棋子合法动作

Question

我和vue.js下了一盘国际象棋，现在我正试着找出每一件棋子的可能动作。我能够修复knight, pawn and bishop的合法动作验证。

在进行主教验证的时候，我遇到了一个问题。它是能够验证，如果有一块站在主教面前。

请看图片，以了解更多。

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你看，红色的数字是如何在典当和继续上升的。它应该停在当铺的数字上。

这是我计算主教的代码。如果有可能的话，如果你也能为女王和其他作品提供验证，那将是非常有帮助的。

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    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循环根本不包括检测冲突的代码。它一直延伸到董事会的边缘。考虑将循环分成四个独立的for循环，每个循环在板的边缘终止，或者当它检测到碰撞时终止。对于同一颜色(非法移动)或相反颜色(捕获)，碰撞需要单独处理。

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);

很难提供确切的冲突检测代码，所以我在那里留下了一些占位符。一些额外的想法：

1. 显然，您应该参数化dx和dy来循环所有+1和-1组合，这样就不会重复上面的代码四次了。
2. 如果内存不是一个问题，您可以在所有四个方面使用一些特殊的值来填充矩阵，这样就不需要每次检查x和y是否正确。例如，如果white要移动，然后用白爪子填充矩阵，那么您可以一起删除onBoard；循环将在samePiece变为真时终止。这确实增加了您的董事会从64平方到100，这可能是重要的，一旦你添加转位表。
3. 更好的是，考虑一下旋转位板，它是一种完全不同、而且速度更快的移动生成方法。