init-commit

internbootcamp/libs/masyu/masyu_z3_solver.py

@@ -0,0 +1,774 @@
+from z3 import *
+import numpy as np
+import time
+import itertools
+import logging
+
+# 配置日志
+logging.basicConfig(
+    level=logging.INFO,
+    format='%(asctime)s - %(levelname)s - %(message)s'
+)
+logger = logging.getLogger(__name__)
+
+
+def solve_masyu_with_z3(puzzle, timeout=None):
+    """
+    使用Z3求解器解决Masyu谜题
+
+    参数:
+        puzzle: 表示谜题的二维数组或列表，0表示空单元格，1表示白珠，2表示黑珠
+        timeout: 求解超时时间（秒）
+
+    返回:
+        解决方案路径，超时或无解则返回None
+    """
+    start_time = time.time()
+
+    # 将列表转换为NumPy数组（如果尚未转换）
+    if not isinstance(puzzle, np.ndarray):
+        # 打印原始谜题数据以进行调试
+        # print("Original puzzle data:", puzzle)
+
+        # 检查谜题格式
+        if isinstance(puzzle, list) and all(isinstance(row, list) for row in puzzle):
+            # 如果是字符表示的谜题，转换为数字表示
+            if isinstance(puzzle[0][0], str):
+                # print("Converting character-based puzzle to numeric format...")
+                numeric_puzzle = []
+                for row in puzzle:
+                    numeric_row = []
+                    for cell in row:
+                        if cell == ' ':
+                            numeric_row.append(0)  # 空单元格
+                        elif cell == 'W':
+                            numeric_row.append(1)  # 白珠
+                        elif cell == 'B':
+                            numeric_row.append(2)  # 黑珠
+                        else:
+                            numeric_row.append(0)  # 默认为空
+                    numeric_puzzle.append(numeric_row)
+                puzzle = np.array(numeric_puzzle)
+            else:
+                # 标准的二维列表格式
+                puzzle = np.array(puzzle)
+        else:
+            # 尝试解析其他格式
+            # print("Attempting to parse non-standard puzzle format...")
+
+            # 如果是字典格式，尝试提取珠子位置
+            if isinstance(puzzle, dict):
+                # 获取白珠和黑珠的位置
+                white_pearls = puzzle.get('white', [])
+                black_pearls = puzzle.get('black', [])
+
+                # 确定谜题尺寸
+                max_row = max([p[0] for p in white_pearls + black_pearls], default=0) + 1
+                max_col = max([p[1] for p in white_pearls + black_pearls], default=0) + 1
+
+                # 创建空谜题
+                size = max(max_row, max_col, 4)  # 至少4x4
+                puzzle_array = np.zeros((size, size), dtype=int)
+
+                # 放置珠子
+                for r, c in white_pearls:
+                    puzzle_array[r, c] = 1
+                for r, c in black_pearls:
+                    puzzle_array[r, c] = 2
+
+                puzzle = puzzle_array
+            else:
+                # 如果格式无法识别，创建一个默认的空谜题
+                # print("Warning: Unrecognized puzzle format. Creating empty 4x4 puzzle.")
+                puzzle = np.zeros((4, 4), dtype=int)
+
+    rows, cols = puzzle.shape
+
+    # 打印谜题信息
+    # print(f"Puzzle dimensions: {rows}x{cols}")
+    # print("Puzzle contents:")
+    for r in range(rows):
+        row_str = ""
+        for c in range(cols):
+            if puzzle[r, c] == 0:
+                row_str += ". "
+            elif puzzle[r, c] == 1:
+                row_str += "○ "
+            elif puzzle[r, c] == 2:
+                row_str += "● "
+        # print(row_str)
+
+    # 计算珠子数量
+    white_pearls = np.sum(puzzle == 1)
+    black_pearls = np.sum(puzzle == 2)
+    # print(f"White pearls: {white_pearls}, Black pearls: {black_pearls}")
+
+    # 如果没有珠子，这可能是一个格式问题
+    if white_pearls == 0 and black_pearls == 0:
+        # print("Warning: No pearls found in the puzzle. This might indicate a format issue.")
