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MTd2
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« on: January 26, 2010, 03:26:03 AM » |
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Hi People! Did anyone try this? I searched for it, but couldn't find anything: Rn = (x,y,z) R0 = (x 0,y 0,z 0) R1 = (x 1,y 1,z 1) Rn+2 <- Rn+1X Rn + R0X denotes the usual cross product: http://en.wikipedia.org/wiki/Cross_product#Computing_the_cross_productThe rules for plotting should be the same for the generic Mandelbrot type of fractal, where the distance from the origin is d = (x 2 + y 2 + z 2) 1/2. I don't know how to plot that in 3D, I hardly know any programing. If someone is interested in that, a plot would be nice. Or at least, a link for a graphic of such set, if was already found. Thank you! |
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« Last Edit: January 26, 2010, 12:44:05 PM by MTd2 »
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makc
Strange Attractor
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« Reply #1 on: January 26, 2010, 12:11:00 PM » |
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doesnt RXR immediately return 0 vector?
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MTd2
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« Reply #2 on: January 26, 2010, 12:43:35 PM » |
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Yeah... But I fixed that!  |
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kram1032
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« Reply #3 on: January 26, 2010, 03:28:19 PM » |
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 already has to be set to  ,  ,  in that case, because ![cross([a,b,c],[0,0,0])](/cgi-bin/mimetex.cgi? cross([a,b,c],[0,0,0])) returns ![[0,0,0]](/cgi-bin/mimetex.cgi?[0,0,0]) too  |
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MTd2
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« Reply #4 on: January 26, 2010, 03:37:07 PM » |
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You just have to set just 2 parameters R1 and R0, R1=/=0. It should work like a Z axis for the 1st and 2nd iteration of all points So that, you have: R2 = R1XR0 + R0 R3 = R2XR1 + R0 R4 = R3X42 + R0 . . . BTW, don't say c, because I am not thinking about complex numbers. It its just a triad of real numbers. |
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makc
Strange Attractor
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« Reply #5 on: January 26, 2010, 03:37:59 PM » |
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I gave this idea a test run in terms of numbers and, indeed, there are areas of convergence and divergence (which means it should be possible to render the border).
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makc
Strange Attractor
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« Reply #6 on: January 26, 2010, 04:11:26 PM » |
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ok, so unless I made an error somewhere, it looks like boring smooth surface:   (filaments look too unlikely to be real, probably bad iteration/bailout limits, or even bugs; this program is very wip) |
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kram1032
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« Reply #7 on: January 26, 2010, 04:14:24 PM » |
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I doubt it's bugs: Bugs usually make it look better, lol j/k Try a higher bailout radius, then |
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MTd2
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« Reply #8 on: January 26, 2010, 04:32:21 PM » |
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I guess it happened because if you think well, having just 2 3d-vectors, it means that there is a cylindrical symmetry around R1. So, although the computation is 3D, you've got just 2 interesting dimensions. And that's why you just saw an ovoid surphace, because that's just the outer shell of the rotation along the cylinder.
Set R1=(0,0,1) and plot just the yz. This 2d symmetry will be enough and everything.
That cylindrical symmetry should be broken, so, I guess this formula would work better:
RN+3=RN+2XRN+1XRN +R0
where
R2=/=R1=/=0
and R0 span all R3
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makc
Strange Attractor
Posts: 272
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« Reply #9 on: January 26, 2010, 04:52:08 PM » |
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so I upped bailout sqrt(10) times, decreased "bail in" sqrt(10) times, upped resolution 4 times, decreased max step 10 times... at this point my PC reminded me it does not qualify for quantum computing competition, so I aborted it at about 34%. never the less, filaments are still there. although a bit different shape, but that might be because of also increased field of view. this has a potential for good image like this one edit: I can see new megathread coming, "The search for FSM fractal" |
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« Last Edit: January 26, 2010, 04:54:45 PM by makc »
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kram1032
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« Reply #10 on: January 26, 2010, 05:13:43 PM » |
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}\times{(d|e|f)} = {(b f-c e|c d-a f|a e-b d)}) ... maybe this could give something interesting after being turned a bit? What would happen if you initialize it with a,b,c being the actual location x0,y0,z0 in the coordinate system and d,e,f being a rotated coordinate system and just iterate as usual? }\times{rot(a|b|c)}+{(x|y|z)}) |
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MTd2
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« Reply #11 on: January 26, 2010, 05:28:18 PM » |
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My motivation here is to try to get fractals by iterating surfacess in the process rather than just numbers. What I mean by surface, in this case, is the cross product, which determines a surface. You can generalize that to any dimensions using exterior algebra. http://en.wikipedia.org/wiki/Exterior_algebra |
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makc
Strange Attractor
Posts: 272
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« Reply #12 on: January 26, 2010, 05:31:12 PM » |
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...never the less, filaments are still there... I rolled back to original settings except fov/resolution to see how that affects filaments, and I believe they are likely to vanish entirely at sufficiently high iterations:  maybe this could give something interesting after being turned a bit? maybe. but I just discovered slight bug in my implementation, R2 = R1XR0 + R0 was not respected. I.e., I had R2 = R1xR0 + location, R3 = R2xR1 + location, etc, with R0 actually != location also, R1 is fixed vector any way, and one could play with that too, before dismissing the idea completely. |
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makc
Strange Attractor
Posts: 272
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« Reply #14 on: January 26, 2010, 05:42:20 PM » |
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...I just discovered slight bug in my implementation, R2 = R1XR0 + R0 was not respected... well fixed it, and guess what we have now   apparently the filaments correspond to vectors that generate 0 cross product in 1st iteration, and so "converge" really quickly. it is funny how that bug created curved shapes, as kram said Bugs usually make it look better, lol |
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