## Almost-uniform tilings?

Higher-dimensional geometry (previously "Polyshapes").

### Almost-uniform tilings?

Here's something that I found out recently.

In hyperbolic plane, we can put, say, pentagons and triangles together in (5,5,5,3) tiling. But that tiling is not uniform -- you can't make it so each vertex looks the same.

Surprisingly, it turns out that when you double this tiling (six pentagons and two triangles per vertex, arranged as (5,5,5,3,5,5,5,3)), there IS a way to make it uniform. The key is that two (5,5,5,3) "halves" that form a vertex do not have to be identical. So, does that mean that there exists an "almost-uniform" version of (5,5,5,3) that is still periodic, just has two distinct classes of vertices instead of one class?
Marek14
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### Re: Almost-uniform tilings?

The only polytope answering to a vertex-cycle of (5,5,5,3,5,5,5,3) is the archiform: "(1) (2) 3(3,4) 5(5,6) 3(7,8)", which reduces to (s,s,s,3,s,5,s,3), where s are snub-faces, here a pentagon.

The reason that i doubt that this could be dismantled into something less by using an unwrap as proposed, is that the double-cycle above is a snub, and is not symmetric on both halves, unless you draw the line through the pentagon (5,6), and the snub face between the digons at (1) and (2).

I an not sure how significant this is, but (5,5,5,3) can be constructed with horocyclic segments, specifically of orders 3, 7, ... [alternating lucas-numbers], whereas anything that is finite, and constructed by conway's archiform rules necessarily can not have such segments. I don't know how this fares in the wrapping.

Conway has given us a very good test for uniform polytopes. We can suppose a 'quasi-uniform' tiling has vertecies with the same ordering, up to reversed, but not a larger scale uniformity. Thus any of the vc = (5,5,5,3) would be quasi-uniform, but not uniform. This lines up quite nicely with quasi-crystals, which have fragments of a symmetry like {5,10/3}, but breaks down on the larger scale.
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wendy
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