Odd spanning trees of a graph

Abstract

A graph $G=(V,E)$ is said to be odd (or even, resp.) if $d_G(v)$ is odd (or even, resp.) for any $v\in V$. Trivially, the order of an odd graph must be even. In this paper, we show that every 4-edge connected graph of even order has a connected odd factor. A spanning tree $T$ of $G$ is called a homeomorphically irreducible spanning tree (HIST by simply) if $T$ contains no vertex of degree two. Trivially, an odd spanning tree must be a HIST. In 1990, Albertson, Berman, Hutchinson, and Thomassen showed that every connected graph of order $n$ with $\delta(G)\geq \min\{\frac n 2, 4\sqrt{2n}\}$ contains a HIST. We show that every complete bipartite graph with both parts being even has no odd spanning tree, thereby for any even integer $n$ divisible by 4, there exists a graph of order $n$ with the minimum degree $\frac n 2$ having no odd spanning tree. Furthermore, we show that every graph of order $n$ with $\delta(G)\geq \frac n 2 +1$ has an odd spanning tree. We also characterize all split graphs having an odd spanning tree. As an application, for any graph $G$ with diameter at least 4, $\overline{G}$ has a spanning odd double star. Finally, we also give a necessary and sufficient condition for a triangle-free graph $G$ whose complement contains an odd spanning tree. A number of related open problems are proposed.