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Any idea how to prove that when 3D space is tiled with truncated octahedra, all vertices can be colored black and white such that no two vertices, sharing the same color, are adjacent?

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Is this homework? – Qiaochu Yuan Dec 3 '10 at 22:26

The graph of a truncated octahedron is bipartite but there are space fillers with regular polygons as faces which are not bipartite.

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What I need is a proof that the graph of the truncated octehedron is indeed bipartite. – user4389 Dec 4 '10 at 0:12
@victor Draw a copy of the graph of the truncated octahedron. Now try to apply the definition of what it means to be bipartite. You might also want to think about the fact that the graph of the truncated octahedron can obtained from the the graph of the regular octahedron. – Joseph Malkevitch Dec 4 '10 at 1:38
What I mean is the graph obtained when tiling space with truncated octahedrons, not just a single one of them. – user4389 Dec 5 '10 at 18:11

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