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Is it possible to tile the plane with regular polygons such that every edge is one of two colours, and no two adjacent edges are the same colour?

Tobias Nash
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    Do you define edges as adjacent if they share a polygon or just a vertex of the tiling? The answer is nearly trivial either way, but it makes the difference between a trivial 'yes' and a trivial 'no'... – Steven Stadnicki Apr 18 '15 at 00:34
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    (Trivial yes: tile the plane with squares and set all vertical edges to be color A and all horizontal edges to be color B. Trivial no: pigeonhole the >=3 edges that must meet at any vertex and note that some two of them must share a color) – Steven Stadnicki Apr 18 '15 at 00:35
  • @StevenStadnicki I don't know, the vertex condition makes it impossible so it has to be the sharing a polygon condition. Are there any other possibilities apart from squares, maybe using multiple polygons? – Tobias Nash Apr 18 '15 at 00:39
  • Tobias: all your polygons need to have an even number of sides (since the colors must alternate around any given polygon) and all of your vertices need to have an even number of edges off of them (for the same reason). None of the regular or semiregular tesselations other than the square tiling satisfy this condition. – Steven Stadnicki Apr 18 '15 at 00:43
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    (Also, an easy proof: the evenness conditions imply that there must be >=4 polygons meeting at each vertex, and that each polygon has >=4 sides (and thus subtends an angle of >= 90 degrees at each vertex). The only way to make the angles fit at a vertex is to have four polygons with 90-degree angles - i.e., the square tiling.) – Steven Stadnicki Apr 18 '15 at 00:50
  • @StevenStadnicki What about tiling the surface of a sphere? – Tobias Nash Apr 18 '15 at 09:35
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    The sphere has less 'room' around each vertex than the plane does, so the same argument goes through and shows that there are none on the sphere. The hyperbolic plane does have such tilings, and they're not hard to come by: just consider the (6,4) tiling of the hyperbolic plane, for instance, and note that alternate edges of each cell can be colored with alternate colors. – Steven Stadnicki Apr 18 '15 at 15:36

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