钻石广场算法固定大小

我试图找出办法有用于固定的规模：钻石广场算法固定大小

https://en.wikipedia.org/wiki/Diamond-square_algorithm

我看到算法要求的2（+1）大小的力量阵列。

我遇到的问题是，我想有不管决议产生相同的高度图。所以如果我有一个512的分辨率，它看起来与分辨率256相同，但是只有更少的细节。我只是无法弄清楚如何做到这一点。

我的初始想法是总是创建在某一尺寸如高度图1024和下采样到我想要的水库。问题是我希望较高的分辨率相当高（比如说4096），并且这会严重降低较低分辨率下的性能，因为我们必须以最高分辨率运行算法。

目前该算法是在JavaScript这里是一个片段：

function Advanced() { 
    var adv = {}, 
    res, max, heightmap, roughness; 

    adv.heightmap = function() { 
     // heightmap has one extra pixel this is ot remove it. 
     var hm = create2DArray(res-1, res-1); 
     for(var x = 0;x< res-1;x++) { 
      for(var y = 0;y< res-1;y++) { 
       hm[x][y] = heightmap[x][y]; 
      } 
     } 
     return hm; 
    } 

    adv.get = function(x,y) { 
     if (x < 0 || x > max || y < 0 || y > max) return -1; 
     return heightmap[x][y]; 
    } 

    adv.set = function(x,y,val) { 
     if(val < 0) { 
      val = 0; 
     } 

     heightmap[x][y] = val; 

    } 

    adv.divide = function(size) { 
     var x, y, half = size/2; 
     var scale = roughness * size; 
     if (half < 1) return; 

     for (y = half; y < max; y += size) { 
      for (x = half; x < max; x += size) { 
       adv.square(x, y, half, Math.random() * scale * 2 - scale); 
      } 
     } 
     for (y = 0; y <= max; y += half) { 
      for (x = (y + half) % size; x <= max; x += size) { 
       adv.diamond(x, y, half, Math.random() * scale * 2 - scale); 
      } 
     } 
     adv.divide(size/2); 
    } 

    adv.average = function(values) { 
     var valid = values.filter(function(val) { 
      return val !== -1; 
     }); 
     var total = valid.reduce(function(sum, val) { 
      return sum + val; 
     }, 0); 
     return total/valid.length; 
    } 

    adv.square = function(x, y, size, offset) { 
     var ave = adv.average([ 
      adv.get(x - size, y - size), // upper left 
      adv.get(x + size, y - size), // upper right 
      adv.get(x + size, y + size), // lower right 
      adv.get(x - size, y + size) // lower left 
     ]); 
     adv.set(x, y, ave + offset); 
    } 

    adv.diamond = function(x, y, size, offset) { 

     var ave = adv.average([ 
      adv.get(x, y - size), // top 
      adv.get(x + size, y), // right 
      adv.get(x, y + size), // bottom 
      adv.get(x - size, y) // left 
     ]); 

     adv.set(x, y, Math.abs(ave + offset)); 
    } 

    adv.generate = function(properties, resolution) { 
     Math.seedrandom(properties.seed); 

     res = resolution + 1; 
     max = res - 1; 
     heightmap = create2DArray(res, res); 

     roughness = properties.roughness; 

     adv.set(0, 0, max); 
     adv.set(max, 0, max/2); 
     adv.set(max, max, 0); 
     adv.set(0, max, max/2); 

     adv.divide(max); 
    } 

    function create2DArray(d1, d2) { 
     var x = new Array(d1), 
     i = 0, 
     j = 0; 

     for (i = 0; i < d1; i += 1) { 
      x[i] = new Array(d2); 
     } 

     for (i=0; i < d1; i += 1) { 
      for (j = 0; j < d2; j += 1) { 
       x[i][j] = 0; 
      } 
     } 

     return x; 
    } 

    return adv; 
}

有谁这样做过？

来源

2016-07-06 Dave3of5

貌似没有再笑！ – Dave3of5