NONCULPABILITY", "extra_args": {}, "output": "原始文本: nonculpability\n处理后的文本(仅大写字母): NONCULPABILITY\n开始逐字符加密:\n字符 N 对应的多击编码是: 6^2\n字符 O 对应的多击编码是: 6^3\n字符 N 对应的多击编码是: 6^2\n字符 C 对应的多击编码是: 2^3\n字符 U 对应的多击编码是: 8^2\n字符 L 对应的多击编码是: 5^3\n字符 P 对应的多击编码是: 7^1\n字符 A 对应的多击编码是: 2^1\n字符 B 对应的多击编码是: 2^2\n字符 I 对应的多击编码是: 4^3\n字符 L 对应的多击编码是: 5^3\n字符 I 对应的多击编码是: 4^3\n字符 T 对应的多击编码是: 8^1\n字符 Y 对应的多击编码是: 9^3\n最终加密结果: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3\n", "ground_truth": "6^26^36^22^38^25^37^12^12^24^35^34^38^19^3"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对非加密信息进行加密\n\n加密规则:\n- 输入:\n    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格。\n- 输出:\n    - 加密信息: 不含标点的字符串。\n- 准备:\n    - 多击编码表\n        | 字母 | 多击编码 |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- 加密步骤:\n    - 对于每个非加密信息字符p:\n        - 如果p是大写字母且存在于多击编码表中:\n            - 用多击编码表中对应的多击编码替换p。\n\n非加密信息: nervii 加密为加密信息: 6^23^27^38^34^34^3\n非加密信息: nervii 加密为加密信息: 6^23^27^38^34^34^3\n非加密信息: nonculpability 加密为加密信息: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3\n非加密信息: pyritic 加密为加密信息是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 246, "source_filename": "icl_with_rule_encode_Kor_rule3_MultiTapPhoneCode_cn.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "PYRITIC", "extra_args": {}, "output": "原始文本: pyritic\n处理后的文本(仅大写字母): PYRITIC\n开始逐字符加密:\n字符 P 对应的多击编码是: 7^1\n字符 Y 对应的多击编码是: 9^3\n字符 R 对应的多击编码是: 7^3\n字符 I 对应的多击编码是: 4^3\n字符 T 对应的多击编码是: 8^1\n字符 I 对应的多击编码是: 4^3\n字符 C 对应的多击编码是: 2^3\n最终加密结果: 7^19^37^34^38^14^32^3\n", "ground_truth": "7^19^37^34^38^14^32^3"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird(All row and column numbers are counted from 0.):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - Located at the first letter of all rows, MKGWB is the row header letter.\n        - Located at the last letter of all columns, BTEHI is the bottom column letter.\n- Encryption Steps:\n    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.\n    - For each letter p in the plaintext.\n        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.\n        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.\n        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.\n    \n    Concatenate all the encrypted messages for the final ciphertext output.\n    \n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Take two letters c1,c2 at a time from the ciphertext.\n        - The letter c1 identifies the row position of the decrypted letter p in the grid, find the row where c1 is the row header.\n        - The letter c2 identifies the column position of the decrypted letter p in the grid, and finds the row with c2 at the bottom of the column.\n        - Find the letter at this row and column position in the grid, which is the c1,c2 decrypted message p.\n        - For example, if c1,c2=BH, B is the header of the row is the 4th row, H is the bottom of the row is the 2nd column, (4,2) at the letter H, so BH decrypted to H.\n    - Decrypt the letters in the ciphertext two by two, and output the connected decrypted message as the final plaintext.\ncipher text: KTGEKTWIGIKHMIGTBBBIKHBIBTWE decode into original information: nonculpability\ncipher text: MIWEWTBIBTBIWI decode into original information: pyritic\ncipher text: MIWEWTBIBTBIWI decode into original information: pyritic\ncipher text: KTBEWTWHBIBI decode into original information is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 247, "source_filename": "icl_with_rule_decode_Kor_rule15_CollonCipher_en.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "KTBEWTWHBIBI", "extra_args": {}, "output": "开始解密过程...\n加密文本: KTBEWTWHBIBI\n\n每次解密两个字符:\n\n处理字符对: KT\n行首字符: K, 列尾字符: T\n-> 在第2行第2列找到原文字符: N\n\n处理字符对: BE\n行首字符: B, 列尾字符: E\n-> 在第5行第3列找到原文字符: E\n\n处理字符对: WT\n行首字符: W, 列尾字符: T\n-> 在第4行第2列找到原文字符: R\n\n处理字符对: WH\n行首字符: W, 列尾字符: H\n-> 在第4行第4列找到原文字符: V\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird(All row and column numbers are counted from 0.):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - Located at the first letter of all rows, MKGWB is the row header letter.\n        - Located at the last letter of all columns, BTEHI is the bottom column letter.\n- Encryption Steps:\n    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.\n    - For each letter p in the plaintext.\n        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.\n        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.\n        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.\n    \n    Concatenate all the encrypted messages for the final ciphertext output.\n    \n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Take two letters c1,c2 at a time from the ciphertext.\n        - The letter c1 identifies the row position of the decrypted letter p in the grid, find the row where c1 is the row header.\n        - The letter c2 identifies the column position of the decrypted letter p in the grid, and finds the row with c2 at the bottom of the column.\n        - Find the letter at this row and column position in the grid, which is the c1,c2 decrypted message p.\n        - For example, if c1,c2=BH, B is the header of the row is the 4th row, H is the bottom of the row is the 2nd column, (4,2) at the letter H, so BH decrypted to H.\n    - Decrypt the letters in the ciphertext two by two, and output the connected decrypted message as the final plaintext.\nencrypted text: KTBEWTWHBIBI decode to plain text: nervii\nencrypted text: KTGEKTWIGIKHMIGTBBBIKHBIBTWE decode to plain text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 248, "source_filename": "icl_with_rule_decode_Kor_rule15_CollonCipher_en.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "KTGEKTWIGIKHMIGTBBBIKHBIBTWE", "extra_args": {}, "output": "开始解密过程...\n加密文本: KTGEKTWIGIKHMIGTBBBIKHBIBTWE\n\n每次解密两个字符:\n\n处理字符对: KT\n行首字符: K, 列尾字符: T\n-> 在第2行第2列找到原文字符: N\n\n处理字符对: GE\n行首字符: G, 列尾字符: E\n-> 在第3行第3列找到原文字符: O\n\n处理字符对: KT\n行首字符: K, 列尾字符: T\n-> 在第2行第2列找到原文字符: N\n\n处理字符对: WI\n行首字符: W, 列尾字符: I\n-> 在第4行第5列找到原文字符: C\n\n处理字符对: GI\n行首字符: G, 列尾字符: I\n-> 在第3行第5列找到原文字符: U\n\n处理字符对: KH\n行首字符: K, 列尾字符: H\n-> 在第2行第4列找到原文字符: L\n\n处理字符对: MI\n行首字符: M, 列尾字符: I\n-> 在第1行第5列找到原文字符: P\n\n处理字符对: GT\n行首字符: G, 列尾字符: T\n-> 在第3行第2列找到原文字符: A\n\n处理字符对: BB\n行首字符: B, 列尾字符: B\n-> 在第5行第1列找到原文字符: B\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: KH\n行首字符: K, 列尾字符: H\n-> 在第2行第4列找到原文字符: L\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: BT\n行首字符: B, 列尾字符: T\n-> 在第5行第2列找到原文字符: T\n\n处理字符对: WE\n行首字符: W, 列尾字符: E\n-> 在第4行第3列找到原文字符: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird(All row and column numbers are counted from 0.):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - Located at the first letter of all rows, MKGWB is the row header letter.\n        - Located at the last letter of all columns, BTEHI is the bottom column letter.\n- Encryption Steps:\n    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.\n    - For each letter p in the plaintext.\n        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.\n        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.\n        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.\n    \n    Concatenate all the encrypted messages for the final ciphertext output.\n    \n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Take two letters c1,c2 at a time from the ciphertext.\n        - The letter c1 identifies the row position of the decrypted letter p in the grid, find the row where c1 is the row header.\n        - The letter c2 identifies the column position of the decrypted letter p in the grid, and finds the row with c2 at the bottom of the column.\n        - Find the letter at this row and column position in the grid, which is the c1,c2 decrypted message p.\n        - For example, if c1,c2=BH, B is the header of the row is the 4th row, H is the bottom of the row is the 2nd column, (4,2) at the letter H, so BH decrypted to H.\n    - Decrypt the letters in the ciphertext two by two, and output the connected decrypted message as the final plaintext.\nencoded text: KTBEWTWHBIBI decode into original information: nervii\nencoded text: KTGEKTWIGIKHMIGTBBBIKHBIBTWE decode into original information: nonculpability\nencoded text: MIWEWTBIBTBIWI decode into original information is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 249, "source_filename": "icl_with_rule_decode_Kor_rule15_CollonCipher_en.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "MIWEWTBIBTBIWI", "extra_args": {}, "output": "开始解密过程...\n加密文本: MIWEWTBIBTBIWI\n\n每次解密两个字符:\n\n处理字符对: MI\n行首字符: M, 列尾字符: I\n-> 在第1行第5列找到原文字符: P\n\n处理字符对: WE\n行首字符: W, 列尾字符: E\n-> 在第4行第3列找到原文字符: Y\n\n处理字符对: WT\n行首字符: W, 列尾字符: T\n-> 在第4行第2列找到原文字符: R\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: BT\n行首字符: B, 列尾字符: T\n-> 在第5行第2列找到原文字符: T\n\n处理字符对: BI\n行首字符: B, 列尾字符: I\n-> 在第5行第5列找到原文字符: I\n\n处理字符对: WI\n行首字符: W, 列尾字符: I\n-> 在第4行第5列找到原文字符: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string without punctuation.\n- Preparation:\n    - Multitap Code Table\n        | Letter | Multitap Code |\n        | --- | --- |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Multitap Code Table:\n            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation: Multitap Code Table (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each given ciphertext Multitap Code c:\n        - If `c` is a Multitap Code from the Multitap Code Table:\n            - Replace `c` with the corresponding uppercase letter from the Multitap Code Table.\ncipher text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decrypt to clear text: nonculpability\ncipher text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decrypt to clear text: nonculpability\ncipher text: 6^23^27^38^34^34^3 decrypt to clear text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 250, "source_filename": "icl_with_rule_decode_Kor_rule3_MultiTapPhoneCode_en.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "6^23^27^38^34^34^3", "extra_args": {}, "output": "加密文本: 6^23^27^38^34^34^3\n开始解密:\n多击编码 6^2 对应的字符是: N\n多击编码 3^2 对应的字符是: E\n多击编码 7^3 对应的字符是: R\n多击编码 8^3 对应的字符是: V\n多击编码 4^3 对应的字符是: I\n多击编码 4^3 对应的字符是: I\n解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string without punctuation.\n- Preparation:\n    - Multitap Code Table\n        | Letter | Multitap Code |\n        | --- | --- |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Multitap Code Table:\n            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation: Multitap Code Table (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each given ciphertext Multitap Code c:\n        - If `c` is a Multitap Code from the Multitap Code Table:\n            - Replace `c` with the corresponding uppercase letter from the Multitap Code Table.\nencoded text: 7^19^37^34^38^14^32^3 decode into original information: pyritic\nencoded text: 6^23^27^38^34^34^3 decode into original information: nervii\nencoded text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decode into original information: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 251, "source_filename": "icl_with_rule_decode_Kor_rule3_MultiTapPhoneCode_en.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "6^26^36^22^38^25^37^12^12^24^35^34^38^19^3", "extra_args": {}, "output": "加密文本: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3\n开始解密:\n多击编码 6^2 对应的字符是: N\n多击编码 6^3 对应的字符是: O\n多击编码 6^2 对应的字符是: N\n多击编码 2^3 对应的字符是: C\n多击编码 8^2 对应的字符是: U\n多击编码 5^3 对应的字符是: L\n多击编码 7^1 对应的字符是: P\n多击编码 2^1 对应的字符是: A\n多击编码 2^2 对应的字符是: B\n多击编码 4^3 对应的字符是: I\n多击编码 5^3 对应的字符是: L\n多击编码 4^3 对应的字符是: I\n多击编码 8^1 对应的字符是: T\n多击编码 9^3 对应的字符是: Y\n解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A string without punctuation.\n- Preparation:\n    - Multitap Code Table\n        | Letter | Multitap Code |\n        | --- | --- |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- Encryption Steps:\n    - For each given plaintext character p:\n        - If `p` is an uppercase letter and exists in the Multitap Code Table:\n            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation: Multitap Code Table (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - For each given ciphertext Multitap Code c:\n        - If `c` is a Multitap Code from the Multitap Code Table:\n            - Replace `c` with the corresponding uppercase letter from the Multitap Code Table.\ncipher text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decode into original information: nonculpability\ncipher text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decode into original information: nonculpability\ncipher text: 7^19^37^34^38^14^32^3 decode into original information: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 252, "source_filename": "icl_with_rule_decode_Kor_rule3_MultiTapPhoneCode_en.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "7^19^37^34^38^14^32^3", "extra_args": {}, "output": "加密文本: 7^19^37^34^38^14^32^3\n开始解密:\n多击编码 7^1 对应的字符是: P\n多击编码 9^3 对应的字符是: Y\n多击编码 7^3 对应的字符是: R\n多击编码 4^3 对应的字符是: I\n多击编码 8^1 对应的字符是: T\n多击编码 4^3 对应的字符是: I\n多击编码 2^3 对应的字符是: C\n解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards:\n        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.\n        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n    - **Keystream Algorithm**\n        \n        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).\n        \n        For example, starting with:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        \n        Perform the following steps to generate a character of the keystream:\n        \n        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21\n        3. Perform a \"triple cut\": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.\n        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26\n        4. Perform a \"count cut\": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.\n        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26\n        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).\n        6. Return the resulting keystream value: 14\n- Encryption Steps:\n    - cards=Initial sequence of cards\n    - For each given plaintext character p:\n        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).\n        - Generate a keystream value y for p using the Initial sequence of cards:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:\n            - z=(y+x) % 26\n        - Use the LETTERS list to return the letter corresponding to position z.\n        - Append it to the ciphertext.\n\n \noriginal information: nonculpability encode to encrypted text: FMMLWKFPPPDEOF\noriginal information: nonculpability encode to encrypted text: FMMLWKFPPPDEOF\noriginal information: nervii encode to encrypted text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 253, "source_filename": "icl_with_rule_encode_Kor_rule6_SolitaireCipher_en.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n开始加密过程:\n\n处理第1个字符 'N':\n字符'N'在字母表中的位置是: 23\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (23 + 14) % 26 = 11\n加密后的字符是: F\n\n处理第2个字符 'E':\n字符'E'在字母表中的位置是: 13\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n加密计算: (13 + 7) % 26 = 20\n加密后的字符是: B\n\n处理第3个字符 'R':\n字符'R'在字母表中的位置是: 5\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n加密计算: (5 + 13) % 26 = 18\n加密后的字符是: P\n\n处理第4个字符 'V':\n字符'V'在字母表中的位置是: 19\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n加密计算: (19 + 17) % 26 = 10\n加密后的字符是: M\n\n处理第5个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (25 + 38) % 26 = 11\n加密后的字符是: F\n\n处理第6个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (25 + 45) % 26 = 18\n加密后的字符是: P\n\n最终加密结果: FBPMFP\n", "ground_truth": "FBPMFP"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards:\n        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.\n        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n    - **Keystream Algorithm**\n        \n        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).\n        \n        For example, starting with:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        \n        Perform the following steps to generate a character of the keystream:\n        \n        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21\n        3. Perform a \"triple cut\": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.\n        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26\n        4. Perform a \"count cut\": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.\n        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26\n        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).\n        6. Return the resulting keystream value: 14\n- Encryption Steps:\n    - cards=Initial sequence of cards\n    - For each given plaintext character p:\n        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).\n        - Generate a keystream value y for p using the Initial sequence of cards:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:\n            - z=(y+x) % 26\n        - Use the LETTERS list to return the letter corresponding to position z.\n        - Append it to the ciphertext.\n\n \noriginal information: nervii encode into encrypted text: FBPMFP\noriginal information: nonculpability encode into encrypted text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 254, "source_filename": "icl_with_rule_encode_Kor_rule6_SolitaireCipher_en.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n开始加密过程:\n\n处理第1个字符 'N':\n字符'N'在字母表中的位置是: 23\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (23 + 14) % 26 = 11\n加密后的字符是: F\n\n处理第2个字符 'O':\n字符'O'在字母表中的位置是: 3\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n加密计算: (3 + 7) % 26 = 10\n加密后的字符是: M\n\n处理第3个字符 'N':\n字符'N'在字母表中的位置是: 23\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n加密计算: (23 + 13) % 26 = 10\n加密后的字符是: M\n\n处理第4个字符 'C':\n字符'C'在字母表中的位置是: 7\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n加密计算: (7 + 17) % 26 = 24\n加密后的字符是: L\n\n处理第5个字符 'U':\n字符'U'在字母表中的位置是: 16\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (16 + 38) % 26 = 2\n加密后的字符是: W\n\n处理第6个字符 'L':\n字符'L'在字母表中的位置是: 24\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (24 + 45) % 26 = 17\n加密后的字符是: K\n\n处理第7个字符 'P':\n字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (18 + 45) % 26 = 11\n加密后的字符是: F\n\n处理第8个字符 'A':\n字符'A'在字母表中的位置是: 6\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (6 + 38) % 26 = 18\n加密后的字符是: P\n\n处理第9个字符 'B':\n字符'B'在字母表中的位置是: 20\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 50\n加密计算: (20 + 50) % 26 = 18\n加密后的字符是: P\n\n处理第10个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (25 + 45) % 26 = 18\n加密后的字符是: P\n\n处理第11个字符 'L':\n字符'L'在字母表中的位置是: 24\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 29\n加密计算: (24 + 29) % 26 = 1\n加密后的字符是: D\n\n处理第12个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (25 + 14) % 26 = 13\n加密后的字符是: E\n\n处理第13个字符 'T':\n字符'T'在字母表中的位置是: 4\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n加密计算: (4 + 25) % 26 = 3\n加密后的字符是: O\n\n处理第14个字符 'Y':\n字符'Y'在字母表中的位置是: 12\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n加密计算: (12 + 25) % 26 = 11\n加密后的字符是: F\n\n最终加密结果: FMMLWKFPPPDEOF\n", "ground_truth": "FMMLWKFPPPDEOF"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards:\n        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.\n        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n    - **Keystream Algorithm**\n        \n        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).\n        \n        For example, starting with:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        \n        Perform the following steps to generate a character of the keystream:\n        \n        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21\n        3. Perform a \"triple cut\": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.\n        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26\n        4. Perform a \"count cut\": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.\n        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26\n        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).\n        6. Return the resulting keystream value: 14\n- Encryption Steps:\n    - cards=Initial sequence of cards\n    - For each given plaintext character p:\n        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).\n        - Generate a keystream value y for p using the Initial sequence of cards:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:\n            - z=(y+x) % 26\n        - Use the LETTERS list to return the letter corresponding to position z.\n        - Append it to the ciphertext.\n\n \nclear text: nervii encrypt into encrypted text: FBPMFP\nclear text: nervii encrypt into encrypted text: FBPMFP\nclear text: nonculpability encrypt into encrypted text: FMMLWKFPPPDEOF\nclear text: pyritic encrypt into encrypted text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 255, "source_filename": "icl_with_rule_encode_Kor_rule6_SolitaireCipher_en.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n开始加密过程:\n\n处理第1个字符 'P':\n字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n加密计算: (18 + 14) % 26 = 6\n加密后的字符是: A\n\n处理第2个字符 'Y':\n字符'Y'在字母表中的位置是: 12\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n加密计算: (12 + 7) % 26 = 19\n加密后的字符是: V\n\n处理第3个字符 'R':\n字符'R'在字母表中的位置是: 5\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n加密计算: (5 + 13) % 26 = 18\n加密后的字符是: P\n\n处理第4个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n加密计算: (25 + 17) % 26 = 16\n加密后的字符是: U\n\n处理第5个字符 'T':\n字符'T'在字母表中的位置是: 4\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n加密计算: (4 + 38) % 26 = 16\n加密后的字符是: U\n\n处理第6个字符 'I':\n字符'I'在字母表中的位置是: 25\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (25 + 45) % 26 = 18\n加密后的字符是: P\n\n处理第7个字符 'C':\n字符'C'在字母表中的位置是: 7\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n加密计算: (7 + 45) % 26 = 0\n加密后的字符是: J\n\n最终加密结果: AVPUUPJ\n", "ground_truth": "AVPUUPJ"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird(All row and column numbers are counted from 0.):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - Located at the first letter of all rows, MKGWB is the row header letter.\n        - Located at the last letter of all columns, BTEHI is the bottom column letter.\n- Encryption Steps:\n    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.\n    - For each letter p in the plaintext.\n        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.\n        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.\n        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.\n    \n    Concatenate all the encrypted messages for the final ciphertext output.\n    \n\n \nclear text: pyritic encode into encrypted text: MIWEWTBIBTBIWI\nclear text: nonculpability encode into encrypted text: KTGEKTWIGIKHMIGTBBBIKHBIBTWE\nclear text: pyritic encode into encrypted text: MIWEWTBIBTBIWI\nclear text: nervii encode into encrypted text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 256, "source_filename": "icl_with_rule_encode_Kor_rule15_CollonCipher_en.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n原始文本: nervii\n预处理后的文本(移除空格标点,转大写,移除J): NERVII\n\n逐字符加密:\n字符 N 位于第2行第2列\n-> 行首字符为K, 列尾字符为T\n-> 加密为: KT\n字符 E 位于第5行第3列\n-> 行首字符为B, 列尾字符为E\n-> 加密为: BE\n字符 R 位于第4行第2列\n-> 行首字符为W, 列尾字符为T\n-> 加密为: WT\n字符 V 位于第4行第4列\n-> 行首字符为W, 列尾字符为H\n-> 加密为: WH\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n\n最终加密结果: KTBEWTWHBIBI\n", "ground_truth": "KTBEWTWHBIBI"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird(All row and column numbers are counted from 0.):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - Located at the first letter of all rows, MKGWB is the row header letter.\n        - Located at the last letter of all columns, BTEHI is the bottom column letter.\n- Encryption Steps:\n    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.\n    - For each letter p in the plaintext.\n        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.\n        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.\n        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.\n    \n    Concatenate all the encrypted messages for the final ciphertext output.\n    \n\n \nplain text: nonculpability encrypt to cipher text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 257, "source_filename": "icl_with_rule_encode_Kor_rule15_CollonCipher_en.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n原始文本: nonculpability\n预处理后的文本(移除空格标点,转大写,移除J): NONCULPABILITY\n\n逐字符加密:\n字符 N 位于第2行第2列\n-> 行首字符为K, 列尾字符为T\n-> 加密为: KT\n字符 O 位于第3行第3列\n-> 行首字符为G, 列尾字符为E\n-> 加密为: GE\n字符 N 位于第2行第2列\n-> 行首字符为K, 列尾字符为T\n-> 加密为: KT\n字符 C 位于第4行第5列\n-> 行首字符为W, 列尾字符为I\n-> 加密为: WI\n字符 U 位于第3行第5列\n-> 行首字符为G, 列尾字符为I\n-> 加密为: GI\n字符 L 位于第2行第4列\n-> 行首字符为K, 列尾字符为H\n-> 加密为: KH\n字符 P 位于第1行第5列\n-> 行首字符为M, 列尾字符为I\n-> 加密为: MI\n字符 A 位于第3行第2列\n-> 行首字符为G, 列尾字符为T\n-> 加密为: GT\n字符 B 位于第5行第1列\n-> 行首字符为B, 列尾字符为B\n-> 加密为: BB\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 L 位于第2行第4列\n-> 行首字符为K, 列尾字符为H\n-> 加密为: KH\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 T 位于第5行第2列\n-> 行首字符为B, 列尾字符为T\n-> 加密为: BT\n字符 Y 位于第4行第3列\n-> 行首字符为W, 列尾字符为E\n-> 加密为: WE\n\n最终加密结果: KTGEKTWIGIKHMIGTBBBIKHBIBTWE\n", "ground_truth": "KTGEKTWIGIKHMIGTBBBIKHBIBTWE"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 5x5 Gird(All row and column numbers are counted from 0.):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - Located at the first letter of all rows, MKGWB is the row header letter.\n        - Located at the last letter of all columns, BTEHI is the bottom column letter.\n- Encryption Steps:\n    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.\n    - For each letter p in the plaintext.\n        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.\n        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.\n        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.\n    \n    Concatenate all the encrypted messages for the final ciphertext output.\n    \n\n \nclear text: nervii encode into cipher text: KTBEWTWHBIBI\nclear text: nervii encode into cipher text: KTBEWTWHBIBI\nclear text: pyritic encode into cipher text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 258, "source_filename": "icl_with_rule_encode_Kor_rule15_CollonCipher_en.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n原始文本: pyritic\n预处理后的文本(移除空格标点,转大写,移除J): PYRITIC\n\n逐字符加密:\n字符 P 位于第1行第5列\n-> 行首字符为M, 列尾字符为I\n-> 加密为: MI\n字符 Y 位于第4行第3列\n-> 行首字符为W, 列尾字符为E\n-> 加密为: WE\n字符 R 位于第4行第2列\n-> 行首字符为W, 列尾字符为T\n-> 加密为: WT\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 T 位于第5行第2列\n-> 行首字符为B, 列尾字符为T\n-> 加密为: BT\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 C 位于第4行第5列\n-> 行首字符为W, 列尾字符为I\n-> 加密为: WI\n\n最终加密结果: MIWEWTBIBTBIWI\n", "ground_truth": "MIWEWTBIBTBIWI"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)\n        \n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'],\n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'],\n        ['6', 'J', 'V', 'O', 'E', 'M']]\n        \n    - Encrypted Character Set\n        - \"ADFGVX\"\n- Encryption Steps:\n    - For each character in the plaintext:\n        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.\n        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.\n            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).\n        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.\n            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters  string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (Same as encryption)\n    - Encrypted Character Set (Same as encryption)\n- Decryption Steps:\n    - Divide the ciphertext into two halves.\n    - Take one letter from the upper half and one letter from the lower half each time as a decryption binary:\n        - The position of each character in the encrypted character set is flagged as 0-6, and the position of the corresponding character in the encrypted character set is used to decrypt the number of rows and columns represented by the decrypted binary.\n            - For example, the decrypted binary is the encrypted binary as (A,F), the position of A is 0 and the position of F is 2, so the number of rows is 0 and the number of columns is 2, obtaining (0, 2).\n        - Use the number of rows and columns obtained to find the letter in the corresponding position in the 6x6 matrix as the character after decryption.\n            - For example, the character at (0, 2) is A, and the character after AF decryption is A.\n    - Concatenate all the decrypted characters to get the final plaintext.\nencoded text: VGAVFVGVAAFVFD decrypt to plain text: pyritic\nencoded text: VXAXVVXVAFFF decrypt to plain text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 259, "source_filename": "icl_with_rule_decode_Kor_rule19_ADFGVX_en.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "VXAXVVXVAFFF", "extra_args": {}, "output": "输入密文: VXAXVVXVAFFF\n\n解密过程:\n1. 将密文分成两半:\n前半部分: VXAXVV\n后半部分: XVAFFF\n\n2. 将前后半部分配对并解密:\n配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N\n配对 (X,V) 对应矩阵位置 (5,4)，解密为字符: E\n配对 (A,A) 对应矩阵位置 (0,0)，解密为字符: R\n配对 (X,F) 对应矩阵位置 (5,2)，解密为字符: V\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n\n最终明文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)\n        \n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'],\n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'],\n        ['6', 'J', 'V', 'O', 'E', 'M']]\n        \n    - Encrypted Character Set\n        - \"ADFGVX\"\n- Encryption Steps:\n    - For each character in the plaintext:\n        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.\n        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.\n            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).\n        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.\n            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters  string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (Same as encryption)\n    - Encrypted Character Set (Same as encryption)\n- Decryption Steps:\n    - Divide the ciphertext into two halves.\n    - Take one letter from the upper half and one letter from the lower half each time as a decryption binary:\n        - The position of each character in the encrypted character set is flagged as 0-6, and the position of the corresponding character in the encrypted character set is used to decrypt the number of rows and columns represented by the decrypted binary.\n            - For example, the decrypted binary is the encrypted binary as (A,F), the position of A is 0 and the position of F is 2, so the number of rows is 0 and the number of columns is 2, obtaining (0, 2).\n        - Use the number of rows and columns obtained to find the letter in the corresponding position in the 6x6 matrix as the character after decryption.\n            - For example, the character at (0, 2) is A, and the character after AF decryption is A.\n    - Concatenate all the decrypted characters to get the final plaintext.\nencrypted text: VXVGAGVAAVGVFGXGXDDXVFXFXFVA decode into clear text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 260, "source_filename": "icl_with_rule_decode_Kor_rule19_ADFGVX_en.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "VXVGAGVAAVGVFGXGXDDXVFXFXFVA", "extra_args": {}, "output": "输入密文: VXVGAGVAAVGVFGXGXDDXVFXFXFVA\n\n解密过程:\n1. 将密文分成两半:\n前半部分: VXVGAGVAAVGVFG\n后半部分: XGXDDXVFXFXFVA\n\n2. 将前后半部分配对并解密:\n配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N\n配对 (X,G) 对应矩阵位置 (5,3)，解密为字符: O\n配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N\n配对 (G,D) 对应矩阵位置 (3,1)，解密为字符: C\n配对 (A,D) 对应矩阵位置 (0,1)，解密为字符: U\n配对 (G,X) 对应矩阵位置 (3,5)，解密为字符: L\n配对 (V,V) 对应矩阵位置 (4,4)，解密为字符: P\n配对 (A,F) 对应矩阵位置 (0,2)，解密为字符: A\n配对 (A,X) 对应矩阵位置 (0,5)，解密为字符: B\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (G,X) 对应矩阵位置 (3,5)，解密为字符: L\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (F,V) 对应矩阵位置 (2,4)，解密为字符: T\n配对 (G,A) 对应矩阵位置 (3,0)，解密为字符: Y\n\n最终明文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)\n        \n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'],\n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'],\n        ['6', 'J', 'V', 'O', 'E', 'M']]\n        \n    - Encrypted Character Set\n        - \"ADFGVX\"\n- Encryption Steps:\n    - For each character in the plaintext:\n        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.\n        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.\n            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).\n        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.\n            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters  string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (Same as encryption)\n    - Encrypted Character Set (Same as encryption)\n- Decryption Steps:\n    - Divide the ciphertext into two halves.\n    - Take one letter from the upper half and one letter from the lower half each time as a decryption binary:\n        - The position of each character in the encrypted character set is flagged as 0-6, and the position of the corresponding character in the encrypted character set is used to decrypt the number of rows and columns represented by the decrypted binary.\n            - For example, the decrypted binary is the encrypted binary as (A,F), the position of A is 0 and the position of F is 2, so the number of rows is 0 and the number of columns is 2, obtaining (0, 2).\n        - Use the number of rows and columns obtained to find the letter in the corresponding position in the 6x6 matrix as the character after decryption.\n            - For example, the character at (0, 2) is A, and the character after AF decryption is A.\n    - Concatenate all the decrypted characters to get the final plaintext.\ncipher text: VGAVFVGVAAFVFD decrypt into clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 261, "source_filename": "icl_with_rule_decode_Kor_rule19_ADFGVX_en.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "VGAVFVGVAAFVFD", "extra_args": {}, "output": "输入密文: VGAVFVGVAAFVFD\n\n解密过程:\n1. 将密文分成两半:\n前半部分: VGAVFVG\n后半部分: VAAFVFD\n\n2. 将前后半部分配对并解密:\n配对 (V,V) 对应矩阵位置 (4,4)，解密为字符: P\n配对 (G,A) 对应矩阵位置 (3,0)，解密为字符: Y\n配对 (A,A) 对应矩阵位置 (0,0)，解密为字符: R\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (F,V) 对应矩阵位置 (2,4)，解密为字符: T\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (G,D) 对应矩阵位置 (3,1)，解密为字符: C\n\n最终明文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n以下是经过整理后的文字：\n\n---\n\n### 5x5 网格布局\n\n共有四个5x5的字符网格，每个网格代表了不同的排列方式。\n\n1. **网格1 原始网格**\n\n这是最初的字符排列，按照特定顺序组织如下：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n2. **网格2 ECHO 网格**\n\n该网格根据\"ECHO\"这个词进行了重新排列：\n\n```\nE C H O A\nB D F G I\nJ K L M N\nP R S T U\nV W X Y Z\n```\n\n3. **网格3 VORTEX 网格**\n\n此网格基于\"VORTEX\"一词进行了独特的字符重组：\n\n```\nV O R T E\nX A B C D\nF G H I J\nK L M N P\nS U W Y Z\n```\n\n4. **网格4 重复原始网格**\n\n最后一个网格与第一个原始网格完全相同，没有进行任何改变：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n每个网格展示了不同主题词下字符的独特排列。\n- 加密步骤:\n    - 清理原文，移除空格和非字母字符，移除字母Q，转换为大写\n    - 如果原文长度为奇数，添加字母'X'使其成为偶数\n    - 将处理后的原文分成两个字母一组\n    - 对于每组两个字母p1,p2:\n        - 在网格1和网格4中找到第一个字母和第二个字母的位置\n        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果\n    - 连接所有加密后的字母组形成最终暗文\n解密算法:\n解密规则:\n- 输入:\n    - 暗文: 大写字母字符串\n- 输出:\n    - 原文: 大写字母字符串\n- 准备:\n    - 四个5x5网格(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 清理暗文，移除空格和非字母字符，转换为大写\n    - 将处理后的暗文分成两个字母一组\n    - 对于每组两个字母c1,c2:\n        - 在网格2和网格3中找到第一个字母和第二个字母的位置\n        - 在网格1和网格4中找到这两个位置对应的字母，用这两个字母作为该组的解密结果\n    - 连接所有解密后的字母组形成最终原文\n暗文: BIDYGYSH 解密成原文: pyritic\n暗文: OPDFYY 解密成原文: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 262, "source_filename": "icl_with_rule_decode_Kor_rule11_FourSquareCipher_cn.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "OPDFYY", "extra_args": {}, "output": "开始解密过程...\n加密文本: OPDFYY\n\n开始逐对字母解密:\n\n处理加密字母对: OP\n第一个字母 O 在ECHO表格中的位置: (0, 3)\n第二个字母 P 在VORTEX表格中的位置: (3, 4)\n解密后的字母对: NE\n\n处理加密字母对: DF\n第一个字母 D 在ECHO表格中的位置: (1, 1)\n第二个字母 F 在VORTEX表格中的位置: (2, 0)\n解密后的字母对: RV\n\n处理加密字母对: YY\n第一个字母 Y 在ECHO表格中的位置: (4, 3)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: II\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n以下是经过整理后的文字：\n\n---\n\n### 5x5 网格布局\n\n共有四个5x5的字符网格，每个网格代表了不同的排列方式。\n\n1. **网格1 原始网格**\n\n这是最初的字符排列，按照特定顺序组织如下：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n2. **网格2 ECHO 网格**\n\n该网格根据\"ECHO\"这个词进行了重新排列：\n\n```\nE C H O A\nB D F G I\nJ K L M N\nP R S T U\nV W X Y Z\n```\n\n3. **网格3 VORTEX 网格**\n\n此网格基于\"VORTEX\"一词进行了独特的字符重组：\n\n```\nV O R T E\nX A B C D\nF G H I J\nK L M N P\nS U W Y Z\n```\n\n4. **网格4 重复原始网格**\n\n最后一个网格与第一个原始网格完全相同，没有进行任何改变：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n每个网格展示了不同主题词下字符的独特排列。\n- 加密步骤:\n    - 清理原始信息，移除空格和非字母字符，移除字母Q，转换为大写\n    - 如果原始信息长度为奇数，添加字母'X'使其成为偶数\n    - 将处理后的原始信息分成两个字母一组\n    - 对于每组两个字母p1,p2:\n        - 在网格1和网格4中找到第一个字母和第二个字母的位置\n        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果\n    - 连接所有加密后的字母组形成最终密文\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 大写字母字符串\n- 输出:\n    - 原始信息: 大写字母字符串\n- 准备:\n    - 四个5x5网格(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 清理密文，移除空格和非字母字符，转换为大写\n    - 将处理后的密文分成两个字母一组\n    - 对于每组两个字母c1,c2:\n        - 在网格2和网格3中找到第一个字母和第二个字母的位置\n        - 在网格1和网格4中找到这两个位置对应的字母，用这两个字母作为该组的解密结果\n    - 连接所有解密后的字母组形成最终原始信息\n密文: BIDYGYSH 解密成原始信息: pyritic\n密文: BIDYGYSH 解密成原始信息: pyritic\n密文: BIDYGYSH 解密成原始信息: pyritic\n密文: OEOMIOBKRYCYGI 解密成原始信息: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 263, "source_filename": "icl_with_rule_decode_Kor_rule11_FourSquareCipher_cn.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "OEOMIOBKRYCYGI", "extra_args": {}, "output": "开始解密过程...