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我正在尝试为井字游戏实现极小极大算法，其中两个玩家都是人类，并且每次计算机都使用极小极大算法建议最佳移动。但它并不是每次都给出正确的建议。例如：它没有为以下场景提供正确的建议：玩家 X：1 玩家 O：2 玩家 X：5。这是我的代码：

#include <stdio.h>
#include <algorithm>  
#include <string>
using namespace std;

#define inf 1<<20
int posmax, posmin;
char board[15];

void print_board()
{
    int i;
    for (i = 1; i <= 9; i++)
    {   
        printf("%c ",board[i]);
        if (i % 3 == 0)
            printf("\n");
    }
    printf("\n");
}

int check_win(char board[])
{
    if ((board[1] == 'X' && board[2] == 'X' && board[3] == 'X') ||
        (board[4] == 'X' && board[5] == 'X' && board[6] == 'X') ||
        (board[7] == 'X' && board[8] == 'X' && board[9] == 'X') ||
        (board[1] == 'X' && board[4] == 'X' && board[7] == 'X') ||
        (board[2] == 'X' && board[5] == 'X' && board[8] == 'X') ||
        (board[3] == 'X' && board[6] == 'X' && board[9] == 'X') ||
        (board[1] == 'X' && board[5] == 'X' && board[9] == 'X') ||
        (board[3] == 'X' && board[5] == 'X' && board[7] == 'X'))
    {
        return 1;
    }
    else if((board[1] == 'O' && board[2] == 'O' && board[3] == 'O') ||
            (board[4] == 'O' && board[5] == 'O' && board[6] == 'O') ||
            (board[7] == 'O' && board[8] == 'O' && board[9] == 'O') ||
            (board[1] == 'O' && board[4] == 'O' && board[7] == 'O') ||
            (board[2] == 'O' && board[5] == 'O' && board[8] == 'O') ||
            (board[3] == 'O' && board[6] == 'O' && board[9] == 'O') ||
            (board[1] == 'O' && board[5] == 'O' && board[9] == 'O') ||
            (board[3] == 'O' && board[5] == 'O' && board[7] == 'O'))
    {
        return -1;
    }
    else return 0;
}

int check_draw(char board[])
{
    if ((check_win(board) == 0) && (board[1] != '_') && (board[2] != '_') &&
        (board[3] != '_') && (board[4] != '_') && (board[5] != '_') &&
        (board[6] != '_') && (board[7] != '_') && (board[8] != '_') &&
        (board[9] != '_'))
    {
        return 1;
    }
    else return 0;
}

int minimax(int player, char board[], int n)
{
    int i, res, j;

    int max = -inf;
    int min = inf;

    int chk = check_win(board);
    if (chk == 1)
        return 1;
    else if (chk == (-1))
        return -1;
    else if (chk = check_draw(board))
        return 0;

    for (i = 1; i <= 9; i++)
    {
        if(board[i] == '_')
        {
            if(player == 2)  
            {
                board[i] = 'O';
                res = minimax(1, board, n + 1);

                board[i] = '_';
                if(res < min)
                {
                    min = res;
                    if (n == 0)
                        posmin = i;
                }
            }
            else if (player == 1)
            {
                board[i] = 'X';
                res = minimax(2, board, n + 1);
                board[i] = '_';
                if (res > max)
                {
                    max = res;
                    if (n == 0)
                        posmax = i;
                }
            }
        }
    }

    if (player == 1)
        return max;
    else return min;    
}


// 1 is X, 2 is O
int main()
{
    int i, j, input, opt;

    for(i = 1; i <= 9; i++)
        board[i] = '_';

    printf("Board:\n");
    print_board();

    for(i = 1; i <= 9; i++)
    {
        if (i % 2 == 0)
            printf("Player O give input:\n");
        else 
            printf("Player X give input:\n");

        scanf("%d", &input);
        if (i % 2 != 0)
            board[input] = 'X';
        else
            board[input] = 'O';

        printf("Board:\n");
        print_board();

        int chk = check_win(board);
        if (chk == 1)
        {
            printf("Player X wins!\n");
            break;
        }
        else if (chk == -1)
        {
            printf("Player O wins!\n");
            break;
        }
        else if ((chk == 0) && (i != 9))
        {
            posmax = -1;
            posmin = -1;
            if(i % 2 == 0)
            {
                opt = minimax(1, board, 0);
                printf("Optimal move for player X is %d\n", posmax);
            }
            else 
            {
            opt = minimax(2, board, 0);
            printf("Optimal move for player O is %d\n", posmin);
            }
        }
        else 
            printf("The game is tied!\n");
    }
    return 0;
}
4

4 回答 4

2

在我看来，您的程序没有给出错误的建议。如果两个玩家都处于最佳状态，Minimax 会计算移动的分数。在您的情况下，得分可以是 +1、-1 和 0，因此，如果一场比赛（例如）不可避免地输掉，那么输掉的深度并没有什么不同。给定以下游戏状态

X O _
X _ _
_ _ _

和玩家 X 的最佳玩法，玩家 O 在哪里下棋并不重要（他在任何一种情况下都输了）：

  • O打7后，X打5，O打6，X打8 --> X赢
  • O 打 3 后，X 打 7 --> X 获胜

玩家 X 获胜。移动 7 与移动 3 和所有其他可玩移动的得分相同。如果你想让你的算法给出这个例子的移动建议 7，你必须在你的评估函数中包含游戏深度。您可以通过将函数的返回值更改为以下内容来做到这一点：

int chk = check_win(board);
if (chk == 1)
    return (10 - n);
else if (chk == (-1))
    return -(10 - n);
else if (chk = check_draw(board))
    return 0;
于 2012-07-30T21:49:37.990 回答
0

我认为这（尽管编码效率低下）是正确的。如果不是，请给出您认为程序错误的移动顺序。

它没有给出最短的移动序列，这可能是你所追求的。然后，您应该重构它以返回给出最短移动序列（如果获胜）或最长移动序列（如果失败）的移动。

于 2012-07-30T21:09:47.793 回答
0

除非我读错 main() ，否则在宣布平局之前，您只需要填充 8 个方格。这可能不是您要寻找的错误，但它是一个开始。

于 2012-07-30T17:05:27.963 回答
0

替换 printf("Optimal move for player X is %d %d\n", posmax); 为 printf("Optimal move for player X is %d\n", posmax);

和

printf("Optimal move for player O is %d %d\n", posmin); 和 printf("Optimal move for player O is %d\n", posmin);

其他一切似乎都是正确的，尽管它并不总是打印出最快的胜利（如果胜利存在的话）。

于 2012-07-30T21:22:50.227 回答