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On packing spanning arborescences with matroid constraint
Let D=(V+s,A) be a digraph with a designated root vertex s. Edmonds’ seminal result (see J. Edmonds [4]) implies that D has a packing of k spanning s‐arborescences if and only if D has a packing of k(s,t)‐paths for all t∈V, where a packing means arc‐disjoint subgraphs. Let M be a matroid on the set...
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Published in: | Journal of graph theory 2020-02, Vol.93 (2), p.230-252 |
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Main Authors: | , , , |
Format: | Article |
Language: | English |
Subjects: | |
Citations: | Items that this one cites |
Online Access: | Get full text |
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Summary: | Let D=(V+s,A) be a digraph with a designated root vertex s. Edmonds’ seminal result (see J. Edmonds [4]) implies that D has a packing of k spanning s‐arborescences if and only if D has a packing of k(s,t)‐paths for all t∈V, where a packing means arc‐disjoint subgraphs. Let M be a matroid on the set of arcs leaving s. A packing of (s,t)‐paths is called M‐based if their arcs leaving s form a base of M while a packing of s‐arborescences is called M‐based if, for all t∈V, the packing of (s,t)‐paths provided by the arborescences is M‐based. Durand de Gevigney, Nguyen, and Szigeti proved in [3] that D has an M‐based packing of s‐arborescences if and only if D has an M‐based packing of (s,t)‐paths for all t∈V. Bérczi and Frank conjectured that this statement can be strengthened in the sense of Edmonds’ theorem such that each s‐arborescence is required to be spanning. Specifically, they conjectured that D has an M‐based packing of spanning s‐arborescences if and only if D has an M‐based packing of (s,t)‐paths for all t∈V. In this paper we disprove this conjecture in its general form and we prove that the corresponding decision problem is NP‐complete. We also prove that the conjecture holds for several fundamental classes of matroids, such as graphic matroids and transversal matroids. For all the results presented in this paper, the undirected counterpart also holds. |
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ISSN: | 0364-9024 1097-0118 |
DOI: | 10.1002/jgt.22484 |