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Author Topic: Metabrot  (Read 1298 times)
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fivex
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« on: August 22, 2013, 11:47:16 AM »

Iterating zn = J(zn-1)j + c where J(x)n = Jn-12 + x and j > 2 produces interesting results.
First, the julia set (or metabrot) for c = (0,0) and j = 3:

Not unexpectedly, it resembles the mandelbrot set.
Increasing j initially has weird effects, before it begins to resemble the mandelbrot set more and more. j = 4, j = 5 and j = 100:



Presumably, if j = infinity, then (0,0) IS the mandelbrot set.
Slight changes in the real part of c have huge effects on the metabrots:
(.01, .01):

(.02, .01):

You're probably wondering what the mandelbrot (or mandelmetabrot?) for this looks like by now. Here is is for j = 3:

This is a close up of the lower half.
Sections of the mandelmetabrot closely resemble the associated metabrot. This is from one of the "asteroids" on the left:

And from the upper left corner of the close up:

Which resembles the antenna.

You could generate similar sets for any escape-time fractal.
« Last Edit: August 31, 2013, 03:49:32 AM by fivex » Logged
KRAFTWERK
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« Reply #1 on: August 22, 2013, 01:20:51 PM »

Interesting stuff fivex!

That (.02, .01) image could have been a great contender for the axolotl challenge in this years competition...  afro
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fivex
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« Reply #2 on: August 24, 2013, 08:55:04 AM »

This is the end of the antenna.
It's amazing how much the julias resemble where they are from:
This is from between the two spikey bulbs (-0.3833333333333333, 0):

And from one of the twisted things deep in the valley(-0.314, -0.017):

From the mess near the top of the twisted bulb(-0.3496666666666667, -0.0608333333333333):

And from one of the satellites on the antenna(-0.5751666666666667, -0.0076666666666667):

The solid parts heavily resemble normal julias. This is the same for all the satellites

I'm going to try to create a "metabox" next.
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fivex
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« Reply #3 on: August 31, 2013, 11:40:40 AM »

Is it possible/normal for a m-set-like function to have multiple critical points?
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element90
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« Reply #4 on: August 31, 2013, 01:42:58 PM »

Quote
Is it possible/normal for a m-set-like function to have multiple critical points?

Yes, it is. The critical points are the roots of f'(z) = 0.

So if

f(z) = z^2 + c

then

f'(z) = 2z = 0, so there is one critical point at 0

Generally for quadratics if a z term is included which is multiplied by a number N then the critical point will be -N/2 (+ve if the z^2 term is -ve), the resulting fractal however always looks like a Mandelbrot just its size and position change.

But if

f(z) = z^3 - 3z + c

f'(z) = 3z^2 - 3z = 0, the critical points are -1 and 1

and

f(z) = z^3 + 3z + c

f'(z) = 3z^2 + 3z = 0, then the critical points are i and -i.

In general for cubics there are two roots, if both roots are 0 then the result will be the standard M3 Mandelbrot set (subject to size and position). If the roots are both of the same magnitude but different signs then the image features M2 Mandelbrots and the image with one critical point will be the mirror image of the other. If the roots are completely different then the images produced will be different.

There are two posts on my blog about cubics and their critical points:

http://element90.wordpress.com/2012/11/02/cubic-observations/
http://element90.wordpress.com/2012/11/25/more-cubic-observations/

The same is true of higher power formulae, the number of critical points goes up as does the resulting number of pictures produced with those critical points. Just by adjusting the factors applied to each term in the formulae a great variety of m-set-type fractals can be produced. The critical points can be real, imaginary and of course complex as indeed can the factors applied to the terms in the formulae.

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Elelemt90 Fractals blog www.element90.wordpress.com
fivex
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« Reply #5 on: September 02, 2013, 06:04:23 AM »

I have figured out a method to combine multiple critical points. For every point on the complex plane, you iterate using all of the critical points. You color a point black if ANY of the iterations escape.
There are multiple ways to color the outside of this. One is to use the escape time of each critical point as RGB values. This is the j=3 set using that method:

Alternatively, you can use the highest escape time of all the critical points to color the point:

As far as I can tell, there are very few malformed ares, and even then the malformed area is covered in normal minis.
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laser blaster
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« Reply #6 on: January 24, 2014, 10:03:51 PM »

I know this is an old topic, but this is some really interesting stuff. I love your idea to try and create a meta Mandelbrot

