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This based on another of my posted codes, titled "Spiral IFS Fractals".

Python, 42 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``` ```# Random Spiral Fractals # FB36 - 20130929 import math import random from collections import deque from PIL import Image imgx = 512; imgy = 512 image = Image.new("RGB", (imgx, imgy)) pixels = image.load() xa = -1.5; xb = 1.5; ya = -1.5; yb = 1.5 # view n = random.randint(2, 9) # of spiral arms a = 2.0 * math.pi / n # angle between arms t = 2.0 * math.pi * random.random() # rotation angle of central copy r1 = 0.1 * random.random() + 0.1 # scale factor of outmost copies of the spiral arms r0 = 1.0 - r1 # scale factor of central copy ts = math.sin(t) * r0; tc = math.cos(t) * r0 maxIt = 64 # max number of iterations allowed for ky in range(imgy): print str(100 * ky / (imgy - 1)).zfill(3) + "%" for kx in range(imgx): x = float(kx) / (imgx - 1) * (xb - xa) + xa y = float(ky) / (imgy - 1) * (yb - ya) + ya queue = deque([]) queue.append((x, y, 0)) while len(queue) > 0: # iterate points until none left (x, y, i) = queue.popleft() # apply all (inverse) IFS transformations for k in range(n + 1): # n outmost copies + central copy if k == n: # central copy # inverse rotation and scaling xnew = (y + x * tc / ts) / (ts + tc * tc / ts) ynew = (y - x / tc * ts) / (tc + ts / tc * ts) else: # outmost copies on the spiral arms c = k * a # angle # inverse scaling and translation xnew = (x - math.cos(c)) / r1 ynew = (y - math.sin(c)) / r1 if xnew >= xa and xnew <= xb and ynew >= ya and ynew <= yb: if i + 1 == maxIt: break queue.append((xnew, ynew, i + 1)) pixels[kx, ky] = (i % 16 * 16 , i % 8 * 32, i % 4 * 64) image.save("RandomSpiralFractal_" + str(n) + ".png", "PNG") ```

#### 1 comment

FB36 (author) 9 years, 8 months ago

Another version using IFS method:

``````# Random Spiral IFS Fractals
# FB - 20130928
import math
import random
from PIL import Image
imgx = 512; imgy = 512
image = Image.new("RGB", (imgx, imgy))
maxIt = imgx * imgy
n = random.randint(2, 9)
a = 2.0 * math.pi / n
t = 2.0 * math.pi * random.random() # rotation angle of central copy
ts = math.sin(t); tc = math.cos(t)
r1 = 0.2 * random.random() + 0.1 # scale factor of outmost copies on the spiral arms
r0 = 1.0 - r1 # scale factor of central copy
p0 = r0 ** 2.0 / (n * r1 ** 2.0 + r0 ** 2.0) # probability of central copy
x = 0.0; y = 0.0
for i in range(maxIt):
if random.random() < p0: # central copy
x *= r0; y *= r0 # scaling
# rotation
h = x * tc - y * ts
y = x * ts + y * tc
x = h
else: # outmost copies on the spiral arms
k = random.randint(0, n - 1) # select an arm
c = k * a # angle
# scaling and translation
x = x * r1 + math.cos(c)
y = y * r1 + math.sin(c)
kx = int((x + 2.0) / 4.0 * (imgx - 1))
ky = int((y + 2.0) / 4.0 * (imgy - 1))
pixels[kx, ky] = (255, 255, 255)
image.save("RandomSpiralIFSFractal.png", "PNG")
``````
 Created by FB36 on Sat, 14 Sep 2013 (MIT)