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Is a "nice" plane tiling possible where each tile has 7 (8, 9, ...) neighbors?

With "nice" I mean:

  • The tiling is (preferably) periodic.
  • The tiles are from a finite set
  • The tiles themselves are "nice" (non-degenerate, no holes, connected). It's OK if the tiles are not convex.

This seem to be a simple question, but I lack the terminology to do a proper search.

Are there general results of tiling possibilities in terms of number of neighbors that I can look at? (For example, if I want to know whether a tiling exist where each cell as $m$, $n$, ..., or $p$ neighbors.)

(I have seen this question Why a tesselation of the plane by a convex polygon of 7 or more sides is not possible?, but this is not quite what I am interested in).

(Background: I am the author of a Grids package that allows programmers to set up various types of grids for game programming. Once customer asked whether we will support octagonal grids in the future, and I wondered whether such a grid is even possible).

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Do you need each tile to share the same number of neighbours? For instance, is a tiling by regular octagons and squares 'filling in the holes' a suitable tiling where tiles have 8 neighbours, or does it only count as 4 (or something else)? – Dan Rust May 18 '14 at 19:53
Yes, I'd like each tile to have 8 neighbors, although I also wonder more generally what neighbor configurations are possible. (I also know about the tiling with octagons and squares, but that would be too easy!) – Herman Tulleken May 18 '14 at 23:55
up vote 1 down vote accepted

Suppose it is possible. Take a large region of tiles. Put a frame around it, and then make a new framing polygon that touches all the outer tiles.

Map this map onto a sphere, and add a point to the framing polygon so that's it's unpunctured.

From there, we have a polyhedron. If all the tiles have 5 or 6 sides, then there will be exactly 12 pentagons via Euler's V+F-E=2. The Fullerenes enumerate varying numbers of hexagons.

Sadly, it's not possible for all the polygons to touch 7 or more others on this sphere. That overloads V+F-E=2. It's possible to use only heptagons on surfaces of higher genus. For example, the klein quartic uses 24 heptagons on a three holed torus.

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Thanks for the answer. I follow the last part (from the polyhedron), but I am a bit unclear about the mapping to a sphere part. (I know the bear minimum of this field and what may be totally obvious is very mysterious to me!) For example, the same argument (as far as I can understand it) must apply to to this tiling, which obviously is not "nice" by my definition, but nevertheless is possible... What bit of information am I missing? – Herman Tulleken Jun 3 '14 at 20:23
For that image from the hyperbolic heptagon tiling, you'll have an outer area with a lot of polygons that don't touch at least 7 others. – Ed Pegg Jun 3 '14 at 20:55
I marked your answer as correct, as it does answer it. I still don;t understand fully why (how exactly the mapping is suppose to work) but I think I need to do some more reading first. At least now I can stop filling my notebooks with futile pictures of heptogonal grids! – Herman Tulleken Jun 4 '14 at 17:02
OK, after giving it some more thought, I convinced myself of the truth of what you say. It's a bit easier to use the dual graph, and then imagine a mapping onto a torus (assuming a periodic tiling, and taking the "unit patch" to map to the torus), and using $V - E + F = 0$. Not much different from your answer (but I still don't understand the sphere mapping properly :P) – Herman Tulleken Jun 6 '14 at 7:24
See also Those are sphere mappings. You need to have one polygon on the outside. from its vertices, you connect to the vertices of polygons on the inner part that expose 2 sides. With Heptagons, you'll always have outer polygons with three sides exposed. – Ed Pegg Jun 6 '14 at 14:15

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