试图在2D中点细分中计算环路的逻辑

Question

我理解这个概念，但我在实现2D细分背后的循环逻辑时遇到了困难。我有一个二维数组代表一个网格的角落播种。我相信我需要3个循环。 1循环进行细分迭代次数。对于行中的每一列第二个，对于每一行第三个。

这就是我所拥有的。它显示了左上角方形细分的结果。这就是为什么行和列只循环一次。如果我得到基本的逻辑，其余的应该是蛋糕。但是循环在第三次迭代中不起作用。我很确定循环需要更复杂。

迭代是手动设置的变量。

 // 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; 
       } 
      } 
     } 

如果这有助于我用于一维细分。

// increase the number of segments 
for(var i = 1; i < size; i *= 2){ 

    // iterate through each segment calculating the center point 
    for(var j = (size/i)/2; j < size; j += size/i){ 
     points[j] = ((points[j - (size/i)/2] + points[j + (size/i)/2])/2); 
    } 

} 

Answer 1

这将是更容易在时间细分一个方向：

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); 
     } 
    } 
} 

结果是：

[[ 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 ]]