a*算法实现八数码python
时间: 2023-08-12 15:28:38 浏览: 115
A*算法解决八数码问题 python实现
下面是使用Python实现A*算法解决八数码问题的示例代码:
```python
import heapq
def manhattan_distance(state):
"""
计算当前状态和目标状态之间的曼哈顿距离
"""
distance = 0
for i in range(3):
for j in range(3):
if state[i][j] != 0:
x, y = divmod(state[i][j]-1, 3)
distance += abs(x-i) + abs(y-j)
return distance
def get_neighbors(state):
"""
获取当前状态的所有邻居
"""
neighbors = []
x, y = next((i, j) for i in range(3) for j in range(3) if state[i][j] == 0)
for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
new_x, new_y = x + dx, y + dy
if 0 <= new_x < 3 and 0 <= new_y < 3:
new_state = [row[:] for row in state]
new_state[x][y], new_state[new_x][new_y] = new_state[new_x][new_y], new_state[x][y]
neighbors.append(new_state)
return neighbors
def a_star(start, goal):
"""
使用A*算法解决八数码问题
"""
heap = [(0, start)]
g = {str(start): 0}
f = {str(start): manhattan_distance(start)}
while heap:
_, current = heapq.heappop(heap)
if current == goal:
path = []
while str(current) in g:
path.append(current)
current = g[str(current)]
path.append(current)
return path[::-1]
for neighbor in get_neighbors(current):
tentative_g = g[str(current)] + 1
if str(neighbor) not in g or tentative_g < g[str(neighbor)]:
g[str(neighbor)] = tentative_g
f[str(neighbor)] = tentative_g + manhattan_distance(neighbor)
heapq.heappush(heap, (f[str(neighbor)], neighbor))
return None
start_state = [[2, 8, 3], [1, 6, 4], [7, 0, 5]]
goal_state = [[1, 2, 3], [8, 0, 4], [7, 6, 5]]
path = a_star(start_state, goal_state)
if path:
for state in path:
print(state)
else:
print("No solution found.")
```
在上述示例代码中,首先定义了一个计算曼哈顿距离的函数manhattan_distance和一个获取当前状态所有邻居的函数get_neighbors。然后定义了A*算法的主函数a_star,其中使用了堆(heapq)来存储待搜索的状态,并使用字典(g和f)来存储已经搜索过的状态的g值和f值。最后,将初始状态和目标状态作为参数调用a_star函数,输出最优路径。
阅读全文