1. From Chinese Postman to Salesman and Beyond: Shortest Tour $\delta$-Covering All Points on All Edges
- Author
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Frei, Fabian, Ghazy, Ahmed, Hartmann, Tim A., Hörsch, Florian, and Marx, Dániel
- Subjects
Computer Science - Data Structures and Algorithms ,Computer Science - Computational Complexity - Abstract
A well-studied continuous model of graphs considers each edge as a continuous unit-length interval of points. For $\delta \geq 0$, we introduce the problem $\delta$-Tour, where the objective is to find the shortest tour that comes within a distance of $\delta$ of every point on every edge. It can be observed that 0-Tour is essentially equivalent to the Chinese Postman Problem, which is solvable in polynomial time. In contrast, 1/2-Tour is essentially equivalent to the graphic Traveling Salesman Problem (TSP), which is NP-hard but admits a constant-factor approximation in polynomial time. We investigate $\delta$-Tour for other values of $\delta$, noting that the problem's behavior and the insights required to understand it differ significantly across various $\delta$ regimes. On one hand, we examine the approximability of the problem for every fixed $\delta > 0$: (1) For every fixed $0 < \delta < 3/2$, the problem $\delta$-Tour admits a constant-factor approximation and is APX-hard, while for every fixed $\delta \geq 3/2$, the problem admits an $O(\log n)$-approximation algorithm and has no polynomial-time $o(\log n)$-approximation, unless P=NP. Our techniques also yield a new APX-hardness result for graphic TSP on cubic bipartite graphs. When parameterizing by tour length, it is relatively easy to show that 3/2 is the threshold of fixed-parameter tractability: (2) For every fixed $0 < \delta < 3/2$, the problem $\delta$-Tour is FPT parameterized by tour length but is W[2]-hard for every fixed $\delta \geq 3/2$. On the other hand, if $\delta$ is part of the input, then an interesting phenomenon occurs when $\delta$ is a constant fraction of n: (3) Here, the problem can be solved in time $f(k) n^{O(k)}$, where $k = \lceil n/\delta \rceil$; however, assuming ETH, there is no algorithm that solves the problem in time $f(k) n^{o(k/\log k)}$., Comment: To appear in ISAAC 2024
- Published
- 2024