Latin squares with no transversals
The electronic journal of combinatorics, Tome 24 (2017) no. 2
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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'.
DOI : 10.37236/6481
Classification : 05B15, 05D15
Mots-clés : Latin square, transversal, plex, triplex

Nicholas J. Cavenagh    ; Ian M. Wanless  1

1 Monash University
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     author = {Nicholas J. Cavenagh and Ian M. Wanless},
     title = {Latin squares with no transversals},
     journal = {The electronic journal of combinatorics},
     year = {2017},
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Nicholas J. Cavenagh; Ian M. Wanless. Latin squares with no transversals. The electronic journal of combinatorics, Tome 24 (2017) no. 2. doi: 10.37236/6481

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