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Author Topic: Inverse Mandelbrot  (Read 7331 times)
Description: Why doesnt it work?
0 Members and 1 Guest are viewing this topic.
makc
Strange Attractor
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Posts: 272



« on: April 09, 2010, 01:03:19 PM »

I was thinking, it must be possible to reverse the formula z(k+1) = z(k)^2 + c to z(k) = sqrt (z(k+1) - c), run k downwards, change bailout condition from r > 2 to r < 2 and still get same Mandelbrot. However, when I try this, I get only two major features plus some weird shape in re>0 area. Why is that?


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makc
Strange Attractor
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Posts: 272



« Reply #1 on: April 09, 2010, 01:18:10 PM »

I get only two major features plus some weird shape in re>0 area.
Never mind two major features, they were due to the code to explicitly skip them; that shape in re>0 area is all there is to it.
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hobold
Fractal Bachius
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Posts: 573


« Reply #2 on: April 09, 2010, 03:24:40 PM »

How do you know which k to start with? And which z(k)?
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makc
Strange Attractor
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Posts: 272



« Reply #3 on: April 09, 2010, 06:32:42 PM »

How do you know which k to start with? And which z(k)?
ah, you're right, I've been starting from c.
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aluminumstudios
Conqueror
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Posts: 135


« Reply #4 on: April 09, 2010, 07:47:22 PM »

If it were possible to run backwards, Buddhabrot's would be a heck of a lot easier to render!
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kram1032
Fractal Senior
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Posts: 1863


« Reply #5 on: April 09, 2010, 07:54:16 PM »

not bad actually.
If the upper half is "true", you could just mirror it to get the whole set... smiley

However isn't that up down the imaginary direction?
So, the strange shape is at Im<0, rather than Re>0?

(Or did I oversee something?Grin with closed eyes)
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makc
Strange Attractor
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Posts: 272



« Reply #6 on: April 09, 2010, 08:11:35 PM »

nah up side is normal mandelbrot to compare it easier.

here is how it looks. I actually wrote that to extend the orbits into the past, and then I spontaneously cam up with idea to render fractal itself using that code.
« Last Edit: April 09, 2010, 08:15:52 PM by makc » Logged
matsoljare
Fractal Lover
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Posts: 215



WWW
« Reply #7 on: April 09, 2010, 11:07:22 PM »

The problem is that for the complex squaring formula, there are two different numbers that give the same result, so you can't know for sure which one of them is the correct "step backwards". For the power of three, there are three different roots, and so on.....
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kram1032
Fractal Senior
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Posts: 1863


« Reply #8 on: April 09, 2010, 11:26:30 PM »

isn't it rather that BOTH are correct in that case?
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