+        return None
+
+    s = Solver()
+
+    if timeout:
+        s.set("timeout", timeout * 1000)  # Z3 timeout in milliseconds
+
+    # 变量定义
+    in_path = {}
+    for r in range(rows):
+        for c in range(cols):
+            in_path[(r, c)] = Bool(f"in_path_{r}_{c}")
+
+    # 边变量
+    edge = {}
+    for r in range(rows):
+        for c in range(cols):
+            # 水平边
+            if c < cols - 1:
+                edge[(r, c, r, c + 1)] = Bool(f"edge_{r}_{c}_{r}_{c + 1}")
+                edge[(r, c + 1, r, c)] = edge[(r, c, r, c + 1)]  # 双向边
+            # 垂直边
+            if r < rows - 1:
+                edge[(r, c, r + 1, c)] = Bool(f"edge_{r}_{c}_{r + 1}_{c}")
+                edge[(r + 1, c, r, c)] = edge[(r, c, r + 1, c)]  # 双向边
+
+    # 约束：路径必须是连续的闭环
+    for r in range(rows):
+        for c in range(cols):
+            # 如果单元格在路径上，它必须恰好有两个相邻的单元格也在路径上
+            neighbors = []
+            if r > 0:
+                neighbors.append((r - 1, c))
+            if r < rows - 1:
+                neighbors.append((r + 1, c))
+            if c > 0:
+                neighbors.append((r, c - 1))
+            if c < cols - 1:
+                neighbors.append((r, c + 1))
+
+            # 如果单元格在路径上，它必须有恰好两个连接的邻居
+            connected_neighbors = [edge[(r, c, nr, nc)] for nr, nc in neighbors]
+
+            # 在路径上的单元格必须有恰好两个连接的邻居
+            s.add(Implies(in_path[(r, c)], PbEq([(conn, 1) for conn in connected_neighbors], 2)))
+
+            # 如果单元格不在路径上，它不能连接到任何邻居
+            s.add(Implies(Not(in_path[(r, c)]), And([Not(edge[(r, c, nr, nc)]) for nr, nc in neighbors])))
+
+            # 确保边的一致性：如果有边连接，则两端的单元格都必须在路径上
+            for nr, nc in neighbors:
+                s.add(Implies(edge[(r, c, nr, nc)], And(in_path[(r, c)], in_path[(nr, nc)])))
+
+    # 添加珠子约束
+    for r in range(rows):
+        for c in range(cols):
+            if puzzle[r, c] == 1:  # 白珠
+                # print(f"Adding white pearl constraints at ({r}, {c})")
+                # 白珠必须在路径上
+                s.add(in_path[(r, c)])
+
+                # 白珠约束：路径必须在白珠处直行
+                horizontal_neighbors = []
+                if c > 0:
+                    horizontal_neighbors.append((r, c - 1))
+                if c < cols - 1:
+                    horizontal_neighbors.append((r, c + 1))
+
+                vertical_neighbors = []
+                if r > 0:
+                    vertical_neighbors.append((r - 1, c))
+                if r < rows - 1:
+                    vertical_neighbors.append((r + 1, c))
+
+                # 如果水平方向有两个邻居，则路径可以水平穿过
+                if len(horizontal_neighbors) == 2:
+                    h_straight = And(edge[(r, c, horizontal_neighbors[0][0], horizontal_neighbors[0][1])],
+                                     edge[(r, c, horizontal_neighbors[1][0], horizontal_neighbors[1][1])])
+                else:
+                    h_straight = False
+
+                # 如果垂直方向有两个邻居，则路径可以垂直穿过
+                if len(vertical_neighbors) == 2:
+                    v_straight = And(edge[(r, c, vertical_neighbors[0][0], vertical_neighbors[0][1])],
+                                     edge[(r, c, vertical_neighbors[1][0], vertical_neighbors[1][1])])
+                else:
+                    v_straight = False
+
+                # 路径必须在白珠处直行（水平或垂直）
+                valid_straights = [x for x in [h_straight, v_straight] if x is not False]
+                if valid_straights:
+                    s.add(Or(*valid_straights))
+
+                # 添加白珠转弯约束（至少一侧必须转弯）
+                if h_straight is not False:
+                    # 检查左侧是否可以转弯
+                    left_r, left_c = horizontal_neighbors[0]
+                    left_turns = []
+                    if left_r > 0:
