用python编写一个 BFS 算法和一个 DFS 算 法来求解八数码问题并添加注释
时间: 2024-09-20 17:09:22 浏览: 46
当然可以,这里我会为你展示如何用 Python 编写广度优先搜索(BFS)和深度优先搜索(DFS)算法来解决经典的八数码问题(又称数独)。我们首先假设已经有一个函数 `check_sudoku` 来检查给定的数组是否满足数独规则。
```python
import collections
# 定义数独的空格
def empty_board():
return [[0] * 9 for _ in range(9)]
# 检查给定的数独布局是否有效
def check_sudoku(board):
# 逻辑省略,你可以自行实现九宫格划分、行、列和宫的遍历以及数字重复性的检查
pass
# 广度优先搜索(BFS)
def bfs_solve(board):
visited = set()
queue = collections.deque([[i // 9, i % 9, board.copy()] for i in range(81) if board[i] == 0]) # 初始化队列
while queue:
row, col, current_board = queue.popleft() # 弹出队首元素
if (row, col) not in visited:
visited.add((row, col))
for num in range(1, 10): # 尝试填入数字
if is_valid_move(row, col, num, current_board):
current_board[row][col] = num
if check_sudoku(current_board):
return current_board # 找到解决方案,返回
else:
queue.append([row, col, current_board]) # 无效则回溯
return None # 没有找到解决方案
# 深度优先搜索(DFS)
def dfs_solve(board):
def backtrack(row, col):
nonlocal found_solution
if row == 9:
found_solution = True
return
for num in range(1, 10):
if is_valid_move(row, col, num, board):
board[row][col] = num
if check_sudoku(board):
solution.append(board.copy())
found_solution = True
else:
backtrack(row + 1, col) # 递归尝试下一个位置
board[row][col] = 0 # 回溯,撤销当前选择
found_solution = False
solution = []
backtrack(0, 0)
return solution if found_solution else None # 返回所有解决方案列表
# 判断当前位置填入num是否合法
def is_valid_move(row, col, num, board):
# 省略对行、列和宫的检查逻辑
pass
# 示例用法
board = empty_board()
solution_bfs = bfs_solve(board)
solution_dfs = dfs_solve(board)
if solution_bfs:
print("BFS Solution:")
print(solution_bfs)
else:
print("BFS couldn't find a solution.")
if solution_dfs:
print("\nDFS Solutions:")
for sol in solution_dfs:
print(sol)
else:
print("DFS couldn't find a solution.")
```
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