Random Spiral Fractals
(Python recipe)
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This based on another of my posted codes, titled "Spiral IFS Fractals".
Python, 42 lines
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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")
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1 comment
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)) pixels = image.load() 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")
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