# Are there higher-dimensional tessellations touching only nearest neighbours?

One property of a hexagonal tiling is that each hexagon only touches its nearest neighbours - in contrast to e.g. a square tiling, where each corner also connects to a second-to-next neighbouring square. But does a higher dimensional generalization of this exist (with only one type of hyper-polygon/polyhedron/poly...?)?

• for 3d I think it's the dodecahedron.. I'd guess yes but idk Feb 28, 2017 at 20:38
• Truncated Octahedra might do what you want ? Feb 28, 2017 at 20:46
• @DonaldSplutterwit Sounds about right in 3d, thanks Mar 1, 2017 at 19:23
• Mar 1, 2017 at 19:23
• (Probably not too related, but a nice find: math.stackexchange.com/questions/36834/…) Mar 1, 2017 at 19:29

Any answer will depend on what you mean by "nearest neighbor". If we assume that the hyper-poly has a well-defined "center" point (which would be for example its center of mass) then yes, there are other examples.

One example in 3-space is to "tile" the volume by hexagonal right prisms, of hexagonal side $1$ and prism height $\sqrt{2}$. The strategy is to tile each plane in the usual hexagonal pattern, and stack the planes. Of course, if you were to stack the planes with the centers vertically below one another, then eax tile would touch not only its two nearest neighbors (distance $\sqrt{2}$ but also its twelve next-nearest neighbors, with which it shares an edge (distance $\sqrt{5}$. But if instead you put the centers of plane $2$ directly above the vertices of plane $1$, then each tile will touch twelve other tiles, each of which is $\sqrt{3}$ in center-to-center distance.

This same idea can be extended to higher dimensions.

By the way, you can't tile the plane with squares such that each only touches its nearest neighbor. But you can with rectangles of side ratio $2:\sqrt{3}$.

You canot tile 3-space (or any higher dimension) with one platonic solid, without some tiels touching non-nearest neighbors.

• Nice idea. Probably not the most efficient one (whatever "efficient" may mean here, maybe something along the lines of another question), but very straightforward to generalized. Meanwhile I think my question is answered by answering the Kissing number problem, which unfortunately isn't solved for many dimensions (but for 24d‽) Mar 1, 2017 at 19:28

I think you are looking for Rhombic dodecahedron. It can fill 3D space and has "similar" properties to the 2D hexagon. You can find out more on Wikipedia.

• Thanks, that's one I also found, but via the corners they touch other non-nearest cells, so it's not exactly what I'm looking for. Mar 1, 2017 at 19:26

In arbitrary dimension -

(1) Here is an example that satisfies your criteria, I think:

The (equilateral, most symmetric) diplo-simplex with 2N+2 vertices tiles space in 2D, 3D, and 7D.

2D diplo-simplex is hexagon, vertices of tesselation correspond to non-lattice hexagonal packing.

3D diplo-simplex is cube, vertices correspond to trivial cubic lattice.. doesn't the cubic lattice in arbitrary dimension satisfy your criteria?

7D diplo-simplex tesselation's vertices correspond to E7* lattice, i.e. the sole Delaunay polytope of the E7* lattice is the 7D diplo-simplex.

(2) Furthermore, in 4D:

The 16-cell and 24-cell tile space in 4D. The vertices of the 16-cell tesselation/honeycomb are the self-dual D4 lattice and the vertices of the 24-cell tesselation/honeycomb are a non-lattice packing, I think. The 24-cell is the sole Voronoi polytope and the 16-cell is the sole Delaunay polytope of the D4 lattice.

(3) In 6D,

the 2_21 Schlafli polytope. 2_21 is the sole Delaunay polytope of the E6 lattice.