Is there any place where I can find a pair of MOLS(mutually orthogonal latin squares) of order 15? I can't seem to find a place where it's spelled out explicitly.
A pair of MOLS of order 15?
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0A finite field $F$ of $q$ elements generates $q-1$ MOLS following the recipe $L_\alpha(x,y)=\alpha x+y$ for all $\alpha\in F^*$, $x,y\in F$. If $F'$ is another finite field with $q'$ elements, then the Cartesian products $L_{\alpha,\alpha'}((x,x'),(y,y'))=(\alpha x+y,\alpha'x'+y'),$ with $x,y$ (resp. $x',y'$) ranging over their respective fields, are Latin squares of size $qq'$. If $q\neq q'$, $\alpha_1\neq\alpha_2$, $\alpha_1'\neq\alpha_2'$, then the squares $L_{\alpha_1,\alpha_1'}$ and $L_{\alpha_2,\alpha_2'}$ are orthogonal. That's all there is to it. – 2012-07-14
3 Answers
For any odd $n$, a diagonally cyclic Latin square construction works. Here's GAP code that implements it:
n:=15; # Diagonally cyclic Latin square L:=List([1..n],i->List([1..n],j->0)); for j in [1..n] do L[1][j]:=((2*(j-1)) mod n)+1; od; for i in [2..n] do for j in [1..n] do L[i][j]:=(L[i-1][((j-2) mod n)+1] mod n)+1; od; od; L; # Fixed diagonal Latin square M:=List([1..n],i->List([1..n],j->((j-i) mod n)+1)); M;
and this is what the output looks like:
gap> L; [ [ 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14 ], [ 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13 ], [ 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12 ], [ 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11 ], [ 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10 ], [ 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9 ], [ 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8 ], [ 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7 ], [ 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6 ], [ 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5 ], [ 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4 ], [ 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3 ], [ 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2 ], [ 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1 ], [ 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15 ] ]
and
gap> M; [ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 ], [ 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 ], [ 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 ], [ 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ], [ 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ], [ 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ], [ 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9 ], [ 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8 ], [ 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7 ], [ 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6 ], [ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5 ], [ 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4 ], [ 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3 ], [ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2 ], [ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1 ] ]
We can readily see that they're orthogonal (the forwards broken diagonals of the first square are $1,2,\ldots,n$ cyclically permuted, whereas the forwards broken diagonals of the second are $x,x,\ldots,x$, for some $x$). The only thing we should check is that the first square is always a Latin square, which comes from its first row being an orthomorphism.
The Wikipedia page on Greco-Latin squares shows squares of order 3 and 5. Gerry Myerson gave an answer to your question /mutually-orthogonal-latin-squares-of-order-mn-from-order-m-and-order-n which shows how to make one. What don't you understand?