python制作一个框架，可用BFS DFS与A*三种方法解决九宫重排问题，要求初始状态和最终状态由自己决定

好的，这里是一个 Python 框架，可以使用 BFS、DFS 和 A* 三种算法来解决九宫重排问题。在这个框架中，您可以自行设定起始状态和目标状态。

from queue import Queue
from collections import deque


class Node:
    def __init__(self, state, parent=None, action=None, cost=0):
        self.state = state
        self.parent = parent
        self.action = action
        self.cost = cost

    def __eq__(self, other):
        return self.state == other.state

    def __hash__(self):
        return hash(str(self.state))

    def __str__(self):
        return str(self.state)

    def expand(self):
        successors = []
        zero_pos = self.state.index(0)
        if zero_pos > 2:
            new_state = self.state.copy()
            new_state[zero_pos], new_state[zero_pos - 3] = new_state[zero_pos - 3], new_state[zero_pos]
            successors.append(Node(new_state, self, 'up', self.cost+1))
        if zero_pos < 6:
            new_state = self.state.copy()
            new_state[zero_pos], new_state[zero_pos + 3] = new_state[zero_pos + 3], new_state[zero_pos]
            successors.append(Node(new_state, self, 'down', self.cost+1))
        if zero_pos % 3 > 0:
            new_state = self.state.copy()
            new_state[zero_pos], new_state[zero_pos - 1] = new_state[zero_pos - 1], new_state[zero_pos]
            successors.append(Node(new_state, self, 'left', self.cost+1))
        if zero_pos % 3 < 2:
            new_state = self.state.copy()
            new_state[zero_pos], new_state[zero_pos + 1] = new_state[zero_pos + 1], new_state[zero_pos]
            successors.append(Node(new_state, self, 'right', self.cost+1))
        return successors

    def is_goal(self, goal):
        return self.state == goal


def bfs(start, goal):
    frontier = Queue()
    frontier.put(start)
    explored = set()
    while not frontier.empty():
        node = frontier.get()
        if node.is_goal(goal):
            return node
        explored.add(node)
        for child in node.expand():
            if child not in explored and child not in frontier.queue:
                frontier.put(child)


def dfs(start, goal):
    frontier = deque()
    frontier.append(start)
    explored = set()
    while frontier:
        node = frontier.pop()
        if node.is_goal(goal):
            return node
        explored.add(node)
        for child in node.expand()[::-1]:
            if child not in explored and child not in frontier:
                frontier.append(child)


def astar(start, goal):
    frontier = []
    frontier.append(start)
    explored = set()
    while frontier:
        frontier.sort(key=lambda x: x.cost + heuristic(x.state, goal))
        node = frontier.pop(0)
        if node.is_goal(goal):
            return node
        explored.add(node)
        for child in node.expand():
            if child not in explored and child not in frontier:
                frontier.append(child)


def heuristic(state, goal):
    return sum(abs(b%3 - g%3) + abs(b//3 - g//3) for b, g in ((state.index(i), goal.index(i)) for i in range(1, 9)))


# Example usage
start_state = Node([1, 2, 3, 4, 5, 6, 0, 7, 8])
goal_state = Node([1, 2, 3, 4, 5, 6, 7, 8, 0])

bfs_result = bfs(start_state, goal_state)
dfs_result = dfs(start_state, goal_state)
astar_result = astar(start_state, goal_state)

print('BFS solution:', [node.action for node in bfs_result.path()])
print('DFS solution:', [node.action for node in dfs_result.path()])
print('A* solution:', [node.action for node in astar_result.path()]))

在这个框架中，我们定义了一个 Node 类来表示每个状态，其中包含以下属性：

- state：表示当前的状态列表。
- parent：表示当前状态的父状态。
- action：表示从父状态到当前状态的操作。
- cost：表示从起始状态到当前状态的代价。

我们还定义了三个函数，分别使用 BFS、DFS 和 A* 算法来搜索状态空间，并找到从起始状态到目标状态的最短路径。最后，我们打印出每个算法的解决方案。

注意：在使用 A* 算法时，我们还定义了一个启发式函数 heuristic，以估计从当前状态到目标状态的距离。在这个框架中，我们使用曼哈顿距离作为启发式函数。您可以根据需要修改启发式函数，以获得更好的性能。