javascript - 协助在 Javascript 中连接四极小极大算法

Question

我正在尝试使用 minimax 算法在我的连连四游戏中创建 AI，但我无法让它发挥作用。我相信我已经很接近了，但我仍然无法正确理解。谁能帮我解决任何错误？我意识到我的代码不是很好，因为我使用 Javascript 的效率还不是很高，这就是我想尝试一下的原因。如果我完全偏离轨道，有人可以告诉我更好的方法吗？提前致谢。

编辑:我已经用更新后的代码替换了下面的代码。就给我 AI Action 而言，它现在“有效”，但问题是它们不是“智能” Action 。我看过许多 minimax 定义，我觉得我已经正确地实现了它。即使当我达到 7 的深度时，一个 5 岁的 child 也能击败它。任何帮助将不胜感激。

    function getBestMove(currBoard,depth,who) {
        var opp;
        //Get opponent for next piece
        if(who == 'a') {
            opp = 'p';
        } else {
            opp = 'a';
        }

        var tBoard = new Array(rows);
        for(var i=0; i<tBoard.length; i++) {
            tBoard[i] = new Array(cols);
        }

        var moves = new Array(aiOpenCols.length);
        //Drop each piece and use minimax function until depth == 0
        for(var i=0; i<aiOpenCols.length; i++) {
            for(var j=0; j<rows; j++) {
                for(var k=0; k<cols; k++) {
                    tBoard[j][k] = currBoard[j][k];
                }
            }
            tBoard = dropPiece(aiOpenCols[i],who,tBoard);
            moves[i] = minimax(tBoard,(+depth - 1),opp,aiOpenCols[i]);
        }

        var bestAlpha = -100000;    //Large negative
        //Use random column if no moves are "good"
        var bestMove = Math.floor(Math.random() * aiOpenCols.length);
        bestMove = +aiOpenCols[bestMove];
        //Get largest value from moves for best move
        for(var i=0; i<aiOpenCols.length; i++) {
            if(+moves[i] > bestAlpha) {
                bestAlpha = moves[i];
                bestMove = aiOpenCols[i];
            }
        }

        bestMove++; //Offset by 1 due to actual drop function
        return bestMove;
    }
    function minimax(currBoard,depth,who,col) {
        //Drop current piece, called from getBestMove function
        currBoard = dropPiece(col,who,currBoard);

        //When depth == 0 return heuristic/eval of board
        if(+depth == 0) {
            var ev = evalMove(currBoard);
            return ev;
        }
        var alpha = -100000;    //Large negative
        var opp;
        //Get opponent for next piece
        if(who == 'a') {
            opp = 'p';
        } else {
            opp = 'a';
        }

        //Loop through all available moves
        for(var i=0; i<aiOpenCols.length; i++) {
            var tBoard = new Array(rows);
            for(var i=0; i<tBoard.length; i++) {
                tBoard[i] = new Array(cols);
            }
            for(var j=0; j<rows; j++) {
                for(var k=0; k<cols; k++) {
                    tBoard[j][k] = currBoard[j][k];
                }
            }
            //Continue recursive minimax until depth == 0
            var next = minimax(tBoard,(+depth - 1),opp,aiOpenCols[i]);
            //Alpha = max(alpha, -minimax()) for negamax
            alpha = Math.max(alpha, (0 - +next));
        }
        return alpha;
    }
    function evalMove(currBoard) {
        //heuristic function
        //AI = # of 4 streaks + # of 3 streaks + # of 2 streaks - # of 3 streaks opp - # of 2 streaks opp           
        var fours = checkFours(currBoard,'b') * 1000;
        var threes = checkThrees(currBoard,'b') * 100;
        var twos = checkTwos(currBoard,'b') * 10;
        var oppThrees = checkThrees(currBoard,'r') * 100;
        var oppTwos = checkTwos(currBoard,'r') * 10;

        var scores = fours + threes + twos - oppThrees - oppTwos;

        //If opponent wins, return large negative
        var oppFours = checkFours(currBoard,'r');
        if(+oppFours > 0) {
            return -100000;
        } else {
            return scores;
        }
    }
    function dropPiece(col,who,currBoard) {
        for(var i=0; i<currBoard.length; i++) {
            if(currBoard[i][col] != 'w') {
                //Make sure column isn't full
                if(i != 0) {
                    if(who == 'p') {
                        currBoard[i-1][col] = 'r';
                    } else {
                        currBoard[i-1][col] = 'b';
                    }
                    break;
                }
            }
            //If column is empty, place in first row
            if(i == (currBoard.length - 1)) {
                if(who == 'p') {
                    currBoard[i][col] = 'r';
                } else {
                    currBoard[i][col] = 'b';
                }
            }
        }
        return currBoard;
    }

Answer 1

您的评估函数正在为没有值(value)的位置特征赋值。连续两个相邻的单元格没有任何值(value)。如果第四格可以被对手占据而不允许在其他地方获胜，那么连续三个也没有任何值(value)。

简化。对于 Connect Four，您只需要关心该位置是赢还是输，或者该位置是否可以在下一步中赢或输。如果赢了，返回一个大的正值。如果是亏损，则返回一个大的负值。如果下一步可以赢或输，则将该分支的搜索深度扩展一层，然后再次调用 minimax()，返回结果。后者将避免 horizon effect所有固定深度极小极大搜索都容易受到影响。否则返回零。