+                        left_turns.append(edge[(left_r, left_c, left_r - 1, left_c)])
+                    if left_r < rows - 1:
+                        left_turns.append(edge[(left_r, left_c, left_r + 1, left_c)])
+
+                    # 检查右侧是否可以转弯
+                    right_r, right_c = horizontal_neighbors[1]
+                    right_turns = []
+                    if right_r > 0:
+                        right_turns.append(edge[(right_r, right_c, right_r - 1, right_c)])
+                    if right_r < rows - 1:
+                        right_turns.append(edge[(right_r, right_c, right_r + 1, right_c)])
+
+                    # 至少一侧必须转弯
+                    all_turns = left_turns + right_turns
+                    if all_turns:
+                        s.add(Implies(h_straight, Or(*all_turns)))
+
+                if v_straight is not False:
+                    # 检查上侧是否可以转弯
+                    top_r, top_c = vertical_neighbors[0]
+                    top_turns = []
+                    if top_c > 0:
+                        top_turns.append(edge[(top_r, top_c, top_r, top_c - 1)])
+                    if top_c < cols - 1:
+                        top_turns.append(edge[(top_r, top_c, top_r, top_c + 1)])
+
+                    # 检查下侧是否可以转弯
+                    bottom_r, bottom_c = vertical_neighbors[1]
+                    bottom_turns = []
+                    if bottom_c > 0:
+                        bottom_turns.append(edge[(bottom_r, bottom_c, bottom_r, bottom_c - 1)])
+                    if bottom_c < cols - 1:
+                        bottom_turns.append(edge[(bottom_r, bottom_c, bottom_r, bottom_c + 1)])
+
+                    # 至少一侧必须转弯
+                    all_turns = top_turns + bottom_turns
+                    if all_turns:
+                        s.add(Implies(v_straight, Or(*all_turns)))
+
+            elif puzzle[r, c] == 2:  # 黑珠
+                # print(f"Adding black pearl constraints at ({r}, {c})")
+                # 黑珠必须在路径上
+                s.add(in_path[(r, c)])
+
+                # 获取四个方向的邻居
+                neighbors = []
+                directions = []  # 存储方向向量
+
+                if r > 0:
+                    neighbors.append((r - 1, c))
+                    directions.append((-1, 0))
+
+                if r < rows - 1:
+                    neighbors.append((r + 1, c))
+                    directions.append((1, 0))
+
+                if c > 0:
+                    neighbors.append((r, c - 1))
+                    directions.append((0, -1))
+
+                if c < cols - 1:
+                    neighbors.append((r, c + 1))
+                    directions.append((0, 1))
+
+                # 黑珠必须连接到恰好两个邻居
+                connected_neighbors = [edge[(r, c, nr, nc)] for nr, nc in neighbors]
+                s.add(PbEq([(conn, 1) for conn in connected_neighbors], 2))
+
+                # 黑珠约束1：路径必须在黑珠处转弯
+                for i, (nr1, nc1) in enumerate(neighbors):
+                    for j, (nr2, nc2) in enumerate(neighbors):
+                        if i < j:
+                            if (nr1 == r and nr2 == r and nc1 != nc2) or (nc1 == c and nc2 == c and nr1 != nr2):
+                                s.add(Not(And(edge[(r, c, nr1, nc1)], edge[(r, c, nr2, nc2)])))
+
+                # 黑珠约束2：路径必须在黑珠两侧至少直行一格
+                for i, ((nr, nc), (dr, dc)) in enumerate(zip(neighbors, directions)):
+                    next_r, next_c = nr + dr, nc + dc
+                    if 0 <= next_r < rows and 0 <= next_c < cols:
+                        s.add(Implies(
+                            edge[(r, c, nr, nc)],
+                            edge[(nr, nc, next_r, next_c)]
+                        ))
+
+    # 添加约束：至少有一个单元格在路径上