I like the name metabrot, but I think that name should be reserved for the "mandelbrot" version of your formula, and not used for the fixed-C "julias". Because your fixed-C sets are actually Julia sets, just using a higher-degree polynomial formula. But the interesting thing is that J(x)_j is equivalent to iterating the mandelbrot formula z=z^2+C (where c = the inital value of x) j times. So when you iterate z = J(z)_j + C, you're basically wrapping the mandelbrot formula into itself. And when j=infinity, the julia set for C = (0,0) would be the original mandelbrot set, and the "mandelbrot" set would be a true meta Mandelbrot! (sorry if I'm just stating the obvious to everyone, but it took me a while to get how this was a meta mandelbrot)

But the question is whether this theoretical j=infinity set would be interesting to look at. The j = 3 "Mandelbrot" set (rendered using your new method of drawing fractals with more than one critical points) looks like 3 mandelbrot-like fractals in some radial arrangement plopped on top of each other, and although it has some interesting features, at the places where the 3 fractals meet, the shape looks discontinuous and uncharacteristic of the mandelbrot set. So when j = infinity, will the infinite amount of overlapping fractals intersect each other to create complex and pleasing shapes, or will they meld into one giant blob? The only way to tell is to start rendering sets with higher and higher values of j (which will get prohibitively expensive as j gets higher, due to the need to iterate multiple points).

Also, maybe there's some better way of rendering fractals with more than one critical point that doesn't create displeasing or discontinuous shapes. Hmmm...
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laser blaster
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« Reply #7 on: January 24, 2014, 11:42:33 PM »

Here's J=5 smiley. It's surrounded by lots of dust.



Here's a closeup of the valley on the left. Some really beautiful detail! cheesy



Stupid photobucket had to convert my pictures to JPEG, increasing the filesize and introducing artifacts.
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kram1032
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« Reply #8 on: January 25, 2014, 02:30:11 AM »

oh I completely missed out on this before, thanks for bringing it back smiley
Looks great!
Makes me wonder what happens if you take this process all the way to infinity yet again. E.g. if you take the meta-meta-meta-(...)-meta-mandelbrot set.

The first of the two proposed coloring techniques could be expanded into the spectral MSet we had going for a while.

Also, if you don't like photobucket, try http://imgur.com/ for image sharing. Maybe that suits you better.

Could you expand the formula out a bit? I was trying to wrap my head around the recursion but didn't quite get it. I'd like to find out what the buddhabrot rendering technique gives with this.
« Last Edit: January 25, 2014, 02:38:52 AM by kram1032 » Logged
laser blaster
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« Reply #9 on: January 25, 2014, 05:00:39 AM »

Sure thing. So we we have our function J(x)j where J(x)n = J(n-1x)2 + x, and J(x)0 = 0
So, starting from J(x)0 and expanding up,
J0 = 0
J1 = J02 + x = x
J2 = J12 + x = x2 + x
J3 = J22 + x = (x2 + x)2 + x

So the final formula when j=3 is:
zn+1 = (z2 + z)2 + z + c

You need to iterate that formula for each critical point to get good results. The problem is that the number of critical points increases exponentially with j. There are 127 critical points for j=8. I tried calculating them all using online math applets, and pasting them into my code, but the result was obviously not right. It looked like it had chunks bitten out of it like the typical results you get when don't start iterating at the critical point. I must've messed up.  sad

Here's the result anyway (by the way, I just discovered I use imageshack, not photobucket, when I tried to log into photobucket and it didn't work  laugh)


The overall shape is interesting, but I wouldn't count on it being even remotely correct.
« Last Edit: January 25, 2014, 05:03:06 AM by laser blaster » Logged
kram1032
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« Reply #10 on: January 25, 2014, 01:50:01 PM »

Ok, so if I got that right, your roots for j=3 would be:

as http://www.wolframalpha.com/input/?i=d%2Fdz+%28z%5E2%2Bz%29%5E2%2Bz tells me.

Thanks for the expansion smiley

One more question:
What exactly do you mean by rendering for each critical point? What is changed from the "normal" standard MSet code to achieve this? Do you simply start from initial (x,y)=the critical  point, rather than it being 0? Or how to achieve that?
Right now, I'm simply rendering a "naive" Buddhabrot and the results are not all too pretty.
For quick reference (j=3):


In your images, could you color the interior by how many of the critical points escape? Perhaps in shades of grey with a single point being white and all being black or something like that.
Or would that make the calculation too slow? (I'd think that, since you stop once any escape, you'll be much faster if you don't have to check that for every single one to the end)

EDIT: turns out, I had an error in the formula. Already looks much better.
EDIT2: this isn't a very long render yet, and the paths aren't all that interesting so far, though it's rendered with random starting point for each path, so it would already include the critical points in the limit, if that's the correct way of doing it:
« Last Edit: January 25, 2014, 03:23:45 PM by kram1032 » Logged
kram1032
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« Reply #11 on: January 26, 2014, 01:21:45 AM »

Ok, I now have a much longer render and I am trying to compare it to fivex' render of j=3.

