No abstract

A latin square of order n is an n × n array of n symbols in which each symbol occurs exactly once in each row and column. A transversal of such a square is a set of n entries such that no two entries share the same row, column or symbol. Transversals are closely related to the notions of complete mappings and orthomorphisms in (quasi)groups, and are fundamental to the concept of mutually orthogonal latin squares.Here we provide a brief survey of the literature on transversals. We cover (1) existence and enumeration results, (2) generalisations of transversals including partial transversals and plexes, (3) the special case when the latin square is a group table, (4) a connection with covering radii of sets of permutations. The survey includes a number of conjectures and open problems.2000 Mathematics Subject Classifications: 05B15 20N05

We (1) determine the number of Latin rectangles with 11 columns and each possible number of rows, including the Latin squares of order 11, (2) answer some questions of Alter by showing that the number of reduced Latin squares of order n is divisible by f ! where f is a particular integer close to 1 2 n, (3) provide a formula for the number of Latin squares in terms of permanents of (+1, −1)-matrices, (4) find the extremal values for the number of 1-factorisations of k-regular bipartite graphs on 2n vertices whenever 1 ≤ k ≤ n ≤ 11, (5) show that the proportion of Latin squares with a non-trivial symmetry group tends quickly to zero as the order increases.

We define a $k$-plex to be a partial latin square of order $n$ containing $kn$ entries such that exactly $k$ entries lie in each row and column and each of $n$ symbols occurs exactly $k$ times. A transversal of a latin square corresponds to the case $k=1$. For $k>n/4$ we prove that not all $k$-plexes are completable to latin squares. Certain latin squares, including the Cayley tables of many groups, are shown to contain no $(2c+1)$-plex for any integer $c$. However, Cayley tables of soluble groups have a $2c$-plex for each possible $c$. We conjecture that this is true for all latin squares and confirm this for orders $n\leq8$. Finally, we demonstrate the existence of indivisible $k$-plexes, meaning that they contain no $c$-plex for $1\leq c < k$.