Question

所以，我一直试图在我的项目中使用Perlin Noise，而且奇怪的是噪声函数输出主要是从[-1,1]边界出来的值。我已通过Perlin's code for his improved noise in Java和another useful implementation in C++ by sol在C中实现了它。我已经完成了至少20次并确保它完全相同，但无法找到问题的原因。以下是它的外观：

//   DEFAULT_PERMUTATION is defined through the preprocessor.
// It is basically an initializer of the same array used in Perlin's
// implementation, except it has been duplicated into a 512 element array


double simpleNoise(double x, double y, double z){

double res;

int p[512] = DEFAULT_PERMUTATION;

int X = (int) floor(x) & 255;
int Y = (int) floor(y) & 255;
int Z = (int) floor(z) & 255;

double u = fade(x);
double v = fade(y);
double w = fade(z);

int A = p[X] + Y;
int AA = p[A] + Z;
int AB = p[A + 1] + Z;
int B = p[X + 1] + Y;
int BA = p[B] + Z;
int BB = p[B + 1] + Z;

x -= floor(x);
y -= floor(y);
z -= floor(z);


res = 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 ))));
return res;
}

其他私人功能也非常直接：

double fade(double t){
return t * t * t * (t * (t * 6 - 15) + 10);
}

double lerp(double t, double a, double b){
return a + t * (b - a); 
}

double grad(int hash, double x, double y, double z){
int h = hash & 15;                     
  double 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);
}

In my project, the noise looks like this，因为颜色只关心0到1之间的值，但是某些位置的测试会给出-15,32等荒谬值，有些值大于数千。至少可以说，提供值（1，-4,1），（2,5,1），（2,0,0）很有意思。

问题是什么？

编辑：出于调查目的，该程序列出了发现的不一致性：

#define DEFAULT_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,\
    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 }
#define SIZE 10
double fade(double t){
return t * t * t * (t * (t * 6 - 15) + 10);
}

double lerp(double t, double a, double b){
return a + t * (b - a); 
}

double grad(int hash, double x, double y, double z){
int h = hash & 15;                     
double 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);
}
double simpleNoise(double x, double y, double z){
double res;
int p[512] = DEFAULT_PERMUTATION;
int X = (int) floor(x) & 255;
int Y = (int) floor(y) & 255;
int Z = (int) floor(z) & 255;
double u = fade(x);
double v = fade(y);
double w = fade(z);
int A = p[X] + Y;
int AA = p[A] + Z;
int AB = p[A + 1] + Z;
int B = p[X + 1] + Y;
int BA = p[B] + Z;
int BB = p[B + 1] + Z;
x -= floor(x);
y -= floor(y);
z -= floor(z);
res = 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 ))));
return res;
}
main(){
int i, j, k, n=0;
double e;
for(i=-SIZE/2; i<SIZE/2; i++){
    for(j=-SIZE/2; j<SIZE/2; j++){
        for(k=-SIZE/2; k<SIZE/2; k++){
            e = simpleNoise(i, j, k);
            if(fabs(e) > 1){
               printf("P(%d,%d,%d): %f\n", i, j, k, e);
               n++;
            }
        }
    }
}
printf("%d inconsistencies found", n);
}

Answer 1

参考实现here确实

x -= floor(x);
y -= floor(y);
z -= floor(z);

之前的

步骤

double u = fade(x);
double v = fade(y);
double w = fade(z);

您的实施在代码中进一步推动了这一步骤。为什么呢？

如果您将其移回原来的位置，请在您的调查＆＃34;代码噪声函数总是预期返回零，因为Perlin噪声函数应该在整数点返回零。

Perlin Noise功能 - 值超出范围

1 个答案: