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Author Topic: True 3D mandelbrot type fractal  (Read 276032 times)
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David Makin
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« Reply #15 on: November 23, 2007, 07:20:18 PM »

Quote
That render doesn't half remind me of this:

<a href="http://www.youtube.com/v/jeVDBvE5Bg0&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/jeVDBvE5Bg0&rel=1&fs=1&hd=1</a>

Hi David, still love to see that quaternion technique at creating the 3D Mandelbrot set. Any attempts so far?


Not yet as I'm concentrating on the 3D IFS/RIFS formula I just released for Ultrafractal at the moment smiley
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twinbee
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« Reply #16 on: December 05, 2007, 12:02:17 AM »

I think we may have something...



Sorry about the low resolution - even this took around 30 hours to render!! Formula and full story coming when I tidy up the code...
« Last Edit: December 05, 2007, 12:20:37 AM by twinbee » Logged
Duncan C
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« Reply #17 on: December 27, 2007, 01:29:18 AM »

Hi all!
I'm relatively new to fractals, but I have searched for hours to find a true 3D mandlebrot type fractal, all in vain. I don't want the raised mountain type of mandlebrot, and I don't want any true (but trivially simple) ones such as the Menger sponge. I instead want a true 3D equivalent of the mandlebrot (or near enough).

The closest I got was the Quasi-fuchsian sphere fractal:
<a href="http://www.youtube.com/v/3lcO9zRCv-4&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/3lcO9zRCv-4&rel=1&fs=1&hd=1</a>

However, that is still mostly self-similar. I want the amazing variety that can be found in the mandlebrot. Anything out there?


Could you take the 4D Mandelbrot/Julia set equation:

(real and imaginary Z and real and imaginary C values) and hold one of the 4 values constant, while running the other 3 thorough a cubic grid of values? I would think this would generate very interesting 3D fractals. It would be a 3D slice through the 4D Mandelbrot/Jullia equation, where normal Mandelbrot and Julia sets are 2D slices.



Duncan
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Regards,

Duncan C
David Makin
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« Reply #18 on: December 27, 2007, 02:46:54 AM »


Could you take the 4D Mandelbrot/Julia set equation:

(real and imaginary Z and real and imaginary C values) and hold one of the 4 values constant, while running the other 3 thorough a cubic grid of values? I would think this would generate very interesting 3D fractals. It would be a 3D slice through the 4D Mandelbrot/Jullia equation, where normal Mandelbrot and Julia sets are 2D slices.

Duncan

2D renders of 3D slices from the 4 dimensional "Julibrot" space are possible from Fractint and some formulas in Ultrafractal but they aren't exactly what is being sought smiley

Here's one view of a Julibrot:

http://makinmagic.deviantart.com/art/Classic-3D-Julibrot-43168532
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twinbee
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« Reply #19 on: December 27, 2007, 10:39:46 AM »

Looks like it's not just us lot that have searched for the 3D Mandelbulb. I found some interesting threads in Google Groups from quite a few who seemed to wonder if such an object could exist. Notable posts include Erick's info about using arbitrary powers of n (instead of the usual n^2 in the main formula) to represent the third dimension. Also Laszlo's rumour indicates that the thing exists, and actually has a name (oops he's lost the reference - there's a surprise!). Finally, Lloyd Mitchell in this thread claims that "3d slices of the 4D Mandelbrot set (let the k part of c = 0) do reveal spheres instead of circular disks.", and precedes that with formulae which maybe someone here can follow better than I can. Of course, I'm am very skeptical he's actually found anything.

I'll come clean about the image I posted on December 04. It's not from any maths at all - I drew it in an art program...... :-} Yeah I know, seemed a fun joke at the time, though I'd be a lot more excited if I really found it. If you look carefully, you should be able to notice some discrepancies which indicate it's not the genuine article heh.

I still think the 3D version of the Mandelbrot would look the best and most fascinating fractal ever (especially if we could make the surface multi-colored - not sure how easy that is, since the surface of the ordinary 2D Mandelbrot is always only a single colour). To find the Mandelbulb, I've tried everything from cylindrical coords to spherical-distance based rotations (instead of angle-based). Unfortunately, nothing's got closer than the one I posted earlier which Lycium rendered.