\n加密文本: OEOMIOBKRYCYGI\n\n开始逐对字母解密:\n\n处理加密字母对: OE\n第一个字母 O 在ECHO表格中的位置: (0, 3)\n第二个字母 E 在VORTEX表格中的位置: (0, 4)\n解密后的字母对: NO\n\n处理加密字母对: OM\n第一个字母 O 在ECHO表格中的位置: (0, 3)\n第二个字母 M 在VORTEX表格中的位置: (3, 2)\n解密后的字母对: NC\n\n处理加密字母对: IO\n第一个字母 I 在ECHO表格中的位置: (1, 4)\n第二个字母 O 在VORTEX表格中的位置: (0, 1)\n解密后的字母对: UL\n\n处理加密字母对: BK\n第一个字母 B 在ECHO表格中的位置: (1, 0)\n第二个字母 K 在VORTEX表格中的位置: (3, 0)\n解密后的字母对: PA\n\n处理加密字母对: RY\n第一个字母 R 在ECHO表格中的位置: (3, 1)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: BI\n\n处理加密字母对: CY\n第一个字母 C 在ECHO表格中的位置: (0, 1)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: LI\n\n处理加密字母对: GI\n第一个字母 G 在ECHO表格中的位置: (1, 3)\n第二个字母 I 在VORTEX表格中的位置: (2, 3)\n解密后的字母对: TY\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n以下是经过整理后的文字：\n\n---\n\n### 5x5 网格布局\n\n共有四个5x5的字符网格，每个网格代表了不同的排列方式。\n\n1. **网格1 原始网格**\n\n这是最初的字符排列，按照特定顺序组织如下：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n2. **网格2 ECHO 网格**\n\n该网格根据\"ECHO\"这个词进行了重新排列：\n\n```\nE C H O A\nB D F G I\nJ K L M N\nP R S T U\nV W X Y Z\n```\n\n3. **网格3 VORTEX 网格**\n\n此网格基于\"VORTEX\"一词进行了独特的字符重组：\n\n```\nV O R T E\nX A B C D\nF G H I J\nK L M N P\nS U W Y Z\n```\n\n4. **网格4 重复原始网格**\n\n最后一个网格与第一个原始网格完全相同，没有进行任何改变：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n每个网格展示了不同主题词下字符的独特排列。\n- 加密步骤:\n    - 清理原始信息，移除空格和非字母字符，移除字母Q，转换为大写\n    - 如果原始信息长度为奇数，添加字母'X'使其成为偶数\n    - 将处理后的原始信息分成两个字母一组\n    - 对于每组两个字母p1,p2:\n        - 在网格1和网格4中找到第一个字母和第二个字母的位置\n        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果\n    - 连接所有加密后的字母组形成最终加密文本\n解密算法:\n解密规则:\n- 输入:\n    - 加密文本: 大写字母字符串\n- 输出:\n    - 原始信息: 大写字母字符串\n- 准备:\n    - 四个5x5网格(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 清理加密文本，移除空格和非字母字符，转换为大写\n    - 将处理后的加密文本分成两个字母一组\n    - 对于每组两个字母c1,c2:\n        - 在网格2和网格3中找到第一个字母和第二个字母的位置\n        - 在网格1和网格4中找到这两个位置对应的字母，用这两个字母作为该组的解密结果\n    - 连接所有解密后的字母组形成最终原始信息\n加密文本: BIDYGYSH 解密为原始信息是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 264, "source_filename": "icl_with_rule_decode_Kor_rule11_FourSquareCipher_cn.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "BIDYGYSH", "extra_args": {}, "output": "开始解密过程...\n加密文本: BIDYGYSH\n\n开始逐对字母解密:\n\n处理加密字母对: BI\n第一个字母 B 在ECHO表格中的位置: (1, 0)\n第二个字母 I 在VORTEX表格中的位置: (2, 3)\n解密后的字母对: PY\n\n处理加密字母对: DY\n第一个字母 D 在ECHO表格中的位置: (1, 1)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: RI\n\n处理加密字母对: GY\n第一个字母 G 在ECHO表格中的位置: (1, 3)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: TI\n\n处理加密字母对: SH\n第一个字母 S 在ECHO表格中的位置: (3, 2)\n第二个字母 H 在VORTEX表格中的位置: (2, 2)\n解密后的字母对: CX\n\n最终解密结果: PYRITICX\n", "ground_truth": "PYRITICX"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n非加密信息: pyritic 密钥或额外参数: {'shift': 1} 编码为加密文本: qzsjujd\n非加密信息: pyritic 密钥或额外参数: {'shift': 1} 编码为加密文本: qzsjujd\n非加密信息: nonculpability 密钥或额外参数: {'shift': 15} 编码为加密文本: cdcrjaepqxaxin\n非加密信息: nonculpability 密钥或额外参数: {'shift': 15} 编码为加密文本: cdcrjaepqxaxin\n非加密信息: nervii 密钥或额外参数: {'shift': 7} 编码为加密文本: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 265, "source_filename": "icl_encode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "nervii", "extra_args": {"shift": 7}, "output": "开始编码过程，文本为: nervii，位移数为: 7\n处理字符: n\n字符 n 向后移动 7 位，变为 u\n处理字符: e\n字符 e 向后移动 7 位，变为 l\n处理字符: r\n字符 r 向后移动 7 位，变为 y\n处理字符: v\n字符 v 向后移动 7 位，变为 c\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n编码完成，结果为: ulycpp\n", "ground_truth": "ulycpp"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd\n明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd\n明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd\n明文: nervii 密钥或额外参数: {'shift': 7} 加密成为隐文: ulycpp\n明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd\n明文: nonculpability 密钥或额外参数: {'shift': 15} 加密成为隐文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 266, "source_filename": "icl_encode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "nonculpability", "extra_args": {"shift": 15}, "output": "开始编码过程，文本为: nonculpability，位移数为: 15\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: o\n字符 o 向后移动 15 位，变为 d\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: c\n字符 c 向后移动 15 位，变为 r\n处理字符: u\n字符 u 向后移动 15 位，变为 j\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: p\n字符 p 向后移动 15 位，变为 e\n处理字符: a\n字符 a 向后移动 15 位，变为 p\n处理字符: b\n字符 b 向后移动 15 位，变为 q\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: t\n字符 t 向后移动 15 位，变为 i\n处理字符: y\n字符 y 向后移动 15 位，变为 n\n编码完成，结果为: cdcrjaepqxaxin\n", "ground_truth": "cdcrjaepqxaxin"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n原文: nonculpability 密钥或额外参数: {'shift': 15} 加密为隐文: cdcrjaepqxaxin\n原文: nonculpability 密钥或额外参数: {'shift': 15} 加密为隐文: cdcrjaepqxaxin\n原文: nervii 密钥或额外参数: {'shift': 7} 加密为隐文: ulycpp\n原文: nonculpability 密钥或额外参数: {'shift': 15} 加密为隐文: cdcrjaepqxaxin\n原文: nervii 密钥或额外参数: {'shift': 7} 加密为隐文: ulycpp\n原文: pyritic 密钥或额外参数: {'shift': 1} 加密为隐文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 267, "source_filename": "icl_encode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "pyritic", "extra_args": {"shift": 1}, "output": "开始编码过程，文本为: pyritic，位移数为: 1\n处理字符: p\n字符 p 向后移动 1 位，变为 q\n处理字符: y\n字符 y 向后移动 1 位，变为 z\n处理字符: r\n字符 r 向后移动 1 位，变为 s\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: t\n字符 t 向后移动 1 位，变为 u\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: c\n字符 c 向后移动 1 位，变为 d\n编码完成，结果为: qzsjujd\n", "ground_truth": "qzsjujd"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将非加密信息文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n解密算法:\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的非加密信息字母。\n4. 输出解密后的文本。\n密文: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为非加密信息: nonculpability\n密文: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为非加密信息: nonculpability\n密文: RV RB RD WX 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解密成为非加密信息: pyritic\n密文: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密成为非加密信息: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 268, "source_filename": "icl_with_rule_decode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "JV KT CA", "extra_args": {"str1": "parametral", "str2": "quiverful"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nP A R M E\nT L B C D\nF G H I J\nK N O S U\nV W X Y Z\n\nM2 矩阵 (根据密钥 str2):\nQ U I V E\nR F L A B\nC D G H J\nK M N O P\nS T W X Y\n\nStep 3: 将密文分成两个字母一组: ['JV', 'KT', 'CA']\n\nStep 4.1: 解密字母对 JV\n  - J 在 M1 中的位置: (2, 4)\n  - V 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 4) 的字母: E\n  - 解密结果为: NE\n\nStep 4.2: 解密字母对 KT\n  - K 在 M1 中的位置: (3, 0)\n  - T 在 M2 中的位置: (4, 1)\n  - 交换列索引后新位置: (3, 1) 和 (4, 0)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (4, 0) 的字母: V\n  - 解密结果为: RV\n\nStep 4.3: 解密字母对 CA\n  - C 在 M1 中的位置: (1, 3)\n  - A 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (1, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: II\n\n解密完成！最终解密结果为: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将原始信息文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n解密算法:\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将加密文本按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的原始信息字母。\n4. 输出解密后的文本。\n加密文本: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密为原始信息: nervii\n加密文本: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密为原始信息: nervii\n加密文本: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密为原始信息: nervii\n加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密为原始信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 269, "source_filename": "icl_with_rule_decode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "JK FI LL KS CE GE NY", "extra_args": {"str1": "trichophytia", "str2": "sphinges"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nT R I C H\nO P Y A B\nD E F G J\nK L M N S\nU V W X Z\n\nM2 矩阵 (根据密钥 str2):\nS P H I N\nG E A B C\nD F J K L\nM O R T U\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['JK', 'FI', 'LL', 'KS', 'CE', 'GE', 'NY']\n\nStep 4.1: 解密字母对 JK\n  - J 在 M1 中的位置: (2, 4)\n  - K 在 M2 中的位置: (2, 3)\n  - 交换列索引后新位置: (2, 3) 和 (2, 4)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (2, 4) 的字母: O\n  - 解密结果为: NO\n\nStep 4.2: 解密字母对 FI\n  - F 在 M1 中的位置: (2, 2)\n  - I 在 M2 中的位置: (0, 3)\n  - 交换列索引后新位置: (2, 3) 和 (0, 2)\n  - 在 M_T 中查找位置 (2, 3) 的字母: N\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 解密结果为: NC\n\nStep 4.3: 解密字母对 LL\n  - L 在 M1 中的位置: (3, 1)\n  - L 在 M2 中的位置: (2, 4)\n  - 交换列索引后新位置: (3, 4) 和 (2, 1)\n  - 在 M_T 中查找位置 (3, 4) 的字母: U\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 解密结果为: UL\n\nStep 4.4: 解密字母对 KS\n  - K 在 M1 中的位置: (3, 0)\n  - S 在 M2 中的位置: (0, 0)\n  - 交换列索引后新位置: (3, 0) 和 (0, 0)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (0, 0) 的字母: A\n  - 解密结果为: PA\n\nStep 4.5: 解密字母对 CE\n  - C 在 M1 中的位置: (0, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (0, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (0, 1) 的字母: B\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: BI\n\nStep 4.6: 解密字母对 GE\n  - G 在 M1 中的位置: (2, 3)\n  - E 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (2, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (2, 1) 的字母: L\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: LI\n\nStep 4.7: 解密字母对 NY\n  - N 在 M1 中的位置: (3, 3)\n  - Y 在 M2 中的位置: (4, 3)\n  - 交换列索引后新位置: (3, 3) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: TY\n\n解密完成！最终解密结果为: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将初始文本文本转换为大写字母，仅保留字母字符并分成两个字母一组。\n3. 对每对字母进行加密：\n   - 查找每个字母在默认矩阵中的位置，交换其y坐标，\n   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。\n4. 输出加密后的文本。\n解密算法:\n解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将加密文本按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的初始文本字母。\n4. 输出解密后的文本。\n加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为初始文本: nonculpability\n加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为初始文本: nonculpability\n加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为初始文本: nonculpability\n加密文本: RV RB RD WX 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解密成为初始文本是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 270, "source_filename": "icl_with_rule_decode_FourSquareCipher_cn.jsonl", "cipher_name": "FourSquareCipher", "input": "RV RB RD WX", "extra_args": {"str1": "sawbill", "str2": "houri"}, "output": "解题方案：\n1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。\n2. 将密文按两个字母一组分组。\n3. 对每对加密字母进行解密：\n   - 在M1和M2中找到加密字母的位置，交换列坐标\n   - 从默认矩阵中获取对应的明文字母。\n4. 输出解密后的文本。\n\nStep 1: 创建默认字母表矩阵 M_T (去除Q)：\nA B C D E\nF G H I J\nK L M N O\nP R S T U\nV W X Y Z\n\nStep 2: 使用密钥构建加密矩阵 M1 和 M2\nM1 矩阵 (根据密钥 str1):\nS A W B I\nL C D E F\nG H J K M\nN O P R T\nU V X Y Z\n\nM2 矩阵 (根据密钥 str2):\nH O U R I\nA B C D E\nF G J K L\nM N P S T\nV W X Y Z\n\nStep 3: 将密文分成两个字母一组: ['RV', 'RB', 'RD', 'WX']\n\nStep 4.1: 解密字母对 RV\n  - R 在 M1 中的位置: (3, 3)\n  - V 在 M2 中的位置: (4, 0)\n  - 交换列索引后新位置: (3, 0) 和 (4, 3)\n  - 在 M_T 中查找位置 (3, 0) 的字母: P\n  - 在 M_T 中查找位置 (4, 3) 的字母: Y\n  - 解密结果为: PY\n\nStep 4.2: 解密字母对 RB\n  - R 在 M1 中的位置: (3, 3)\n  - B 在 M2 中的位置: (1, 1)\n  - 交换列索引后新位置: (3, 1) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 1) 的字母: R\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: RI\n\nStep 4.3: 解密字母对 RD\n  - R 在 M1 中的位置: (3, 3)\n  - D 在 M2 中的位置: (1, 3)\n  - 交换列索引后新位置: (3, 3) 和 (1, 3)\n  - 在 M_T 中查找位置 (3, 3) 的字母: T\n  - 在 M_T 中查找位置 (1, 3) 的字母: I\n  - 解密结果为: TI\n\nStep 4.4: 解密字母对 WX\n  - W 在 M1 中的位置: (0, 2)\n  - X 在 M2 中的位置: (4, 2)\n  - 交换列索引后新位置: (0, 2) 和 (4, 2)\n  - 在 M_T 中查找位置 (0, 2) 的字母: C\n  - 在 M_T 中查找位置 (4, 2) 的字母: X\n  - 解密结果为: CX\n\n解密完成！最终解密结果为: PYRITICX\n", "ground_truth": "PYRITICX"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\ncipher text: OCAX..C decode to clear text: pyritic\ncipher text: NFJBXZUHXEJCCY decode to clear text: nonculpability\ncipher text: NFJBXZUHXEJCCY decode to clear text: nonculpability\ncipher text: MSQIQR decode to clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 271, "source_filename": "icl_decode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "MSQIQR", "extra_args": {}, "output": "步骤1: 将密文转换为大写: MSQIQR\n步骤2: 将每个字母转换为坐标: 212311223133223233\n步骤3: 计算列数: 6\n步骤4: 将坐标组织成3xN的矩阵:\n212311\n223133\n223233\n步骤5: 转置矩阵:\n222\n122\n233\n312\n133\n133\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (1, 2, 2), (2, 3, 3), (3, 1, 2), (1, 3, 3), (1, 3, 3)]\n步骤7: 将坐标转换回字母: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\nencoded text: MSQIQR decode to clear text: nervii\nencoded text: MSQIQR decode to clear text: nervii\nencoded text: MSQIQR decode to clear text: nervii\nencoded text: OCAX..C decode to clear text: pyritic\nencoded text: NFJBXZUHXEJCCY decode to clear text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 272, "source_filename": "icl_decode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "NFJBXZUHXEJCCY", "extra_args": {}, "output": "步骤1: 将密文转换为大写: NFJBXZUHXEJCCY\n步骤2: 将每个字母转换为坐标: 222132211121332323331123332122211131131313\n步骤3: 计算列数: 14\n步骤4: 将坐标组织成3xN的矩阵:\n22213221112133\n23233311233321\n22211131131313\n步骤5: 转置矩阵:\n222\n232\n222\n131\n331\n231\n213\n111\n121\n133\n231\n133\n321\n313\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 2, 2), (2, 3, 2), (2, 2, 2), (1, 3, 1), (3, 3, 1), (2, 3, 1), (2, 1, 3), (1, 1, 1), (1, 2, 1), (1, 3, 3), (2, 3, 1), (1, 3, 3), (3, 2, 1), (3, 1, 3)]\n步骤7: 将坐标转换回字母: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\nencoded text: MSQIQR decrypt to original information: nervii\nencoded text: MSQIQR decrypt to original information: nervii\nencoded text: MSQIQR decrypt to original information: nervii\nencoded text: MSQIQR decrypt to original information: nervii\nencoded text: MSQIQR decrypt to original information: nervii\nencoded text: OCAX..C decrypt to original information: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 273, "source_filename": "icl_decode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "OCAX..C", "extra_args": {}, "output": "步骤1: 将密文转换为大写: OCAX..C\n步骤2: 将每个字母转换为坐标: 232131111332333333131\n步骤3: 计算列数: 7\n步骤4: 将坐标组织成3xN的矩阵:\n2321311\n1133233\n3333131\n步骤5: 转置矩阵:\n213\n313\n233\n133\n321\n133\n131\n立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\n步骤6: 组合转置后的坐标: [(2, 1, 3), (3, 1, 3), (2, 3, 3), (1, 3, 3), (3, 2, 1), (1, 3, 3), (1, 3, 1)]\n步骤7: 将坐标转换回字母: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).\n- Preparation:\n    - Key(SHA hash value)\n        - Performing SHA-256 on \"SECRET_KEY\" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.\n- Encryption Steps:\n    - Converts a plaintext string into a sequence of bytes (ASCII encoding).\n    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.\n    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Hexadecimal string (and letters a-e in lower case).\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Key (Same as encryption,is a hashed hash obtained by SHA_256)\n- Decryption Steps:\n    - Converts the ciphertext string into a sequence of bytes.\n    - Each byte is decrypted (the decryption process is the same as the encryption process) with the corresponding byte of the key using a bitwise different-or (XOR) operation.\n    - The decrypted byte sequence is converted to a plaintext string.\nencrypted text: 3da0640d88362ffe45b41326fbc0 decode to plain text: nonculpability\nencrypted text: 23b6780789333c decode to plain text: pyritic\nencrypted text: 3daa78189433 decode to plain text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 274, "source_filename": "icl_with_rule_decode_Kor_rule25_SHACipher_en.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "3daa78189433", "extra_args": {}, "output": "开始解密过程...\n加密的十六进制文本: 3daa78189433\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n十六进制转换为字节序列: b'=\\xaax\\x18\\x943'\n开始XOR解密...\n解密后的字节序列: b'NERVII'\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).\n- Preparation:\n    - Key(SHA hash value)\n        - Performing SHA-256 on \"SECRET_KEY\" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.\n- Encryption Steps:\n    - Converts a plaintext string into a sequence of bytes (ASCII encoding).\n    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.\n    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Hexadecimal string (and letters a-e in lower case).\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Key (Same as encryption,is a hashed hash obtained by SHA_256)\n- Decryption Steps:\n    - Converts the ciphertext string into a sequence of bytes.\n    - Each byte is decrypted (the decryption process is the same as the encryption process) with the corresponding byte of the key using a bitwise different-or (XOR) operation.\n    - The decrypted byte sequence is converted to a plaintext string.\ncipher text: 23b6780789333c decrypt into clear text: pyritic\ncipher text: 3da0640d88362ffe45b41326fbc0 decrypt into clear text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 275, "source_filename": "icl_with_rule_decode_Kor_rule25_SHACipher_en.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "3da0640d88362ffe45b41326fbc0", "extra_args": {}, "output": "开始解密过程...\n加密的十六进制文本: 3da0640d88362ffe45b41326fbc0\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n十六进制转换为字节序列: b'=\\xa0d\\r\\x886/\\xfeE\\xb4\\x13&\\xfb\\xc0'\n开始XOR解密...\n解密后的字节序列: b'NONCULPABILITY'\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).\n- Preparation:\n    - Key(SHA hash value)\n        - Performing SHA-256 on \"SECRET_KEY\" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.\n- Encryption Steps:\n    - Converts a plaintext string into a sequence of bytes (ASCII encoding).\n    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.\n    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Hexadecimal string (and letters a-e in lower case).\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Key (Same as encryption,is a hashed hash obtained by SHA_256)\n- Decryption Steps:\n    - Converts the ciphertext string into a sequence of bytes.\n    - Each byte is decrypted (the decryption process is the same as the encryption process) with the corresponding byte of the key using a bitwise different-or (XOR) operation.\n    - The decrypted byte sequence is converted to a plaintext string.\nencoded text: 3da0640d88362ffe45b41326fbc0 decrypt to original information: nonculpability\nencoded text: 3da0640d88362ffe45b41326fbc0 decrypt to original information: nonculpability\nencoded text: 3da0640d88362ffe45b41326fbc0 decrypt to original information: nonculpability\nencoded text: 23b6780789333c decrypt to original information: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 276, "source_filename": "icl_with_rule_decode_Kor_rule25_SHACipher_en.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "23b6780789333c", "extra_args": {}, "output": "开始解密过程...\n加密的十六进制文本: 23b6780789333c\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n十六进制转换为字节序列: b'#\\xb6x\\x07\\x893<'\n开始XOR解密...\n解密后的字节序列: b'PYRITIC'\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Maximum number of characters per line: 5\n- Encryption Steps:\n    - Lines are counted from 1.\n    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.\n        - For example, for the plaintext \"LIDAHELLOWORLD\", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows\n        LIDAH\n        WOLLE\n        ORLD\n    - Then read by column, each column is read from the top down, each read a column plus a \"#\", read out the contents of the final ciphertext.\n        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.\n\n \nplain text: nonculpability encode to encrypted text: NIL#OBI#NAT#CPY#UL#\nplain text: nonculpability encode to encrypted text: NIL#OBI#NAT#CPY#UL#\nplain text: nervii encode to encrypted text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 277, "source_filename": "icl_with_rule_encode_Kor_rule17_PathCipher_en.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "NERVII", "extra_args": {}, "output": "1. 输入文本转换为大写并去除非字母字符: NERVII\n2. 确定网格大小: 2行 x 5列\n3. 开始填充网格:\n   第1行从左到右填充: N E R V I \n   第2行从右到左填充: I \n4. 按列读取并添加#号:\n   第1列: N##\n   第2列: E##\n   第3列: R##\n   第4列: V##\n   第5列: II#\n5. 最终密文: N#E#R#V#II#\n", "ground_truth": "N#E#R#V#II#"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Maximum number of characters per line: 5\n- Encryption Steps:\n    - Lines are counted from 1.\n    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.\n        - For example, for the plaintext \"LIDAHELLOWORLD\", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows\n        LIDAH\n        WOLLE\n        ORLD\n    - Then read by column, each column is read from the top down, each read a column plus a \"#\", read out the contents of the final ciphertext.\n        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.\n\n \nplain text: nonculpability encode into encrypted text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 278, "source_filename": "icl_with_rule_encode_Kor_rule17_PathCipher_en.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "1. 输入文本转换为大写并去除非字母字符: NONCULPABILITY\n2. 确定网格大小: 3行 x 5列\n3. 开始填充网格:\n   第1行从左到右填充: N O N C U \n   第2行从右到左填充: L P A B I \n   第3行从左到右填充: L I T Y \n4. 按列读取并添加#号:\n   第1列: NIL#\n   第2列: OBI#\n   第3列: NAT#\n   第4列: CPY#\n   第5列: UL##\n5. 最终密文: NIL#OBI#NAT#CPY#UL#\n", "ground_truth": "NIL#OBI#NAT#CPY#UL#"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Maximum number of characters per line: 5\n- Encryption Steps:\n    - Lines are counted from 1.\n    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.\n        - For example, for the plaintext \"LIDAHELLOWORLD\", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows\n        LIDAH\n        WOLLE\n        ORLD\n    - Then read by column, each column is read from the top down, each read a column plus a \"#\", read out the contents of the final ciphertext.\n        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.\n\n \nplain text: nervii encode into encoded text: N#E#R#V#II#\nplain text: pyritic encode into encoded text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 279, "source_filename": "icl_with_rule_encode_Kor_rule17_PathCipher_en.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "PYRITIC", "extra_args": {}, "output": "1. 输入文本转换为大写并去除非字母字符: PYRITIC\n2. 确定网格大小: 2行 x 5列\n3. 开始填充网格:\n   第1行从左到右填充: P Y R I T \n   第2行从右到左填充: I C \n4. 按列读取并添加#号:\n   第1列: P##\n   第2列: Y##\n   第3列: R##\n   第4列: IC#\n   第5列: TI#\n5. 最终密文: P#Y#R#IC#TI#\n", "ground_truth": "P#Y#R#IC#TI#"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 加密步骤:\n    - 对非加密信息中的每个字母p:\n    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。\n    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。\n    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的暗文字母。例如，如果经过步骤(2)得到字母D，则映射为J。\n解密算法:\n解密规则:\n- 输入:\n    - 暗文: 大写字母字符串\n- 输出:\n    - 非加密信息: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 解密步骤(与加密步骤完全相反):\n    - (1) 对暗文中的每个字母c，在替换字母表中找到其位置，用标准字母表中对应位置的字母替换。\n    - (2) 将步骤(1)得到的字母按标准字母表顺序向后移动4位。\n    - (3) 将步骤(2)得到的字母，在反转字母表中找到其位置，然后用标准字母表中对应位置的字母替换。例如，Z映射为A，Y映射为B。\n暗文: ONOFEQMJDVQVGH 解密成非加密信息: nonculpability\n暗文: OZKCVV 解密成非加密信息是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 280, "source_filename": "icl_with_rule_decode_Kor_rule1_InverseShiftSubstitutionCipher_cn.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "OZKCVV", "extra_args": {}, "output": "标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n解密字符 O:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q\n2. 将得到的字母向后移动4位: Q -> M\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N\n\n解密字符 Z:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Z -> Z\n2. 将得到的字母向后移动4位: Z -> V\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: V -> E\n\n解密字符 K:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: K -> M\n2. 将得到的字母向后移动4位: M -> I\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: I -> R\n\n解密字符 C:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: C -> I\n2. 将得到的字母向后移动4位: I -> E\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: E -> V\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 加密步骤:\n    - 对初始文本中的每个字母p:\n    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。\n    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。\n    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的密文字母。例如，如果经过步骤(2)得到字母D，则映射为J。\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 大写字母字符串\n- 输出:\n    - 初始文本: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 解密步骤(与加密步骤完全相反):\n    - (1) 对密文中的每个字母c，在替换字母表中找到其位置，用标准字母表中对应位置的字母替换。\n    - (2) 将步骤(1)得到的字母按标准字母表顺序向后移动4位。\n    - (3) 将步骤(2)得到的字母，在反转字母表中找到其位置，然后用标准字母表中对应位置的字母替换。例如，Z映射为A，Y映射为B。\n密文: MHKVGVF 解码为初始文本: pyritic\n密文: ONOFEQMJDVQVGH 解码为初始文本: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 281, "source_filename": "icl_with_rule_decode_Kor_rule1_InverseShiftSubstitutionCipher_cn.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "ONOFEQMJDVQVGH", "extra_args": {}, "output": "标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n解密字符 O:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q\n2. 将得到的字母向后移动4位: Q -> M\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N\n\n解密字符 N:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: N -> P\n2. 将得到的字母向后移动4位: P -> L\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: L -> O\n\n解密字符 O:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q\n2. 将得到的字母向后移动4位: Q -> M\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N\n\n解密字符 F:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: F -> B\n2. 将得到的字母向后移动4位: B -> X\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: X -> C\n\n解密字符 E:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: E -> J\n2. 将得到的字母向后移动4位: J -> F\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: F -> U\n\n解密字符 Q:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Q -> S\n2. 将得到的字母向后移动4位: S -> O\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: O -> L\n\n解密字符 M:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: M -> O\n2. 将得到的字母向后移动4位: O -> K\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: K -> P\n\n解密字符 J:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: J -> D\n2. 将得到的字母向后移动4位: D -> Z\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: Z -> A\n\n解密字符 D:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: D -> C\n2. 将得到的字母向后移动4位: C -> Y\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: Y -> B\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 Q:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Q -> S\n2. 将得到的字母向后移动4位: S -> O\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: O -> L\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 G:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: G -> K\n2. 将得到的字母向后移动4位: K -> G\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: G -> T\n\n解密字符 H:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: H -> F\n2. 将得到的字母向后移动4位: F -> B\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: B -> Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 加密步骤:\n    - 对原始信息中的每个字母p:\n    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。\n    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。\n    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的加密文本字母。例如，如果经过步骤(2)得到字母D，则映射为J。\n解密算法:\n解密规则:\n- 输入:\n    - 加密文本: 大写字母字符串\n- 输出:\n    - 原始信息: 大写字母字符串\n- 准备:\n    - 标准字母表: \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    - 反转字母表: \"ZYXWVUTSRQPONMLKJIHGFEDCBA\"\n    - 替换字母表: \"RFDJUHABCEGIKLMNOPQSTVWXYZ\"\n- 解密步骤(与加密步骤完全相反):\n    - (1) 对加密文本中的每个字母c，在替换字母表中找到其位置，用标准字母表中对应位置的字母替换。\n    - (2) 将步骤(1)得到的字母按标准字母表顺序向后移动4位。\n    - (3) 将步骤(2)得到的字母，在反转字母表中找到其位置，然后用标准字母表中对应位置的字母替换。例如，Z映射为A，Y映射为B。\n加密文本: OZKCVV 解密成为原始信息: nervii\n加密文本: MHKVGVF 解密成为原始信息: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 282, "source_filename": "icl_with_rule_decode_Kor_rule1_InverseShiftSubstitutionCipher_cn.jsonl", "cipher_name": "Kor_rule1_InverseShiftSubstitutionCipher", "input": "MHKVGVF", "extra_args": {}, "output": "标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ\n反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA\n替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ\n\n解密字符 M:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: M -> O\n2. 将得到的字母向后移动4位: O -> K\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: K -> P\n\n解密字符 H:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: H -> F\n2. 将得到的字母向后移动4位: F -> B\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: B -> Y\n\n解密字符 K:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: K -> M\n2. 将得到的字母向后移动4位: M -> I\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: I -> R\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 G:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: G -> K\n2. 将得到的字母向后移动4位: K -> G\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: G -> T\n\n解密字符 V:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V\n2. 将得到的字母向后移动4位: V -> R\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I\n\n解密字符 F:\n1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: F -> B\n2. 将得到的字母向后移动4位: B -> X\n3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: X -> C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Reels\n        \n        Reel1:   \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n        Reel2:   \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n        Reel3:   \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n        Reel4:   \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n        Reel5:   \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n        Reel6:   \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n        Reel7:    \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n        Reel8:    \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n        Reel9:    \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n        Reel10:   \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n        Reel11:    \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n        Reel12:    \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n        Reel13:    \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n        Reel14:    \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n        Reel15:    \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n        Reel16:    \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n        Reel17:    \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n        Reel18:    \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n        Reel19:    \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n        Reel20:    \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n        Reel21:     \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n        Reel22:    \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n        Reel23:    \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n        Reel24:    \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n        Reel25:    \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        \n- Encryption Steps:\n    - Initially select the Reel1.\n    - For each character p in the plaintext:\n        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.\n        - If the current character is at the end of the Reel, go around to the beginning of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.\n\n \nclear text: pyritic encode into encoded text: WGTFJXF\nclear text: nonculpability encode into encoded text: AUXBKUVIXCTSSR\nclear text: pyritic encode into encoded text: WGTFJXF\nclear text: nervii encode into encoded text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 283, "source_filename": "icl_with_rule_encode_Kor_rule10_JeffersonCipher_en.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n初始化Jefferson密码轮...\n开始加密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 N 替换为下一个字符 A\n  * 在第2个密码轮上，将字符 E 替换为下一个字符 H\n  * 在第3个密码轮上，将字符 R 替换为下一个字符 T\n  * 在第4个密码轮上，将字符 V 替换为下一个字符 P\n  * 在第5个密码轮上，将字符 I 替换为下一个字符 T\n  * 在第6个密码轮上，将字符 I 替换为下一个字符 X\n加密完成，结果: AHTPTX\n", "ground_truth": "AHTPTX"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Reels\n        \n        Reel1:   \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n        Reel2:   \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n        Reel3:   \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n        Reel4:   \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n        Reel5:   \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n        Reel6:   \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n        Reel7:    \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n        Reel8:    \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n        Reel9:    \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n        Reel10:   \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n        Reel11:    \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n        Reel12:    \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n        Reel13:    \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n        Reel14:    \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n        Reel15:    \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n        Reel16:    \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n        Reel17:    \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n        Reel18:    \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n        Reel19:    \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n        Reel20:    \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n        Reel21:     \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n        Reel22:    \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n        Reel23:    \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n        Reel24:    \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n        Reel25:    \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        \n- Encryption Steps:\n    - Initially select the Reel1.\n    - For each character p in the plaintext:\n        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.\n        - If the current character is at the end of the Reel, go around to the beginning of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.\n\n \noriginal information: nervii encrypt into encoded text: AHTPTX\noriginal information: pyritic encrypt into encoded text: WGTFJXF\noriginal information: nonculpability encrypt into encoded text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 284, "source_filename": "icl_with_rule_encode_Kor_rule10_JeffersonCipher_en.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n初始化Jefferson密码轮...\n开始加密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 N 替换为下一个字符 A\n  * 在第2个密码轮上，将字符 O 替换为下一个字符 U\n  * 在第3个密码轮上，将字符 N 替换为下一个字符 X\n  * 在第4个密码轮上，将字符 C 替换为下一个字符 B\n  * 在第5个密码轮上，将字符 U 替换为下一个字符 K\n  * 在第6个密码轮上，将字符 L 替换为下一个字符 U\n  * 在第7个密码轮上，将字符 P 替换为下一个字符 V\n  * 在第8个密码轮上，将字符 A 替换为下一个字符 I\n  * 在第9个密码轮上，将字符 B 替换为下一个字符 X\n  * 在第10个密码轮上，将字符 I 替换为下一个字符 C\n  * 在第11个密码轮上，将字符 L 替换为下一个字符 T\n  * 在第12个密码轮上，将字符 I 替换为下一个字符 S\n  * 在第13个密码轮上，将字符 T 替换为下一个字符 S\n  * 在第14个密码轮上，将字符 Y 替换为下一个字符 R\n加密完成，结果: AUXBKUVIXCTSSR\n", "ground_truth": "AUXBKUVIXCTSSR"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Reels\n        \n        Reel1:   \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n        Reel2:   \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n        Reel3:   \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n        Reel4:   \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n        Reel5:   \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n        Reel6:   \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n        Reel7:    \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n        Reel8:    \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n        Reel9:    \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n        Reel10:   \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n        Reel11:    \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n        Reel12:    \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n        Reel13:    \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n        Reel14:    \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n        Reel15:    \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n        Reel16:    \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n        Reel17:    \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n        Reel18:    \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n        Reel19:    \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n        Reel20:    \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n        Reel21:     \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n        Reel22:    \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n        Reel23:    \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n        Reel24:    \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n        Reel25:    \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        \n- Encryption Steps:\n    - Initially select the Reel1.\n    - For each character p in the plaintext:\n        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.\n        - If the current character is at the end of the Reel, go around to the beginning of the Reel.\n        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.\n\n \noriginal information: nonculpability encrypt to encoded text: AUXBKUVIXCTSSR\noriginal information: nonculpability encrypt to encoded text: AUXBKUVIXCTSSR\noriginal information: nervii encrypt to encoded text: AHTPTX\noriginal information: pyritic encrypt to encoded text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 285, "source_filename": "icl_with_rule_encode_Kor_rule10_JeffersonCipher_en.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n初始化Jefferson密码轮...