Python中Tictactoe的Minimax算法

Question

我最近参加了CS50 AI python课程，其中一个项目是为Tictatcoe游戏实现一个minimax算法。我寻求帮助并搜索了stackoverflow，但没有找到对我有帮助的答案。它的图形部分已经实现了，您所需要做的就是对模板的给定函数进行编程，我相信我正确地实现了这些函数，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（棋盘）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”（可以是两个中的任意一个）来计算可能的最佳移动，“X”玩家尝试最大化分数，“O”应该利用工具（棋盘）将其最小化函数，该函数返回1表示X赢、-1表示“O”赢或0表示平局。 到目前为止，人工智能的移动不是最优的，我可以轻松地在我不应该的时候战胜它，因为在最好的情况下，我应该得到的是一个平局，因为人工智能应该在这一点上计算每一个可能的移动。但我不知道怎么了

Answer 1

@harsh kothari，{a1}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 2

首先谈谈调试：如果要打印递归调用中完成的计算，可以跟踪问题的执行情况并快速找到答案

但是，你的问题似乎是最重要的。在minimax调用中，如果当前播放器为X，则调用该状态的每个子级的max_值，然后取该结果的max。但是，这会在树的顶部应用max函数两次。游戏中的下一个玩家是O，所以你应该为下一个玩家调用minu值函数

因此，在minimax调用中，如果当前_播放器为X，则应调用minu value；如果当前_播放器为O，则应调用max_value

Answer 3

改变 此操作的操作（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