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Fractal Math, Chaos Theory & Research => Theory => Topic started by: Prokofiev on July 28, 2010, 12:20:44 PM




Title: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on July 28, 2010, 12:20:44 PM
Hello,
I am a new member in fractalforums, but not quite new in fracal geometry.
I am the guy who commited this : http://en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension (http://en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension), well, most of it. Among other things.
I followed the fascinating efforts towards the discovery of the mandelbulb last year.
The Mandelbulb certainly deserves its place in this list, but where ?

On first thought, its fractal dimension should be 3, very probably, since it adds a full dimension to the boundary of the Mandelbrot set, which has already dimension 2 (Mitsuhiro Shishikura 1991). And this is certainly true for all powers.

Has anyone studied the subject seriously ? Or at least calculated its box-counting dimension ?

Thanks for your help !


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: ker2x on July 28, 2010, 12:58:39 PM
I have the excellent book by benoit mandelbrot "Les objets fractals. forme, hasard et dimension." which describe clearly the non-integer fractal dimensions.
It's odd that i never seen any topic about it here.

i'm glad you wrote one.
Unfortunally, i can't help. :)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Sockratease on July 28, 2010, 01:15:40 PM
Hello and Welcome to the forum!

Glad you joined in rather than just lurking.

But sorry, I too am unable to answer your question.  However, there are folks here who surely can.  Give them time to spot this thread...


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: msltoe on July 28, 2010, 02:07:40 PM
One of the two Mandelbulb inventors, twinbee, did some preliminary box counting on his web page:
http://www.skytopia.com/project/fractal/2mandelbulb.html#stats
He sees a value of ~2.945 for the 8th order Mandelbulb. If you know of a formal method of calculating the Hausdorff dimension, you may be to able modify the source code of a program such as Mandelbulber by Buddhi.
We also have another intriguing 3-D fractal called the Mandelbox that could be analyzed.

-mike


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: kram1032 on July 28, 2010, 02:12:55 PM
I guess, K-IFS change their dimensionality, huh?
Because their Hausdorff Dimension would be interesting too :)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on July 28, 2010, 02:31:46 PM
He sees a value of ~2.945 for the 8th order Mandelbulb.
-mike
Thanks Mike, but that was Daniel White's estimation of the volume, if I read correctly.
And, as he adds just underneath, he has "no idea about its Hausdorff or fractal dimension". So far ?

I do agree the Mandelbox should also be studied.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Nahee_Enterprises on July 28, 2010, 05:54:57 PM
    Hello,   I am a new member in FractalForums, but not quite new in fracal geometry.
    The Mandelbulb certainly deserves its place in this list, but where ?
    Has anyone studied the subject seriously ?  Or at least calculated its box-counting dimension ?

Greetings, and Welcome to this particular Forum !!!    :)

Good questions, but not something I can answer for you nor point to a location to acquire the answers.  Most likely there are none at present.  Sounds like a good project that you might wish to take on, if you so desire.    ;D
 


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Calcyman on July 31, 2010, 06:03:11 PM
I assume everyone is discussing the boundary of the Mandelbulb here -- the boundary is fractal, whereas the interior is solid, with a dimension of exactly 3.

The boundary of the ordinary Mandelbrot set has a Hausdorff dimension of 2 (and thus could even have an area!). The Mandelbulb does not look convoluted enough to have a Hausdorff dimension of 3 -- if it falls short, then it is definitely not the ideal 3D extrapolation of the Mandelbrot Set.

Because the triplex algebra is horribly complex (no pun intended!) to analyse, determining the fractal dimension analytically may be intractable.


If I remember correctly, the Mandelbulb is multiply connected (it is topologically inequivalent to a 3-ball). In fact, coupled with its quasi-self-similarity, the Mandelbulb is genus-infinity, and therefore has an infinite Euler characteristic. An ideal 3D Mandelbrot Set should be genus-0.


The Mandelbox is another thing entirely. Again, assuming I have interpreted it correctly, it doesn't use an underlying 'algebra', but instead the fractal is determined by the bounding planes. A Mandel-dodecahedron should be equally possible, or indeed any convex polyhedron.



