Distance Estimated Menger Sponge Normal Calculation

[makc]

Like that is something hard to do

Using code from original thread:

Code:

bool mengerSponge(double x2d, double y2d) {
/*
 * 1) cutting plane goes through 0.5,0.5,0.5
 * 2) plane normal is 1/√3,1/√3,1/√3, therefore vectors 1,-1,0 and 0,1,-1 are good 2D basis
 */
double x = 0.5 + x2d;
double y = 0.5 + y2d - x2d;
double z = 0.5 - y2d;

if(x<0 || x>1 || y<0 || y>1 || z<0 || z>1 )     { return 0; } // point is not part of menger sponge

int iterations=8;

double depth=3;

for(int iter=0; iter<iterations; iter++) {
int holex=1;
while(holex<depth) {
int holey=1;
while(holey<depth) {
if(
((x > holex/depth && x< (holex+1) /depth) && (y > holey/depth && y< (holey+1) /depth)) ||
((y > holex/depth && y< (holex+1) /depth) && (z > holey/depth && z< (holey+1) /depth)) ||
((x > holex/depth && x< (holex+1) /depth) && (z > holey/depth && z< (holey+1) /depth))
) { return 0; } // point is not part of menger sponge
holey+=3;
}
holex+=3;
}
depth*=3;
}

return 1; // point is part of menger sponge
}

p.s. I happened to have this swf that renders random menger sponge cuts as you move your mouse around:
<a href="http://swf.wonderfl.net/swf/usercode/5/53/536d/536dfb5d15b061017191a4ec6bdcb2f40512deff.swf" target="_blank">http://swf.wonderfl.net/swf/usercode/5/53/536d/536dfb5d15b061017191a4ec6bdcb2f40512deff.swf</a>

[cKleinhuis]

try this, in 3d view you can rotate the axis, it is from jesse the formula combination, i changed it to menger3, can be useful for other (fractal) cuttings for sure

 

Code:

Mandelbulb3Dv18{
f.....S....O/...w....2....EKakjwGSH/.zhCqh4vvQ1Ew95Ldb3wOznIZrAY9mNazgiF6yJP9zuj
................................nmSHqnHisz1........Y./..................y.2...wD
...Uz./....0..../M.0/......T....n/....E3.....g873aGCHSqD/..........m/dkpXmH....U
z.....kD12../..........wz.................................U0.....y1...sD...../..
.z1...sDu9tznVdl5wf45nb/8QPLz25PQTOsA7qjpHkC9E7ZowvnE66HjfWMzEiwry.eOqpD3uwfwraP
bx1Kxs7KrRRAzwLG4yE4QJmDU.....Y/..............sD.6....sD.6E.....................
.............oAnAt1...sD....zw1.........................................A....k1.
.....m7lkz1.......kz.wzzz1.U..6.P....U5...EB....m....c3....F....6/...I1.....SF52
...U.qFG9yzb2zzzRYoWzz7lz16.pc..zXCc..kvrEtMc7xD6ocyFE0ujz1..........2.28.kFrA0.
.Ub96aAIVz9.1se7Umvxz0........../EU0.wzzz1...........s/...................E.2c..
zzzz.............0...................2./8.kzzzD............8....................
/EU0.wzzz1...................................2CcN/UvNPcveeWCNq0.yRiibHJJUk1f..XR
SvBmx3CcN/UvNPcvQsLsUa3.ibhVi1bTV1OK.sSq4uCly3CcN/UvNPcvMwLsUa3.ibhVinqTV1OK.sSq
4uCkz3CcN/UvNPcv..EsUa3feeWCNqGQIJ36wk8EwyLsUa3f................................
E....AU6V2E.....I....A....kLGx4RVFLN..kPrJaQ.........................MU/4.U/4M..
..................nO./......UV4E.......U4/2.....................................
................................................................................
.....................2.....3....0....wJRkFKMoJqE................................
..............................zD................................................
................................................................................
................................/....E/...U0....BJaPbJaQn..IjRLNm/..............
................4MU/4MU/................6.2........wz.........zD........kz1.....
................................................................................
............................................}
{Titel: Mandelbulb 3DckJRotate}

[makc]

Wait

Code:

plane normal is 1/√3,1/√3,1/√3, therefore vectors 1,-1,0 and 0,1,-1 are good 2D basis

It's not, vectors are at right angle to normal but not to each other So let's keep 1,-1,0 then 2nd vector must be cross with normal = -1,-1,2. Now we have vectors of different length, √2 and √6, so we also need to scale one: 1,-1,0 now turns into √3,-√3,0. In the end it gives
x = 0.5 + 1.732 x2d - y2d
y = 0.5 - 1.732 x2d - y2d
z = 0.5 + 2 y2d

I am sleepy, so I could still mess it up, but you get the idea, right?

