Talk:Latin Squares in reduced form/Randomizing using Jacobson and Matthews’ Technique
https://brainwagon.org/2016/05/17/code-for-generating-a-random-latin-square/ some good news and some bad news
Quoting from Drizen the following is this task:
Theorem 4. Let X0 be an arbitrarily distributed order-n Latin square that starts a Markov chain of (proper and improper) squares: to each square, apply a move chosen uniformly at random from the permissible ±1-moves [n2(n-1) from a proper square, 8 from an improper square]. Let Xᵜ ≡ (X1;X2;X3; : : :) be the subsequence of proper squares we encounter; then Xᵜ is a Markov chain with a (unique) stationary distribution that is uniform over the set of order-n Latin squares.--Nigel Galloway (talk) 11:57, 12 August 2019 (UTC)
some bad news
The C code above has defined a meaning for random which is not suitable for this task, and in my view not suitable for its own purpose. It performs permissible ±1-moves in a loop DIM*DIM*DIM times, where DIM is the order of the required Latin Square. In effect it generates a Markov chain and returns X(DIM*DIM*DIM) as its Random Latin Square. If one uses this as X1 it is: not this task; and slow. To generate X5 for Latin Square order 5 the C-code generates 42*42*42*5 proper squares. The task is only asking for 750. As observed you will wait a very long time for X1 of order 256 while the C-code generates 256*256*256 proper squares.--Nigel Galloway (talk) 11:57, 12 August 2019 (UTC)
some good news
- Thanks Nigel.
- If I remove the outer loop (i.e. perform 1 iteration rather than DIM*DIM*DIM iterations), then parts 3 and 4 complete in about 7 seconds on my elderly Linux machine. If we assume a modern machine would be 3 or 4 times faster and rewriting the whole thing in C rather than Go would be twice as fast again, then your estimate of around 1 second looks about right.
- They both seem to be looping n ^ 3 times so it may be that the brainwagon.org implementation has worked from these.