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?
-
1-1 Please think about previous answers before asking new questions on the same subject. What does this have to do with finite fields? There are mutually orthogonal Latin squares of orders that have no fields (10, for example). – 2012-07-14
-
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?