by **wendy** » Wed Nov 28, 2018 9:03 am

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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