claude
Fractal Bachius
Posts: 563
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« on: February 27, 2015, 07:16:50 PM » |
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Here's an attempt to explain the relationship between external angles and binary decomposition rendering for quadratic polynomials: http://mathr.co.uk/mandelbrot/2015-02-27_morph_demo.mkv (3.3MB, 30secs) Need to re-do the first morph section once I figure out (1) the conformal mapping between the complement of the Mandelbrot set and the complement of the unit disc (2) how to interpolate smoothlybetween the identity mapping and that mapping. And I also want to add the ray-drawing parts on the Mandelbrot set too.
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Adam Majewski
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« Reply #1 on: February 27, 2015, 07:58:03 PM » |
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Nice video. I wait for source code and detailed description of the algorithm. Do you think that : - it would be easier to understand the algorithm first on the dynamic plane then on the parameter plane ?
- you can add point ont the ray to show how ray is created ?
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« Last Edit: March 01, 2015, 10:31:40 AM by Adam Majewski, Reason: list »
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claude
Fractal Bachius
Posts: 563
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« Reply #2 on: March 03, 2015, 12:15:08 PM » |
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Fragmentarium source code for the video attached, it's very messy with lots of unnecessary code... I rendered 30 seconds at 25 fps making 750 frames, then converted with avconv -i out.%05d.png -crf 18 out.mkv It would probably be easier to understand if the video was reversed - good idea! Adding points on the ray could be useful too. Also adding shading (so the local position in each cell is visible, rather than just binarily decomposed). For full effect I'd need to add the binary digits, luckily rendering ".01" characters shouldn't be too difficult in a fragment shader.
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Chris Thomasson
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« Reply #3 on: March 04, 2015, 01:05:58 AM » |
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Fragmentarium source code for the video attached, it's very messy with lots of unnecessary code... I rendered 30 seconds at 25 fps making 750 frames, then converted with avconv -i out.%05d.png -crf 18 out.mkv It would probably be easier to understand if the video was reversed - good idea! Adding points on the ray could be useful too. Also adding shading (so the local position in each cell is visible, rather than just binarily decomposed). For full effect I'd need to add the binary digits, luckily rendering ".01" characters shouldn't be too difficult in a fragment shader. Great work! I am wondering if you gave my suggestion a try wrt the trouble you had with my escape condition for field lines: http://www.fractalforums.com/new-theories-and-research/plotting-field-lines-during-iteration
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billtavis
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« Reply #4 on: March 04, 2015, 07:09:58 PM » |
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cool video!
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claude
Fractal Bachius
Posts: 563
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« Reply #5 on: March 04, 2015, 08:48:31 PM » |
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Here's one showing how the bits are collected when tracing a ray outwards:
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Adam Majewski
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« Reply #6 on: March 04, 2015, 09:49:14 PM » |
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cool. Thx.
Next questions arise : - When rays are not straight lines then ... ? - When you are tracing inwards then ... ? - "as you pass each dwell band " - What does pass on parameter plane : choosing next point c with grater radius = abs(c) ?
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claude
Fractal Bachius
Posts: 563
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« Reply #7 on: March 04, 2015, 09:58:50 PM » |
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cool. Thx.
Next questions arise : - When rays are not straight lines then ... ? - When you are tracing inwards then ... ? - "as you pass each dwell band " - What does pass on parameter plane : choosing next point c with grater radius = abs(c) ?
The lines can be made straight by conformal mapping to the complemnt of the unit disc (at least for Mandelbrot set and connected Julia sets). When tracing inwards you start with the bits and use them to decide whether to go left or right at each step. Removing the first bit is the same as doubling mod 1. Dwell bands are regions where the integer iteration count is constant, when the iteration count decreases (increases) by 1 then you have passed a dwell band going outwards (inwards). Continuous iteration count can give a measure of how far through the dwell band you are - further out the first escaped |z| will be larger.
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DarkBeam
Global Moderator
Fractal Senior
Posts: 2512
Fragments of the fractal -like the tip of it
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« Reply #8 on: March 04, 2015, 11:09:19 PM » |
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No sweat, guardian of wisdom!
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billtavis
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« Reply #9 on: March 07, 2015, 11:19:56 PM » |
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Adam Majewski
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« Reply #10 on: March 17, 2015, 04:31:33 PM » |
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So when tracing : * inwards I add bit at the end * outwards I add the bit at the beginning
Is it correct ?
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3dickulus
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« Reply #11 on: June 07, 2015, 09:01:40 PM » |
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Escher fits
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tit_toinou
Iterator
Posts: 192
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« Reply #12 on: June 08, 2015, 12:40:38 AM » |
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Nice video but that doesn't give or explain the formula for the conformal map ! The last image with Escher was generated with a Julia ? The map doesn't seem conformal at all....
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laser blaster
Iterator
Posts: 178
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« Reply #13 on: June 08, 2015, 01:40:59 PM » |
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Nice video but that doesn't give or explain the formula for the conformal map ! The last image with Escher was generated with a Julia ? The map doesn't seem conformal at all....
I agree, it doesn't look conformal. Any hyperbolic tiling should be possible to map conformally to the exterior of the mandelbrot set, but this mapping doesn't seem right.
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claude
Fractal Bachius
Posts: 563
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« Reply #14 on: June 08, 2015, 03:01:22 PM » |
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Nice video but that doesn't give or explain the formula for the conformal map !
Thanks. It's not an explicit formula (though I think there does exist a complicated formula based on power series that is very slow converging - I still haven't looked into it in more detail). Rather the grid is generated implicitly by iterations of z→z²+c, see http://www.mrob.com/pub/muency/binarydecomposition.html for the basic idea, you can get "local coordinates" within each cell by taking the fractional part of continuous escape time and the argument of the first iterate to escape. Some hyperbolic tilings fit the binary decomposition structure, so can be mapped using these local coordinates, but in general you need global coordinates, which is much harder (needs the full external angle).
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