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# Flafla2/Perlin_Tiled.cs

Last active September 14, 2024 03:58
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A slightly modified implementation of Ken Perlin's improved noise that allows for tiling the noise arbitrarily.
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### MisterE123 commented Feb 19, 2024

What is the license of this code?

This does some weird stuff when y is negative.... the output goes way outside 0 and 1.

Should `double xf = x-(int)x;` be.... `double xf = Math.abs(x-(int)x)`? Or somehow reverse into another unit square?

Fix - the xi and xf variables need to be using 'floor' instead of int casting, which will take -2.4 to -2 instead of -3 (there may be a faster way, please let me know if you find it 😄 )

``````    let xi = Math.floor(x) & 255;
let yi = Math.floor(y) & 255;
let zi = Math.floor(z) & 255;
let xf = x - Math.floor(x);
let yf = y - Math.floor(y);
let zf = z - Math.floor(z);
``````

### sakura-ice commented Sep 14, 2024

Fixed the bug in the code.

``````public class Perlin {

public int repeat;

public Perlin(int repeat = -1) {
this.repeat = repeat;
}

public double OctavePerlin(double x, double y, double z, int octaves, double persistence) {
double total = 0;
double frequency = 1;
double amplitude = 1;
double maxValue = 0;			// Used for normalizing result to 0.0 - 1.0
for(int i=0;i<octaves;i++) {
total += perlin(x * frequency, y * frequency, z * frequency) * amplitude;

maxValue += amplitude;

amplitude *= persistence;
frequency *= 2;
}

}

private static readonly int[] permutation = { 151,160,137,91,90,15,					// Hash lookup table as defined by Ken Perlin.  This is a randomly
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,	// arranged array of all numbers from 0-255 inclusive.
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
};

private static readonly int[] p; 													// Doubled permutation to avoid overflow

static Perlin() {
p = new int[512];
for(int x=0;x<512;x++) {
p[x] = permutation[x%256];
}
}

public double perlin(double x, double y, double z) {
if(repeat > 0) {									// If we have any repeat on, change the coordinates to their "local" repetitions
x = x%repeat;
y = y%repeat;
z = z%repeat;
}

int xi = (int)Math.Floor(x) & 255;					// Calculate the "unit cube" that the point asked will be located in
int yi = (int)Math.Floor(y) & 255;					// The left bound is ( |_x_|,|_y_|,|_z_| ) and the right bound is that
int zi = (int)Math.Floor(z) & 255;					// plus 1.  Next we calculate the location (from 0.0 to 1.0) in that cube.
double xf = x - Math.Floor(x);						// We also fade the location to smooth the result.
double yf = y - Math.Floor(y);
double zf = z - Math.Floor(z);

int aaa, aba, aab, abb, baa, bba, bab, bbb;
aaa = p[p[p[    xi ]+    yi ]+    zi ];
aba = p[p[p[    xi ]+inc(yi)]+    zi ];
aab = p[p[p[    xi ]+    yi ]+inc(zi)];
abb = p[p[p[    xi ]+inc(yi)]+inc(zi)];
baa = p[p[p[inc(xi)]+    yi ]+    zi ];
bba = p[p[p[inc(xi)]+inc(yi)]+    zi ];
bab = p[p[p[inc(xi)]+    yi ]+inc(zi)];
bbb = p[p[p[inc(xi)]+inc(yi)]+inc(zi)];

double x1, x2, y1, y2;
x1 = lerp(	grad (aaa, xf  , yf  , zf),				// The gradient function calculates the dot product between a pseudorandom
grad (baa, xf-1, yf  , zf),				// gradient vector and the vector from the input coordinate to the 8
u);										// surrounding points in its unit cube.
x2 = lerp(	grad (aba, xf  , yf-1, zf),				// This is all then lerped together as a sort of weighted average based on the faded (u,v,w)
u);
y1 = lerp(x1, x2, v);

x1 = lerp(	grad (aab, xf  , yf  , zf-1),
grad (bab, xf-1, yf  , zf-1),
u);
x2 = lerp(	grad (abb, xf  , yf-1, zf-1),
u);
y2 = lerp (x1, x2, v);

return (lerp (y1, y2, w)+1)/2;						// For convenience we bound it to 0 - 1 (theoretical min/max before is -1 - 1)
}

public int inc(int num) {
num++;
if (repeat > 0) num %= repeat;

return num;
}

public static double grad(int hash, double x, double y, double z) {
int h = hash & 15;									// Take the hashed value and take the first 4 bits of it (15 == 0b1111)
double u = h < 8 /* 0b1000 */ ? x : y;				// If the most significant bit (MSB) of the hash is 0 then set u = x.  Otherwise y.

double v;											// In Ken Perlin's original implementation this was another conditional operator (?:).  I

if(h < 4 /* 0b0100 */)								// If the first and second significant bits are 0 set v = y
v = y;
else if(h == 12 /* 0b1100 */ || h == 14 /* 0b1110*/)// If the first and second significant bits are 1 set v = x
v = x;
else 												// If the first and second significant bits are not equal (0/1, 1/0) set v = z
v = z;

return ((h&1) == 0 ? u : -u)+((h&2) == 0 ? v : -v); // Use the last 2 bits to decide if u and v are positive or negative.  Then return their addition.
}

public static double fade(double t) {
// Fade function as defined by Ken Perlin.  This eases coordinate values
// so that they will "ease" towards integral values.  This ends up smoothing
// the final output.
return t * t * t * (t * (t * 6 - 15) + 10);			// 6t^5 - 15t^4 + 10t^3
}

public static double lerp(double a, double b, double x) {
return a + x * (b - a);
}
``````

}