Finally, one glimmer of hope that such a thing may exist after all. I quote from this excellent page:

"Also Mandelbrot curves have been discovered in cross-sections of magnetic field borders, implying there is a 3-D mandelbrot equivalent that is closely tied to electromagnetism and therefore a deep structural and fundamental aspect of life, and physical space/time."
« Last Edit: December 27, 2007, 12:46:04 PM by twinbee » Logged
lkmitch
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« Reply #20 on: December 29, 2007, 05:34:33 AM »

actually has a name (oops he's lost the reference - there's a surprise!). Finally, Lloyd Mitchell in this thread claims that "3d slices of the 4D Mandelbrot set (let the k part of c = 0) do reveal spheres instead of circular disks.", and precedes that with formulae which maybe someone here can follow better than I can. Of course, I'm am very skeptical he's actually found anything.

Wow--it's not often that something I wrote 15 years ago comes back to life!  :-)

I think that the search for a 3D Mandelbrot analog depends on a suitable definition of a 3D number system.  As has probably been pointed out, the more useful extensions of the real numbers have 2^n components, so there isn't a standard 3D system.  Thus, you're free to define your own and see how it works with the Mandelbrot set.

(Lloyd) Kerry Mitchell
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glbn
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« Reply #21 on: January 09, 2008, 02:59:53 PM »

Hi,

Here's a little program I did long time ago to explore the quaternions fractals:

http://www.cse.yorku.ca/~gilles/fractal.zip

I got some interesting results:


I can't remember exactly which multiplication I used, but it wasn't the quaternion one.

Gilles
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twinbee
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« Reply #22 on: January 12, 2008, 02:31:52 PM »

Quote
Wow--it's not often that something I wrote 15 years ago comes back to life!  :-)

Lol smiley I take it what you said back then didn't produce what we're looking for (a Mandelbulb with bulbs on all axis) ?

Hey glbn, that looks pretty cool! Obviously, it's not quite what we're after, but it looks quite nice regardless and although the chance is small, there are certainly areas which could hold more and more detail as you zoom in.

Nice renders Lycium!

As for me, I've almost given up (but then I always say that don't I? wink). I'm exploring one last idea involving toroidal coords, which could be interesting.
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Karl131058
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« Reply #23 on: November 19, 2008, 04:19:12 PM »

Oh, my deeply sleeping thread!
I revive Thee from the dead...  angel


Ok, seriously, after staying away from FractalForums for a while because of RealLife(TM), I came back and (obviously) found this thread. I had a little exchange of mails with twinbee about the formula involved and he has asked me to post the results here. So, here it comes:


twinbee defined (a "few" posts back)

double r    = sqrt(x*x + y*y + z*z );   
double yang = atan2(sqrt(x*x + y*y) , z  ) // that would be theta in std polar coordinates
double zang = atan2(y , x);                      // that would be phi  in std polar coordinates

so I would suppose he has implicitly

x = r*sin(yang)*cos(zang)
y = r*sin(yang)*sin(zang)
z = r*cos(yang)

and I would have expected (doubling the angles!)

newx = (r*r) * sin(yang*2)*cos(zang*2)
newy = (r*r) * sin(yang*2)*sin(zang*2)
newz = (r*r) * cos(yang*2)

but he defines

newx = (r*r) * sin( yang*2 + 0.5*pi ) * cos(zang*2 +pi);
newy = (r*r) * sin( yang*2 + 0.5*pi ) * sin(zang*2 +pi);
newz = (r*r) * cos( yang*2 + 0.5*pi );

which can be simplified by taking into account the symmetries of sin() and cos() to

newx = - (r*r) * cos(yang*2) * cos(zang*2)
newy = - (r*r) * cos(yang*2) * sin(zang*2)
newz = - (r*r) * sin(yang*2)

which is not exactly equal to doubling the angles. smiley
... but it leads to interesting pictures.
One DOES NOT need atan2(), sin() and cos() to implement these formulae,
because they can be simplified a lot by using the following identities for any
angle a :
cos(a)*cos(a)+sin(a)*sin(a)=1  (well known, I suppose)
cos(2*a) = cos(a)*cos(a)-sin(a)*sin(a)   (less well known, it seems smiley)
sin(2*a) = 2*cos(a)*sin(a)

