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Because of Euler we know that for polyhedra without holes we have

V+F=E+2. As a result we can't have a polyhedron made entirely of

hexagons, and need at least 12 pentagons or an equivalent set of

shapes. Etc.

But things change if we allow intersecting faces. Clearly we have

a lot of scope for variations, but we can always retain limits to

make things reasonable.

But ...

Can we make a "polyhedron" consisting only of hexagonal "faces"?

@bremner I didn't know that, thanks. Someone has also pointed me at this:

@bremner Yeah, we know that with flat, convex, non-interesting faces, and something without holes, you can't have all hexagons. So the question is, what conditions do we relax? What is the *minimum* amount of relaxation we can do?

It's an odd question, but there's something there.

David Bremner@bremner@mathstodon.xyz@ColinTheMathmo I don't know the answer, but I remember that Branko Gruenbaum was very interested in polyhedra where regularity conditions are maintained at the potential expense of allowing faces to self intersect.

Maybe you know all this...