Odd spanning trees of a graph
Abstract
A graph is said to be odd (or even, resp.) if is odd (or even, resp.) for any . 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 of is called a homeomorphically irreducible spanning tree (HIST by simply) if 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 with 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 divisible by 4, there exists a graph of order with the minimum degree having no odd spanning tree. Furthermore, we show that every graph of order with has an odd spanning tree. We also characterize all split graphs having an odd spanning tree. As an application, for any graph with diameter at least 4, has a spanning odd double star. Finally, we also give a necessary and sufficient condition for a triangle-free graph whose complement contains an odd spanning tree. A number of related open problems are proposed.
Keywords
Cite
@article{arxiv.2503.17676,
title = {Odd spanning trees of a graph},
author = {Jingyu Zheng and Baoyindureng Wu},
journal= {arXiv preprint arXiv:2503.17676},
year = {2025}
}
Comments
19 pages, 11 figures