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Author Topic: Is there any shape that can not be found in the Mandelbrot?  (Read 5313 times)
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grobblewobble
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« on: May 09, 2013, 11:16:39 PM »

A question for the mathematicians. Is there any shape that is not present, at least approximately, in the Mandelbrot set?

I am slowly starting to understand better how shapes stack when you zoom in, and I suspect you could in principle "draw" any sort of picture by stacking shapes, if you had unlimited computational power to zoom. But that's just speculation.

Then I have a slightly related question. I heard somewhere on the internet (unfortunately I don't remember where it was) that there are special mathematical problems in understanding what happens at the point where a baby Mandelbrot is connected to its "mother". Is there any truth in that, or is this just a rumour?

I did have a look at this region and noticed unusual patterns, that look almost like tears or faultlines in the fractal. Here is a youtube link to the zoom (the pattern at the end is especially strange):

<a href="http://www.youtube.com/v/a1To14ilGzA&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/a1To14ilGzA&rel=1&fs=1&hd=1</a>
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Levi
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« Reply #1 on: May 10, 2013, 01:31:52 AM »

Define 'approximately'.

The perimeter of the Mandelbrot Set is fractal everywhere. Therefore no shape with a non-fractal perimeter can truly occur in it. No circles, no triangles, no rectangles - nothing with a finite number of sides, no curves whose derivatives will eventually reach 0 or e.

Now what might 'approximately' occur is not something that is well defined in mathematics. For the most part, similarity is really just an illusion - either two shapes are the same (with or without the appropriate transformations applied) or they aren't. That said, I don't know if there have been any investigations into the occurence of approximate shapes in the Mandelbrot, but approximation would have to be well defined for that to even be possible.

I was thinking perhaps using limits as you traverse the perimeter of the set, but the problem with that is--I don't think you can traverse the perimeter of a dense fractal like the Mandelbrot because I believe the distance between any two points on the edge, if you're travelling along the edge, is infinity. Someone let me know if I'm mistaken.

There might be some definitions of 'approximate' that would make this an analytical problem, but none that I can think of.
« Last Edit: May 10, 2013, 01:33:51 AM by Levi » Logged

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Tglad
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« Reply #2 on: May 10, 2013, 05:25:04 AM »

Grobblewobble...
if you choose a colour spectrum that wraps around, and contains every one of the 256x256x256 rgb colours... then it is almost certain that any rgb image is possible somewhere in the mandelbrot set.

One way to think of this is to imagine an extremely dusty area of the set... in these areas the colour of each pixel is very very sensitive to the exact location of the pixel, so pretty much any colour is possible by a slight movement of the pixel location or a slight variation of the dusty area.
So even if the image looks non-fractal, e.g. a perfect white square... that just means that the centre of each pixel is the correct number of iterations for the colour white. It doesn't mean that the mandelbrot set really has a square in that location.
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grobblewobble
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« Reply #3 on: May 10, 2013, 09:15:02 AM »

then it is almost certain that any rgb image is possible somewhere in the mandelbrot set.
Thanks, that answers my question. It is just the definition of "approximate" I had in mind. Let me try to make it more precise:

Say we have a finite grid of size M times N and assign a color code to each grid node. Is it possible to find an area of the Mandelbrot and a mapping of color codes to numbers of iterations, such that if we map the number of iterations it takes at each grid point before the absolute value of the complex number exceeds 2 corresponds to its color code?
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Levi
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« Reply #4 on: May 10, 2013, 09:24:41 AM »

Ah I see. In that case it's more of an aliasing issue than a case of the actual set. I suppose it would be possible to get any pixel-image from the set with the right parameters.
I've actually seen some really interesting things happen as a result of aliasing in the Mandelbrot. For example...

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grobblewobble
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« Reply #5 on: May 10, 2013, 09:36:31 AM »

Ah I see. In that case it's more of an aliasing issue than a case of the actual set.

Haha, awesome image. smiley I don't know if is even really possible to define "approximate" in a way that avoids aliasing? I do like the idea of basing the definition of "approximate" on grids. Our eyes and brains can only handle a finite number of points in an image, so grids cover every distinct shape we could ever see.
« Last Edit: May 10, 2013, 09:40:42 AM by grobblewobble » Logged
Dinkydau
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« Reply #6 on: May 10, 2013, 02:27:10 PM »

I did have a look at this region and noticed unusual patterns, that look almost like tears or faultlines in the fractal. Here is a youtube link to the zoom (the pattern at the end is especially strange):

<a href="http://www.youtube.com/v/a1To14ilGzA&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/a1To14ilGzA&rel=1&fs=1&hd=1</a>
The thing at the end is just a higher order julia set as I call them for personal use. They're the result of zooming close to the border of a very small mandelbrot set, just like regular julia sets are the result of zooming close to the border of a "somewhat" small mandelbrot set.

Grobblewobble...
if you choose a colour spectrum that wraps around, and contains every one of the 256x256x256 rgb colours... then it is almost certain that any rgb image is possible somewhere in the mandelbrot set.

