Javascript 尝试在二维中点细分中描绘循环的逻辑

我理解这个概念，但在实现2D细分后的循环逻辑时遇到困难。我有一个2d数组，表示一个带有角点种子的网格。我想我需要3圈。1来循环细分迭代次数。第二个用于行中的每列，第三个用于每行

这就是我所拥有的。它显示左上角方形细分的结果。这就是为什么行和列只循环一次。如果我得到了基本的逻辑，剩下的应该是蛋糕。但是，循环在第三次迭代中不起作用。我很确定这个循环需要更加复杂

迭代是一个手动设置的变量

       // iterate though subdivision levels
       for(i = 1; i <= iterations; i++) {                   // iteration

            // iterate through each row
            for(row = 1; row <= 1; row += size / i ) {      // row

                // iterate through each column and subdivide
                for(col = 1; col <= 1; col += size / i) {   // column

                    //*** ONLY SHOWING THE FIRST SUBDIVIDE OF EACH ITERATION ***//

                    // Math.pow(2, iterations) / 2 / i = 1
                    // i = 1, iteration = 1
                    heightmap[0][1] = (heightmap[0][0] + heightmap[0][2]) / 2;

                    // Math.pow(2, iterations) / 2 / i = 2
                    // i = 1, iterations = 2
                    heightmap[0][2] = (heightmap[0][0] + heightmap[0][4]) / 2;

                    // Math.pow(2, iterations) / 2 / i = 4
                    // i = 1, iterations = 3
                    heightmap[0][4] = (heightmap[0][0] + heightmap[0][8]) / 2;

                    // if iterations == 1, i != 2

                    // Math.pow(2, iterations) / 2 / i = 1
                    // i = 2, iterations = 2
                    heightmap[0][1] = (heightmap[0][0] + heightmap[0][2]) / 2;

                    // Math.pow(2, iterations) / 2 / i = 2
                    // i = 2, iterations = 3
                    heightmap[0][2] = (heightmap[0][0] + heightmap[0][4]) / 2;

                    // if iterations == 1 or 2, i != 3

                    // Math.pow(2, iterations) / 2 / i = 4/3 != 1
                    // i = 3, iterations = 3
                    heightmap[0][1] = (heightmap[0][0] + heightmap[0][2]) / 2;
                }
            }
        } 

//迭代细分级别
对于（i=1；i而言，一次细分一个方向会更容易：

var points = [[ 0,-1,-1,-1, 1],
              [-1,-1,-1,-1,-1],
              [-1,-1,-1,-1,-1],
              [-1,-1,-1,-1,-1],
              [ 2,-1,-1,-1, 0]];
var size = 4;

for (var iteration = 1; iteration < size; iteration *= 2) {
    var step = size / iteration;

    // * . * . *
    // x . x . x
    // * . * . *
    for (var row = step / 2; row < size; row += step) {
        for (var col = 0; col <= size; col += step) {
            points[row][col] = ((points[row - step / 2][col] + points[row + step / 2][col]) / 2);
        }
    }

    // * x * x *
    // * x * x *
    // * x * x *
    for (var row = 0; row <= size; row += step / 2) {
        for (var col = step / 2; col < size; col += step) {
            points[row][col] = ((points[row][col - step / 2] + points[row][col + step / 2]) / 2);
        }
    }
}

var points = [[ 0,-1,-1,-1, 1],
              [-1,-1,-1,-1,-1],
              [-1,-1,-1,-1,-1],
              [-1,-1,-1,-1,-1],
              [ 2,-1,-1,-1, 0]];
var size = 4;

for (var iteration = 1; iteration < size; iteration *= 2) {
    var step = size / iteration;

    // * . * . *
    // x . x . x
    // * . * . *
    for (var row = step / 2; row < size; row += step) {
        for (var col = 0; col <= size; col += step) {
            points[row][col] = ((points[row - step / 2][col] + points[row + step / 2][col]) / 2);
        }
    }

    // * x * x *
    // * x * x *
    // * x * x *
    for (var row = 0; row <= size; row += step / 2) {
        for (var col = step / 2; col < size; col += step) {
            points[row][col] = ((points[row][col - step / 2] + points[row][col + step / 2]) / 2);
        }
    }
}

[[ 0,   0.25,   0.5,   0.75,   1    ],
 [ 0.5, 0.5625, 0.625, 0.6875, 0.75 ],
 [ 1,   0.875,  0.75,  0.625,  0.5  ],
 [ 1.5, 1.1875, 0.875, 0.5625, 0.25 ],
 [ 2,   1.5,    1,     0.5,    0    ]]