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Author Topic: intro: gandreas  (Read 2349 times)
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David Makin
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« Reply #15 on: December 14, 2006, 09:57:11 PM »

Just on the escape time method, I'm not sure but I thought that there will be some functions that are expanding (like an expanding affine transform) but have no definable inverse and in these cases you can get an attractor using the escape-time method where you can't using the chaos game or deterministic methods that rely on using a contracting function.
I don't think I explained that too well but I hope the idea gets across:
Basically there will be attractors that can be found using the escape-time method that cannot be found/calculated using the standard methods.
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David Makin
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« Reply #16 on: December 15, 2006, 01:20:52 AM »

Also in the beta formula the non-linear bits can be applied as modifiers mixed with the affine transforms, so you can get escape-time renders that are as "rich" as using the purely convergent methods - it just takes judicious use of the modifier mixing.

Here's an example:

"Undersea Illuminations"



Full-size:
http://www.deviantart.com/deviation/44737262/
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gandreas
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« Reply #17 on: December 15, 2006, 05:36:26 AM »

Definitely cool!  I'd wouldn't say "as rich" rather "differently rich" (where the magnitude is similar, but the direction is orthogonal - it's seriously different than what you'd expect from IFS images).

What happens when you try to use a higher order polynomial instead of an affine transform?
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David Makin
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« Reply #18 on: December 15, 2006, 11:36:53 AM »

If you use the plain inverse of the convergent version of say z^2+c then you end up computing (-z)^2+c and (+z)^2+c which makes the inverse IFS method simply a very slow way of calculating the inside of a standard escape-time fractal - the beauty is that you can change it so that (for example) the two transforms are z^2+c1 and z^2+c2 which results in attractors that wouldn't be possible any other way - obviously you could use any polynomial instead of z^2+c. Alternatively you could use the two transforms as two different polynomials say z^2+c and z^3+c for example - obviously the possibilities are endless as with anyything fractal related !
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lycium
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« Reply #19 on: December 15, 2006, 11:45:03 AM »

Alternatively you could use the two transforms as two different polynomials say z^2+c and z^3+c for example

that specific case results in this btw (minus crazy stuff in the middle): http://www.deviantart.com/deviation/35013449/
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