Python 中 Tictactoe 的 Minimax 算法

Question

我最近注册了 CS50 AI python 课程，其中一个项目是为 tictactoe 游戏实现 minimax 算法。我寻求帮助并搜索了 Whosebug，但没有找到可以帮助我的答案。它的图形部分已经实现了，你需要做的就是编写一个模板的给定函数，我相信我做对了，除了算法部分，函数如下：

import math
import copy

X = "X"
O = "O"
EMPTY = None


def initial_state():
    """
    Returns starting state of the board.
    """
    return [[EMPTY, EMPTY, EMPTY],
            [EMPTY, EMPTY, EMPTY],
            [EMPTY, EMPTY, EMPTY]]


def player(board):
    """
    Returns player who has the next turn on a board.
    """
    if board == initial_state():
        return X

    xcounter = 0
    ocounter = 0
    for row in board:
        xcounter += row.count(X)
        ocounter += row.count(O)

    if xcounter == ocounter:
        return X
    else:
        return O


def actions(board):
    """
    Returns set of all possible actions (i, j) available on the board.
    """
    possible_moves = []
    for i in range(3):
        for j in range(3):
            if board[i][j] == EMPTY:
                possible_moves.append([i, j])
    return possible_moves


def result(board, action):
    """
    Returns the board that results from making move (i, j) on the board.
    """
    boardcopy = copy.deepcopy(board)
    try:
        if boardcopy[action[0]][action[1]] != EMPTY:
            raise IndexError
        else:
            boardcopy[action[0]][action[1]] = player(boardcopy)
            return boardcopy
    except IndexError:
        print('Spot already occupied')


def winner(board):
    """
    Returns the winner of the game, if there is one.
    """
    columns = []
    # Checks rows
    for row in board:
        xcounter = row.count(X)
        ocounter = row.count(O)
        if xcounter == 3:
            return X
        if ocounter == 3:
            return O

    # Checks columns
    for j in range(len(board)):
        column = [row[j] for row in board]
        columns.append(column)

    for j in columns:
        xcounter = j.count(X)
        ocounter = j.count(O)
        if xcounter == 3:
            return X
        if ocounter == 3:
            return O

    # Checks diagonals
    if board[0][0] == O and board[1][1] == O and board[2][2] == O:
        return O
    if board[0][0] == X and board[1][1] == X and board[2][2] == X:
        return X
    if board[0][2] == O and board[1][1] == O and board[2][0] == O:
        return O
    if board[0][2] == X and board[1][1] == X and board[2][0] == X:
        return X

    # No winner/tie
    return None


def terminal(board):
    """
    Returns True if game is over, False otherwise.
    """
    # Checks if board is full or if there is a winner
    empty_counter = 0
    for row in board:
        empty_counter += row.count(EMPTY)
    if empty_counter == 0:
        return True
    elif winner(board) is not None:
        return True
    else:
        return False


def utility(board):
    """
    Returns 1 if X has won the game, -1 if O has won, 0 otherwise.
    """
    if winner(board) == X:
        return 1
    elif winner(board) == O:
        return -1
    else:
        return 0


def minimax(board):
    current_player = player(board)

    if current_player == X:
        v = -math.inf
        for action in actions(board):
            k = min_value(result(board, action))    #FIXED
            if k > v:
                v = k
                best_move = action
    else:
        v = math.inf
        for action in actions(board):
            k = max_value(result(board, action))    #FIXED
            if k < v:
                v = k
                best_move = action
    return best_move

def max_value(board):
    if terminal(board):
        return utility(board)
    v = -math.inf
    for action in actions(board):
        v = max(v, min_value(result(board, action)))
    return v    #FIXED

def min_value(board):
    if terminal(board):
        return utility(board)
    v = math.inf
    for action in actions(board):
        v = min(v, max_value(result(board, action)))
    return v    #FIXED

最后一部分是 minimax(board) 函数所在的位置，它应该获取棋盘的当前状态并根据 AI 是玩家 'X' 还是 'O'（可以是两者中的任何一个），'X' 玩家尝试最大化分数，而 'O' 应该利用 returns a 1 的效用（棋盘）函数将其最小化X 获胜，-1 代表 'O' 获胜，0 代表平局。到目前为止，AI 的动作并不是最佳的，我可以在不应该的时候轻松地战胜它，因为在最好的情况下，我应该得到的只是平局，因为 AI 应该计算那一点上的每一个可能的动作。但是不知道怎么回事...

Answer 1

首先谈谈调试：如果您要打印在递归调用中完成的计算，您可以跟踪问题的执行并快速找到答案。

但是，您的问题似乎排在首位。在您的 minimax 调用中，如果当前玩家是 X，您将在状态的每个 children 上调用 max_value，然后取该结果的最大值。但是，这会在树的顶部应用两次 max 函数。游戏中的下一位玩家是 O，因此您应该为下一位玩家调用 min_value 函数。

所以，在 minimax 调用中，如果 current_player 是 X，你应该调用 min_value，如果 current_player 是 O，你应该调用 max_value。

Answer 2

@harsh-kothari，cs50 project page 说

Importantly, the original board should be left unmodified: since Minimax will ultimately require considering many different board states during its computation. This means that simply updating a cell in board itself is not a correct implementation of the result function. You’ll likely want to make a deep copy of the board first before making any changes.

为了避免子列表被修改，我们使用deepcopy而不是copy

Read more about copy() and deepcopy() here

Answer 3

改变对此的操作（板）代码

possibleActions = set()

for i in range(0, len(board)):
    for j in range(0, len(board[0])):
        if board[i][j] == EMPTY:
            possibleActions.add((i, j))

return possibleActions

Python 中 Tictactoe 的 Minimax 算法

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