Are there any academic papers relating to these new objects, or the triplex algebra itself?


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on August 01, 2010, 12:50:16 PM
Hi Prokofiev, and excellent work on that wikipaedia page I have looked at it often, its a great reference.
I can't give a value for the hausdorff dimension of either of these two objects but I am almost certain that they both are 'multi-fractals' so their hausdorff, box-counting dimension etc probably aren't the same, and the hausdorff dimension itself I think would change depending on where you look on the object... but maybe an overall average can be calculated.

The reason why I think they are multi-fractal is because the scale-factor in the iteraction varies with location, e.g. larger near the poles on the mandelbulb and on the mandelbox it is the ball-fold that gives varying scale factor.
For this reason I think that hausdorff dimension for the kaleidoscopic fractals should be much easier to calculate, they aren't multi-fractal.

By the way, I know that article is on Hausdorff dimension, but it would be also cool to know the 'size' of each of those fractals, e.g. is the unit moore curve 'larger' than the unit peano curve or Hilbert curve?
Hausdorff dimension is a bit like saying a disk area is order(r^2), it would be nice to know that it is in fact pi*r^2, here's a ref: http://www.fractalforums.com/mathematics/how-long-is-the-coastline-of-great-britain/ (http://www.fractalforums.com/mathematics/how-long-is-the-coastline-of-great-britain/)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Calcyman on August 01, 2010, 01:25:25 PM
Quote
By the way, I know that article is on Hausdorff dimension, but it would be also cool to know the 'size' of each of those fractals

One measure I have devised (probably re-invented) is as follows:

* Divide the space into hypercubes of side length x units.
* Count the number of hypercubes that contain at least one point of the set (call this value n), and calculate n*x^D, for some value of D.
* As x becomes arbitrarily small, n*x^D will either approach 0 (if D is greater than the Hausdorff dimension), infinity (if D is less than the Hausdorff dimension), or a finite value (if D is the Hausdorff dimension).
* This finite value, the limit of n*x^D as x --> 0, is the size of the set, in units^D. So, for example, this could be used to calculate the size of a Sierpinski triangle, in metre^(ln(3)/ln(2)).


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: kram1032 on August 02, 2010, 04:29:37 PM
would it have any sense to do those calculations over the whole parameter space? (Which of course would be way more than 3D)
Maybe even the generalized multifractal dimension and lacunarity. It could end up to somehow highlight nice parameter sets :)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Paolo Bonzini on August 02, 2010, 04:48:22 PM
Are there any academic papers relating to these new objects, or the triplex algebra itself?

I would not call it academic at all, but there's

http://github.com/bonzini/mbulb/raw/master/mbulb.pdf

Paolo


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on August 03, 2010, 02:08:11 AM
Calcyman, that sounds right.
 My own guess is that they both have an average hausdorff dimension of around 2.8. And yes, this is the dimension of the mandelbulb surface, but the dimension of the whole mandelbox.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on August 18, 2010, 11:27:00 AM
Hello,
Interesting comments. :)
Apparently nobody has attempted this calculation.
Calcyman, I'm affraid your measure will always give 1 for any self-similar fractal pattern of side 1 (at least for strictly self-similar patterns such as the Sierpinski triangle, your definition derives from the box-counting dimension).
Now, a point to keep in mind : The Hausdorff Dimension of a set is the highest dimension encountered in any of its subsets (Falconer, Mandebrot...). There is no such thing as an "average Hausdorff dimension". The boundary of the Mandelbulb, as a whole, "seems" to have a dimension around 2.7 - 2.8, but there could be somewhere a subset, no matter how small, more dense, with a dimension closer to 3. (The boundary of the Mandelbrot set has dimension 2, in spite of what we could guess, because of such subsets).
I agree it must be really difficult to get an analytic result of the Hausdorff dimension with such a difficult algebric complexity. :death: I would not know where to start.
At least we could try to look for a "dense-looking subset" and measure its box-counting dimension. That would give a reasonable lower bound.
Actually we should search in the neighbourhood of the xy-plane, the highly dense regions of the corresponding Mandelbrot Set.




Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on August 24, 2010, 04:47:23 AM
I'm not sure that is right about the mandelbrot set, the boundary looks equally rough everywhere to me, and I'd be surprised if you can find a subset of the boundary with a dimension < 2.
I also don't think it is right to say it will give 1 for a pattern of side 1, for example, a Koch curve (from a to b) has side 1, but a double koch curve (from a to be then back) is twice as much, at every stage of box counting for instance there will be twice as many boxes filled with the curve. In fact I looked it up and its called the Hausdorff measure of the fractal.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on August 25, 2010, 05:15:24 AM
I tried box counting the scale -2 mandelbox today, and if my code is right then its box counting dimension is about 2.8.
The code is below so let me know if you see a mistake. I measured the dimension by estimating what the dimension tends to as count increases.
I use a kind of distance estimate by keeping track of the radius of each box as it is transformed in the iterations. The whole box is only outside if the centre of the box is more than 6 + radius away from 0,0,0. (I'm using 6 as the escape distance, even though 4 may be sufficient.)
 
Code:
   int count = 10;
  uint64 numBoxes[100];
  float dimension[100];
  for (int b = 0; b<100; b++)
  {
    count = count * 2;
    numBoxes[b] = 0;
    float boxDiameter = 2.f / (float)count;
    for (uint64 x = 0; x<count; x++)
    {
      float X = (0.5f+(float)x) * boxDiameter;
      for (uint64 y = 0; y<count; y++)
      {
        float Y = (0.5f+(float)y) * boxDiameter;
        for (uint64 z = x + y%2; z<count; z++) // starting at x means we only do half the work by exploiting symmetry
        {
          float Z = (0.5f+(float)z) * boxDiameter;
          Vector3 start(X,Y,Z);
          Vector3 point = start;
          float radius = 0.5f*boxDiameter;
          numBoxes[b]++; // in anticipation of it being inside
          for (int i = 0; radius<4.f; i++)
          {
            if (point.x > 1.f)
              point.x = 2.f - point.x;
            else if (point.x < -1.f)
              point.x = -2.f - point.x;
            if (point.y > 1.f)
              point.y = 2.f - point.y;
            else if (point.y < -1.f)
              point.y = -2.f - point.y;
            if (point.z > 1.f)
              point.z = 2.f - point.z;
            else if (point.z < -1.f)
              point.z = -2.f - point.z;
            float size = point.magnitudeSquared();
            if (size < 0.25f)
            {
              point *= 4.f;
              radius *= 4.f;
            }
            else if (size < 1)
            {
              point /= size;
              radius /= size;
            }
            point *= -2.f; // scale 2
            radius *= 2.f;
            point += start;
            float mag = max(abs(point.x), max(abs(point.y), abs(point.z)));
            if (mag > 6.f + radius)
            {
              numBoxes[b]--; // wasn't inside
              break;
            }
          }
        }
      }
    }
    dimension[b] = logf(2.f*(float)numBoxes[b]) / logf((float)count);
    printf("dimension %f", dimension[b]);
  }
so it is more solid than a Menger sponge. (sequence that limit estimated from: 2.9861, 2.9518, 2.92, 2.8916, 2.871, 2.857)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on August 26, 2010, 09:59:30 AM
Nice result, Tom !  :).
Although I'm not quite sure I understand your algorithm, here  :hmh: Are your boxes cubic or spheric ? Why should "the radius of each box be transformed in each iteration" ?
Could you give more explanation ? More details on the parameters and intermediary results ?
How many transformations were applied ? what is the range in sizes of boxes ? Does the log-log plot look linear ?

It would be interesting to test your code on the Sierpinski teraedron, or the Menger sponge, to be sure.

Now what about the scale 2 Mandelbox (standard one?) ?


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on August 26, 2010, 10:10:58 AM
I see your point here:
but a double koch curve (from a to be then back) is twice as much, at every stage of box counting for instance there will be twice as many boxes filled with the curve.
It makes sense. So we could say:
- The size of a 2-meters Sierpinki trangle is 3 m^d
- The size of a 3-meters Sierpinski triangle equals 5.7 m^d, that is 5.7 times the unit Sierpinski triangle.
- we can also compare fractals with similar dimensions and say "this one is "larger" than that one".
- Or choose a unit fractal and say "this coast equals 1535 unit Koch curves".

With this measure I don't see why one could not estimate the "area" of the boundary of the mandelbrot set in square meters.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on August 27, 2010, 01:56:41 AM
Yes I think that's right. You can also compare the size of fractals with different dimensions by determining the crossover resolution where they are the same size. Here's a recent discussion- http://www.fractalforums.com/mathematics/how-long-is-the-coastline-of-great-britain/ (http://www.fractalforums.com/mathematics/how-long-is-the-coastline-of-great-britain/)
"I don't see why one could not estimate the "area" of the boundary of the mandelbrot set in square meters"
According to wikipedia no-one knows yet whether it has a lebesque measure (an area in the traditional sense)
I also found this:
Quote
Shishikura proved that, excluding certain points on the boundary, the boundary's two-dimensional area itself is zero. "This means the boundary is as thick as it could possibly be without occupying an area," Milnor says. "He's pinned it down quite precisely."
Whether that is just the lebesque measure of area, or the hausdorff measure or something else I don't know.

I have updated the code and results above to be a bit more accurate, and fixed the radius which was twice what it should have been. I am not sure how valid the algorithm is, but looks ok to me. The boxes are cubic, if I just count individual points in the set then the number of points (and indeed the dimension) becomes dependent on the number of iterations since like a menger sponge it is nowhere dense. We want to know if any point in each little box could be inside the set, so I approximate the transformation of the whole box by storing it as a centre + radius, and scaling the radius whenever the point gets scaled. This allows you to determine when a box is partially inside, since its centre can grow at max *2 + 2 when its radius is around 6, this means any radius of >4 can not escape from a distance of 6 away.
The numbers tend towards a value but the convergence is slow so I use this site to find an exponential fit (from http://zunzun.com/ (http://zunzun.com/)) so giving an estimated limit.
The equivalent for the menger sponge gives this sequence 2.86406, 2.82157, 2.811604, 2.80002, 2.792933, 2.786642, 2.782381 so looks roughly on course to give the right result around 2.72. But as with the previous, convergence is incredibly slow with this method. Its version of the code is:
Code:
  int count = 10;
  uint64 numBoxes[100];
  float dimension[100];
  for (int b = 0; b<100; b++)
  {
    count = count * 2;
    numBoxes[b] = 0;
    float boxDiameter = 1.5f / (float)count;
    for (uint64 x = 0; x<count; x++)
    {
      float X = 1.5f + (0.5f+(float)x) * boxDiameter;
      for (uint64 y = 0; y<count; y++)
      {
        float Y = 1.5f + (0.5f+(float)y) * boxDiameter;
        for (uint64 z = x + y%2; z<count; z++) // starting at x means we only do half the work by exploiting symmetry
        {
          float Z = 1.5f + (0.5f+(float)z) * boxDiameter;
          Vector3 start(X,Y,Z);
          Vector3 point = start;
          float radius = 0.5f*boxDiameter;
          numBoxes[b]++; // in anticipation of it being inside
          for (int i = 0; ; i++)
          {
            if (point.x > 1+radius && point.x < 2-radius && point.y>1+radius && point.y < 2-radius)
            {
              numBoxes[b]--;
              break;
            }
            if (point.x > 1+radius && point.x < 2-radius && point.z>1+radius && point.z < 2-radius)
            {
              numBoxes[b]--;
              break;
            }
            if (point.z > 1+radius && point.z < 2-radius && point.y>1+radius && point.y < 2-radius)
            {
              numBoxes[b]--;
              break;
            }
            if (radius > 0.5f) // obviously the above can never be true in this case
            {
              break;
            }
            point *= 3.f;
            radius *= 3.f;
            point.x = fmodf(point.x, 3.f);
            point.y = fmodf(point.y, 3.f);
            point.z = fmodf(point.z, 3.f);
          }
        }
      }
    }
    dimension[b] = logf(2.f*(float)numBoxes[b]) / logf((float)count);
    printf("dimension %f", dimension[b]);
  }


