An iterative algorithm to eliminate edges for traveling salesman problem based on a new binomial distribution

Abstract

Traveling salesman problem (TSP) is one of the extensively studied NP-hard problems. The recent research showed that the TSP on sparse graphs could be resolved in the relatively shorter computation time than that on the complete graph $K_{n}$. This paper updates a previous probability model for the optimal Hamiltonian cycle edges according to the frequency quadrilaterals in $K_{n}$. A new binomial distribution for TSP is rebuilt to show the probability that an edge e has the frequency 5 in a frequency quadrilateral. Based on the binomial distribution, an iterative algorithm is designed to compute the sparse graphs for TSP. There are two steps at each computation cycle. Firstly, N frequency quadrilaterals containing an edge e in the input graph is chosen to compute the average frequency $\bar {f}(e)$ with the frequency quadrilaterals where e has the frequency 5. Secondly, half edges with the small values $\bar {f}(e)$ are eliminated. The two steps are repeated until a sparse graph is computed. The computation time of the algorithm is $O(Nn^{2})$. For the TSP instances in the TSPLIB, the experimental results illustrated that the sparse graphs with the $O(n\log _{2} n)$ edges are computed and the original optimal solution is preserved. The experiments means the optimal Hamiltonian cycle edges have the bigger average frequency $\bar {f}(e)$ in $K_{n}$ and the subgraphs of $K_{n}$ so they are preserved in the computation process.

Appendix

Theorem 1

Given the $p_{5}(e)=\frac {1}{4}+ \epsilon _{n}$ is the minimum probability that the OHC edge e has the frequency 5 in a frequency quadrilateral in $K_{n}$, the probability $p_{5}(e)=\frac {1}{4} + \epsilon _{n}(or\ \epsilon _{n})$ tends to 1 (or 3/4) as n is big enough.

Proof

In the following, we prove that the $p_{5}(e)\to 1$ holds for $e\in OHC$ as n is big enough. We assume that the bigger graph $K_{n + 1}$ originates from a smaller graph $K_{n}$. Some notations are given first for proof.

Based on the axiom of choice, we will construct such two graphs $K_{n}$ and $K_{n + 1}$ which have the two probabilities ${p_{5}^{n}}(e)$ and $p_{5}^{n + 1}(e^{\prime })$.

${p_{5}^{n}}(e)=\frac {1}{4} + \epsilon _{n}$::

The minimum probability that $e\in OHC$ has the frequency $f = 5$ in a frequency quadrilateral in $K_{n}$ where $\epsilon _{n}\to 0$.

$p_{5}^{n + 1}(e^{\prime })=\frac {1}{3} + \epsilon _{n + 1}$::

The minimum probability that $e^{\prime }\in OHC$ has the frequency $f = 5$ in a frequency quadrilateral in $K_{n + 1}$ where $\epsilon _{n + 1}\to 0$.

$M_{n}={p_{5}^{n}}(e){{n-2}\choose {2}}$::

The minimum number of the frequency quadrilaterals where $e\in OHC$ has the frequency $f = 5$ in $K_{n}$.

$M_{n + 1}=p_{5}^{n + 1}(e^{\prime }){{n-1}\choose {2}}$::

The minimum number of the frequency quadrilaterals where $e^{\prime }\in OHC$ has the frequency $f = 5$ in $K_{n + 1}$.

Since $\epsilon _{n + 1}\to 0$ and $\epsilon _{n}\to 0$, we have the limitations $p_{5}^{n + 1}(e^{\prime }) \to \frac {1}{4}$ and ${p_{5}^{n}}(e) \to \frac {1}{4}$. Because $M_{n + 1} \geq M_{n}$, we derive the inequality

$$\frac{p_{5}^{n + 1}(e^{\prime})}{{p_{5}^{n}}(e)} \geq \frac{{{n-2}\choose{2}}}{{{n-1}\choose{2}}} = 1-\frac{2}{n-1}.$$

We assume a big number M to ensure the limitation $\lim _{n\to M}\frac {2}{n-1} = 0$. For any $n > M$, we have $p_{5}^{n + 1}(e^{\prime }) \geq {p_{5}^{n}}(e)$. It means the minimum probability $p_{5}(e)$ that the OHC edge e has the frequency $f = 5$ in a frequency quadrilateral in $K_{n}$ increases according to n until it reaches the maximum value 1. Thus, the OHC edges have the big probability $p_{5}(e)$ in any subgraphs containing e of $K_{n}$ as n is big enough. □