I can recognize the antenna and the "major bulb" - note that my render is tilted to look like it's standing up.
The middle body is completely missing. But the left/right orange stuff looks a bit like Labrador heads (or some other dog like it) looking up to the major bulb. That is somewhat like the big top and bottom big bulbs in fivex' render. However I'm not quite sure if the proportions match up.

Note, that red has the highest bailout of the three in this render, so that means, that the values I chose may not be high enough to bring out a lot of the detail yet.
I chose 200, 2000 and 20000 for blue, green and red respectively.
All samples are sampled across the entire plane, so the critical points should at least be (almost) represented in the render too. (Since it's random sampling, the chances that I hit the critical points as starting position exactly are zero, just like for any point.)
I'll try zooming out a bit to see if anything appears outside the boundaries the render shows here, since it seems like the buddhabrot and mandelbrot variant do not match up in proportions, although I can't quite tell whether this is just an illusion based on the lack in the buddhabrot set of the middle body or it's truly narrower.

I'll also look into zooms to find out whether there is some detail to be found in this rather smooth overall image.

« Last Edit: January 26, 2014, 01:24:38 AM by kram1032 » Logged
kram1032
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« Reply #12 on: January 26, 2014, 11:43:58 PM »

j=3 seems to not have great details and there also isn't anything outside the part I already posted, as far as I could tell. Just blackness.

However, j=4 seems to result in much more interesting shapes which makes me think that I might have done something wrong for j=3. At the same time, though, it also renders much slower. It hasn't even come close to clearing up.

Here is what I've got so far:


Edit: Ok, I tried it now and I'm most certain that I indeed made some errors with the previous j=3. Not sure what went wrong though, since I simply expanded out the polynomial via wolfram alpha and took that as my j=3. Now I'm using a simpler method in which I don't have to expand the polynomials. Rendering goes faster that way and I'm less likely to err.

Edit: I'm now cooking up a j=20. Though because it is incredibly slow, I confined it to sample only from (x,y)=(0,0), as well as really low bailout values (20, 200, 2000), though despite this possibly not being the complete set (I don't quite know how the buddhabrot style rendering is affected by not sampling all the critical points), it already looks really interesting. Right now it's just, basically, a sparse pixel cloud. However you can already recognize the shape. It looks like this will turn into a Buddhabrot set that looks precisely like the Mandelbrot set, with, apparently, some additional structures around it. (Though, because of the very low SNR, it's really hard to tell right now).
This might end up becoming an alternate way of coloring the interior of the Mandelbrot set. An incredibly slow way, but if the render keeps up the appearance  it seems to converge to, a great way none-the-less.
Perhaps somebody with a better optimized buddhabrot-style renderer than me could test it more effectively.

This is an obviously heavily adjusted version. I only did so because of the incredibly low sampling. I don't have time right now to sample it more.
« Last Edit: January 27, 2014, 01:11:44 AM by kram1032 » Logged
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« Reply #13 on: January 27, 2014, 11:19:15 AM »

Love this topic. And beautiful results kram!
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element90
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« Reply #14 on: January 27, 2014, 04:14:05 PM »

Combining the images of Mandelbrot algorithm fractals based on the critical points produces interesting results. The construction of the j = 3 version of the 'Metabrot' can be shown by combining the images using its 3 critical points, the resulting image of the overlap is produced by layering the images with the opacity of the top two images adjusted to show the images below.

The pictures were produced using Gnofract4d and GIMP for the final layered version.






This technique is primitive and limited in scope. Better control of combining the individual critical point images can be achieved by writing some software. I'm currently testing version 4.1.0 of Saturn and Titan and I'll have a go at producing some software to produce "Multiple Critical Point" images after 4.1.0 has been released. The inner area could be selected based on whether the orbit for a location is captive for at least one critical point, or all critical points. "Grey areas" that are both inner and outer could have their colours combined such that the features of both critical points can represented.  

The method can be applied to any "Mandelbrot Algorithm" fractal with two or more critical points.

For example the "quartic" z = z^4 + z^2 + c has three critical points (0, +1/(sqrt*2) and -1/(sqrt(2)), however the are only two images to layer as two of the critical points produce the same image.





Note: the final three images were produced using Saturn and are hence a bit rough as high resolution versions haven't been produced using Titan. The images were combined using GIMP simply by reducing the opacity of one of the layers.
« Last Edit: May 26, 2014, 10:34:51 AM by element90, Reason: Replaced file links. » Logged

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