Wikipedia:Reference desk/Archives/Mathematics/2006 August 8

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Domino tiling[edit]

Dominos are black tiles of size 2 by 1. I know that the number of ways to tile an area of size 2 by n with these dominos is the (n+1)th Fibonnacci Number (depending on the starting point) and this can be proven quite easily using recursions. However, how many different ways are there to tile an m by n rectangle with dominoes? --AMorris (talk)●(contribs) 08:08, 8 August 2006 (UTC)[reply]

For every fixed value of m you can give a system of recurrence relations for varying n, just like for the 2 by n case (but increasingly more complicated as m increases), and use that to give a closed formula depending on n. As far as I know, no-one has succeeded in doing something similar with both m and n as parameters. --Lambiam ^Talk 09:38, 8 August 2006 (UTC)[reply]

This is called the "dimer problem" (which I can't find in Wikipedia). There is no simple closed solution, but Kasteleyn found a determinant formula that leads to a pretty double product formula. This survey of the dimer problem has the formula at the start of section 5. McKay 10:42, 8 August 2006 (UTC)[reply]

However, the pretty double product formula that McKay referenced only applies to cases where M and N are even. StevoX 12:55, 8 August 2006 (UTC)[reply]

Why do you say that? Experimentally it seems to work for all sizes. McKay 03:26, 9 August 2006 (UTC)[reply]

Oops- I misread the formula in this paper/lecture by Richard P. Stanley and Federico Ardila to only work with even numbers due to the statement that it solves for tilings of 2M x 2N. I wrongly assumed that M and N had to be integers. --StevoX 04:15, 9 August 2006 (UTC)[reply]

I read that part of the paper but i can't see how it would work for a rectangle of dimensions 3 by 4, because in this case m = 1.5 and how do u take the product from k=1 to 1.5? And if both 2m and 2n are odd, does the answer come out as zero? Otherwise it's a pretty damn interesting formula especially the fact that it yields an integer value with all those cosines. Thanks. --AMorris (talk)●(contribs) 06:49, 9 August 2006 (UTC)[reply]

The formula given by Stanley-Ardila is less general. Look at the version I referenced--it works for all sizes and even correctly gives the value 0 if both dimensions are odd. McKay 11:58, 9 August 2006 (UTC)[reply]

The OP will no doubt be interested in the work of Jim Propp and others on Aztec diamonds and other topics connected with random domino tilings; see this page. Propp is well known as coinventor with David Wilson of an algorithm for perfect sampling, e.g. for a simulation of a near-critical Ising model. Hmm.. lotta red links there, which seems strange since the Propp-Wilson algorithm is more or less the centerpiece of a recent undergraduate textbook in the highly regarded London Mathematical Society student text series (volume 52):

Häggström, Olle (2002). Finite Markov Chains and Algorithmic Applications. Cambridge: Cambridge University Press. ISBN 0-521-89001-2..

I'd agree with the implied suggestion that the dimer problem deserves a Wikipedia article. (In the interests of ridiculously complete disclosure: I know Jim Propp, but then the theory of tiling is still a pretty small world, so this is hardly surprising!)---CH 02:20, 12 August 2006 (UTC)[reply]