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Let $n\in\mathbb{N^+}, A $ be a matrix where $(a_{ij}) \in$ ${\{0,1\}}^{n\times n} $ so that the sum of each row or column of $A$ is $x$. For which $x \geq 1$ does a permutation $\sigma \in S_n$ exist, so $a_{i\sigma(i)} = 1 \; \forall i \in \{1,\cdots, n\}$?

I first thought about how such an $x$ would look like. It is clear that such a matrix could be built up for all $x \leq n$, and for each row $i$ of $A$ the permutation need to map $i$ to the position of a one. But now I'm stuck, can you please help me to go on?

muffel
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1 Answers1

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If $A$ is any $n\times n$ matrix with non-negative integer entries, and if all the row sums and all the column sums are equal and not zero, then there is a permutation matrix $P$ (a matrix with a single 1 in each row and in each column, and zero everywhere else) such that $A-P$ has non-negative entries. David Leep and I wrote a paper about this and realted matters: Marriage, Magic, and Solitaire, American Mathematical Monthly 106 (1999) 419-429. It's not our result; it goes back many decades, to Konig. It can also be proved easily using Hall's Marriage Theorem. It follows that $A$ can be written as a positive integer linear combination of permutation matrices.

The question here is the special case where the entries of $A$ are restricted to be zeros and ones.

Gerry Myerson
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