+    s.add(Or([in_path[(r, c)] for r in range(rows) for c in range(cols)]))
+
+    # 添加子路径消除约束
+    # 使用一个辅助变量表示单元格的访问顺序
+    visit_order = {}
+    for r in range(rows):
+        for c in range(cols):
+            visit_order[(r, c)] = Int(f"visit_{r}_{c}")
+            # 如果单元格不在路径上，其访问顺序为0
+            s.add(Implies(Not(in_path[(r, c)]), visit_order[(r, c)] == 0))
+            # 如果单元格在路径上，其访问顺序为正数
+            s.add(Implies(in_path[(r, c)], visit_order[(r, c)] > 0))
+            # 访问顺序不能超过单元格总数
+            s.add(visit_order[(r, c)] <= rows * cols)
+
+    # 确保访问顺序形成一个有效的序列
+    for r in range(rows):
+        for c in range(cols):
+            neighbors = []
+            if r > 0:
+                neighbors.append((r - 1, c))
+            if r < rows - 1:
+                neighbors.append((r + 1, c))
+            if c > 0:
+                neighbors.append((r, c - 1))
+            if c < cols - 1:
+                neighbors.append((r, c + 1))
+
+            # 如果单元格在路径上，它的访问顺序必须比其中一个连接的邻居大1，比另一个小1
+            # 或者是首尾连接的特殊情况
+            neighbor_constraints = []
+            for nr, nc in neighbors:
+                neighbor_constraints.append(
+                    And(
+                        edge[(r, c, nr, nc)],
+                        Or(
+                            visit_order[(r, c)] == visit_order[(nr, nc)] + 1,
+                            visit_order[(r, c)] == visit_order[(nr, nc)] - 1,
+                            # 特殊情况：处理环的首尾连接
+                            And(
+                                visit_order[(r, c)] == 1,
+                                visit_order[(nr, nc)] == rows * cols
+                            ),
+                            And(
+                                visit_order[(r, c)] == rows * cols,
+                                visit_order[(nr, nc)] == 1
+                            )
+                        )
+                    )
+                )
+
+            # 如果单元格在路径上，必须满足至少两个邻居约束
+            s.add(Implies(
+                in_path[(r, c)],
+                PbEq([(constraint, 1) for constraint in neighbor_constraints], 2)
+            ))
+
+    # 求解
+    # print("Starting Z3 solver...")
+    result = s.check()
+    solve_time = time.time() - start_time
+    # print(f"Z3 result: {result}")
+
+    if result == sat:
+        model = s.model()
+        # print("Solution found!")
+
+        # 构建有序路径
+        path_with_order = []
+        for r in range(rows):
+            for c in range(cols):
+                if model.evaluate(in_path[(r, c)]):
+                    order = model.evaluate(visit_order[(r, c)]).as_long()
+                    if order > 0:  # 只考虑在路径上的单元格
+                        path_with_order.append((order, (r, c)))
+
+        # 按访问顺序排序
+        path_with_order.sort()
+        ordered_path = [pos for _, pos in path_with_order]
+
+        # 确保路径是闭环
+        if ordered_path and ordered_path[0] != ordered_path[-1]:
+            ordered_path.append(ordered_path[0])
+
+            # 验证路径连续性
+            if not is_continuous_path(ordered_path):
+                # print("Warning: Z3 solution is not a continuous path. Attempting to fix...")
+                fixed_path = construct_continuous_path(ordered_path, model, edge, rows, cols)
+                if fixed_path:
+                    ordered_path = fixed_path
+                    # print(f"Fixed path: {ordered_path}")
+                else:
+                    # print("Failed to fix path. Using fallback solver...")
+                    ordered_path = solve_with_backtracking(puzzle)
+
+            # 验证路径是否满足所有规则
+            if ordered_path and not verify_complete_path(puzzle, ordered_path):
+                # print("Warning: Z3 solution does not satisfy all rules. Using fallback solver...")
