JavaScript单工/珀林噪声
JavaScript单工/珀林噪声
提问于 2011-12-07 03:21:54
我尝试在JavaScript中创建一个perlin / simplex / value噪声函数，它将产生类似于以下结果的结果：
(注意:此图像已经应用了treshold。我想要不带发夹的。)
我已经在网上找了两天了。有很多解释噪音和大量代码的链接，但在JavaScript中没有一个，每当我尝试转换它时，我都会得到奇怪的结果。以下是我的一些尝试：
- http://jsfiddle.net/GZCye/
- http://jsfiddle.net/Lk56f/ (based on this)
- http://pastebin.com/nMA40SrL (由于某种原因不能在based on this
中运行
我认为部分问题可能源于在JavaScript中使用按位运算符，在这种情况下很难强制数字的类型。
我正在寻找的是一个工作的例子，JavaScript noise (任何类型)。
我也愿意奖励任何与参数有关的信息，这些信息可能导致类似于我发布的图像的模式(阈值之前)或任何优化提示，因为我需要它尽可能快地运行(尽管如此，在JavaScript中)。
回答 3
Stack Overflow用户
发布于 2013-03-22 08:38:53
我知道这很古老，但也许它对其他人仍然有用。我已经构建了一个javascript应用程序，可以将Perlin和单工噪声渲染到HTML5画布上，在这里查看它：http://lencinhaus.github.com/canvas-noise
该应用程序允许调整噪声计算和渲染中涉及的每个参数，并保存生成的纹理。它还会将参数添加到URL，以便可以共享。例如，this configuration会生成与您显示的纹理非常相似的纹理。
希望这能有所帮助！
Stack Overflow用户
发布于 2012-10-19 06:45:22
这是我为2D / 3D Perlin噪声而结束的代码。请注意，它使用的是RequireJS的AMD模块语法，但如果您不使用AMD加载器，则可以将其去掉。
define(
[],
function()
{
function fade(t)
{
return t * t * t * (t * (t * 6 - 15) + 10);
}
function lerp(t, a, b)
{
return a + t * (b - a);
}
function grad(hash, x, y, z)
{
// Convert lo 4 bits of hash code into 12 gradient directions.
var h = hash & 15,
u = h < 8 ? x : y,
v = h < 4 ? y : h == 12 || h == 14 ? x : z;
return ((h & 1) == 0 ? u : -u) + ((h & 2) == 0 ? v : -v);
}
function scale(n)
{
return (1 + n) / 2;
}
var p = new Array(512);
var permutation = [
151, 160, 137, 91, 90, 15, 131, 13, 201, 95, 96, 53, 194, 233, 7, 225, 140, 36, 103, 30, 69, 142, 8, 99,
37, 240, 21, 10, 23, 190, 6, 148, 247, 120, 234, 75, 0, 26, 197, 62, 94, 252, 219, 203, 117, 35, 11, 32,
57, 177, 33, 88, 237, 149, 56, 87, 174, 20, 125, 136, 171, 168, 68, 175, 74, 165, 71, 134, 139, 48, 27, 166,
77, 146, 158, 231, 83, 111, 229, 122, 60, 211, 133, 230, 220, 105, 92, 41, 55, 46, 245, 40, 244, 102, 143,
54, 65, 25, 63, 161, 1, 216, 80, 73, 209, 76, 132, 187, 208, 89, 18, 169, 200, 196, 135, 130, 116, 188, 159,
86, 164, 100, 109, 198, 173, 186, 3, 64, 52, 217, 226, 250, 124, 123, 5, 202, 38, 147, 118, 126, 255, 82,
85, 212, 207, 206, 59, 227, 47, 16, 58, 17, 182, 189, 28, 42, 223, 183, 170, 213, 119, 248, 152, 2, 44,
154, 163, 70, 221, 153, 101, 155, 167, 43, 172, 9, 129, 22, 39, 253, 19, 98, 108, 110, 79, 113, 224, 232,
178, 185, 112, 104, 218, 246, 97, 228, 251, 34, 242, 193, 238, 210, 144, 12, 191, 179, 162, 241, 81, 51,
145, 235, 249, 14, 239, 107, 49, 192, 214, 31, 181, 199, 106, 157, 184, 84, 204, 176, 115, 121, 50, 45,
127, 4, 150, 254, 138, 236, 205, 93, 222, 114, 67, 29, 24, 72, 243, 141, 128, 195, 78, 66, 215, 61, 156, 180
];
for (var i = 0; i < 256; i++) {
p[256 + i] = p[i] = permutation[i];
}
return {
/** Returns a number between 0 and 1. */
noise3d: function(x, y, z)
{
// Find unit cube that contains point.
var X = Math.floor(x) & 255,
Y = Math.floor(y) & 255,
Z = Math.floor(z) & 255;
// Find relative x,y,z of point in cube.
x -= Math.floor(x);
y -= Math.floor(y);
z -= Math.floor(z);
// Compute fade curves for each of x,y,z.
var u = fade(x),
v = fade(y),
w = fade(z);
// Hash coordinates of the corners.
var A = p[X ] + Y, AA = p[A] + Z, AB = p[A + 1] + Z,
B = p[X + 1] + Y, BA = p[B] + Z, BB = p[B + 1] + Z;
// Add blended results from 8 corners of cube.
return scale(
lerp(
w,
lerp(
v,
lerp(
u,
grad(p[AA], x, y, z),
grad(p[BA], x - 1, y, z)
),
lerp(
u,
grad(p[AB], x, y - 1, z),
grad(p[BB], x - 1, y - 1, z)
)
),
lerp(
v,
lerp(
u,
grad(p[AA + 1], x, y, z - 1),
grad(p[BA + 1], x - 1, y, z - 1)
),
lerp(
u,
grad(p[AB + 1], x, y - 1, z - 1),
grad(p[BB + 1], x - 1, y - 1, z - 1)
)
)
)
);
},
/** Returns a number between 0 and 1. */
noise2d: function(x, y)
{
// Find unit square that contains point.
var X = Math.floor(x) & 255,
Y = Math.floor(y) & 255;
// Find relative x,y of point in square.
x -= Math.floor(x);
y -= Math.floor(y);
// Compute fade curves for each of x,y.
var u = fade(x),
v = fade(y);
// Hash coordinates of the corners.
var A = p[X ] + Y, AA = p[A], AB = p[A + 1],
B = p[X + 1] + Y, BA = p[B], BB = p[B + 1];
// Add blended results from the corners.
return scale(
lerp(
v,
lerp(
u,
grad(p[AA], x, y, 0),
grad(p[BA], x - 1, y, 0)
),
lerp(
u,
grad(p[AB], x, y - 1, 0),
grad(p[BB], x - 1, y - 1, 0)
)
)
);
}
};
}
);Stack Overflow用户
发布于 2018-11-09 06:08:35
我想出了这个解决方案，它为我的用例提供了更好的结果：
http://asserttrue.blogspot.com/2011/12/perlin-noise-in-javascript_31.html
它使用了一个很大的排列矩阵：
var permutation = [ 151,160,137,91,90,15,
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180
];将其用作：
pn = PerlinNoise()
n = pn.noise( random_x, random_y, .8 )页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/8405526
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