给出使用DFS实现八数码问题的代码
时间: 2024-05-14 22:14:03 浏览: 14
以下是使用DFS实现八数码问题的Python代码:
```
class Node:
def __init__(self, state, parent=None, action=None):
self.state = state
self.parent = parent
self.action = action
def __eq__(self, other):
return self.state == other.state
def __hash__(self):
return hash(self.state)
def expand(self):
"""返回当前状态下所有可能的扩展节点"""
children = []
state = self.state
i = state.index(0)
if i not in [0, 1, 2]:
new_state = state[:]
new_state[i], new_state[i - 3] = new_state[i - 3], new_state[i]
children.append(Node(new_state, self, "Up"))
if i not in [6, 7, 8]:
new_state = state[:]
new_state[i], new_state[i + 3] = new_state[i + 3], new_state[i]
children.append(Node(new_state, self, "Down"))
if i not in [0, 3, 6]:
new_state = state[:]
new_state[i], new_state[i - 1] = new_state[i - 1], new_state[i]
children.append(Node(new_state, self, "Left"))
if i not in [2, 5, 8]:
new_state = state[:]
new_state[i], new_state[i + 1] = new_state[i + 1], new_state[i]
children.append(Node(new_state, self, "Right"))
return children
def dfs(start_state, goal_state):
"""使用深度优先搜索查找从起始状态到目标状态的路径"""
start_node = Node(start_state)
goal_node = Node(goal_state)
visited = set()
stack = [start_node]
while stack:
node = stack.pop()
if node == goal_node:
path = []
while node.parent is not None:
path.append(node.action)
node = node.parent
return path[::-1]
visited.add(node)
for child in node.expand():
if child not in visited:
stack.append(child)
return None
# 示例用法
start_state = [1, 2, 3, 4, 0, 5, 6, 7, 8]
goal_state = [1, 2, 3, 4, 5, 8, 6, 0, 7]
path = dfs(start_state, goal_state)
print(path)
```
解释:
- `Node` 类表示八数码问题中的一个节点,包括状态、父节点和到达该节点的动作。
- `expand()` 方法返回当前状态下所有可能的扩展节点。
- `dfs()` 函数使用深度优先搜索查找从起始状态到目标状态的路径,返回一个包含到达目标状态的动作序列的列表。如果找不到路径,则返回 `None`。
- 示例中使用的起始状态和目标状态为经典的八数码问题。
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