\n开始加密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 P 替换为下一个字符 W\n  * 在第2个密码轮上，将字符 Y 替换为下一个字符 G\n  * 在第3个密码轮上，将字符 R 替换为下一个字符 T\n  * 在第4个密码轮上，将字符 I 替换为下一个字符 F\n  * 在第5个密码轮上，将字符 T 替换为下一个字符 J\n  * 在第6个密码轮上，将字符 I 替换为下一个字符 X\n  * 在第7个密码轮上，将字符 C 替换为下一个字符 F\n加密完成，结果: WGTFJXF\n", "ground_truth": "WGTFJXF"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\nplain text: nonculpability encrypt into cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\nplain text: nonculpability encrypt into cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\nplain text: pyritic encrypt into cipher text: [112, 121, 114, 105, 116, 105, 99]\nplain text: nervii encrypt into cipher text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 286, "source_filename": "icl_encode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": "nervii", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   e, ASCII Code: 101\nCharacter:   r, ASCII Code: 114\nCharacter:   v, ASCII Code: 118\nCharacter:   i, ASCII Code: 105\nCharacter:   i, ASCII Code: 105\nASCII List: [110, 101, 114, 118, 105, 105]\n", "ground_truth": [110, 101, 114, 118, 105, 105]}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\nplain text: nervii encrypt to encrypted text: [110, 101, 114, 118, 105, 105]\nplain text: nervii encrypt to encrypted text: [110, 101, 114, 118, 105, 105]\nplain text: nonculpability encrypt to encrypted text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 287, "source_filename": "icl_encode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": "nonculpability", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   o, ASCII Code: 111\nCharacter:   n, ASCII Code: 110\nCharacter:   c, ASCII Code: 99\nCharacter:   u, ASCII Code: 117\nCharacter:   l, ASCII Code: 108\nCharacter:   p, ASCII Code: 112\nCharacter:   a, ASCII Code: 97\nCharacter:   b, ASCII Code: 98\nCharacter:   i, ASCII Code: 105\nCharacter:   l, ASCII Code: 108\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   y, ASCII Code: 121\nASCII List: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n", "ground_truth": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\nclear text: nervii encode to cipher text: [110, 101, 114, 118, 105, 105]\nclear text: nervii encode to cipher text: [110, 101, 114, 118, 105, 105]\nclear text: nonculpability encode to cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\nclear text: nervii encode to cipher text: [110, 101, 114, 118, 105, 105]\nclear text: pyritic encode to cipher text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 288, "source_filename": "icl_encode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": "pyritic", "extra_args": {}, "output": "Character:   p, ASCII Code: 112\nCharacter:   y, ASCII Code: 121\nCharacter:   r, ASCII Code: 114\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   i, ASCII Code: 105\nCharacter:   c, ASCII Code: 99\nASCII List: [112, 121, 114, 105, 116, 105, 99]\n", "ground_truth": [112, 121, 114, 105, 116, 105, 99]}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\nPlease decrypt the password according to the encryption and decryption algorithms.\n\nEncryption Algorithm:\nThe Atbash cipher encoding rule is to replace each letter with its reverse corresponding letter in the alphabet. For example, 'A' becomes 'Z', 'B' becomes 'Y', and so on. Non-letter characters remain unchanged.\n\nDecryption Algorithm:\nThe Atbash cipher decoding rule is to replace each letter with its reverse corresponding letter in the alphabet. Since the Atbash cipher is symmetric, the decoding process is the same as the encoding process. For example, 'Z' becomes 'A', 'Y' becomes 'B', and so on. Non-letter characters remain unchanged.\ncipher text: mlmxfokzyrorgb decrypt to plain text: nonculpability\ncipher text: kbirgrx decrypt to plain text: pyritic\ncipher text: kbirgrx decrypt to plain text: pyritic\ncipher text: mvierr decrypt to plain text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 289, "source_filename": "icl_with_rule_decode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "mvierr", "extra_args": {}, "output": "解码过程:\n转换单词: mvierr\n  'm' -> 'n'\n  'v' -> 'e'\n  'i' -> 'r'\n  'e' -> 'v'\n  'r' -> 'i'\n  'r' -> 'i'\n解码后的文本: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\nPlease decrypt the password according to the encryption and decryption algorithms.\n\nEncryption Algorithm:\nThe Atbash cipher encoding rule is to replace each letter with its reverse corresponding letter in the alphabet. For example, 'A' becomes 'Z', 'B' becomes 'Y', and so on. Non-letter characters remain unchanged.\n\nDecryption Algorithm:\nThe Atbash cipher decoding rule is to replace each letter with its reverse corresponding letter in the alphabet. Since the Atbash cipher is symmetric, the decoding process is the same as the encoding process. For example, 'Z' becomes 'A', 'Y' becomes 'B', and so on. Non-letter characters remain unchanged.\ncipher text: mvierr decrypt into clear text: nervii\ncipher text: kbirgrx decrypt into clear text: pyritic\ncipher text: mlmxfokzyrorgb decrypt into clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 290, "source_filename": "icl_with_rule_decode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "mlmxfokzyrorgb", "extra_args": {}, "output": "解码过程:\n转换单词: mlmxfokzyrorgb\n  'm' -> 'n'\n  'l' -> 'o'\n  'm' -> 'n'\n  'x' -> 'c'\n  'f' -> 'u'\n  'o' -> 'l'\n  'k' -> 'p'\n  'z' -> 'a'\n  'y' -> 'b'\n  'r' -> 'i'\n  'o' -> 'l'\n  'r' -> 'i'\n  'g' -> 't'\n  'b' -> 'y'\n解码后的文本: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "Please decrypt the password according to the encryption and decryption algorithms.\n\nEncryption Algorithm:\nThe Atbash cipher encoding rule is to replace each letter with its reverse corresponding letter in the alphabet. For example, 'A' becomes 'Z', 'B' becomes 'Y', and so on. Non-letter characters remain unchanged.\n\nDecryption Algorithm:\nThe Atbash cipher decoding rule is to replace each letter with its reverse corresponding letter in the alphabet. Since the Atbash cipher is symmetric, the decoding process is the same as the encoding process. For example, 'Z' becomes 'A', 'Y' becomes 'B', and so on. Non-letter characters remain unchanged.\nencoded text: kbirgrx decrypt to plain text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 291, "source_filename": "icl_with_rule_decode_AtbashCipher_en.jsonl", "cipher_name": "AtbashCipher", "input": "kbirgrx", "extra_args": {}, "output": "解码过程:\n转换单词: kbirgrx\n  'k' -> 'p'\n  'b' -> 'y'\n  'i' -> 'r'\n  'r' -> 'i'\n  'g' -> 't'\n  'r' -> 'i'\n  'x' -> 'c'\n解码后的文本: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards:\n        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.\n        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n    - **Keystream Algorithm**\n        \n        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).\n        \n        For example, starting with:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        \n        Perform the following steps to generate a character of the keystream:\n        \n        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21\n        3. Perform a \"triple cut\": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.\n        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26\n        4. Perform a \"count cut\": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.\n        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26\n        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).\n        6. Return the resulting keystream value: 14\n- Encryption Steps:\n    - cards=Initial sequence of cards\n    - For each given plaintext character p:\n        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).\n        - Generate a keystream value y for p using the Initial sequence of cards:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:\n            - z=(y+x) % 26\n        - Use the LETTERS list to return the letter corresponding to position z.\n        - Append it to the ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards(Same as encryption)\n    - Keystream Algorithm(Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - cards=Initial sequence of cards\n    - For each ciphertext character c:\n        - Convert c to its corresponding positional value z using the LETTERS (starting from 0).\n        - Generate the keystream value y for c:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - Calculate the original positional value x from ciphertext character c:\n            - x=(z-y) mod 26\n        - Use the LETTERS list to return the letter corresponding to position x.\n        - Append it to the decrypted plaintext.\ncipher text: FMMLWKFPPPDEOF decode into clear text: nonculpability\ncipher text: FMMLWKFPPPDEOF decode into clear text: nonculpability\ncipher text: AVPUUPJ decode into clear text: pyritic\ncipher text: FBPMFP decode into clear text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 292, "source_filename": "icl_with_rule_decode_Kor_rule6_SolitaireCipher_en.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "FBPMFP", "extra_args": {}, "output": "开始解密密文: FBPMFP\n\n处理第1个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (11 - 14) % 26 = 23\n解密后的字符是: N\n\n处理第2个字符 'B':\n密文字符'B'在字母表中的位置是: 20\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n解密计算: (20 - 7) % 26 = 13\n解密后的字符是: E\n\n处理第3个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n解密计算: (18 - 13) % 26 = 5\n解密后的字符是: R\n\n处理第4个字符 'M':\n密文字符'M'在字母表中的位置是: 10\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n解密计算: (10 - 17) % 26 = 19\n解密后的字符是: V\n\n处理第5个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (11 - 38) % 26 = 25\n解密后的字符是: I\n\n处理第6个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (18 - 45) % 26 = 25\n解密后的字符是: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards:\n        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.\n        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n    - **Keystream Algorithm**\n        \n        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).\n        \n        For example, starting with:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        \n        Perform the following steps to generate a character of the keystream:\n        \n        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21\n        3. Perform a \"triple cut\": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.\n        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26\n        4. Perform a \"count cut\": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.\n        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26\n        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).\n        6. Return the resulting keystream value: 14\n- Encryption Steps:\n    - cards=Initial sequence of cards\n    - For each given plaintext character p:\n        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).\n        - Generate a keystream value y for p using the Initial sequence of cards:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:\n            - z=(y+x) % 26\n        - Use the LETTERS list to return the letter corresponding to position z.\n        - Append it to the ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards(Same as encryption)\n    - Keystream Algorithm(Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - cards=Initial sequence of cards\n    - For each ciphertext character c:\n        - Convert c to its corresponding positional value z using the LETTERS (starting from 0).\n        - Generate the keystream value y for c:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - Calculate the original positional value x from ciphertext character c:\n            - x=(z-y) mod 26\n        - Use the LETTERS list to return the letter corresponding to position x.\n        - Append it to the decrypted plaintext.\ncipher text: FBPMFP decode to original information: nervii\ncipher text: FMMLWKFPPPDEOF decode to original information is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 293, "source_filename": "icl_with_rule_decode_Kor_rule6_SolitaireCipher_en.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "FMMLWKFPPPDEOF", "extra_args": {}, "output": "开始解密密文: FMMLWKFPPPDEOF\n\n处理第1个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (11 - 14) % 26 = 23\n解密后的字符是: N\n\n处理第2个字符 'M':\n密文字符'M'在字母表中的位置是: 10\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n解密计算: (10 - 7) % 26 = 3\n解密后的字符是: O\n\n处理第3个字符 'M':\n密文字符'M'在字母表中的位置是: 10\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n解密计算: (10 - 13) % 26 = 23\n解密后的字符是: N\n\n处理第4个字符 'L':\n密文字符'L'在字母表中的位置是: 24\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n解密计算: (24 - 17) % 26 = 7\n解密后的字符是: C\n\n处理第5个字符 'W':\n密文字符'W'在字母表中的位置是: 2\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (2 - 38) % 26 = 16\n解密后的字符是: U\n\n处理第6个字符 'K':\n密文字符'K'在字母表中的位置是: 17\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (17 - 45) % 26 = 24\n解密后的字符是: L\n\n处理第7个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (11 - 45) % 26 = 18\n解密后的字符是: P\n\n处理第8个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (18 - 38) % 26 = 6\n解密后的字符是: A\n\n处理第9个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 50\n解密计算: (18 - 50) % 26 = 20\n解密后的字符是: B\n\n处理第10个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (18 - 45) % 26 = 25\n解密后的字符是: I\n\n处理第11个字符 'D':\n密文字符'D'在字母表中的位置是: 1\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 29\n解密计算: (1 - 29) % 26 = 24\n解密后的字符是: L\n\n处理第12个字符 'E':\n密文字符'E'在字母表中的位置是: 13\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (13 - 14) % 26 = 25\n解密后的字符是: I\n\n处理第13个字符 'O':\n密文字符'O'在字母表中的位置是: 3\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n解密计算: (3 - 25) % 26 = 4\n解密后的字符是: T\n\n处理第14个字符 'F':\n密文字符'F'在字母表中的位置是: 11\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 25\n解密计算: (11 - 25) % 26 = 12\n解密后的字符是: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards:\n        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.\n        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]\n    - **Keystream Algorithm**\n        \n        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).\n        \n        For example, starting with:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        \n        Perform the following steps to generate a character of the keystream:\n        \n        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21\n        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:\n        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21\n        3. Perform a \"triple cut\": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.\n        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26\n        4. Perform a \"count cut\": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.\n        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26\n        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).\n        6. Return the resulting keystream value: 14\n- Encryption Steps:\n    - cards=Initial sequence of cards\n    - For each given plaintext character p:\n        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).\n        - Generate a keystream value y for p using the Initial sequence of cards:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:\n            - z=(y+x) % 26\n        - Use the LETTERS list to return the letter corresponding to position z.\n        - Append it to the ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',\n    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']\n    - Associate each letter with its position in the LETTERS (starting from 0):\n        \n        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,\n        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,\n        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25\n        \n    - Initial sequence of cards(Same as encryption)\n    - Keystream Algorithm(Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - cards=Initial sequence of cards\n    - For each ciphertext character c:\n        - Convert c to its corresponding positional value z using the LETTERS (starting from 0).\n        - Generate the keystream value y for c:\n            - y, cards = Keystream Algorithm (cards)\n            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.\n        - Calculate the original positional value x from ciphertext character c:\n            - x=(z-y) mod 26\n        - Use the LETTERS list to return the letter corresponding to position x.\n        - Append it to the decrypted plaintext.\nencrypted text: FBPMFP decrypt into clear text: nervii\nencrypted text: FBPMFP decrypt into clear text: nervii\nencrypted text: FMMLWKFPPPDEOF decrypt into clear text: nonculpability\nencrypted text: AVPUUPJ decrypt into clear text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 294, "source_filename": "icl_with_rule_decode_Kor_rule6_SolitaireCipher_en.jsonl", "cipher_name": "Kor_rule6_SolitaireCipher", "input": "AVPUUPJ", "extra_args": {}, "output": "开始解密密文: AVPUUPJ\n\n处理第1个字符 'A':\n密文字符'A'在字母表中的位置是: 6\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 14\n解密计算: (6 - 14) % 26 = 18\n解密后的字符是: P\n\n处理第2个字符 'V':\n密文字符'V'在字母表中的位置是: 19\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 7\n解密计算: (19 - 7) % 26 = 12\n解密后的字符是: Y\n\n处理第3个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 13\n解密计算: (18 - 13) % 26 = 5\n解密后的字符是: R\n\n处理第4个字符 'U':\n密文字符'U'在字母表中的位置是: 16\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 17\n解密计算: (16 - 17) % 26 = 25\n解密后的字符是: I\n\n处理第5个字符 'U':\n密文字符'U'在字母表中的位置是: 16\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 38\n解密计算: (16 - 38) % 26 = 4\n解密后的字符是: T\n\n处理第6个字符 'P':\n密文字符'P'在字母表中的位置是: 18\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (18 - 45) % 26 = 25\n解密后的字符是: I\n\n处理第7个字符 'J':\n密文字符'J'在字母表中的位置是: 0\n生成密钥流值:\n1. 移动A王牌下移一位\n2. 移动B王牌下移两位\n3. 以王牌为界进行三切\n4. 根据底牌进行计数切牌\n生成的密钥流值是: 45\n解密计算: (0 - 45) % 26 = 7\n解密后的字符是: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids:\n        \n        Grid0:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid1:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid2:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n        Grid3:\n        \n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        \n        Grid4:\n        \n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        \n        Grid5:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid6:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid7:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n- Encryption Steps:\n    - The plaintext is grouped into blocks of 5 characters, numbered from 0.\n    - For blocks of 5 characters:\n        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.\n        - For each character in the current block:\n            - If the character is \"J\", it is not encrypted and is appended directly to the encrypted block.\n            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.\n            - Add the encrypted letter to the encryption block.\n    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.\n\n \nplain text: nonculpability encrypt to cipher text: OWOMVEBGKDEDZS\nplain text: nonculpability encrypt to cipher text: OWOMVEBGKDEDZS\nplain text: nervii encrypt to cipher text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 295, "source_filename": "icl_with_rule_encode_Kor_rule7_PhillipsFigureCipher_en.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n\n处理第0个块: NERVI\n使用第0号网格\n字母N在位置(2,4)，向右下移动一格得到O\n字母E在位置(1,4)，向右下移动一格得到F\n字母R在位置(3,2)，向右下移动一格得到Y\n字母V在位置(4,0)，向右下移动一格得到H\n字母I在位置(0,2)，向右下移动一格得到D\n加密后的块: OFYHD\n\n处理第1个块: I\n使用第1号网格\n字母I在位置(1,2)，向右下移动一格得到D\n加密后的块: D\n\n最终加密结果: OFYHDD\n", "ground_truth": "OFYHDD"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids:\n        \n        Grid0:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid1:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid2:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n        Grid3:\n        \n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        \n        Grid4:\n        \n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        \n        Grid5:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid6:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid7:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n- Encryption Steps:\n    - The plaintext is grouped into blocks of 5 characters, numbered from 0.\n    - For blocks of 5 characters:\n        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.\n        - For each character in the current block:\n            - If the character is \"J\", it is not encrypted and is appended directly to the encrypted block.\n            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.\n            - Add the encrypted letter to the encryption block.\n    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.\n\n \nclear text: pyritic encrypt to cipher text: BSYDZDM\nclear text: pyritic encrypt to cipher text: BSYDZDM\nclear text: pyritic encrypt to cipher text: BSYDZDM\nclear text: nonculpability encrypt to cipher text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 296, "source_filename": "icl_with_rule_encode_Kor_rule7_PhillipsFigureCipher_en.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n\n处理第0个块: NONCU\n使用第0号网格\n字母N在位置(2,4)，向右下移动一格得到O\n字母O在位置(3,0)，向右下移动一格得到W\n字母N在位置(2,4)，向右下移动一格得到O\n字母C在位置(1,2)，向右下移动一格得到M\n字母U在位置(3,4)，向右下移动一格得到V\n加密后的块: OWOMV\n\n处理第1个块: LPABI\n使用第1号网格\n字母L在位置(1,3)，向右下移动一格得到E\n字母P在位置(1,0)，向右下移动一格得到B\n字母A在位置(2,0)，向右下移动一格得到G\n字母B在位置(2,1)，向右下移动一格得到K\n字母I在位置(1,2)，向右下移动一格得到D\n加密后的块: EBGKD\n\n处理第2个块: LITY\n使用第2号网格\n字母L在位置(2,3)，向右下移动一格得到E\n字母I在位置(2,2)，向右下移动一格得到D\n字母T在位置(0,3)，向右下移动一格得到Z\n字母Y在位置(1,3)，向右下移动一格得到S\n加密后的块: EDZS\n\n最终加密结果: OWOMVEBGKDEDZS\n", "ground_truth": "OWOMVEBGKDEDZS"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)\n    - 8 Grids:\n        \n        Grid0:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid1:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid2:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n        Grid3:\n        \n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        \n        Grid4:\n        \n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        \n        Grid5:\n        \n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        \n        Grid6:\n        \n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        | O | Q | R | T | U |\n        \n        Grid7:\n        \n        | O | Q | R | T | U |\n        | V | W | X | Y | Z |\n        | P | H | I | L | S |\n        | A | B | C | D | E |\n        | F | G | K | M | N |\n        \n- Encryption Steps:\n    - The plaintext is grouped into blocks of 5 characters, numbered from 0.\n    - For blocks of 5 characters:\n        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.\n        - For each character in the current block:\n            - If the character is \"J\", it is not encrypted and is appended directly to the encrypted block.\n            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.\n            - Add the encrypted letter to the encryption block.\n    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.\n\n \nclear text: nonculpability encrypt into encoded text: OWOMVEBGKDEDZS\nclear text: nervii encrypt into encoded text: OFYHDD\nclear text: nervii encrypt into encoded text: OFYHDD\nclear text: pyritic encrypt into encoded text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 297, "source_filename": "icl_with_rule_encode_Kor_rule7_PhillipsFigureCipher_en.jsonl", "cipher_name": "Kor_rule7_PhillipsFigureCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n\n处理第0个块: PYRIT\n使用第0号网格\n字母P在位置(0,0)，向右下移动一格得到B\n字母Y在位置(4,3)，向右下移动一格得到S\n字母R在位置(3,2)，向右下移动一格得到Y\n字母I在位置(0,2)，向右下移动一格得到D\n字母T在位置(3,3)，向右下移动一格得到Z\n加密后的块: BSYDZ\n\n处理第1个块: IC\n使用第1号网格\n字母I在位置(1,2)，向右下移动一格得到D\n字母C在位置(2,2)，向右下移动一格得到M\n加密后的块: DM\n\n最终加密结果: BSYDZDM\n", "ground_truth": "BSYDZDM"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\nclear text: pyritic secret key: {'shift': 1} encrypt into cipher text: qzsjujd\nclear text: pyritic secret key: {'shift': 1} encrypt into cipher text: qzsjujd\nclear text: nonculpability secret key: {'shift': 15} encrypt into cipher text: cdcrjaepqxaxin\nclear text: nonculpability secret key: {'shift': 15} encrypt into cipher text: cdcrjaepqxaxin\nclear text: nervii secret key: {'shift': 7} encrypt into cipher text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 298, "source_filename": "icl_encode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "nervii", "extra_args": {"shift": 7}, "output": "开始编码过程，文本为: nervii，位移数为: 7\n处理字符: n\n字符 n 向后移动 7 位，变为 u\n处理字符: e\n字符 e 向后移动 7 位，变为 l\n处理字符: r\n字符 r 向后移动 7 位，变为 y\n处理字符: v\n字符 v 向后移动 7 位，变为 c\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n编码完成，结果为: ulycpp\n", "ground_truth": "ulycpp"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\nclear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd\nclear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd\nclear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd\nclear text: nervii secret key: {'shift': 7} encode into encrypted text: ulycpp\nclear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd\nclear text: nonculpability secret key: {'shift': 15} encode into encrypted text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 299, "source_filename": "icl_encode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "nonculpability", "extra_args": {"shift": 15}, "output": "开始编码过程，文本为: nonculpability，位移数为: 15\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: o\n字符 o 向后移动 15 位，变为 d\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: c\n字符 c 向后移动 15 位，变为 r\n处理字符: u\n字符 u 向后移动 15 位，变为 j\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: p\n字符 p 向后移动 15 位，变为 e\n处理字符: a\n字符 a 向后移动 15 位，变为 p\n处理字符: b\n字符 b 向后移动 15 位，变为 q\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: t\n字符 t 向后移动 15 位，变为 i\n处理字符: y\n字符 y 向后移动 15 位，变为 n\n编码完成，结果为: cdcrjaepqxaxin\n", "ground_truth": "cdcrjaepqxaxin"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\nplain text: nonculpability secret key: {'shift': 15} encrypt to cipher text: cdcrjaepqxaxin\nplain text: nonculpability secret key: {'shift': 15} encrypt to cipher text: cdcrjaepqxaxin\nplain text: nervii secret key: {'shift': 7} encrypt to cipher text: ulycpp\nplain text: nonculpability secret key: {'shift': 15} encrypt to cipher text: cdcrjaepqxaxin\nplain text: nervii secret key: {'shift': 7} encrypt to cipher text: ulycpp\nplain text: pyritic secret key: {'shift': 1} encrypt to cipher text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 300, "source_filename": "icl_encode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "pyritic", "extra_args": {"shift": 1}, "output": "开始编码过程，文本为: pyritic，位移数为: 1\n处理字符: p\n字符 p 向后移动 1 位，变为 q\n处理字符: y\n字符 y 向后移动 1 位，变为 z\n处理字符: r\n字符 r 向后移动 1 位，变为 s\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: t\n字符 t 向后移动 1 位，变为 u\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: c\n字符 c 向后移动 1 位，变为 d\n编码完成，结果为: qzsjujd\n", "ground_truth": "qzsjujd"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\nTap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。\n解密算法:\n要解码 Tap Code，需要计算每一行和每一列中的点数。通过查找表格中的位置，可以确定对应的字母。\n暗码: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解码为明文: pyritic\n暗码: ... ...  . .....  .... ..  ..... .  .. ....  .. .... 解码为明文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 301, "source_filename": "icl_with_rule_decode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "... ...  . .....  .... ..  ..... .  .. ....  .. ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '. .....'，它表示第 1 行第 5 列，对应的字母是 'E'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '..... .'，它表示第 5 行第 1 列，对应的字母是 'V'\n步骤 5：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n最终步骤：解码完成，解码后的消息是：NERVII \n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\nTap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。\n解密算法:\n要解码 Tap Code，需要计算每一行和每一列中的点数。通过查找表格中的位置，可以确定对应的字母。\n加密信息: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密为原文是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 302, "source_filename": "icl_with_rule_decode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '... ....'，它表示第 3 行第 4 列，对应的字母是 'O'\n步骤 3：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 4：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n步骤 5：正在解码 '.... .....'，它表示第 4 行第 5 列，对应的字母是 'U'\n步骤 6：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 7：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 8：正在解码 '. .'，它表示第 1 行第 1 列，对应的字母是 'A'\n步骤 9：正在解码 '. ..'，它表示第 1 行第 2 列，对应的字母是 'B'\n步骤 10：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 11：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 12：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 13：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 14：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n最终步骤：解码完成，解码后的消息是：NONCULPABILITY \n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\nTap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。\n解密算法:\n要解码 Tap Code，需要计算每一行和每一列中的点数。通过查找表格中的位置，可以确定对应的字母。\n暗码: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密成初始文本是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 303, "source_filename": "icl_with_rule_decode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...", "extra_args": {}, "output": "步骤 1：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 2：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 5：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 7：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n最终步骤：解码完成，解码后的消息是：PYRITIC \n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\nencrypted text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into clear text: nonculpability\nencrypted text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into clear text: nonculpability\nencrypted text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into clear text: nonculpability\nencrypted text: qzsjujd secret key: {'shift': 1} decrypt into clear text: pyritic\nencrypted text: ulycpp secret key: {'shift': 7} decrypt into clear text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 304, "source_filename": "icl_decode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "ulycpp", "extra_args": {"shift": 7}, "output": "开始解码过程，文本为: ulycpp，位移数为: 7\n处理字符: u\n字符 u 向前移动 7 位，变为 n\n处理字符: l\n字符 l 向前移动 7 位，变为 e\n处理字符: y\n字符 y 向前移动 7 位，变为 r\n处理字符: c\n字符 c 向前移动 7 位，变为 v\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n解码完成，结果为: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "encoded text: qzsjujd extra parameter: {'shift': 1} decode into clear text: pyritic\nencoded text: qzsjujd extra parameter: {'shift': 1} decode into clear text: pyritic\nencoded text: cdcrjaepqxaxin extra parameter: {'shift': 15} decode into clear text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 305, "source_filename": "icl_decode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "cdcrjaepqxaxin", "extra_args": {"shift": 15}, "output": "开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: d\n字符 d 向前移动 15 位，变为 o\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: r\n字符 r 向前移动 15 位，变为 c\n处理字符: j\n字符 j 向前移动 15 位，变为 u\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: e\n字符 e 向前移动 15 位，变为 p\n处理字符: p\n字符 p 向前移动 15 位，变为 a\n处理字符: q\n字符 q 向前移动 15 位，变为 b\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: i\n字符 i 向前移动 15 位，变为 t\n处理字符: n\n字符 n 向前移动 15 位，变为 y\n解码完成，结果为: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\ncipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability\ncipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability\ncipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability\ncipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability\ncipher text: qzsjujd secret key: {'shift': 1} decrypt into plain text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 306, "source_filename": "icl_decode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "qzsjujd", "extra_args": {"shift": 1}, "output": "开始解码过程，文本为: qzsjujd，位移数为: 1\n处理字符: q\n字符 q 向前移动 1 位，变为 p\n处理字符: z\n字符 z 向前移动 1 位，变为 y\n处理字符: s\n字符 s 向前移动 1 位，变为 r\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: u\n字符 u 向前移动 1 位，变为 t\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: d\n字符 d 向前移动 1 位，变为 c\n解码完成，结果为: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则：\n1. 将输入文本中的每个字母向后移动指定的位数（shift）。\n2. 输出加密后的文本。\n解密算法:\n解码规则：\n1. 将输入文本中的每个字母向前移动指定的位数（shift）。\n2. 输出解密后的文本。\n暗文: qzsjujd 密钥: {'shift': 1} 解密成初始文本: pyritic\n暗文: qzsjujd 密钥: {'shift': 1} 解密成初始文本: pyritic\n暗文: qzsjujd 密钥: {'shift': 1} 解密成初始文本: pyritic\n暗文: ulycpp 密钥: {'shift': 7} 解密成初始文本: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 307, "source_filename": "icl_with_rule_decode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "ulycpp", "extra_args": {"shift": 7}, "output": "开始解码过程，文本为: ulycpp，位移数为: 7\n处理字符: u\n字符 u 向前移动 7 位，变为 n\n处理字符: l\n字符 l 向前移动 7 位，变为 e\n处理字符: y\n字符 y 向前移动 7 位，变为 r\n处理字符: c\n字符 c 向前移动 7 位，变为 v\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n解码完成，结果为: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则：\n1. 将输入文本中的每个字母向后移动指定的位数（shift）。\n2. 输出加密后的文本。\n解密算法:\n解码规则：\n1. 将输入文本中的每个字母向前移动指定的位数（shift）。\n2. 输出解密后的文本。\n暗码: ulycpp 额外参数: {'shift': 7} 解密成初始文本: nervii\n暗码: qzsjujd 额外参数: {'shift': 1} 解密成初始文本: pyritic\n暗码: ulycpp 额外参数: {'shift': 7} 解密成初始文本: nervii\n暗码: cdcrjaepqxaxin 额外参数: {'shift': 15} 解密成初始文本是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 308, "source_filename": "icl_with_rule_decode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "cdcrjaepqxaxin", "extra_args": {"shift": 15}, "output": "开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: d\n字符 d 向前移动 15 位，变为 o\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: r\n字符 r 向前移动 15 位，变为 c\n处理字符: j\n字符 j 向前移动 15 位，变为 u\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: e\n字符 e 向前移动 15 位，变为 p\n处理字符: p\n字符 p 向前移动 15 位，变为 a\n处理字符: q\n字符 q 向前移动 15 位，变为 b\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: i\n字符 i 向前移动 15 位，变为 t\n处理字符: n\n字符 n 向前移动 15 位，变为 y\n解码完成，结果为: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则：\n1. 将输入文本中的每个字母向后移动指定的位数（shift）。\n2. 输出加密后的文本。\n解密算法:\n解码规则：\n1. 将输入文本中的每个字母向前移动指定的位数（shift）。\n2. 输出解密后的文本。\n加密文本: qzsjujd 额外参数: {'shift': 1} 解码为非加密信息是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 309, "source_filename": "icl_with_rule_decode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "qzsjujd", "extra_args": {"shift": 1}, "output": "开始解码过程，文本为: qzsjujd，位移数为: 1\n处理字符: q\n字符 q 向前移动 1 位，变为 p\n处理字符: z\n字符 z 向前移动 1 位，变为 y\n处理字符: s\n字符 s 向前移动 1 位，变为 r\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: u\n字符 u 向前移动 1 位，变为 t\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: d\n字符 d 向前移动 1 位，变为 c\n解码完成，结果为: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):\n        - U  B  I  L  A\n        N  T  C  D  E\n        F  G  H  K  M\n        O  P  Q  R  S\n        V  W  X  Y  Z\n- Encryption Steps:\n    - Remove punctuation, spaces, and the letter J from plaintext.\n    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).\n    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.\n    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.\n\n \nclear text: nervii encrypt into encrypted text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 310, "source_filename": "icl_with_rule_encode_Kor_rule13_BifidCipher_en.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "NERVII", "extra_args": {}, "output": "加密步骤:\n1. 去除标点和空格,将J替换为I后的文本: NERVII\n2. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n3. 获取每个字母的行列坐标: [1, 1, 3, 4, 0, 0, 0, 4, 3, 0, 2, 2]\n4. 最终加密结果: TSUAOH\n", "ground_truth": "TSUAOH"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):\n        - U  B  I  L  A\n        N  T  C  D  E\n        F  G  H  K  M\n        O  P  Q  R  S\n        V  W  X  Y  Z\n- Encryption Steps:\n    - Remove punctuation, spaces, and the letter J from plaintext.\n    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).\n    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.\n    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.\n\n \nplain text: nervii encode to encrypted text: TSUAOH\nplain text: pyritic encode to encrypted text: SONTRGH\nplain text: pyritic encode to encrypted text: SONTRGH\nplain text: nonculpability encode to encrypted text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 311, "source_filename": "icl_with_rule_encode_Kor_rule13_BifidCipher_en.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "加密步骤:\n1. 去除标点和空格,将J替换为I后的文本: NONCULPABILITY\n2. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n3. 获取每个字母的行列坐标: [1, 3, 1, 1, 0, 0, 3, 0, 0, 0, 0, 0, 1, 4, 0, 0, 0, 2, 0, 3, 1, 4, 1, 2, 3, 2, 1, 3]\n4. 最终加密结果: DTUOUUEUILECQD\n", "ground_truth": "DTUOUUEUILECQD"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):\n        - U  B  I  L  A\n        N  T  C  D  E\n        F  G  H  K  M\n        O  P  Q  R  S\n        V  W  X  Y  Z\n- Encryption Steps:\n    - Remove punctuation, spaces, and the letter J from plaintext.\n    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).\n    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.\n    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.\n\n \nclear text: nervii encrypt into cipher text: TSUAOH\nclear text: nervii encrypt into cipher text: TSUAOH\nclear text: pyritic encrypt into cipher text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 312, "source_filename": "icl_with_rule_encode_Kor_rule13_BifidCipher_en.