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Crossposted at Theoretical Computer Science SE


A problem I study reduces to a system of linear equations $A\mathbf{x}=\mathbf{1}$ where $A$ is an $m\times n$ matrix with each entry $a_{ij}\in\{0,1\}$. $\mathbf{1}$ is the $m$-dimensional vector with all its coordinates being 1. The nature of the problem I study guarantees that there exists at least a solution $\mathbf{x}\in[0,1]^n$ for this system, which means that the polyhedron $P=\{\mathbf{x}\mid A\mathbf{x}=\mathbf{1}, \mathbf{x}\geq0\}$ is nonempty. I want to know what condition guarantees the existence of a 0-1 solution $\mathbf{x}\in\{0,1\}^n$ to this system.

I know that the unimodularity of $A$ is a sufficient condition, which has been useful for me. I also know the TDI condition in integer programming is sufficient, whereas I do not know how to make use of this condition. Both unimodularity and TDI guarantee that all vertices of polyhedron $P$ to be integral, thus I know there is at least a 0-1 solution (vertex), since the polyhedron is nonempty. However, all vertices being integral is more than I want, because I only care about whether there exists an integral vertex. I wonder whether there are more specialized results for this problem.

As pointed out by the helpful comment below, this problem is essentially equivalent to the EXACT SET COVER problem. I would now ask whether there is any condition or structure that guarantees the existence of an exact cover.

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    $\begingroup$ This is essentially equivalent to the EXACT SET COVER problem (which is NP-complete). So we would not expect there to be any simple criterion that is necessary and sufficient for the existence of a 0/1 solution. There are a few very special cases where you can deduce that there is a solution, and a few others where you can deduce that there isn't, but my feeling is that these are extremely rare. But look up the literature on EXACT SET COVER and see how you go. Disclaimer: I am not an expert on combinatorial optimization, but I recently had to solve a few exact set cover problems. $\endgroup$ Apr 20, 2021 at 5:28
  • $\begingroup$ @GordonRoyle Very good advice! Thank you! $\endgroup$ Apr 20, 2021 at 11:13

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