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Author Topic: 3D Mandelbrot Formula based on the Hopf Map  (Read 12832 times)
Description: 3D Mandelbrot Formula based on the Hopf Map
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bugman
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« on: December 04, 2009, 05:41:07 PM »

In search of the ideal 3D Mandelbrot set, a couple of people on this forum have mentioned the hairy ball theorem. Basically, the problem with the Mandelbulb formula has something to do with the poles of the spherical coordinate system. Everything gets distorted near the poles. An ideal formula would have no poles, but this is only possible in 4 dimensions (because of the hairy ball theorem). This got me thinking, maybe we could create the ideal 4D Mandelbrot using the Hopf map (a mapping for the surface of the 4D sphere with no poles), I'll call it the "Hopfbrot". So I came up with the following formula and I tried rendering it. Unfortunately, it doesn't look as I had hoped, although it does contain the Mandelbulb. But perhaps this will provide some more food for thought.


* Presentation.jpg (194.15 KB, 713x852 - viewed 1193 times.)
« Last Edit: December 04, 2009, 05:59:58 PM by bugman » Logged
bib
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« Reply #1 on: December 04, 2009, 06:05:53 PM »

Very nice! The quest goes on... smiley

As I posted somewhere else, I think everybody "imagines" the ideal 3D Mandelbrot, it would look like this famous picture: http://www.renderosity.com/mod/gallery/index.php?image_id=1308487&member

So why not try a completely different approach, and of course completely wrong and purely artistic, for example by using a 3D software to draw a kind of apple (a 3D cardioid) and then plug spheres all around, etc... ??
« Last Edit: December 04, 2009, 06:07:57 PM by bib » Logged

Between order and disorder reigns a delicious moment. (Paul ValÚry)
BradC
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« Reply #2 on: December 04, 2009, 06:49:03 PM »

Coolness! Even though it has whipped cream / taffy, I like this idea. Thanks for the lesson. smiley

In general, I think the taffy problem happens because the power function isn't conformal. The 3D White-Nylander power functions are least conformal near the poles. In experimenting with 2D escape time fractals over the years, I think I've noticed that fractals that look all stretched out come from formulas that aren't conformal, and fractals that look "nice" invariably come from formulas that are. I think this makes sense because that's what a non-conformal map does--it "stretches out" space.

As pointed out by Tglad on another thread, Liouville's theorem says that the only smooth conformal maps possible in > 2 dimensions are Moebius transformations, which unfortunately don't become complicated when iterated since they form a closed group. I suspect that if one could find a 3D map that was somehow "close" to being conformal, it still might not be good enough because any deviations from conformality might get magnified when iterated lots of times. I think completely getting rid of the taffy is likely a hopeless task, but there may be other nice ways to compromise. In my opinion, the high-degree Mandelbulb fractals are nice ones because they "spread the taffy around" so that there are lots of areas that aren't dominated by it. It seems kinda hard to find much variety in them though from what I've seen so far.
« Last Edit: December 04, 2009, 06:50:48 PM by BradC » Logged
kram1032
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« Reply #3 on: December 04, 2009, 08:32:06 PM »

Funny that you come with that idea...

I had an other idea which is kinda similar probably...

Yesterday I did some tries of slight variations of the riemann's surface mapping and got some interesting (yet not at all Mandelbrot-like) results. (Actually that was clear as I didn't do any variation which would include anything close to z^n+c)

The direct transformation didn't work as the third dimension is based on just two input values.

So my idea right now is, to use

Stereographic Projection but basically ignoring that the whole thing usually is said to be done on the hypersphere's surface.
If you use outranged values and use the convertion formulas as usually, you still get points inside the space....

That might allow for interesting projections... - only [0.0] can't be mapped afaik... - though an equivalent of the Mandelbrot would stay at 0 at that value, anyway....

The results might be mathematically senselessish but they should work anyway smiley
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bugman
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« Reply #4 on: December 05, 2009, 07:28:00 PM »

Here are some more slices of the Hopfbrot. The slice at zc = 0 looks like it might be worth further investigation. I think the slice at yc = 0 looks like a Christmas tree. And the slice at xc = 0 is just, weird.


* Hopfbrot.jpg (168.07 KB, 563x563 - viewed 1080 times.)
« Last Edit: December 05, 2009, 09:10:34 PM by bugman » Logged
stigomaster
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« Reply #5 on: December 05, 2009, 09:32:32 PM »

Wow! The zc = 0 one actually looks a little holy grail-ish. That's a lot like how I would imagine the 3d m-set.
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JosLeys
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« Reply #6 on: December 05, 2009, 10:35:43 PM »

Hi Paul,

Although I made a movie some time ago about the Hopf fibration, I don't understand what you are doing here.

Here are some things I know about it, and I don't really see how it relates to you formula.

Let z1 and z2 be complex numbers, related as z2=a.z1 (a is a complex number also). The intersection of the plane z2=a.z1 with the three-sphere gives a circle in R^3 under stereographic projection from S^3 to R^3. As there are an infinite number of complex numbers a this gives an infinity of circles (and every one of those circles is linked with all the other ones)

See http://www.dimensions-math.org/Dim_reg_E.htm (see the films Dimensions_7 and Dimensions_8)

Can you clarify?
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kram1032
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« Reply #7 on: December 06, 2009, 12:46:02 AM »

Sorry for OT but can you tell me what piece is that nice main music used?
(Used all the way through in the preview and also a lot in the others...)
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JosLeys
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« Reply #8 on: December 06, 2009, 12:42:24 PM »

It's "Sheep may safely graze" by J.S.Bach
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kram1032
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« Reply #9 on: December 07, 2009, 03:15:11 PM »

thanks a lot cheesy
I like it very much smiley
- generally, the music you chose is nice smiley
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Paolo Bonzini
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« Reply #10 on: December 07, 2009, 11:24:23 PM »

What does the zc=0 slice look like for power 8?
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Daniel_P
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« Reply #11 on: December 08, 2009, 02:46:51 PM »

Paul - great work, I think this is a very interesting direction.