One way to think of this is to imagine an extremely dusty area of the set... in these areas the colour of each pixel is very very sensitive to the exact location of the pixel, so pretty much any colour is possible by a slight movement of the pixel location or a slight variation of the dusty area.
So even if the image looks non-fractal, e.g. a perfect white square... that just means that the centre of each pixel is the correct number of iterations for the colour white. It doesn't mean that the mandelbrot set really has a square in that location.
The problem with this idea, I think, is that the traditional way of using such a color spectrum is to give points that escape at the same iteration the same color. That means that you are restricted to the (approximately) symmetrical shapes of the mandelbrot set. But I may be wrong. The mandelbrot set surprised me countless times already.
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lkmitch
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« Reply #7 on: May 10, 2013, 05:33:56 PM »

It seems to me that if the color coding is based solely on the number of iterations, then only a fairly small subset of the rgb image space might be actually realized in the Mandelbrot set.  For example, I find it very difficult to image a location that would produce the Mona Lisa or a solid square.
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Levi
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« Reply #8 on: May 10, 2013, 06:56:14 PM »

It seems to me that if the color coding is based solely on the number of iterations, then only a fairly small subset of the rgb image space might be actually realized in the Mandelbrot set.  For example, I find it very difficult to image a location that would produce the Mona Lisa or a solid square.

I was thinking that too. However, I'm not sure. Have you ever zoomed into an area that was so fractal that the background appeared to become just static? I know I have. To prevent this you have to decrease the frequency of color changes (if you're using the smooth coloration algorithm).  Likewise, by increasing the color frequency you can make patterns show up in relatively non-fractal places, as shown in the image I posted. But if you increase frequency in a location that is already highly fractal, static occurs instead of patterns. I'm sure you could find such images where the escape time of every pixel is unique (assuming that all visible points are escaping), and from there you could adjust your coloration rules in such a way that it would create any image you want.
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Dinkydau
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« Reply #9 on: May 10, 2013, 07:59:29 PM »

So does a location exist where, if you take a grid of points, all have a unique escape-time?

To start with a more simple task: make the grid 2×2. That's easy. There are plenty of locations where that happens to be possible. 15×15? maybe. This is a 15×15 render of a strong spiral in the elephant valley. I have been able to verify that this grid contains at least two points with escape time 2295. To find a place where indeed all escape times are unique, I think that could be plausible, especially considering how dense some areas are where we cannot even (yet) go.


2000×1500 high-quality pictures then?
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Levi
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« Reply #10 on: May 10, 2013, 08:24:39 PM »

I see no reason why a unique picture of any size m*n would be impossible. But I have a feeling it is an NP problem--given a grid of Mandelbrot coordinates of any size it is easy to determine that each square is unique, but there is no way to know where such a grid will be before you find it.

However, if you use normalized iteration count the problem becomes much easier - the chance of two points not only escaping on the same turn but also having exactly the same magnitude after escaping is much slimmer. So then you could find an m*n grid that contains only escaping points, normalize their iteration count to get a unique number for each point, and then customize your coloring rule based on those new values.  

And given a set of unique iteration counts and their desired colors, you would simply have to perform some kind of interpolation to create a function that would satisfy this color scheme while also creating a continuous color palette. You could make a cool video by finding one of these locations and its interpolated color scheme, then showing a zoom to that point - it would look like a strangely colored Mandelbrot zoom until you reach the appropriate location then - voila! - it's a picture of Mona Lisa. wink

Only problem is, this would be very computationally expensive - to create a polynomial fit of m*n points, the polynomial would be of degree (m*n - 1)...which for a 700*700 viewport is 489,999. Then you would have to consult this giant polynomial for every pixel of every frame of your entire video.
« Last Edit: May 10, 2013, 08:30:00 PM by Levi » Logged

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grobblewobble
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« Reply #11 on: May 12, 2013, 06:12:39 PM »

The thing at the end is just a higher order julia set as I call them for personal use. They're the result of zooming close to the border of a very small mandelbrot set, just like regular julia sets are the result of zooming close to the border of a "somewhat" small mandelbrot set.
Okay.. could you please explain more precisely what you mean by "higher order Julia set"? I think I have an idea, but not sure.
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Dinkydau
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« Reply #12 on: May 12, 2013, 06:44:11 PM »

If you go past a small mandelbrot set there will be julia sets that, as you zoom further, increase in symmetry infinitely and approach the border of a smaller mandelbrot set. If then you also go past that second small mandelbrot set, you will see the same julia sets increasing in symmetry infinitely and approaching the border of what would then be a second order julia set. Those second order julia sets increase in symmetry and approach the border of an even smaller mandelbrot set. Go past that mandelbrot set, and you will see the second order julia sets increase in symmetry approaching a 3rd order julia set. This can go on infinitely.

"order" is not a commonly used name, though. I don't know if there is any official or de-facto name in use for them.
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grobblewobble
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« Reply #13 on: May 12, 2013, 07:38:56 PM »

Ahh.. like, if you zoom in on a minibrot inside a spiral, the border consists of infinitely many small spirals, and so on? I think this is called "shape stacking" sometimes?
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Dinkydau
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« Reply #14 on: May 12, 2013, 10:03:29 PM »

That's a nice name because it describes what happens more carefully. I'll use in the future
« Last Edit: May 12, 2013, 10:07:58 PM by Dinkydau » Logged

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