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on August 27, 2010, 11:29:30 AM
Applying transformations to boxes seem a bit risky to me  :hmh:.  And just rescaling the boxes might too approximative, there are foldings and inversions in your transformation.  I don’t know exactly if it over-estimates or under-estimates the reallity. It is not the definition of the box-counting.
Why not just iterate all the transforms first and then divide into smaller and smaller boxes afterwards, in the usual way  ? It is the most simple way, to me, especially if you choose to divide the size of the boxes by 2. We would then be sure.

The quote about Shishikura is interesting, Where did you find it ?

by the way, I created the Mandelbox article for the french Wikipedia, here http://fr.wikipedia.org/wiki/Mandelbox (http://fr.wikipedia.org/wiki/Mandelbox). With a link to your site, of course.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on August 27, 2010, 01:44:57 PM
"Why not just iterate all the transforms first and then divide into smaller and smaller boxes afterwards, in the usual way"
I tried this but the number of iterations I used was affecting the resulting dimension... hmmm tricky, maybe someone else here on fractalforums could give it a try. The mandelbulb sounds trickier still, since to count the border you need do decide which are the border boxes.

Nice article in french, its better than the english version :)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on August 27, 2010, 03:31:26 PM
I tried this but the number of iterations I used was affecting the resulting dimension... hmmm tricky, maybe someone else here on fractalforums could give it a try.
I am not suprized by that. The dimension must be calculated with the highest possible number of iterations. That's the only way to get as close as possible to the dimension on the "real thing".

Pity i am such an awful programmer (I used to program a very long time ago). :-\


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on August 27, 2010, 04:05:17 PM
The mandelbulb sounds trickier still, since to count the border you need do decide which are the border boxes.
A voxel of the boundary has less than 6 neighbours. The best way to get rid of the inside.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on September 01, 2010, 06:42:26 PM
I have been doing some research in the papers about our problem and asked several questions around. 
I give you some news:

1) The Hausdorf dimension of any subset of the boundary of the Mandelbrot set is 2. Tglad, you were right and I was wrong. Strangely, even the tip of the antenna has dimension 2  :surprised:. It comes from Shishikura's paper : "for any open set U which intersects ∂M, we have H-dim(∂M ∩ U) = 2". Whatever the subset, you always find a dense subset of points around which the dimension is 2. (because of embedded copies of the Mandelbrot set all over the place?).

2) The boundary of the power-8 mandelbrot set has also dimension 2. It was confirmed to me by Jacques Carette, (he worked under Douaddy and Hubbard). Shishikura's result goes for any "rational map f0  of degree d (>1)  which has a parabolic fixed point ζ with multiplier exp(2πip/q) (p,q∈Z,gcd(p,q)=1) and that the immediate parabolic basin of ζ contains only one critical point of f0". This also applies to the map z^8+c. And any subset of its boundary has dimension 2, for the same reason explained in 1).

so: The entire boundary of the cross-section of the power-8 mandelbulb in the xy-plane has dimension 2.
We're getting somewhere  :dink:

Now if you give some "thickness" to this boundary in the z direction, for example a cartesian product with a segment (dim=1), whatever small, (a cylinder is the cartesian product of a circle and a segment), then you definitively have a set with Hausdorff dimension 2+1 = 3.
And if the mandelbulb has dimension 3 in any neighbourhood of the xy-plane, whatever small, then the entire set has dimension 3.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: fractower on September 01, 2010, 09:29:32 PM

The final step seems a valid way to argue that HD = 3 for the power 8 bulb. For example an extrusion of any object with HD = N by an orthogonal axis results in an object with HD=N+1. So the extrusion of a 2D Mandelbrot will have HD=3 without really increasing the entropy (physicist not mathematician.)

Another way to say this is stretched taffy is qualitatively different from bulby goodness. Integer HDs seem to miss this distinction.



Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Tglad on September 02, 2010, 12:33:15 AM
Maybe that is why you can't have a real 3d mandelbrot... since a simple extruded mandelbrot (mandelbrot based prism) already has a full 3 dimensional boundary, the surface simply can't be any rougher than that. Or maybe you need to go to the next level down...
So a mandelbrot has area 2d and edge 2d
Any extruded mandelbrot has volume 3d and surface 3d
But I think the pursuit that lead to the mandelbulb was looking for: volume 3d, surface 3d and all lines over its surface being 2d. i.e. not just a fractal surface but fractal paths over every surface.

I wonder if there is a shape with volume 3d, surface 3d and all circumference lines being 3d? That would be nuts.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: fractower on September 02, 2010, 06:50:26 AM
I don't want to give up on the holy bulb just yet. Prokofiev was able to demonstrate that the HD=3. I was just pointing out that that is not quite the complete answer. I suspect Tglad's path defination is a good metric to consider. For example the distance between any two points on the boundery of the Mandelbrot is infinite. Carette's result would seem to indicate the same for the xy plane (assuming a connectivity) of the power 8 bulb. If the same could be said for all paths in some small volume we would be closer.

I have not done a lot of bulb diving. Are there taffy regions in the power 8 bulb? It probably only takes one to propigate to all scales.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on September 02, 2010, 11:31:02 AM
An interesting result is that we didn't make any hypothesis on the formula (except that it equals Mandelbrot's one in the xy-plane), so this result is valid for any variant of a 3D Mandelbrot set such as the ones mentionned in  Daniel White's page (http://www.skytopia.com/project/fractal/mandelbrot.html) or in Nylander's page (http://www.bugman123.com/Hypercomplex/) (and post (http://www.fractalforums.com/theory/summary-of-3d-mandelbrot-set-formulas/)).

Now, of course it would be interesting to see if any cross-section by any plane has dimension 2,  and if any subset of these cross-sections have dimension 2.
I'd like to think so but I have doubts, particularly when I see smooth surfaces such as these ones :
1) The bottom of the "caves" (cross-section along the xy-plane)
(http://lh4.ggpht.com/_R1baPul19Kw/TH9rmLhknMI/AAAAAAAABEY/sEHOR3Xf4eg/s576/smoothsurface.jpg)
2) The "toruses" surrounding the set (cross section along the xz-plane).
(http://lh3.ggpht.com/_R1baPul19Kw/TH9rmIijW8I/AAAAAAAABEc/Qx0afOvivBE/s640/smoothsurface2.jpg)

But, again, we must not trust our eyes on that matter.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on September 02, 2010, 03:41:14 PM
Actually, why not make a small test ?
When you zoom on the tip of the antenna of the Mandelbrot set, everything looks smooth and straight. But, as you keep on zooming, at some stage, small copies of the Mandelbrot set appear.
Could it be the same on the Mandelbulb ? So I tried this on a smooth part of the Mandelbulb, just to see if a minute and hidden rough pattern of some sort appears on the screen. I zoomed on the point of coordinates x=0.45, y=-0.5427 and z=0 which is on the “wall” of a “cave”  (see left image).
(http://lh6.ggpht.com/_R1baPul19Kw/TH-m_KQyllI/AAAAAAAABEg/z8gXF876CI8/testsmooth.jpg)
The result is disapointing  :-\. You see the classic spiral patterns unravel on the xy-plane, but the surface along the z axis always looks desperately smooth, even with a high degree of iterations. 
I zoomed up to  10^-14 (see right image), which is the best I could get..

Of course that is just a quick test and we must not draw conclusions from that.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: M Benesi on September 03, 2010, 09:39:22 PM
   Silly idea,  brain stopped after creation of it, so need another to glance at it:

  Idea:  Circumference of a circle has a Hausdorff of 1.   
           Area of the surface of a sphere has a Hausdorff of 2.
           Area of sphere is a product of the rotation of a circle around a perpendicular circle with the same center.

   Similarly:

          Mandelbrot outer edge has a Hausdorff of 2.
          Bulb outer edge would have a Hausdorff of ???   
          Area of surface of bulb is the product of the rotation of a 2d Mandelbrot around a 2d Mandelbrot (both with Hausdorff of 2).

