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/*
* A speed-improved simplex noise algorithm for 2D, 3D and 4D in Java.
*
* Based on example code by Stefan Gustavson ([email protected]).
* Optimisations by Peter Eastman ([email protected]).
* Better rank ordering method by Stefan Gustavson in 2012.
*
* This could be speeded up even further, but it's useful as it is.
*
* Version 2012-03-09
*
* This code was placed in the public domain by its original author,
* Stefan Gustavson. You may use it as you see fit, but
* attribution is appreciated.
*
*/
package se.liu.itn.stegu;
import java.util.Random;
/**
* Simplex noise in 2D, 3D and 4D
*/
public final class SimplexNoise {
public SimplexNoise(long seed) {
short[] permutation = new short[256];
for (short i = 0; i < 256; i++) {
permutation[i] = i;
}
Random random = new Random(seed);
for (int i = 256; i > 1; i--) {
swap(permutation, i - 1, random.nextInt(i));
}
for (int i = 0; i < 256; i++) {
perm[256 + i] = perm[i] = permutation[i];
}
for(int i=0; i<512; i++)
{
permMod12[i] = (short)(perm[i] % 12);
}
}
/**
* 2D simplex noise
*/
public double noise(double xin, double yin) {
double n0, n1, n2; // Noise contributions from the three corners
// Skew the input space to determine which simplex cell we're in
double s = (xin+yin)*F2; // Hairy factor for 2D
int i = fastfloor(xin+s);
int j = fastfloor(yin+s);
double t = (i+j)*G2;
double X0 = i-t; // Unskew the cell origin back to (x,y) space
double Y0 = j-t;
double x0 = xin-X0; // The x,y distances from the cell origin
double y0 = yin-Y0;
// For the 2D case, the simplex shape is an equilateral triangle.
// Determine which simplex we are in.
int i1, j1; // Offsets for second (middle) corner of simplex in (i,j) coords
if(x0>y0) {i1=1; j1=0;} // lower triangle, XY order: (0,0)->(1,0)->(1,1)
else {i1=0; j1=1;} // upper triangle, YX order: (0,0)->(0,1)->(1,1)
// A step of (1,0) in (i,j) means a step of (1-c,-c) in (x,y), and
// a step of (0,1) in (i,j) means a step of (-c,1-c) in (x,y), where
// c = (3-sqrt(3))/6
double x1 = x0 - i1 + G2; // Offsets for middle corner in (x,y) unskewed coords
double y1 = y0 - j1 + G2;
double x2 = x0 - 1.0 + 2.0 * G2; // Offsets for last corner in (x,y) unskewed coords
double y2 = y0 - 1.0 + 2.0 * G2;
// Work out the hashed gradient indices of the three simplex corners
int ii = i & 255;
int jj = j & 255;
int gi0 = permMod12[ii+perm[jj]];
int gi1 = permMod12[ii+i1+perm[jj+j1]];
int gi2 = permMod12[ii+1+perm[jj+1]];
// Calculate the contribution from the three corners
double t0 = 0.5 - x0*x0-y0*y0;
if(t0<0) n0 = 0.0;
else {
t0 *= t0;
n0 = t0 * t0 * dot(grad3[gi0], x0, y0); // (x,y) of grad3 used for 2D gradient
}
double t1 = 0.5 - x1*x1-y1*y1;
if(t1<0) n1 = 0.0;
else {
t1 *= t1;
n1 = t1 * t1 * dot(grad3[gi1], x1, y1);
}
double t2 = 0.5 - x2*x2-y2*y2;
if(t2<0) n2 = 0.0;
else {
t2 *= t2;
n2 = t2 * t2 * dot(grad3[gi2], x2, y2);
}
// Add contributions from each corner to get the final noise value.