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "PYRITIC", "extra_args": {}, "output": "加密步骤:\n1. 去除标点和空格,将J替换为I后的文本: PYRITIC\n2. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n3. 获取每个字母的行列坐标: [3, 4, 3, 0, 1, 0, 1, 1, 3, 3, 2, 1, 2, 2]\n4. 最终加密结果: SONTRGH\n", "ground_truth": "SONTRGH"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - k_q_x : 12\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x corresponding to the ascii code of p.\n        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.\n    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.\n\n \nclear text: pyritic encrypt to encoded text: 960,1068,984,876,1008,876,804\nclear text: pyritic encrypt to encoded text: 960,1068,984,876,1008,876,804\nclear text: nonculpability encrypt to encoded text: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068\nclear text: nervii encrypt to encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 313, "source_filename": "icl_with_rule_encode_Kor_rule24_ECCCipher_en.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NERVII\n使用预设的k_q_x值: 12\n\n处理字符: N\n对应的ASCII值: 78\n加密后的值: 936\n\n处理字符: E\n对应的ASCII值: 69\n加密后的值: 828\n\n处理字符: R\n对应的ASCII值: 82\n加密后的值: 984\n\n处理字符: V\n对应的ASCII值: 86\n加密后的值: 1032\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n最终加密结果: 936,828,984,1032,876,876\n", "ground_truth": "936,828,984,1032,876,876"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - k_q_x : 12\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x corresponding to the ascii code of p.\n        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.\n    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.\n\n \noriginal information: pyritic encrypt to encrypted text: 960,1068,984,876,1008,876,804\noriginal information: pyritic encrypt to encrypted text: 960,1068,984,876,1008,876,804\noriginal information: pyritic encrypt to encrypted text: 960,1068,984,876,1008,876,804\noriginal information: nonculpability encrypt to encrypted text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 314, "source_filename": "icl_with_rule_encode_Kor_rule24_ECCCipher_en.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: NONCULPABILITY\n使用预设的k_q_x值: 12\n\n处理字符: N\n对应的ASCII值: 78\n加密后的值: 936\n\n处理字符: O\n对应的ASCII值: 79\n加密后的值: 948\n\n处理字符: N\n对应的ASCII值: 78\n加密后的值: 936\n\n处理字符: C\n对应的ASCII值: 67\n加密后的值: 804\n\n处理字符: U\n对应的ASCII值: 85\n加密后的值: 1020\n\n处理字符: L\n对应的ASCII值: 76\n加密后的值: 912\n\n处理字符: P\n对应的ASCII值: 80\n加密后的值: 960\n\n处理字符: A\n对应的ASCII值: 65\n加密后的值: 780\n\n处理字符: B\n对应的ASCII值: 66\n加密后的值: 792\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: L\n对应的ASCII值: 76\n加密后的值: 912\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: T\n对应的ASCII值: 84\n加密后的值: 1008\n\n处理字符: Y\n对应的ASCII值: 89\n加密后的值: 1068\n\n最终加密结果: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068\n", "ground_truth": "936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A sequence of numbers separated by ,. For example, \"y1,y2,…\".\n- Preparation:\n    - k_q_x : 12\n- Encryption Steps:\n    - For each letter p in the plaintext:\n        - Get the decimal number x corresponding to the ascii code of p.\n        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.\n    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.\n\n \noriginal information: nervii encrypt into cipher text: 936,828,984,1032,876,876\noriginal information: pyritic encrypt into cipher text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 315, "source_filename": "icl_with_rule_encode_Kor_rule24_ECCCipher_en.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n处理后的输入文本: PYRITIC\n使用预设的k_q_x值: 12\n\n处理字符: P\n对应的ASCII值: 80\n加密后的值: 960\n\n处理字符: Y\n对应的ASCII值: 89\n加密后的值: 1068\n\n处理字符: R\n对应的ASCII值: 82\n加密后的值: 984\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: T\n对应的ASCII值: 84\n加密后的值: 1008\n\n处理字符: I\n对应的ASCII值: 73\n加密后的值: 876\n\n处理字符: C\n对应的ASCII值: 67\n加密后的值: 804\n\n最终加密结果: 960,1068,984,876,1008,876,804\n", "ground_truth": "960,1068,984,876,1008,876,804"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密文本: 不含标点和空格的字符串\n- 准备:\n    - 行数: 5\n- 加密步骤:\n    - 在第一行第一个位置填入第一个非加密信息字母\n    - 两种填充方式:\n        - 向下填充: 在第一行填入\"#\"(除去第一列，因为第一个位置已经填入非加密信息字母)，然后从第二行到最后一行(第五行)向下填充非加密信息\n        - 向上填充: 从最后一行(第五行)到第二行向上填充非加密信息，然后在第一行填入\"#\"\n    - 对于非加密信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入\n    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终加密文本\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 加密文本: 不含标点和空格的字符串\n- 输出:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 行数: 5\n- 解密步骤:\n    - 根据加密文本中的*(不包括*号)，可以分成五组，依次填入五行。得到恢复的五行数据\n    - 然后按照先向下读再向上读交替的方式读取所有列。得到未清理的消息\n    - 从未清理的消息中删除#，得到最终非加密信息\n\n加密文本: N*E*R*VI*I#* 解密成为非加密信息是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 316, "source_filename": "icl_with_rule_decode_Kor_rule16_RedefenceFigureCipher_cn.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "N*E*R*VI*I#*", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的密文: N*E*R*VI*I#*\n2. 根据*号分割密文\n  分割得到6个部分\n3. 将分割后的内容填入矩阵:\n  第1行: N\n  第2行: E\n  第3行: R\n  第4行: VI\n  第5行: I#\n  第6行: \n\n4. 交替读取矩阵内容:\n  向下读取第1列\n  向上读取第2列\n  向下读取第3列\n  向上读取第4列\n  向下读取第5列\n  向上读取第6列\n  向下读取第7列\n  向上读取第8列\n  向下读取第9列\n  向上读取第10列\n  向下读取第11列\n  向上读取第12列\n5. 移除#号获得最终明文\n解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 不含标点和空格的字符串\n- 准备:\n    - 行数: 5\n- 加密步骤:\n    - 在第一行第一个位置填入第一个非加密信息字母\n    - 两种填充方式:\n        - 向下填充: 在第一行填入\"#\"(除去第一列，因为第一个位置已经填入非加密信息字母)，然后从第二行到最后一行(第五行)向下填充非加密信息\n        - 向上填充: 从最后一行(第五行)到第二行向上填充非加密信息，然后在第一行填入\"#\"\n    - 对于非加密信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入\n    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终密文\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 密文: 不含标点和空格的字符串\n- 输出:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 行数: 5\n- 解密步骤:\n    - 根据密文中的*(不包括*号)，可以分成五组，依次填入五行。得到恢复的五行数据\n    - 然后按照先向下读再向上读交替的方式读取所有列。得到未清理的消息\n    - 从未清理的消息中删除#，得到最终非加密信息\n\n密文: N*E*R*VI*I#* 解密成非加密信息: nervii\n密文: N*E*R*VI*I#* 解密成非加密信息: nervii\n密文: N*E*R*VI*I#* 解密成非加密信息: nervii\n密文: NB#*OAI*NPL*CLIY*U#T#* 解密成非加密信息是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 317, "source_filename": "icl_with_rule_decode_Kor_rule16_RedefenceFigureCipher_cn.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "NB#*OAI*NPL*CLIY*U#T#*", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的密文: NB#*OAI*NPL*CLIY*U#T#*\n2. 根据*号分割密文\n  分割得到6个部分\n3. 将分割后的内容填入矩阵:\n  第1行: NB#\n  第2行: OAI\n  第3行: NPL\n  第4行: CLIY\n  第5行: U#T#\n  第6行: \n\n4. 交替读取矩阵内容:\n  向下读取第1列\n  向上读取第2列\n  向下读取第3列\n  向上读取第4列\n  向下读取第5列\n  向上读取第6列\n  向下读取第7列\n  向上读取第8列\n  向下读取第9列\n  向上读取第10列\n  向下读取第11列\n  向上读取第12列\n  向下读取第13列\n  向上读取第14列\n  向下读取第15列\n  向上读取第16列\n  向下读取第17列\n  向上读取第18列\n  向下读取第19列\n  向上读取第20列\n  向下读取第21列\n  向上读取第22列\n5. 移除#号获得最终明文\n解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密文本: 不含标点和空格的字符串\n- 准备:\n    - 行数: 5\n- 加密步骤:\n    - 在第一行第一个位置填入第一个原始信息字母\n    - 两种填充方式:\n        - 向下填充: 在第一行填入\"#\"(除去第一列，因为第一个位置已经填入原始信息字母)，然后从第二行到最后一行(第五行)向下填充原始信息\n        - 向上填充: 从最后一行(第五行)到第二行向上填充原始信息，然后在第一行填入\"#\"\n    - 对于原始信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入\n    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终加密文本\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 加密文本: 不含标点和空格的字符串\n- 输出:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 行数: 5\n- 解密步骤:\n    - 根据加密文本中的*(不包括*号)，可以分成五组，依次填入五行。得到恢复的五行数据\n    - 然后按照先向下读再向上读交替的方式读取所有列。得到未清理的消息\n    - 从未清理的消息中删除#，得到最终原始信息\n\n加密文本: NB#*OAI*NPL*CLIY*U#T#* 解密为原始信息: nonculpability\n加密文本: NB#*OAI*NPL*CLIY*U#T#* 解密为原始信息: nonculpability\n加密文本: N*E*R*VI*I#* 解密为原始信息: nervii\n加密文本: P*Y*RC*II*T#* 解密为原始信息是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 318, "source_filename": "icl_with_rule_decode_Kor_rule16_RedefenceFigureCipher_cn.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "P*Y*RC*II*T#*", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的密文: P*Y*RC*II*T#*\n2. 根据*号分割密文\n  分割得到6个部分\n3. 将分割后的内容填入矩阵:\n  第1行: P\n  第2行: Y\n  第3行: RC\n  第4行: II\n  第5行: T#\n  第6行: \n\n4. 交替读取矩阵内容:\n  向下读取第1列\n  向上读取第2列\n  向下读取第3列\n  向上读取第4列\n  向下读取第5列\n  向上读取第6列\n  向下读取第7列\n  向上读取第8列\n  向下读取第9列\n  向上读取第10列\n  向下读取第11列\n  向上读取第12列\n  向下读取第13列\n5. 移除#号获得最终明文\n解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\nPlease decrypt the password according to the encryption algorithm and decryption algorithm.\nEncryption Algorithm:\nEncryption Rules:\n1. Shift each letter in the input text forward by a specified number of positions (shift).\n2. Output the encrypted text.\nDecryption Algorithm:\nDecryption Rules:\n1. Shift each letter in the input text backward by a specified number of positions (shift).\n2. Output the decrypted text.\nencoded text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt to original information: pyritic\nencoded text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt to original information: pyritic\nencoded text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt to original information: pyritic\nencoded text: ulycpp secret key or extra parameter: {'shift': 7} decrypt to original information: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 319, "source_filename": "icl_with_rule_decode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "ulycpp", "extra_args": {"shift": 7}, "output": "开始解码过程，文本为: ulycpp，位移数为: 7\n处理字符: u\n字符 u 向前移动 7 位，变为 n\n处理字符: l\n字符 l 向前移动 7 位，变为 e\n处理字符: y\n字符 y 向前移动 7 位，变为 r\n处理字符: c\n字符 c 向前移动 7 位，变为 v\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n解码完成，结果为: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\nPlease decrypt the password according to the encryption algorithm and decryption algorithm.\nEncryption Algorithm:\nEncryption Rules:\n1. Shift each letter in the input text forward by a specified number of positions (shift).\n2. Output the encrypted text.\nDecryption Algorithm:\nDecryption Rules:\n1. Shift each letter in the input text backward by a specified number of positions (shift).\n2. Output the decrypted text.\ncipher text: ulycpp secret key or extra parameter: {'shift': 7} decrypt into original information: nervii\ncipher text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt into original information: pyritic\ncipher text: ulycpp secret key or extra parameter: {'shift': 7} decrypt into original information: nervii\ncipher text: cdcrjaepqxaxin secret key or extra parameter: {'shift': 15} decrypt into original information is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 320, "source_filename": "icl_with_rule_decode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "cdcrjaepqxaxin", "extra_args": {"shift": 15}, "output": "开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: d\n字符 d 向前移动 15 位，变为 o\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: r\n字符 r 向前移动 15 位，变为 c\n处理字符: j\n字符 j 向前移动 15 位，变为 u\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: e\n字符 e 向前移动 15 位，变为 p\n处理字符: p\n字符 p 向前移动 15 位，变为 a\n处理字符: q\n字符 q 向前移动 15 位，变为 b\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: i\n字符 i 向前移动 15 位，变为 t\n处理字符: n\n字符 n 向前移动 15 位，变为 y\n解码完成，结果为: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\nPlease decrypt the password according to the encryption algorithm and decryption algorithm.\nEncryption Algorithm:\nEncryption Rules:\n1. Shift each letter in the input text forward by a specified number of positions (shift).\n2. Output the encrypted text.\nDecryption Algorithm:\nDecryption Rules:\n1. Shift each letter in the input text backward by a specified number of positions (shift).\n2. Output the decrypted text.\nencrypted text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt into plain text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 321, "source_filename": "icl_with_rule_decode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "qzsjujd", "extra_args": {"shift": 1}, "output": "开始解码过程，文本为: qzsjujd，位移数为: 1\n处理字符: q\n字符 q 向前移动 1 位，变为 p\n处理字符: z\n字符 z 向前移动 1 位，变为 y\n处理字符: s\n字符 s 向前移动 1 位，变为 r\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: u\n字符 u 向前移动 1 位，变为 t\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: d\n字符 d 向前移动 1 位，变为 c\n解码完成，结果为: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 大写字母字符串,不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - 5x5网格(不含字母J,行列坐标在0到4之间):\n        - U B I L A\n        N T C D E\n        F G H K M\n        O P Q R S\n        V W X Y Z\n- 加密步骤:\n    - 去除原始信息中的标点、空格和字母J\n    - 对原始信息中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)\n    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142\n    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成暗码。例如对于0142,暗码对应(0,1)是B,(4,2)是X。最终暗码是BX\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 大写字母字符串,不含标点和空格\n- 输出:\n    - 原始信息: 大写字母字符串,不含标点和空格\n- 准备:\n    - 5x5网格(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 对暗码中的每个字母,找到其在网格中对应的行列坐标(0-4)得到坐标序列\n    - 将坐标序列分成两半,前半部分为所有原始行坐标,后半部分为所有原始列坐标,从原始行坐标读一个,从原始列坐标读一个,根据原始行列坐标在网格中找到对应字母形成原始信息\n        - 例如[0,1,4,2],前半部分作为行坐标:[0,1],后半部分作为列坐标:[4,2]\n        - 交替读取行坐标和列坐标。得到(0,4),(1,2),分别在网格中找到对应字母为AC\n暗码: DTUOUUEUILECQD 解密成为原始信息: nonculpability\n暗码: TSUAOH 解密成为原始信息是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 322, "source_filename": "icl_with_rule_decode_Kor_rule13_BifidCipher_cn.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "TSUAOH", "extra_args": {}, "output": "解密步骤:\n1. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n2. 获取密文字母的坐标: [1, 3, 0, 0, 3, 2, 1, 4, 0, 4, 0, 2]\n3. 还原原始坐标: ['1', '0', '1', '4', '3', '3', '4', '0', '0', '2', '0', '2']\n4. 最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串,不含标点和空格\n- 输出:\n    - 加密信息: 大写字母字符串\n- 准备:\n    - 5x5网格(不含字母J,行列坐标在0到4之间):\n        - U B I L A\n        N T C D E\n        F G H K M\n        O P Q R S\n        V W X Y Z\n- 加密步骤:\n    - 去除原文中的标点、空格和字母J\n    - 对原文中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)\n    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142\n    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成加密信息。例如对于0142,加密信息对应(0,1)是B,(4,2)是X。最终加密信息是BX\n解密算法:\n解密规则:\n- 输入:\n    - 加密信息: 大写字母字符串,不含标点和空格\n- 输出:\n    - 原文: 大写字母字符串,不含标点和空格\n- 准备:\n    - 5x5网格(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 对加密信息中的每个字母,找到其在网格中对应的行列坐标(0-4)得到坐标序列\n    - 将坐标序列分成两半,前半部分为所有原始行坐标,后半部分为所有原始列坐标,从原始行坐标读一个,从原始列坐标读一个,根据原始行列坐标在网格中找到对应字母形成原文\n        - 例如[0,1,4,2],前半部分作为行坐标:[0,1],后半部分作为列坐标:[4,2]\n        - 交替读取行坐标和列坐标。得到(0,4),(1,2),分别在网格中找到对应字母为AC\n加密信息: SONTRGH 解密为原文: pyritic\n加密信息: DTUOUUEUILECQD 解密为原文: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 323, "source_filename": "icl_with_rule_decode_Kor_rule13_BifidCipher_cn.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "DTUOUUEUILECQD", "extra_args": {}, "output": "解密步骤:\n1. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n2. 获取密文字母的坐标: [1, 1, 0, 3, 0, 0, 1, 0, 0, 0, 1, 1, 3, 1, 3, 1, 0, 0, 0, 0, 4, 0, 2, 3, 4, 2, 2, 3]\n3. 还原原始坐标: ['1', '0', '3', '0', '1', '0', '1', '2', '0', '0', '0', '3', '3', '1', '0', '4', '0', '1', '0', '2', '0', '3', '0', '2', '1', '1', '4', '3']\n4. 最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串,不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 5x5网格(不含字母J,行列坐标在0到4之间):\n        - U B I L A\n        N T C D E\n        F G H K M\n        O P Q R S\n        V W X Y Z\n- 加密步骤:\n    - 去除原文中的标点、空格和字母J\n    - 对原文中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)\n    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142\n    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成加密文本。例如对于0142,加密文本对应(0,1)是B,(4,2)是X。最终加密文本是BX\n解密算法:\n解密规则:\n- 输入:\n    - 加密文本: 大写字母字符串,不含标点和空格\n- 输出:\n    - 原文: 大写字母字符串,不含标点和空格\n- 准备:\n    - 5x5网格(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 对加密文本中的每个字母,找到其在网格中对应的行列坐标(0-4)得到坐标序列\n    - 将坐标序列分成两半,前半部分为所有原始行坐标,后半部分为所有原始列坐标,从原始行坐标读一个,从原始列坐标读一个,根据原始行列坐标在网格中找到对应字母形成原文\n        - 例如[0,1,4,2],前半部分作为行坐标:[0,1],后半部分作为列坐标:[4,2]\n        - 交替读取行坐标和列坐标。得到(0,4),(1,2),分别在网格中找到对应字母为AC\n加密文本: DTUOUUEUILECQD 解码为原文: nonculpability\n加密文本: SONTRGH 解码为原文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 324, "source_filename": "icl_with_rule_decode_Kor_rule13_BifidCipher_cn.jsonl", "cipher_name": "Kor_rule13_BifidCipher", "input": "SONTRGH", "extra_args": {}, "output": "解密步骤:\n1. 生成的5x5矩阵:\n['U', 'B', 'I', 'L', 'A']\n['N', 'T', 'C', 'D', 'E']\n['F', 'G', 'H', 'K', 'M']\n['O', 'P', 'Q', 'R', 'S']\n['V', 'W', 'X', 'Y', 'Z']\n2. 获取密文字母的坐标: [3, 3, 1, 1, 3, 2, 2, 4, 0, 0, 1, 3, 1, 2]\n3. 还原原始坐标: ['3', '1', '4', '3', '3', '3', '0', '2', '1', '1', '0', '2', '1', '2']\n4. 最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules**\n\n- Input:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Output:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Preparation:\n    - Fixed Key: A binary string of length 8 (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Encryption Steps:\n    1. Convert Each Character to Binary:\n        - For each character in the plaintext, convert it to its ASCII value.\n        - Convert this ASCII value to an 8-bit binary string.\n    2. XOR Operation:\n        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.\n        - Ensure the key is repeated to match the length of the binary representation if needed.\n    3. Permutation:\n        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.\n    4. Combine Binary Strings:\n        - Concatenate the binary strings of all characters to form the final ciphertext.\n\n\noriginal information: pyritic encode to cipher text: 11111100111101011111100011010101111111101101010111011001\noriginal information: nonculpability encode to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\noriginal information: nervii encode to cipher text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 325, "source_filename": "icl_with_rule_encode_Kor_rule21_XORCipher_en.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "NERVII", "extra_args": {}, "output": "1. 输入文本处理为: NERVII\n\n处理字符: N\n2. 转换为8位二进制: 01001110\n3. 与密钥进行XOR运算: 11100100\n4. 应用置换表: 11010010\n\n处理字符: E\n2. 转换为8位二进制: 01000101\n3. 与密钥进行XOR运算: 11101111\n4. 应用置换表: 11011111\n\n处理字符: R\n2. 转换为8位二进制: 01010010\n3. 与密钥进行XOR运算: 11111000\n4. 应用置换表: 11111000\n\n处理字符: V\n2. 转换为8位二进制: 01010110\n3. 与密钥进行XOR运算: 11111100\n4. 应用置换表: 11111010\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n最终加密结果: 110100101101111111111000111110101101010111010101\n", "ground_truth": "110100101101111111111000111110101101010111010101"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules**\n\n- Input:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Output:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Preparation:\n    - Fixed Key: A binary string of length 8 (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Encryption Steps:\n    1. Convert Each Character to Binary:\n        - For each character in the plaintext, convert it to its ASCII value.\n        - Convert this ASCII value to an 8-bit binary string.\n    2. XOR Operation:\n        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.\n        - Ensure the key is repeated to match the length of the binary representation if needed.\n    3. Permutation:\n        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.\n    4. Combine Binary Strings:\n        - Concatenate the binary strings of all characters to form the final ciphertext.\n\n\nclear text: nervii encrypt to encoded text: 110100101101111111111000111110101101010111010101\nclear text: nervii encrypt to encoded text: 110100101101111111111000111110101101010111010101\nclear text: nervii encrypt to encoded text: 110100101101111111111000111110101101010111010101\nclear text: nonculpability encrypt to encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 326, "source_filename": "icl_with_rule_encode_Kor_rule21_XORCipher_en.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "1. 输入文本处理为: NONCULPABILITY\n\n处理字符: N\n2. 转换为8位二进制: 01001110\n3. 与密钥进行XOR运算: 11100100\n4. 应用置换表: 11010010\n\n处理字符: O\n2. 转换为8位二进制: 01001111\n3. 与密钥进行XOR运算: 11100101\n4. 应用置换表: 11010011\n\n处理字符: N\n2. 转换为8位二进制: 01001110\n3. 与密钥进行XOR运算: 11100100\n4. 应用置换表: 11010010\n\n处理字符: C\n2. 转换为8位二进制: 01000011\n3. 与密钥进行XOR运算: 11101001\n4. 应用置换表: 11011001\n\n处理字符: U\n2. 转换为8位二进制: 01010101\n3. 与密钥进行XOR运算: 11111111\n4. 应用置换表: 11111111\n\n处理字符: L\n2. 转换为8位二进制: 01001100\n3. 与密钥进行XOR运算: 11100110\n4. 应用置换表: 11010110\n\n处理字符: P\n2. 转换为8位二进制: 01010000\n3. 与密钥进行XOR运算: 11111010\n4. 应用置换表: 11111100\n\n处理字符: A\n2. 转换为8位二进制: 01000001\n3. 与密钥进行XOR运算: 11101011\n4. 应用置换表: 11011101\n\n处理字符: B\n2. 转换为8位二进制: 01000010\n3. 与密钥进行XOR运算: 11101000\n4. 应用置换表: 11011000\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: L\n2. 转换为8位二进制: 01001100\n3. 与密钥进行XOR运算: 11100110\n4. 应用置换表: 11010110\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: T\n2. 转换为8位二进制: 01010100\n3. 与密钥进行XOR运算: 11111110\n4. 应用置换表: 11111110\n\n处理字符: Y\n2. 转换为8位二进制: 01011001\n3. 与密钥进行XOR运算: 11110011\n4. 应用置换表: 11110101\n\n最终加密结果: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\n", "ground_truth": "1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules**\n\n- Input:\n    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.\n- Output:\n    - Ciphertext: A binary string containing only 0s and 1s.\n- Preparation:\n    - Fixed Key: A binary string of length 8 (e.g., '10101010').\n    - Permutation Table:\n        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)\n        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)\n- Encryption Steps:\n    1. Convert Each Character to Binary:\n        - For each character in the plaintext, convert it to its ASCII value.\n        - Convert this ASCII value to an 8-bit binary string.\n    2. XOR Operation:\n        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.\n        - Ensure the key is repeated to match the length of the binary representation if needed.\n    3. Permutation:\n        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.\n    4. Combine Binary Strings:\n        - Concatenate the binary strings of all characters to form the final ciphertext.\n\n\nplain text: nonculpability encrypt to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\nplain text: nonculpability encrypt to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\nplain text: nonculpability encrypt to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101\nplain text: pyritic encrypt to cipher text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 327, "source_filename": "icl_with_rule_encode_Kor_rule21_XORCipher_en.jsonl", "cipher_name": "Kor_rule21_XORCipher", "input": "PYRITIC", "extra_args": {}, "output": "1. 输入文本处理为: PYRITIC\n\n处理字符: P\n2. 转换为8位二进制: 01010000\n3. 与密钥进行XOR运算: 11111010\n4. 应用置换表: 11111100\n\n处理字符: Y\n2. 转换为8位二进制: 01011001\n3. 与密钥进行XOR运算: 11110011\n4. 应用置换表: 11110101\n\n处理字符: R\n2. 转换为8位二进制: 01010010\n3. 与密钥进行XOR运算: 11111000\n4. 应用置换表: 11111000\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: T\n2. 转换为8位二进制: 01010100\n3. 与密钥进行XOR运算: 11111110\n4. 应用置换表: 11111110\n\n处理字符: I\n2. 转换为8位二进制: 01001001\n3. 与密钥进行XOR运算: 11100011\n4. 应用置换表: 11010101\n\n处理字符: C\n2. 转换为8位二进制: 01000011\n3. 与密钥进行XOR运算: 11101001\n4. 应用置换表: 11011001\n\n最终加密结果: 11111100111101011111100011010101111111101101010111011001\n", "ground_truth": "11111100111101011111100011010101111111101101010111011001"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\nPlease encrypt the plaintext according to the encryption algorithm.\nEncryption rules:\n1. Shift each letter in the input text forward by a specified number of positions (shift).\n2. Output the encrypted text.\nclear text: nervii secret key: {'shift': 7} encrypt into encoded text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 328, "source_filename": "icl_with_rule_encode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "nervii", "extra_args": {"shift": 7}, "output": "开始编码过程，文本为: nervii，位移数为: 7\n处理字符: n\n字符 n 向后移动 7 位，变为 u\n处理字符: e\n字符 e 向后移动 7 位，变为 l\n处理字符: r\n字符 r 向后移动 7 位，变为 y\n处理字符: v\n字符 v 向后移动 7 位，变为 c\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n处理字符: i\n字符 i 向后移动 7 位，变为 p\n编码完成，结果为: ulycpp\n", "ground_truth": "ulycpp"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\nPlease encrypt the plaintext according to the encryption algorithm.\nEncryption rules:\n1. Shift each letter in the input text forward by a specified number of positions (shift).\n2. Output the encrypted text.\noriginal information: nervii secret key: {'shift': 7} encrypt into encoded text: ulycpp\noriginal information: nervii secret key: {'shift': 7} encrypt into encoded text: ulycpp\noriginal information: nonculpability secret key: {'shift': 15} encrypt into encoded text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 329, "source_filename": "icl_with_rule_encode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "nonculpability", "extra_args": {"shift": 15}, "output": "开始编码过程，文本为: nonculpability，位移数为: 15\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: o\n字符 o 向后移动 15 位，变为 d\n处理字符: n\n字符 n 向后移动 15 位，变为 c\n处理字符: c\n字符 c 向后移动 15 位，变为 r\n处理字符: u\n字符 u 向后移动 15 位，变为 j\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: p\n字符 p 向后移动 15 位，变为 e\n处理字符: a\n字符 a 向后移动 15 位，变为 p\n处理字符: b\n字符 b 向后移动 15 位，变为 q\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: l\n字符 l 向后移动 15 位，变为 a\n处理字符: i\n字符 i 向后移动 15 位，变为 x\n处理字符: t\n字符 t 向后移动 15 位，变为 i\n处理字符: y\n字符 y 向后移动 15 位，变为 n\n编码完成，结果为: cdcrjaepqxaxin\n", "ground_truth": "cdcrjaepqxaxin"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\nPlease encrypt the plaintext according to the encryption algorithm.\nEncryption rules:\n1. Shift each letter in the input text forward by a specified number of positions (shift).\n2. Output the encrypted text.\nclear text: nonculpability extra parameter: {'shift': 15} encode to encoded text: cdcrjaepqxaxin\nclear text: nonculpability extra parameter: {'shift': 15} encode to encoded text: cdcrjaepqxaxin\nclear text: pyritic extra parameter: {'shift': 1} encode to encoded text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 330, "source_filename": "icl_with_rule_encode_Caesar_Cipher_en.jsonl", "cipher_name": "Caesar_Cipher", "input": "pyritic", "extra_args": {"shift": 1}, "output": "开始编码过程，文本为: pyritic，位移数为: 1\n处理字符: p\n字符 p 向后移动 1 位，变为 q\n处理字符: y\n字符 y 向后移动 1 位，变为 z\n处理字符: r\n字符 r 向后移动 1 位，变为 s\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: t\n字符 t 向后移动 1 位，变为 u\n处理字符: i\n字符 i 向后移动 1 位，变为 j\n处理字符: c\n字符 c 向后移动 1 位，变为 d\n编码完成，结果为: qzsjujd\n", "ground_truth": "qzsjujd"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - k_q_x值: 12\n- 加密步骤:\n    - 对明文中的每个字母p:\n        - 获取字母p对应的ASCII码值x\n        - 计算x * k_q_x作为该字母对应的密文数字y\n    - 最后将所有y值用逗号连接得到最终密文\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 输出:\n    - 明文: 不含标点和空格的大写字母字符串\n- 准备:\n    - k_q_x值: 12\n- 解密步骤:\n    - 对密文中的每个数字c:\n        - 计算z = c // k_q_x，其中//表示整数除法运算，返回商的整数部分\n        - 根据z对应的ASCII码值找到对应的字母作为明文字母p\n    - 最后将所有p连接得到最终明文\n密文: 960,1068,984,876,1008,876,804 解密成为明文: pyritic\n密文: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 解密成为明文: nonculpability\n密文: 960,1068,984,876,1008,876,804 解密成为明文: pyritic\n密文: 936,828,984,1032,876,876 解密成为明文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 331, "source_filename": "icl_with_rule_decode_Kor_rule24_ECCCipher_cn.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "936,828,984,1032,876,876", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 936,828,984,1032,876,876\n使用预设的k_q_x值: 12\n\n处理数字: 936\n除以k_q_x后的值: 78\n对应的字符: N\n\n处理数字: 828\n除以k_q_x后的值: 69\n对应的字符: E\n\n处理数字: 984\n除以k_q_x后的值: 82\n对应的字符: R\n\n处理数字: 1032\n除以k_q_x后的值: 86\n对应的字符: V\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - k_q_x值: 12\n- 加密步骤:\n    - 对原始信息中的每个字母p:\n        - 获取字母p对应的ASCII码值x\n        - 计算x * k_q_x作为该字母对应的暗码数字y\n    - 最后将所有y值用逗号连接得到最终暗码\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 输出:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - k_q_x值: 12\n- 解密步骤:\n    - 对暗码中的每个数字c:\n        - 计算z = c // k_q_x，其中//表示整数除法运算，返回商的整数部分\n        - 根据z对应的ASCII码值找到对应的字母作为原始信息字母p\n    - 最后将所有p连接得到最终原始信息\n暗码: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 解密成原始信息: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 332, "source_filename": "icl_with_rule_decode_Kor_rule24_ECCCipher_cn.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068\n使用预设的k_q_x值: 12\n\n处理数字: 936\n除以k_q_x后的值: 78\n对应的字符: N\n\n处理数字: 948\n除以k_q_x后的值: 79\n对应的字符: O\n\n处理数字: 936\n除以k_q_x后的值: 78\n对应的字符: N\n\n处理数字: 804\n除以k_q_x后的值: 67\n对应的字符: C\n\n处理数字: 1020\n除以k_q_x后的值: 85\n对应的字符: U\n\n处理数字: 912\n除以k_q_x后的值: 76\n对应的字符: L\n\n处理数字: 960\n除以k_q_x后的值: 80\n对应的字符: P\n\n处理数字: 780\n除以k_q_x后的值: 65\n对应的字符: A\n\n处理数字: 792\n除以k_q_x后的值: 66\n对应的字符: B\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 912\n除以k_q_x后的值: 76\n对应的字符: L\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 1008\n除以k_q_x后的值: 84\n对应的字符: T\n\n处理数字: 1068\n除以k_q_x后的值: 89\n对应的字符: Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 准备:\n    - k_q_x值: 12\n- 加密步骤:\n    - 对非加密信息中的每个字母p:\n        - 获取字母p对应的ASCII码值x\n        - 计算x * k_q_x作为该字母对应的暗码数字y\n    - 最后将所有y值用逗号连接得到最终暗码\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 由逗号分隔的数字序列，例如\"y1,y2,...\"\n- 输出:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - k_q_x值: 12\n- 解密步骤:\n    - 对暗码中的每个数字c:\n        - 计算z = c // k_q_x，其中//表示整数除法运算，返回商的整数部分\n        - 根据z对应的ASCII码值找到对应的字母作为非加密信息字母p\n    - 最后将所有p连接得到最终非加密信息\n暗码: 936,828,984,1032,876,876 解密成为非加密信息: nervii\n暗码: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 解密成为非加密信息: nonculpability\n暗码: 960,1068,984,876,1008,876,804 解密成为非加密信息: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 333, "source_filename": "icl_with_rule_decode_Kor_rule24_ECCCipher_cn.jsonl", "cipher_name": "Kor_rule24_ECCCipher", "input": "960,1068,984,876,1008,876,804", "extra_args": {}, "output": "开始解密过程...\n收到的加密文本: 960,1068,984,876,1008,876,804\n使用预设的k_q_x值: 12\n\n处理数字: 960\n除以k_q_x后的值: 80\n对应的字符: P\n\n处理数字: 1068\n除以k_q_x后的值: 89\n对应的字符: Y\n\n处理数字: 984\n除以k_q_x后的值: 82\n对应的字符: R\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 1008\n除以k_q_x后的值: 84\n对应的字符: T\n\n处理数字: 876\n除以k_q_x后的值: 73\n对应的字符: I\n\n处理数字: 804\n除以k_q_x后的值: 67\n对应的字符: C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: A string of uppercase letters and spaces.\n- Output:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - S_BOX\n        - S_BOX = {\n        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n        }\n- Encryption Steps:\n    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \\x00 (null character) to make its length a multiple of 8 bytes.\n    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.\n    3. Block Encryption:\n        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.\n        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.\n        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.\n    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.\n    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.\n\n**Decryption Rules:**\n\n- Input:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Output:\n    - Plaintext: A string of uppercase letters and spaces.\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - INV_S_BOX\n        - INV_S_BOX = {\n        0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,\n        0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,\n        0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,\n        0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F\n        }\n- Decryption Steps:\n    1. Block Splitting: Divide the ciphertext into 16-character (8-byte) blocks (each block is 8 bytes long).\n    2. Block Decryption:\n        - Convert Hex to Bytes: Convert each block from hexadecimal string to bytes.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Inverse Permute: Perform an inverse permutation by rotating each byte right by 1 bit.\n        - Substitute: Use INV_S_BOX to replace the high four and low four bits in a block of bytes and splice them together.\n        - XOR with Key: Apply XOR between the byte block and the fixed key again.\n    3. Convert Bytes to Text: Convert the decrypted byte blocks back to text using ASCII decoding.\n    4. Remove Padding: Remove any padding characters (\\x00) from the end of the decrypted plaintext.\n    5. Concatenate: Concatenate all the decrypted blocks to form the final plaintext.\nencoded text: 293637348A3E96283B3E2B30808F9784 decrypt to plain text: nonculpability\nencoded text: 293637348A3E96283B3E2B30808F9784 decrypt to plain text: nonculpability\nencoded text: 2932869B372E9784 decrypt to plain text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 334, "source_filename": "icl_with_rule_decode_Kor_rule22_SBOXCipher_en.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "2932869B372E9784", "extra_args": {}, "output": "开始解密过程...\n1. 接收到的密文: 2932869B372E9784\n2. 开始分块处理...\n\n处理第1个块: 2932869B372E9784\n  - 转换为字节: 2932869B372E9784\n  - 与密钥XOR后: 1800B5AF0218A0BC\n  - 右移置换后: 0C00DAD7010C505E\n  - 逆S盒替换后: 7F7761627C7F3738\n  - 最终与密钥XOR后: 4E45525649490000\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: A string of uppercase letters and spaces.\n- Output:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - S_BOX\n        - S_BOX = {\n        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n        }\n- Encryption Steps:\n    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \\x00 (null character) to make its length a multiple of 8 bytes.\n    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.\n    3. Block Encryption:\n        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.\n        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.\n        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.\n    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.\n    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.\n\n**Decryption Rules:**\n\n- Input:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Output:\n    - Plaintext: A string of uppercase letters and spaces.\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - INV_S_BOX\n        - INV_S_BOX = {\n        0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,\n        0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,\n        0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,\n        0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F\n        }\n- Decryption Steps:\n    1. Block Splitting: Divide the ciphertext into 16-character (8-byte) blocks (each block is 8 bytes long).\n    2. Block Decryption:\n        - Convert Hex to Bytes: Convert each block from hexadecimal string to bytes.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Inverse Permute: Perform an inverse permutation by rotating each byte right by 1 bit.\n        - Substitute: Use INV_S_BOX to replace the high four and low four bits in a block of bytes and splice them together.\n        - XOR with Key: Apply XOR between the byte block and the fixed key again.\n    3. Convert Bytes to Text: Convert the decrypted byte blocks back to text using ASCII decoding.\n    4. Remove Padding: Remove any padding characters (\\x00) from the end of the decrypted plaintext.\n    5. Concatenate: Concatenate all the decrypted blocks to form the final plaintext.\nencoded text: 2932869B372E9784 decode to clear text: nervii\nencoded text: 2932869B372E9784 decode to clear text: nervii\nencoded text: 849F8630802E2584 decode to clear text: pyritic\nencoded text: 293637348A3E96283B3E2B30808F9784 decode to clear text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 335, "source_filename": "icl_with_rule_decode_Kor_rule22_SBOXCipher_en.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "293637348A3E96283B3E2B30808F9784", "extra_args": {}, "output": "开始解密过程...\n1. 接收到的密文: 293637348A3E96283B3E2B30808F9784\n2. 开始分块处理...\n\n处理第1个块: 293637348A3E9628\n  - 转换为字节: 293637348A3E9628\n  - 与密钥XOR后: 18040400BF08A110\n  - 右移置换后: 0C020200DF04D008\n  - 逆S盒替换后: 7F7D7D77607A6779\n  - 最终与密钥XOR后: 4E4F4E43554C5041\n\n处理第2个块: 3B3E2B30808F9784\n  - 转换为字节: 3B3E2B30808F9784\n  - 与密钥XOR后: 0A0C1804B5B9A0BC\n  - 右移置换后: 05060C02DADC505E\n  - 逆S盒替换后: 737B7F7D616F3738\n  - 最终与密钥XOR后: 42494C4954590000\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: A string of uppercase letters and spaces.\n- Output:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - S_BOX\n        - S_BOX = {\n        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n        }\n- Encryption Steps:\n    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \\x00 (null character) to make its length a multiple of 8 bytes.\n    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.\n    3. Block Encryption:\n        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.\n        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.\n        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.\n    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.\n    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.\n\n**Decryption Rules:**\n\n- Input:\n    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).\n- Output:\n    - Plaintext: A string of uppercase letters and spaces.\n- Preparation:\n    - Fixed key\n        - KEY = b'1234567890ABCDEF'\n    - INV_S_BOX\n        - INV_S_BOX = {\n        0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,\n        0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,\n        0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,\n        0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F\n        }\n- Decryption Steps:\n    1. Block Splitting: Divide the ciphertext into 16-character (8-byte) blocks (each block is 8 bytes long).\n    2. Block Decryption:\n        - Convert Hex to Bytes: Convert each block from hexadecimal string to bytes.\n        - XOR with Key: Apply XOR between the byte block and the fixed key.\n        - Inverse Permute: Perform an inverse permutation by rotating each byte right by 1 bit.