Client LuaCsForBarotrauma
PerlinNoise.cs
1 using Microsoft.Xna.Framework;
2 using System;
3 using System.Collections.Generic;
4 using System.Text;
5 
6 namespace Barotrauma
7 {
8  //By Adrian Biagioli (Flafla2)
9  //under a Creative Commons Attribution 4.0 International License.
10 
11  public static class PerlinNoise
12  {
13  public static double OctavePerlin(double x, double y, double z, double frequency, int octaves, double persistence)
14  {
15  double total = 0;
16  double amplitude = 3;
17  for (int i = 0; i < octaves; i++)
18  {
19  total += CalculatePerlin(x * frequency, y * frequency, z * frequency) * amplitude;
20 
21  amplitude *= persistence;
22  frequency *= 2;
23  }
24 
25  return total;
26  }
27 
28  // Hash lookup table as defined by Ken Perlin. This is a randomly
29  // arranged array of all numbers from 0-255 inclusive.
30  private static readonly int[] permutation =
31  {
32  151,160,137,91,90,15,
33  131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
34  190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
35  88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
36  77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
37  102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
38  135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
39  5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
40  223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
41  129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
42  251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
43  49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
44  138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180
45  };
46 
47  private static readonly int[] p; // Doubled permutation to avoid overflow
48 
49  private static readonly float[] cachedNoise;
50 
51  private const int CacheResolution = 256;
52 
53  static PerlinNoise()
54  {
55  p = new int[512];
56  for (int x = 0; x < 512; x++)
57  {
58  p[x] = permutation[x % 256];
59  }
60 
61  float minValue = float.MaxValue;
62  float maxValue = float.MinValue;
63  cachedNoise = new float[CacheResolution * CacheResolution];
64  for (int x = 0; x < CacheResolution; x++)
65  {
66  for (int y = 0; y < CacheResolution; y++)
67  {
68  cachedNoise[x + CacheResolution * y] = (float)OctavePerlin(x / (double)CacheResolution, y / (double)CacheResolution, 0.5, 10, 4, 0.5f);
69  }
70  }
71 
72  for (int i = 0; i < CacheResolution * CacheResolution; i++)
73  {
74  minValue = Math.Min(cachedNoise[i], minValue);
75  maxValue = Math.Max(cachedNoise[i], maxValue);
76  }
77 
78  //normalize to 0-1 range
79  for (int i = 0; i < CacheResolution * CacheResolution; i++)
80  {
81  cachedNoise[i] = (cachedNoise[i] - minValue) / (maxValue - minValue);
82  }
83  }
84 
91  public static float GetPerlin(float x, float y)
92  {
93  x = Math.Abs(x) % 1.0f;
94  y = Math.Abs(y) % 1.0f;
95 
96  float xIndex = x < 0.5f ? (x * 2.0f * CacheResolution) : CacheResolution - ((x - 0.5f) * 2.0f * CacheResolution);
97  xIndex = Math.Min(xIndex, CacheResolution - 1);
98 
99  float yIndex = y < 0.5f ? (y * 2.0f * CacheResolution) : CacheResolution - ((y - 0.5f) * 2.0f * CacheResolution);
100  yIndex = Math.Min(yIndex, CacheResolution - 1);
101 
102  int minX = (int)xIndex, maxX = (int)Math.Ceiling(xIndex);
103  int minY = (int)yIndex, maxY = (int)Math.Ceiling(yIndex);
104 
105  return MathHelper.Lerp(
106  MathHelper.Lerp(cachedNoise[minX + minY * CacheResolution], cachedNoise[maxX + minY * CacheResolution], xIndex % 1.0f),
107  MathHelper.Lerp(cachedNoise[minX + maxY * CacheResolution], cachedNoise[maxX + maxY * CacheResolution], xIndex % 1.0f),
108  yIndex % 1.0f);
109  }
110 
111  public static double CalculatePerlin(double x, double y, double z)
112  {
113  int xi = (int)x & 255; // Calculate the "unit cube" that the point asked will be located in
114  int yi = (int)y & 255; // The left bound is ( |_x_|,|_y_|,|_z_| ) and the right bound is that
115  int zi = (int)z & 255; // plus 1. Next we calculate the location (from 0.0 to 1.0) in that cube.
