Toughness and spanning trees in K4‐minor‐free graphs

For an integer k, a k‐tree is a tree with maximum degree at most k. More generally, if f is an integer‐valued function on vertices, an f‐tree is a tree in which each vertex v has degree at most f(v). Let c(G) denote the number of components of a graph G. We show that if G is a connected K4‐minor‐fre...

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Bibliographic Details
Published inJournal of graph theory Vol. 96; no. 3; pp. 379 - 402
Main Authors Ellingham, M. N., Shan, Songling, Ye, Dong, Zha, Xiaoya
Format Journal Article
LanguageEnglish
Published Hoboken Wiley Subscription Services, Inc 01.03.2021
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Summary:For an integer k, a k‐tree is a tree with maximum degree at most k. More generally, if f is an integer‐valued function on vertices, an f‐tree is a tree in which each vertex v has degree at most f(v). Let c(G) denote the number of components of a graph G. We show that if G is a connected K4‐minor‐free graph and c(G−S)≤∑v∈S(f(v)−1)for all S⊆V(G) with S≠∅, then G has a spanning f‐tree. Consequently, if G is a 1k−1‐tough K4‐minor‐free graph, then G has a spanning k‐tree. These results are stronger than results for general graphs due to Win (for k‐trees) and Ellingham, Nam, and Voss (for f‐trees). The K4‐minor‐free graphs form a subclass of planar graphs, and are identical to graphs of treewidth at most 2, and also to graphs whose blocks are series‐parallel. We provide examples to show that the inequality above cannot be relaxed by adding 1 to the right‐hand side, and also to show that our result does not hold for general planar graphs. Our proof uses a technique where we incorporate toughness‐related information into weights associated with vertices and cutsets.
ISSN:0364-9024
1097-0118
DOI:10.1002/jgt.22620