Latin transversals of rectangular arrays

Let m and n be integers, $2 \leq m \leq n$. An m by n array consists of mn cells, arranged in m rows and n columns, and each cell contains exactly one symbol. A transversal of an array consists of m cells, one from each row and no two from the same column. A latin transversal is a transversal in which no symbol appears more than once. We will establish a sufficient condition that a 3 by n array has a latin transversal.Comment: Theorem 4 has been added, which provides a lower bound on L(m,n

Cross-intersecting sub-families of hereditary families

Families $\mathcal{A}_1, \mathcal{A}_2, ..., \mathcal{A}_k$ of sets are said to be \emph{cross-intersecting} if for any $i$ and $j$ in $\{1, 2, ..., k\}$ with $i \neq j$, any set in $\mathcal{A}_i$ intersects any set in $\mathcal{A}_j$. For a finite set $X$, let $2^X$ denote the \emph{power set of $X$} (the family of all subsets of $X$). A family $\mathcal{H}$ is said to be \emph{hereditary} if all subsets of any set in $\mathcal{H}$ are in $\mathcal{H}$; so $\mathcal{H}$ is hereditary if and only if it is a union of power sets. We conjecture that for any non-empty hereditary sub-family $\mathcal{H} \neq \{\emptyset\}$ of $2^X$ and any $k \geq |X|+1$, both the sum and product of sizes of $k$ cross-intersecting sub-families $\mathcal{A}_1, \mathcal{A}_2, ..., \mathcal{A}_k$ (not necessarily distinct or non-empty) of $\mathcal{H}$ are maxima if $\mathcal{A}_1 = \mathcal{A}_2 = ... = \mathcal{A}_k = \mathcal{S}$ for some largest \emph{star $\mathcal{S}$ of $\mathcal{H}$} (a sub-family of $\mathcal{H}$ whose sets have a common element). We prove this for the case when $\mathcal{H}$ is \emph{compressed with respect to an element $x$ of $X$}, and for this purpose we establish new properties of the usual \emph{compression operation}. For the product, we actually conjecture that the configuration $\mathcal{A}_1 = \mathcal{A}_2 = ... = \mathcal{A}_k = \mathcal{S}$ is optimal for any hereditary $\mathcal{H}$ and any $k \geq 2$, and we prove this for a special case too.Comment: 13 page

Decomposing almost complete graphs by random trees

An old conjecture of Ringel states that every tree with m edges decomposes the complete graph K2m+1. The best known lower bound for the order of a complete graph which admits a decomposition by every given tree with m edges is O(m3). We show that asymptotically almost surely a random tree with m edges and p=2m+1 a prime decomposes K2m+1(r) for every r=2, the graph obtained from the complete graph K2m+1 by replacing each vertex by a coclique of order r. Based on this result we show, among other results, that a random tree with m+1 edges a.a.s. decomposes the compete graph K6m+5 minus one edge.Peer ReviewedPostprint (author's final draft