// instead of adding another permutation array, just use hash table defined above
public static float newPerlin(float x, float y, float z) {
int A, AA, AB, B, BA, BB;
float floorX = (float) Math.floor(x),
floorY = (float) Math.floor(y),
floorZ = (float) Math.floor(z);
int intX = (int) floorX & 0xFF, intY = (int) floorY & 0xFF, intZ = (int) floorZ & 0xFF;
x -= floorX;
y -= floorY;
z -= floorZ;
// computing fading curves
floorX = NoiseMath.npfade(x);
floorY = NoiseMath.npfade(y);
floorZ = NoiseMath.npfade(z);
A = hash[intX] + intY;
AA = hash[A] + intZ;
AB = hash[A + 1] + intZ;
B = hash[intX + 1] + intY;
BA = hash[B] + intZ;
BB = hash[B + 1] + intZ;
return NoiseMath.lerp(
floorZ,
NoiseMath.lerp(
floorY,
NoiseMath.lerp(
floorX, NoiseMath.grad(hash[AA], x, y, z), NoiseMath.grad(hash[BA], x - 1, y, z)),
NoiseMath.lerp(
floorX,
NoiseMath.grad(hash[AB], x, y - 1, z),
NoiseMath.grad(hash[BB], x - 1, y - 1, z))),
NoiseMath.lerp(
floorY,
NoiseMath.lerp(
floorX,
NoiseMath.grad(hash[AA + 1], x, y, z - 1),
NoiseMath.grad(hash[BA + 1], x - 1, y, z - 1)),
NoiseMath.lerp(
floorX,
NoiseMath.grad(hash[AB + 1], x, y - 1, z - 1),
NoiseMath.grad(hash[BB + 1], x - 1, y - 1, z - 1))));
}
public static float noise3Perlin(float x, float y, float z) {
float t = x + 10000.0f;
int bx0 = (int) t & 0xFF;
int bx1 = bx0 + 1 & 0xFF;
float rx0 = t - (int) t;
float rx1 = rx0 - 1.0f;
t = y + 10000.0f;
int by0 = (int) t & 0xFF;
int by1 = by0 + 1 & 0xFF;
float ry0 = t - (int) t;
float ry1 = ry0 - 1.0f;
t = z + 10000.0f;
int bz0 = (int) t & 0xFF;
int bz1 = bz0 + 1 & 0xFF;
float rz0 = t - (int) t;
float rz1 = rz0 - 1.0f;
int i = p[bx0];
int j = p[bx1];
int b00 = p[i + by0];
int b10 = p[j + by0];
int b01 = p[i + by1];
int b11 = p[j + by1];
float sx = NoiseMath.surve(rx0);
float sy = NoiseMath.surve(ry0);
float sz = NoiseMath.surve(rz0);
float[] q = g[b00 + bz0];
float u = NoiseMath.at(rx0, ry0, rz0, q);
q = g[b10 + bz0];
float v = NoiseMath.at(rx1, ry0, rz0, q);
float a = NoiseMath.lerp(sx, u, v);
q = g[b01 + bz0];
u = NoiseMath.at(rx0, ry1, rz0, q);
q = g[b11 + bz0];
v = NoiseMath.at(rx1, ry1, rz0, q);
float b = NoiseMath.lerp(sx, u, v);
float c = NoiseMath.lerp(sy, a, b);
q = g[b00 + bz1];
u = NoiseMath.at(rx0, ry0, rz1, q);
q = g[b10 + bz1];
v = NoiseMath.at(rx1, ry0, rz1, q);
a = NoiseMath.lerp(sx, u, v);
q = g[b01 + bz1];
u = NoiseMath.at(rx0, ry1, rz1, q);
q = g[b11 + bz1];
v = NoiseMath.at(rx1, ry1, rz1, q);
b = NoiseMath.lerp(sx, u, v);
float d = NoiseMath.lerp(sy, a, b);
return 1.5f * NoiseMath.lerp(sz, c, d);
}
/**
* Not 'pure' Worley, but the results are virtually the same. Returns distances in da and point
* coords in pa
*/
public static void voronoi(
float x,
float y,
float z,
float[] da,
float[] pa,
float distanceExponent,
int distanceType) {
float xd, yd, zd, d, p[] = new float[3];
DistanceFunction distanceFunc = distanceFunctions.get(Integer.valueOf(distanceType));
if (distanceFunc == null) {
distanceFunc = distanceFunctions.get(Integer.valueOf(0));
}
int xi = (int) FastMath.floor(x);
int yi = (int) FastMath.floor(y);
int zi = (int) FastMath.floor(z);
da[0] = da[1] = da[2] = da[3] = Float.MAX_VALUE; // 1e10f;
for (int i = xi - 1; i <= xi + 1; ++i) {
for (int j = yi - 1; j <= yi + 1; ++j) {
for (int k = zi - 1; k <= zi + 1; ++k) {
NoiseMath.hash(i, j, k, p);
xd = x - (p[0] + i);
yd = y - (p[1] + j);
zd = z - (p[2] + k);
d = distanceFunc.execute(xd, yd, zd, distanceExponent);
if (d < da[0]) {
da[3] = da[2];
da[2] = da[1];
da[1] = da[0];
da[0] = d;
pa[9] = pa[6];
pa[10] = pa[7];
pa[11] = pa[8];
pa[6] = pa[3];
pa[7] = pa[4];
pa[8] = pa[5];
pa[3] = pa[0];
pa[4] = pa[1];
pa[5] = pa[2];
pa[0] = p[0] + i;
pa[1] = p[1] + j;
pa[2] = p[2] + k;
} else if (d < da[1]) {
da[3] = da[2];
da[2] = da[1];
da[1] = d;
pa[9] = pa[6];
pa[10] = pa[7];
pa[11] = pa[8];
pa[6] = pa[3];
pa[7] = pa[4];
pa[8] = pa[5];
pa[3] = p[0] + i;
pa[4] = p[1] + j;
pa[5] = p[2] + k;
} else if (d < da[2]) {
da[3] = da[2];
da[2] = d;
pa[9] = pa[6];
pa[10] = pa[7];
pa[11] = pa[8];
pa[6] = p[0] + i;
pa[7] = p[1] + j;
pa[8] = p[2] + k;
} else if (d < da[3]) {
da[3] = d;
pa[9] = p[0] + i;
pa[10] = p[1] + j;
pa[11] = p[2] + k;
}
}
}
}
}