Latin Squares with No Transversals
Nicholas J. Cavenagh, Ian M. Wanless
Abstract
Nicholas J. Cavenagh, Ian M. Wanless
Abstract
A $k$-plex in a latin square of order $n$ is a selection of $kn$ entries that includes $k$ representatives from each row and column and $k$ occurrences of each symbol. A $1$-plex is also known as a transversal.It is well known that if $n$ is even then $B_n$, the addition table for the integers modulo $n$, possesses no transversals. We show that there are a great many latin squares that are similar to $B_n$ and have no transversal. As a consequence, the number of species of transversal-free latin squares is shown to be at least $n^{n^{3/2}(1/2-o(1))}$ for even $n\rightarrow\infty$.We also produce various constructions for latin squares that have no transversal but do have a $k$-plex for some odd $k>1$. We prove a 2002 conjecture of the second author that for all even orders $n>4$ there is a latin square of order $n$ that contains a $3$-plex but no transversal. We also show that for odd $k$ and $m\geq 2$, there exists a latin square of order $2km$ with a $k$-plex but no $k'$-plex for odd $k'
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A $k$-plex in a latin square of order $n$ is a selection of $kn$ entries that includes $k$ representatives from each row and column and $k$ occurrences of each symbol. A $1$-plex is also known as a transversal.It is well known that if $n$ is even then $B_n$, the addition table for the integers modulo $n$, possesses no transversals. We show that there are a great many latin squares that are similar to $B_n$ and have no transversal. As a consequence, the number of species of transversal-free latin squares is shown to be at least $n^{n^{3/2}(1/2-o(1))}$ for even $n\rightarrow\infty$.We also produce various constructions for latin squares that have no transversal but do have a $k$-plex for some odd $k>1$. We prove a 2002 conjecture of the second author that for all even orders $n>4$ there is a latin square of order $n$ that contains a $3$-plex but no transversal. We also show that for odd $k$ and $m\geq 2$, there exists a latin square of order $2km$ with a $k$-plex but no $k'$-plex for odd $k'
Key concepts: Transversal (combinatorics), Latin square, Combinatorics, Mathematics, Conjecture, Order (exchange), Modulo, Square (algebra)