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Author Topic: Buddhabrot fractals  (Read 9010 times)
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FractalWoman
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« Reply #30 on: April 26, 2010, 02:00:22 AM »

Quote
experimentation, you'll only find them analytically because experimentally the number of irrational attractive values is orders of infinity more than the number of rationals.

Yes, you're right. The set of irrational numbers is "dense" compared to rational numbers and so experimentally, I will probably always pick an point that tends toward an irrational number rather than a rational one. Thanks for clarifying that for me.
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reesej2
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« Reply #31 on: April 26, 2010, 02:30:11 AM »

Whoa whoa whoa! Hold on a minute. The number of transcendental numbers between 0 and 1 is orders of infinity larger than the number of algebraic numbers, but because of our descriptive limitations, if you "pick" a number in that interval at "random" it's almost guaranteed algebraic. What I'm saying is that the choices you have to make in experimentation (e. g. only using rational initial points) will skew the probabilities infinitely in one direction.

Also, David: Perhaps I'm mistaken, but by that last formula, z is rational in exactly the case that sqrt(1 - 4c) is rational, which is only the case when c is rational. Of course, there are many choices of c for which z isn't rational, but I think I'm correct in claiming that whenever z is rational, c is too.
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David Makin
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« Reply #32 on: April 26, 2010, 12:23:20 PM »

Whoa whoa whoa! Hold on a minute. The number of transcendental numbers between 0 and 1 is orders of infinity larger than the number of algebraic numbers, but because of our descriptive limitations, if you "pick" a number in that interval at "random" it's almost guaranteed algebraic. What I'm saying is that the choices you have to make in experimentation (e. g. only using rational initial points) will skew the probabilities infinitely in one direction.

Also, David: Perhaps I'm mistaken, but by that last formula, z is rational in exactly the case that sqrt(1 - 4c) is rational, which is only the case when c is rational. Of course, there are many choices of c for which z isn't rational, but I think I'm correct in claiming that whenever z is rational, c is too.

In the first case you are correct that experimentally the values of c used will always effectively be rational, but the relative infinities are such that the number of rational attractors for those rational values of c will be comparatively very limited.
In the second case you are also correct, but again there are of course more cases where c is rational and z is not than there are of z (both) being rational.
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lkmitch
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« Reply #33 on: April 26, 2010, 05:43:37 PM »

Following up on Dave's ideas, the equation for attractor points inside the main cardioid is:

z2 + C = z,

or, rearranging,

c = z - z2.

So, one could create a complex rational z value with magnitude less than 0.5 and find the corresponding c point inside the cardioid.  Since z is rational (by construction), c will be, as well.  For example, using z = (1 + i) / (2 + 3i), I found c = (-1 + 3i) / (-5 + 12i), or about 0.2426-0.01775i.

By this method, there are infinite rational (c, z) pairs.  It seems to me that since the iteration only involves squaring and adding, using a rational c value should guarantee rational z values, but now I'm thinking like an engineer, not a mathematician, as my complex analysis professor once told me.  smiley

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lkmitch
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« Reply #34 on: April 26, 2010, 05:59:55 PM »

It seems to me that since the iteration only involves squaring and adding, using a rational c value should guarantee rational z values, but now I'm thinking like an engineer, not a mathematician, as my complex analysis professor once told me.  smiley

Scratch that.  If that were true, then all power series expansions of functions with rational arguments would be rationally-valued, but that's obviously not the case.  Nonetheless, there do seem to be infinite pairs of rational c & final z values.
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FractalWoman
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« Reply #35 on: April 27, 2010, 10:00:09 PM »

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You need to ignore the actual value and consider the value that it's tending towards, I mean if you take the point (0.25,0) and you use infinite precision then your orbit value will never actually reach (0.5,0) even though that is the attractor.

I wanted to address this comment, because I guess it wasn't clear that I am in fact watching where the point is tending to go. What I do is I start with an initial (rational) point, then I iterate zillions of times until the digits begin to stabilize. The stabilized digits are the final digits that I am analyzing, not the in between digits during the iteration process. As you can see from the orbit that I showed earlier (with two spiral arms), the points are spiraling into a particular point near the center of this dynamic, and this is the point I am trying to resolve. With infinite digits of precision, this point would never resolve. My current experiment uses 500 digits of precision and when all 500 digits finally stabilize, what I find is that the "digits" are statistically identical to the digits of PI (or any other irrational number of this kind).

NOTE: in a period three bulb, there would be three orbits, and each of these orbits would tend toward a single point (whether irrational or not) which can only be resolved absolutely given infinite number of iterations.

