Constrained ear decompositions in graphs and digraphs
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 whi...
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| Published in: | Discrete Mathematics and Theoretical Computer Science Vol. 21 no. 4; no. Graph Theory; pp. COV2 - 19 |
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| Format: | Journal Article |
| Language: | English |
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02.09.2019
Discrete Mathematics & Theoretical Computer Science |
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| ISSN: | 1365-8050, 1462-7264, 1365-8050 |
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| Abstract | 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}. |
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| AbstractList | 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 NP-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 Lovasz 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 l is polynomial-time solvable for all fixed positive integer l. On the other hand, deciding whether a graph admits an ear decomposition without ears of length in F is NP-complete for any finite set F of positive integers. We also prove that, for any k [greater than or equal to] 2, deciding whether a graph admits an ear decomposition with all ears of length 0 mod k is NP-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 l is polynomial-time solvable for all positive integer l; deciding whether a digraph admits a handle decomposition without handles of length in F is NP -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 - [epsilon])); for any k [greater than or equal to] 2, deciding whether a digraph admits a handle decomposition with all handles of length 0 mod k is NP-complete. Also, in contrast with the result of Lovasz, we prove that deciding whether a digraph admits a handle decomposition with all handles of odd length is NP-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 NP-complete, unless there exists h [member of] N such that A = 1, ***, h. 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}. 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 ЛЛР-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 1. On the other hand, deciding whether a graph admits an ear decomposition without ears of length in У is AiP-complete for any finite set У of positive integers. We also prove that, for any к > 2, deciding whether a graph admits an ear decomposition with all ears of length 0 mod к is AfP-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 1; deciding whether a digraph admits a handle decomposition without handles of length in У is AP-complete for any finite set У of positive integers (and minimizing the number of handles of length in У is not approximable up to n(l - e)); for any к > 2, deciding whether a digraph admits a handle decomposition with all handles of length 0 mod k is AfP-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 AfP-complete. Finally, we conjecture that, for every set Л of integers, deciding whether a digraph has a handle decomposition with all handles of length in A is AfP-complete, unless there exists h E N such that A = {1, * * * , h\. 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 NP-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 Lovasz 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 l is polynomial-time solvable for all fixed positive integer l. On the other hand, deciding whether a graph admits an ear decomposition without ears of length in F is NP-complete for any finite set F of positive integers. We also prove that, for any k [greater than or equal to] 2, deciding whether a graph admits an ear decomposition with all ears of length 0 mod k is NP-complete. |
| Audience | Academic |
| Author | Havet, Frédéric Nisse, Nicolas |
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| SubjectTerms | [info.info-cc]computer science [cs]/computational complexity [cs.cc] [info.info-dm]computer science [cs]/discrete mathematics [cs.dm] Computational Complexity Computer Science Decomposition Decomposition (Mathematics) Discrete Mathematics Ear Graph theory Graphs Handles Integers Mathematical research Polynomials Traveling salesman problem |
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| Title | Constrained ear decompositions in graphs and digraphs |
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