\n        - Substitute: Use INV_S_BOX to replace the high four and low four bits in a block of bytes and splice them together.\n        - XOR with Key: Apply XOR between the byte block and the fixed key again.\n    3. Convert Bytes to Text: Convert the decrypted byte blocks back to text using ASCII decoding.\n    4. Remove Padding: Remove any padding characters (\\x00) from the end of the decrypted plaintext.\n    5. Concatenate: Concatenate all the decrypted blocks to form the final plaintext.\nencoded text: 2932869B372E9784 decode to clear text: nervii\nencoded text: 293637348A3E96283B3E2B30808F9784 decode to clear text: nonculpability\nencoded text: 849F8630802E2584 decode to clear text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 336, "source_filename": "icl_with_rule_decode_Kor_rule22_SBOXCipher_en.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "849F8630802E2584", "extra_args": {}, "output": "开始解密过程...\n1. 接收到的密文: 849F8630802E2584\n2. 开始分块处理...\n\n处理第1个块: 849F8630802E2584\n  - 转换为字节: 849F8630802E2584\n  - 与密钥XOR后: B5ADB504B51812BC\n  - 右移置换后: DAD6DA02DA0C095E\n  - 逆S盒替换后: 616B617D617F7438\n  - 最终与密钥XOR后: 5059524954494300\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 隐文: 不含标点和空格的字符串\n- 准备:\n    - 每行最大字符数: 5\n- 加密步骤:\n    - 行数从1开始计数\n    - 原文按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母\n        - 例如，对于原文\"LIDAHELLOWORLD\"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按列读取，每列从上到下读取，每读完一列加一个\"#\"，读出的内容即为最终隐文\n        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#\n解密算法:\n解密规则:\n- 输入:\n    - 隐文: 不含标点和空格的字符串\n- 输出:\n    - 原文: 大写字母字符串，不含标点和空格\n- 准备:\n    - 行数: 5\n- 解密步骤(与加密步骤完全相反):\n    - 对隐文中的每个字符，从上到下写入；如果遇到#，则切换到下一列继续写入，直到整个隐文写完\n        - 例如，对于隐文LWO#IOR#DLL#ALD#HE#，写出如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按照奇数行从左到右读，偶数行从右到左读的顺序依次读取，最终结果即为解密后的原文\n        - 例如，对于上述写好的内容，第一行从左到右读为LIDAH，第二行从右到左读为ELLOW，第三行从左到右读为ORLD，最终全部内容连接起来，解密原文为LIDAHELLOWORLD\n隐文: N#E#R#V#II# 解密成为原文: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 337, "source_filename": "icl_with_rule_decode_Kor_rule17_PathCipher_cn.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "N#E#R#V#II#", "extra_args": {}, "output": "1. 收到密文: N#E#R#V#II#\n2. 将密文分割成5列: ['N', 'E', 'R', 'V', 'II']\n3. 按列填充网格:\n   第1列: N \n   第2列: E \n   第3列: R \n   第4列: V \n   第5列: I I \n4. 按特定顺序读取:\n   第1行从左到右读取: N E R V I \n   第2行从右到左读取: I \n5. 最终明文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗文: 不含标点和空格的字符串\n- 准备:\n    - 每行最大字符数: 5\n- 加密步骤:\n    - 行数从1开始计数\n    - 原文按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母\n        - 例如，对于原文\"LIDAHELLOWORLD\"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按列读取，每列从上到下读取，每读完一列加一个\"#\"，读出的内容即为最终暗文\n        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#\n解密算法:\n解密规则:\n- 输入:\n    - 暗文: 不含标点和空格的字符串\n- 输出:\n    - 原文: 大写字母字符串，不含标点和空格\n- 准备:\n    - 行数: 5\n- 解密步骤(与加密步骤完全相反):\n    - 对暗文中的每个字符，从上到下写入；如果遇到#，则切换到下一列继续写入，直到整个暗文写完\n        - 例如，对于暗文LWO#IOR#DLL#ALD#HE#，写出如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按照奇数行从左到右读，偶数行从右到左读的顺序依次读取，最终结果即为解密后的原文\n        - 例如，对于上述写好的内容，第一行从左到右读为LIDAH，第二行从右到左读为ELLOW，第三行从左到右读为ORLD，最终全部内容连接起来，解密原文为LIDAHELLOWORLD\n暗文: NIL#OBI#NAT#CPY#UL# 解密成为原文: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 338, "source_filename": "icl_with_rule_decode_Kor_rule17_PathCipher_cn.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "NIL#OBI#NAT#CPY#UL#", "extra_args": {}, "output": "1. 收到密文: NIL#OBI#NAT#CPY#UL#\n2. 将密文分割成5列: ['NIL', 'OBI', 'NAT', 'CPY', 'UL']\n3. 按列填充网格:\n   第1列: N I L \n   第2列: O B I \n   第3列: N A T \n   第4列: C P Y \n   第5列: U L \n4. 按特定顺序读取:\n   第1行从左到右读取: N O N C U \n   第2行从右到左读取: L P A B I \n   第3行从左到右读取: L I T Y \n5. 最终明文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密信息: 不含标点和空格的字符串\n- 准备:\n    - 每行最大字符数: 5\n- 加密步骤:\n    - 行数从1开始计数\n    - 明文按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母\n        - 例如，对于明文\"LIDAHELLOWORLD\"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按列读取，每列从上到下读取，每读完一列加一个\"#\"，读出的内容即为最终加密信息\n        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#\n解密算法:\n解密规则:\n- 输入:\n    - 加密信息: 不含标点和空格的字符串\n- 输出:\n    - 明文: 大写字母字符串，不含标点和空格\n- 准备:\n    - 行数: 5\n- 解密步骤(与加密步骤完全相反):\n    - 对加密信息中的每个字符，从上到下写入；如果遇到#，则切换到下一列继续写入，直到整个加密信息写完\n        - 例如，对于加密信息LWO#IOR#DLL#ALD#HE#，写出如下\n        LIDAH\n        WOLLE\n        ORLD\n    - 然后按照奇数行从左到右读，偶数行从右到左读的顺序依次读取，最终结果即为解密后的明文\n        - 例如，对于上述写好的内容，第一行从左到右读为LIDAH，第二行从右到左读为ELLOW，第三行从左到右读为ORLD，最终全部内容连接起来，解密明文为LIDAHELLOWORLD\n加密信息: N#E#R#V#II# 解密成明文: nervii\n加密信息: NIL#OBI#NAT#CPY#UL# 解密成明文: nonculpability\n加密信息: P#Y#R#IC#TI# 解密成明文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 339, "source_filename": "icl_with_rule_decode_Kor_rule17_PathCipher_cn.jsonl", "cipher_name": "Kor_rule17_PathCipher", "input": "P#Y#R#IC#TI#", "extra_args": {}, "output": "1. 收到密文: P#Y#R#IC#TI#\n2. 将密文分割成5列: ['P', 'Y', 'R', 'IC', 'TI']\n3. 按列填充网格:\n   第1列: P \n   第2列: Y \n   第3列: R \n   第4列: I C \n   第5列: T I \n4. 按特定顺序读取:\n   第1行从左到右读取: P Y R I T \n   第2行从右到左读取: I C \n5. 最终明文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Encryption Steps:\n    - For each given plaintext character p:\n        - Let  x  denote its position in the affine alphabet.\n        - Apply the affine coding function to compute y:\n            - y = (Ax + B) mod 26\n        - Find the corresponding letter in the affine alphabet at position  y , forming the encrypted message.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Decryption Steps (exact opposite of encryption steps):\n    - For each given ciphertext character c:\n        - Let y denote its position in the affine alphabet.\n        - Calculate x:\n            - x = A_inv * (y - B) % n.\n        - Replace c with the letter at position x  in the affine alphabet to form the decrypted message.\nencrypted text: XEXAKJQMSLJLVH decode into original information: nonculpability\nencrypted text: QHOLVLA decode into original information: pyritic\nencrypted text: XEXAKJQMSLJLVH decode into original information: nonculpability\nencrypted text: XYOFLL decode into original information: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 340, "source_filename": "icl_with_rule_decode_Kor_rule5_AffineCipher_en.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "XYOFLL", "extra_args": {}, "output": "加密文本: XYOFLL\n开始解密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5, A的逆元=9\n字符 X -> 位置 0 -> 计算9*(0-5)%26=7 -> 解密为 N\n字符 Y -> 位置 22 -> 计算9*(22-5)%26=23 -> 解密为 E\n字符 O -> 位置 6 -> 计算9*(6-5)%26=9 -> 解密为 R\n字符 F -> 位置 15 -> 计算9*(15-5)%26=12 -> 解密为 V\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n解密完成，结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Encryption Steps:\n    - For each given plaintext character p:\n        - Let  x  denote its position in the affine alphabet.\n        - Apply the affine coding function to compute y:\n            - y = (Ax + B) mod 26\n        - Find the corresponding letter in the affine alphabet at position  y , forming the encrypted message.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Decryption Steps (exact opposite of encryption steps):\n    - For each given ciphertext character c:\n        - Let y denote its position in the affine alphabet.\n        - Calculate x:\n            - x = A_inv * (y - B) % n.\n        - Replace c with the letter at position x  in the affine alphabet to form the decrypted message.\nencrypted text: XYOFLL decode to clear text: nervii\nencrypted text: QHOLVLA decode to clear text: pyritic\nencrypted text: XYOFLL decode to clear text: nervii\nencrypted text: XEXAKJQMSLJLVH decode to clear text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 341, "source_filename": "icl_with_rule_decode_Kor_rule5_AffineCipher_en.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "XEXAKJQMSLJLVH", "extra_args": {}, "output": "加密文本: XEXAKJQMSLJLVH\n开始解密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5, A的逆元=9\n字符 X -> 位置 0 -> 计算9*(0-5)%26=7 -> 解密为 N\n字符 E -> 位置 23 -> 计算9*(23-5)%26=6 -> 解密为 O\n字符 X -> 位置 0 -> 计算9*(0-5)%26=7 -> 解密为 N\n字符 A -> 位置 16 -> 计算9*(16-5)%26=21 -> 解密为 C\n字符 K -> 位置 17 -> 计算9*(17-5)%26=4 -> 解密为 U\n字符 J -> 位置 2 -> 计算9*(2-5)%26=25 -> 解密为 L\n字符 Q -> 位置 3 -> 计算9*(3-5)%26=8 -> 解密为 P\n字符 M -> 位置 1 -> 计算9*(1-5)%26=16 -> 解密为 A\n字符 S -> 位置 18 -> 计算9*(18-5)%26=13 -> 解密为 B\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 J -> 位置 2 -> 计算9*(2-5)%26=25 -> 解密为 L\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 V -> 位置 12 -> 计算9*(12-5)%26=11 -> 解密为 T\n字符 H -> 位置 19 -> 计算9*(19-5)%26=22 -> 解密为 Y\n解密完成，结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Encryption Steps:\n    - For each given plaintext character p:\n        - Let  x  denote its position in the affine alphabet.\n        - Apply the affine coding function to compute y:\n            - y = (Ax + B) mod 26\n        - Find the corresponding letter in the affine alphabet at position  y , forming the encrypted message.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Decryption Steps (exact opposite of encryption steps):\n    - For each given ciphertext character c:\n        - Let y denote its position in the affine alphabet.\n        - Calculate x:\n            - x = A_inv * (y - B) % n.\n        - Replace c with the letter at position x  in the affine alphabet to form the decrypted message.\ncipher text: XYOFLL decrypt to clear text: nervii\ncipher text: XYOFLL decrypt to clear text: nervii\ncipher text: XYOFLL decrypt to clear text: nervii\ncipher text: QHOLVLA decrypt to clear text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 342, "source_filename": "icl_with_rule_decode_Kor_rule5_AffineCipher_en.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "QHOLVLA", "extra_args": {}, "output": "加密文本: QHOLVLA\n开始解密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5, A的逆元=9\n字符 Q -> 位置 3 -> 计算9*(3-5)%26=8 -> 解密为 P\n字符 H -> 位置 19 -> 计算9*(19-5)%26=22 -> 解密为 Y\n字符 O -> 位置 6 -> 计算9*(6-5)%26=9 -> 解密为 R\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 V -> 位置 12 -> 计算9*(12-5)%26=11 -> 解密为 T\n字符 L -> 位置 25 -> 计算9*(25-5)%26=24 -> 解密为 I\n字符 A -> 位置 16 -> 计算9*(16-5)%26=21 -> 解密为 C\n解密完成，结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密方案概述：将字符串转换为ASCII编码。\n解密算法:\n解密方案概述：遍历ASCII编码字符串,返回转换后的字符串\n暗码: [112, 121, 114, 105, 116, 105, 99] 解密成为非加密信息: pyritic\n暗码: [112, 121, 114, 105, 116, 105, 99] 解密成为非加密信息: pyritic\n暗码: [110, 101, 114, 118, 105, 105] 解密成为非加密信息: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 343, "source_filename": "icl_with_rule_decode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": [110, 101, 114, 118, 105, 105], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 101, Character: e\nASCII   Code: 114, Character: r\nASCII   Code: 118, Character: v\nASCII   Code: 105, Character: i\nASCII   Code: 105, Character: i\nResult: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密方案概述：将字符串转换为ASCII编码。\n解密算法:\n解密方案概述：遍历ASCII编码字符串,返回转换后的字符串\n加密文本: [110, 101, 114, 118, 105, 105] 解密成为初始文本: nervii\n加密文本: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] 解密成为初始文本: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 344, "source_filename": "icl_with_rule_decode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 111, Character: o\nASCII   Code: 110, Character: n\nASCII   Code: 99, Character: c\nASCII   Code: 117, Character: u\nASCII   Code: 108, Character: l\nASCII   Code: 112, Character: p\nASCII   Code: 97, Character: a\nASCII   Code: 98, Character: b\nASCII   Code: 105, Character: i\nASCII   Code: 108, Character: l\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 121, Character: y\nResult: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密方案概述：将字符串转换为ASCII编码。\n解密算法:\n解密方案概述：遍历ASCII编码字符串,返回转换后的字符串\n暗文: [110, 101, 114, 118, 105, 105] 解密成为原文: nervii\n暗文: [110, 101, 114, 118, 105, 105] 解密成为原文: nervii\n暗文: [112, 121, 114, 105, 116, 105, 99] 解密成为原文: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 345, "source_filename": "icl_with_rule_decode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": [112, 121, 114, 105, 116, 105, 99], "extra_args": {}, "output": "ASCII   Code: 112, Character: p\nASCII   Code: 121, Character: y\nASCII   Code: 114, Character: r\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 105, Character: i\nASCII   Code: 99, Character: c\nResult: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n    - 25个密码轮，每个密码轮包含26个字母的不同排列\n    [\n            \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n            \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n            \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n            \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n            \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n            \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n            \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n            \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n            \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n            \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n            \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n            \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n            \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n            \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n            \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n            \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n            \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n            \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n            \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n            \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n            \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n            \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n            \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n            \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n            \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        ]\n- 加密步骤:\n    - 初始选择第1个密码轮\n    - 对原始信息中的每个字符p:\n        - 在当前密码轮上找到字符p，用其后一个字符替换得到密文字符\n        - 如果当前字符在密码轮末尾，则回到密码轮开头\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续加密过程\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 大写字母字符串\n- 输出:\n    - 原始信息: 大写字母字符串\n- 准备:\n    - 25个密码轮，与加密相同\n- 解密步骤(与加密步骤相反):\n    - 初始选择第1个密码轮\n    - 对密文中的每个字符c:\n        - 在当前密码轮上找到字符c，用其前一个字符替换得到原始信息字符\n        - 如果当前字符在密码轮开头，则回到密码轮末尾\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续解密过程\n密文: WGTFJXF 解密为原始信息: pyritic\n密文: WGTFJXF 解密为原始信息: pyritic\n密文: AHTPTX 解密为原始信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 346, "source_filename": "icl_with_rule_decode_Kor_rule10_JeffersonCipher_cn.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "AHTPTX", "extra_args": {}, "output": "初始化Jefferson密码轮...\n开始解密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 A 替换为前一个字符 N\n  * 在第2个密码轮上，将字符 H 替换为前一个字符 E\n  * 在第3个密码轮上，将字符 T 替换为前一个字符 R\n  * 在第4个密码轮上，将字符 P 替换为前一个字符 V\n  * 在第5个密码轮上，将字符 T 替换为前一个字符 I\n  * 在第6个密码轮上，将字符 X 替换为前一个字符 I\n解密完成，结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 25个密码轮，每个密码轮包含26个字母的不同排列\n    [\n            \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n            \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n            \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n            \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n            \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n            \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n            \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n            \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n            \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n            \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n            \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n            \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n            \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n            \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n            \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n            \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n            \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n            \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n            \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n            \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n            \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n            \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n            \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n            \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n            \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        ]\n- 加密步骤:\n    - 初始选择第1个密码轮\n    - 对初始文本中的每个字符p:\n        - 在当前密码轮上找到字符p，用其后一个字符替换得到加密文本字符\n        - 如果当前字符在密码轮末尾，则回到密码轮开头\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续加密过程\n解密算法:\n解密规则:\n- 输入:\n    - 加密文本: 大写字母字符串\n- 输出:\n    - 初始文本: 大写字母字符串\n- 准备:\n    - 25个密码轮，与加密相同\n- 解密步骤(与加密步骤相反):\n    - 初始选择第1个密码轮\n    - 对加密文本中的每个字符c:\n        - 在当前密码轮上找到字符c，用其前一个字符替换得到初始文本字符\n        - 如果当前字符在密码轮开头，则回到密码轮末尾\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续解密过程\n加密文本: AUXBKUVIXCTSSR 解密为初始文本: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 347, "source_filename": "icl_with_rule_decode_Kor_rule10_JeffersonCipher_cn.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "AUXBKUVIXCTSSR", "extra_args": {}, "output": "初始化Jefferson密码轮...\n开始解密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 A 替换为前一个字符 N\n  * 在第2个密码轮上，将字符 U 替换为前一个字符 O\n  * 在第3个密码轮上，将字符 X 替换为前一个字符 N\n  * 在第4个密码轮上，将字符 B 替换为前一个字符 C\n  * 在第5个密码轮上，将字符 K 替换为前一个字符 U\n  * 在第6个密码轮上，将字符 U 替换为前一个字符 L\n  * 在第7个密码轮上，将字符 V 替换为前一个字符 P\n  * 在第8个密码轮上，将字符 I 替换为前一个字符 A\n  * 在第9个密码轮上，将字符 X 替换为前一个字符 B\n  * 在第10个密码轮上，将字符 C 替换为前一个字符 I\n  * 在第11个密码轮上，将字符 T 替换为前一个字符 L\n  * 在第12个密码轮上，将字符 S 替换为前一个字符 I\n  * 在第13个密码轮上，将字符 S 替换为前一个字符 T\n  * 在第14个密码轮上，将字符 R 替换为前一个字符 Y\n解密完成，结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - 25个密码轮，每个密码轮包含26个字母的不同排列\n    [\n            \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n            \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n            \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n            \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n            \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n            \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n            \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n            \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n            \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n            \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n            \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n            \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n            \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n            \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n            \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n            \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n            \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n            \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n            \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n            \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n            \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n            \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n            \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n            \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n            \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        ]\n- 加密步骤:\n    - 初始选择第1个密码轮\n    - 对明文中的每个字符p:\n        - 在当前密码轮上找到字符p，用其后一个字符替换得到加密文本字符\n        - 如果当前字符在密码轮末尾，则回到密码轮开头\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续加密过程\n解密算法:\n解密规则:\n- 输入:\n    - 加密文本: 大写字母字符串\n- 输出:\n    - 明文: 大写字母字符串\n- 准备:\n    - 25个密码轮，与加密相同\n- 解密步骤(与加密步骤相反):\n    - 初始选择第1个密码轮\n    - 对加密文本中的每个字符c:\n        - 在当前密码轮上找到字符c，用其前一个字符替换得到明文字符\n        - 如果当前字符在密码轮开头，则回到密码轮末尾\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续解密过程\n加密文本: AHTPTX 解密为明文: nervii\n加密文本: AHTPTX 解密为明文: nervii\n加密文本: AUXBKUVIXCTSSR 解密为明文: nonculpability\n加密文本: WGTFJXF 解密为明文: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 348, "source_filename": "icl_with_rule_decode_Kor_rule10_JeffersonCipher_cn.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "WGTFJXF", "extra_args": {}, "output": "初始化Jefferson密码轮...\n开始解密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 W 替换为前一个字符 P\n  * 在第2个密码轮上，将字符 G 替换为前一个字符 Y\n  * 在第3个密码轮上，将字符 T 替换为前一个字符 R\n  * 在第4个密码轮上，将字符 F 替换为前一个字符 I\n  * 在第5个密码轮上，将字符 J 替换为前一个字符 T\n  * 在第6个密码轮上，将字符 X 替换为前一个字符 I\n  * 在第7个密码轮上，将字符 F 替换为前一个字符 C\n解密完成，结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Encryption Steps:\n    - Fill in the first plaintext in the first position of the first line.\n    - Two ways to fill:\n        - Downward Fill: Fill the first row with a \"#\" (removing the first column, since the first position is already filled with a plaintext letter), then fill the plaintext downwards from the second row to the last row (fifth row).\n        - Upwards: the plaintext is filled upwards from the last line (fifth line) to the second line, and then the first line is filled with a \"#\".\n    - For each letter in the plaintext (except the first letter which has been filled in the first position), perform downward padding first, fill one column, then turn to upward padding, then turn to downward padding, and so on alternately, until all the letters have been filled.\n        - For example, HELLOWORLD.The content of the five lines after filling is\n        H L #\n        E R D\n        L O\n        L W\n        O #\n    - Filled out, read line by line, first read the contents of the first line, after reading the contents of each line are added a * sign, marking the end of the line; and then read the contents of the second line, and so on, read all the lines, the formation of the final password text.\n        H L #\n        E R D\n        L O\n        L W\n        O #\n        - Read the final cipher text line by line as \"HL#*ERD*LO*LW*O#*\" (i.e., there must be five *'s in the final cipher text to indicate that there are five lines)\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Decryption Steps (exact opposite of encryption steps):\n    - According to the * in the ciphertext (excluding the * sign), it can be divided into five groups, which are filled into five rows in turn. Get the recovered data in five rows.\n        - For example, like \"HL#*ERD*LO*LW*O#*\" ,according to the division of * and then fill in the five lines to get the recovered five lines of data D:\n            H L #\n            E R D\n            L O\n            L W\n            O #\n    - Then read all the columns alternately according to read down first and then read up. Get the uncleared messages.\n        - For example, for the five rows of message D in the previous step, read alternately to get \"HELLO#WORL#D\"\n    - Remove the # from the uncleaned message to get the final plaintext.\nencoded text: N*E*R*VI*I#* decode to plain text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 349, "source_filename": "icl_with_rule_decode_Kor_rule16_RedefenceFigureCipher_en.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "N*E*R*VI*I#*", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的密文: N*E*R*VI*I#*\n2. 根据*号分割密文\n  分割得到6个部分\n3. 将分割后的内容填入矩阵:\n  第1行: N\n  第2行: E\n  第3行: R\n  第4行: VI\n  第5行: I#\n  第6行: \n\n4. 交替读取矩阵内容:\n  向下读取第1列\n  向上读取第2列\n  向下读取第3列\n  向上读取第4列\n  向下读取第5列\n  向上读取第6列\n  向下读取第7列\n  向上读取第8列\n  向下读取第9列\n  向上读取第10列\n  向下读取第11列\n  向上读取第12列\n5. 移除#号获得最终明文\n解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Please decode the ciphertext according to the examples and the given information.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Encryption Steps:\n    - Fill in the first plaintext in the first position of the first line.\n    - Two ways to fill:\n        - Downward Fill: Fill the first row with a \"#\" (removing the first column, since the first position is already filled with a plaintext letter), then fill the plaintext downwards from the second row to the last row (fifth row).\n        - Upwards: the plaintext is filled upwards from the last line (fifth line) to the second line, and then the first line is filled with a \"#\".\n    - For each letter in the plaintext (except the first letter which has been filled in the first position), perform downward padding first, fill one column, then turn to upward padding, then turn to downward padding, and so on alternately, until all the letters have been filled.\n        - For example, HELLOWORLD.The content of the five lines after filling is\n        H L #\n        E R D\n        L O\n        L W\n        O #\n    - Filled out, read line by line, first read the contents of the first line, after reading the contents of each line are added a * sign, marking the end of the line; and then read the contents of the second line, and so on, read all the lines, the formation of the final password text.\n        H L #\n        E R D\n        L O\n        L W\n        O #\n        - Read the final cipher text line by line as \"HL#*ERD*LO*LW*O#*\" (i.e., there must be five *'s in the final cipher text to indicate that there are five lines)\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Decryption Steps (exact opposite of encryption steps):\n    - According to the * in the ciphertext (excluding the * sign), it can be divided into five groups, which are filled into five rows in turn. Get the recovered data in five rows.\n        - For example, like \"HL#*ERD*LO*LW*O#*\" ,according to the division of * and then fill in the five lines to get the recovered five lines of data D:\n            H L #\n            E R D\n            L O\n            L W\n            O #\n    - Then read all the columns alternately according to read down first and then read up. Get the uncleared messages.\n        - For example, for the five rows of message D in the previous step, read alternately to get \"HELLO#WORL#D\"\n    - Remove the # from the uncleaned message to get the final plaintext.\nencoded text: N*E*R*VI*I#* decrypt into clear text: nervii\nencoded text: N*E*R*VI*I#* decrypt into clear text: nervii\nencoded text: N*E*R*VI*I#* decrypt into clear text: nervii\nencoded text: NB#*OAI*NPL*CLIY*U#T#* decrypt into clear text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 350, "source_filename": "icl_with_rule_decode_Kor_rule16_RedefenceFigureCipher_en.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "NB#*OAI*NPL*CLIY*U#T#*", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的密文: NB#*OAI*NPL*CLIY*U#T#*\n2. 根据*号分割密文\n  分割得到6个部分\n3. 将分割后的内容填入矩阵:\n  第1行: NB#\n  第2行: OAI\n  第3行: NPL\n  第4行: CLIY\n  第5行: U#T#\n  第6行: \n\n4. 交替读取矩阵内容:\n  向下读取第1列\n  向上读取第2列\n  向下读取第3列\n  向上读取第4列\n  向下读取第5列\n  向上读取第6列\n  向下读取第7列\n  向上读取第8列\n  向下读取第9列\n  向上读取第10列\n  向下读取第11列\n  向上读取第12列\n  向下读取第13列\n  向上读取第14列\n  向下读取第15列\n  向上读取第16列\n  向下读取第17列\n  向上读取第18列\n  向下读取第19列\n  向上读取第20列\n  向下读取第21列\n  向上读取第22列\n5. 移除#号获得最终明文\n解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  A string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Encryption Steps:\n    - Fill in the first plaintext in the first position of the first line.\n    - Two ways to fill:\n        - Downward Fill: Fill the first row with a \"#\" (removing the first column, since the first position is already filled with a plaintext letter), then fill the plaintext downwards from the second row to the last row (fifth row).\n        - Upwards: the plaintext is filled upwards from the last line (fifth line) to the second line, and then the first line is filled with a \"#\".\n    - For each letter in the plaintext (except the first letter which has been filled in the first position), perform downward padding first, fill one column, then turn to upward padding, then turn to downward padding, and so on alternately, until all the letters have been filled.\n        - For example, HELLOWORLD.The content of the five lines after filling is\n        H L #\n        E R D\n        L O\n        L W\n        O #\n    - Filled out, read line by line, first read the contents of the first line, after reading the contents of each line are added a * sign, marking the end of the line; and then read the contents of the second line, and so on, read all the lines, the formation of the final password text.\n        H L #\n        E R D\n        L O\n        L W\n        O #\n        - Read the final cipher text line by line as \"HL#*ERD*LO*LW*O#*\" (i.e., there must be five *'s in the final cipher text to indicate that there are five lines)\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: A string without punctuation and spaces.\n- Output:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Preparation:\n    - Number of lines: 5\n- Decryption Steps (exact opposite of encryption steps):\n    - According to the * in the ciphertext (excluding the * sign), it can be divided into five groups, which are filled into five rows in turn. Get the recovered data in five rows.\n        - For example, like \"HL#*ERD*LO*LW*O#*\" ,according to the division of * and then fill in the five lines to get the recovered five lines of data D:\n            H L #\n            E R D\n            L O\n            L W\n            O #\n    - Then read all the columns alternately according to read down first and then read up. Get the uncleared messages.\n        - For example, for the five rows of message D in the previous step, read alternately to get \"HELLO#WORL#D\"\n    - Remove the # from the uncleaned message to get the final plaintext.\ncipher text: NB#*OAI*NPL*CLIY*U#T#* decode into clear text: nonculpability\ncipher text: NB#*OAI*NPL*CLIY*U#T#* decode into clear text: nonculpability\ncipher text: N*E*R*VI*I#* decode into clear text: nervii\ncipher text: P*Y*RC*II*T#* decode into clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 351, "source_filename": "icl_with_rule_decode_Kor_rule16_RedefenceFigureCipher_en.jsonl", "cipher_name": "Kor_rule16_RedefenceFigureCipher", "input": "P*Y*RC*II*T#*", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的密文: P*Y*RC*II*T#*\n2. 根据*号分割密文\n  分割得到6个部分\n3. 将分割后的内容填入矩阵:\n  第1行: P\n  第2行: Y\n  第3行: RC\n  第4行: II\n  第5行: T#\n  第6行: \n\n4. 交替读取矩阵内容:\n  向下读取第1列\n  向上读取第2列\n  向下读取第3列\n  向上读取第4列\n  向下读取第5列\n  向上读取第6列\n  向下读取第7列\n  向上读取第8列\n  向下读取第9列\n  向上读取第10列\n  向下读取第11列\n  向上读取第12列\n  向下读取第13列\n5. 移除#号获得最终明文\n解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)\n        \n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'],\n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'],\n        ['6', 'J', 'V', 'O', 'E', 'M']]\n        \n    - Encrypted Character Set\n        - \"ADFGVX\"\n- Encryption Steps:\n    - For each character in the plaintext:\n        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.\n        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.\n            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).\n        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.\n            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.\n\n \nclear text: nervii encode into cipher text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 352, "source_filename": "icl_with_rule_encode_Kor_rule19_ADFGVX_en.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n\n加密过程:\n1. 获取每个字符的行列位置并转换为加密字符:\n字符 N 在矩阵中的位置是 (4,5)，转换为加密二元组 (V,X)\n字符 E 在矩阵中的位置是 (5,4)，转换为加密二元组 (X,V)\n字符 R 在矩阵中的位置是 (0,0)，转换为加密二元组 (A,A)\n字符 V 在矩阵中的位置是 (5,2)，转换为加密二元组 (X,F)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n\n2. 合并所有加密二元组的行和列:\n先读取所有行: VXAXVV\n再读取所有列: XVAFFF\n最终密文: VXAXVVXVAFFF\n", "ground_truth": "VXAXVVXVAFFF"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)\n        \n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'],\n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'],\n        ['6', 'J', 'V', 'O', 'E', 'M']]\n        \n    - Encrypted Character Set\n        - \"ADFGVX\"\n- Encryption Steps:\n    - For each character in the plaintext:\n        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.\n        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.\n            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).\n        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.\n            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.\n\n \nclear text: nervii encode into encrypted text: VXAXVVXVAFFF\nclear text: nonculpability encode into encrypted text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 353, "source_filename": "icl_with_rule_encode_Kor_rule19_ADFGVX_en.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n\n加密过程:\n1. 获取每个字符的行列位置并转换为加密字符:\n字符 N 在矩阵中的位置是 (4,5)，转换为加密二元组 (V,X)\n字符 O 在矩阵中的位置是 (5,3)，转换为加密二元组 (X,G)\n字符 N 在矩阵中的位置是 (4,5)，转换为加密二元组 (V,X)\n字符 C 在矩阵中的位置是 (3,1)，转换为加密二元组 (G,D)\n字符 U 在矩阵中的位置是 (0,1)，转换为加密二元组 (A,D)\n字符 L 在矩阵中的位置是 (3,5)，转换为加密二元组 (G,X)\n字符 P 在矩阵中的位置是 (4,4)，转换为加密二元组 (V,V)\n字符 A 在矩阵中的位置是 (0,2)，转换为加密二元组 (A,F)\n字符 B 在矩阵中的位置是 (0,5)，转换为加密二元组 (A,X)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 L 在矩阵中的位置是 (3,5)，转换为加密二元组 (G,X)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 T 在矩阵中的位置是 (2,4)，转换为加密二元组 (F,V)\n字符 Y 在矩阵中的位置是 (3,0)，转换为加密二元组 (G,A)\n\n2. 合并所有加密二元组的行和列:\n先读取所有行: VXVGAGVAAVGVFG\n再读取所有列: XGXDDXVFXFXFVA\n最终密文: VXVGAGVAAVGVFGXGXDDXVFXFXFVA\n", "ground_truth": "VXVGAGVAAVGVFGXGXDDXVFXFXFVA"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Uppercase letters  string without punctuation and spaces.\n- Preparation:\n    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)\n        \n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'],\n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'],\n        ['6', 'J', 'V', 'O', 'E', 'M']]\n        \n    - Encrypted Character Set\n        - \"ADFGVX\"\n- Encryption Steps:\n    - For each character in the plaintext:\n        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.\n        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.\n            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).\n        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.\n            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.\n\n \nplain text: nonculpability encrypt into encrypted text: VXVGAGVAAVGVFGXGXDDXVFXFXFVA\nplain text: nervii encrypt into encrypted text: VXAXVVXVAFFF\nplain text: pyritic encrypt into encrypted text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 354, "source_filename": "icl_with_rule_encode_Kor_rule19_ADFGVX_en.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n\n加密过程:\n1. 获取每个字符的行列位置并转换为加密字符:\n字符 P 在矩阵中的位置是 (4,4)，转换为加密二元组 (V,V)\n字符 Y 在矩阵中的位置是 (3,0)，转换为加密二元组 (G,A)\n字符 R 在矩阵中的位置是 (0,0)，转换为加密二元组 (A,A)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 T 在矩阵中的位置是 (2,4)，转换为加密二元组 (F,V)\n字符 I 在矩阵中的位置是 (4,2)，转换为加密二元组 (V,F)\n字符 C 在矩阵中的位置是 (3,1)，转换为加密二元组 (G,D)\n\n2. 合并所有加密二元组的行和列:\n先读取所有行: VGAVFVG\n再读取所有列: VAAFVFD\n最终密文: VGAVFVGVAAFVFD\n", "ground_truth": "VGAVFVGVAAFVFD"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\nPlease encrypt the plaintext according to the encryption algorithm.\nOverview of the encryption scheme: Convert the string to ASCII encoding.\nclear text: nonculpability encode to cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\nclear text: nervii encode to cipher text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 355, "source_filename": "icl_with_rule_encode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": "nervii", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   e, ASCII Code: 101\nCharacter:   r, ASCII Code: 114\nCharacter:   v, ASCII Code: 118\nCharacter:   i, ASCII Code: 105\nCharacter:   i, ASCII Code: 105\nASCII List: [110, 101, 114, 118, 105, 105]\n", "ground_truth": [110, 101, 114, 118, 105, 105]}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nPlease encrypt the plaintext according to the encryption algorithm.\nOverview of the encryption scheme: Convert the string to ASCII encoding.\noriginal information: nervii encrypt into cipher text: [110, 101, 114, 118, 105, 105]\noriginal information: nonculpability encrypt into cipher text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 356, "source_filename": "icl_with_rule_encode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": "nonculpability", "extra_args": {}, "output": "Character:   n, ASCII Code: 110\nCharacter:   o, ASCII Code: 111\nCharacter:   n, ASCII Code: 110\nCharacter:   c, ASCII Code: 99\nCharacter:   u, ASCII Code: 117\nCharacter:   l, ASCII Code: 108\nCharacter:   p, ASCII Code: 112\nCharacter:   a, ASCII Code: 97\nCharacter:   b, ASCII Code: 98\nCharacter:   i, ASCII Code: 105\nCharacter:   l, ASCII Code: 108\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   y, ASCII Code: 121\nASCII List: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]\n", "ground_truth": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nPlease encrypt the plaintext according to the encryption algorithm.\nOverview of the encryption scheme: Convert the string to ASCII encoding.