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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).

RavenclawPrefect
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Herman Tulleken
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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

1 Answers1

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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.

Ed Pegg
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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 https://i.sstatic.net/IOnhq.jpg, which obviously is not "nice" by my definition, but nevertheless is possible... What bit of information am I missing? – Herman Tulleken Jun 03 '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 03 '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 04 '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 06 '14 at 07:24
  • See also http://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 06 '14 at 14:15
  • Ah thanks for the link; the images immediately make it clear. (I totally had the wrong visualisation, and feel a bit stupid now). Thanks again for your answer too (I see now it is trivial), it has set me on the right track. (I discovered many interesting and useful pages armed with better search terms!). BTW the reason for me not upvoting is because of my low cred. – Herman Tulleken Jun 07 '14 at 17:04
  • I don't see why this necessarily forms a polyhedron? Certainly there are some polygon configurations which don't map onto the sphere. (I'm also not sure if "adding a point" is supposed to turn the extra faces into triangles meeting at a point or not.) – RavenclawPrefect Jun 05 '21 at 19:05