2016•Monash University Research Portal (Monash University)Open access

Latin squares with no transversals

Nicholas J. Cavenagh, Ian M. Wanless

Open full text 0 citations

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'

About this research paper

What this paper is about

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'

Why it matters

A significance statement is not available in the OpenAlex record.

Key contribution

A contribution statement is not available in the OpenAlex record.

Method / approach

Method details are not available in the OpenAlex metadata.

Main findings

Findings are not separately available in the OpenAlex metadata.

Limitations

Limitations are not available in the OpenAlex metadata.

Applications

Application details are not available in the OpenAlex metadata.

Available 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'

Key concepts: Transversal (combinatorics), Latin square, Combinatorics, Conjecture, Mathematics, Order (exchange), Modulo, Square (algebra)

Related papers

Back to paper searchBrowse research topicsOriginal source
Latin squares with no transversals — Research Paper | ScholarLens