+                ordered_path = solve_with_backtracking(puzzle)
+
+            # print(f"Final solution path: {ordered_path}")
+            # print(f"Solved in {solve_time:.2f} seconds.")
+            return ordered_path
+
+    elif result == unknown:
+        # print(f"Z3 timed out after {solve_time:.2f} seconds")
+        # print("Using fallback backtracking solver...")
+        return solve_with_backtracking(puzzle)
+    else:
+        # print(f"Z3 determined puzzle is unsatisfiable in {solve_time:.2f} seconds")
+        return None
+
+
+def is_continuous_path(path):
+    """检查路径是否连续"""
+    if not path or len(path) < 2:
+        return False
+
+    for i in range(len(path) - 1):
+        r1, c1 = path[i]
+        r2, c2 = path[i + 1]
+        if not ((abs(r1 - r2) == 1 and c1 == c2) or (r1 == r2 and abs(c1 - c2) == 1)):
+            # print(f"Path discontinuity at {path[i]} to {path[i + 1]}")
+            return False
+
+    return True
+
+
+def construct_continuous_path(cells, model, edge, rows, cols):
+    """
+    从Z3解中构建连续路径
+
+    参数:
+        cells: 路径上的单元格集合
+        model: Z3模型
+        edge: 边变量字典
+        rows, cols: 网格尺寸
+
+    返回:
+        连续路径
+    """
+    if not cells:
+        return None
+
+    # 构建邻接图
+    graph = {cell: [] for cell in cells}
+    for r1 in range(rows):
+        for c1 in range(cols):
+            for dr, dc in [(0, 1), (1, 0)]:
+                r2, c2 = r1 + dr, c1 + dc
+                if r2 < rows and c2 < cols:
+                    if (r1, c1) in graph and (r2, c2) in graph:
+                        if model.evaluate(edge[(r1, c1, r2, c2)]):
+                            graph[(r1, c1)].append((r2, c2))
+                            graph[(r2, c2)].append((r1, c1))
+
+    # 从第一个单元格开始DFS
+    start = cells[0]
+    path = [start]
+    visited = {start}
+
+    # DFS构建路径
+    current = start
+    while len(visited) < len(cells):
+        found_next = False
+        for neighbor in graph[current]:
+            if neighbor not in visited:
+                path.append(neighbor)
+                visited.add(neighbor)
+                current = neighbor
+                found_next = True
+                break
+
+        if not found_next:
+            # 如果无法继续，尝试从其他已访问点继续
+            for i, node in enumerate(path):
+                for neighbor in graph[node]:
+                    if neighbor not in visited:
+                        # 重新排序路径
+                        new_path = path[i:] + path[:i]
+                        path = new_path
+                        current = node
+                        found_next = True
+                        break
+                if found_next:
+                    break
+
+            # 如果仍然无法继续，可能是无效解
+            if not found_next:
+                return None
+
+    # 检查是否可以闭合路径
+    if start in graph[current]:
+        path.append(start)  # 闭合路径
+
+    return path
+
+
+def solve_with_backtracking(puzzle):
+    """
+    使用回溯算法解决Masyu谜题
+
+    参数:
+        puzzle: 谜题网格
+
+    返回:
+        solution_path: 解决方案路径，如果无解则返回None
+    """
+    # print("Using improved backtracking algorithm to solve Masyu puzzle...")