// The result is scaled to return values in the interval [-1,1].
return 70.0 * (n0 + n1 + n2);
}
/**
* 3D simplex noise
*/
public double noise(double xin, double yin, double zin) {
double n0, n1, n2, n3; // Noise contributions from the four corners
// Skew the input space to determine which simplex cell we're in
double s = (xin+yin+zin)*F3; // Very nice and simple skew factor for 3D
int i = fastfloor(xin+s);
int j = fastfloor(yin+s);
int k = fastfloor(zin+s);
double t = (i+j+k)*G3;
double X0 = i-t; // Unskew the cell origin back to (x,y,z) space
double Y0 = j-t;
double Z0 = k-t;
double x0 = xin-X0; // The x,y,z distances from the cell origin
double y0 = yin-Y0;
double z0 = zin-Z0;
// For the 3D case, the simplex shape is a slightly irregular tetrahedron.
// Determine which simplex we are in.
int i1, j1, k1; // Offsets for second corner of simplex in (i,j,k) coords
int i2, j2, k2; // Offsets for third corner of simplex in (i,j,k) coords
if(x0>=y0) {
if(y0>=z0)
{ i1=1; j1=0; k1=0; i2=1; j2=1; k2=0; } // X Y Z order
else if(x0>=z0) { i1=1; j1=0; k1=0; i2=1; j2=0; k2=1; } // X Z Y order
else { i1=0; j1=0; k1=1; i2=1; j2=0; k2=1; } // Z X Y order
}
else { // x0 y0) rankx++; else ranky++;
if(x0 > z0) rankx++; else rankz++;
if(x0 > w0) rankx++; else rankw++;
if(y0 > z0) ranky++; else rankz++;
if(y0 > w0) ranky++; else rankw++;
if(z0 > w0) rankz++; else rankw++;
int i1, j1, k1, l1; // The integer offsets for the second simplex corner
int i2, j2, k2, l2; // The integer offsets for the third simplex corner
int i3, j3, k3, l3; // The integer offsets for the fourth simplex corner
// simplex[c] is a 4-vector with the numbers 0, 1, 2 and 3 in some order.
// Many values of c will never occur, since e.g. x>y>z>w makes x= 3 ? 1 : 0;
j1 = ranky >= 3 ? 1 : 0;
k1 = rankz >= 3 ? 1 : 0;
l1 = rankw >= 3 ? 1 : 0;
// Rank 2 denotes the second largest coordinate.
i2 = rankx >= 2 ? 1 : 0;
j2 = ranky >= 2 ? 1 : 0;
k2 = rankz >= 2 ? 1 : 0;
l2 = rankw >= 2 ? 1 : 0;
// Rank 1 denotes the second smallest coordinate.
i3 = rankx >= 1 ? 1 : 0;
j3 = ranky >= 1 ? 1 : 0;
k3 = rankz >= 1 ? 1 : 0;
l3 = rankw >= 1 ? 1 : 0;
// The fifth corner has all coordinate offsets = 1, so no need to compute that.
double x1 = x0 - i1 + G4; // Offsets for second corner in (x,y,z,w) coords
double y1 = y0 - j1 + G4;
double z1 = z0 - k1 + G4;
double w1 = w0 - l1 + G4;
double x2 = x0 - i2 + 2.0*G4; // Offsets for third corner in (x,y,z,w) coords
double y2 = y0 - j2 + 2.0*G4;
double z2 = z0 - k2 + 2.0*G4;
double w2 = w0 - l2 + 2.0*G4;
double x3 = x0 - i3 + 3.0*G4; // Offsets for fourth corner in (x,y,z,w) coords
double y3 = y0 - j3 + 3.0*G4;
double z3 = z0 - k3 + 3.0*G4;
double w3 = w0 - l3 + 3.0*G4;
double x4 = x0 - 1.0 + 4.0*G4; // Offsets for last corner in (x,y,z,w) coords
double y4 = y0 - 1.0 + 4.0*G4;
double z4 = z0 - 1.0 + 4.0*G4;
double w4 = w0 - 1.0 + 4.0*G4;
