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arXiv · 2504.19608

The frequency $K_i$s for symmetrical traveling salesman problem

Abstract

The frequency $K_i$s ($i\in[4,n]$) are studied for symmetric traveling salesman problem ($TSP$) to characterize the structure properties of the edges in optimal Hamiltonian cycle ($OHC$). For a given $K_i$ in the complete graph $K_n$, the frequency $K_i$ is computed with the set of ${{i}\choose{2}}$ optimal $i$-vertex paths with fixed endpoints (optimal $i$-vertex paths) in the $K_i$. Given an $OHC$ edge related to $K_i$, it has certain frequency bigger than $\frac{1}{2}{{i}\choose{2}}$ in the frequency $K_i$, and that of an ordinary edge not in $OHC$ is smaller than $2(n-3)$. Moreover, given a frequency $K_i$ containing an $OHC$ edge related to $K_n$, the frequency of the $OHC$ edge is bigger than $\frac{1}{2}{{i}\choose{2}}$ in the average case. It also found that the probability that an $OHC$ edge is contained in the optimal $i$-vertex paths increases according to $i\in [4, n]$ or keeps stable if it decreases from $i$ to $i+1\leq n$. As the frequency $K_i$s are used to compute the frequency of an edge, each $OHC$ edge reaches its own peak frequency at $i=P_0$ where $P_0=\frac{n}{2} + 2$ for even $n$ or $\frac{n+1}{2} + 1$ for odd $n$. For each ordinary edge out of $OHC$, the probability that they are contained in the optimal $i$-vertex paths decreases according to $i$, respectively, in the average case. Moreover, the probability of an ordinary edge definitely decreases if $i \geq i_d$ where $i_d = O(n^{\frac{4}{7}})$ is the smallest number meeting the inequality $\frac{(n-2)(n-3) - (i_d-2)(i_d-3)}{(n-2)(n-3) - (i_d-1)(i_d-2)} \geq \sqrt{1 + \frac{2}{i_d(i_d+1)}}$. Based on these findings, an algorithm is presented to find $OHC$ in $O(n^2i_d^42^{i_d})$ time using dynamic programming. The experiments are executed to verify these findings with various $TSP$ instances.

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BibTeXRIS

Yong Wang. 2026-04-03. The frequency $K_i$s for symmetrical traveling salesman problem. https://arxiv.org/abs/2504.19608

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