Parity of transversals of Latin squares

TL;DR

This paper explores the parity properties of transversals in Latin squares, establishing new divisibility results and congruences, especially for Latin squares of order 2 mod 4, and introduces related combinatorial and algebraic findings.

Contribution

It introduces a novel parity notion for transversals, proves divisibility and congruence properties for Latin squares and related structures, and conjectures new modular relations.

Findings

Number of transversals in Latin squares of order 2 mod 4 is divisible by 4.

Transversal counts modulo 2 are consistent across fixed positions.

Divisibility results for perfect matchings in certain bipartite graphs and permanents of matrices.

Abstract

We introduce a notion of parity for transversals, and use it to show that in Latin squares of order $2 \bmod 4$, the number of transversals is a multiple of 4. We also demonstrate a number of relationships (mostly congruences modulo 4) involving $E_1,\dots, E_n$, where $E_i$ is the number of diagonals of a given Latin square that contain exactly $i$ different symbols. Let $A(i\mid j)$ denote the matrix obtained by deleting row $i$ and column $j$ from a parent matrix $A$. Define $t_{ij}$ to be the number of transversals in $L(i\mid j)$, for some fixed Latin square $L$. We show that $t_{ab}\equiv t_{cd}\bmod2$ for all $a,b,c,d$ and $L$. Also, if $L$ has odd order then the number of transversals of $L$ equals $t_{ab}$ mod 2. We conjecture that $t_{ac} + t_{bc} + t_{ad} + t_{bd} \equiv 0 \bmod 4$ for all $a,b,c,d$. In the course of our investigations we prove several results that could be of interest in other contexts. For example, we show that the number of perfect matchings in a $k$-regular bipartite graph on $2n$ vertices is divisible by $4$ when $n$ is odd and $k\equiv0\bmod 4$. We also show that $${\rm per}\, A(a \mid c)+{\rm per}\, A(b \mid c)+{\rm per}\, A(a \mid d)+{\rm per}\, A(b \mid d) \equiv 0 \bmod 4$$ for all $a,b,c,d$, when $A$ is an integer matrix of odd order with all row and columns sums equal to $k\equiv2\bmod4$.