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Kalles Fraktaler
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« Reply #90 on: March 11, 2016, 12:57:09 PM » |
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Not too surprising though. Each "feature" in the progression has twice the complexity of the last. The 1024 X formation has 4096 arms, and the 2048X formation will have 8192 arms. Each arm also has a specific number of dendrites coming off it. The length of said arms, and period of the spirals therin are determined early on in the zoom sequence.
You don't need to write "will have" since the parameters I gave is indeed the 2048X image  You can try it in Mandel Machine. It renders in 640x360 in 5.5 seconds... (13 times faster than KF for that location) |
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stardust4ever
Fractal Bachius
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« Reply #91 on: March 11, 2016, 01:41:24 PM » |
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I added some info to my previous post. 164864 = 1024 x 23 x 7. Zoom way out, and I suspect the mini within the initial 2X formation might have a period of 161. How do I check for periodicity in Kalles Fraktaler? Not too surprising though. Each "feature" in the progression has twice the complexity of the last. The 1024 X formation has 4096 arms, and the 2048X formation will have 8192 arms. Each arm also has a specific number of dendrites coming off it. The length of said arms, and period of the spirals therin are determined early on in the zoom sequence.
In fact, your period number, 164864, is divisible by 1024 x 161. And 161 = 7 x 23. No coincidence each "arm" have 7 segments. So I will postulize that if you zoomed all the way out to the original little "X" formation I zoomed sideways from, the period of said minibrot contained within would be 161. Or perhaps the 161 refers to the minibrot that lies in the center of the initial 2X formation. Every chromosome in the series has a pair of little chromosomes inside it.
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Kalles Fraktaler
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« Reply #92 on: March 11, 2016, 02:27:36 PM » |
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I added some info to my previous post. 164864 = 1024 x 23 x 7. Zoom way out, and I suspect the mini within the initial 2X formation might have a period of 161. How do I check for periodicity in Kalles Fraktaler?
Yep, the first 2X formation is located at 1.91E18 and has a period of 161, just as you predicted I have already made a new version with the Newton-Raphson method, just haven't announced it yet |
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hapf
Fractal Lover
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« Reply #93 on: March 11, 2016, 03:02:48 PM » |
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Yep, the first 2X formation is located at 1.91E18 and has a period of 161, just as you predicted The keys to the kingdom are the periods...  |
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stardust4ever
Fractal Bachius
Posts: 513
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« Reply #94 on: March 11, 2016, 03:07:58 PM » |
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The keys to the kingdom are the periods... Was gonna insert joke about my fiance holding the keys; decided it would be best not to.  |
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hapf
Fractal Lover
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« Reply #96 on: March 11, 2016, 04:36:42 PM » |
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Zoom between the outside arms of the outermost chromosome. More specifically, there are seven sideways facing dendrites along the edge of each chromosomal arm counting from center to outer edge. To continue the pattern such that all chromosomes are equidistant to each other, zoom into the diamond-shaped julia midway between the endpoints of the third dendrites on the outer arms of the outer chromosome. There are several of these shapes one could zoom into but the distance from the center of the chromosome will determine the exact spacing between the center two chromosomes in the next target formation.
It's easier to take a screengrab but I'm away from the PC right now and don't feel like booting it as it is late.
EDIT: I appear to have misinterpreted the post. Hapf was referring to the automated process by which one finds the central minibrot (which also reveal the location of the next formation without manual progression)
So you are only interested in one of the minibrots with period 164864 and not the others? |
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stardust4ever
Fractal Bachius
Posts: 513
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« Reply #97 on: March 11, 2016, 05:57:42 PM » |
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So you are only interested in one of the minibrots with period 164864 and not the others?
