I’m looking for algorithms to generate randomized instances of Latin squares.

I found only one paper:

M. T. Jacobson and P. Matthews,

Generating uniformly distributed random Latin squares, J. Combinatorial Design 4 (1996), 405-437Is there any other method known which generates truly random instances not just isomorphic instances of other instances?

**Answer**

In the combinatorics community, the Jacobson and Matthews approach is widely considered the best approach to obtain random Latin squares (approximately) uniformly at random from the set of all Latin squares. A practical non-MCMC approach that samples uniformly would be extremely well-received (but seems beyond current techniques).

Other uniform sampling methods are:

- Generating Latin squares row-by-row by appending random permutations and restarting whenever their is a clash gives the uniform distribution. [Or equivalently, uniformly sampling from the set of row-Latin squares, then restarting if there is a clash.]
- Generating a list of all Latin squares, and picking one at random. Storage requirements could be reduced by (a) storing only a list of normalised Latin squares [i.e. first row in order] then randomly permuting the columns after sampling, and (b) storing only the differences between subsequent Latin squares (i.e. Latin trades) [although, this makes the algorithm more complicated].

In either case, for not particularly large n [maybe n>5], these approaches are either impractically slow, or requires impractically large storage space. However, for some applications, we don’t need to sample n \times n Latin squares for n>5, in which case, this is not a problem.

Moreover, for statistical applications, sampling from all possible Latin squares is often not necessary, in which case we just apply a random isotopism to any given Latin square (i.e. we pick a Latin square, then permute its rows, columns and symbols randomly).

Any attempt to sample via extending a Latin rectangle (or partial Latin square) to a Latin square without restarting from scratch after a clash occurs will almost certainly result in a non-uniform distribution. [I suppose theoretically you could add weights to your intermediate choices.] Different Latin rectangles admit a different number of completions, so if we don’t restart from scratch, we will favour Latin squares that have Latin rectangles that admit fewer completions (i.e. there’s less competition for those Latin squares).

This non-uniformity might seem like a subtle difference, but consider a (n-2) \times n Latin rectangle. The number of completions is always a power of 2. If the number of completions is 1 (which can happen: take a cyclic group’s Cayley table and delete the last two rows), then its completion is guaranteed to be generated from that point on. If the number of completions is 2^{n/2} (which can also happen: take the elementary abelian 2-group’s Cayley table and delete the last two rows), then the probability of it being generated from that point on could be 2^{-n/2} (depending on how things are implemented). So, the difference in probabilities can be at least exponential in n.

Even if you don’t care too much about the uniform distribution, the Jacobson and Matthews is still reasonable: it is quite fast and simple to implement (there’s also implementations for GAP (“loops”) and SAGE available, and probably others I’m unaware of).

**Attribution***Source : Link , Question Author : Community , Answer Author : Douglas S. Stones*