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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.

  • 0
    A 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 3

4

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.

2

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?