我使用了perlin噪声来生成2D高度图。起初我手动尝试了一些参数,并找到了振幅,持久性......的良好组合,用于我的工作。
现在我正在开发程序,我添加了用户更改地图参数并为自己制作新地图的功能,但现在我看到,对于某些参数(大多数是八度音和频率),值不在我常常看到的范围。我认为如果设置幅度= 20,我得到的值(高度)将在例如[0,20]或[-10,10]或[-20,20]范围内,但现在我看到幅度是不是控制输出范围的唯一参数。
我的问题是:是否有精确的数学公式(幅度,八度,频率和持久性的函数)来计算范围,或者我应该采取大量样本(如100,000)并检查它们的最小值和最大值是否可以猜测近似范围?
注意:以下代码是perlin噪声的一个实现,其中一个stackoverflow人员在C中对它进行了操作,并将其移植到java。
PerlinNoiseParameters.java
public class PerlinNoiseParameters {
public double persistence;
public double frequency;
public double amplitude;
public int octaves;
public int randomseed;
public PerlinNoiseParameters(double persistence, double frequency, double amplitude, int octaves, int randomseed) {
this.ChangeParameters(persistence, frequency, amplitude, octaves, randomseed);
}
public void ChangeParameters(double persistence, double frequency, double amplitude, int octaves, int randomseed) {
this.persistence = persistence;
this.frequency = frequency;
this.amplitude = amplitude;
this.octaves = octaves;
this.randomseed = 2 + randomseed * randomseed;
}
}
PerlinNoiseGenerator.java
public class PerlinNoiseGenerator {
PerlinNoiseParameters parameters;
public PerlinNoiseGenerator() {
}
public PerlinNoiseGenerator(PerlinNoiseParameters parameters) {
this.parameters = parameters;
}
public void ChangeParameters(double persistence, double frequency, double amplitude, int octaves, int randomseed) {
parameters.ChangeParameters(persistence, frequency, amplitude, octaves, randomseed);
}
public void ChangeParameters(PerlinNoiseParameters newParams) {
parameters = newParams;
}
public double get(double x, double y) {
return parameters.amplitude * Total(x, y);
}
private double Total(double i, double j) {
double t = 0.0f;
double _amplitude = 1;
double freq = parameters.frequency;
for (int k = 0; k < parameters.octaves; k++) {
t += GetValue(j * freq + parameters.randomseed, i * freq + parameters.randomseed)
* _amplitude;
_amplitude *= parameters.persistence;
freq *= 2;
}
return t;
}
private double GetValue(double x, double y) {
int Xint = (int) x;
int Yint = (int) y;
double Xfrac = x - Xint;
double Yfrac = y - Yint;
double n01 = Noise(Xint - 1, Yint - 1);
double n02 = Noise(Xint + 1, Yint - 1);
double n03 = Noise(Xint - 1, Yint + 1);
double n04 = Noise(Xint + 1, Yint + 1);
double n05 = Noise(Xint - 1, Yint);
double n06 = Noise(Xint + 1, Yint);
double n07 = Noise(Xint, Yint - 1);
double n08 = Noise(Xint, Yint + 1);
double n09 = Noise(Xint, Yint);
double n12 = Noise(Xint + 2, Yint - 1);
double n14 = Noise(Xint + 2, Yint + 1);
double n16 = Noise(Xint + 2, Yint);
double n23 = Noise(Xint - 1, Yint + 2);
double n24 = Noise(Xint + 1, Yint + 2);
double n28 = Noise(Xint, Yint + 2);
double n34 = Noise(Xint + 2, Yint + 2);
double x0y0 = 0.0625 * (n01 + n02 + n03 + n04) + 0.1250
* (n05 + n06 + n07 + n08) + 0.2500 * n09;
double x1y0 = 0.0625 * (n07 + n12 + n08 + n14) + 0.1250
* (n09 + n16 + n02 + n04) + 0.2500 * n06;
double x0y1 = 0.0625 * (n05 + n06 + n23 + n24) + 0.1250
* (n03 + n04 + n09 + n28) + 0.2500 * n08;
double x1y1 = 0.0625 * (n09 + n16 + n28 + n34) + 0.1250
* (n08 + n14 + n06 + n24) + 0.2500 * n04;
double v1 = Interpolate(x0y0, x1y0, Xfrac);
double v2 = Interpolate(x0y1, x1y1, Xfrac);
double fin = Interpolate(v1, v2, Yfrac);
return fin;
}
private double Interpolate(double x, double y, double a) {
double negA = 1.0 - a;
double negASqr = negA * negA;
double fac1 = 3.0 * (negASqr) - 2.0 * (negASqr * negA);
double aSqr = a * a;
double fac2 = 3.0 * aSqr - 2.0 * (aSqr * a);
return x * fac1 + y * fac2;
}
private double Noise(int x, int y) {
int n = x + y * 57;
n = (n << 13) ^ n;
int t = (n * (n * n * 15731 + 789221) + 1376312589) & 0x7fffffff;
return 1.0 - (double) t * 0.931322574615478515625e-9;
}
}
答案 0 :(得分:1)
这对于八度音阶和频率影响振幅不是问题,至少不是直接影响振幅。这是整数溢出的问题。因为你通过将随机种子添加到x和y坐标中来引入它(这是不寻常的,我不认为这是通常的暗示)
t += GetValue(j * freq + parameters.randomseed, i * freq + parameters.randomseed)* _amplitude;
随机种子可能很大(可能是int的接近完整大小),因为
this.randomseed = 2 + randomseed * randomseed;
因此,如果您为j和i输入较大的值,则最终会在GetValue(double x, double y)
处传递的双倍大于int的最大大小,此时您调用
int Xint = (int) x;
int Yint = (int) y;
Xint和YInt不会像x和y那样(因为x和y可能很大!)所以
double Xfrac = x - Xint;
double Yfrac = y - Yint;
可能比1大得多,允许返回介于-1和1之间的值。
使用合理的小值,使用代码的范围介于-1和1之间(幅度1)
作为助手,在java中,方法名称通常是methodName
,而不是MethodName
如果它有用请在这里查找perlin噪声的另一个java实现: http://mrl.nyu.edu/~perlin/noise/
答案 1 :(得分:0)
单个Perlin噪声阶跃的范围为: http://digitalfreepen.com/2017/06/20/range-perlin-noise.html
-sqrt(N/4), sqrt(N/4)
以N为尺寸数。 2您的情况。
八度,余辉和振幅附加于此:
double range = 0.0;
double _amplitude = parameters.;
for (int k = 0; k < parameters.octaves; k++) {
range += sqrt(N/4) * _amplitude;
_amplitude *= parameters.persistence;
}
return range;
可能有一些方法可以将其作为单个数学表达式来执行。涉及pow(),但由于大脑的作用,我现在无法接受。