[kram1032]

makc, that's really nice, but you're still using a 3D Menger sponge that you then cut with a plane.
I was wondering about a formulation that never even builds a 3D Version.

[makc]

I don't get it.

Suppose there is (squared) length of 3D vector, L1 = x*x+y*y+z*z. It's 3D, no problem here.

Suppose there is (squared) length of 2D vector, L2 = a*a+b*b. It's 2D, no problem here.

Now, I want to calculate L1, where x = a, y = b and z = 0. Is it 3D or 2D?

From my perspective, there is no difference, I mean I just substitute variables in some expression. What could be a problem with that?

[tit_toinou]

There is two fractals hiding in this image.
The first one is an hexagon that divides into 6 hexagons and 6 equilateral triangles.
The second one is an equilateral triangle that divides into 9 hexagons and 15 equilateral triangles (see my "Triangle de Menger" preview image).

[Geometrian]

So, I've gotten a new render (direct link) (Fractal Forums art thread), but it still has a few problems.

In particular, it is using the distance estimator to calculate shadows, which mostly works. However, there's still some nasty banding artifacts in some places that I'm pretty sure are erroneous. Ideas as to what these might be caused by?

[cKleinhuis]

Hey kram i think the pure 2d version should not be too hard to implement
i have made a 2d version of the kifs in ultrafractal

but for this cut i think it should be extended to a 3d cut slice with 3d rotation possibility right now i am busy with other stuff but i plan to make a review of the kifs soon

you can find my kifs 2d variant in the uf formula database

[kram1032]

makc, well, the 2D-sierpinsky carpet is built from squares. The 3D-menger sponge is built from cubes.
This cut-plane seems to be built from triangles with 6-pointed star shapes (that is, 12 smaller triangles in a certain configuration) being taken out of them.
I'm wondering about having a way to build it without using cubes that you then cut through at all.

[makc]

kram, well:



the whole thing is just a triangle without corners

[tit_toinou]

There is two fractals hiding in this image.
The first one is an hexagon that divides into 6 hexagons and 6 equilateral triangles.
The second one is an equilateral triangle that divides into 9 hexagons and 15 equilateral triangles (see my "Triangle de Menger" preview image).

Oops I was wrong (too complicated), an equilateral triangle divides into 1 hexagon and 3 little equilateral triangles, thx makc for the image.
@kram1032: I think my description is enough to create the fractal..

[cKleinhuis]

the triangle is split into 1 hexagon and 3 triangles, but into what is the hexagon split
i mean, it would be nice to have a nice formulation of this for the hexagon as well, right now i do not see it easily  
the rule seems to only rely on the star shape,

- start with a hexagon
- cutout the big star shape in the middle
- repeat with the six remaining hexagons

.... but i thing it is missing the "triangle sections ... there must be a 2 way rule...

[kram1032]

Well, you essentially fill a hexagon with smaller hexagons that touch each other and are alligned with the starting hexagon's corners.
However, alongside each edge, there will be a missing triangle that you need to fill in too.
Repeat as before for the hexagons but
note that the edge-triangles also has a new hexagon inside them, that you need to repeat the process with too, however
you again need to fill the new holes around the triangles in that iteration.

It's a bit more complicated than your typical Sierpinsky-type fractal because it needs some extra filling. I wonder if that's actually necessary though.

Equivalently you could probably also start from a triangle of which you cut away the corners to form a hexagon.
In either case, it seems to be a two-step iteration loop, rather than a single-step one. In that sense, it's more complicated than a typical 2D sierpinsky-type fractal...
Unless there is some way (other than going 3D and just cutting out the final result from an arbitrarily iterated menger sponge) to merge the two steps into one.
(Even cutting the final form out of an iterated 3D object seems like an additional step)

If you don't fill the missing triangles, you'll instead end up with something that looks like a Koch-flake- with Kock-flake-holes inside each Koch-flake-part:


On a different note:
Would anything change if you tried this with a menger-hypercube? Are there other interesting special cut-planes in that case?
What about interesting cut 3-planes, e.g. 3D cut-outs from 4D-structures?
Notably, that problem would be related to Quasicrystal-structures: If you cut a 3D space from an n-D periodic structure (n>3), you'll get all possible crystal structures if the 3D space is rotated with rational angles.
quasi-crystals happen if the angle is irrational (for instance the golden angle, which corresponds to a Penrose-tiling)
In case of a high-dimensional menger-cube it would obviously be a regular n-cube-lattice, cut in arbitrary angles. But other regular 4D structures might be tried just as well.