\nclear text: nervii encode into cipher text: [110, 101, 114, 118, 105, 105]\nclear text: pyritic encode into cipher text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 357, "source_filename": "icl_with_rule_encode_ASCII_en.jsonl", "cipher_name": "ASCII", "input": "pyritic", "extra_args": {}, "output": "Character:   p, ASCII Code: 112\nCharacter:   y, ASCII Code: 121\nCharacter:   r, ASCII Code: 114\nCharacter:   i, ASCII Code: 105\nCharacter:   t, ASCII Code: 116\nCharacter:   i, ASCII Code: 105\nCharacter:   c, ASCII Code: 99\nASCII List: [112, 121, 114, 105, 116, 105, 99]\n", "ground_truth": [112, 121, 114, 105, 116, 105, 99]}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板:\n        - 准备一个空白网格和一个带孔的模板(栅栏)\n        - 使用的模板是:\n            ▮ ▮ ▮ ▮\n            ▮ ▮ ▯ ▯\n            ▮ ▮ ▮ ▯\n            ▯ ▮ ▮ ▮\n            其中白色的是孔，将模板放在空白网格上，只通过白色孔洞，在网格的对应位置填入字符。\n- 加密步骤:\n    - 将初始文本逐个分成16个字母的块(如果初始文本少于16个长度则为一个块)\n    - 对每个块执行以下加密操作:\n        - 将带孔的模板放在空白网格上\n        - 通过模板中的孔按顺序填入初始文本字母\n        - 模板总共有四个孔，所以填完四个字母后，需要将模板逆时针旋转90度\n        - 重复填充可见孔中的初始文本下一个字母并旋转模板，直到整个网格完全填满。这将执行4次填充+旋转，最终模板会转回原始模板。如果消息不足以填满整个网格，用填充字符(如'#')补充\n        - 网格填满后，按行读取网格内容作为该块的加密消息\n        - 进入下一个块时，清空网格内容并重做整个填充和旋转操作\n    最后，将所有块的加密消息连接在一起作为最终暗文。\n解密算法:\n解密规则:\n- 输入:\n    - 暗文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 将暗文逐个分成16个字母的块\n    - 对每个块执行以下操作:\n        - 将16个字母按行写入填满网格\n        - 将带孔的模板放在填满的网格上\n        - 读取通过孔可见的字母获得部分初始文本消息\n        - 由于只有四个孔，此时需要将模板逆时针旋转90度读取下一组字母\n        - 重复读取步骤四次以获得这块解密消息\n    - 连接所有块的解密消息得到最终初始文本。\n暗文: TULBIPNOLIYNC##A 解密成初始文本: nonculpability\n暗文: #TI##CPY###RI### 解密成初始文本: pyritic\n暗文: #TI##CPY###RI### 解密成初始文本: pyritic\n暗文: #II###NE###RV### 解密成初始文本: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 358, "source_filename": "icl_with_rule_decode_Kor_rule18_RotatingGridCipher_cn.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "#II###NE###RV###", "extra_args": {}, "output": "使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始解密过程:\n\n处理第1个块:\n当前块的密文: #II###NE###RV###\n\n重构的网格:\n[['#' 'I' 'I' '#']\n ['#' '#' 'N' 'E']\n ['#' '#' '#' 'R']\n ['V' '#' '#' '#']]\n\n第1次旋转后读取:\n从位置(1,2)读取字符: N\n从位置(1,3)读取字符: E\n从位置(2,3)读取字符: R\n从位置(3,0)读取字符: V\n\n第2次旋转后读取:\n从位置(0,1)读取字符: I\n从位置(0,2)读取字符: I\n从位置(1,1)读取字符: #\n从位置(3,3)读取字符: #\n\n第3次旋转后读取:\n从位置(0,3)读取字符: #\n从位置(1,0)读取字符: #\n从位置(2,0)读取字符: #\n从位置(2,1)读取字符: #\n\n第4次旋转后读取:\n从位置(0,0)读取字符: #\n从位置(2,2)读取字符: #\n从位置(3,1)读取字符: #\n从位置(3,2)读取字符: #\n当前块解密结果: NERVII##########\n\n最终明文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板:\n        - 准备一个空白网格和一个带孔的模板(栅栏)\n        - 使用的模板是:\n            ▮ ▮ ▮ ▮\n            ▮ ▮ ▯ ▯\n            ▮ ▮ ▮ ▯\n            ▯ ▮ ▮ ▮\n            其中白色的是孔，将模板放在空白网格上，只通过白色孔洞，在网格的对应位置填入字符。\n- 加密步骤:\n    - 将原始信息逐个分成16个字母的块(如果原始信息少于16个长度则为一个块)\n    - 对每个块执行以下加密操作:\n        - 将带孔的模板放在空白网格上\n        - 通过模板中的孔按顺序填入原始信息字母\n        - 模板总共有四个孔，所以填完四个字母后，需要将模板逆时针旋转90度\n        - 重复填充可见孔中的原始信息下一个字母并旋转模板，直到整个网格完全填满。这将执行4次填充+旋转，最终模板会转回原始模板。如果消息不足以填满整个网格，用填充字符(如'#')补充\n        - 网格填满后，按行读取网格内容作为该块的加密消息\n        - 进入下一个块时，清空网格内容并重做整个填充和旋转操作\n    最后，将所有块的加密消息连接在一起作为最终加密信息。\n解密算法:\n解密规则:\n- 输入:\n    - 加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 将加密信息逐个分成16个字母的块\n    - 对每个块执行以下操作:\n        - 将16个字母按行写入填满网格\n        - 将带孔的模板放在填满的网格上\n        - 读取通过孔可见的字母获得部分原始信息消息\n        - 由于只有四个孔，此时需要将模板逆时针旋转90度读取下一组字母\n        - 重复读取步骤四次以获得这块解密消息\n    - 连接所有块的解密消息得到最终原始信息。\n加密信息: #TI##CPY###RI### 解密成为原始信息: pyritic\n加密信息: TULBIPNOLIYNC##A 解密成为原始信息是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 359, "source_filename": "icl_with_rule_decode_Kor_rule18_RotatingGridCipher_cn.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "TULBIPNOLIYNC##A", "extra_args": {}, "output": "使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始解密过程:\n\n处理第1个块:\n当前块的密文: TULBIPNOLIYNC##A\n\n重构的网格:\n[['T' 'U' 'L' 'B']\n ['I' 'P' 'N' 'O']\n ['L' 'I' 'Y' 'N']\n ['C' '#' '#' 'A']]\n\n第1次旋转后读取:\n从位置(1,2)读取字符: N\n从位置(1,3)读取字符: O\n从位置(2,3)读取字符: N\n从位置(3,0)读取字符: C\n\n第2次旋转后读取:\n从位置(0,1)读取字符: U\n从位置(0,2)读取字符: L\n从位置(1,1)读取字符: P\n从位置(3,3)读取字符: A\n\n第3次旋转后读取:\n从位置(0,3)读取字符: B\n从位置(1,0)读取字符: I\n从位置(2,0)读取字符: L\n从位置(2,1)读取字符: I\n\n第4次旋转后读取:\n从位置(0,0)读取字符: T\n从位置(2,2)读取字符: Y\n从位置(3,1)读取字符: #\n从位置(3,2)读取字符: #\n当前块解密结果: NONCULPABILITY##\n\n最终明文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板:\n        - 准备一个空白网格和一个带孔的模板(栅栏)\n        - 使用的模板是:\n            ▮ ▮ ▮ ▮\n            ▮ ▮ ▯ ▯\n            ▮ ▮ ▮ ▯\n            ▯ ▮ ▮ ▮\n            其中白色的是孔，将模板放在空白网格上，只通过白色孔洞，在网格的对应位置填入字符。\n- 加密步骤:\n    - 将初始文本逐个分成16个字母的块(如果初始文本少于16个长度则为一个块)\n    - 对每个块执行以下加密操作:\n        - 将带孔的模板放在空白网格上\n        - 通过模板中的孔按顺序填入初始文本字母\n        - 模板总共有四个孔，所以填完四个字母后，需要将模板逆时针旋转90度\n        - 重复填充可见孔中的初始文本下一个字母并旋转模板，直到整个网格完全填满。这将执行4次填充+旋转，最终模板会转回原始模板。如果消息不足以填满整个网格，用填充字符(如'#')补充\n        - 网格填满后，按行读取网格内容作为该块的加密消息\n        - 进入下一个块时，清空网格内容并重做整个填充和旋转操作\n    最后，将所有块的加密消息连接在一起作为最终暗码。\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 输出:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板(与加密相同)\n- 解密步骤(与加密步骤完全相反):\n    - 将暗码逐个分成16个字母的块\n    - 对每个块执行以下操作:\n        - 将16个字母按行写入填满网格\n        - 将带孔的模板放在填满的网格上\n        - 读取通过孔可见的字母获得部分初始文本消息\n        - 由于只有四个孔，此时需要将模板逆时针旋转90度读取下一组字母\n        - 重复读取步骤四次以获得这块解密消息\n    - 连接所有块的解密消息得到最终初始文本。\n暗码: TULBIPNOLIYNC##A 解码为初始文本: nonculpability\n暗码: #II###NE###RV### 解码为初始文本: nervii\n暗码: #TI##CPY###RI### 解码为初始文本: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 360, "source_filename": "icl_with_rule_decode_Kor_rule18_RotatingGridCipher_cn.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "#TI##CPY###RI###", "extra_args": {}, "output": "使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始解密过程:\n\n处理第1个块:\n当前块的密文: #TI##CPY###RI###\n\n重构的网格:\n[['#' 'T' 'I' '#']\n ['#' 'C' 'P' 'Y']\n ['#' '#' '#' 'R']\n ['I' '#' '#' '#']]\n\n第1次旋转后读取:\n从位置(1,2)读取字符: P\n从位置(1,3)读取字符: Y\n从位置(2,3)读取字符: R\n从位置(3,0)读取字符: I\n\n第2次旋转后读取:\n从位置(0,1)读取字符: T\n从位置(0,2)读取字符: I\n从位置(1,1)读取字符: C\n从位置(3,3)读取字符: #\n\n第3次旋转后读取:\n从位置(0,3)读取字符: #\n从位置(1,0)读取字符: #\n从位置(2,0)读取字符: #\n从位置(2,1)读取字符: #\n\n第4次旋转后读取:\n从位置(0,0)读取字符: #\n从位置(2,2)读取字符: #\n从位置(3,1)读取字符: #\n从位置(3,2)读取字符: #\n当前块解密结果: PYRITIC#########\n\n最终明文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - 25个密码轮，每个密码轮包含26个字母的不同排列\n    [\n            \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n            \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n            \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n            \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n            \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n            \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n            \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n            \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n            \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n            \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n            \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n            \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n            \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n            \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n            \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n            \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n            \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n            \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n            \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n            \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n            \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n            \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n            \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n            \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n            \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        ]\n- 加密步骤:\n    - 初始选择第1个密码轮\n    - 对原始信息中的每个字符p:\n        - 在当前密码轮上找到字符p，用其后一个字符替换得到暗码字符\n        - 如果当前字符在密码轮末尾，则回到密码轮开头\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续加密过程\n原始信息: pyritic 加密成暗码: WGTFJXF\n原始信息: nonculpability 加密成暗码: AUXBKUVIXCTSSR\n原始信息: pyritic 加密成暗码: WGTFJXF\n原始信息: nervii 加密成暗码: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 361, "source_filename": "icl_with_rule_encode_Kor_rule10_JeffersonCipher_cn.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n初始化Jefferson密码轮...\n开始加密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 N 替换为下一个字符 A\n  * 在第2个密码轮上，将字符 E 替换为下一个字符 H\n  * 在第3个密码轮上，将字符 R 替换为下一个字符 T\n  * 在第4个密码轮上，将字符 V 替换为下一个字符 P\n  * 在第5个密码轮上，将字符 I 替换为下一个字符 T\n  * 在第6个密码轮上，将字符 I 替换为下一个字符 X\n加密完成，结果: AHTPTX\n", "ground_truth": "AHTPTX"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - 25个密码轮，每个密码轮包含26个字母的不同排列\n    [\n            \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n            \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n            \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n            \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n            \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n            \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n            \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n            \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n            \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n            \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n            \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n            \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n            \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n            \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n            \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n            \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n            \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n            \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n            \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n            \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n            \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n            \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n            \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n            \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n            \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        ]\n- 加密步骤:\n    - 初始选择第1个密码轮\n    - 对明文中的每个字符p:\n        - 在当前密码轮上找到字符p，用其后一个字符替换得到暗码字符\n        - 如果当前字符在密码轮末尾，则回到密码轮开头\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续加密过程\n明文: nervii 加密成暗码: AHTPTX\n明文: pyritic 加密成暗码: WGTFJXF\n明文: nonculpability 加密成暗码: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 362, "source_filename": "icl_with_rule_encode_Kor_rule10_JeffersonCipher_cn.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n初始化Jefferson密码轮...\n开始加密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 N 替换为下一个字符 A\n  * 在第2个密码轮上，将字符 O 替换为下一个字符 U\n  * 在第3个密码轮上，将字符 N 替换为下一个字符 X\n  * 在第4个密码轮上，将字符 C 替换为下一个字符 B\n  * 在第5个密码轮上，将字符 U 替换为下一个字符 K\n  * 在第6个密码轮上，将字符 L 替换为下一个字符 U\n  * 在第7个密码轮上，将字符 P 替换为下一个字符 V\n  * 在第8个密码轮上，将字符 A 替换为下一个字符 I\n  * 在第9个密码轮上，将字符 B 替换为下一个字符 X\n  * 在第10个密码轮上，将字符 I 替换为下一个字符 C\n  * 在第11个密码轮上，将字符 L 替换为下一个字符 T\n  * 在第12个密码轮上，将字符 I 替换为下一个字符 S\n  * 在第13个密码轮上，将字符 T 替换为下一个字符 S\n  * 在第14个密码轮上，将字符 Y 替换为下一个字符 R\n加密完成，结果: AUXBKUVIXCTSSR\n", "ground_truth": "AUXBKUVIXCTSSR"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格\n- 输出:\n    - 隐文: 大写字母字符串\n- 准备:\n    - 25个密码轮，每个密码轮包含26个字母的不同排列\n    [\n            \"ABCEIGDJFVUYMHTQKZOLRXSPWN\",\n            \"ACDEHFIJKTLMOUVYGZNPQXRWSB\",\n            \"ADKOMJUBGEPHSCZINXFYQRTVWL\",\n            \"AEDCBIFGJHLKMRUOQVPTNWYXZS\",\n            \"AFNQUKDOPITJBRHCYSLWEMZVXG\",\n            \"AGPOCIXLURNDYZHWBJSQFKVMET\",\n            \"AHXJEZBNIKPVROGSYDULCFMQTW\",\n            \"AIHPJOBWKCVFZLQERYNSUMGTDX\",\n            \"AJDSKQOIVTZEFHGYUNLPMBXWCR\",\n            \"AKELBDFJGHONMTPRQSVZUXYWIC\",\n            \"ALTMSXVQPNOHUWDIZYCGKRFBEJ\",\n            \"AMNFLHQGCUJTBYPZKXISRDVEWO\",\n            \"ANCJILDHBMKGXUZTSWQYVORPFE\",\n            \"AODWPKJVIUQHZCTXBLEGNYRSMF\",\n            \"APBVHIYKSGUENTCXOWFQDRLJZM\",\n            \"AQJNUBTGIMWZRVLXCSHDEOKFPY\",\n            \"ARMYOFTHEUSZJXDPCWGQIBKLNV\",\n            \"ASDMCNEQBOZPLGVJRKYTFUIWXH\",\n            \"ATOJYLFXNGWHVCMIRBSEKUPDZQ\",\n            \"AUTRZXQLYIOVBPESNHJWMDGFCK\",\n            \"AVNKHRGOXEYBFSJMUDQCLZWTIP\",\n            \"AWVSFDLIEBHKNRJQZGMXPUCOTY\",\n            \"AXKWREVDTUFOYHMLSIQNJCPGBZ\",\n            \"AYJPXMVKBQWUGLOSTECHNZFRID\",\n            \"AZDNBUHYFWJLVGRCQMPSOEXTKI\"\n        ]\n- 加密步骤:\n    - 初始选择第1个密码轮\n    - 对原始信息中的每个字符p:\n        - 在当前密码轮上找到字符p，用其后一个字符替换得到隐文字符\n        - 如果当前字符在密码轮末尾，则回到密码轮开头\n        - 移动到下一个密码轮处理下一个字符，当到达最后一个密码轮时，返回第一个密码轮继续加密过程\n原始信息: nonculpability 加密成隐文: AUXBKUVIXCTSSR\n原始信息: nonculpability 加密成隐文: AUXBKUVIXCTSSR\n原始信息: nervii 加密成隐文: AHTPTX\n原始信息: pyritic 加密成隐文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 363, "source_filename": "icl_with_rule_encode_Kor_rule10_JeffersonCipher_cn.jsonl", "cipher_name": "Kor_rule10_JeffersonCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n初始化Jefferson密码轮...\n开始加密过程:\n- 从第1个密码轮开始\n- 对每个字符:\n  * 在第1个密码轮上，将字符 P 替换为下一个字符 W\n  * 在第2个密码轮上，将字符 Y 替换为下一个字符 G\n  * 在第3个密码轮上，将字符 R 替换为下一个字符 T\n  * 在第4个密码轮上，将字符 I 替换为下一个字符 F\n  * 在第5个密码轮上，将字符 T 替换为下一个字符 J\n  * 在第6个密码轮上，将字符 I 替换为下一个字符 X\n  * 在第7个密码轮上，将字符 C 替换为下一个字符 F\n加密完成，结果: WGTFJXF\n", "ground_truth": "WGTFJXF"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\nplain text: pyritic encrypt into cipher text: OCAX..C\nplain text: nervii encrypt into cipher text is: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 364, "source_filename": "icl_encode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "NERVII", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NERVII\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  E -> 层: 1, 列: 2, 行: 2\n  R -> 层: 2, 列: 3, 行: 3\n  V -> 层: 3, 列: 1, 行: 2\n  I -> 层: 1, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\nStep 3: 合并坐标:\n  层坐标: 212311\n  列坐标: 223133\n  行坐标: 223233\nStep 4: 重组坐标: ['212', '311', '223', '133', '223', '233']\nStep 5: 根据新坐标获取密文:\n  212 -> M\n  311 -> S\n  223 -> Q\n  133 -> I\n  223 -> Q\n  233 -> R\n", "ground_truth": "MSQIQR"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\nclear text: nervii encrypt to encrypted text: MSQIQR\nclear text: pyritic encrypt to encrypted text: OCAX..C\nclear text: nonculpability encrypt to encrypted text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 365, "source_filename": "icl_encode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NONCULPABILITY\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  O -> 层: 2, 列: 3, 行: 2\n  N -> 层: 2, 列: 2, 行: 2\n  C -> 层: 1, 列: 3, 行: 1\n  U -> 层: 3, 列: 3, 行: 1\n  L -> 层: 2, 列: 3, 行: 1\n  P -> 层: 2, 列: 1, 行: 3\n  A -> 层: 1, 列: 1, 行: 1\n  B -> 层: 1, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  L -> 层: 2, 列: 3, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  Y -> 层: 3, 列: 1, 行: 3\nStep 3: 合并坐标:\n  层坐标: 22213221112133\n  列坐标: 23233311233321\n  行坐标: 22211131131313\nStep 4: 重组坐标: ['222', '132', '211', '121', '332', '323', '331', '123', '332', '122', '211', '131', '131', '313']\nStep 5: 根据新坐标获取密文:\n  222 -> N\n  132 -> F\n  211 -> J\n  121 -> B\n  332 -> X\n  323 -> Z\n  331 -> U\n  123 -> H\n  332 -> X\n  122 -> E\n  211 -> J\n  131 -> C\n  131 -> C\n  313 -> Y\n", "ground_truth": "NFJBXZUHXEJCCY"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\noriginal information: nonculpability encrypt to encoded text: NFJBXZUHXEJCCY\noriginal information: nonculpability encrypt to encoded text: NFJBXZUHXEJCCY\noriginal information: nonculpability encrypt to encoded text: NFJBXZUHXEJCCY\noriginal information: pyritic encrypt to encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 366, "source_filename": "icl_encode_TrifidCipher_en.jsonl", "cipher_name": "TrifidCipher", "input": "PYRITIC", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: PYRITIC\nStep 2: 获取每个字符的坐标:\n  P -> 层: 2, 列: 1, 行: 3\n  Y -> 层: 3, 列: 1, 行: 3\n  R -> 层: 2, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  C -> 层: 1, 列: 3, 行: 1\nStep 3: 合并坐标:\n  层坐标: 2321311\n  列坐标: 1133233\n  行坐标: 3333131\nStep 4: 重组坐标: ['232', '131', '111', '332', '333', '333', '131']\nStep 5: 根据新坐标获取密文:\n  232 -> O\n  131 -> C\n  111 -> A\n  332 -> X\n  333 -> .\n  333 -> .\n  131 -> C\n", "ground_truth": "OCAX..C"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (矩阵中的行和列从0开始计数)\n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'], \n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'], \n        ['6', 'J', 'V', 'O', 'E', 'M']]\n    - 加密字符集: \"ADFGVX\"\n- 加密步骤:\n    - 对初始文本中的每个字符:\n        - 在6X6矩阵中找到该字符的行数和列数。例如，A的行数为0，列数为2。\n        - 加密字符集中的字符位置标记为0-6。使用加密字符集中对应位置的字符替换行数和列数得到加密二元组。\n            - 例如，A的行数为0对应加密字符集中的A，列数为2对应加密字符集中的F，所以A的加密二元组为(A,F)。\n        - 读取所有加密二元组的行，然后读取所有加密二元组的列得到最终暗码。\n            - 例如，加密二元组为(A,F)(X,V)，最后读取为AXFV，所以最终暗码为AXFV。\n解密算法:\n解密规则:\n- 输入:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 输出:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (与加密相同)\n    - 加密字符集 (与加密相同)\n- 解密步骤:\n    - 将暗码分成两半\n    - 每次从上半部分和下半部分各取一个字母作为解密二元组:\n        - 加密字符集中的字符位置标记为0-6，使用加密字符集中对应字符的位置解密出解密二元组代表的行数和列数。\n            - 例如，解密二元组为加密二元组(A,F)，A的位置为0，F的位置为2，所以行数为0，列数为2，得到(0,2)。\n        - 使用得到的行数和列数在6x6矩阵中找到对应位置的字母作为解密后的字符。\n            - 例如，(0,2)位置的字符为A，所以AF解密后的字符为A。\n    - 连接所有解密后的字符得到最终初始文本。\n暗码: VGAVFVGVAAFVFD 解密为初始文本: pyritic\n暗码: VXAXVVXVAFFF 解密为初始文本: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 367, "source_filename": "icl_with_rule_decode_Kor_rule19_ADFGVX_cn.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "VXAXVVXVAFFF", "extra_args": {}, "output": "输入密文: VXAXVVXVAFFF\n\n解密过程:\n1. 将密文分成两半:\n前半部分: VXAXVV\n后半部分: XVAFFF\n\n2. 将前后半部分配对并解密:\n配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N\n配对 (X,V) 对应矩阵位置 (5,4)，解密为字符: E\n配对 (A,A) 对应矩阵位置 (0,0)，解密为字符: R\n配对 (X,F) 对应矩阵位置 (5,2)，解密为字符: V\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n\n最终明文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 隐文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (矩阵中的行和列从0开始计数)\n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'], \n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'], \n        ['6', 'J', 'V', 'O', 'E', 'M']]\n    - 加密字符集: \"ADFGVX\"\n- 加密步骤:\n    - 对初始文本中的每个字符:\n        - 在6X6矩阵中找到该字符的行数和列数。例如，A的行数为0，列数为2。\n        - 加密字符集中的字符位置标记为0-6。使用加密字符集中对应位置的字符替换行数和列数得到加密二元组。\n            - 例如，A的行数为0对应加密字符集中的A，列数为2对应加密字符集中的F，所以A的加密二元组为(A,F)。\n        - 读取所有加密二元组的行，然后读取所有加密二元组的列得到最终隐文。\n            - 例如，加密二元组为(A,F)(X,V)，最后读取为AXFV，所以最终隐文为AXFV。\n解密算法:\n解密规则:\n- 输入:\n    - 隐文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (与加密相同)\n    - 加密字符集 (与加密相同)\n- 解密步骤:\n    - 将隐文分成两半\n    - 每次从上半部分和下半部分各取一个字母作为解密二元组:\n        - 加密字符集中的字符位置标记为0-6，使用加密字符集中对应字符的位置解密出解密二元组代表的行数和列数。\n            - 例如，解密二元组为加密二元组(A,F)，A的位置为0，F的位置为2，所以行数为0，列数为2，得到(0,2)。\n        - 使用得到的行数和列数在6x6矩阵中找到对应位置的字母作为解密后的字符。\n            - 例如，(0,2)位置的字符为A，所以AF解密后的字符为A。\n    - 连接所有解密后的字符得到最终初始文本。\n隐文: VXVGAGVAAVGVFGXGXDDXVFXFXFVA 解密成初始文本是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 368, "source_filename": "icl_with_rule_decode_Kor_rule19_ADFGVX_cn.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "VXVGAGVAAVGVFGXGXDDXVFXFXFVA", "extra_args": {}, "output": "输入密文: VXVGAGVAAVGVFGXGXDDXVFXFXFVA\n\n解密过程:\n1. 将密文分成两半:\n前半部分: VXVGAGVAAVGVFG\n后半部分: XGXDDXVFXFXFVA\n\n2. 将前后半部分配对并解密:\n配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N\n配对 (X,G) 对应矩阵位置 (5,3)，解密为字符: O\n配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N\n配对 (G,D) 对应矩阵位置 (3,1)，解密为字符: C\n配对 (A,D) 对应矩阵位置 (0,1)，解密为字符: U\n配对 (G,X) 对应矩阵位置 (3,5)，解密为字符: L\n配对 (V,V) 对应矩阵位置 (4,4)，解密为字符: P\n配对 (A,F) 对应矩阵位置 (0,2)，解密为字符: A\n配对 (A,X) 对应矩阵位置 (0,5)，解密为字符: B\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (G,X) 对应矩阵位置 (3,5)，解密为字符: L\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (F,V) 对应矩阵位置 (2,4)，解密为字符: T\n配对 (G,A) 对应矩阵位置 (3,0)，解密为字符: Y\n\n最终明文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n加密规则:\n- 输入:\n    - 原文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (矩阵中的行和列从0开始计数)\n        [['R', 'U', 'A', '0', 'Q', 'B'], \n        ['D', '2', 'W', 'K', 'S', '1'], \n        ['H', '4', '5', 'F', 'T', 'Z'], \n        ['Y', 'C', 'G', 'X', '7', 'L'], \n        ['9', '8', 'I', '3', 'P', 'N'], \n        ['6', 'J', 'V', 'O', 'E', 'M']]\n    - 加密字符集: \"ADFGVX\"\n- 加密步骤:\n    - 对原文中的每个字符:\n        - 在6X6矩阵中找到该字符的行数和列数。例如，A的行数为0，列数为2。\n        - 加密字符集中的字符位置标记为0-6。使用加密字符集中对应位置的字符替换行数和列数得到加密二元组。\n            - 例如，A的行数为0对应加密字符集中的A，列数为2对应加密字符集中的F，所以A的加密二元组为(A,F)。\n        - 读取所有加密二元组的行，然后读取所有加密二元组的列得到最终密文。\n            - 例如，加密二元组为(A,F)(X,V)，最后读取为AXFV，所以最终密文为AXFV。\n解密算法:\n解密规则:\n- 输入:\n    - 密文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 原文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 6x6矩阵 (与加密相同)\n    - 加密字符集 (与加密相同)\n- 解密步骤:\n    - 将密文分成两半\n    - 每次从上半部分和下半部分各取一个字母作为解密二元组:\n        - 加密字符集中的字符位置标记为0-6，使用加密字符集中对应字符的位置解密出解密二元组代表的行数和列数。\n            - 例如，解密二元组为加密二元组(A,F)，A的位置为0，F的位置为2，所以行数为0，列数为2，得到(0,2)。\n        - 使用得到的行数和列数在6x6矩阵中找到对应位置的字母作为解密后的字符。\n            - 例如，(0,2)位置的字符为A，所以AF解密后的字符为A。\n    - 连接所有解密后的字符得到最终原文。\n密文: VGAVFVGVAAFVFD 解密为原文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 369, "source_filename": "icl_with_rule_decode_Kor_rule19_ADFGVX_cn.jsonl", "cipher_name": "Kor_rule19_ADFGVX", "input": "VGAVFVGVAAFVFD", "extra_args": {}, "output": "输入密文: VGAVFVGVAAFVFD\n\n解密过程:\n1. 将密文分成两半:\n前半部分: VGAVFVG\n后半部分: VAAFVFD\n\n2. 将前后半部分配对并解密:\n配对 (V,V) 对应矩阵位置 (4,4)，解密为字符: P\n配对 (G,A) 对应矩阵位置 (3,0)，解密为字符: Y\n配对 (A,A) 对应矩阵位置 (0,0)，解密为字符: R\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (F,V) 对应矩阵位置 (2,4)，解密为字符: T\n配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I\n配对 (G,D) 对应矩阵位置 (3,1)，解密为字符: C\n\n最终明文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n暗码: cdcrjaepqxaxin 额外参数: {'shift': 15} 解密为原文: nonculpability\n暗码: cdcrjaepqxaxin 额外参数: {'shift': 15} 解密为原文: nonculpability\n暗码: cdcrjaepqxaxin 额外参数: {'shift': 15} 解密为原文: nonculpability\n暗码: qzsjujd 额外参数: {'shift': 1} 解密为原文: pyritic\n暗码: ulycpp 额外参数: {'shift': 7} 解密为原文: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 370, "source_filename": "icl_decode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "ulycpp", "extra_args": {"shift": 7}, "output": "开始解码过程，文本为: ulycpp，位移数为: 7\n处理字符: u\n字符 u 向前移动 7 位，变为 n\n处理字符: l\n字符 l 向前移动 7 位，变为 e\n处理字符: y\n字符 y 向前移动 7 位，变为 r\n处理字符: c\n字符 c 向前移动 7 位，变为 v\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n处理字符: p\n字符 p 向前移动 7 位，变为 i\n解码完成，结果为: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n隐文: qzsjujd 密钥或额外参数: {'shift': 1} 解密为原文: pyritic\n隐文: qzsjujd 密钥或额外参数: {'shift': 1} 解密为原文: pyritic\n隐文: cdcrjaepqxaxin 密钥或额外参数: {'shift': 15} 解密为原文是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 371, "source_filename": "icl_decode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "cdcrjaepqxaxin", "extra_args": {"shift": 15}, "output": "开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: d\n字符 d 向前移动 15 位，变为 o\n处理字符: c\n字符 c 向前移动 15 位，变为 n\n处理字符: r\n字符 r 向前移动 15 位，变为 c\n处理字符: j\n字符 j 向前移动 15 位，变为 u\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: e\n字符 e 向前移动 15 位，变为 p\n处理字符: p\n字符 p 向前移动 15 位，变为 a\n处理字符: q\n字符 q 向前移动 15 位，变为 b\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: a\n字符 a 向前移动 15 位，变为 l\n处理字符: x\n字符 x 向前移动 15 位，变为 i\n处理字符: i\n字符 i 向前移动 15 位，变为 t\n处理字符: n\n字符 n 向前移动 15 位，变为 y\n解码完成，结果为: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n加密文本: cdcrjaepqxaxin 密钥或额外参数: {'shift': 15} 解密成为原始信息: nonculpability\n加密文本: cdcrjaepqxaxin 密钥或额外参数: {'shift': 15} 解密成为原始信息: nonculpability\n加密文本: cdcrjaepqxaxin 密钥或额外参数: {'shift': 15} 解密成为原始信息: nonculpability\n加密文本: cdcrjaepqxaxin 密钥或额外参数: {'shift': 15} 解密成为原始信息: nonculpability\n加密文本: qzsjujd 密钥或额外参数: {'shift': 1} 解密成为原始信息: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 372, "source_filename": "icl_decode_Caesar_Cipher_cn.jsonl", "cipher_name": "Caesar_Cipher", "input": "qzsjujd", "extra_args": {"shift": 1}, "output": "开始解码过程，文本为: qzsjujd，位移数为: 1\n处理字符: q\n字符 q 向前移动 1 位，变为 p\n处理字符: z\n字符 z 向前移动 1 位，变为 y\n处理字符: s\n字符 s 向前移动 1 位，变为 r\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: u\n字符 u 向前移动 1 位，变为 t\n处理字符: j\n字符 j 向前移动 1 位，变为 i\n处理字符: d\n字符 d 向前移动 1 位，变为 c\n解码完成，结果为: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds\n        - Gird1\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n        - Gird2\n            - E  C  H  O  A\n            B  D  F  G  I\n            J  K  L  M  N\n            P  R  S  T  U\n            V  W  X  Y  Z\n        - Gird3\n            - V  O  R  T  E\n            X  A  B  C  D\n            F  G  H  I  J\n            K  L  M  N  P\n            S  U  W  Y  Z\n        - Gird4\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n- Encryption Steps:\n    - Cleans up plaintext, removing spaces and non-alphabetic characters, removing the letter Q, and converting all letters to uppercase.\n    - If the plaintext is of odd length, add a letter 'X' to make it of even length.\n    - Split the cleaned plaintext into two-letter groups.\n    - For each double-letter group p1,p2 of the plaintext:\n        - In Gird1 and Gird4, find the position of the first letter and the second letter.\n        - In Gird2 and Gird3, find the letters corresponding to these two positions, use these two letters as the result of encrypting the double letter combination.\n    - Concatenate all the encrypted double letter groups to form the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Cleans up ciphertext, removing spaces and non-alphabetic characters, and converting all letters to uppercase.\n    - Split the cleaned ciphertext into two-letter groups.\n    - For each double-letter group c1,c2 of the ciphertext:\n        - In Gird2 and Gird3, find the position of the first letter and the second letter.\n        - In Gird1 and Gird4, find the letters corresponding to these two positions, use these two letters as the result of decrypting the double letter combination.\n    - Concatenate all the decrypted double letter groups to form the final plaintext.\nencoded text: BIDYGYSH decode to clear text: pyritic\nencoded text: OPDFYY decode to clear text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 373, "source_filename": "icl_with_rule_decode_Kor_rule11_FourSquareCipher_en.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "OPDFYY", "extra_args": {}, "output": "开始解密过程...\n加密文本: OPDFYY\n\n开始逐对字母解密:\n\n处理加密字母对: OP\n第一个字母 O 在ECHO表格中的位置: (0, 3)\n第二个字母 P 在VORTEX表格中的位置: (3, 4)\n解密后的字母对: NE\n\n处理加密字母对: DF\n第一个字母 D 在ECHO表格中的位置: (1, 1)\n第二个字母 F 在VORTEX表格中的位置: (2, 0)\n解密后的字母对: RV\n\n处理加密字母对: YY\n第一个字母 Y 在ECHO表格中的位置: (4, 3)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: II\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds\n        - Gird1\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n        - Gird2\n            - E  C  H  O  A\n            B  D  F  G  I\n            J  K  L  M  N\n            P  R  S  T  U\n            V  W  X  Y  Z\n        - Gird3\n            - V  O  R  T  E\n            X  A  B  C  D\n            F  G  H  I  J\n            K  L  M  N  P\n            S  U  W  Y  Z\n        - Gird4\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n- Encryption Steps:\n    - Cleans up plaintext, removing spaces and non-alphabetic characters, removing the letter Q, and converting all letters to uppercase.\n    - If the plaintext is of odd length, add a letter 'X' to make it of even length.\n    - Split the cleaned plaintext into two-letter groups.\n    - For each double-letter group p1,p2 of the plaintext:\n        - In Gird1 and Gird4, find the position of the first letter and the second letter.\n        - In Gird2 and Gird3, find the letters corresponding to these two positions, use these two letters as the result of encrypting the double letter combination.\n    - Concatenate all the encrypted double letter groups to form the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Cleans up ciphertext, removing spaces and non-alphabetic characters, and converting all letters to uppercase.\n    - Split the cleaned ciphertext into two-letter groups.\n    - For each double-letter group c1,c2 of the ciphertext:\n        - In Gird2 and Gird3, find the position of the first letter and the second letter.\n        - In Gird1 and Gird4, find the letters corresponding to these two positions, use these two letters as the result of decrypting the double letter combination.\n    - Concatenate all the decrypted double letter groups to form the final plaintext.\nencoded text: BIDYGYSH decode into plain text: pyritic\nencoded text: BIDYGYSH decode into plain text: pyritic\nencoded text: BIDYGYSH decode into plain text: pyritic\nencoded text: OEOMIOBKRYCYGI decode into plain text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 374, "source_filename": "icl_with_rule_decode_Kor_rule11_FourSquareCipher_en.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "OEOMIOBKRYCYGI", "extra_args": {}, "output": "开始解密过程...\n加密文本: OEOMIOBKRYCYGI\n\n开始逐对字母解密:\n\n处理加密字母对: OE\n第一个字母 O 在ECHO表格中的位置: (0, 3)\n第二个字母 E 在VORTEX表格中的位置: (0, 4)\n解密后的字母对: NO\n\n处理加密字母对: OM\n第一个字母 O 在ECHO表格中的位置: (0, 3)\n第二个字母 M 在VORTEX表格中的位置: (3, 2)\n解密后的字母对: NC\n\n处理加密字母对: IO\n第一个字母 I 在ECHO表格中的位置: (1, 4)\n第二个字母 O 在VORTEX表格中的位置: (0, 1)\n解密后的字母对: UL\n\n处理加密字母对: BK\n第一个字母 B 在ECHO表格中的位置: (1, 0)\n第二个字母 K 在VORTEX表格中的位置: (3, 0)\n解密后的字母对: PA\n\n处理加密字母对: RY\n第一个字母 R 在ECHO表格中的位置: (3, 1)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: BI\n\n处理加密字母对: CY\n第一个字母 C 在ECHO表格中的位置: (0, 1)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: LI\n\n处理加密字母对: GI\n第一个字母 G 在ECHO表格中的位置: (1, 3)\n第二个字母 I 在VORTEX表格中的位置: (2, 3)\n解密后的字母对: TY\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds\n        - Gird1\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n        - Gird2\n            - E  C  H  O  A\n            B  D  F  G  I\n            J  K  L  M  N\n            P  R  S  T  U\n            V  W  X  Y  Z\n        - Gird3\n            - V  O  R  T  E\n            X  A  B  C  D\n            F  G  H  I  J\n            K  L  M  N  P\n            S  U  W  Y  Z\n        - Gird4\n            - K  L  M  N  O\n            P  R  S  T  U\n            V  W  X  Y  Z\n            A  B  C  D  E\n            F  G  H  I  J\n- Encryption Steps:\n    - Cleans up plaintext, removing spaces and non-alphabetic characters, removing the letter Q, and converting all letters to uppercase.\n    - If the plaintext is of odd length, add a letter 'X' to make it of even length.\n    - Split the cleaned plaintext into two-letter groups.\n    - For each double-letter group p1,p2 of the plaintext:\n        - In Gird1 and Gird4, find the position of the first letter and the second letter.\n        - In Gird2 and Gird3, find the letters corresponding to these two positions, use these two letters as the result of encrypting the double letter combination.\n    - Concatenate all the encrypted double letter groups to form the final ciphertext.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - Four 5x5 Girds (Same as encryption)\n- Decryption Steps (exact opposite of encryption steps):\n    - Cleans up ciphertext, removing spaces and non-alphabetic characters, and converting all letters to uppercase.\n    - Split the cleaned ciphertext into two-letter groups.\n    - For each double-letter group c1,c2 of the ciphertext:\n        - In Gird2 and Gird3, find the position of the first letter and the second letter.\n        - In Gird1 and Gird4, find the letters corresponding to these two positions, use these two letters as the result of decrypting the double letter combination.\n    - Concatenate all the decrypted double letter groups to form the final plaintext.\ncipher text: BIDYGYSH decrypt to original information is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 375, "source_filename": "icl_with_rule_decode_Kor_rule11_FourSquareCipher_en.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "BIDYGYSH", "extra_args": {}, "output": "开始解密过程...\n加密文本: BIDYGYSH\n\n开始逐对字母解密:\n\n处理加密字母对: BI\n第一个字母 B 在ECHO表格中的位置: (1, 0)\n第二个字母 I 在VORTEX表格中的位置: (2, 3)\n解密后的字母对: PY\n\n处理加密字母对: DY\n第一个字母 D 在ECHO表格中的位置: (1, 1)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: RI\n\n处理加密字母对: GY\n第一个字母 G 在ECHO表格中的位置: (1, 3)\n第二个字母 Y 在VORTEX表格中的位置: (4, 3)\n解密后的字母对: TI\n\n处理加密字母对: SH\n第一个字母 S 在ECHO表格中的位置: (3, 2)\n第二个字母 H 在VORTEX表格中的位置: (2, 2)\n解密后的字母对: CX\n\n最终解密结果: PYRITICX\n", "ground_truth": "PYRITICX"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对明文进行加密\nTap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。\n明文: nonculpability 加密为密文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n明文: nonculpability 加密为密文: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n明文: nervii 加密为密文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 376, "source_filename": "icl_with_rule_encode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "NERVII", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'E'，它位于第 1 行第 5 列，编码为 '. .....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'V'，它位于第 5 行第 1 列，编码为 '..... .'\n步骤 5：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n最终步骤：加密完成，加密后的消息是：... ...  . .....  .... ..  ..... .  .. ....  .. ....\n", "ground_truth": "... ...  . .....  .... ..  ..... .  .. ....  .. ...."}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对明文进行加密\nTap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。\n明文: pyritic 编码为加密信息: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n明文: nervii 编码为加密信息: ... ...  . .....  .... ..  ..... .  .. ....  .. ....\n明文: nervii 编码为加密信息: ... ...  . .....  .... ..  ..... .  .. ....  .. ....\n明文: nonculpability 编码为加密信息: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 377, "source_filename": "icl_with_rule_encode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "NONCULPABILITY", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'O'，它位于第 3 行第 4 列，编码为 '... ....'\n步骤 3：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 4：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n步骤 5：正在加密字母 'U'，它位于第 4 行第 5 列，编码为 '.... .....'\n步骤 6：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 7：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 8：正在加密字母 'A'，它位于第 1 行第 1 列，编码为 '. .'\n步骤 9：正在加密字母 'B'，它位于第 1 行第 2 列，编码为 '. ..'\n步骤 10：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 11：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 12：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 13：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 14：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n最终步骤：加密完成，加密后的消息是：... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n", "ground_truth": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ...."}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对原始信息进行加密\nTap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。\n原始信息: pyritic 加密为密文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 378, "source_filename": "icl_with_rule_encode_TapCode_cn.jsonl", "cipher_name": "TapCode", "input": "PYRITIC", "extra_args": {}, "output": "步骤 1：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 2：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 5：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 7：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n最终步骤：加密完成，加密后的消息是：... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n", "ground_truth": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ..."}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原始信息进行加密\n加密方案概述：\n        1. 过滤掉非字母字符并转换为大写。\n        2. 将每个字母转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成暗码。\n        5. 按照加密逻辑，将暗码输出。\n        \n原始信息: pyritic 编码为暗码: OCAX..C\n原始信息: nonculpability 编码为暗码: NFJBXZUHXEJCCY\n原始信息: nervii 编码为暗码是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 379, "source_filename": "icl_with_rule_encode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "NERVII", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NERVII\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  E -> 层: 1, 列: 2, 行: 2\n  R -> 层: 2, 列: 3, 行: 3\n  V -> 层: 3, 列: 1, 行: 2\n  I -> 层: 1, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\nStep 3: 合并坐标:\n  层坐标: 212311\n  列坐标: 223133\n  行坐标: 223233\nStep 4: 重组坐标: ['212', '311', '223', '133', '223', '233']\nStep 5: 根据新坐标获取密文:\n  212 -> M\n  311 -> S\n  223 -> Q\n  133 -> I\n  223 -> Q\n  233 -> R\n", "ground_truth": "MSQIQR"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对原始信息进行加密\n加密方案概述：\n        1. 过滤掉非字母字符并转换为大写。\n        2. 将每个字母转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成密文。\n        5. 按照加密逻辑，将密文输出。\n        \n原始信息: pyritic 编码为密文: OCAX..C\n原始信息: nonculpability 编码为密文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 380, "source_filename": "icl_with_rule_encode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: NONCULPABILITY\nStep 2: 获取每个字符的坐标:\n  N -> 层: 2, 列: 2, 行: 2\n  O -> 层: 2, 列: 3, 行: 2\n  N -> 层: 2, 列: 2, 行: 2\n  C -> 层: 1, 列: 3, 行: 1\n  U -> 层: 3, 列: 3, 行: 1\n  L -> 层: 2, 列: 3, 行: 1\n  P -> 层: 2, 列: 1, 行: 3\n  A -> 层: 1, 列: 1, 行: 1\n  B -> 层: 1, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  L -> 层: 2, 列: 3, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  Y -> 层: 3, 列: 1, 行: 3\nStep 3: 合并坐标:\n  层坐标: 22213221112133\n  列坐标: 23233311233321\n  行坐标: 22211131131313\nStep 4: 重组坐标: ['222', '132', '211', '121', '332', '323', '331', '123', '332', '122', '211', '131', '131', '313']\nStep 5: 根据新坐标获取密文:\n  222 -> N\n  132 -> F\n  211 -> J\n  121 -> B\n  332 -> X\n  323 -> Z\n  331 -> U\n  123 -> H\n  332 -> X\n  122 -> E\n  211 -> J\n  131 -> C\n  131 -> C\n  313 -> Y\n", "ground_truth": "NFJBXZUHXEJCCY"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对初始文本进行加密\n加密方案概述：\n        1. 