  Wouldn't this make the Hausdorff 4?

    Consider a 2d Mandelbrot spun in a circle (I know I've seen a few :D):

    Mandelbrot Hausdorff of 2.
   Circle is Hausdorff of 1.
   Is the Hausdorff of the spun Mandelbrot 3?



Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Prokofiev on September 03, 2010, 11:21:38 PM
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          Area of surface of bulb is the product of the rotation of a 2d Mandelbrot around a 2d Mandelbrot (both with Hausdorff of 2).
The mandelbulb is not the product of 2 Mandelbrot sets.
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Wouldn't this make the Hausdorff 4?
Certainly not, you can't just add dimensions. And, it can't be more than the space in which it lies.
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Consider a 2d Mandelbrot spun in a circle (I know I've seen a few :D):
    Mandelbrot Hausdorff of 2.
   Circle is Hausdorff of 1.
   Is the Hausdorff of the spun Mandelbrot 3?
Yes. Absolutely  :). If you are interested in the subject I would advise you to read Falconer's "Fractal Geometry" (There is an entire chapter about the product of fractal sets)


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: fractower on September 04, 2010, 05:36:03 AM
M Benesi was not entirely off base when she/he described the product of two Mandelbrot sets as a 4D object. One just needs to use the outer product rather than some kind of rotation (There may be an outer product rotation as well, but I will ignore that for now since it hurts my head). The outer product of two column vectors is a 2D object known as a matrix. The outer produce of two matrices is a 4D tensor. We are not use to thinking of 2D objects such as the M set as a matrix, but it can be expressed as the sum of outer products of sins and cosins (Fourier transform). Granted an infiinite number would be required, but we should be use to infinite sums by now.

Try graphing sin(Ax)*cos(By) to get an idea of what an outer product of 2 continuous functions looks like.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: Calcyman on September 04, 2010, 09:00:35 PM
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M Benesi was not entirely off base when she/he described the product of two Mandelbrot sets as a 4D object.

The Cartesian product of two orthogonal Mandelbrot sets is four-dimensional. But the Mandelbulb, or any other 3D fractal, is certainly not the M-set˛; such a set would lie in four-dimensional Euclidean space.

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We are not used to thinking of 2D objects such as the M set as a matrix

That's because the Mandelbrot Set is not a matrix, and it is physically impossible to express the Mandelbrot Set as a matrix -- there are only Beth One matrices, and Beth Two sets over the complex plane. Placing them in one-to-one correspondence would conflict with Cantor.


Title: Re: Hausdorff dimension of the Mandelbulb
Post by: M Benesi on September 05, 2010, 12:38:08 AM
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          Area of surface of bulb is the product of the rotation of a 2d Mandelbrot around a 2d Mandelbrot (both with Hausdorff of 2).
The mandelbulb is not the product of 2 Mandelbrot sets. 
  Here is some code  (should make it clear, since this shows how the bulb works ... like the regular 2d set):
Code:
  sx,sy,sz are starting x,y, and z components; nx, ny, and nz are calculated components before pixel component addition;
r1=sqrt(sqr(sy)+sqr(sz));
r3=r1^-n;   //take it to the negative power to avoid division by zero error!

victor=complex(sx,r1);   // complex (x,y) is ChaosPro's command to create a complex number out of 2 real components x and y
bravo=complex(sy,sz);  //  the second part (1st,2nd) is the imaginary component

victor=victor^n;
bravo=bravo^n;

nx=real (victor);
ny=imag (victor) * real (bravo) *r3;  //  you have to divide out the magnitude (r3=r1^n) because it has already been applied
nz=imag (victor) * imag (bravo) *r3;  //  twice, once computing victor, once computing bravo

  then add in your pixel values..

  So you can see we have something that resembles two 2d sets influencing one another, almost like it's a product of 2 2d sets.... 
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Wouldn't this make the Hausdorff 4?
Certainly not, you can't just add dimensions. And, it can't be more than the space in which it lies.
  So it's maximum hausdorff is 3, and it has to be 3.. although it has a higher apparent complexity than a spun brot...