Fine-Grained Complexity Analysis of Two Classic TSP Variants
July 10, 2016 Β· Declared Dead Β· π International Colloquium on Automata, Languages and Programming
"No code URL or promise found in abstract"
Evidence collected by the PWNC Scanner
Authors
Mark de Berg, Kevin Buchin, Bart M. P. Jansen, Gerhard Woeginger
arXiv ID
1607.02725
Category
cs.DS: Data Structures & Algorithms
Cross-listed
cs.CG
Citations
32
Venue
International Colloquium on Automata, Languages and Programming
Last Checked
3 months ago
Abstract
We analyze two classic variants of the Traveling Salesman Problem using the toolkit of fine-grained complexity. Our first set of results is motivated by the Bitonic TSP problem: given a set of $n$ points in the plane, compute a shortest tour consisting of two monotone chains. It is a classic dynamic-programming exercise to solve this problem in $O(n^2)$ time. While the near-quadratic dependency of similar dynamic programs for Longest Common Subsequence and Discrete Frechet Distance has recently been proven to be essentially optimal under the Strong Exponential Time Hypothesis, we show that bitonic tours can be found in subquadratic time. More precisely, we present an algorithm that solves bitonic TSP in $O(n \log^2 n)$ time and its bottleneck version in $O(n \log^3 n)$ time. Our second set of results concerns the popular $k$-OPT heuristic for TSP in the graph setting. More precisely, we study the $k$-OPT decision problem, which asks whether a given tour can be improved by a $k$-OPT move that replaces $k$ edges in the tour by $k$ new edges. A simple algorithm solves $k$-OPT in $O(n^k)$ time for fixed $k$. For 2-OPT, this is easily seen to be optimal. For $k=3$ we prove that an algorithm with a runtime of the form $\tilde{O}(n^{3-Ξ΅})$ exists if and only if All-Pairs Shortest Paths in weighted digraphs has such an algorithm. The results for $k=2,3$ may suggest that the actual time complexity of $k$-OPT is $Ξ(n^k)$. We show that this is not the case, by presenting an algorithm that finds the best $k$-move in $O(n^{\lfloor 2k/3 \rfloor + 1})$ time for fixed $k \geq 3$. This implies that 4-OPT can be solved in $O(n^3)$ time, matching the best-known algorithm for 3-OPT. Finally, we show how to beat the quadratic barrier for $k=2$ in two important settings, namely for points in the plane and when we want to solve 2-OPT repeatedly.
Community Contributions
Found the code? Know the venue? Think something is wrong? Let us know!
π Similar Papers
In the same crypt β Data Structures & Algorithms
π
π
The Cartographer
R.I.P.
π»
Ghosted
Route Planning in Transportation Networks
R.I.P.
π»
Ghosted
Near-linear time approximation algorithms for optimal transport via Sinkhorn iteration
R.I.P.
π»
Ghosted
Hierarchical Clustering: Objective Functions and Algorithms
R.I.P.
π»
Ghosted
Graph Isomorphism in Quasipolynomial Time
π
π
The Cartographer
Simulation optimization: A review of algorithms and applications
Died the same way β π» Ghosted
R.I.P.
π»
Ghosted
Federated Learning: Strategies for Improving Communication Efficiency
R.I.P.
π»
Ghosted
In-Datacenter Performance Analysis of a Tensor Processing Unit
R.I.P.
π»
Ghosted
Deep Convolutional Neural Networks for Computer-Aided Detection: CNN Architectures, Dataset Characteristics and Transfer Learning
R.I.P.
π»
Ghosted