过滤掉非字母字符并转换为大写。\n        2. 将每个字母转换为对应的3D坐标。\n        3. 将3D坐标重新组合，每三个字符一组。\n        4. 根据新的坐标组在3D立方体中找到对应的字母，生成暗文。\n        5. 按照加密逻辑，将暗文输出。\n        \n初始文本: nonculpability 加密成为暗文: NFJBXZUHXEJCCY\n初始文本: nonculpability 加密成为暗文: NFJBXZUHXEJCCY\n初始文本: nonculpability 加密成为暗文: NFJBXZUHXEJCCY\n初始文本: pyritic 加密成为暗文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 381, "source_filename": "icl_with_rule_encode_TrifidCipher_cn.jsonl", "cipher_name": "TrifidCipher", "input": "PYRITIC", "extra_args": {}, "output": "立方体布局:\n第1层:\nA B C\nD E F\nG H I\n\n第2层:\nJ K L\nM N O\nP Q R\n\n第3层:\nS T U\nV W X\nY Z .\n\nStep 1: 过滤掉非字母字符并转换为大写: PYRITIC\nStep 2: 获取每个字符的坐标:\n  P -> 层: 2, 列: 1, 行: 3\n  Y -> 层: 3, 列: 1, 行: 3\n  R -> 层: 2, 列: 3, 行: 3\n  I -> 层: 1, 列: 3, 行: 3\n  T -> 层: 3, 列: 2, 行: 1\n  I -> 层: 1, 列: 3, 行: 3\n  C -> 层: 1, 列: 3, 行: 1\nStep 3: 合并坐标:\n  层坐标: 2321311\n  列坐标: 1133233\n  行坐标: 3333131\nStep 4: 重组坐标: ['232', '131', '111', '332', '333', '333', '131']\nStep 5: 根据新坐标获取密文:\n  232 -> O\n  131 -> C\n  111 -> A\n  332 -> X\n  333 -> .\n  333 -> .\n  131 -> C\n", "ground_truth": "OCAX..C"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原始信息进行加密\n加密规则:\n- 输入:\n    - 原始信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(所有行和列号从0开始计数):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - 位于所有行第一个字母的MKGWB是行首字母\n        - 位于所有列最后一个字母的PQUCL是列尾字母\n- 加密步骤:\n    - 移除原始信息中的空格、标点和字母J，并将所有字母转换为大写\n    - 对原始信息中的每个字母p:\n        - 在网格中找到字母p的位置，然后找到相应的行首和列尾字符\n        - 将行首和列尾字符连接成二元组作为该字母p的加密消息\n        - 例如，如果字母p是H，它在第4行，行首字符是B；它在第3列，列尾字符是H，所以加密消息是BH\n    \n    连接所有加密消息作为最终暗文输出\n原始信息: pyritic 编码为暗文: MIWEWTBIBTBIWI\n原始信息: nonculpability 编码为暗文: KTGEKTWIGIKHMIGTBBBIKHBIBTWE\n原始信息: pyritic 编码为暗文: MIWEWTBIBTBIWI\n原始信息: nervii 编码为暗文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 382, "source_filename": "icl_with_rule_encode_Kor_rule15_CollonCipher_cn.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n原始文本: nervii\n预处理后的文本(移除空格标点,转大写,移除J): NERVII\n\n逐字符加密:\n字符 N 位于第2行第2列\n-> 行首字符为K, 列尾字符为T\n-> 加密为: KT\n字符 E 位于第5行第3列\n-> 行首字符为B, 列尾字符为E\n-> 加密为: BE\n字符 R 位于第4行第2列\n-> 行首字符为W, 列尾字符为T\n-> 加密为: WT\n字符 V 位于第4行第4列\n-> 行首字符为W, 列尾字符为H\n-> 加密为: WH\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n\n最终加密结果: KTBEWTWHBIBI\n", "ground_truth": "KTBEWTWHBIBI"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对非加密信息进行加密\n加密规则:\n- 输入:\n    - 非加密信息: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(所有行和列号从0开始计数):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - 位于所有行第一个字母的MKGWB是行首字母\n        - 位于所有列最后一个字母的PQUCL是列尾字母\n- 加密步骤:\n    - 移除非加密信息中的空格、标点和字母J，并将所有字母转换为大写\n    - 对非加密信息中的每个字母p:\n        - 在网格中找到字母p的位置，然后找到相应的行首和列尾字符\n        - 将行首和列尾字符连接成二元组作为该字母p的加密消息\n        - 例如，如果字母p是H，它在第4行，行首字符是B；它在第3列，列尾字符是H，所以加密消息是BH\n    \n    连接所有加密消息作为最终暗文输出\n非加密信息: nonculpability 编码为暗文: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 383, "source_filename": "icl_with_rule_encode_Kor_rule15_CollonCipher_cn.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n原始文本: nonculpability\n预处理后的文本(移除空格标点,转大写,移除J): NONCULPABILITY\n\n逐字符加密:\n字符 N 位于第2行第2列\n-> 行首字符为K, 列尾字符为T\n-> 加密为: KT\n字符 O 位于第3行第3列\n-> 行首字符为G, 列尾字符为E\n-> 加密为: GE\n字符 N 位于第2行第2列\n-> 行首字符为K, 列尾字符为T\n-> 加密为: KT\n字符 C 位于第4行第5列\n-> 行首字符为W, 列尾字符为I\n-> 加密为: WI\n字符 U 位于第3行第5列\n-> 行首字符为G, 列尾字符为I\n-> 加密为: GI\n字符 L 位于第2行第4列\n-> 行首字符为K, 列尾字符为H\n-> 加密为: KH\n字符 P 位于第1行第5列\n-> 行首字符为M, 列尾字符为I\n-> 加密为: MI\n字符 A 位于第3行第2列\n-> 行首字符为G, 列尾字符为T\n-> 加密为: GT\n字符 B 位于第5行第1列\n-> 行首字符为B, 列尾字符为B\n-> 加密为: BB\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 L 位于第2行第4列\n-> 行首字符为K, 列尾字符为H\n-> 加密为: KH\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 T 位于第5行第2列\n-> 行首字符为B, 列尾字符为T\n-> 加密为: BT\n字符 Y 位于第4行第3列\n-> 行首字符为W, 列尾字符为E\n-> 加密为: WE\n\n最终加密结果: KTGEKTWIGIKHMIGTBBBIKHBIBTWE\n", "ground_truth": "KTGEKTWIGIKHMIGTBBBIKHBIBTWE"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 暗码: 不含标点和空格的大写字母字符串\n- 准备:\n    - 5x5网格(所有行和列号从0开始计数):\n        - M Z S D P\n        K N F L Q\n        G A O X U\n        W R Y V C\n        B T E H I\n        - 位于所有行第一个字母的MKGWB是行首字母\n        - 位于所有列最后一个字母的PQUCL是列尾字母\n- 加密步骤:\n    - 移除明文中的空格、标点和字母J，并将所有字母转换为大写\n    - 对明文中的每个字母p:\n        - 在网格中找到字母p的位置，然后找到相应的行首和列尾字符\n        - 将行首和列尾字符连接成二元组作为该字母p的加密消息\n        - 例如，如果字母p是H，它在第4行，行首字符是B；它在第3列，列尾字符是H，所以加密消息是BH\n    \n    连接所有加密消息作为最终暗码输出\n明文: nervii 编码为暗码: KTBEWTWHBIBI\n明文: nervii 编码为暗码: KTBEWTWHBIBI\n明文: pyritic 编码为暗码: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 384, "source_filename": "icl_with_rule_encode_Kor_rule15_CollonCipher_cn.jsonl", "cipher_name": "Kor_rule15_CollonCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n原始文本: pyritic\n预处理后的文本(移除空格标点,转大写,移除J): PYRITIC\n\n逐字符加密:\n字符 P 位于第1行第5列\n-> 行首字符为M, 列尾字符为I\n-> 加密为: MI\n字符 Y 位于第4行第3列\n-> 行首字符为W, 列尾字符为E\n-> 加密为: WE\n字符 R 位于第4行第2列\n-> 行首字符为W, 列尾字符为T\n-> 加密为: WT\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 T 位于第5行第2列\n-> 行首字符为B, 列尾字符为T\n-> 加密为: BT\n字符 I 位于第5行第5列\n-> 行首字符为B, 列尾字符为I\n-> 加密为: BI\n字符 C 位于第4行第5列\n-> 行首字符为W, 列尾字符为I\n-> 加密为: WI\n\n最终加密结果: MIWEWTBIBTBIWI\n", "ground_truth": "MIWEWTBIBTBIWI"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对原文进行加密\n加密规则:\n- 输入:\n    - 原文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 加密文本: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板:\n        - 准备一个空白网格和一个带孔的模板(栅栏)\n        - 使用的模板是:\n            ▮ ▮ ▮ ▮\n            ▮ ▮ ▯ ▯\n            ▮ ▮ ▮ ▯\n            ▯ ▮ ▮ ▮\n            其中白色的是孔，将模板放在空白网格上，只通过白色孔洞，在网格的对应位置填入字符。\n- 加密步骤:\n    - 将原文逐个分成16个字母的块(如果原文少于16个长度则为一个块)\n    - 对每个块执行以下加密操作:\n        - 将带孔的模板放在空白网格上\n        - 通过模板中的孔按顺序填入原文字母\n        - 模板总共有四个孔，所以填完四个字母后，需要将模板逆时针旋转90度\n        - 重复填充可见孔中的原文下一个字母并旋转模板，直到整个网格完全填满。这将执行4次填充+旋转，最终模板会转回原始模板。如果消息不足以填满整个网格，用填充字符(如'#')补充\n        - 网格填满后，按行读取网格内容作为该块的加密消息\n        - 进入下一个块时，清空网格内容并重做整个填充和旋转操作\n    最后，将所有块的加密消息连接在一起作为最终加密文本。\n原文: nervii 编码为加密文本是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 385, "source_filename": "icl_with_rule_encode_Kor_rule18_RotatingGridCipher_cn.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "NERVII", "extra_args": {}, "output": "处理后的输入文本: NERVII\n使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始加密过程:\n\n处理第1个块:\n当前块的文本: NERVII\n文本长度不足16，用#补充\n补充后的文本: NERVII##########\n\n填充后的网格:\n[['#' 'I' 'I' '#']\n ['#' '#' 'N' 'E']\n ['#' '#' '#' 'R']\n ['V' '#' '#' '#']]\n当前块的密文: #II###NE###RV###\n\n最终密文: #II###NE###RV###\n", "ground_truth": "#II###NE###RV###"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。\n请根据加密算法对明文进行加密\n加密规则:\n- 输入:\n    - 明文: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板:\n        - 准备一个空白网格和一个带孔的模板(栅栏)\n        - 使用的模板是:\n            ▮ ▮ ▮ ▮\n            ▮ ▮ ▯ ▯\n            ▮ ▮ ▮ ▯\n            ▯ ▮ ▮ ▮\n            其中白色的是孔，将模板放在空白网格上，只通过白色孔洞，在网格的对应位置填入字符。\n- 加密步骤:\n    - 将明文逐个分成16个字母的块(如果明文少于16个长度则为一个块)\n    - 对每个块执行以下加密操作:\n        - 将带孔的模板放在空白网格上\n        - 通过模板中的孔按顺序填入明文字母\n        - 模板总共有四个孔，所以填完四个字母后，需要将模板逆时针旋转90度\n        - 重复填充可见孔中的明文下一个字母并旋转模板，直到整个网格完全填满。这将执行4次填充+旋转，最终模板会转回原始模板。如果消息不足以填满整个网格，用填充字符(如'#')补充\n        - 网格填满后，按行读取网格内容作为该块的加密消息\n        - 进入下一个块时，清空网格内容并重做整个填充和旋转操作\n    最后，将所有块的加密消息连接在一起作为最终密文。\n明文: nonculpability 加密成密文: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 386, "source_filename": "icl_with_rule_encode_Kor_rule18_RotatingGridCipher_cn.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "处理后的输入文本: NONCULPABILITY\n使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始加密过程:\n\n处理第1个块:\n当前块的文本: NONCULPABILITY\n文本长度不足16，用#补充\n补充后的文本: NONCULPABILITY##\n\n填充后的网格:\n[['T' 'U' 'L' 'B']\n ['I' 'P' 'N' 'O']\n ['L' 'I' 'Y' 'N']\n ['C' '#' '#' 'A']]\n当前块的密文: TULBIPNOLIYNC##A\n\n最终密文: TULBIPNOLIYNC##A\n", "ground_truth": "TULBIPNOLIYNC##A"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 不含标点和空格的大写字母字符串\n- 输出:\n    - 密文: 不含标点和空格的大写字母字符串\n- 准备:\n    - 网格和模板:\n        - 准备一个空白网格和一个带孔的模板(栅栏)\n        - 使用的模板是:\n            ▮ ▮ ▮ ▮\n            ▮ ▮ ▯ ▯\n            ▮ ▮ ▮ ▯\n            ▯ ▮ ▮ ▮\n            其中白色的是孔，将模板放在空白网格上，只通过白色孔洞，在网格的对应位置填入字符。\n- 加密步骤:\n    - 将初始文本逐个分成16个字母的块(如果初始文本少于16个长度则为一个块)\n    - 对每个块执行以下加密操作:\n        - 将带孔的模板放在空白网格上\n        - 通过模板中的孔按顺序填入初始文本字母\n        - 模板总共有四个孔，所以填完四个字母后，需要将模板逆时针旋转90度\n        - 重复填充可见孔中的初始文本下一个字母并旋转模板，直到整个网格完全填满。这将执行4次填充+旋转，最终模板会转回原始模板。如果消息不足以填满整个网格，用填充字符(如'#')补充\n        - 网格填满后，按行读取网格内容作为该块的加密消息\n        - 进入下一个块时，清空网格内容并重做整个填充和旋转操作\n    最后，将所有块的加密消息连接在一起作为最终密文。\n初始文本: nervii 编码为密文: #II###NE###RV###\n初始文本: nonculpability 编码为密文: TULBIPNOLIYNC##A\n初始文本: nonculpability 编码为密文: TULBIPNOLIYNC##A\n初始文本: pyritic 编码为密文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 387, "source_filename": "icl_with_rule_encode_Kor_rule18_RotatingGridCipher_cn.jsonl", "cipher_name": "Kor_rule18_RotatingGridCipher", "input": "PYRITIC", "extra_args": {}, "output": "处理后的输入文本: PYRITIC\n使用的模板:\n▮ ▮ ▮ ▮\n▮ ▮ ▯ ▯\n▮ ▮ ▮ ▯\n▯ ▮ ▮ ▮\n\n开始加密过程:\n\n处理第1个块:\n当前块的文本: PYRITIC\n文本长度不足16，用#补充\n补充后的文本: PYRITIC#########\n\n填充后的网格:\n[['#' 'T' 'I' '#']\n ['#' 'C' 'P' 'Y']\n ['#' '#' '#' 'R']\n ['I' '#' '#' '#']]\n当前块的密文: #TI##CPY###RI###\n\n最终密文: #TI##CPY###RI###\n", "ground_truth": "#TI##CPY###RI###"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对原始信息进行加密\n加密规则:\n    - 输入:\n        - 原始信息: 大写字母和空格组成的字符串\n    - 输出:\n        - 隐文: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - S盒\n            - S_BOX = {\n            0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n            0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n            0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n            0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n            }\n    - 加密步骤:\n        1. 填充: 如果原始信息长度不是8字节的倍数，用\\x00(空字符)填充使其长度成为8字节的倍数\n        2. 分块: 将填充后的原始信息分成8字节的块\n        3. 块加密:\n            - 转换为字节: 使用ASCII编码将每个块转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 替换: 使用S盒替换每个字节的高4位和低4位并拼接\n            - 置换: 通过将每个字节左移1位进行简单置换\n            - 与密钥XOR: 置换后的字节块再次与固定密钥进行XOR运算\n        4. 十六进制编码: 将加密后的字节块转换为十六进制字符串\n        5. 拼接: 将所有加密块的十六进制字符串拼接形成最终隐文\n原始信息: nervii 加密成为隐文是什么？ 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 388, "source_filename": "icl_with_rule_encode_Kor_rule22_SBOXCipher_cn.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n1. 规范化输入文本: NERVII\n2. 开始分块处理...\n\n处理第1个块: NERVII\n  - 填充后的块: NERVII\u0000\u0000\n  - 与密钥XOR后: 7F7761627C7F3738\n  - S盒替换后: 0C00DAD7010C505E\n  - 左移置换后: 1800B5AF0218A0BC\n  - 最终与密钥XOR后: 2932869B372E9784\n\n最终加密结果: 2932869B372E9784\n", "ground_truth": "2932869B372E9784"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对非加密信息进行加密\n加密规则:\n    - 输入:\n        - 非加密信息: 大写字母和空格组成的字符串\n    - 输出:\n        - 加密信息: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - S盒\n            - S_BOX = {\n            0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n            0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n            0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n            0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n            }\n    - 加密步骤:\n        1. 填充: 如果非加密信息长度不是8字节的倍数，用\\x00(空字符)填充使其长度成为8字节的倍数\n        2. 分块: 将填充后的非加密信息分成8字节的块\n        3. 块加密:\n            - 转换为字节: 使用ASCII编码将每个块转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 替换: 使用S盒替换每个字节的高4位和低4位并拼接\n            - 置换: 通过将每个字节左移1位进行简单置换\n            - 与密钥XOR: 置换后的字节块再次与固定密钥进行XOR运算\n        4. 十六进制编码: 将加密后的字节块转换为十六进制字符串\n        5. 拼接: 将所有加密块的十六进制字符串拼接形成最终加密信息\n非加密信息: nervii 加密成加密信息: 2932869B372E9784\n非加密信息: nervii 加密成加密信息: 2932869B372E9784\n非加密信息: nonculpability 加密成加密信息是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 389, "source_filename": "icl_with_rule_encode_Kor_rule22_SBOXCipher_cn.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n1. 规范化输入文本: NONCULPABILITY\n2. 开始分块处理...\n\n处理第1个块: NONCULPA\n  - 填充后的块: NONCULPA\n  - 与密钥XOR后: 7F7D7D77607A6779\n  - S盒替换后: 0C020200DF04D008\n  - 左移置换后: 18040400BF08A110\n  - 最终与密钥XOR后: 293637348A3E9628\n\n处理第2个块: BILITY\n  - 填充后的块: BILITY\u0000\u0000\n  - 与密钥XOR后: 737B7F7D616F3738\n  - S盒替换后: 05060C02DADC505E\n  - 左移置换后: 0A0C1804B5B9A0BC\n  - 最终与密钥XOR后: 3B3E2B30808F9784\n\n最终加密结果: 293637348A3E96283B3E2B30808F9784\n", "ground_truth": "293637348A3E96283B3E2B30808F9784"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原文进行加密\n加密规则:\n    - 输入:\n        - 原文: 大写字母和空格组成的字符串\n    - 输出:\n        - 暗文: 表示加密数据的十六进制字符串(其中A-E需要大写)\n    - 准备:\n        - 固定密钥\n            - KEY = b'1234567890ABCDEF'\n        - S盒\n            - S_BOX = {\n            0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,\n            0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,\n            0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,\n            0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C\n            }\n    - 加密步骤:\n        1. 填充: 如果原文长度不是8字节的倍数，用\\x00(空字符)填充使其长度成为8字节的倍数\n        2. 分块: 将填充后的原文分成8字节的块\n        3. 块加密:\n            - 转换为字节: 使用ASCII编码将每个块转换为字节\n            - 与密钥XOR: 字节块与固定密钥进行XOR运算\n            - 替换: 使用S盒替换每个字节的高4位和低4位并拼接\n            - 置换: 通过将每个字节左移1位进行简单置换\n            - 与密钥XOR: 置换后的字节块再次与固定密钥进行XOR运算\n        4. 十六进制编码: 将加密后的字节块转换为十六进制字符串\n        5. 拼接: 将所有加密块的十六进制字符串拼接形成最终暗文\n原文: pyritic 加密为暗文是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 390, "source_filename": "icl_with_rule_encode_Kor_rule22_SBOXCipher_cn.jsonl", "cipher_name": "Kor_rule22_SBOXCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n1. 规范化输入文本: PYRITIC\n2. 开始分块处理...\n\n处理第1个块: PYRITIC\n  - 填充后的块: PYRITIC\u0000\n  - 与密钥XOR后: 616B617D617F7438\n  - S盒替换后: DAD6DA02DA0C095E\n  - 左移置换后: B5ADB504B51812BC\n  - 最终与密钥XOR后: 849F8630802E2584\n\n最终加密结果: 849F8630802E2584\n", "ground_truth": "849F8630802E2584"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\nPlease decrypt the password based on the encryption and decryption algorithms.\nEncryption Algorithm:\nTap Code is a simple cryptographic technique where each letter is represented by a dot pattern. The pattern is encoded by counting the number of dots in the rows and columns. For example, the letter 'A' is represented by a single dot in the first row and the first column.\nDecryption Algorithm:\nTo decode Tap Code, you need to count the number of dots in each row and column. By finding the position in the table, you can determine the corresponding letter.\nencrypted text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decrypt into plain text: pyritic\nencrypted text: ... ...  . .....  .... ..  ..... .  .. ....  .. .... decrypt into plain text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 391, "source_filename": "icl_with_rule_decode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "... ...  . .....  .... ..  ..... .  .. ....  .. ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '. .....'，它表示第 1 行第 5 列，对应的字母是 'E'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '..... .'，它表示第 5 行第 1 列，对应的字母是 'V'\n步骤 5：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n最终步骤：解码完成，解码后的消息是：NERVII \n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "Please decrypt the password based on the encryption and decryption algorithms.\nEncryption Algorithm:\nTap Code is a simple cryptographic technique where each letter is represented by a dot pattern. The pattern is encoded by counting the number of dots in the rows and columns. For example, the letter 'A' is represented by a single dot in the first row and the first column.\nDecryption Algorithm:\nTo decode Tap Code, you need to count the number of dots in each row and column. By finding the position in the table, you can determine the corresponding letter.\ncipher text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... decrypt into original information: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 392, "source_filename": "icl_with_rule_decode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....", "extra_args": {}, "output": "步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 2：正在解码 '... ....'，它表示第 3 行第 4 列，对应的字母是 'O'\n步骤 3：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'\n步骤 4：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n步骤 5：正在解码 '.... .....'，它表示第 4 行第 5 列，对应的字母是 'U'\n步骤 6：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 7：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 8：正在解码 '. .'，它表示第 1 行第 1 列，对应的字母是 'A'\n步骤 9：正在解码 '. ..'，它表示第 1 行第 2 列，对应的字母是 'B'\n步骤 10：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 11：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'\n步骤 12：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 13：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 14：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n最终步骤：解码完成，解码后的消息是：NONCULPABILITY \n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.\nPlease decrypt the password based on the encryption and decryption algorithms.\nEncryption Algorithm:\nTap Code is a simple cryptographic technique where each letter is represented by a dot pattern. The pattern is encoded by counting the number of dots in the rows and columns. For example, the letter 'A' is represented by a single dot in the first row and the first column.\nDecryption Algorithm:\nTo decode Tap Code, you need to count the number of dots in each row and column. By finding the position in the table, you can determine the corresponding letter.\ncipher text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... decrypt to plain text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 393, "source_filename": "icl_with_rule_decode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...", "extra_args": {}, "output": "步骤 1：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'\n步骤 2：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'\n步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'\n步骤 4：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 5：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'\n步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'\n步骤 7：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'\n最终步骤：解码完成，解码后的消息是：PYRITIC \n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n\n输入:\n- 初始文本: 大写字母字符串，不含标点和空格\n- period: 定义内圈多久旋转一次。周期性表示在加密过程中每处理指定数量的字符后，内圈将根据增量值旋转一次\n- increment: 定义内圈每次旋转的字符数。在每个周期结束时，内圈将根据增量值向右旋转相应数量的字符\n\n输出:\n- 隐文: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n加密步骤:\n- 对初始文本中的每个字符p:\n    - 在外圈找到该字符\n    - 用内圈对应位置的字符替换它\n    - 每加密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n解密算法:\n\n解密规则:\n\n输入:\n- 隐文: 大写字母字符串\n- period (与加密相同)\n- increment (与加密相同)\n\n输出:\n- 初始文本: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n解密步骤 (与加密步骤完全相反):\n- 对隐文中的每个字符c:\n    - 在内圈找到该字符\n    - 用外圈对应位置的字符替换它\n    - 每解密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n隐文: YLZRBS 解密成初始文本是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 394, "source_filename": "icl_with_rule_decode_Kor_rule9_AlertiCipher_cn.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "YLZRBS", "extra_args": {}, "output": "初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n解密过程:\n字符 Y 在内圈位置 18 对应外圈字符 N\n字符 L 在内圈位置 24 对应外圈字符 E\n字符 Z 在内圈位置 25 对应外圈字符 R\n字符 R 在内圈位置 16 对应外圈字符 V\n字符 B 在内圈位置 3 对应外圈字符 I\n已处理5个字符，内圈向右旋转4个位置\n字符 S 在内圈位置 7 对应外圈字符 I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n\n输入:\n- 初始文本: 大写字母字符串，不含标点和空格\n- period: 定义内圈多久旋转一次。周期性表示在加密过程中每处理指定数量的字符后，内圈将根据增量值旋转一次\n- increment: 定义内圈每次旋转的字符数。在每个周期结束时，内圈将根据增量值向右旋转相应数量的字符\n\n输出:\n- 密文: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n加密步骤:\n- 对初始文本中的每个字符p:\n    - 在外圈找到该字符\n    - 用内圈对应位置的字符替换它\n    - 每加密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n解密算法:\n\n解密规则:\n\n输入:\n- 密文: 大写字母字符串\n- period (与加密相同)\n- increment (与加密相同)\n\n输出:\n- 初始文本: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n解密步骤 (与加密步骤完全相反):\n- 对密文中的每个字符c:\n    - 在内圈找到该字符\n    - 用外圈对应位置的字符替换它\n    - 每解密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n密文: YNYEVZFGOSBHDF 解码为初始文本是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 395, "source_filename": "icl_with_rule_decode_Kor_rule9_AlertiCipher_cn.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "YNYEVZFGOSBHDF", "extra_args": {}, "output": "初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n解密过程:\n字符 Y 在内圈位置 14 对应外圈字符 N\n字符 N 在内圈位置 0 对应外圈字符 O\n字符 Y 在内圈位置 14 对应外圈字符 N\n字符 E 在内圈位置 11 对应外圈字符 C\n字符 V 在内圈位置 24 对应外圈字符 U\n已处理5个字符，内圈向右旋转4个位置\n字符 Z 在内圈位置 21 对应外圈字符 L\n字符 F 在内圈位置 5 对应外圈字符 P\n字符 G 在内圈位置 6 对应外圈字符 A\n字符 O 在内圈位置 17 对应外圈字符 B\n字符 S 在内圈位置 3 对应外圈字符 I\n已处理5个字符，内圈向右旋转4个位置\n字符 B 在内圈位置 25 对应外圈字符 L\n字符 H 在内圈位置 7 对应外圈字符 I\n字符 D 在内圈位置 4 对应外圈字符 T\n字符 F 在内圈位置 5 对应外圈字符 Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n\n输入:\n- 原文: 大写字母字符串，不含标点和空格\n- period: 定义内圈多久旋转一次。周期性表示在加密过程中每处理指定数量的字符后，内圈将根据增量值旋转一次\n- increment: 定义内圈每次旋转的字符数。在每个周期结束时，内圈将根据增量值向右旋转相应数量的字符\n\n输出:\n- 暗码: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n加密步骤:\n- 对原文中的每个字符p:\n    - 在外圈找到该字符\n    - 用内圈对应位置的字符替换它\n    - 每加密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n解密算法:\n\n解密规则:\n\n输入:\n- 暗码: 大写字母字符串\n- period (与加密相同)\n- increment (与加密相同)\n\n输出:\n- 原文: 大写字母字符串\n\n准备:\n- outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n- inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n\n解密步骤 (与加密步骤完全相反):\n- 对暗码中的每个字符c:\n    - 在内圈找到该字符\n    - 用外圈对应位置的字符替换它\n    - 每解密period个字符后，将内圈向右旋转increment个字符。例如，将'ZXCVBNMASDFGHJKLQWERTYUIOP'旋转4位得到'BNMASDFGHJKLQWERTYUIOPZXCV'\n        \n暗码: YLZRBS 解密为原文: nervii\n暗码: YNYEVZFGOSBHDF 解密为原文: nonculpability\n暗码: YLZRBS 解密为原文: nervii\n暗码: MCZBXSU 解密为原文: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 396, "source_filename": "icl_with_rule_decode_Kor_rule9_AlertiCipher_cn.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "MCZBXSU", "extra_args": {}, "output": "初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n解密过程:\n字符 M 在内圈位置 5 对应外圈字符 P\n字符 C 在内圈位置 1 对应外圈字符 Y\n字符 Z 在内圈位置 25 对应外圈字符 R\n字符 B 在内圈位置 3 对应外圈字符 I\n字符 X 在内圈位置 0 对应外圈字符 T\n已处理5个字符，内圈向右旋转4个位置\n字符 S 在内圈位置 7 对应外圈字符 I\n字符 U 在内圈位置 19 对应外圈字符 C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "请根据加密算法对明文进行加密\n\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密文本: 大写字母字符串\n- 准备:\n    - Polybius方阵:\n    1  2  3  4  5\n1   R  T  X  F  S\n2   W  C  M  V  H\n3   Z  J  A  P  B\n4   L  Q  Y  G  K\n5   N  E  U  D  I\n\n- 加密步骤:\n    - 对于每个明文字符p:\n        - 如果p是存在于Polybius方阵中的大写字母:\n            - 用字符在方阵中的行号和列号(都从1开始计数)替换该字符\n        - 特别地，字母O不存在于方阵中，用66替换\n    \n明文: pyritic 编码为加密文本: 34431155125522\n明文: nervii 编码为加密文本: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 397, "source_filename": "icl_with_rule_encode_Kor_rule4_PolybiusSquareCipher_cn.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密文本: nervii\n预处理后的文本: NERVII\n开始逐字符加密:\n处理字符: N\n字符 N 在方阵中的位置是: 第5行第1列，替换为: 51\n处理字符: E\n字符 E 在方阵中的位置是: 第5行第2列，替换为: 52\n处理字符: R\n字符 R 在方阵中的位置是: 第1行第1列，替换为: 11\n处理字符: V\n字符 V 在方阵中的位置是: 第2行第4列，替换为: 24\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n最终加密结果: 515211245555\n", "ground_truth": "515211245555"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对原文进行加密\n\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n    - Polybius方阵:\n    1  2  3  4  5\n1   R  T  X  F  S\n2   W  C  M  V  H\n3   Z  J  A  P  B\n4   L  Q  Y  G  K\n5   N  E  U  D  I\n\n- 加密步骤:\n    - 对于每个原文字符p:\n        - 如果p是存在于Polybius方阵中的大写字母:\n            - 用字符在方阵中的行号和列号(都从1开始计数)替换该字符\n        - 特别地，字母O不存在于方阵中，用66替换\n    \n原文: pyritic 编码为暗码: 34431155125522\n原文: nonculpability 编码为暗码是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 398, "source_filename": "icl_with_rule_encode_Kor_rule4_PolybiusSquareCipher_cn.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密文本: nonculpability\n预处理后的文本: NONCULPABILITY\n开始逐字符加密:\n处理字符: N\n字符 N 在方阵中的位置是: 第5行第1列，替换为: 51\n处理字符: O\n字符 O 不在Polybius方阵中，替换为: 66\n处理字符: N\n字符 N 在方阵中的位置是: 第5行第1列，替换为: 51\n处理字符: C\n字符 C 在方阵中的位置是: 第2行第2列，替换为: 22\n处理字符: U\n字符 U 在方阵中的位置是: 第5行第3列，替换为: 53\n处理字符: L\n字符 L 在方阵中的位置是: 第4行第1列，替换为: 41\n处理字符: P\n字符 P 在方阵中的位置是: 第3行第4列，替换为: 34\n处理字符: A\n字符 A 在方阵中的位置是: 第3行第3列，替换为: 33\n处理字符: B\n字符 B 在方阵中的位置是: 第3行第5列，替换为: 35\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: L\n字符 L 在方阵中的位置是: 第4行第1列，替换为: 41\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: T\n字符 T 在方阵中的位置是: 第1行第2列，替换为: 12\n处理字符: Y\n字符 Y 在方阵中的位置是: 第4行第3列，替换为: 43\n最终加密结果: 5166512253413433355541551243\n", "ground_truth": "5166512253413433355541551243"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法对明文进行加密\n\n加密规则:\n- 输入:\n    - 明文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 隐文: 大写字母字符串\n- 准备:\n    - Polybius方阵:\n    1  2  3  4  5\n1   R  T  X  F  S\n2   W  C  M  V  H\n3   Z  J  A  P  B\n4   L  Q  Y  G  K\n5   N  E  U  D  I\n\n- 加密步骤:\n    - 对于每个明文字符p:\n        - 如果p是存在于Polybius方阵中的大写字母:\n            - 用字符在方阵中的行号和列号(都从1开始计数)替换该字符\n        - 特别地，字母O不存在于方阵中，用66替换\n    \n明文: nervii 加密为隐文: 515211245555\n明文: nervii 加密为隐文: 515211245555\n明文: nervii 加密为隐文: 515211245555\n明文: pyritic 加密为隐文: ? 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 399, "source_filename": "icl_with_rule_encode_Kor_rule4_PolybiusSquareCipher_cn.jsonl", "cipher_name": "Kor_rule4_PolybiusSquareCipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密文本: pyritic\n预处理后的文本: PYRITIC\n开始逐字符加密:\n处理字符: P\n字符 P 在方阵中的位置是: 第3行第4列，替换为: 34\n处理字符: Y\n字符 Y 在方阵中的位置是: 第4行第3列，替换为: 43\n处理字符: R\n字符 R 在方阵中的位置是: 第1行第1列，替换为: 11\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: T\n字符 T 在方阵中的位置是: 第1行第2列，替换为: 12\n处理字符: I\n字符 I 在方阵中的位置是: 第5行第5列，替换为: 55\n处理字符: C\n字符 C 在方阵中的位置是: 第2行第2列，替换为: 22\n最终加密结果: 34431155125522\n", "ground_truth": "34431155125522"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 原始信息: 仅包含大写字母的字符串，不含标点和空格。\n- 输出:\n    - 暗文: 不含标点的字符串。\n- 准备:\n    - 多击编码表\n        | 字母 | 多击编码 |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- 加密步骤:\n    - 对于每个原始信息字符p:\n        - 如果p是大写字母且存在于多击编码表中:\n            - 用多击编码表中对应的多击编码替换p。\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 暗文: 不含标点的字符串。\n- 输出:\n    - 原始信息: 大写字母字符串。\n- 准备: 多击编码表(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 对于每个暗文中的多击编码c:\n        - 如果c是多击编码表中的编码:\n            - 用多击编码表中对应的大写字母替换c。\n\n暗文: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 解码为原始信息: nonculpability\n暗文: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 解码为原始信息: nonculpability\n暗文: 6^23^27^38^34^34^3 解码为原始信息: ? 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 400, "source_filename": "icl_with_rule_decode_Kor_rule3_MultiTapPhoneCode_cn.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "6^23^27^38^34^34^3", "extra_args": {}, "output": "加密文本: 6^23^27^38^34^34^3\n开始解密:\n多击编码 6^2 对应的字符是: N\n多击编码 3^2 对应的字符是: E\n多击编码 7^3 对应的字符是: R\n多击编码 8^3 对应的字符是: V\n多击编码 4^3 对应的字符是: I\n多击编码 4^3 对应的字符是: I\n解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 明文: 仅包含大写字母的字符串，不含标点和空格。\n- 输出:\n    - 暗文: 不含标点的字符串。\n- 准备:\n    - 多击编码表\n        | 字母 | 多击编码 |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- 加密步骤:\n    - 对于每个明文字符p:\n        - 如果p是大写字母且存在于多击编码表中:\n            - 用多击编码表中对应的多击编码替换p。\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 暗文: 不含标点的字符串。\n- 输出:\n    - 明文: 大写字母字符串。\n- 准备: 多击编码表(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 对于每个暗文中的多击编码c:\n        - 如果c是多击编码表中的编码:\n            - 用多击编码表中对应的大写字母替换c。\n\n暗文: 7^19^37^34^38^14^32^3 解密为明文: pyritic\n暗文: 6^23^27^38^34^34^3 解密为明文: nervii\n暗文: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 解密为明文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 401, "source_filename": "icl_with_rule_decode_Kor_rule3_MultiTapPhoneCode_cn.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "6^26^36^22^38^25^37^12^12^24^35^34^38^19^3", "extra_args": {}, "output": "加密文本: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3\n开始解密:\n多击编码 6^2 对应的字符是: N\n多击编码 6^3 对应的字符是: O\n多击编码 6^2 对应的字符是: N\n多击编码 2^3 对应的字符是: C\n多击编码 8^2 对应的字符是: U\n多击编码 5^3 对应的字符是: L\n多击编码 7^1 对应的字符是: P\n多击编码 2^1 对应的字符是: A\n多击编码 2^2 对应的字符是: B\n多击编码 4^3 对应的字符是: I\n多击编码 5^3 对应的字符是: L\n多击编码 4^3 对应的字符是: I\n多击编码 8^1 对应的字符是: T\n多击编码 9^3 对应的字符是: Y\n解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 原文: 仅包含大写字母的字符串，不含标点和空格。\n- 输出:\n    - 密文: 不含标点的字符串。\n- 准备:\n    - 多击编码表\n        | 字母 | 多击编码 |\n        | A | 2^1 |\n        | B | 2^2 |\n        | C | 2^3 |\n        | D | 3^1 |\n        | E | 3^2 |\n        | F | 3^3 |\n        | G | 4^1 |\n        | H | 4^2 |\n        | I | 4^3 |\n        | J | 5^1 |\n        | K | 5^2 |\n        | L | 5^3 |\n        | M | 6^1 |\n        | N | 6^2 |\n        | O | 6^3 |\n        | P | 7^1 |\n        | Q | 7^2 |\n        | R | 7^3 |\n        | S | 7^4 |\n        | T | 8^1 |\n        | U | 8^2 |\n        | V | 8^3 |\n        | W | 9^1 |\n        | X | 9^2 |\n        | Y | 9^3 |\n        | Z | 9^4 |\n- 加密步骤:\n    - 对于每个原文字符p:\n        - 如果p是大写字母且存在于多击编码表中:\n            - 用多击编码表中对应的多击编码替换p。\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 密文: 不含标点的字符串。\n- 输出:\n    - 原文: 大写字母字符串。\n- 准备: 多击编码表(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 对于每个密文中的多击编码c:\n        - 如果c是多击编码表中的编码:\n            - 用多击编码表中对应的大写字母替换c。\n\n密文: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 解码为原文: nonculpability\n密文: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 解码为原文: nonculpability\n密文: 7^19^37^34^38^14^32^3 解码为原文是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 402, "source_filename": "icl_with_rule_decode_Kor_rule3_MultiTapPhoneCode_cn.jsonl", "cipher_name": "Kor_rule3_MultiTapPhoneCode", "input": "7^19^37^34^38^14^32^3", "extra_args": {}, "output": "加密文本: 7^19^37^34^38^14^32^3\n开始解密:\n多击编码 7^1 对应的字符是: P\n多击编码 9^3 对应的字符是: Y\n多击编码 7^3 对应的字符是: R\n多击编码 4^3 对应的字符是: I\n多击编码 8^1 对应的字符是: T\n多击编码 4^3 对应的字符是: I\n多击编码 2^3 对应的字符是: C\n解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "运用您的专业技能，将提供的数据通过加密算法转换为安全的密文形式，是您的主要职责。\n请根据加密算法对原文进行加密\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 密文: 大写字母字符串\n- 准备:\n以下是经过整理后的文字：\n\n---\n\n### 5x5 网格布局\n\n共有四个5x5的字符网格，每个网格代表了不同的排列方式。\n\n1. **网格1 原始网格**\n\n这是最初的字符排列，按照特定顺序组织如下：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n2. **网格2 ECHO 网格**\n\n该网格根据\"ECHO\"这个词进行了重新排列：\n\n```\nE C H O A\nB D F G I\nJ K L M N\nP R S T U\nV W X Y Z\n```\n\n3. **网格3 VORTEX 网格**\n\n此网格基于\"VORTEX\"一词进行了独特的字符重组：\n\n```\nV O R T E\nX A B C D\nF G H I J\nK L M N P\nS U W Y Z\n```\n\n4. **网格4 重复原始网格**\n\n最后一个网格与第一个原始网格完全相同，没有进行任何改变：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n每个网格展示了不同主题词下字符的独特排列。\n- 加密步骤:\n    - 清理原文，移除空格和非字母字符，移除字母Q，转换为大写\n    - 如果原文长度为奇数，添加字母'X'使其成为偶数\n    - 将处理后的原文分成两个字母一组\n    - 对于每组两个字母p1,p2:\n        - 在网格1和网格4中找到第一个字母和第二个字母的位置\n        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果\n    - 连接所有加密后的字母组形成最终密文\n原文: pyritic 加密成密文: BIDYGYSH\n原文: nervii 加密成密文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 403, "source_filename": "icl_with_rule_encode_Kor_rule11_FourSquareCipher_cn.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n原始文本: nervii\n清理后的文本(移除非字母字符并转大写，移除Q): NERVII\n\n开始逐对字母加密:\n\n处理字母对: NE\n第一个字母 N 在原始表格中的位置: (0, 3)\n第二个字母 E 在原始表格中的位置: (3, 4)\n加密后的字母对: OP\n\n处理字母对: RV\n第一个字母 R 在原始表格中的位置: (1, 1)\n第二个字母 V 在原始表格中的位置: (2, 0)\n加密后的字母对: DF\n\n处理字母对: II\n第一个字母 I 在原始表格中的位置: (4, 3)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: YY\n\n最终加密结果: OPDFYY\n", "ground_truth": "OPDFYY"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对初始文本进行加密\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗码: 大写字母字符串\n- 准备:\n以下是经过整理后的文字：\n\n---\n\n### 5x5 网格布局\n\n共有四个5x5的字符网格，每个网格代表了不同的排列方式。\n\n1. **网格1 原始网格**\n\n这是最初的字符排列，按照特定顺序组织如下：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n2. **网格2 ECHO 网格**\n\n该网格根据\"ECHO\"这个词进行了重新排列：\n\n```\nE C H O A\nB D F G I\nJ K L M N\nP R S T U\nV W X Y Z\n```\n\n3. **网格3 VORTEX 网格**\n\n此网格基于\"VORTEX\"一词进行了独特的字符重组：\n\n```\nV O R T E\nX A B C D\nF G H I J\nK L M N P\nS U W Y Z\n```\n\n4. **网格4 重复原始网格**\n\n最后一个网格与第一个原始网格完全相同，没有进行任何改变：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n每个网格展示了不同主题词下字符的独特排列。\n- 加密步骤:\n    - 清理初始文本，移除空格和非字母字符，移除字母Q，转换为大写\n    - 如果初始文本长度为奇数，添加字母'X'使其成为偶数\n    - 将处理后的初始文本分成两个字母一组\n    - 对于每组两个字母p1,p2:\n        - 在网格1和网格4中找到第一个字母和第二个字母的位置\n        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果\n    - 连接所有加密后的字母组形成最终暗码\n初始文本: nervii 编码为暗码: OPDFYY\n初始文本: pyritic 编码为暗码: BIDYGYSH\n初始文本: nervii 编码为暗码: OPDFYY\n初始文本: nonculpability 编码为暗码: ? 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 404, "source_filename": "icl_with_rule_encode_Kor_rule11_FourSquareCipher_cn.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n原始文本: nonculpability\n清理后的文本(移除非字母字符并转大写，移除Q): NONCULPABILITY\n\n开始逐对字母加密:\n\n处理字母对: NO\n第一个字母 N 在原始表格中的位置: (0, 3)\n第二个字母 O 在原始表格中的位置: (0, 4)\n加密后的字母对: OE\n\n处理字母对: NC\n第一个字母 N 在原始表格中的位置: (0, 3)\n第二个字母 C 在原始表格中的位置: (3, 2)\n加密后的字母对: OM\n\n处理字母对: UL\n第一个字母 U 在原始表格中的位置: (1, 4)\n第二个字母 L 在原始表格中的位置: (0, 1)\n加密后的字母对: IO\n\n处理字母对: PA\n第一个字母 P 在原始表格中的位置: (1, 0)\n第二个字母 A 在原始表格中的位置: (3, 0)\n加密后的字母对: BK\n\n处理字母对: BI\n第一个字母 B 在原始表格中的位置: (3, 1)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: RY\n\n处理字母对: LI\n第一个字母 L 在原始表格中的位置: (0, 1)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: CY\n\n处理字母对: TY\n第一个字母 T 在原始表格中的位置: (1, 3)\n第二个字母 Y 在原始表格中的位置: (2, 3)\n加密后的字母对: GI\n\n最终加密结果: OEOMIOBKRYCYGI\n", "ground_truth": "OEOMIOBKRYCYGI"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n请根据加密算法对原文进行加密\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗文: 大写字母字符串\n- 准备:\n以下是经过整理后的文字：\n\n---\n\n### 5x5 网格布局\n\n共有四个5x5的字符网格，每个网格代表了不同的排列方式。\n\n1. **网格1 原始网格**\n\n这是最初的字符排列，按照特定顺序组织如下：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n2. **网格2 ECHO 网格**\n\n该网格根据\"ECHO\"这个词进行了重新排列：\n\n```\nE C H O A\nB D F G I\nJ K L M N\nP R S T U\nV W X Y Z\n```\n\n3. **网格3 VORTEX 网格**\n\n此网格基于\"VORTEX\"一词进行了独特的字符重组：\n\n```\nV O R T E\nX A B C D\nF G H I J\nK L M N P\nS U W Y Z\n```\n\n4. **网格4 重复原始网格**\n\n最后一个网格与第一个原始网格完全相同，没有进行任何改变：\n\n```\nK L M N O\nP R S T U\nV W X Y Z\nA B C D E\nF G H I J\n```\n\n每个网格展示了不同主题词下字符的独特排列。\n- 加密步骤:\n    - 清理原文，移除空格和非字母字符，移除字母Q，转换为大写\n    - 如果原文长度为奇数，添加字母'X'使其成为偶数\n    - 将处理后的原文分成两个字母一组\n    - 对于每组两个字母p1,p2:\n        - 在网格1和网格4中找到第一个字母和第二个字母的位置\n        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果\n    - 连接所有加密后的字母组形成最终暗文\n原文: nervii 加密为暗文: OPDFYY\n原文: pyritic 加密为暗文: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 405, "source_filename": "icl_with_rule_encode_Kor_rule11_FourSquareCipher_cn.jsonl", "cipher_name": "Kor_rule11_FourSquareCipher", "input": "PYRITICX", "extra_args": {}, "output": "开始加密过程...\n原始文本: pyritic\n清理后的文本(移除非字母字符并转大写，移除Q): PYRITIC\n文本长度为奇数，添加X: PYRITICX\n\n开始逐对字母加密:\n\n处理字母对: PY\n第一个字母 P 在原始表格中的位置: (1, 0)\n第二个字母 Y 在原始表格中的位置: (2, 3)\n加密后的字母对: BI\n\n处理字母对: RI\n第一个字母 R 在原始表格中的位置: (1, 1)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: DY\n\n处理字母对: TI\n第一个字母 T 在原始表格中的位置: (1, 3)\n第二个字母 I 在原始表格中的位置: (4, 3)\n加密后的字母对: GY\n\n处理字母对: CX\n第一个字母 C 在原始表格中的位置: (3, 2)\n第二个字母 X 在原始表格中的位置: (2, 2)\n加密后的字母对: SH\n\n最终加密结果: BIDYGYSH\n", "ground_truth": "BIDYGYSH"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Encryption Steps:\n    - For each given plaintext character p:\n        - Let  x  denote its position in the affine alphabet.\n        - Apply the affine coding function to compute y:\n            - y = (Ax + B) mod 26\n        - Find the corresponding letter in the affine alphabet at position  y , forming the encrypted message.\n\n \noriginal information: nonculpability encode to encrypted text: XEXAKJQMSLJLVH\noriginal information: nonculpability encode to encrypted text: XEXAKJQMSLJLVH\noriginal information: nervii encode to encrypted text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 406, "source_filename": "icl_with_rule_encode_Kor_rule5_AffineCipher_en.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "NERVII", "extra_args": {}, "output": "原始文本: NERVII\n开始加密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5\n字符 N -> 位置 7 -> 计算(3*7+5)%26=0 -> 加密为 X\n字符 E -> 位置 23 -> 计算(3*23+5)%26=22 -> 加密为 Y\n字符 R -> 位置 9 -> 计算(3*9+5)%26=6 -> 加密为 O\n字符 V -> 位置 12 -> 计算(3*12+5)%26=15 -> 加密为 F\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n加密完成，结果: XYOFLL\n", "ground_truth": "XYOFLL"}, "data_source": "Cipher"}
