Re: IFS inverse algoritm
From: Stewart Robert Hinsley ({$news$}_at_meden.demon.co.uk)
Date: 06/01/04
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Date: Tue, 1 Jun 2004 21:34:34 +0100
In article <8oyuc.840$9n5.786@amstwist00>, Suncode
<sol_developments@hotmail.com> writes
>
>IFS's are made from randomly choose which index to use when selecting
>transformation-parameters from a table, now if we try to do this the outher
>way around and test for which index that was used to get to the current
>position and then inverse transform from there. Then render a picture of how
>many iteriations it takes to get to a defined area (ex: the outside of a
>surface or a circle). The same as when doing the Mr: Julia's and Mr:
>Mandelbrot's things and likewise. The points that still is left out (not
>exclueded) when the iteriations has ran out is considered as IFS! Here I'm
>trying to explain how this can be made :
A few minor points.
1) The random or chaos game algorithm isn't the only means of drawing an
IFS. OTOH, I think your technique works equal well with the xerox or
copy machine and deterministic algorithms.
2) There are Iterated Function Systems in which there is not a unique
index that gets you to a point, so your technique is not universally
applicable, at least without modification.
3) Strictly speaking the IFS is those points where an infinite number of
iterations is required (assuming that the target area does not intersect
with the attractor). But, then, the other algorithms also produce
approximations.
However, you've produced an ("escape time") algorithm for IFSs that I
haven't encountered before.
-- Stewart Robert Hinsley
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