Question

大家好，我真的需要帮助，我的计算能力很差，而且我真的被困在这里。

所以我用vue.js制作了一个国际象棋游戏，现在我正试图找出每个棋子的可能动作，并且能够修正knight, pawn and bishop的合法动作验证

在进行主教验证时遇到一个问题，就是要能够验证主教面前是否有一块东西

请查看图片以了解更多信息。

您会看到红色数字越过典当并继续上升。它应该停止在典当行。

这是我用于主教计算的代码。如果可能的话，如果您还可以提供对皇后区的验证，那么其他部分将真的很有帮助。

    var el = {  };
    // sample Data
    el.whiteMoves = [{"x":1,"y":1,"type":"Rook1","name":"1A","cleanType":"rook"},{"x":8,"y":1,"type":"Rook2","name":"1H","cleanType":"rook"},{"x":2,"y":1,"type":"Knight1","name":"1B","cleanType":"knight"},{"x":7,"y":1,"type":"Knight2","name":"1G","cleanType":"knight"},{"x":3,"y":1,"type":"Bishop1","name":"1C","cleanType":"bishop"},{"x":6,"y":1,"type":"Bishop2","name":"1F","cleanType":"bishop"},{"x":4,"y":1,"type":"Queen","name":"1D","cleanType":"queen"},{"x":5,"y":1,"type":"King","name":"1E","cleanType":"king"},{"x":1,"y":2,"type":"Pawn1","name":"2A","cleanType":"pawn"},{"x":2,"y":2,"type":"Pawn2","name":"2B","cleanType":"pawn"},{"x":3,"y":2,"type":"Pawn3","name":"2C","cleanType":"pawn"},{"x":4,"y":2,"type":"Pawn4","name":"2D","cleanType":"pawn"},{"x":5,"y":2,"type":"Pawn5","name":"2E","cleanType":"pawn"},{"x":6,"y":2,"type":"Pawn6","name":"2F","cleanType":"pawn"},{"x":7,"y":2,"type":"Pawn7","name":"2G","cleanType":"pawn"},{"x":8,"y":2,"type":"Pawn8","name":"2H","cleanType":"pawn"}];
    
    el.blackMoves = [{"x":1,"y":8,"type":"Rook1","name":"8A","cleanType":"rook"},{"x":8,"y":8,"type":"Rook2","name":"8H","cleanType":"rook"},{"x":2,"y":8,"type":"Knight1","name":"8B","cleanType":"knight"},{"x":7,"y":8,"type":"Knight2","name":"8G","cleanType":"knight"},{"x":3,"y":8,"type":"Bishop1","name":"8C","cleanType":"bishop"},{"x":6,"y":8,"type":"Bishop2","name":"8F","cleanType":"bishop"},{"x":4,"y":8,"type":"Queen","name":"8D","cleanType":"queen"},{"x":5,"y":8,"type":"King","name":"8E","cleanType":"king"},{"x":1,"y":7,"type":"Pawn1","name":"7A","cleanType":"pawn"},{"x":2,"y":7,"type":"Pawn2","name":"7B","cleanType":"pawn"},{"x":3,"y":7,"type":"Pawn3","name":"7C","cleanType":"pawn"},{"x":4,"y":7,"type":"Pawn4","name":"7D","cleanType":"pawn"},{"x":5,"y":7,"type":"Pawn5","name":"7E","cleanType":"pawn"},{"x":6,"y":7,"type":"Pawn6","name":"7F","cleanType":"pawn"},{"x":7,"y":7,"type":"Pawn7","name":"7G","cleanType":"pawn"},{"x":8,"y":7,"type":"Pawn8","name":"7H","cleanType":"pawn"}]

    el.rank = ["A", "B", "C", "D", "E", "F", "G", "H"];
    var result = []
    type = "white";
    piece= "bishop";
    var x = 5;
    var y = 1;
    var v = {
    // the validation methods
    bishop: function () {
        var offSet = [];

        for (var i = 1; i <= 8; i++) {
          if (x + i < 8 && y + i < 8)
            offSet.push({ x: x + i, y: y + i });

          if (x + i < 8 && y - i < 8)
            offSet.push({ x: x + i, y: y - i });

          if (x - i < 8 && y + i < 8)
            offSet.push({ x: x - i, y: y + i });

          if (x - i < 8 && y - i < 8)
            offSet.push({ x: x - i, y: y - i });
        }
                       
                  
  if (type == "white") 
    result = offSet.flatMap((item) => item.y + el.rank[item.x]).filter((item) => el.whiteMoves.filter((x) => x.name == item).length <= 0);
    else result = offSet.flatMap((item) => item.y + el.rank[item.x]).filter((item) => el.blackMoves.filter((x) => x.name == item).length <= 0);

  return result;
   }
  }
  v[piece]();
  // there is some invalid values like -5A or NaN but its not a problem these will be removed later on
  console.log(result);

Answer 1

您的for循环根本不包含检测冲突的代码。它一直延伸到板的边缘。考虑将循环分成四个独立的for循环，每个循环终止于电路板的边缘，或者在检测到碰撞时终止。对于相同颜色（非法移动）或相反颜色（捕获）的一块，需要分别处理碰撞。

var dx = +1, dy = +1;
do {
  x += dx;
  y += dy;

  // Running into any color piece terminates the loop.
  // However, running into an opposite color piece adds one last legal move.
  var onBoard = (x >= 0) && (x < 8) && (y >= 0) && (y < 8);
  var samePiece = onBoard ? (detect_collision_with_same_color_piece) : false;
  var oppPiece = onBoard ? (detect_collision_with_opp_color_piece) : false;

  if (onBoard && !samePiece) {
    offSet.push({ x: x, y: y });
  }
} while (onBoard && !samePiece && !oppPiece);

很难提供确切的碰撞检测代码，所以我在那儿留了一些占位符。其他一些想法：

很显然，您应该对dx和dy进行参数化，以遍历所有+1和-1组合，以免重复执行上述代码四次。
如果内存不是问题，则可以在矩阵的四个侧面上都填充一些特殊值，这样就不必每次都检查x和y的正确性。例如，如果要移动白色，并且用白色棋子填充矩阵，则可以一起删除onBoard； samePiece为真时，循环将终止。确实会使您的电路板从64平方增加到100平方，一旦添加了转置表，这将非常重要。
更好的是，考虑一下rotated bitboards，这是一种完全不同且更快的移动生成方法。

国际象棋棋子的合法举动

1 个答案: