八数码问题a*算法python

以下是使用python语言实现八数码问题的a*算法代码：

from queue import PriorityQueue

class Node:
    def __init__(self, state, parent=None, move=None):
        self.state = state
        self.parent = parent
        self.move = move
        self.g = 0
        self.h = 0
    
    def __lt__(self, other):
        return (self.g + self.h) < (other.g + other.h)
    
    def __eq__(self, other):
        return self.state == other.state
    
    def __str__(self):
        return str(self.state)
    
    def __repr__(self):
        return str(self.state)

def get_blank_pos(state):
    for i in range(3):
        for j in range(3):
            if state[i][j] == 0:
                return i, j

def get_successors(state):
    i, j = get_blank_pos(state)
    successors = []
    if i > 0:
        new_state = [row[:] for row in state]
        new_state[i][j], new_state[i-1][j] = new_state[i-1][j], new_state[i][j]
        successors.append(Node(new_state))
    if i < 2:
        new_state = [row[:] for row in state]
        new_state[i][j], new_state[i+1][j] = new_state[i+1][j], new_state[i][j]
        successors.append(Node(new_state))
    if j > 0:
        new_state = [row[:] for row in state]
        new_state[i][j], new_state[i][j-1] = new_state[i][j-1], new_state[i][j]
        successors.append(Node(new_state))
    if j < 2:
        new_state = [row[:] for row in state]
        new_state[i][j], new_state[i][j+1] = new_state[i][j+1], new_state[i][j]
        successors.append(Node(new_state))
    return successors

def manhattan_distance(state):
    dist = 0
    for i in range(3):
        for j in range(3):
            if state[i][j] != 0:
                row, col = divmod(state[i][j]-1, 3)
                dist += abs(row-i) + abs(col-j)
    return dist

def a_star(initial_state):
    initial_node = Node(initial_state)
    initial_node.h = manhattan_distance(initial_state)
    frontier = PriorityQueue()
    frontier.put(initial_node)
    explored = set()
    while not frontier.empty():
        node = frontier.get()
        explored.add(node.state)
        if node.h == 0:
            path = []
            while node.parent != None:
                path.append(node.move)
                node = node.parent
            path.reverse()
            return path
        for successor in get_successors(node.state):
            if successor.state not in explored:
                successor.g = node.g + 1
                successor.h = manhattan_distance(successor.state)
                successor.parent = node
                successor.move = successor.state[node.state.index(0)][node.state[node.state.index(0)].index(0)] # move is the number moved to the blank position
                frontier.put(successor)
    return None

# example usage
initial_state = [[1, 2, 3], [4, 5, 6], [0, 7, 8]]
path = a_star(initial_state)
print(path)

该代码实现了a*算法来解决八数码问题，其中Node类表示搜索树中的节点，包括节点状态、父节点、移动操作、g值和h值等信息。get_blank_pos()函数用于获取状态中空白位置的行和列。get_successors()函数用于生成当前状态的所有可能后继状态。manhattan_distance()函数用于计算当前状态与目标状态之间的曼哈顿距离。a_star()函数是主要的搜索函数，使用优先队列来管理搜索树中的所有节点，同时使用explored集合来记录已经访问过的状态，以避免重复搜索。最终返回的是从初始状态到目标状态的一系列移动操作。