+
+    rows, cols = puzzle.shape
+
+    # 获取所有珠子位置
+    pearls = []
+    for r in range(rows):
+        for c in range(cols):
+            if puzzle[r, c] in [1, 2]:  # 白珠或黑珠
+                pearls.append((r, c))
+
+    # 如果没有珠子，返回None
+    if not pearls:
+        return None
+
+    # 尝试从每个珠子开始
+    for start in pearls:
+        # 初始化路径和访问状态
+        path = [start]
+        visited = {start}
+
+        # 尝试构建路径
+        if improved_backtrack(puzzle, path, visited, pearls, start):
+            # 确保路径是闭环
+            if path[0] != path[-1]:
+                path.append(path[0])
+            return path
+
+    return None
+
+
+def improved_backtrack(puzzle, path, visited, pearls, start_pos):
+    """
+    改进的回溯算法构建路径
+
+    参数:
+        puzzle: 谜题网格
+        path: 当前路径
+        visited: 已访问的单元格
+        pearls: 所有珠子位置
+        start_pos: 起始位置
+
+    返回:
+        是否找到有效路径
+    """
+    # 如果已经访问了所有珠子，尝试闭合路径
+    if all(p in visited for p in pearls):
+        # 检查是否可以回到起点
+        r, c = path[-1]
+        start_r, start_c = start_pos
+
+        # 如果可以直接回到起点，路径有效
+        if abs(r - start_r) + abs(c - start_c) == 1:
+            # 验证整个路径是否满足所有规则
+            if verify_complete_path(puzzle, path + [start_pos]):
+                return True
+
+        return False
+
+    # 获取当前位置
+    r, c = path[-1]
+    rows, cols = puzzle.shape
+
+    # 尝试四个方向
+    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
+
+    # 如果当前位置是珠子，需要特殊处理方向优先级
+    if puzzle[r, c] == 1:  # 白珠
+        # 如果路径长度大于1，确保直行
+        if len(path) > 1:
+            prev_r, prev_c = path[-2]
+            dr, dc = r - prev_r, c - prev_c
+            # 优先直行方向
+            directions = [(dr, dc)] + [d for d in directions if d != (dr, dc)]
+    elif puzzle[r, c] == 2:  # 黑珠
+        # 如果路径长度大于1，确保转弯
+        if len(path) > 1:
+            prev_r, prev_c = path[-2]
+            dr, dc = r - prev_r, c - prev_c
+            # 优先转弯方向
+            turn_directions = [(-dc, dr), (dc, -dr)]
+            directions = turn_directions + [d for d in directions if d not in turn_directions]
+
+    for dr, dc in directions:
+        nr, nc = r + dr, c + dc
+
+        # 检查是否在网格内且未访问
+        if 0 <= nr < rows and 0 <= nc < cols and (nr, nc) not in visited:
+            # 检查移动是否符合规则
+            if improved_is_valid_move(puzzle, path, r, c, nr, nc):
+                # 添加到路径
+                path.append((nr, nc))
+                visited.add((nr, nc))
+
+                # 递归搜索
+                if improved_backtrack(puzzle, path, visited, pearls, start_pos):
+                    return True
+
+                # 回溯
+                path.pop()
+                visited.remove((nr, nc))
+
+    return False
+
+
+def improved_is_valid_move(puzzle, path, r1, c1, r2, c2):
+    """
+    改进的移动有效性检查
+
+    参数:
+        puzzle: 谜题网格
+        path: 当前路径
+        r1, c1: 当前位置
+        r2, c2: 目标位置
+
+    返回:
+        移动是否有效
+    """
+    rows, cols = puzzle.shape
+
+    # 检查白珠规则
+    if puzzle[r1, c1] == 1:  # 白珠
+        # 获取前一个位置（如果有）
+        if len(path) >= 2:
+            prev_r, prev_c = path[-2]
+
+            # 检查是否直行
+            if not ((prev_r == r1 == r2) or (prev_c == c1 == c2)):
+                return False
+
+    # 检查黑珠规则
+    if puzzle[r1, c1] == 2:  # 黑珠
+        # 获取前一个位置（如果有）
+        if len(path) >= 2:
+            prev_r, prev_c = path[-2]
+
+            # 检查是否转弯
+            if (prev_r == r1 == r2) or (prev_c == c1 == c2):
+                return False
+
+            # 检查后方是否可以直行
+            dr, dc = r2 - r1, c2 - c1
+            next_r, next_c = r2 + dr, c2 + dc
+
+            # 检查下一个位置是否在网格内
+            if not (0 <= next_r < rows and 0 <= next_c < cols):
+                return False
+
+    # 检查目标位置是否为白珠
