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From all the published images, it seems that the Mandelbox is a solid object. If we slice through it, we see complicated patterns of cavities, but they seem to be surrounded by solid material.
My point is that all this is not true at higher iterations.

The inside of the classic Mandelbrot set is "solid". Any point in any area of the inside of the set will stay bounded, even at an infinite number of iterations. For the Mandelbox , this is not the case.

The film below shows a series of 100 slices through the box, so from z=-6 to 6, with iterations set at 13.
http://www.josleys.com/gfx/Mandelbox2D-04.mov (http://www.josleys.com/gfx/Mandelbox2D-04.mov)  (18Mb, be patient)
You will recognise familiar patterns of the "standard" box ( scale=2)

Now the same 100 slices at 10.000 iterations:
http://www.josleys.com/gfx/Mandelbox2D-05.mov (http://www.josleys.com/gfx/Mandelbox2D-05.mov) (277kb, this will load quickly)

There now only appear to be 'inside' points between -3 and 3. ( with, strangely enough, a set of points at -6 and +6)

In none of the slices do there appear to be contiguous areas of inside points. We see nothing but collections of 'dust'.
It is to be noted also that without the inversion step, the limit set becomes a cube half as large as the 'box' (vertices at <+/-3,+/-3,+/-3> instead of  <+/-6,+/-6,+/-6>) (see the image below : the area between 3 and 6 gets more and more tenuous: the spacing between the lines appears to be Cantor dust)

It would be nice if somebody could confirm these results. A voxel render at high iterations should confirm this.

The second mov doesn't run for me, but it sounds wrong, what is your bailout distance? I think it needs to be at least 12 (144 in UF), I set it to 1024 to be safe.
Whole sections shouldn't disappear with higher iterations, however, I think you are right that any surface that appears solid will probably have small cracks and gaps and arches in it if you zoom close enough.

Both films run without problems from here.

The bailout was 15, but I get the same result with a bailout of 15.000..
Note that there is a huge difference between a surface with pits and cracks, and a cloud of dust..

I took Dave Makin's UF implementation, which has an option to demand  a minimum number of iterations before drawing something solid.
In the images below, the first has no such demand, and the second demands a minimum of 100 iterations.
You see that of the original cube, there remains something dusty, at exactly half the size..

I just don't have quicktime :)
I think its a bug in the minimum iterations option on Mandelbox. But did the videos use this min iterations?

The videos us a specially written program in UF to show only 2D, so no distance estimation is involved. I take a pixel and iterate it through the folding and inversion. If it does not bailout it turns blue, if not, white..

Just to add that my remarks about the Mandelbox are not meant to be disparaging in any way. I love the pictures!
I'm just trying to dissect the animal mathematically.
What I think is that the Mandelbox has properties like the Cantor set : the higher the iterations, the more dust..

To be honest just looking at the transforms that make up the Mandelbox formula I would expect it to be made up of Cantor Dust.
As to the appearance of renders using my formula but with a high "min. iterations" value, I think the result is more of an issue of very small points being missed by the render - i.e. it is Cantor Dust but there should be a lot more of it.
The "Min. iterations" option was only really designed to skip small iteration counts for on Mandelboxes where there are discontinuities in the smoothness of the smoth iteration colouring - here for example a value of say 8.5 is immediately followed by one of say 8.4 which looks to my renderer as if we've hit solid but the point is still "outside" and the driop in value is simply due to a discontinuity in the smooth iteration values, actually the option is obsolete if you enable "Double check for solid".

Agree, but "seeing" the dust particles becomes virtually impossible at high iterations because they become smaller and smaller and smaller...

A while ago, I too had played with these slices. The only things I remember was how at z=0, z=+/-2, z=+/-4, you get a "full" slice; something that appears to fill a whole plane. On the other hand, z=2.00001, with sufficient iterations, gave cantor dust. I assume this is due to some very fragile equilibrium between the folding and spherical inversion; I am curious what would happen if the folding and inversions are tested with < and <=, since these might "break" that one slice.

I do not remember any unusual dust, though. Do you think you could try rendering a video of any random slice (say, z=2.3 or whatever) with increasing iterations? This might show us how the differences came about.

Interesting find, for sure!  ^-^

To Timeroot:

See below the slice at z=2.3 at 15,20,35 and 75 iterations.

A solid square is to be found at z=+/-1, +/-3, and +/-5, even at high iterations.(and of course also at the same values for x and y, due to symmetry)

JosLeys I think you're on to something here. It does make sense that it would be like dense Cantor dust.
Thinking about it, even if the cracks are tiny each iteration, you still end up with 0 volume... if you have a line and you remove 1% as the biggest crack.. then all the remaining percent segments also have tiny cracks 1% of their own length, then the total line length is .99 * .99.. continuing you get .99^infinity which is 0 total length.
You get the same result with a Sierpinski tetrahedron and with a Menger sponge, and the renders would also disappear as sparse dust because you are rendering with a probe that is very thin compared to the pixel size.

A Menger sponge or carpet renders better if the probe is about the width of each pixel, ie colour it black if you get within half a pixel radius of an inside point.

I agree with Tglad, it seems that the fractal is still "correct", but it becomes so sparse that nothing happens to hit upon some of the dust. I'm sure that if you rendered it in 2D with distance estimation, it would stay the same regardless of the iterations. (Even with high minimum iterations).

..hmm
If a fractal turns to dust, distance estimation is not a good method I think..
My 2D images are made by iterating the pixel through the folding and inversion , and either it bails out or it does not.
If it bails out then we have empty space on that spot..

but just whether or not it is on that spot is not a good indicator of whether the fractal is actually "there". Distance estimation can tell you, "is there fractal anywhere within this pixel?", and if so, color it for you. If you iterate a whole buncha times, Almost All points will be excluded, and so Almost No points would be plotted on screen. Try look at the Mandelbrot set. It's not even dust. But unless you're looking at some bulb or some minibrot, you won't really get to see any inside points. The high iteration value converging there just tell you there's a point right in the middle; but chances are none of the pixels landed exactly on the scraggly little line.

OK, I see what you mean, but if the fractal has turned to disconnected specks of very, very small particles, then I'm not sure that distance estimation will tell us more. The Mandelbrot set is connected and distance estimation allows to find the fine tendrils between minibrots, but I think this is different..

But it's not different! Look at the Julia set (-1.5, 0.8) - it has an infinite number of points. Chances are, not a single pixel on your screen is "inside". On the other hand, the iteration values - or Distance Estimation, if you prefer, tell us where the points are, even if we don't land on them.

Timeroot is quite right I think, but I think it is great JosLeys that you are looking into what this shape is. I think your post has lead us to a collective conjecture that the Mandelbox is nowhere dense, so has a volume of 0. I wouldn't even like to think how that could be proved though :)

And such a fractal does have some weird properties, as you saw 0 radius probes will miss the set 100% of the time with enough iterations. That doesn't mean that there is nothing there, it would miss the edge of a circle 100% of the time too. It means we need to be careful how to render a 2d Mandelbox... one problem with just colouring black if there exists a point inside the radius is that there is no anti-aliasing, but too much anti-aliasing and the points will fade away again, a similar problem to rendering the smooth edge of a circle.

It is actually easier in 3d, strangely a 0 width ray of light will bounce off the Mandelbox even though it has no volume, it can be demonstrated on small-hole cantor dust. So should always look solid/opaque.

I guess another question about this object, what do you measure as the fractal? Is it the surface area that is fractal, or is it just the set of points that never escape? What is its topological dimension I wonder.

I believe that, because it is nowhere dense, it's surface is the whole fractal (that is, the set of points that don't escape). And even if that's false, the fractal would be considered the points that don't escape. We say the Mandelbrot Set is even the fatty insides, even though only the border is truly fractal.

I think that, even in 3D, rays would generally miss the solid. DE is implemented. .... - No, wait - there are actually the full 2D slices which are always reoccurring, giving it something to hit. But with ray tracing, there's the chance that it would skip right over that very thin slice, so, yeah, I guess you actually would need DE to see anything.  :sad1:

Make a 3D print of it?  :D

It would be pretty cheap if you are right about zero volume  O0

cheap and unstable  :evil1:

Well, if it's nowhere dense it's kind of hard to say that it has a "surface" at all... But when it comes to "seeing" the fractal, why not just use rays with nonzero width? A zero-radius ray is almost guaranteed to miss, if it's nowhere dense, but a "thick" ray would hit whenever the zero-radius ray would have come close. It has a physical analog, I guess, since light beams do have thickness... Anyway, distance estimate would help hugely with actually displaying that.

Question: The hypothesis that the Mandelbox is nowhere dense entails that, as the iteration count gets higher and higher, the image should gradually vanish (as long as your rays have zero width). Has anyone actually noticed this? Or does it just exhibit the "solidifying" behavior of the Mandelbrot or Mandelbulb?

"Well, if it's nowhere dense it's kind of hard to say that it has a "surface" at all.."
True, it has no boundary between an inside volume and the outside.. it might be better to think of the Mandelbox as being just a set of points in space.

In 2d the box will vanish with higher and higher resolution, just like a circle will vanish with higher and higher resolution.
But just as a sphere shell is visible regardless of the resolution (despite it having no volume and being nowhere dense) I'm most certain that the 3d Mandelbox will not vapourise at higher resolutions.

It is very similar to cantor dust, and this picture shows how diagonal light rays (red) hitting cantor dust do not penetrate regardless of the resolution... the maximum penetration just gets closer and closer to reaching the middle.
0 volume but 100% opacity.

But as Timeroot said, in practice the 3d box might start to vanish as the rays will accidentally skip over the inside points.. so non-zero ray width would help with this I expect :)

That's only true when it's from that exact angle, though....

It only starts to penetrate when the box angle gets closer to horizontal... and the smaller the gap in the dust the closer to horizontal it can get without penetrating. The Mandelbox has very small gaps compared to standard cantor dust. Additionally the sphere inversion mixes up the crack orientations so gaps won't ever line up sufficiently allow this sort of penetration.

Not a bulletproof arguement but its enough to convince me that the Mandelbox would look much the same with 0 width rays (or infinite resolution) as it does with the renders we do now.

I think I've been thinking about fractals too much, my girlfriend heard me talk in my sleep yesterday, mumbling "its not very fractally" !

I'm convinced. Admittedly for certain angles, light would pass through perfectly, but an observer never views every point with the same angle. So the Mandelbox might look "thin" in some places but I don't think it would ever disappear. Regardless, though, the density poses a problem for rendering, because if these isolated points are missed in the steps (almost guaranteed without some sort of dynamic step size) then the Box WOULD disappear.

[q]I think I've been thinking about fractals too much, my girlfriend heard me talk in my sleep yesterday, mumbling "its not very fractally" ![/q]

:P Know the feeling...

Hi, here you can see a test showing mandelbox's slices.
A part of the same data set was used to create a .dxf file (with low-iterations).
I used mandelbulb3D and imageJ; using higher iterations creates more problems with the isosurface
(maybe because I'm using a 32 bit system).

http://www.youtube.com/watch?v=MDVhQJLPIm8 (http://www.youtube.com/watch?v=MDVhQJLPIm8)

From what i have seen in renderings with high iterations, it looked like the scale 2 and minR 0.5 box just
is no more solid, but if you decrease the scale and/or increase the minR value, it "looked" like getting more
and more solid in certain areas.
I cant proof this mathematical yet, but looking at the case for the vectorlength to be between minR and 1,
the spherical folding becomes scale/vectorlength, what leads to nonlinear gradients with the lowest value at 1,
beeing -1?
If this gradient is getting below 1, there might be possible areas that dont get stretched... but these are only
some first thoughts of what i discovered so far.

Edit:
Duh, i think i made a mistake in the scaling of the graphic for my tests, it seems that the gradient does not
goes below the scale value!
That does not affects the conclusion from what i have seen with high iteration counts, but i am totally out of
doing an analytical explanation.