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Birkhoff-von Neumann Graphs that are PM-compact
Marcelo H. de Carvalho, Nishad Kothari, Xiumei Wang, Yixun Lin · 2018-07-19 · via math.CO updates on arXiv.org

A well-studied geometric object in combinatorial optimization is the perfect matching polytope of a graph $G$. In any investigation concerning the perfect matching polytope, one may assume that $G$ is matching covered --- that is, it is a connected graph (of order at least two) and each edge lies in some perfect matching. A graph $G$ is Birkhoff-von Neumann (BvN) if its perfect matching polytope is characterized solely by non-negativity and degree constraints. A result of Balas (1981) implies that $G$ is BvN if and only if $G$ does not contain a pair of vertex-disjoint odd cycles $(C_1,C_2)$ such that $G-V(C_1)-V(C_2)$ has a perfect matching. It follows immediately that the corresponding decision problem is in co-NP. However, it is not known to be in NP. The problem is in P if the input graph is planar --- due to a result of Carvalho, Lucchesi and Murty (2004). These authors, along with Kothari (2018), have shown that this problem is equivalent to the seemingly unrelated problem of deciding whether a given graph is $\overline{C_6}$-free. The combinatorial diameter of a polytope is the diameter of its $1$-skeleton graph. A graph $G$ is PM-compact (PMc) if the combinatorial diameter of its perfect matching polytope equals one. A result of Chvátal (1975) implies that $G$ is PMc if and only if $G$ does not contain a pair of vertex-disjoint even cycles $(C_1,C_2)$ such that $G-V(C_1)-V(C_2)$ has a perfect matching. Once again the corresponding decision problem is in co-NP, but it is not known to be in NP. The problem is in P if the input graph is bipartite or is near-bipartite --- due to a result of Wang, Lin, Carvalho, Lucchesi, Sanjith and Little (2013). In this paper, we consider the "intersection" of the aforementioned problems. We give a complete characterization of matching covered graphs that are BvN as well as PMc. (Thus the corresponding decision problem is in P.)