Are there higher-dimensional tessellations touching only nearest neighbours?

Question

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...?)?

Answer 1

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.

Answer 2

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.