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* feat(run_eval): add checkpoint resume functionality and update example documentation; - update new bootcamp benchmark dataset * refactor(data_pipeline): optimize data generation pipeline; add multiple preset configurations for data generation * docs: update bootcamp list and add new scripts - Update Fulllist_InternBootcamp.md with new bootcamps and categories - Add new scripts to .gitignore: - examples/pipelines/filter_autogen_configs.py - examples/pipelines/quickgen_data_configs_from_eval_meta.py - Update dependencies in setup.py: - Add scipy and scikit-learn * refactor(internbootcamp): update bootcamp modules and improve error handling - Update import statements in __init__.py files - Add timestamp to target directory name in verl_data_preprocess.py - Improve error handling and scoring logic in bootcamp_judger.py - Remove unnecessary comments and update puzzle descriptions in multiple files
206 lines
7.5 KiB
Python
Executable file
206 lines
7.5 KiB
Python
Executable file
"""### 谜题描述
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The objective of the Bridges puzzle (Hashiwokakero) is to connect all numbered \"islands\" on a grid using horizontal/vertical bridges, adhering to these principles:
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1. **Island Numbers**: Each island (node) displays a number (1-8) indicating how many bridges must connect to it.
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- Example: A \"3\" island must have exactly 3 bridges linked to it.
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2. **Bridge Placement**:
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- Bridges connect **two adjacent islands** horizontally/vertically.
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- Bridges cannot cross islands, other bridges, or \"turn\" mid-connection.
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3. **Bridge Limits**:
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- A maximum of **2 bridges** can connect any pair of islands.
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- Bridges may overlap in straight lines if they connect different island pairs (no crossing).
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4. **Connectivity**:
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- All islands must be interconnected into a **single continuous network** via bridges.
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Key Constraints:
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- Bridges cannot be placed diagonally.
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- Islands cannot have fewer/more bridges than their number.
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- Overlapping bridges (parallel lines) must align in the same direction without intersecting.
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import re
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import random
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from collections import defaultdict
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class Bridgesbootcamp(Basebootcamp):
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def __init__(self, width=5, height=5):
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self.width = width
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self.height = height
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def case_generator(self):
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direction = random.choice(['horizontal', 'vertical'])
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islands = []
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bridges = []
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if direction == 'horizontal':
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x = random.randint(0, self.width-1)
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y1 = random.randint(0, self.height-3)
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y2 = y1 + 2
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islands = [
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{'x': x, 'y': y1, 'num': 2},
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{'x': x, 'y': y2, 'num': 2},
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]
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bridges = [{'from': (x, y1), 'to': (x, y2), 'count': 2}]
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else:
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y = random.randint(0, self.height-1)
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x1 = random.randint(0, self.width-3)
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x2 = x1 + 2
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islands = [
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{'x': x1, 'y': y, 'num': 2},
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{'x': x2, 'y': y, 'num': 2},
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]
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bridges = [{'from': (x1, y), 'to': (x2, y), 'count': 2}]
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return {
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'islands': islands,
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'bridges': bridges
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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islands = question_case['islands']
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islands_desc = []
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for island in islands:
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islands_desc.append(f"坐标({island['x']}, {island['y']})的岛屿数字为{island['num']}。")
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islands_text = '\n'.join(islands_desc)
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prompt = f"""你是Hashiwokakero谜题的解题专家，请根据以下规则连接所有岛屿：
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规则：
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1. 每个岛屿上的数字表示必须连接的桥梁数目。
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2. 桥梁必须水平或竖直连接相邻的岛屿，中间不能有其他岛屿或桥梁阻挡。
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3. 每对岛屿之间最多可以建造两座桥梁。
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4. 所有岛屿必须通过桥梁连通成一个单一网络。
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5. 桥梁不能交叉或转弯。
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当前的岛屿分布如下：
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{islands_text}
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请建造桥梁以满足所有条件，并将答案按照以下格式放置于[answer]和[/answer]之间。每个桥梁的格式为：(x1,y1)-(x2,y2):数量，多个桥梁用换行分隔。
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示例：
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[answer]
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(0,0)-(0,2):2
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(1,3)-(3,3):1
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[/answer]"""
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return prompt
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@staticmethod
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def extract_output(output):
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pattern = re.compile(r'\[answer\](.*?)\[/answer\]', re.DOTALL)
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matches = pattern.findall(output)
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if not matches:
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return None
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return matches[-1].strip()
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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bridges = []
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pattern = re.compile(r'\((\d+)\s*,\s*(\d+)\)\s*-\s*\((\d+)\s*,\s*(\d+)\)\s*:\s*(\d+)')
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matches = pattern.findall(solution)
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for m in matches:
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x1, y1, x2, y2, cnt = map(int, m)
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if cnt not in (1, 2):
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return False
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if (x1, y1) > (x2, y2):
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x1, x2, y1, y2 = x2, x1, y2, y1
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bridges.append({'from': (x1, y1), 'to': (x2, y2), 'count': cnt})
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island_coords = {(i['x'], i['y']) for i in identity['islands']}
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for bridge in bridges:
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if bridge['from'] not in island_coords or bridge['to'] not in island_coords:
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return False
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x1, y1 = bridge['from']
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x2, y2 = bridge['to']
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if not (x1 == x2 or y1 == y2):
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return False
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if x1 == x2:
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y_min, y_max = sorted([y1, y2])
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for y in range(y_min+1, y_max):
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if (x1, y) in island_coords:
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return False
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else:
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x_min, x_max = sorted([x1, x2])
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for x in range(x_min+1, x_max):
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if (x, y1) in island_coords:
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return False
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bridge_counts = defaultdict(int)
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for bridge in bridges:
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pair = (bridge['from'], bridge['to'])
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bridge_counts[pair] += bridge['count']
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if any(v > 2 for v in bridge_counts.values()):
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return False
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island_num = {(i['x'], i['y']): i['num'] for i in identity['islands']}
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usage = defaultdict(int)
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for bridge in bridges:
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usage[bridge['from']] += bridge['count']
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usage[bridge['to']] += bridge['count']
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for coord, num in island_num.items():
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if usage.get(coord, 0) != num:
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return False
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bridges_path = []
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for bridge in bridges:
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x1, y1 = bridge['from']
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x2, y2 = bridge['to']
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if x1 == x2:
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y_start, y_end = sorted([y1, y2])
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bridges_path.append(('vertical', x1, y_start, y_end))
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else:
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x_start, x_end = sorted([x1, x2])
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bridges_path.append(('horizontal', y1, x_start, x_end))
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for i in range(len(bridges_path)):
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ti, ai, si, ei = bridges_path[i]
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for j in range(i+1, len(bridges_path)):
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tj, aj, sj, ej = bridges_path[j]
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if ti == tj:
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continue
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if ti == 'horizontal':
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y_h = ai
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xh_s, xh_e = si, ej
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x_v = aj
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yv_s, yv_e = sj, ej
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else:
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x_v = ai
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yv_s, yv_e = si, ei
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y_h = aj
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xh_s, xh_e = sj, ej
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if (xh_s <= x_v <= xh_e) and (yv_s <= y_h <= yv_e):
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return False
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coord_to_id = {(i['x'], i['y']): idx for idx, i in enumerate(identity['islands'])}
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parent = list(range(len(coord_to_id)))
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def find(u):
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while parent[u] != u:
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parent[u] = parent[parent[u]]
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u = parent[u]
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return u
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def union(u, v):
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pu, pv = find(u), find(v)
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if pu != pv:
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parent[pu] = pv
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for bridge in bridges:
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u = coord_to_id[bridge['from']]
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v = coord_to_id[bridge['to']]
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union(u, v)
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roots = {find(i) for i in range(len(parent))}
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return len(roots) == 1
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except Exception as e:
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return False
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