I'll spare you the details, but you end on:

newx = ( x*x + y*y - z*z )*( x*x - y*y) / ( x*x + y*y )
newy = 2 * ( x*x + y*y - z*z )*x*y / ( x*x + y*y )
newz = - 2 * z * sqrt( x*x + y*y )

no trigonometric functions at all, just additions, multiplications, divisions and a squareroot.
There is NO pole on the z-axis, BUT there might be numerical problems because of the
denominator, solvable e.g. by taking

if( abs(y) < really_small_value )
newx = x*x-z*z
newy = 0
newz = -2*z*sqrt(x*x)
else (view above)

(end of simplification)

Interesting side effect: when I first read about "doubling both angles" I wanted
to try that for myself. When doing geometry instead of math, I prefer measuring
the angle theta not against the z-axis, but against the x-y-plane, so in that case

phi = atan2( y, x ) // just like before
theta = atan2( z, sqrt(x*x + y*y) ) // exchange of arguments, angle is positive above, negative below x-y-Plane

then
x = r*cos(theta)*cos(phi)
y = r*cos(theta)*sin(phi)
z = r*sin(theta)

and simply doubling the angles

newx = r*r*cos(2*theta)*cos(2*phi)
newy = r*r*cos(2*theta)*sin(2*phi)
newz = r*r*sin(2*theta)

simplifying this analogous to above gives

newx = ( x*x + y*y - z*z )*( x*x - y*y) / ( x*x + y*y )
newy = 2 * ( x*x + y*y - z*z )*x*y / ( x*x + y*y )
newz = 2 * z * sqrt( x*x + y*y )

IDENTICAL to twinbee's stuff except for the sign in z !

Since I'm sitting at an OLD Mac (350MHz) and try to use POV-Ray to produce pictures I can't show any yet; that takes TIME! But the first little "thumbnails" I produced show that this change of sign dramatically changes the resulting set!

Forgive my ranting - I hope somebody might find these "simplifications" useful - twinbee thought so, at least!

Happy iterating...
Karl
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twinbee
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« Reply #24 on: February 09, 2009, 12:25:57 AM »

Hi Karl,

I've just recently 'modularized' my drawing/mandelbrot algorithms, and so I've been trying to get some results out of your simplified functions such as:

newx = ( x*x + y*y - z*z )*( x*x - y*y) / ( x*x + y*y )
newy = 2 * ( x*x + y*y - z*z )*x*y / ( x*x + y*y )
newz = - 2 * z * sqrt( x*x + y*y )

Pictures are turning up blank so far... any idea why? I'll keep trying anyway...
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twinbee
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« Reply #25 on: February 09, 2009, 01:21:04 PM »

Yay, cracked it! I just added Karl's "if( abs(y) < really_small_value )" section, and that seemed to do the trick. The (orthographic) image seems practically identical to the old method (except a difference in 'pixel noise' which is expected really).

Can't wait to see the new 3D renders btw! 100x may be a bit optimistic however - I only managed a 4-5x increase on my setup. However that may be different when working with proper 3D, or it could be because of my own SSC (super sucky code) wink
« Last Edit: February 09, 2009, 01:46:59 PM by twinbee » Logged
Karl131058
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« Reply #26 on: February 10, 2009, 12:14:29 PM »

Hi twinbee,

sorry, haven't been here for a while - my new part time job puts quite a strain on my capabilities to look at a computer screen  wink
I'm quite happy that someone made use of my mathematical gymnastics - finally, I might add.

I only managed a 4-5x increase on my setup.

OK - that's not THAT much, but it's SOMETHING. I would seriously love a speedup like that, but I'm looking forward to the next pay check from aforementioned new job; that just MIGHT become a faster computer (crosses fingers)

My pictures came up very fast (and empty!) too, before I worked around that non-singularity with that "really_small_value"-trick...

