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Author Topic: Buddhabrot fractals  (Read 8957 times)
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FractalWoman
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« Reply #15 on: April 23, 2010, 02:42:03 PM »

Here is a star field image generated from the M-Set.

See below for details on how I generated them:

http://www.butterflyeffect.ca/Close/Pages/SpaceTimeFluctuations.html


* Another Star Field.jpg (129.43 KB, 990x792 - viewed 486 times.)
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Pauldelbrot
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« Reply #16 on: April 23, 2010, 08:23:26 PM »

Pauldelbrot, you mentioned "Brent's algorithm"? I've never heard the term, any pointers on where I could find that info?

Wikipedia.

Basically it works like this. You have some sequence defined by a recurrence, a function to get the next element from the previous. You also have a starting point. Brent's algorithm works like this: first you calculate the function to get element two and compare to the starting point. Then you save element two and calculate element three, compare it to two. If still no cycle, calculate element four and compare to two. Save element four now and for points five through eight compare to that. Then save eight.

That is, you maintain a saved point from earlier in the sequence, compare every new element to the saved point, and every time the orbit length reaches a power of two you update the saved point to the current one.

Cycles of any length can be detected and there's no need to perform extra tandem iterations or anything like that; just store and compare values. So the overhead is fairly low, about the same as checking for a known finite attractor (the difference from that is logarithmic in the number of iterations).

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reesej2
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« Reply #17 on: April 23, 2010, 09:04:28 PM »

Oh, awesome, thanks--that algorithm looks very nicely effective! cheesy
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hobold
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« Reply #18 on: April 23, 2010, 10:46:07 PM »

BTW, to quickly find out if an integer N is a power of two, you can test

((N - 1) & N) == 0

On most processors, the comparison to zero is optimized away completely, so this takes just two simple computational instructions and a branch.


(The above expression clears the rightmost 1 bit. The result is zero if and only if there was at most one 1 bit, i.e. for powers of two and N == 0.)
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kram1032
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« Reply #19 on: April 24, 2010, 01:17:57 PM »

I guess, it's this...
http://mathworld.wolfram.com/BrentsMethod.html
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FractalWoman
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« Reply #20 on: April 24, 2010, 05:02:19 PM »

Great explanation of Brent's algorithm. I'm going to try that for sure. What I do is I save the whole sequence and periodically check for cycles. This is useful if you want to measure the length of the cycle and/or find the starting point of the cycle. The length of the cycle seems to correlate to the number of "spiral arms" in the galaxy-like shapes generated by the orbits. For example:

http://www.butterflyeffect.ca/Close/Pages/SpaceTimeFluctuations.html

The dynamic or orbit this web page has 5 spiral arms so the cycle count is five. This is really interesting because the cycle only happens because of the limited resolution of the floating point processor in the computer. If we had an infinite number of bits or digits of precision, then the cycle would never happen. NOTE: the cycle we are talking about here is different than the orbital periods within the primary bulbs of the M-Set. For instance, a period 3 orbit might have a cycle count of 10 or 100 depending on how close to the "event horizon" you are.

Hope this makes sense...
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FractalWoman
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« Reply #21 on: April 24, 2010, 05:40:45 PM »

For example, in this image, I selected a point from the main bulb. This should represent a period one orbit. However, you can clearly see that it has two spiral arms. This is what I call a loop-2 singularity because it ends up in a cycle of length two. However, it is clearly spiraling onto one point and thus has a period one orbit.


* Loop-2 Singularity.JPG (21.31 KB, 527x587 - viewed 294 times.)
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kram1032
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« Reply #22 on: April 25, 2010, 10:43:29 AM »

In what order do they appear?
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FractalWoman
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« Reply #23 on: April 25, 2010, 04:08:23 PM »

The red dot is the first point. The opposing point at the other end is the second point. The point next to the red point is third etc... so it spirals into the center point but, with infinite digits of precision, would never get there. Essentially, it is spiraling into a complex irrational number. I'm working on a paper now that proves the irrationality of these numbers.
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kram1032
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« Reply #24 on: April 25, 2010, 06:02:09 PM »

So, it's visually a double spiral, but if you connect the points in order of their appearace, they actually are on a single spiral. smiley
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David Makin
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« Reply #25 on: April 25, 2010, 06:19:14 PM »

I'm working on a paper now that proves the irrationality of these numbers.

But not all of them are irrational !
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David Makin
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« Reply #26 on: April 25, 2010, 07:07:48 PM »

I'm working on a paper now that proves the irrationality of these numbers.

But not all of them are irrational !

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.
« Last Edit: April 25, 2010, 07:14:31 PM by David Makin » Logged

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FractalWoman
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« Reply #27 on: April 25, 2010, 11:41:00 PM »

Actually, I believe that all the points on the INSIDE of the Mandelbrot Set are "falling" into one or more complex irrational numbers. The reason I believe this is because I created my own infinite precision floating point library and have followed these points to more than 500 decimal places and have found no end to the number of digits that eventually stabilize. Each point from the INSIDE of the M-Set seems to go this way. In the period 3 bulbs, the points are "falling" toward three irrational points or singularities. In the period 5 bulbs, the points are spiraling around five singularities as seen in this picture. Given infinite digits of precision, these points would fall forever toward these singular points. The closer the starting point is to the "event horizon" the longer it takes to fall into these irrational singularities. At least that's what I find.


* FractalDynamicField 4.JPG (6.79 KB, 589x530 - viewed 194 times.)
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David Makin
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« Reply #28 on: April 26, 2010, 12:29:57 AM »

Actually, I believe that all the points on the INSIDE of the Mandelbrot Set are "falling" into one or more complex irrational numbers. The reason I believe this is because I created my own infinite precision floating point library and have followed these points to more than 500 decimal places and have found no end to the number of digits that eventually stabilize. Each point from the INSIDE of the M-Set seems to go this way. In the period 3 bulbs, the points are "falling" toward three irrational points or singularities. In the period 5 bulbs, the points are spiraling around five singularities as seen in this picture. Given infinite digits of precision, these points would fall forever toward these singular points. The closer the starting point is to the "event horizon" the longer it takes to fall into these irrational singularities. At least that's what I find.

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.
Also apart from the "obvious" points such as (0,0) or (0.25,0) etc. then you'll (probably) never be able to find a rational one by 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.
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David Makin
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« Reply #29 on: April 26, 2010, 01:41:46 AM »

If you consider the main cardioid then all the point attractors are given by:

  z^2 + c = z

Re-arranging:

  z^2 - z + c = 0

Which yields:

  z = (-1 +/- sqrt(1 - 4*c))/2

And I think I'm correct in saying that all the attractors in the main cardioid have a magnitude <=0.5 (magnitude exactly 0.5 being the boundary points) though I don't have a proof of this to hand.

So for the attractors in the main cardioid we have:

  z = (-1 +/- sqrt(1 - 4*c))/2

and cabs(z)<=0.5

I think it's fairly obvious that there will be many rational values for z that satisy the above - though far fewer cases where both z *and* c are rational.
« Last Edit: April 26, 2010, 01:44:07 AM by David Makin » Logged

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