// Work out the hashed gradient indices of the five simplex corners
int ii = i & 255;
int jj = j & 255;
int kk = k & 255;
int ll = l & 255;
int gi0 = perm[ii+perm[jj+perm[kk+perm[ll]]]] % 32;
int gi1 = perm[ii+i1+perm[jj+j1+perm[kk+k1+perm[ll+l1]]]] % 32;
int gi2 = perm[ii+i2+perm[jj+j2+perm[kk+k2+perm[ll+l2]]]] % 32;
int gi3 = perm[ii+i3+perm[jj+j3+perm[kk+k3+perm[ll+l3]]]] % 32;
int gi4 = perm[ii+1+perm[jj+1+perm[kk+1+perm[ll+1]]]] % 32;
// Calculate the contribution from the five corners
double t0 = 0.6 - x0*x0 - y0*y0 - z0*z0 - w0*w0;
if(t0<0) n0 = 0.0;
else {
t0 *= t0;
n0 = t0 * t0 * dot(grad4[gi0], x0, y0, z0, w0);
}
double t1 = 0.6 - x1*x1 - y1*y1 - z1*z1 - w1*w1;
if(t1<0) n1 = 0.0;
else {
t1 *= t1;
n1 = t1 * t1 * dot(grad4[gi1], x1, y1, z1, w1);
}
double t2 = 0.6 - x2*x2 - y2*y2 - z2*z2 - w2*w2;
if(t2<0) n2 = 0.0;
else {
t2 *= t2;
n2 = t2 * t2 * dot(grad4[gi2], x2, y2, z2, w2);
}
double t3 = 0.6 - x3*x3 - y3*y3 - z3*z3 - w3*w3;
if(t3<0) n3 = 0.0;
else {
t3 *= t3;
n3 = t3 * t3 * dot(grad4[gi3], x3, y3, z3, w3);
}
double t4 = 0.6 - x4*x4 - y4*y4 - z4*z4 - w4*w4;
if(t4<0) n4 = 0.0;
else {
t4 *= t4;
n4 = t4 * t4 * dot(grad4[gi4], x4, y4, z4, w4);
}
// Sum up and scale the result to cover the range [-1,1]
return 27.0 * (n0 + n1 + n2 + n3 + n4);
}
private static Grad grad3[] = {new Grad(1,1,0),new Grad(-1,1,0),new Grad(1,-1,0),new Grad(-1,-1,0),
new Grad(1,0,1),new Grad(-1,0,1),new Grad(1,0,-1),new Grad(-1,0,-1),
new Grad(0,1,1),new Grad(0,-1,1),new Grad(0,1,-1),new Grad(0,-1,-1)};
private static Grad grad4[]= {new Grad(0,1,1,1),new Grad(0,1,1,-1),new Grad(0,1,-1,1),new Grad(0,1,-1,-1),
new Grad(0,-1,1,1),new Grad(0,-1,1,-1),new Grad(0,-1,-1,1),new Grad(0,-1,-1,-1),
new Grad(1,0,1,1),new Grad(1,0,1,-1),new Grad(1,0,-1,1),new Grad(1,0,-1,-1),
new Grad(-1,0,1,1),new Grad(-1,0,1,-1),new Grad(-1,0,-1,1),new Grad(-1,0,-1,-1),
new Grad(1,1,0,1),new Grad(1,1,0,-1),new Grad(1,-1,0,1),new Grad(1,-1,0,-1),
new Grad(-1,1,0,1),new Grad(-1,1,0,-1),new Grad(-1,-1,0,1),new Grad(-1,-1,0,-1),
new Grad(1,1,1,0),new Grad(1,1,-1,0),new Grad(1,-1,1,0),new Grad(1,-1,-1,0),
new Grad(-1,1,1,0),new Grad(-1,1,-1,0),new Grad(-1,-1,1,0),new Grad(-1,-1,-1,0)};
// To remove the need for index wrapping, double the permutation table length
private final short perm[] = new short[512];
private final short permMod12[] = new short[512];
// Skewing and unskewing factors for 2, 3, and 4 dimensions
private static final double F2 = 0.5*(Math.sqrt(3.0)-1.0);
private static final double G2 = (3.0-Math.sqrt(3.0))/6.0;
private static final double F3 = 1.0/3.0;
private static final double G3 = 1.0/6.0;
private static final double F4 = (Math.sqrt(5.0)-1.0)/4.0;
private static final double G4 = (5.0-Math.sqrt(5.0))/20.0;
// This method is a *lot* faster than using (int)Math.floor(x)
private static int fastfloor(double x) {
int xi = (int)x;
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