a*八数码问题python代码
时间: 2023-11-27 18:04:47 浏览: 104
以下是使用 A* 算法解决八数码问题的 Python 代码:
```python
import heapq
class PuzzleState:
def __init__(self, board, parent=None, move=""):
self.board = board
self.parent = parent
self.move = move
self.depth = 0
if parent:
self.depth = parent.depth + 1
def __str__(self):
return "\n".join([ " ".join(map(str, row)) for row in self.board ])
@property
def is_solved(self):
return str(self) == "1 2 3\n4 5 6\n7 8 0"
@property
def cost(self):
return self.depth + self.heuristic_cost
@property
def heuristic_cost(self):
return sum([ 1 if self.board[i][j] != 0 and self.board[i][j] != (3*i + j + 1) else 0 for i in range(3) for j in range(3) ])
@property
def possible_moves(self):
row, col = next((i, j) for i in range(3) for j in range(3) if self.board[i][j] == 0)
moves = []
if row > 0:
moves.append("U")
if row < 2:
moves.append("D")
if col > 0:
moves.append("L")
if col < 2:
moves.append("R")
return moves
def move_to(self, move):
row, col = next((i, j) for i in range(3) for j in range(3) if self.board[i][j] == 0)
new_board = [[val for val in row] for row in self.board]
if move == "U":
new_board[row][col] = new_board[row-1][col]
new_board[row-1][col] = 0
elif move == "D":
new_board[row][col] = new_board[row+1][col]
new_board[row+1][col] = 0
elif move == "L":
new_board[row][col] = new_board[row][col-1]
new_board[row][col-1] = 0
elif move == "R":
new_board[row][col] = new_board[row][col+1]
new_board[row][col+1] = 0
return PuzzleState(new_board, parent=self, move=move)
def solve(puzzle):
queue = []
heapq.heappush(queue, (puzzle.cost, puzzle))
visited = set()
while queue:
_, current = heapq.heappop(queue)
if current.is_solved:
moves = []
while current.parent:
moves.append(current.move)
current = current.parent
return moves[::-1]
visited.add(str(current))
for move in current.possible_moves:
next_state = current.move_to(move)
if str(next_state) not in visited:
heapq.heappush(queue, (next_state.cost, next_state))
return None
puzzle = PuzzleState([[2, 3, 6], [1, 5, 0], [7, 8, 4]])
print("\n".join([ " ".join(map(str, row)) for row in puzzle.board ]))
print()
for move in solve(puzzle):
print(move)
puzzle = puzzle.move_to(move)
print("\n".join([ " ".join(map(str, row)) for row in puzzle.board ]))
print()
```
这个代码使用了 `PuzzleState` 类来表示八数码问题的状态,其中包括了当前的棋盘布局、父状态、移动方向和深度。`is_solved` 方法用于判断当前状态是否已经达到目标状态,`cost` 方法用于计算当前状态的总代价,包括深度和启发式代价,`heuristic_cost` 方法则用于计算当前状态的启发式代价,这里采用了曼哈顿距离作为启发式函数。`possible_moves` 方法返回当前状态可以进行的移动方向,`move_to` 方法则根据指定的移动方向生成新的状态。
`solve` 函数使用 A* 算法来求解八数码问题。它使用了一个优先队列来维护待扩展的状态,每次从队列中取出代价最小的状态进行扩展。如果扩展得到的状态已经是目标状态,就返回从初始状态到达该状态的移动方向序列。否则,将该状态加入已访问的集合中,并根据可能的移动方向生成新的状态加入优先队列中。
这个代码使用了 `heapq` 模块来实现优先队列,它采用了小根堆的方式来维护队列。在每个状态中,我们使用字符串形式的棋盘布局来表示状态,这样可以方便地进行比较和哈希。
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