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Mar 10, 2021 at 21:23 vote accept Hans-Peter Stricker
Aug 18, 2020 at 17:54 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 16:15 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 15:46 comment added M. Winter @Hans-PeterStricker Okay, uniqueness is a lie. Imagine that you can build finite graphs of type $(3.3)^3$ by wrapping around the tiling to a torus as you did with $(4)^4$. The Local Theorm assumes that the "topology is simply connected" (e.g. sphere, Euclidean/hyperbolic plane). Similar wrapping might be possible in the spherical and hyperbolic case. I am not sure what this does to vertex-transitivity, but I am optimistic that it is preserved. I might edit the answer later.
Aug 18, 2020 at 15:46 comment added Hans-Peter Stricker Note that $(3.3)^3 = (3)^6$. Note further that $(6)^3$ gives rise to the hexagonal tiling (which you know for sure).
Aug 18, 2020 at 15:41 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 15:36 comment added M. Winter @Hans-PeterStricker The graph with configuration $(3.n)^m$ exists, is unique and vertex-transitive. I edited my answer.
Aug 18, 2020 at 15:35 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 14:33 comment added Hans-Peter Stricker And thanks a lot for the counter example! This was my impression, too: that it is tricky to check a configuration for realizability. By the way: I always tended to consider only "tame" configurations of the form $(3.n)^m$ - these should work. Or do some of them don't pass your rule-of-thumb test?
Aug 18, 2020 at 13:49 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 13:38 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 13:31 comment added Hans-Peter Stricker Thanks for the hint to the Dolbilin-Schulte paper!
Aug 18, 2020 at 13:30 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 13:26 comment added Hans-Peter Stricker As I do .......;-)
Aug 18, 2020 at 13:25 comment added M. Winter @Hans-PeterStricker Sadly no, I always do it by hand.
Aug 18, 2020 at 13:25 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 13:20 comment added Hans-Peter Stricker Do you have a tool at hand that draws graphs for given configurations?
Aug 18, 2020 at 13:19 history edited M. Winter CC BY-SA 4.0
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Aug 18, 2020 at 13:15 comment added Hans-Peter Stricker Thanks, your answer is relevant in two respects: First, it shows that a graph that realizes a configuration doesn't have to be vertex-transitive. (That's important for me to know, and it answers question 3) Second it shows that there are (of course) configurations that tile only the sphere (defining a polyhedron) but not a plane. This case was excluded in the remark to question 1.
Aug 18, 2020 at 13:12 history answered M. Winter CC BY-SA 4.0