Happy iterating
Karl
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twinbee
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« Reply #27 on: February 10, 2009, 01:05:56 PM »

If you do get a new CPU, you might want to treat yourself to a quadcore Q6600 from Intel. I've never loved a CPU as much as this, and it's been getting tons of 5 stars from Amazon - it's amazing for multitasking lots of progs simultanesously too + it runs quite cool (at least with the G0 stepping version).

Quote
OK - that's not THAT much, but it's SOMETHING.
Haha, I'm just spoilt wink I'm guessing you got similar results too?

I wonder if there's a program to simplify all the sine/cosines like you've done. It can't be easy - probably comes under advanced algebra I'm guessing. Thanks again for the results!
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monguin61
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« Reply #28 on: March 17, 2009, 03:06:06 AM »

Hi twinbee, I'm a brand new member here. I stumbled upon your page (http://www.skytopia.com/project/fractal/mandelbrot.html) on my quest to understand how to render the "whipped cream" Julia quaternions. I've always liked how they look... but there's something missing... I was so glad to find your page documenting your attempt to find this object. Ever since I wrote my first crappy Mandelbrot program (probably ten years ago), I've had the same question you do, about a "true" 3D Mandelbrot fractal solid. I just wanted to post to let you know that I understand EXACTLY what you're looking for, and why its so enticing. I briefly searched for this thing years ago, but gave up pretty quickly. Some of the images I've seen here and on your page make me think that it does exist though.

Also, while looking over the description of your 3D multiplication, I wondered why you doubled both angles - that has kind of a weird effect. Wouldn't it be better to double the angle between your number, and the number 1 (for simplicity, say 1=(0,0,1))? This is similar to your approach, but not exactly the same thing. It makes the multiplication operation a little more uniform, in that all points on the circle of constant r and phi, but varying theta, will end up on another circle of constant, but different, r and phi. Did you try this at all? (sorry if that's hard to follow - I can clarify with a diagram if necessary)

lyc, it seems that most of your renders attached to your older posts are gone, are those still available anywhere?

Quote
I wonder if there's a program to simplify all the sine/cosines like you've done. It can't be easy - probably comes under advanced algebra I'm guessing. Thanks again for the results!
I use my TI89 frequently (it uses a program called Derive). There's also Mathematica, and Maple. Matlab has a symbolic manipulation package that just uses Maple.

[sorry for reviving a dying thread, but I was totally astounded to find another person in the world searching for this]
« Last Edit: March 17, 2009, 03:48:24 AM by monguin61 » Logged
twinbee
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« Reply #29 on: March 17, 2009, 07:11:44 PM »

Hey there, welcome to the forum!

Great to see someone else interested in the search too. It's almost like hunting for the yeti, loch ness, or UFOs, except at least this has a chance of existing wink Kinda weird to see this kind of thing in abstract maths, rather than on a material level too. And yeah, the reward is definitely enticing.

There was a thread recently at povray.org that you may find interesting. Goes into a little bit more detail about whether it exists, and how one would render such a beast (in terms of lighting, iteration level etc.):
http://news.povray.org/povray.off-topic/thread/%3C48d7d290%241%40news.povray.org%3E/?mtop=9

I'd be happy to try out your suggestion and post the results here, but I can't quite follow your description so a diagram would be nice. Feel free to to use my 'sphere' as a template:
http://www.skytopia.com/project/fractal/sphere.png

Actually, even better, if you know a little programming, you could alter the function below. I've kept it as the 'slow' long way for clarity:

Code:
// A triplex number is a '3D' number
triplex multiply(triplex a, triplex b) {
    triplex n;
    double pi=3.14159265;
    double r    = pow(a.x*b.x + a.y*b.y + a.z*b.z , 0.5);
    double yang = atan2( pow(a.x*b.x + a.y*b.y, 0.5) ,      a.z  )  ;
    double zang = atan2(a.y , b.x);
    n.x = (r*r) * sin( yang*2 + 0.5*pi ) * cos(zang*2 +pi);
    n.y = (r*r) * sin( yang*2 + 0.5*pi ) * sin(zang*2 +pi);
    n.z = (r*r) * cos( yang*2 + 0.5*pi );
return n;
}
« Last Edit: March 17, 2009, 07:15:58 PM by twinbee » Logged
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