{"prompt": "Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Encryption Steps:\n    - For each given plaintext character p:\n        - Let  x  denote its position in the affine alphabet.\n        - Apply the affine coding function to compute y:\n            - y = (Ax + B) mod 26\n        - Find the corresponding letter in the affine alphabet at position  y , forming the encrypted message.\n\n \noriginal information: pyritic encode to encoded text: QHOLVLA\noriginal information: nonculpability encode to encoded text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 407, "source_filename": "icl_with_rule_encode_Kor_rule5_AffineCipher_en.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "原始文本: NONCULPABILITY\n开始加密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5\n字符 N -> 位置 7 -> 计算(3*7+5)%26=0 -> 加密为 X\n字符 O -> 位置 6 -> 计算(3*6+5)%26=23 -> 加密为 E\n字符 N -> 位置 7 -> 计算(3*7+5)%26=0 -> 加密为 X\n字符 C -> 位置 21 -> 计算(3*21+5)%26=16 -> 加密为 A\n字符 U -> 位置 4 -> 计算(3*4+5)%26=17 -> 加密为 K\n字符 L -> 位置 25 -> 计算(3*25+5)%26=2 -> 加密为 J\n字符 P -> 位置 8 -> 计算(3*8+5)%26=3 -> 加密为 Q\n字符 A -> 位置 16 -> 计算(3*16+5)%26=1 -> 加密为 M\n字符 B -> 位置 13 -> 计算(3*13+5)%26=18 -> 加密为 S\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 L -> 位置 25 -> 计算(3*25+5)%26=2 -> 加密为 J\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 T -> 位置 11 -> 计算(3*11+5)%26=12 -> 加密为 V\n字符 Y -> 位置 22 -> 计算(3*22+5)%26=19 -> 加密为 H\n加密完成，结果: XEXAKJQMSLJLVH\n", "ground_truth": "XEXAKJQMSLJLVH"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - affine alphabet = \"XMJQUDONPRGTVBWFAKSHZCYEIL\"\n    - Associate each letter with its position in the affine alphabet (starting from 0):\n        \n        X -> 0, M -> 1, J -> 2, Q -> 3, U -> 4, D -> 5, O -> 6, N -> 7,\n        P -> 8, R -> 9, G -> 10, T -> 11, V -> 12, B -> 13, W -> 14, F -> 15,\n        A -> 16, K -> 17, S -> 18, H -> 19, Z -> 20, C -> 21, Y -> 22, E -> 23, I -> 24, L -> 25\n        \n    - A: 3\n    - B: 5\n    - A_inv: 9\n- Encryption Steps:\n    - For each given plaintext character p:\n        - Let  x  denote its position in the affine alphabet.\n        - Apply the affine coding function to compute y:\n            - y = (Ax + B) mod 26\n        - Find the corresponding letter in the affine alphabet at position  y , forming the encrypted message.\n\n \nclear text: pyritic encode to encrypted text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 408, "source_filename": "icl_with_rule_encode_Kor_rule5_AffineCipher_en.jsonl", "cipher_name": "Kor_rule5_AffineCipher", "input": "PYRITIC", "extra_args": {}, "output": "原始文本: PYRITIC\n开始加密过程...\n使用仿射字母表: XMJQUDONPRGTVBWFAKSHZCYEIL\n参数 A=3, B=5\n字符 P -> 位置 8 -> 计算(3*8+5)%26=3 -> 加密为 Q\n字符 Y -> 位置 22 -> 计算(3*22+5)%26=19 -> 加密为 H\n字符 R -> 位置 9 -> 计算(3*9+5)%26=6 -> 加密为 O\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 T -> 位置 11 -> 计算(3*11+5)%26=12 -> 加密为 V\n字符 I -> 位置 24 -> 计算(3*24+5)%26=25 -> 加密为 L\n字符 C -> 位置 21 -> 计算(3*21+5)%26=16 -> 加密为 A\n加密完成，结果: QHOLVLA\n", "ground_truth": "QHOLVLA"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Preparation:\n    - 3 Girds(All row and column numbers are counted from 0.):\n        - Grid 1 (3x9):\n            \n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n            \n        - Grid 2 (9x3):\n            \n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n            \n        - Grid 3 (3x3):\n            \n            1 2 3\n            4 5 6\n            7 8 9\n            \n- Encryption Steps:\n    - Remove all spaces and punctuation from the plain text and convert them to capital letters.\n    - Cut the plaintext into groups of 6 characters, and if a group is less than 6 characters long, fill it with `#`.\n    - Split each group of 6 characters into 3 binary groups.\n    - For each binary group `(L1, L2)` do the following:\n        - Determine the row and column numbers  of `L1` in `grid1` as `(l1_row, l1_col)`.\n        - Determine the row and column numbers of `L2` in `grid2` as `(l2_row, l2_col)`.\n        - Find the corresponding number `num3` in `grid3` from `l1_row` and `l2_col`.\n        - Output the triple `(l1_col, num3, l2_row)` of the three numbers.\n            - For example, for the binary (T,H), the row and column numbers of T in `grid1` are (0,4), the row and column numbers of H in `grid2` are (5,0).\n            - Based on `l1_row` (0) and `l2_col`(0) , the corresponding number `num3` found in `grid3` is 1, and the final ternary is (4, 1, 5).\n    - After performing the previous step multiple times and converting all the binary groups to triples, read them in turn to form a string of numbers that can be used as encrypted information. For example (4, 1, 5), (2, 3, 3), (8, 1, 8), the final encrypted message is 415233818.\n\n \noriginal information: pyritic encode to encoded text: 061322422398898898\noriginal information: nervii encode to encoded text: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 409, "source_filename": "icl_with_rule_encode_Kor_rule14_DigrafidCipher_en.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "NERVII", "extra_args": {}, "output": "\n开始加密过程:\n原始文本: nervii\n预处理后的文本: NERVII\n补充#后的文本: NERVII\n分割成二元组: ['NE', 'RV', 'II']\n处理字符对 NE:\n- 在grid1中找到N的位置: 行=2, 列=6\n- 在grid2中找到E的位置: 行=0, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (6, 9, 0)\n处理字符对 RV:\n- 在grid1中找到R的位置: 行=0, 列=3\n- 在grid2中找到V的位置: 行=7, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (3, 2, 7)\n处理字符对 II:\n- 在grid1中找到I的位置: 行=0, 列=7\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (7, 2, 2)\n\n最终加密结果: 690327722\n", "ground_truth": "690327722"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Preparation:\n    - 3 Girds(All row and column numbers are counted from 0.):\n        - Grid 1 (3x9):\n            \n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n            \n        - Grid 2 (9x3):\n            \n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n            \n        - Grid 3 (3x3):\n            \n            1 2 3\n            4 5 6\n            7 8 9\n            \n- Encryption Steps:\n    - Remove all spaces and punctuation from the plain text and convert them to capital letters.\n    - Cut the plaintext into groups of 6 characters, and if a group is less than 6 characters long, fill it with `#`.\n    - Split each group of 6 characters into 3 binary groups.\n    - For each binary group `(L1, L2)` do the following:\n        - Determine the row and column numbers  of `L1` in `grid1` as `(l1_row, l1_col)`.\n        - Determine the row and column numbers of `L2` in `grid2` as `(l2_row, l2_col)`.\n        - Find the corresponding number `num3` in `grid3` from `l1_row` and `l2_col`.\n        - Output the triple `(l1_col, num3, l2_row)` of the three numbers.\n            - For example, for the binary (T,H), the row and column numbers of T in `grid1` are (0,4), the row and column numbers of H in `grid2` are (5,0).\n            - Based on `l1_row` (0) and `l2_col`(0) , the corresponding number `num3` found in `grid3` is 1, and the final ternary is (4, 1, 5).\n    - After performing the previous step multiple times and converting all the binary groups to triples, read them in turn to form a string of numbers that can be used as encrypted information. For example (4, 1, 5), (2, 3, 3), (8, 1, 8), the final encrypted message is 415233818.\n\n \noriginal information: nervii encrypt into encrypted text: 690327722\noriginal information: pyritic encrypt into encrypted text: 061322422398898898\noriginal information: nonculpability encrypt into encrypted text is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 410, "source_filename": "icl_with_rule_encode_Kor_rule14_DigrafidCipher_en.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "\n开始加密过程:\n原始文本: nonculpability\n预处理后的文本: NONCULPABILITY\n补充#后的文本: NONCULPABILITY####\n分割成二元组: ['NO', 'NC', 'UL', 'PA', 'BI', 'LI', 'TY', '##', '##']\n处理字符对 NO:\n- 在grid1中找到N的位置: 行=2, 列=6\n- 在grid2中找到O的位置: 行=2, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (6, 9, 2)\n处理字符对 NC:\n- 在grid1中找到N的位置: 行=2, 列=6\n- 在grid2中找到C的位置: 行=7, 列=0\n- 在grid3中找到对应数字: 7 (使用grid1的行2和grid2的列0)\n- 生成三元组: (6, 7, 7)\n处理字符对 UL:\n- 在grid1中找到U的位置: 行=0, 列=6\n- 在grid2中找到L的位置: 行=6, 列=0\n- 在grid3中找到对应数字: 1 (使用grid1的行0和grid2的列0)\n- 生成三元组: (6, 1, 6)\n处理字符对 PA:\n- 在grid1中找到P的位置: 行=1, 列=0\n- 在grid2中找到A的位置: 行=3, 列=1\n- 在grid3中找到对应数字: 5 (使用grid1的行1和grid2的列1)\n- 生成三元组: (0, 5, 3)\n处理字符对 BI:\n- 在grid1中找到B的位置: 行=2, 列=5\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 8 (使用grid1的行2和grid2的列1)\n- 生成三元组: (5, 8, 2)\n处理字符对 LI:\n- 在grid1中找到L的位置: 行=2, 列=0\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 8 (使用grid1的行2和grid2的列1)\n- 生成三元组: (0, 8, 2)\n处理字符对 TY:\n- 在grid1中找到T的位置: 行=0, 列=4\n- 在grid2中找到Y的位置: 行=1, 列=2\n- 在grid3中找到对应数字: 3 (使用grid1的行0和grid2的列2)\n- 生成三元组: (4, 3, 1)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n\n最终加密结果: 692677616053582082431898898\n", "ground_truth": "692677616053582082431898898"}, "data_source": "Cipher"}
{"prompt": "Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext: A numeric string without punctuation and spaces.\n- Preparation:\n    - 3 Girds(All row and column numbers are counted from 0.):\n        - Grid 1 (3x9):\n            \n            Q W E R T Y U I O\n            P A S D F G H J K\n            L Z X C V B N M #\n            \n        - Grid 2 (9x3):\n            \n            Q W E\n            R T Y\n            U I O\n            P A S\n            D F G\n            H J K\n            L Z X\n            C V B\n            N M #\n            \n        - Grid 3 (3x3):\n            \n            1 2 3\n            4 5 6\n            7 8 9\n            \n- Encryption Steps:\n    - Remove all spaces and punctuation from the plain text and convert them to capital letters.\n    - Cut the plaintext into groups of 6 characters, and if a group is less than 6 characters long, fill it with `#`.\n    - Split each group of 6 characters into 3 binary groups.\n    - For each binary group `(L1, L2)` do the following:\n        - Determine the row and column numbers  of `L1` in `grid1` as `(l1_row, l1_col)`.\n        - Determine the row and column numbers of `L2` in `grid2` as `(l2_row, l2_col)`.\n        - Find the corresponding number `num3` in `grid3` from `l1_row` and `l2_col`.\n        - Output the triple `(l1_col, num3, l2_row)` of the three numbers.\n            - For example, for the binary (T,H), the row and column numbers of T in `grid1` are (0,4), the row and column numbers of H in `grid2` are (5,0).\n            - Based on `l1_row` (0) and `l2_col`(0) , the corresponding number `num3` found in `grid3` is 1, and the final ternary is (4, 1, 5).\n    - After performing the previous step multiple times and converting all the binary groups to triples, read them in turn to form a string of numbers that can be used as encrypted information. For example (4, 1, 5), (2, 3, 3), (8, 1, 8), the final encrypted message is 415233818.\n\n \nclear text: nonculpability encrypt into encrypted text: 692677616053582082431898898\nclear text: nonculpability encrypt into encrypted text: 692677616053582082431898898\nclear text: nervii encrypt into encrypted text: 690327722\nclear text: pyritic encrypt into encrypted text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 411, "source_filename": "icl_with_rule_encode_Kor_rule14_DigrafidCipher_en.jsonl", "cipher_name": "Kor_rule14_DigrafidCipher", "input": "PYRITIC", "extra_args": {}, "output": "\n开始加密过程:\n原始文本: pyritic\n预处理后的文本: PYRITIC\n补充#后的文本: PYRITIC#####\n分割成二元组: ['PY', 'RI', 'TI', 'C#', '##', '##']\n处理字符对 PY:\n- 在grid1中找到P的位置: 行=1, 列=0\n- 在grid2中找到Y的位置: 行=1, 列=2\n- 在grid3中找到对应数字: 6 (使用grid1的行1和grid2的列2)\n- 生成三元组: (0, 6, 1)\n处理字符对 RI:\n- 在grid1中找到R的位置: 行=0, 列=3\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (3, 2, 2)\n处理字符对 TI:\n- 在grid1中找到T的位置: 行=0, 列=4\n- 在grid2中找到I的位置: 行=2, 列=1\n- 在grid3中找到对应数字: 2 (使用grid1的行0和grid2的列1)\n- 生成三元组: (4, 2, 2)\n处理字符对 C#:\n- 在grid1中找到C的位置: 行=2, 列=3\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (3, 9, 8)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n处理字符对 ##:\n- 在grid1中找到#的位置: 行=2, 列=8\n- 在grid2中找到#的位置: 行=8, 列=2\n- 在grid3中找到对应数字: 9 (使用grid1的行2和grid2的列2)\n- 生成三元组: (8, 9, 8)\n\n最终加密结果: 061322422398898898\n", "ground_truth": "061322422398898898"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n加密信息: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] 解密为明文: nonculpability\n加密信息: [112, 121, 114, 105, 116, 105, 99] 解密为明文: pyritic\n加密信息: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] 解密为明文: nonculpability\n加密信息: [112, 121, 114, 105, 116, 105, 99] 解密为明文: pyritic\n加密信息: [112, 121, 114, 105, 116, 105, 99] 解密为明文: pyritic\n加密信息: [110, 101, 114, 118, 105, 105] 解密为明文是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 412, "source_filename": "icl_decode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": [110, 101, 114, 118, 105, 105], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 101, Character: e\nASCII   Code: 114, Character: r\nASCII   Code: 118, Character: v\nASCII   Code: 105, Character: i\nASCII   Code: 105, Character: i\nResult: nervii\n", "ground_truth": "nervii"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n暗文: [110, 101, 114, 118, 105, 105] 解码为非加密信息: nervii\n暗文: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121] 解码为非加密信息: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 413, "source_filename": "icl_decode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121], "extra_args": {}, "output": "ASCII   Code: 110, Character: n\nASCII   Code: 111, Character: o\nASCII   Code: 110, Character: n\nASCII   Code: 99, Character: c\nASCII   Code: 117, Character: u\nASCII   Code: 108, Character: l\nASCII   Code: 112, Character: p\nASCII   Code: 97, Character: a\nASCII   Code: 98, Character: b\nASCII   Code: 105, Character: i\nASCII   Code: 108, Character: l\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 121, Character: y\nResult: nonculpability\n", "ground_truth": "nonculpability"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n暗码: [110, 101, 114, 118, 105, 105] 解密成为初始文本: nervii\n暗码: [110, 101, 114, 118, 105, 105] 解密成为初始文本: nervii\n暗码: [110, 101, 114, 118, 105, 105] 解密成为初始文本: nervii\n暗码: [112, 121, 114, 105, 116, 105, 99] 解密成为初始文本是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 414, "source_filename": "icl_decode_ASCII_cn.jsonl", "cipher_name": "ASCII", "input": [112, 121, 114, 105, 116, 105, 99], "extra_args": {}, "output": "ASCII   Code: 112, Character: p\nASCII   Code: 121, Character: y\nASCII   Code: 114, Character: r\nASCII   Code: 105, Character: i\nASCII   Code: 116, Character: t\nASCII   Code: 105, Character: i\nASCII   Code: 99, Character: c\nResult: pyritic\n", "ground_truth": "pyritic"}, "data_source": "Cipher"}
{"prompt": "Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.\nplain text: nonculpability encode into encrypted text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\nplain text: nonculpability encode into encrypted text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\nplain text: pyritic encode into encrypted text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\nplain text: pyritic encode into encrypted text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\nplain text: nonculpability encode into encrypted text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\nplain text: nervii encode into encrypted text: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 415, "source_filename": "icl_encode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "NERVII", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'E'，它位于第 1 行第 5 列，编码为 '. .....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'V'，它位于第 5 行第 1 列，编码为 '..... .'\n步骤 5：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n最终步骤：加密完成，加密后的消息是：... ...  . .....  .... ..  ..... .  .. ....  .. ....\n", "ground_truth": "... ...  . .....  .... ..  ..... .  .. ....  .. ...."}, "data_source": "Cipher"}
{"prompt": "plain text: pyritic encrypt to encoded text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\nplain text: pyritic encrypt to encoded text: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\nplain text: nonculpability encrypt to encoded text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 416, "source_filename": "icl_encode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "NONCULPABILITY", "extra_args": {}, "output": "步骤 1：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 2：正在加密字母 'O'，它位于第 3 行第 4 列，编码为 '... ....'\n步骤 3：正在加密字母 'N'，它位于第 3 行第 3 列，编码为 '... ...'\n步骤 4：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n步骤 5：正在加密字母 'U'，它位于第 4 行第 5 列，编码为 '.... .....'\n步骤 6：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 7：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 8：正在加密字母 'A'，它位于第 1 行第 1 列，编码为 '. .'\n步骤 9：正在加密字母 'B'，它位于第 1 行第 2 列，编码为 '. ..'\n步骤 10：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 11：正在加密字母 'L'，它位于第 3 行第 1 列，编码为 '... .'\n步骤 12：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 13：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 14：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n最终步骤：加密完成，加密后的消息是：... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\n", "ground_truth": "... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ...."}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\noriginal information: nonculpability encrypt to encoded text: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... ....\noriginal information: nervii encrypt to encoded text: ... ...  . .....  .... ..  ..... .  .. ....  .. ....\noriginal information: pyritic encrypt to encoded text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 417, "source_filename": "icl_encode_TapCode_en.jsonl", "cipher_name": "TapCode", "input": "PYRITIC", "extra_args": {}, "output": "步骤 1：正在加密字母 'P'，它位于第 3 行第 5 列，编码为 '... .....'\n步骤 2：正在加密字母 'Y'，它位于第 5 行第 4 列，编码为 '..... ....'\n步骤 3：正在加密字母 'R'，它位于第 4 行第 2 列，编码为 '.... ..'\n步骤 4：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 5：正在加密字母 'T'，它位于第 4 行第 4 列，编码为 '.... ....'\n步骤 6：正在加密字母 'I'，它位于第 2 行第 4 列，编码为 '.. ....'\n步骤 7：正在加密字母 'C'，它位于第 1 行第 3 列，编码为 '. ...'\n最终步骤：加密完成，加密后的消息是：... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ...\n", "ground_truth": "... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ..."}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对非加密信息进行加密\n\n加密规则:\n- 输入:\n    - 非加密信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 隐文: 字符串\n- 准备:\n    - 数字索引映射表\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - 摩斯密码表\n        - A: '.-',     B: '-...',   C: '-.-.',   D: '-..',\n        - E: '.',      F: '..-.',   G: '--.',    H: '....',\n        - I: '..',     J: '.---',   K: '-.-',    L: '.-..',\n        - M: '--',     N: '-.',     O: '---',    P: '.--.',\n        - Q: '--.-',   R: '.-.',    S: '...',    T: '-',\n        - U: '..-',    V: '...-',   W: '.--',    X: '-..-',\n        - Y: '-.--',   Z: '--..'\n- 加密步骤:\n    - 根据摩斯密码表将非加密信息中的每个字符转换为摩斯密码，用/分隔每个字符，例如AB对应'.-/-...'\n    - 将摩斯密码分成两个字符一组。如果摩斯密码长度为奇数，最后一个字符不进行映射，直接添加到隐文末尾\n    - 根据数字索引映射表将每组字符转换为对应的数字字符串\n    - 加密后的消息用字符串表示\n\n非加密信息: nonculpability 加密成为隐文: 485834472743247641732173645\n非加密信息: pyritic 加密成为隐文: 2482623187344\n非加密信息: nervii 加密成为隐文: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 418, "source_filename": "icl_with_rule_encode_Kor_rule12_MorbitCipher_cn.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "NERVII", "extra_args": {}, "output": "加密步骤开始:\n1. 将输入文本标准化为大写字母: NERVII\n2. 将文本转换为摩斯密码: -././.-./...-/../..\n3. 根据数字索引映射表将摩斯密码对转换为数字: 477471617.\n", "ground_truth": "477471617."}, "data_source": "Cipher"}
{"prompt": "作为编码器，您的任务是依据给出的案例中的加密算法，将明文加密为密文。\n请根据加密算法对原文进行加密\n\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 暗文: 字符串\n- 准备:\n    - 数字索引映射表\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - 摩斯密码表\n        - A: '.-',     B: '-...',   C: '-.-.',   D: '-..',\n        - E: '.',      F: '..-.',   G: '--.',    H: '....',\n        - I: '..',     J: '.---',   K: '-.-',    L: '.-..',\n        - M: '--',     N: '-.',     O: '---',    P: '.--.',\n        - Q: '--.-',   R: '.-.',    S: '...',    T: '-',\n        - U: '..-',    V: '...-',   W: '.--',    X: '-..-',\n        - Y: '-.--',   Z: '--..'\n- 加密步骤:\n    - 根据摩斯密码表将原文中的每个字符转换为摩斯密码，用/分隔每个字符，例如AB对应'.-/-...'\n    - 将摩斯密码分成两个字符一组。如果摩斯密码长度为奇数，最后一个字符不进行映射，直接添加到暗文末尾\n    - 根据数字索引映射表将每组字符转换为对应的数字字符串\n    - 加密后的消息用字符串表示\n\n原文: nonculpability 加密成暗文是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 419, "source_filename": "icl_with_rule_encode_Kor_rule12_MorbitCipher_cn.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "加密步骤开始:\n1. 将输入文本标准化为大写字母: NONCULPABILITY\n2. 将文本转换为摩斯密码: -./---/-./-.-./..-/.-../.--./.-/-.../../.-../../-/-.--\n3. 根据数字索引映射表将摩斯密码对转换为数字: 485834472743247641732173645\n", "ground_truth": "485834472743247641732173645"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。\n请根据加密算法对原文进行加密\n\n加密规则:\n- 输入:\n    - 原文: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密文本: 字符串\n- 准备:\n    - 数字索引映射表\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - 摩斯密码表\n        - A: '.-',     B: '-...',   C: '-.-.',   D: '-..',\n        - E: '.',      F: '..-.',   G: '--.',    H: '....',\n        - I: '..',     J: '.---',   K: '-.-',    L: '.-..',\n        - M: '--',     N: '-.',     O: '---',    P: '.--.',\n        - Q: '--.-',   R: '.-.',    S: '...',    T: '-',\n        - U: '..-',    V: '...-',   W: '.--',    X: '-..-',\n        - Y: '-.--',   Z: '--..'\n- 加密步骤:\n    - 根据摩斯密码表将原文中的每个字符转换为摩斯密码，用/分隔每个字符，例如AB对应'.-/-...'\n    - 将摩斯密码分成两个字符一组。如果摩斯密码长度为奇数，最后一个字符不进行映射，直接添加到加密文本末尾\n    - 根据数字索引映射表将每组字符转换为对应的数字字符串\n    - 加密后的消息用字符串表示\n\n原文: pyritic 加密为加密文本是什么？ \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 420, "source_filename": "icl_with_rule_encode_Kor_rule12_MorbitCipher_cn.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "PYRITIC", "extra_args": {}, "output": "加密步骤开始:\n1. 将输入文本标准化为大写字母: PYRITIC\n2. 将文本转换为摩斯密码: .--./-.--/.-./../-/../-.-.\n3. 根据数字索引映射表将摩斯密码对转换为数字: 2482623187344\n", "ground_truth": "2482623187344"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施加密处理。\n按照以下规则将原始文本进行编码转换:使用Atbash加密方式,该方式要求把每个英文字母替换成字母表中相对应的反向字母,比如把'A'换成'Z','B'换成'Y'等。对于文本中的非字母内容,则保留原样不做改变。\n非加密信息: pyritic 加密成为加密信息: kbirgrx\n非加密信息: nonculpability 加密成为加密信息: mlmxfokzyrorgb\n非加密信息: nervii 加密成为加密信息: ? 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 421, "source_filename": "icl_with_rule_encode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "nervii", "extra_args": {}, "output": "编码过程:\n转换单词: nervii\n  'n' -> 'm'\n  'e' -> 'v'\n  'r' -> 'i'\n  'v' -> 'e'\n  'i' -> 'r'\n  'i' -> 'r'\n编码后的文本: mvierr\n", "ground_truth": "mvierr"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n按照以下方式对原文进行编码:使用Atbash加密方法,该方法将每个英文字母替换为字母表中相对位置相反的字母。具体来说,A对应Z,B对应Y,C对应X等。注意,对于非字母的字符(如数字、标点符号等),应保持原样不变。\n原始信息: nervii 加密成暗文: mvierr\n原始信息: nervii 加密成暗文: mvierr\n原始信息: nonculpability 加密成暗文是什么？ 精心完成每一步。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 422, "source_filename": "icl_with_rule_encode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "nonculpability", "extra_args": {}, "output": "编码过程:\n转换单词: nonculpability\n  'n' -> 'm'\n  'o' -> 'l'\n  'n' -> 'm'\n  'c' -> 'x'\n  'u' -> 'f'\n  'l' -> 'o'\n  'p' -> 'k'\n  'a' -> 'z'\n  'b' -> 'y'\n  'i' -> 'r'\n  'l' -> 'o'\n  'i' -> 'r'\n  't' -> 'g'\n  'y' -> 'b'\n编码后的文本: mlmxfokzyrorgb\n", "ground_truth": "mlmxfokzyrorgb"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n按照以下方式对原文进行编码:使用Atbash加密方法,该方法将每个英文字母替换为字母表中相对位置相反的字母。具体来说,A对应Z,B对应Y,C对应X等。注意,对于非字母的字符(如数字、标点符号等),应保持原样不变。\n非加密信息: pyritic 加密为密文是什么？ 一步一步完成\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 423, "source_filename": "icl_with_rule_encode_AtbashCipher_cn.jsonl", "cipher_name": "AtbashCipher", "input": "pyritic", "extra_args": {}, "output": "编码过程:\n转换单词: pyritic\n  'p' -> 'k'\n  'y' -> 'b'\n  'r' -> 'i'\n  'i' -> 'r'\n  't' -> 'g'\n  'i' -> 'r'\n  'c' -> 'x'\n编码后的文本: kbirgrx\n", "ground_truth": "kbirgrx"}, "data_source": "Cipher"}
{"prompt": "您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 非加密信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密信息: 字符串\n- 准备:\n    - 数字索引映射表\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - 摩斯密码表\n        - A: '.-',     B: '-...',   C: '-.-.',   D: '-..',\n        - E: '.',      F: '..-.',   G: '--.',    H: '....',\n        - I: '..',     J: '.---',   K: '-.-',    L: '.-..',\n        - M: '--',     N: '-.',     O: '---',    P: '.--.',\n        - Q: '--.-',   R: '.-.',    S: '...',    T: '-',\n        - U: '..-',    V: '...-',   W: '.--',    X: '-..-',\n        - Y: '-.--',   Z: '--..'\n- 加密步骤:\n    - 根据摩斯密码表将非加密信息中的每个字符转换为摩斯密码，用/分隔每个字符，例如AB对应'.-/-...'\n    - 将摩斯密码分成两个字符一组。如果摩斯密码长度为奇数，最后一个字符不进行映射，直接添加到加密信息末尾\n    - 根据数字索引映射表将每组字符转换为对应的数字字符串\n    - 加密后的消息用字符串表示\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 加密信息: 数字字符串\n- 输出:\n    - 非加密信息: 大写字母字符串\n- 准备:\n    - 数字索引映射表(与加密相同)\n    - 摩斯密码表(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 根据数字索引映射表将加密信息中的每个数字转换为对应的字符对。如果加密信息末尾有非数字字符，则不处理。此时获得完整的摩斯密码\n    - 通过/分隔摩斯密码获得每个字符的摩斯密码\n    - 根据摩斯密码表将每个字符的摩斯密码转换为对应的非加密信息字符\n    - 最终非加密信息字符为大写字符串\n\n加密信息: 485834472743247641732173645 解密为非加密信息: nonculpability\n加密信息: 485834472743247641732173645 解密为非加密信息: nonculpability\n加密信息: 2482623187344 解密为非加密信息: pyritic\n加密信息: 477471617. 解密为非加密信息: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 424, "source_filename": "icl_with_rule_decode_Kor_rule12_MorbitCipher_cn.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "477471617.", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的加密文本: 477471617.\n2. 将数字转换回摩斯密码: -././.-./...-/../..\n3. 将摩斯密码转换为原文: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 初始文本: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密文本: 字符串\n- 准备:\n    - 数字索引映射表\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - 摩斯密码表\n        - A: '.-',     B: '-...',   C: '-.-.',   D: '-..',\n        - E: '.',      F: '..-.',   G: '--.',    H: '....',\n        - I: '..',     J: '.---',   K: '-.-',    L: '.-..',\n        - M: '--',     N: '-.',     O: '---',    P: '.--.',\n        - Q: '--.-',   R: '.-.',    S: '...',    T: '-',\n        - U: '..-',    V: '...-',   W: '.--',    X: '-..-',\n        - Y: '-.--',   Z: '--..'\n- 加密步骤:\n    - 根据摩斯密码表将初始文本中的每个字符转换为摩斯密码，用/分隔每个字符，例如AB对应'.-/-...'\n    - 将摩斯密码分成两个字符一组。如果摩斯密码长度为奇数，最后一个字符不进行映射，直接添加到加密文本末尾\n    - 根据数字索引映射表将每组字符转换为对应的数字字符串\n    - 加密后的消息用字符串表示\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 加密文本: 数字字符串\n- 输出:\n    - 初始文本: 大写字母字符串\n- 准备:\n    - 数字索引映射表(与加密相同)\n    - 摩斯密码表(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 根据数字索引映射表将加密文本中的每个数字转换为对应的字符对。如果加密文本末尾有非数字字符，则不处理。此时获得完整的摩斯密码\n    - 通过/分隔摩斯密码获得每个字符的摩斯密码\n    - 根据摩斯密码表将每个字符的摩斯密码转换为对应的初始文本字符\n    - 最终初始文本字符为大写字符串\n\n加密文本: 2482623187344 解密成初始文本: pyritic\n加密文本: 2482623187344 解密成初始文本: pyritic\n加密文本: 485834472743247641732173645 解密成初始文本: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 425, "source_filename": "icl_with_rule_decode_Kor_rule12_MorbitCipher_cn.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "485834472743247641732173645", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的加密文本: 485834472743247641732173645\n2. 将数字转换回摩斯密码: -./---/-./-.-./..-/.-../.--./.-/-.../../.-../../-/-.--\n3. 将摩斯密码转换为原文: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。\n请根据加密算法和解密算法对密码进行解密\n加密算法:\n\n加密规则:\n- 输入:\n    - 原始信息: 大写字母字符串，不含标点和空格\n- 输出:\n    - 加密信息: 字符串\n- 准备:\n    - 数字索引映射表\n        - '..' : 5\n        - '.-' : 4\n        - './' : 9\n        - '-.' : 8\n        - '--' : 6\n        - '-/' : 7\n        - '/.' : 3\n        - '/-' : 1\n        - '//' : 2\n    - 摩斯密码表\n        - A: '.-',     B: '-...',   C: '-.-.',   D: '-..',\n        - E: '.',      F: '..-.',   G: '--.',    H: '....',\n        - I: '..',     J: '.---',   K: '-.-',    L: '.-..',\n        - M: '--',     N: '-.',     O: '---',    P: '.--.',\n        - Q: '--.-',   R: '.-.',    S: '...',    T: '-',\n        - U: '..-',    V: '...-',   W: '.--',    X: '-..-',\n        - Y: '-.--',   Z: '--..'\n- 加密步骤:\n    - 根据摩斯密码表将原始信息中的每个字符转换为摩斯密码，用/分隔每个字符，例如AB对应'.-/-...'\n    - 将摩斯密码分成两个字符一组。如果摩斯密码长度为奇数，最后一个字符不进行映射，直接添加到加密信息末尾\n    - 根据数字索引映射表将每组字符转换为对应的数字字符串\n    - 加密后的消息用字符串表示\n\n解密算法:\n\n解密规则:\n- 输入:\n    - 加密信息: 数字字符串\n- 输出:\n    - 原始信息: 大写字母字符串\n- 准备:\n    - 数字索引映射表(与加密相同)\n    - 摩斯密码表(与加密相同)\n- 解密步骤(与加密步骤相反):\n    - 根据数字索引映射表将加密信息中的每个数字转换为对应的字符对。如果加密信息末尾有非数字字符，则不处理。此时获得完整的摩斯密码\n    - 通过/分隔摩斯密码获得每个字符的摩斯密码\n    - 根据摩斯密码表将每个字符的摩斯密码转换为对应的原始信息字符\n    - 最终原始信息字符为大写字符串\n\n加密信息: 485834472743247641732173645 解码为原始信息: nonculpability\n加密信息: 2482623187344 解码为原始信息: ? 请一步一步完成，确保过程详细严谨，结果正确。\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 426, "source_filename": "icl_with_rule_decode_Kor_rule12_MorbitCipher_cn.jsonl", "cipher_name": "Kor_rule12_MorbitCipher", "input": "2482623187344", "extra_args": {}, "output": "解密步骤开始:\n1. 接收到的加密文本: 2482623187344\n2. 将数字转换回摩斯密码: .--./-.--/.-./../-/../-.-.\n3. 将摩斯密码转换为原文: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - period: Defines how often the inner disc rotates. Periodicity indicates that after every number of characters processed in the encryption process, the inner disc will rotate once according to the incremental value.\n    - increment: Defines the number of characters the inner disc rotates each time. At the end of each cycle, the inner disc will rotate to the right by the corresponding number of characters based on the increment value. For example, if the increment is 4, the inner disc will rotate 4 characters to the right for each cycle that passes (e.g., 5 characters are processed).\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Encryption Steps:\n    - For each character p in the plaintext:\n        - Find the character in the outer_disk.\n        - Replace it with the character at the corresponding position on the inner_disk.\n        - After encrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n    - period (Same as encryption)\n    - increment (Same as encryption)\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Decryption Steps (exact opposite of encryption steps):\n    - For each character c in the ciphertext:\n        - Find the character in the inner_disk.\n        - Replace the character with the corresponding position in the outer_disk.\n        - After decrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\nencoded text: YLZRBS decrypt to plain text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 427, "source_filename": "icl_with_rule_decode_Kor_rule9_AlertiCipher_en.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "YLZRBS", "extra_args": {}, "output": "初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n解密过程:\n字符 Y 在内圈位置 18 对应外圈字符 N\n字符 L 在内圈位置 24 对应外圈字符 E\n字符 Z 在内圈位置 25 对应外圈字符 R\n字符 R 在内圈位置 16 对应外圈字符 V\n字符 B 在内圈位置 3 对应外圈字符 I\n已处理5个字符，内圈向右旋转4个位置\n字符 S 在内圈位置 7 对应外圈字符 I\n\n最终解密结果: NERVII\n", "ground_truth": "NERVII"}, "data_source": "Cipher"}
{"prompt": "As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - period: Defines how often the inner disc rotates. Periodicity indicates that after every number of characters processed in the encryption process, the inner disc will rotate once according to the incremental value.\n    - increment: Defines the number of characters the inner disc rotates each time. At the end of each cycle, the inner disc will rotate to the right by the corresponding number of characters based on the increment value. For example, if the increment is 4, the inner disc will rotate 4 characters to the right for each cycle that passes (e.g., 5 characters are processed).\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Encryption Steps:\n    - For each character p in the plaintext:\n        - Find the character in the outer_disk.\n        - Replace it with the character at the corresponding position on the inner_disk.\n        - After encrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n    - period (Same as encryption)\n    - increment (Same as encryption)\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Decryption Steps (exact opposite of encryption steps):\n    - For each character c in the ciphertext:\n        - Find the character in the inner_disk.\n        - Replace the character with the corresponding position in the outer_disk.\n        - After decrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\ncipher text: YNYEVZFGOSBHDF decrypt to original information is: ?  Carefully complete each step.\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 428, "source_filename": "icl_with_rule_decode_Kor_rule9_AlertiCipher_en.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "YNYEVZFGOSBHDF", "extra_args": {}, "output": "初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n解密过程:\n字符 Y 在内圈位置 14 对应外圈字符 N\n字符 N 在内圈位置 0 对应外圈字符 O\n字符 Y 在内圈位置 14 对应外圈字符 N\n字符 E 在内圈位置 11 对应外圈字符 C\n字符 V 在内圈位置 24 对应外圈字符 U\n已处理5个字符，内圈向右旋转4个位置\n字符 Z 在内圈位置 21 对应外圈字符 L\n字符 F 在内圈位置 5 对应外圈字符 P\n字符 G 在内圈位置 6 对应外圈字符 A\n字符 O 在内圈位置 17 对应外圈字符 B\n字符 S 在内圈位置 3 对应外圈字符 I\n已处理5个字符，内圈向右旋转4个位置\n字符 B 在内圈位置 25 对应外圈字符 L\n字符 H 在内圈位置 7 对应外圈字符 I\n字符 D 在内圈位置 4 对应外圈字符 T\n字符 F 在内圈位置 5 对应外圈字符 Y\n\n最终解密结果: NONCULPABILITY\n", "ground_truth": "NONCULPABILITY"}, "data_source": "Cipher"}
{"prompt": "**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n    - period: Defines how often the inner disc rotates. Periodicity indicates that after every number of characters processed in the encryption process, the inner disc will rotate once according to the incremental value.\n    - increment: Defines the number of characters the inner disc rotates each time. At the end of each cycle, the inner disc will rotate to the right by the corresponding number of characters based on the increment value. For example, if the increment is 4, the inner disc will rotate 4 characters to the right for each cycle that passes (e.g., 5 characters are processed).\n- Output:\n    - Ciphertext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Encryption Steps:\n    - For each character p in the plaintext:\n        - Find the character in the outer_disk.\n        - Replace it with the character at the corresponding position on the inner_disk.\n        - After encrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\n\n **Decryption Rules:**\n\n- Input:\n    - Ciphertext: Uppercase letters string.\n    - period (Same as encryption)\n    - increment (Same as encryption)\n- Output:\n    - Plaintext: Uppercase letters string.\n- Preparation:\n    - outer_disk = \"QWERTYUIOPASDFGHJZXCVBNMKL\"\n    - inner_disk = \"JKLZXCVBNMASDFGHJQWERTYUIO\"\n- Decryption Steps (exact opposite of encryption steps):\n    - For each character c in the ciphertext:\n        - Find the character in the inner_disk.\n        - Replace the character with the corresponding position in the outer_disk.\n        - After decrypting every `period` characters, rotate the inner_disk to the right by `increment` characters.For example, rotating 'ZXCVBNMASDFGHJKLQWERTYUIOP' by 4 gives 'BNMASDFGHJKLQWERTYUIOPZXCV'.\nencrypted text: YLZRBS decrypt to clear text: nervii\nencrypted text: YNYEVZFGOSBHDF decrypt to clear text: nonculpability\nencrypted text: YLZRBS decrypt to clear text: nervii\nencrypted text: MCZBXSU decrypt to clear text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 429, "source_filename": "icl_with_rule_decode_Kor_rule9_AlertiCipher_en.jsonl", "cipher_name": "Kor_rule9_AlertiCipher", "input": "MCZBXSU", "extra_args": {}, "output": "初始化Alberti密码盘:\n外圈: QWERTYUIOPASDFGHJZXCVBNMKL\n内圈: JKLZXCVBNMASDFGHJQWERTYUIO\n周期: 5 (每处理5个字符后旋转内圈)\n增量: 4 (每次旋转4个位置)\n\n解密过程:\n字符 M 在内圈位置 5 对应外圈字符 P\n字符 C 在内圈位置 1 对应外圈字符 Y\n字符 Z 在内圈位置 25 对应外圈字符 R\n字符 B 在内圈位置 3 对应外圈字符 I\n字符 X 在内圈位置 0 对应外圈字符 T\n已处理5个字符，内圈向右旋转4个位置\n字符 S 在内圈位置 7 对应外圈字符 I\n字符 U 在内圈位置 19 对应外圈字符 C\n\n最终解密结果: PYRITIC\n", "ground_truth": "PYRITIC"}, "data_source": "Cipher"}
{"prompt": "As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).\n- Preparation:\n    - Key(SHA hash value)\n        - Performing SHA-256 on \"SECRET_KEY\" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.\n- Encryption Steps:\n    - Converts a plaintext string into a sequence of bytes (ASCII encoding).\n    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.\n    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.\n\n \nplain text: nonculpability encrypt to encrypted text: 3da0640d88362ffe45b41326fbc0\nplain text: nonculpability encrypt to encrypted text: 3da0640d88362ffe45b41326fbc0\nplain text: pyritic encrypt to encrypted text: 23b6780789333c\nplain text: nervii encrypt to encrypted text is: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 430, "source_filename": "icl_with_rule_encode_Kor_rule25_SHACipher_en.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "NERVII", "extra_args": {}, "output": "开始加密过程...\n原始输入文本: nervii\n处理后的输入文本: NERVII\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n文本转换为字节序列: b'NERVII'\n开始XOR加密...\n加密后的字节序列: b'=\\xaax\\x18\\x943'\n最终加密结果(十六进制): 3daa78189433\n", "ground_truth": "3daa78189433"}, "data_source": "Cipher"}
{"prompt": "By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).\n- Preparation:\n    - Key(SHA hash value)\n        - Performing SHA-256 on \"SECRET_KEY\" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.\n- Encryption Steps:\n    - Converts a plaintext string into a sequence of bytes (ASCII encoding).\n    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.\n    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.\n\n \noriginal information: nonculpability encode to cipher text: ? \n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 431, "source_filename": "icl_with_rule_encode_Kor_rule25_SHACipher_en.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "NONCULPABILITY", "extra_args": {}, "output": "开始加密过程...\n原始输入文本: nonculpability\n处理后的输入文本: NONCULPABILITY\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n文本转换为字节序列: b'NONCULPABILITY'\n开始XOR加密...\n加密后的字节序列: b'=\\xa0d\\r\\x886/\\xfeE\\xb4\\x13&\\xfb\\xc0'\n最终加密结果(十六进制): 3da0640d88362ffe45b41326fbc0\n", "ground_truth": "3da0640d88362ffe45b41326fbc0"}, "data_source": "Cipher"}
{"prompt": "You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.\n**Encryption Rules:**\n\n- Input:\n    - Plaintext: Uppercase letters string without punctuation and spaces.\n- Output:\n    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).\n- Preparation:\n    - Key(SHA hash value)\n        - Performing SHA-256 on \"SECRET_KEY\" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.\n- Encryption Steps:\n    - Converts a plaintext string into a sequence of bytes (ASCII encoding).\n    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.\n    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.\n\n \noriginal information: nonculpability encode into cipher text: 3da0640d88362ffe45b41326fbc0\noriginal information: nonculpability encode into cipher text: 3da0640d88362ffe45b41326fbc0\noriginal information: nonculpability encode into cipher text: 3da0640d88362ffe45b41326fbc0\noriginal information: pyritic encode into cipher text is: ? step by step\n\nLet's think step by step and output the final answer with an example markdown formatting:\nFinal-answer: ```text\nBTWTBIGTKTGBGIKHGTBTBEME\n```", "ground_truth": {"id": 432, "source_filename": "icl_with_rule_encode_Kor_rule25_SHACipher_en.jsonl", "cipher_name": "Kor_rule25_SHACipher", "input": "PYRITIC", "extra_args": {}, "output": "开始加密过程...\n原始输入文本: pyritic\n处理后的输入文本: PYRITIC\n生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3\n文本转换为字节序列: b'PYRITIC'\n开始XOR加密...\n加密后的字节序列: b'#\\xb6x\\x07\\x893<'\n最终加密结果(十六进制): 23b6780789333c\n", "ground_truth": "23b6780789333c"}, "data_source": "Cipher"}