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I read that among the 24 heptiamonds there is one piece that does not tile the Euclidean plane. My question is the following, given a particular polyiamond how do you prove that the piece does tile the Euclidean plane? ( I thought about arranging the pieces and possibly stretching the piece to a form of which it is known that it tiles the plane, i.e. triangle, square, parallelogram or hexagon. )

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The V heptiamond will not tile the plane. All others will.

The proof of tiling is usually to find a tiling. That's often very simple for the smaller iamonds, polyhexes, and polyominoes.

For larger polyforms, such as the wheelbarrow polyhex, finding the tiling can be quite tricky. See the tiling database for more.

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Can you be more precise about when a tiling has been found. What is the exact criterium for that? Imagine an algorithm doing so, what would be the stop rule? –  ndroock1 Jan 23 at 10:35
    
Octiamond tilings -- When a repeated tiling with translation is found, you've divided the plane into identical cells with identical orientation. Each cell will have some number of of polyforms. –  Ed Pegg Jan 23 at 14:27
    
That's a nice page, there is one about the heptiamonds too. Anyway, I had hoped that there was a way to a priori say, from its properties, that the V-shape does not tile the plane. –  ndroock1 Jan 23 at 15:07
    
With the V shape, lay down one. To fill the hole, the next piece is forced. To fill that hole, the next move is forced. A few more forced moves gives unfillable holes. Generally, the non-tilers cannot surround themselves, the Heesch problem, but there are many non-tilers that are Heesch-2. –  Ed Pegg Jan 23 at 17:22

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