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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)? –  Daniel 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

1 Answer 1

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 i.stack.imgur.com/IOnhq.jpg, 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 mathworld.wolfram.com/TruncatedIcosahedralGraph.html 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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