Instead of rendering the flat slice where W=constant, perhaps you could try the intersection of the set with a 3-sphere, projected down stereographically ?

I described something along these lines here:
http://www.fractalforums.com/3d-fractal-generation/true-3d-mandlebrot-type-fractal/msg8472/#msg8472

Here is the stereographic projection from the unit 3-sphere down to flat 3-space:

X = x / (1 - w)
Y = y / (1 - w)
Z = z / (1 - w)

And here is the reverse projection from flat 3-space up to the 3-sphere:

 X = 2 * x / (1 + x ^ 2 + y ^ 2 + z ^ 2)
 Y = 2 * y / (1 + x ^ 2 + y ^ 2 + z ^ 2)
 Z = 2 * z / (1 + x ^ 2 + y ^ 2 + z ^ 2)
 W = (-1 + x ^ 2 + y ^ 2 + z ^ 2) / (1 + x ^ 2 + y ^ 2 + z ^ 2)

These are the equations I used (in the opposite order and with a 4D rotation in between) to create the animations on these pages:
http://spacesymmetrystructure.wordpress.com/2008/12/11/4-dimensional-rotations/.


Under this projection/rotation/projection any point in 3D sweeps out one of the circles of the Hopf fibration as described here:
http://spacesymmetrystructure.wordpress.com/links/4-dimensional-rotations-page4/
<a href="http://vimeo.com/moogaloop.swf?clip_id=3945328&amp;server=vimeo.com&amp;fullscreen=1&amp;show_title=1&amp;show_byline=1&amp;show_portrait=0&amp;color=01AAEA" target="_blank">http://vimeo.com/moogaloop.swf?clip_id=3945328&amp;server=vimeo.com&amp;fullscreen=1&amp;show_title=1&amp;show_byline=1&amp;show_portrait=0&amp;color=01AAEA</a>
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kram1032
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« Reply #12 on: December 08, 2009, 04:29:44 PM »

great videos cheesy

That way, both the quaternionean (or offspring of that) Mandelbrot as well as the full Mandelbrot set, where also the c-plane changes, could be downprojected during rotations...

I'm not totally sure but I *think* I really never saw a stereographic projection of the full 2+2iD Mandelbrot object... - only inversed Mandelbrots, I think...
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bugman
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« Reply #13 on: December 08, 2009, 06:42:51 PM »

Hi Paul,

Although I made a movie some time ago about the Hopf fibration, I don't understand what you are doing here.

Here are some things I know about it, and I don't really see how it relates to you formula.

Let z1 and z2 be complex numbers, related as z2=a.z1 (a is a complex number also). The intersection of the plane z2=a.z1 with the three-sphere gives a circle in R^3 under stereographic projection from S^3 to R^3. As there are an infinite number of complex numbers a this gives an infinity of circles (and every one of those circles is linked with all the other ones)

See http://www.dimensions-math.org/Dim_reg_E.htm (see the films Dimensions_7 and Dimensions_8)

Can you clarify?

I used slightly different notation, but I got the formula for my HopfInverse function from this Mathematica notebook by Srdjan Vukmirovic:
http://library.wolfram.com/infocenter/MathSource/5193/

I also posted some sample code here:
http://bugman123.com/Math/Math.html#HopfFibration

If the sphere is parameterized by angles phi and theta, then the HopfInverse function returns the 4D point from the fiber over phi and theta. Psi describes a particular point in the fiber:
HopfInverse(theta, phi, psi) = {cos(phi/2)*cos(psi), cos(phi/2)*sin(psi), cos(theta+psi)*sin(phi/2), sin(theta+psi)*sin(phi/2)}

I also defined the radius of the hypersphere:
r = sqrt(x▓+y▓+z▓+w▓).

So the complete transformation becomes:
{x, y, z, w} = r*HopfInverse(theta, phi, psi)

First I solved the inverse of this transformation for {r, theta, phi, and psi} in terms of {x, y, z, w}:
theta = acos(z/rzw) - psi
phi = 2*acos(rxy/r)
psi = acos(x/rxy)
where rxy = sqrt(x▓+y▓), rzw = sqrt(z▓+w▓)

Then I multiplied these angles by my constant power n and raised r to the nth power before feeding the new angles back into the HopfInverse function. So this becomes my final power function:
{x, y, z, w}^n = r^n HopfInverse(n*theta, n*phi, n*psi)
« Last Edit: December 09, 2009, 12:24:35 AM by bugman » Logged
Paolo Bonzini
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« Reply #14 on: December 08, 2009, 07:57:57 PM »

I think you're right in eliminating the Abs.  It is just an artefact of using ArcCos (which has a range of only PI, not 2PI).  Just to be sure, if you feed Mathematica something based on arctan(y/x) it should produce formulas without Abs.

Have you taken a look at my quaternion thing? Can you try feeding it to Mathematica and see what it simplifies to?...
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