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Parity of transversals of Latin squares

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 Added by Ian Wanless
 Publication date 2019
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and research's language is English




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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 $A(imid j)$ 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 $L(imid j)$, 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 $kequiv0bmod 4$. 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 $kequiv2bmod4$.

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