Constrained ear decompositions in graphs and digraphs

Description

Ear decompositions of graphs are a standard concept related to several major problems in graph theory like the Traveling Salesman Problem. For example, the Hamiltonian Cycle Problem, which is notoriously N P-complete, is equivalent to deciding whether a given graph admits an ear decomposition in which all ears except one are trivial (i.e. of length 1). On the other hand, a famous result of Lovász states that deciding whether a graph admits an ear decomposition with all ears of odd length can be done in polynomial time. In this paper, we study the complexity of deciding whether a graph admits an ear decomposition with prescribed ear lengths. We prove that deciding whether a graph admits an ear decomposition with all ears of length at most is polynomial-time solvable for all fixed positive integer. On the other hand, deciding whether a graph admits an ear decomposition without ears of length in F is N P-complete for any finite set F of positive integers. We also prove that, for any k ≥ 2, deciding whether a graph admits an ear decomposition with all ears of length 0 mod k is N P-complete. We also consider the directed analogue to ear decomposition, which we call handle decomposition, and prove analogous results : deciding whether a digraph admits a handle decomposition with all handles of length at most is polynomial-time solvable for all positive integer ; deciding whether a digraph admits a handle decomposition without handles of length in F is N P-complete for any finite set F of positive integers (and minimizing the number of handles of length in F is not approximable up to n(1 −)); for any k ≥ 2, deciding whether a digraph admits a handle decomposition with all handles of length 0 mod k is N P-complete. Also, in contrast with the result of Lovász, we prove that deciding whether a digraph admits a handle decomposition with all handles of odd length is N P-complete. Finally, we conjecture that, for every set A of integers, deciding whether a digraph has a handle decomposition with all handles of length in A is N P-complete, unless there exists h ∈ N such that A = {1, · · · , h}.

Abstract

International audience

Additional details