I hope this clarifies what I am doing.

FW
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David Makin
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« Reply #36 on: April 27, 2010, 10:57:21 PM »

Quote
You need to ignore the actual value and consider the value that it's tending towards, I mean if you take the point (0.25,0) and you use infinite precision then your orbit value will never actually reach (0.5,0) even though that is the attractor.

I wanted to address this comment,

OK, I guess I was just trying to point out 2 things, first that even using infinite precision you'll never reach the true value/s of the attractor except in special cases - like c=(-2, 0) - even if the attractor is rational, and secondly that using less than infinite precision means that the final value stabilising (or not) doesn't tell you whether the true value is rational or not because to know that from the stabilised value itself inherently requires infinite precision.
This issue is essentially what makes it so difficult to accurately do "smooth iteration" or distance estimation for inside points since it's very difficult to tell apart for example a point attractor that's actually oscillating between two values but very slowly tending towards a single point and a period 2 attractor that's oscillating and tending to the 2 periodic values.

I should add that if you have a statistically random distribution to 10,000 decimal places then this could simply mean that the number is rational as x/10000 smiley

Edit: Forgive the schoolboy error, I meant x/(10^10000) smiley


« Last Edit: April 27, 2010, 11:22:53 PM by David Makin » Logged

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Pauldelbrot
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pderbyshire2
« Reply #37 on: April 28, 2010, 04:34:54 AM »

Actually, convergent smoothed iterations is fairly easy. If the attractor is superattracting, use the divergent smoothed iterations algorithm conjugated by an inversion about a point of the attractor (and looking only at every nth iterate where n is the attractor's period). If it's not superattracting, the method described in this post of mine works wonders.
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David Makin
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« Reply #38 on: April 28, 2010, 03:59:31 PM »

Actually, convergent smoothed iterations is fairly easy. If the attractor is superattracting, use the divergent smoothed iterations algorithm conjugated by an inversion about a point of the attractor (and looking only at every nth iterate where n is the attractor's period). If it's not superattracting, the method described in this post of mine works wonders.

Well, I know the various algorithms are fairly easy, it's just the number of iterations required that's the problem, especially if writing generically for a general Mandelbrot rather than for a general Julia (where you can pre-compute the attractor value/s) - for instance if you want to get within say 1e-7 units of the Set boundary smiley
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Pauldelbrot
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pderbyshire2
« Reply #39 on: April 28, 2010, 10:19:00 PM »

There's always a lot of iterations required to render points near the set boundary correctly. (Periodicity detection helps.)
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ker2x
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« Reply #40 on: May 04, 2010, 11:32:08 PM »

i posted some (cheap) optimisation tricks i use, here : http://www.fractalforums.com/programming/buddhabrot-in-fortran-%29/15/
You can also find a link to the full source code at the same place.
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
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FractalWoman
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« Reply #41 on: May 06, 2010, 05:47:21 PM »

Quote
Apologies, I just realised that you probably meant proving that all the truly complex attractors (i.e. non-zero real and non-zero imaginary) are irrational ? Which I guess may be true though I'd be surprised if it was, my reasoning being that the attractors vary smoothly from one point to another (given infinitessimal steps) and I suspect most/all values within a given 2D range are covered hence the likelihood that some values are rational complex.

Yes, this is what I mean. Both real and imaginary components must be non-zero. So far, all the points I have tested at seem to "fall" in the same manner where the final result (up to 500 decimal places) look a lot like an irrational number where the digits are fully randomized with no repeats. It's a difficult thing to prove though because it's impossible to test all cases.
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Timeroot
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« Reply #42 on: May 18, 2010, 02:16:41 AM »

(I'm back!  grin)

I'm fairly certain that many rational numbers would eventually lead to rational attractors. For instance, I think that c=-0.3125=-5/16 has a rational attractor. The set of rational numbers with rational attractors is equal in cardinality to the set of perfect square rationals, which are in turn equal in cardinality to the set of all rationals. The set of rationals with irrational attractors would have the same cardinality.
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« Reply #43 on: May 20, 2010, 02:22:06 AM »

A large colored 8000x8000 render made with the lastest version:
http://telin.ugent.be/~jdebock/BuddhaBrotMT/BuddhaBrotMT-8000x8000-001.png

Took about 21 hours to render.
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ker2x
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« Reply #44 on: May 20, 2010, 02:08:06 PM »

A large colored 8000x8000 render made with the lastest version:
http://telin.ugent.be/~jdebock/BuddhaBrotMT/BuddhaBrotMT-8000x8000-001.png

Took about 21 hours to render.

Very nice smiley
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
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