# Fractal Tiling of Rhombic Dodecahedra

Hello, this is my first question on Math Overflow...

Rhombic dodecahedra can be tiled in 3-space, leaving no gaps. This tiling corresponds to the close-packing of spheres.

Consider a "nucleus" rhombic dodecahedron ("rd") surrounded by 12 rd's, such that each of the faces of the nucleus rd is shared with a face of one of the 12 outer rd's. This becomes a composite polyhedron (call it "rd1") which can be described as the union of 13 rd's.

My question is this: can rd1 then be used as a tile for a similar tiling of 13 rd1's? (call it "rd2") - leaving no gaps and no overlapping? (You can think of this as the 3D version of tiling 7 regular hexagons, and then tiling those 7 to form a shape comprised of 49 hexagons.

The next question is: can this process be applied forever? (rdn). If rdn is scaled during each iteration such that it has the same volume as the original rd, then at the limit the resulting shape could be described as a "fractal rhombic dodecahedron" - having a fractal surface. But I don't know if such a thing can exist.

I am wondering if there is a way to arrive at a proof or a strong conjecture that such a fractal tiling ("pertiling") can exist. What kinds of geometry tools should I use to arrive at an answer?

Thanks!

-Jeffrey

-
Actually I do not want several rd's to tile a larger rd. Nor do I want rd's to tile to create a convex body, as you mention. I fully expect the boundary to be fractal, and highly irregular, with infinitely many convex and concave regions, if done in the way I describe in my question. Check out some examples of regular hexagons tiling progressively to make a "Gosper Island". This will give you an example of what I am looking for in 3 dimensions. I would love some kind of proof that what I want to achieve is either possible or impossible ('possible' would be most awesome :) -Jeffrey –  Jeffrey Ventrella Jun 5 '12 at 4:02
I think $\langle (3,-1,0),(0,3,-1),(-1,0,3) \rangle$ works. –  Douglas Zare Jun 5 '12 at 7:22
The tile in that paper has dodecahedral symmetry not the symmetry of a cube. The rhombic dodecahedron has the symmetry of a cube. –  Douglas Zare Jun 5 '12 at 20:58
The way I usually think of a rhombic dodecahedron is that it is between the convex stellated cubes and the nonconvex stellated cubes. –  Douglas Zare Jun 6 '12 at 1:20

Yes, although the process can't be quite as symmetric as it is in 2D.

Let's first try a simpler version of the question to warm up:

Cubes tile 3-space, leaving no gaps.

Consider a "nucleus" cube surrounded by 6 cubes, such that each of the faces of the nucleus cube is shared with a face of one of the 6 outer cubes. This becomes a composite polyhedron (call it "cb1") which can be described as the union of 7 cubes, or as a 3D plus sign.

We can see that this cb1 object is roughly octahedral in some sense. (The 2D cases of squares or hexagons was simpler in that regular n-gons are their own duals.) Unfortunately, unlike the approach used in the 2D cases of 5 squares or 7 hexagons, in 3D we cannot align a slightly twisted octahedron so that its vertices move onto vertices of the cb1. Even if we could, it wouldn't help, since octahedra don't tile space.

Nevertheless the cb1 can tile space in a regular lattice with 6 "nearest" neighbors offset by the cyclic permutations of $\pm(0,2,-1)$. (Actually, lattice member $(1,1,1)$ is closer, but I want to emphasize that the lattice of cb1 centers is a cubic lattice again, just squished in the (1,1,1) direction by a factor of $\sqrt{7}$.) Each layer looks like this: (+/- indicates center on next/previous layer)

+ - . * * * . + - . * * * .
. * + - * . . . * + - * . .
* * * . + - . * * * . + - .
- * . . . * + - * . . . * +
. + - . * * * . + - . * * *
. . * + - * . . . * + - * .
. * * * . + - . * * * . + -
+ - * . . . * + - * . . . *
* . + - . * * * . + - . * *
. . . * + - * . . . * + - *
- . * * * . + - . * * * . +
* + - * . . . * + - * . . .
* * . + - . * * * . + - . *
* . . . * + - * . . . * + -


Now we want to combine these into some kind of "cb2" clusters. If we just consider the cb1 centers, we see (as mentioned above) that we have a cubic lattice, squished in the (1,1,1) direction. So we can make clusters in this lattice just like we did in the initial lattice, attaching 6 neighbor cb1s to a central 7th cb1. The process can be repeated ad infinitum, yielding the desired type of structure, except that the (1,1,1) direction gets progressively more squished at each stage. To avoid this, you could pick a different direction to squish at each level, so that the eccentricity of the object never gets too bad.

So with cubes it works, with the caveat that the eccentricity changes at each level.
So how about with rhombic dodecahedra?

In this case, the rd1 is something like the dual of a rhombic dodecahedron, which doesn't tile space, and again, since we are in 3D, we won't be able to "twist" a tiling-capable polyhedron so its vertices fall onto vertices of the rd1, one per attached rd0. So again, we can't do an exact analog of the 2D case.

But as we saw with the cubes, if we can find some tiling of space by rd1 objects, then we can hopefully look at the lattice of centers of those objects as a cubic or rhombododecahedral lattice (the difference between the two being only a factor of 2 in the (1,1,1) direction), and use that to repeat the original tiling at higher and higher levels.

Here is one simple way to do it: $\ \ \$ (+/T indicate center on next layer, -/= indicate center on previous)

+   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .
+   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .
-   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .
=   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T
.   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T
*   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T
.   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =
*   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -
T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *
+   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .
-   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *
-   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .
-   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +
.   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T
.   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =
.   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =
T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =
T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *
T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *
-   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *
-   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +
-   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +
.   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +
.   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =
.   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =


In this pattern each rd1 cluster touches 14 others, but 2 of them just barely. Leaving those two out, the 12 neighbors are at $\pm(3,-1,0), \pm(0,-3,1), \pm(3,2,-1), \pm(2,2,2), \pm(-1,3,2), \pm(-1,0,3)$. The lattice of rd1 centers is indeed the sort of lattice we were looking for, with basis vectors $2\vec{x}=(2,5,1), 2\vec{y}=(2,-1,3), 2\vec{z}=(-4,1,3)$. Using these basis vectors, we can draw the lattice of cluster centers as a cubic lattice, and make "rd2" clusters of clusters using the same pattern as above, and so on up the hierarchical levels.

The resulting objects do have the flavor of a Gosper Island, although in 3D I would call it an asteroid.

-

Thanks for the ideas, folks!

Douglas: exploring the lattice of index 26 is a good idea. I can certainly draw pictures (or make 3D models) to determine if 13 rd1's can indeed tile to form a rd2 in the same way that 13 rd's tile to form a rd1.

If this is NOT possible, then I have my answer But if it is possible, then I would need to determine whether this process would work for the next level. And the next, and the next.

Also, the following illustration of the formation of the Gosper Island gives an example of the process I am thinking of - only I am extending it into 3D. http://mathworld.wolfram.com/GosperIsland.html

Thanks! -j

-