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112 lines
No EOL
3.5 KiB
Python
Executable file
112 lines
No EOL
3.5 KiB
Python
Executable file
from .BaseCipherEnvironment import BaseCipherEnvironment
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polybius_square = [
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['R', 'T', 'X', 'F', 'S'],
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['W', 'C', 'M', 'V', 'H'],
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['Z', 'J', 'A', 'P', 'B'],
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['L', 'Q', 'Y', 'G', 'K'],
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['N', 'E', 'U', 'D', 'I']
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]
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def find_position(char):
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for i in range(len(polybius_square)):
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for j in range(len(polybius_square[i])):
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if polybius_square[i][j] == char:
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return i + 1, j + 1
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return None
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class KorPolybiusSquareCipherEnvironment(BaseCipherEnvironment):
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def __init__(self, *args, **kwargs):
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problem_description = "Polybius Square Cipher from KOR-bench"
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super().__init__(problem_description, *args, **kwargs)
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@property
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def cipher_name(self) -> str:
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return "Kor_rule4_PolybiusSquareCipher"
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def encode(self, text, **kwargs):
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print(f"开始加密文本: {text}")
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# 转换为大写并只保留字母
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text = ''.join([c.upper() for c in text if c.isalpha()])
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print(f"预处理后的文本: {text}")
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encrypted_text = ""
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print("开始逐字符加密:")
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for char in text:
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print(f"处理字符: {char}")
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if char == 'O':
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encrypted_text += '66'
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print(f"字符 {char} 不在Polybius方阵中，替换为: 66")
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else:
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row, col = find_position(char)
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if row and col:
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encrypted_text += f"{row}{col}"
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print(f"字符 {char} 在方阵中的位置是: 第{row}行第{col}列，替换为: {row}{col}")
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print(f"最终加密结果: {encrypted_text}")
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return encrypted_text
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def decode(self, text, **kwargs):
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print(f"开始解密文本: {text}")
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decrypted_text = ""
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i = 0
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print("开始逐对数字解密:")
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while i < len(text):
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if text[i:i+2] == '66':
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decrypted_text += 'O'
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print(f"遇到数字对: 66，解密为: O")
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i += 2
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elif text[i].isdigit():
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row = int(text[i])
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col = int(text[i+1])
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char = polybius_square[row-1][col-1]
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decrypted_text += char
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print(f"遇到数字对: {row}{col}，对应方阵位置: 第{row}行第{col}列，解密为: {char}")
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i += 2
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print(f"最终解密结果: {decrypted_text}")
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return decrypted_text
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def get_decode_rule(self, ):
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decode_rule = """
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解密规则:
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- 输入:
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- 密文: 字符串
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- 输出:
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- 明文: 大写字母字符串
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- 准备:
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- Polybius方阵(与加密相同)
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- 解密步骤(与加密步骤相反):
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- 对于密文中的每对数字CrCc:
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- 根据CrCc表示的行列号在Polybius方阵中找到对应字母
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- 如果CrCc=66，替换为"O"
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"""
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return decode_rule
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def get_encode_rule(self, ):
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encode_rule = """
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加密规则:
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- 输入:
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- 明文: 大写字母字符串，不含标点和空格
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- 输出:
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- 密文: 大写字母字符串
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- 准备:
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- Polybius方阵:
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1 2 3 4 5
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1 R T X F S
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2 W C M V H
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3 Z J A P B
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4 L Q Y G K
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5 N E U D I
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- 加密步骤:
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- 对于每个明文字符p:
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- 如果p是存在于Polybius方阵中的大写字母:
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- 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
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- 特别地，字母O不存在于方阵中，用66替换
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"""
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return encode_rule
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