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If I have a grid made of equilateral triangles, I can easily form larger convex polygons as a set of tiles in that grid. I believe this holds for some (but not all) tilings of non-equilateral triangles.

The same for quadrilaterals - it's obvious for squares, rectangles and parallelograms, but I believe it also holds for some other (but not all) tilings of quadrilaterals.

I'm pretty sure tiling a flat area purely with convex pentagons is impossible, so skip that case.

With hexagons, a tiled grid is possible, but the only way to make a convex polygon from those convex hexagonal tiles seems to be to use only one tile - no multi-tile convex polygons seem possible.

My speculation is that it's only possible to form a multi-tile convex polygon from convex polygon tiles if all those tiles have interior angles at all vertices of 90 degrees or less, at least for those vertices that are at a vertex or edge of that larger convex polygon.

Is that speculation correct? Is there a proof or disproof?

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2 Answers 2

Are you looking for items like the Hirschhorn Medallion?

Hirschhorn Medallion

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Very nice, but no. The idea started when I was reading a blog post about pathfinding in square grids, and I wondered how far I could generalise to different grids. I thought about an idea I had some time ago about defining cells for the pathfinding to work on where each cell was as large as possible while being convex (so any point within a cell could be reached from any other in one step - I also had a variant where any point could be reached from any other in two steps via a centre point). –  Steve314 May 23 '13 at 2:27
    
The two pathfinding ideas are separate anyway. Tthe whole point of defining convex cells is that there are no pre-defined cells, and the whole point of the two-step non-convex cells is to cope with e.g. circular barriers - if everything is defined on a square/whatever grid these should be non-issues. Basically, I was just going off on a random tangent. maybe thinking about optimising pathfinding on a grid by combining cells. Though this certainly shows a case where larger convex cells are possible, but can't be found by growing incrementally larger convex cells. –  Steve314 May 23 '13 at 2:37
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Sorry - I figured it out within a minute of posting. Should have thought longer before asking.

If two polygons are joined such that such that a vertex and edge are on the same point and line, there are two cases...

  • If the sum of the two interior angles is greater than 180 degrees, this is a concavity. It's only valid as part of a larger convex polygon if that concavity is filled by more tiles.

  • If the sum of the two interior angles is less than or equal to 180 degrees, there is no concavity. This vertex may be at the vertex or edge of a larger convex polygon formed from these tiles.

So the issue is the sum of the angles, not the interior angle at the vertex of one particular tile.

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