{"title":"图和有向图中的约束耳分解","authors":"F. Havet, N. Nisse","doi":"10.23638/DMTCS-21-4-3","DOIUrl":null,"url":null,"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 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 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 Lovasz, 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}.","PeriodicalId":110830,"journal":{"name":"Discret. Math. Theor. Comput. Sci.","volume":"111 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"2019-09-02","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"Constrained ear decompositions in graphs and digraphs\",\"authors\":\"F. Havet, N. Nisse\",\"doi\":\"10.23638/DMTCS-21-4-3\",\"DOIUrl\":null,\"url\":null,\"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 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 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 Lovasz, 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}.\",\"PeriodicalId\":110830,\"journal\":{\"name\":\"Discret. Math. Theor. Comput. Sci.\",\"volume\":\"111 1\",\"pages\":\"0\"},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2019-09-02\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Discret. Math. Theor. Comput. 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引用次数: 0
摘要
图的耳分解是图论中几个主要问题的标准概念,比如旅行商问题。例如,众所周知的np完备的哈密顿循环问题,等价于判定一个给定的图是否允许除一个耳朵以外的所有耳朵都是平凡的(即长度为1)的耳分解。另一方面,Lovasz的一个著名结果表明,判定一个图是否允许所有耳朵长度为奇数的耳分解可以在多项式时间内完成。在本文中,我们研究了在给定耳长下图是否允许耳分解的复杂性。证明了对于所有固定正整数,判定一个图是否允许最大长度为所有耳朵的耳朵分解是多项式时间可解的。另一方面,对于任意正整数的有限集合F,判定图是否允许在F中没有耳的耳分解是np完全的。我们还证明了,对于任意k≥2,判定一个图是否允许所有耳长为0 mod k的耳分解是np完全的。我们还考虑了对耳分解的有向模拟,我们称之为柄分解,并证明了类似的结果:判定一个有向图是否允许柄分解,且所有柄的长度最多为正整数,是多项式时间可解的;判定一个有向图是否允许无长度为F的句柄分解对任何正整数有限集F是np完全的(且最小化长度为F的句柄的数量不逼近到N(1−));对于任意k≥2,判定一个有向图是否允许所有句柄长度为0 mod k的句柄分解为np完全。同时,与Lovasz的结果相反,我们证明了判定一个有向图是否允许所有柄长为奇数的柄分解是np完全的。最后,我们推测,对于每一个整数集合A,判定一个有向图是否有一个句柄分解,且所有句柄的长度都在A中,是np完全的,除非存在h∈N使得A ={1,···,h}。
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 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 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 Lovasz, 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}.