Koch Snowflake and Sierpinski Triangle Combination Fractal using recursion.
Python, 68 lines
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# FB - 201003265
from PIL import Image, ImageDraw
import math
imgx = 512
imgy = 512
image = Image.new("L", (imgx, imgy))
draw = ImageDraw.Draw(image)
def tf (x0, y0, x1, y1, x2, y2):
global draw
a = math.sqrt((x1 - x0) ** 2 + (y1 - y0) ** 2)
b = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)
c = math.sqrt((x0 - x2) ** 2 + (y0 - y2) ** 2)
if (a < 2) or (b < 2) or (c < 2): return
x3 = (x0 + x1) / 2
y3 = (y0 + y1) / 2
x4 = (x1 + x2) / 2
y4 = (y1 + y2) / 2
x5 = (x2 + x0) / 2
y5 = (y2 + y0) / 2
draw.line ([(int(x3), int(y3)),(int(x4), int(y4))], 255)
draw.line ([(int(x4), int(y4)),(int(x5), int(y5))], 255)
draw.line ([(int(x5), int(y5)),(int(x3), int(y3))], 255)
tf(x0, y0, x3, y3, x5, y5)
tf(x3, y3, x1, y1, x4, y4)
tf(x5, y5, x4, y4, x2, y2)
def sf (ax, ay, bx, by):
f = math.sqrt((bx - ax) ** 2 + (by - ay) ** 2)
if f < 1: return
f3 = f / 3
cs = (bx - ax) / f
sn = (by - ay) / f
cx = ax + cs * f3
cy = ay + sn * f3
h = f3 * math.sqrt(3) / 2
dx = (ax + bx) / 2 + sn * h
dy = (ay + by) / 2 - cs * h
ex = bx - cs * f3
ey = by - sn * f3
tf(cx, cy, dx, dy, ex, ey)
sf(ax, ay, cx, cy)
sf(cx, cy, dx, dy)
sf(dx, dy, ex, ey)
sf(ex, ey, bx, by)
# main
mx2 = imgx / 2
my2 = imgy / 2
r = my2
a = 2 * math.pi / 3
for k in range(3):
x0 = mx2 + r * math.cos(a * k)
y0 = my2 + r * math.sin(a * k)
x1 = mx2 + r * math.cos(a * (k + 1))
y1 = my2 + r * math.sin(a * (k + 1))
sf(x0, y0, x1, y1)
x2 = mx2 + r * math.cos(a)
y2 = my2 + r * math.sin(a)
tf(x0, y0, x1, y1, x2, y2)
image.save("sfplustf.png", "PNG")
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