Do K33-Free Latin Squares Exist?

We discuss the problem of existence of latin squares without a substructure consisting of six elements $(r_1,c_2,l_3)$, $(r_2,c_3,l_1)$, $(r_3,c_1,l_2)$, $(r_2,c_1,l_3)$, $(r_3,c_2,l_1)$, $(r_1,c_3,l_2)$. Equivalently, the corresponding latin square graph does not have an induced subgraph isomorphic to $K_{3,3}$. The exhaustive search [Brouwer, Wanless. Universally noncommutative loops. 2011] says that there are no such latin squares of order $3$--$7$, $9$--$11$ and there are only two $K_{3,3}$-free latin squares of order $8$, up to equivalence. We repeat the search, establishing also the number of latin $m$-by-$n$ rectangles for each $m$ and $n$ less than or equal to $11$. As a switched combination of two orthogonal latin squares of order $8$, we construct a $K_{3,3}$-free (universally noncommutative) latin square of order $16$. We also consider a similar problem for orthogonal latin squares, proving that there are both $K_{4,4}$-free and non-$K_{4,4}$-free linear pairs of orthogonal latin squares for each odd prime-power order larger than~$5$. Keywords: latin square; transversal; trade; pattern avoiding; eigenfunction; universally noncommutative loop.