状态空间法野人与修道士Python

您好，状态空间法是一种常用于解决搜索问题的方法，而野人与修道士问题是一种搜索问题，因此可以用状态空间法来解决。下面是一个用 Python 实现的野人与修道士问题的代码示例：

# 野人与修道士问题的状态表示：(左岸野人数, 左岸修道士数, 船的位置)
start_state = (3, 3, 1)  # 初始状态
goal_state = (0, 0, 0)  # 目标状态

def is_valid(state):
    """判断当前状态是否合法"""
    left_m, left_c, boat = state
    right_m, right_c = 3 - left_m, 3 - left_c
    if left_m < 0 or left_c < 0 or right_m < 0 or right_c < 0:
        return False
    if left_m > 0 and left_c > left_m:
        return False
    if right_m > 0 and right_c > right_m:
        return False
    return True

def get_next_states(state):
    """获取当前状态下所有可能的下一步状态"""
    left_m, left_c, boat = state
    right_m, right_c = 3 - left_m, 3 - left_c
    next_states = []
    if boat == 1:  # 船在左岸
        for i in range(3):
            for j in range(3):
                if i + j > 2 or i + j == 0:  # 船上至少要有一人，最多只能带两人
                    continue
                next_state = (left_m - i, left_c - j, 0)
                if is_valid(next_state):
                    next_states.append(next_state)
    else:  # 船在右岸
        for i in range(3):
            for j in range(3):
                if i + j > 2 or i + j == 0:
                    continue
                next_state = (left_m + i, left_c + j, 1)
                if is_valid(next_state):
                    next_states.append(next_state)
    return next_states

def bfs(start_state, goal_state):
    """使用广度优先搜索算法查找从起始状态到目标状态的最短路径"""
    queue = [[start_state]]
    visited = set([start_state])
    while queue:
        path = queue.pop(0)
        current_state = path[-1]
        if current_state == goal_state:
            return path
        for next_state in get_next_states(current_state):
            if next_state not in visited:
                queue.append(path + [next_state])
                visited.add(next_state)
    return None

# 测试代码
path = bfs(start_state, goal_state)
if path:
    print("找到了解决方案：")
    for state in path:
        print(state)
else:
    print("无法找到解决方案。")

这段代码使用了广度优先搜索算法，可以找到从起始状态到目标状态的最短路径。您可以将其保存为一个.py文件，在命令行中运行即可。