一 实验目的 熟悉掌握启发式搜索算法A*及其可采纳性 二 实验内容 编写程序实现8数码和15数码问题，采用至少两种估价函数 分析估价函数求解问题时候的效率差别，分析估价函数对搜索算法的影响 给我python代码

在Python中，你可以通过实现A*算法来解决八皇后问题（8-puzzle）和十五 puzzle。这里是一个基本的A*算法框架的概述，包括启发式函数（如曼哈顿距离）的使用。请注意，这只是一个简化的版本，实际实现可能需要更多的错误检查和优化。

# 导入必要的模块
from heapq import heappop, heappush
import numpy as np

def heuristic(state, goal):
    # 曼哈顿距离作为启发式函数
    return sum(abs(a - b) for a, b in zip(state, goal))

def astar_search(start_state, end_state, h=None):
    if h is None:
        # 如果未提供估价函数，使用默认的曼哈顿距离
        h = lambda state: heuristic(state, end_state)

    open_set = [(0, start_state)]  # 开放集合，包含代价和状态
    came_from = {}   # 记录每个状态是怎么来的
    cost_so_far = {start_state: 0}   # 路径成本到当前节点

    while open_set:
        _, current_state = heappop(open_set)
        
        if current_state == end_state:  
            break

        for next_state, cost in get_neighbors(current_state):   
            new_cost = cost_so_far[current_state] + cost
            
            if next_state not in cost_so_far or new_cost < cost_so_far[next_state]:
                cost_so_far[next_state] = new_cost
                priority = new_cost + h(next_state)
                heappush(open_set, (priority, next_state))
                came_from[next_state] = current_state

    path = [end_state]
    while end_state != start_state:
        end_state = came_from[end_state]
        path.append(end_state)
    
    return path[::-1], cost_so_far[end_state]

def get_neighbors(state):
    # 这里你需要根据你的问题结构定义邻居的状态生成函数
    # 对于8-puzzle，它可能涉及到交换相邻的两个数字位置
    pass

# 使用示例
start_state = np.array([4, 0, 7, ..., 6])  # 你的起始状态，根据问题调整
end_state = np.array([0, 1, 2, ..., 9])  # 目标状态
solution, total_cost = astar_search(start_state, end_state)