0xDE ([info]11011110) wrote,
@ 2006-10-28 00:34:00
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Entry tags:color, geometry, symmetry, tiling

A partial answer to my periodic tiling coloring question
A few weeks ago I asked whether periodic tilings of the plane have periodic colorings with four colors, and if so what the relation is between the periodicity of the coloring and of the tiling. As discussed in the comments there, another way of asking the same question is, if a graph is embedded in the torus, is there a finite cover of the torus such that the lift of the graph to the cover needs only four colors?

I still don't know the answer to the question for four colors, but the answer turns out to be yes for five colors. Albertson and Stromquist [PAMS 84:449-456, 1972] showed that if a torus graph has no noncontractable cycles of length less than eight, it's five-colorable. And it's easy to find a finite cover of a torus in which any noncontractable cycle has to pass across eight copies of the original torus, hence have length at least eight. Or, put another way, for any periodic tiling of the plane, if you make the pattern of repetitions big enough so that each tile in the tiling is eight or more tiles away from its nearest repeated copy, you can use only five colors in a periodic coloring.



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finite and infinite tilings
(Anonymous)
2008-02-20 08:53 pm UTC (link)
Hi Prof. Eppstein,

Scott Aaronson and I have a question re tilings. We know that for discrete-lattice tilings of the plane by a fixed finite tile set (i.e. squares with edge compatibility-constraints), an infinite tiling is possible iff arbitrarily large squares can be tiled.

Is this still true if we move to tiling by general polygons, or even more general tiles? The original argument based on compactness (or completeness or Konig's lemma) seems to break down.

OK if you don't know... Thanks a lot!

-Andy Drucker

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Re: finite and infinite tilings
[info]11011110
2008-02-20 09:09 pm UTC (link)
If you allow more than finitely many tiles in your set, it's not true. The König's lemma based proof breaks down (because the tree can be infinite by having infinitely many children at a node rather than by having an infinite path), and there are sets of tiles, even sets of tiles with diameter bounded above and below uniformly, such that arbitrarily large squares may be covered but the whole plane may not.

For instance, let S = {regular heptagons with side length 2^i}. Clearly, S covers any finite square, but cannot cover the whole plane because there's no way to surround a single vertex of a heptagon. This doesn't have bounded diameter, but let S' be formed by partitioning each heptagon of S into bounded-diameter pieces that interlock in such a way that they can only be rejoined to form S again...

If there are finitely many polygonal tiles (or possibly more general shapes, I'm not sure) then the König's lemma argument still works. Make a tree in which the children of the root are the partial tilings with some vertex of one of the tiles at some fixed orientation at the origin, and in which the children of any other node are the ways of placing one more tile so that it covers the points along one of the edges of the previous partial tiling at the nearest not-completely-covered point of the plane. It has finite degree (there are only finitely many choices of which polygon to place to cover that edge) but infinitely many nodes (from the assumption that arbitrarily large patches may be tiled), hence an infinite path. An argument about area of tiles vs the area of a disk within any fixed radius shows that an infinite path must correspond to a tiling of the whole plane.

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(Anonymous)
2008-02-20 09:45 pm UTC (link)
Thanks... I wondered along these lines, but I don't see why the next tile placed on the tiling-so-far has to completely cover any edge. Or have any of its edges completely covered. For instance, I don't see how this finite-branching process could 'discover' the standard tiling of bricks. (though it's true that bricks have other tesselations.)

I do agree that something like this should prove the result, I just don't see it.

-Andy

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[info]11011110
2008-02-20 09:48 pm UTC (link)
You have a point. At least it does cover tilings that are edge-to-edge and vertex-to-vertex, though.

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