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arXiv 2602.16532cs.DS
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S-Hamiltonian循环问题

The S-Hamiltonian Cycle Problem

Univ. Lille, Inria, CNRS, Centrale Lille(里尔大学、法国国家信息与自动化研究所、法国国家科学研究中心、里尔中央理工学院);DI ENS, ENS, PSL University, CNRS, Inria, Paris(巴黎高等师范学院计算机科学研究所、巴黎高等师范学院、巴黎文理研究大学、法国国家科学研究中心、法国国家信息与自动化研究所)

Antoine Amarilli, Arthur Lombardo, Mikaël Monet

AI总结:

研究S-Hamiltonian循环问题的复杂性,发现不同S集合下的计算难度,部分情况下为NP难,部分情况下可处理。

摘要:

确定给定的无向图是否是Hamiltonian图,即是否存在一个循环经过每个顶点恰好一次,是其中最著名的NP难问题之一。我们考虑Hamiltonian循环的一个推广:对于一个固定的自然数集合S,我们希望每个图G的顶点恰好被访问一次,并且任何两个连续的顶点可以连接k步,其中k是S中的某个选择。形式上,S-Hamiltonian循环是图G顶点的一个排列(v₀,…,v_{n-1}),使得对于0 ≤ i ≤ n-1,存在一条从v_i到v_{i+1 mod n}的路径,其长度在S中。(我们不限制走路径中顶点被访问的次数。) 当然,标准意义上的Hamiltonian循环对应于S={1}。我们研究决定输入图G是否具有S-Hamiltonian循环的S-Hamiltonian循环问题。我们的目标是根据固定的集合S确定该问题的复杂性。已知对于S={1,2},该问题仍然NP难,而对于S={1,2,3}则很简单,因为任何连通图都包含一个{1,2,3}-Hamiltonian循环。我们的工作对大多数类型的S集分类了该问题的复杂性,关键的新结果包括:对于S={2}和S={2,4},问题仍然是NP难的,但对于S={1,2,4}、S={2,4,6}、这些可处理情况的任何超集以及S为所有奇数的无限集合,问题都是可处理的。剩下的开放情况是非单元素的有限奇数集合,特别是S={1,3}。除了循环外,我们还讨论了寻找S-Hamiltonian路径的复杂性,并显示我们的问题在具有有界cliquewidth的图上都是可处理的。

英文摘要:

Determining if an input undirected graph is Hamiltonian, i.e., if it has a cycle that visits every vertex exactly once, is one of the most famous NP-complete problems. We consider the following generalization of Hamiltonian cycles: for a fixed set $S$ of natural numbers, we want to visit each vertex of a graph $G$ exactly once and ensure that any two consecutive vertices can be joined in $k$ hops for some choice of $k \in S$. Formally, an $S$-Hamiltonian cycle is a permutation $(v_0,\ldots,v_{n-1})$ of the vertices of $G$ such that, for $0 \leq i \leq n-1$, there exists a walk between $v_i$ and $v_{i+1 \bmod n}$ whose length is in $S$. (We do not impose any constraints on how many times vertices can be visited as intermediate vertices of walks.) Of course Hamiltonian cycles in the standard sense correspond to $S=\{1\}$. We study the $S$-Hamiltonian cycle problem of deciding whether an input graph $G$ has an $S$-Hamiltonian cycle. Our goal is to determine the complexity of this problem depending on the fixed set $S$. It is already known that the problem remains NP-complete for $S=\{1,2\}$, whereas it is trivial for $S=\{1,2,3\}$ because any connected graph contains a $\{1,2,3\}$-Hamiltonian cycle. Our work classifies the complexity of this problem for most kinds of sets $S$, with the key new results being the following: we have NP-completeness for $S = \{2\}$ and for $S = \{2, 4\}$, but tractability for $S = \{1, 2, 4\}$, for $S = \{2, 4, 6\}$, for any superset of these two tractable cases, and for $S$ the infinite set of all odd integers. The remaining open cases are the non-singleton finite sets of odd integers, in particular $S = \{1, 3\}$. Beyond cycles, we also discuss the complexity of finding $S$-Hamiltonian paths, and show that our problems are all tractable on graphs of bounded cliquewidth.

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