More smoothed-diagonal honeycomb thoughts

[sonata_green]

I've installed Emacs, git, and wc-mode. My $100 Windows laptop is gradually learning to pretend to be a real computer.

A cubic honeycomb meets four cells around an edge, and eight at a vertex. I think (I might be wrong) that a rhombic dodecahedral honeycomb meets three at an edge, six around one type of vertex, and four around the other type of vertex.

If I want an odd number of cells around a vertex, then it seems to me that the arrangement of cells must be described by a polyhedron with an odd number of faces (each face corresponding to the vertex figure of the relevant vertex of one of the cells). (This is not the same operation as forming the dual honeycomb; the cubic honeycomb has a 🐛vertex polyhedron🐛 of an octahedron, while its dual honeycomb is another cubic honeycomb that's syncopated relative to the original.)

What about hexagonal prisms, arranged in a face-centered cubic lattice? No, those meet four (or more) at a vertex. In particular, three hexagons are met by a fourth on the next layer. Syncopated cubes give five to a vertex by the same logic, but where edges cross we necessarily get four cells meeting, so the syncopated-prism strategy in general is out.

Can we make a honeycomb with a vertex polyhedron of a triangular prism? This requires three cells with square vertices and two with triangular vertices.

We've come a long way from the original problem, though. We're only interested in vertices in the context of a half-honeycomb that meshes with its rotation or mirror image. Furthermore, the problematic case with the square tessellation was opposite corners matching across a diagonal (▚); if there's a two-and-two with matching corners being adjacent, the result is fine.

In the rhombic dodecahedral honeycomb, the four-cell vertex has a tetrahedral arrangement, where any two cells are necessarily adjacent. (You can cut a tetrahedron in half by bisecting four of its six edges, making a square cut-face. Each half has two of the original four vertices.) The six-cell vertex's vertex polyhedron is a cube; for three-and-three, there are two possible arrangements: each three can meet each other around a single vertex, or they can make a C shape (the Cs are making out sloppy style). Each of these lends itself to a fairly straightforward choice of plane of bisection, though in the latter case we bisect four of the cube's faces, so it's not as simple as "this corner of the rhombic dodecahedron is a certain color".

[440/250 words, running total 758/500.]

Comments

[ilzolende]

I'd love to hear more about your cheap Windows laptop, if you'd like to post more about it.

[sonata_green]

I don't know that there's much to tell. I got it refurbished; it's an Ideapad 1i with 114G disk space, 4G ram, 1.11 GHz cpu, and Windows 11. It can run Steam, KSP, The Sims 4, and Uplink (maybe not all at once). I'd have to go digging through old emails to figure out where I bought it.