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To draw an IFS fractal probabilities of each transformation must be given normally. This code calculates the probabilities using a heuristic instead. The bounding rectangle of the fractal also calculated automatically.

Python, 103 lines
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103``` ```# IFS fractals w/ automatic probability distribution # http://en.wikipedia.org/wiki/Iterated_function_system # http://en.wikipedia.org/wiki/Chaos_game # FB - 20120311 import random from PIL import Image imgx = 512 imgy = 512 # will be auto-re-adjusted according to aspect ratio of the fractal maxIt = imgx * imgy * 2 ##fractalName = "Barnsley Fern" ##mat=[[0.0, 0.0, 0.0, 0.16, 0.0, 0.0], ## [0.85, 0.04, -0.04, 0.85, 0.0, 1.6], ## [0.2, -0.26, 0.23, 0.22, 0.0, 1.6], ## [-0.15, 0.28, 0.26, 0.24, 0.0, 0.44]] fractalName = "Centipede" mat = [[0.824074, 0.281482, -0.212346, 0.864198, -1.882290, -0.110607], [0.088272, 0.520988, -0.463889, -0.377778, 0.785360, 8.095795]] ##fractalName = "Levy C curve" ##mat = [[0.5, -0.5, 0.5, 0.5, 0.0, 0.0], ## [0.5, 0.5, -0.5, 0.5, 0.5, 0.5]] ##fractalName = "Dragon curve" ##mat = [[0.5, -0.5, 0.5, 0.5, 0.0, 0.0], ## [-0.5, -0.5, 0.5, -0.5, 1.0, 0.0]] ##fractalName = "Sierpinski Triangle" ##mat = [[0.5, 0.0, 0.0, 0.5, 0.0, 0.0], ## [0.5, 0.0, 0.0, 0.5, 0.5, 0.0], ## [0.5, 0.0, 0.0, 0.5, 0.0, 0.5]] # Area of Polygon using Shoelace formula # http://en.wikipedia.org/wiki/Shoelace_formula # corners must be ordered in clockwise or counter-clockwise direction def PolygonArea(corners): n = len(corners) # of corners area = 0.0 for i in range(n): j = (i + 1) % n area += corners[i][0] * corners[j][1] area -= corners[j][0] * corners[i][1] area = abs(area) / 2.0 return area def IFS(x, y, i): # apply ith transformation to given point x0 = x * mat[i][0] + y * mat[i][1] + mat[i][4] y = x * mat[i][2] + y * mat[i][3] + mat[i][5] x = x0 return (x, y) m = len(mat) # number of IFS transformations # calculate probabilities of the transformations areas = [] # areas of transformed rectangles for j in range(m): area = PolygonArea([IFS(1, 1, j), IFS(-1, 1, j), IFS(-1, -1, j), IFS(1, -1, j)]) areas.append(area) totalArea = sum(areas) pArr = [] for j in range(m): pArr.append(areas[j] / totalArea) # find bounding rectangle of the fractal using Chaos Game algorithm x = mat[0][4] y = mat[0][5] xa = x xb = x ya = y yb = y for k in range(maxIt): i = random.randint(0, m - 1) if random.random() <= pArr[i]: (x, y) = IFS(x, y, i) if x < xa: xa = x if x > xb: xb = x if y < ya: ya = y if y > yb: yb = y imgy = int(imgy * (yb - ya) / (xb - xa)) # re-adjust the aspect ratio image = Image.new("RGB", (imgx, imgy)) pixels = image.load() # drawing using Chaos Game algorithm theColor = (255, 255, 255) x=0.0 y=0.0 for k in range(maxIt): i = random.randint(0, m - 1) if random.random() <= pArr[i]: (x, y) = IFS(x, y, i) jx = int((x - xa) / (xb - xa) * (imgx - 1)) jy = (imgy - 1) - int((y - ya) / (yb - ya) * (imgy - 1)) if jx >= 0 and jx < imgx and jy >= 0 and jy < imgy: pixels[jx, jy] = theColor image.save(fractalName + " fractal.png", "PNG") print "Fractal Name: " + fractalName print "Probabilities: " + str(pArr) ```
 Created by FB36 on Sun, 11 Mar 2012 (MIT)