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Author Topic: IFS to Julia set  (Read 2936 times)
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Posts: 81

« on: May 27, 2009, 07:08:39 PM »

Does anybody know how to draw trees with Julia sets? The starting point definitely seems to be IFS code, but it would be nice if they had curvy contours and the delicacy of Julia set over them.
Posts: 81

« Reply #1 on: May 28, 2009, 07:40:56 PM »

One example that comes to mind is the inverse Julia set. We can use some Julia map as an iterative formula to map random points onto Julia set. Procedure works approximately and we have to wait for many points to accumulate onto Julia set (that would be the most inner contour - lake itself of Julia set). Now, trees in my opinion, are best made with recursive programming with fixed two dimensions, root base, branches always connected to each other... The classical "chaos game" based method has a lot of numbers (matrices), for example check Xenodream trees. Numbers are a little bit sensitive. They turn to total mess if you move the defining "holons". Anyhow, sticking to the point, IFS is a set of formulas calculated via random walk. Clearly, they are the "inverse Julias" of some Julia set, I'd say.  police
« Reply #2 on: May 19, 2010, 02:46:55 PM »

using Julia Set to draw Tree is too difficult for me
can you explain how to do it?
Posts: 38

« Reply #3 on: June 09, 2010, 08:34:37 AM »

but it would be nice if they had curvy contours and the delicacy of Julia set over them.

In an ordinary Julia set, a quadratic function (z = z▓ + c) is iterated. We can generalise this to z = az▓ + bz + c.

In an ordinary iterated function system, a number (at least 2) of linear transformations are iterated, by selecting one at random each iteration.

The way to combine these elegantly would be to have several quadratic functions, where one is selected at random each iteration. In other words, it's like chaos game, but with quadratic, rather than affine, transformations.
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