116  double xf = x - (int)x; // We also fade the location to smooth the result.
117  double yf = y - (int)y;
118  double zf = z - (int)z;
119  double u = Fade(xf);
120  double v = Fade(yf);
121  double w = Fade(zf);
122 
123  int a = p[xi] + yi; // This here is Perlin's hash function. We take our x value (remember,
124  int aa = p[a] + zi; // between 0 and 255) and get a random value (from our p[] array above) between
125  int ab = p[a + 1] + zi; // 0 and 255. We then add y to it and plug that into p[], and add z to that.
126  int b = p[xi + 1] + yi; // Then, we get another random value by adding 1 to that and putting it into p[]
127  int ba = p[b] + zi; // and add z to it. We do the whole thing over again starting with x+1. Later
128  int bb = p[b + 1] + zi; // we plug aa, ab, ba, and bb back into p[] along with their +1's to get another set.
129  // in the end we have 8 values between 0 and 255 - one for each vertex on the unit cube.
130  // These are all interpolated together using u, v, and w below.
131 
132  double x1, x2, y1, y2;
133  x1 = Lerp(Grad(p[aa], xf, yf, zf), // This is where the "magic" happens. We calculate a new set of p[] values and use that to get
134  Grad(p[ba], xf - 1, yf, zf), // our final gradient values. Then, we interpolate between those gradients with the u value to get
135  u); // 4 x-values. Next, we interpolate between the 4 x-values with v to get 2 y-values. Finally,
136  x2 = Lerp(Grad(p[ab], xf, yf - 1, zf), // we interpolate between the y-values to get a z-value.
137  Grad(p[bb], xf - 1, yf - 1, zf),
138  u); // When calculating the p[] values, remember that above, p[a+1] expands to p[xi]+yi+1 -- so you are
139  y1 = Lerp(x1, x2, v); // essentially adding 1 to yi. Likewise, p[ab+1] expands to p[p[xi]+yi+1]+zi+1] -- so you are adding
140  // to zi. The other 3 parameters are your possible return values (see grad()), which are actually
141  x1 = Lerp(Grad(p[aa + 1], xf, yf, zf - 1), // the vectors from the edges of the unit cube to the point in the unit cube itself.
142  Grad(p[ba + 1], xf - 1, yf, zf - 1),
143  u);
144  x2 = Lerp(Grad(p[ab + 1], xf, yf - 1, zf - 1),
145  Grad(p[bb + 1], xf - 1, yf - 1, zf - 1),
146  u);
147  y2 = Lerp(x1, x2, v);
148 
149  return (Lerp(y1, y2, w) + 1) / 2; // For convenience we bound it to 0 - 1 (theoretical min/max before is -1 - 1)
150  }
151 
152  public static double Grad(int hash, double x, double y, double z)
153  {
154  int h = hash & 15; // Take the hashed value and take the first 4 bits of it (15 == 0b1111)
155  double u = h < 8 /* 0b1000 */ ? x : y; // If the most signifigant bit (MSB) of the hash is 0 then set u = x. Otherwise y.
156 
157  double v; // In Ken Perlin's original implementation this was another conditional operator (?:). I
158  // expanded it for readability.
159 
160  if (h < 4 /* 0b0100 */) // If the first and second signifigant bits are 0 set v = y
161  v = y;
162  else if (h == 12 /* 0b1100 */ || h == 14 /* 0b1110*/)// If the first and second signifigant bits are 1 set v = x
163  v = x;
164  else // If the first and second signifigant bits are not equal (0/1, 1/0) set v = z
165  v = z;
166 
167  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.
168  }
169 
170  public static double Fade(double t)
171  {
172  // Fade function as defined by Ken Perlin. This eases coordinate values
173  // so that they will "ease" towards integral values. This ends up smoothing
174  // the final output.
175  return t * t * t * (t * (t * 6 - 15) + 10); // 6t^5 - 15t^4 + 10t^3
176  }
177 
178  public static double Lerp(double a, double b, double x)
179  {
180  return a + x * (b - a);
181  }
182  }
183 }