No, I am interested in using your minibrotlocator algorithm to explore features of the mandelbrot set without spending hours manually scrolling through thousands of zoom levels to reach them. It's like setting the fractal explorer on auto-pilot and walking away while your PC crunches the numbers. I published the extreme aspect feature I purposely dug up because lining up multiple objects in a row tends to break many existing centroid locator algorithms. Your period checking algorithm appears to be much more accurate in that it relies on calculating periodicity rather than using patterns within the iteration bands to guide the centroid. Not only does this method appear faster, but not prone to errors like pattern matching of iteration bands. |
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hapf
Fractal Lover
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« Reply #98 on: March 11, 2016, 06:17:15 PM » |
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Ok. The polygon approach has found you the lowest (?) period there it seems. The one stardust4ever was after and which seems to be central. There are infinitely many minibrots in that polygon. If you don't abort once you found one some others are also reported? (164867, 164870, 164876, 164882...) The whole thing should not take 55 minutes, though. You seem to have a bottleneck somewhere. Finding period 164864 and computing the location is a matter of minutes at most. |
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hapf
Fractal Lover
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« Reply #99 on: March 11, 2016, 06:27:09 PM » |
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No, I am interested in using your minibrotlocator algorithm to explore features of the mandelbrot set without spending hours manually scrolling through thousands of zoom levels to reach them. It's like setting the fractal explorer on auto-pilot and walking away while your PC crunches the numbers.
I asked because there are tons of period 164864 in that location. My algorithm uses Newton, period identification of pixels (simple search for minimal iteration value) and nothing else. So Kalle has all he needs to give you what your heart desires.  If you are after high periods from the top level it's not so easy to find something specific like period x in one go automatically. But in principle it's doable. |
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claude
Fractal Bachius
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« Reply #100 on: March 11, 2016, 07:42:27 PM » |
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Ok. The polygon approach has found you the lowest (?) period there it seems. ... If you don't abort once you found one some others are also reported?
It finds the lowest period only, continuing thereafter doesn't work because the box gets folded (wrapped by z^2) in half by the next iteration. See also the second-to-last paragraph of http://www.mrob.com/pub/muency/synchronousorbitalgorithm.htmlEDITED to add: note that this works even in the case where there are no pixels with the dominant period (because the dominant region is too small and falls between pixels), so it can be used from way outside the embedded julia set, see fior examples the images here http://www.mrob.com/pub/muency/secondorderembeddedjuliase.html |
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« Last Edit: March 11, 2016, 08:26:12 PM by claude »
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Kalles Fraktaler
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« Reply #101 on: March 11, 2016, 08:15:27 PM » |
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Ok. The polygon approach has found you the lowest (?) period there it seems. The one stardust4ever was after and which seems to be central.
There are infinitely many minibrots in that polygon. If you don't abort once you found one some others are also reported?
(164867, 164870, 164876, 164882...) The whole thing should not take 55 minutes, though. You seem to have a bottleneck somewhere.
Finding period 164864 and computing the location is a matter of minutes at most.
Finding the minibrot of period 82432 in the center takes "only" some 20 minutes. But the time grows exponentially, more precision is required and period is equal to the number of iterations made. 55 minutes was to find the minibrot on the left of the leftmost X formation, with double period of 164864. I estimate next doubling, i.e. the 4096X image of period 329728, to take about 5 hours. I'll make a test-run of this tonight |
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hapf
Fractal Lover
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« Reply #102 on: March 11, 2016, 08:31:39 PM » |
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Finding the minibrot of period 82432 in the center takes "only" some 20 minutes. But the time grows exponentially, more precision is required and period is equal to the number of iterations made. 55 minutes was to find the minibrot on the left of the leftmost X formation, with double period of 164864. I estimate next doubling, i.e. the 4096X image of period 329728, to take about 5 hours. I'll make a test-run of this tonight I find such periods as 164864 in less than 2 minutes. What arbitrary precision library are you using? |
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hapf
Fractal Lover
Posts: 219
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« Reply #103 on: March 11, 2016, 08:40:37 PM » |
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EDITED to add: note that this works even in the case where there are no pixels with the dominant period (because the dominant region is too small and falls between pixels), so it can be used from way outside the embedded julia set, see fior examples the images here http://www.mrob.com/pub/muency/secondorderembeddedjuliase.htmlYes, that seems the main advantage. |
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Kalles Fraktaler
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« Reply #104 on: March 11, 2016, 08:45:38 PM » |
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I find such periods as 164864 in less than 2 minutes. What arbitrary precision library are you using?
Does that also include the newton-raphson iterations to find the exact location parameters, and the size of the minibrot? For this, that requires high precision division, I use http://speleotrove.com/decimal/decnumber.html |
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March 26, 2007, 06:03:34 PM
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