By Bjorner A., Stanley R.P.

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In this figure the torus is cut up and flattened out — to get back the original torus one has to roll this flattened version up and glue together the two sides marked 1-2-3-1, and then wrap around the cylinder obtained and glue together the two end-circles marked 1-4-5-1. Note that the two circles 1-2-3-1 and 1-4-5-1 in Figure 8 correspond to the circles marked a and b that are drawn with dashed lines on the torus in Figure 7. 1 2 3 5 1 5 6 7 4 4 1 2 3 1 Figure 8: A triangulated torus Having thus cut the torus apart we now have a collection of 14 triangles.

In other words, given a tiling of AZn+1 , we can reconstruct which of the dominos were shuffled from a tiling of AZn and thus also the n + 1 2 × 2 squares that were left over. Since there are exactly 2n+1 tilings of AZn+1 associated with each tiling of AZn , we obtain the recurrence az(n + 1) = 2n+1 az(n). The unique solution to this recurrence satisfying az(1) = 2 is easily seen (for instance by mathematical induction) to be 1 az(n) = 2 2 n(n+1) , proving equation (24). 49 Figure 6: The hexagonal board H(2, 3, 3) 8 Tilings and plane partitions.

1966), Gregory John Kuperberg (b. 1967), Michael Jeffrey Larsen (b. 1962), and James Gary Propp (b. 1960). Their work has stimulated a flurry of activity on exact and approximate enumeration of domino tilings, as well as related questions such as the appearance of a “typical” domino tiling of a given region. , up to 2n squares in the nth row. Then reflect the diagram created so far about the bottom edge and adjoin this reflected diagram to the original. For instance, the Aztec diamond AZ3 looks as follows: Let az(n) be the number of domino tilings of the Aztec diamond AZn .

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