Another possible way to accelerate MB set deep zooming

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Offline knighty

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« Reply #30 on: February 02, 2018, 06:35:12 PM »
Doing derivative of a multi-variate function is a little bit tricky!
This is what I get for the distance estimate:


Offline knighty

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« Reply #31 on: February 02, 2018, 06:42:39 PM »
Instead of a "bailout radius" maybe it's better to think of it as a "stop criterion" like the methods for usual SA?
Agree :) ...Or maybe "super bailout" because we have "super iterations"?

If you write the expansion as \( \sum_{ij}a_{ij}z^i (c-c_0)^j \), the \( a_{0j} \) is just the good old SA. Could that observation be used to integrate this method seamlessly with conventional SA? Sort of like far away from the mini it becomes usual SA but closer you can skip more if you go into the higher "super" iteration bands?
I think there is another way by using roots that are farther away: Those that appear at lower zoom levels. I still don't have a clear picture of how to do it.

Offline gerrit

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« Reply #32 on: February 02, 2018, 07:47:24 PM »
Great, looks perfect. :)

Offline gerrit

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« Reply #33 on: February 04, 2018, 04:15:19 AM »
Have you considered estimating truncation error the usual way and using that as super-bailout criterion similar to SA?
I worked out the formula's which are not too complicated but some work to type in which I can do if it helps.

Offline knighty

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« Reply #34 on: February 04, 2018, 08:18:17 PM »
Have you considered estimating truncation error the usual way and using that as super-bailout criterion similar to SA?
I worked out the formula's which are not too complicated but some work to type in which I can do if it helps.
Yes please.
For now, I use superMB to get the super-bailout: zoom out until the target miniMB's period is not detected (preferably at SA order 4) then use that zoom level to define the super-bailout. I think it is possible to compute it similarly to the atom domain size. The later is already a good super-bailout (but a little bit conservative).

Offline gerrit

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« Reply #35 on: February 04, 2018, 09:56:04 PM »
OK will do. Do you have a feeling yet on computational saving with the super SA compared to regular?

Offline gerrit

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« Reply #36 on: February 05, 2018, 03:05:45 AM »
I added a section on the super SA: http://persianney.com/fractal/fractalNotes.pdf.
Edit: (46f) is wrong, should have all terms with  m>K OR q>M, not AND, same change in (48) of course.
« Last Edit: February 05, 2018, 08:31:31 AM by gerrit »

Offline gerrit

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« Reply #37 on: February 05, 2018, 06:24:59 PM »
Updated pdf, also corrected another error (last formula).

Offline knighty

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« Reply #38 on: February 05, 2018, 07:48:56 PM »
Thanks!

Do you have a feeling yet on computational saving with the super SA compared to regular?
For minibrots with sufficiently big period, say around 1000, the speedup is about one order of magnitude (20 - 30x). Of course the bigger the period is, the faster. As Pauldelbrot said one of the immediate benefits is that one can use a big max iterations number to have a well defined boundary of the minibrot without doing a lot of work. There are other possibilities like early stop criterion for interior points and interior coloring...

When zooming out, it is slower. The relative speed (w.r.t. SA) varies because the number of super-iterations is integer so sometimes it is almost the same and sometimes 4 or 5 times slower (or something like that).

For virtually all realistic examples an order of (4,4) is sufficient: No visible difference with SA. The "deformed" minis which are found near the cusp of main hyperbolic components need more terms or smaller super-bailout. There is little benefit to use super-iteration at those. It could even slow down the computation. One can use the super-iteration at their parent mini instead.

Offline gerrit

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« Reply #39 on: February 06, 2018, 01:41:34 AM »
I find (after fixing up my code a bit) that the atom domain size is too small as super escape radius by varying factors (30 to 1e9) just to get the super iterations to not escape inside the mini. True or bug again?

I tried implementing the truncation error formula for SSA but found it blows up quickly (even after correcting some more typos).

Offline knighty

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« Reply #40 on: February 07, 2018, 01:24:50 PM »
How and when does it blow?

Offline gerrit

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« Reply #41 on: February 07, 2018, 07:59:51 PM »
How and when does it blow?
Eq (48) blows up very quickly. The bound \( |w_{max}| = |x| \) is way too loose, w stays very small in reality, not sure how to get a better (smaller) bound on |w|.
Setting \( w_{max} = R_s \) perhaps, that gives reasonable results, but not sure how to justify it...
Edit: Hang on, at iteration period x(p)=0, so wmax<Rs.
Trying some examples, gradually increasing Rs, pictures seem to break up when r_p gets big. So you'd have to find Rs such that r_p stays small from eq (48). As M,K are small maybe not too hard. For some deep examples I get something like Rs = 1e12*atom_domain_size.
I haven't tried yet propagating r_p into the super iterations cf  eq (53).
« Last Edit: February 07, 2018, 08:24:14 PM by gerrit »

Offline gerrit

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« Reply #42 on: February 08, 2018, 02:56:04 AM »
This example rendered only 3X slower in my bad MATLAB SSA code compared to KF (both without glitch correction).
For comparison my PT MATLAB code is 200X slower here, so this is good.
Still haven't found a way to set the truncation error and/or escape radius, optimal settings (as determined visually) varies hugely depending on the image (not just on depth).

Offline knighty

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« Reply #43 on: February 11, 2018, 08:01:06 PM »
One way to estimate the "super-escape" radius is this:
We suppose a01==0 or is very small.
set r=0;
iterate for i>1:
r = |a01|/|sum(a0i * ri-2)|

Doing (for i==2):
r = |a01|/|a02|
already gives a good estimate. In reality one have to use a fraction of r instead. Say: r/10.

If one takes the SA part of the SSA, that is the part that doesn't depend on w, r is an estimation of the distance from the reference point to the nearest root.
« Last Edit: February 13, 2018, 11:52:44 AM by knighty, Reason: Mistake in the iteration formula. »

Offline gerrit

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« Reply #44 on: February 11, 2018, 10:19:34 PM »
One way to estimate the "super-escape" radius is this:
We suppose a01==0 or is very small.
set r=0;
iterate:
r = |a01|/|sum(a0i * ri-1)|

Doing:
r = |a01|/|a02|
already gives a good estimate. In reality one have to use a fraction of r instead. Say: r/10.

If one takes the SA part of the SSA, that is the part that doesn't depend on w, r is an estimation of the distance from the reference point to the nearest root.
I don't get it. \( |a_{01}/a_{02}| \) gets very large. (7e10 for attached.)
Are we talking about the same a?


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