+    if puzzle[r2, c2] == 1:  # 白珠
+        # 白珠必须能够直行通过
+        dr, dc = r2 - r1, c2 - c1
+        next_r, next_c = r2 + dr, c2 + dc
+
+        # 检查下一个位置是否在网格内
+        if not (0 <= next_r < rows and 0 <= next_c < cols):
+            return False
+
+    # 检查目标位置是否为黑珠
+    if puzzle[r2, c2] == 2:  # 黑珠
+        # 黑珠必须能够转弯
+        dr, dc = r2 - r1, c2 - c1
+
+        # 可能的转弯方向
+        turn_directions = [(-dc, dr), (dc, -dr)]
+
+        valid_turn = False
+        for turn_dr, turn_dc in turn_directions:
+            next_r, next_c = r2 + turn_dr, c2 + turn_dc
+
+            # 检查转弯后的位置是否在网格内
+            if 0 <= next_r < rows and 0 <= next_c < cols:
+                valid_turn = True
+                break
+
+        if not valid_turn:
+            return False
+
+    return True
+
+
+def verify_complete_path(puzzle, path):
+    """
+    验证完整路径是否满足所有规则
+
+    参数:
+        puzzle: 谜题网格
+        path: 完整路径
+
+    返回:
+        路径是否有效
+    """
+    rows, cols = puzzle.shape
+
+    # 检查白珠规则
+    for r in range(rows):
+        for c in range(cols):
+            if puzzle[r, c] == 1:  # 白珠
+                # 找到白珠在路径中的索引
+                try:
+                    idx = path.index((r, c))
+                except ValueError:
+                    return False
+
+                # 获取前后的点
+                prev_idx = (idx - 1) % (len(path) - 1)
+                next_idx = (idx + 1) % (len(path) - 1)
+
+                prev_r, prev_c = path[prev_idx]
+                next_r, next_c = path[next_idx]
+
+                # 检查是否直行
+                is_straight = (prev_r == next_r) or (prev_c == next_c)
+                if not is_straight:
+                    return False
+
+                # 检查至少一侧是否转弯
+                # 获取前前和后后的点
+                prev_prev_idx = (prev_idx - 1) % (len(path) - 1)
+                next_next_idx = (next_idx + 1) % (len(path) - 1)
+
+                prev_prev_r, prev_prev_c = path[prev_prev_idx]
+                next_next_r, next_next_c = path[next_next_idx]
+
+                # 检查前一侧是否转弯
+                prev_turn = not ((prev_prev_r == prev_r == r) or (prev_prev_c == prev_c == c))
+
+                # 检查后一侧是否转弯
+                next_turn = not ((next_r == next_next_r == r) or (next_c == next_next_c == c))
+
+                if not (prev_turn or next_turn):
+                    return False
+
+    # 检查黑珠规则
+    for r in range(rows):
+        for c in range(cols):
+            if puzzle[r, c] == 2:  # 黑珠
+                # 找到黑珠在路径中的索引
+                try:
+                    idx = path.index((r, c))
+                except ValueError:
+                    return False
+
+                # 获取前后的点
+                prev_idx = (idx - 1) % (len(path) - 1)
+                next_idx = (idx + 1) % (len(path) - 1)
+
+                prev_r, prev_c = path[prev_idx]
+                next_r, next_c = path[next_idx]
+
+                # 检查是否转弯
+                is_turn = not ((prev_r == next_r) or (prev_c == next_c))
+                if not is_turn:
+                    return False
+
+                # 检查两侧是否直行至少一格
+                # 获取前前和后后的点
+                prev_prev_idx = (prev_idx - 1) % (len(path) - 1)
+                next_next_idx = (next_idx + 1) % (len(path) - 1)
+
+                prev_prev_r, prev_prev_c = path[prev_prev_idx]
+                next_next_r, next_next_c = path[next_next_idx]
+
+                # 检查前一侧是否直行
+                prev_straight = (prev_prev_r == prev_r) or (prev_prev_c == prev_c)
+
+                # 检查后一侧是否直行
+                next_straight = (next_r == next_next_r) or (next_c == next_next_c)
+
+                if not (prev_straight and next_straight):
+                    return False
+
+    return True
+