Friday, July 12, 2013

Hopalong fractals

Have you ever wondered what happen if you pick a point (x0,y0) and compute hundreds of point using these equations?

Well, you get a hopalong fractal.
Let's plot this fractal using Pylab. The following function computes n points using the equations above:
from __future__ import division
from numpy import sqrt,power

def hopalong(x0,y0,n,a=-55,b=-1,c=-42):
 def update(x,y):
  x1 = y-x/abs(x)*sqrt(abs(b*x+c))
  y1 = a-x
  return x1,y1
 xx = []
 yy = []
 for _ in xrange(n):
  x0,y0 = update(x0,y0) 
 return xx,yy
and this snippet computes 40000 points starting from (-1,10):
from pylab import scatter,show, cm, axis
from numpy import array,mean
x = -1
y = 10
n = 40000
xx,yy = hopalong(x,y,n)
cr = sqrt(power(array(xx)-mean(xx),2)+power(array(yy)-mean(yy),2))
scatter(xx, yy, marker='.', c=cr/max(cr), 
        edgecolor='w', cmap=cm.Dark2, s=50)
Here we have one of the possible hopalong fractals:

Varying the starting point and the values of a, b and c we have different fractals. Here are some of them:




  1. Thank you for sharing your method. it,s great. Tesla and Elon Musk

  2. Hello! Nice code, works fine!

    However, the first equation typeset is written in the wrong way. That is, it does not match the code, which is right.


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