From Chinese Postman to Salesman and Beyond II: Inapproximability and Parameterized Complexity
February 25, 2025 Β· Declared Dead Β· π arXiv.org
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Authors
Fabian Frei, Ahmed Ghazy, Tim A. Hartmann, Florian HΓΆrsch, DΓ‘niel Marx
arXiv ID
2502.18541
Category
cs.DS: Data Structures & Algorithms
Cross-listed
cs.CC
Citations
0
Venue
arXiv.org
Last Checked
5 months ago
Abstract
A well-studied continuous model of graphs considers each edge as a continuous unit-length interval of points. In the problem $Ξ΄$-Tour defined within this model, the objective to find a shortest tour that comes within a distance of $Ξ΄$ of every point on every edge. This parameterized problem was introduced in the predecessor to this article and shown to be essentially equivalent to the Chinese Postman problem for $Ξ΄= 0$, to the graphic Travel Salesman Problem (TSP) for $Ξ΄= 1/2$, and close to first Vertex Cover and then Dominating Set for even larger $Ξ΄$. Moreover, approximation algorithms for multiple parameter ranges were provided. In this article, we provide complementing inapproximability bounds and examine the fixed-parameter tractability of the problem. On the one hand, we show the following: (1) For every fixed $0 < Ξ΄< 3/2$, the problem $Ξ΄$-Tour is APX-hard, while for every fixed $Ξ΄\geq 3/2$, the problem has no polynomial-time $o(\log{n})$-approximation unless P = NP. Our techniques also yield the new result that TSP remains APX-hard on cubic (and even cubic bipartite) graphs. (2) For every fixed $0 < Ξ΄< 3/2$, the problem $Ξ΄$-Tour is fixed-parameter tractable (FPT) when parameterized by the length of a shortest tour, while it is W[2]-hard for every fixed $Ξ΄\geq 3/2$ and para-NP-hard for $Ξ΄$ being part of the input. On the other hand, if $Ξ΄$ is considered to be part of the input, then an interesting nontrivial phenomenon occurs when $Ξ΄$ is a constant fraction of the number of vertices: (3) If $Ξ΄$ is part of the input, then the problem can be solved in time $f(k)n^{O(k)}$, where $k = \lceil n/Ξ΄\rceil$; however, assuming the Exponential-Time Hypothesis (ETH), there is no algorithm that solves the problem and runs in time $f(k)n^{o(k/\log k)}$.
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