The Kth Traveling Salesman Problem is Pseudopolynomial when TSP is polynomial

TL;DR

This paper demonstrates that the Kth Traveling Salesman Problem becomes pseudopolynomial solvable when the classical TSP can be solved in polynomial time, highlighting a nuanced complexity relationship.

Contribution

It establishes that KthTSP is pseudopolynomial under the condition that TSP is polynomial, providing new insights into the problem's complexity.

Findings

KthTSP is NP-hard for fixed K.

KthTSP is pseudopolynomial if TSP is polynomial.

The result links TSP complexity to the KthTSP variant.

Abstract

Given an undirected graph $G=(V, E)$ with a weight function $c\in R^E$, and a positive integer $K$, the Kth Traveling Salesman Problem (KthTSP) is to find $K$ Hamilton cycles $H_1, H_2, , ..., H_K$ such that, for any Hamilton cycle $H\not \in \{H_1, H_2, , ..., H_K \}$, we have $c(H)\geq c(H_i), i=1, 2, ..., K$. This problem is NP-hard even for $K$ fixed. We prove that KthTSP is pseudopolynomial when TSP is polynomial.

Equations

$$x_j\in \bigcup _{i=1}^{j-1}N(x_i)\backslash \{x_1,x_2,...,x_{j-1}\}.$$

Full text

2010 Mathematics Subject Classification: Primary 90C27, Secondary 90C57.

Key words and phrases: K best solutions, Traveling Salesman Problem, Kth best Traveling Salesman Problem, pseudopolynomial.

1. Introduction

Sets and their characterisitic vectors will not be distinguished. We refer to Bondy and Murty [1] and Schrijver [17] about, respectively, graph theory and polyhedra terminolgy and facts.

Given an undirected graph $G=(V,E)$ with a weight function $c\in R^{E}$, and a positive integer $K$, the Kth Traveling Salesman Problem (KthTSP) is to find $K$ distinct Hamilton cycles $H_{1},H_{2},,...,H_{K}$ such that, for any Hamilton cycle $H\not\in\{H_{1},H_{2},,...,H_{K}\}$, we have $c(H)\geq c(H_{i}),i=1,2,...,K$. Since KthTSP is the famous TSP for $K=1$, then KthTSP is NP-hard even for $K$ fixed. KthTSP is motivated by searching near optimal solutions with some special properties: when in addition of the TSP comstraints, ”there are some other wich might be difficult to consider explicitly in a mathematical model, or if considered, would increase largely the size of the model. By finding the best, second best, …, Kth best solution, we are able to sequentially verify these solutions with respect to the additional constraints and stop when a solution that satisfies all of them is found” [19]. Another motivation is that if, for any reason, the route of the best solution is unavailable, then alternate solutions (routes) are desirable [16].

Finding K best solutions of an optimization problem in general has been studied by few authors [11, 14, 15, 18] and almost the same situation happened for particular problems [2, 3, 4, 5, 6, 9, 10, 12, 13, 15].

The remainder of the paper is organized as follows: in section 2, we give an algorithm for finding K best solutions for a general model containing KthTSP, then, in section 3, we apply this algorithm to KthTSP and deduce that it is polynomial on $K$ and $|E|$ when TSP is polynomial. And we conclude in section 4.

2. An Algorithm for Finding K Best Solutions of a Large Class of Combinatorial Optimization Problems

Let $P\subseteq R^{m}$ be a polyhedra, $f(m)$ be the number of its facets, $N(x)$ be the set of all neighbors of an extreme point $x\in P$, $x_{K}$ the Kth best solution in $P$, regarding to a given weight function and a given positive integer $K$.

Based on the following property, an algorithm has been used for particular problems [5, 6].

Proposition 2.1**.**

For any positive integer $j$ such that $2\leq j\leq K$,

[TABLE]

Since selecting $K$ best numbers from a list of $n$ numbers requires a running time complexity of $O(n+KlogK)$ [7], solving an $n\times n$ system of linear equations is $O(n^{3})$ [8], and if $C(m)$ is the running time complexity for finding the best solution on $P$, then we have the following two consequences.

Corollary 2.2**.**

The running time complexity for finding K best solutions of $P$ regarding to a given weight function is $O(C(m)+KNm^{3}+KlogK)$ where $N$ is the maximum cardinality of all $N(x_{i}),i=1,...,K-1$.

Since $N$ can be bounded by $mf(m)-m^{2}$ then:

Corollary 2.3**.**

The running time complexity for finding K best solutions of $P$ regarding to a given weight function is $O(C(m)+Km^{4}f(m)+KlogK)$.

Corollary 2.4**.**

If $C(m)$ and $f(m)$ are polynomial on $m$ then finding K best solutions of $P$ is pseudopolynomial, i.e., polynomial on $m$ and $K$.

We will propose now a new algorithm which generalizes one used in [11] for the Kth Best Base of a Matroid (KBBM).

Let us give a general model of combinatorial objects containing Hamilton cycles.

Let $E$ be a finite set and $\mathcal{X}=\subseteq\{0,1\}^{E}$. We say that $\mathcal{X}$ is an $\alpha$-bases system, where $\alpha$ is a positive integer, if the following conditions hold:

1. (1)

$\alpha=Min\{x\backslash y$ such that $(x,y)\in\mathcal{X}^{2},x\neq y\}$; 2. (2)

there exists a positive integer $r$ such that $x(E)=r$, for any $x\in\mathcal{X}$; 3. (3)

for any $(x,x^{\prime})\in\mathcal{X}^{2}$, there exist $t\in N$, $F^{\prime}_{i}\subseteq x^{\prime}\backslash x$, and $F_{i}\subseteq x\backslash x^{\prime},i=1,2,...,t$ such that $\alpha\leq|F_{i}|=|F^{\prime}_{i}|\leq\alpha+1$, $x^{\prime}=x\backslash(\bigcup_{i=1}^{t}F_{i})\cup(\bigcup_{i=1}^{t}F^{\prime}_{i})$ and $x\backslash(\bigcup_{i\in I\subseteq\{1,...,t\}}F_{i})\cup(\bigcup_{i\in I\subseteq\{1,...,t\}}F^{\prime}_{i})\in\mathcal{X}$.

Such pair $(F_{i},F^{\prime}_{i})$ verifying the condition (3) is called an $x$-exchangeable pair.

Note that bases of a matroid form a 1-bases system and we will prove that Hamilton cycles of a complete graph form a 2-bases system.

We have then the following property for K best solutions of $\alpha$-bases system.

Theorem 2.5**.**

Given a weight function $c\in R^{E}$ and a jth $c$-best solution (of $\mathcal{X}$) $x$. If $(F_{0},F^{\prime}_{0})$ is an $x$-exchangeable pair such that $c(F_{0})-c(F^{\prime}_{0})=Maximum\{c(F)-c(F^{\prime})$ such that $(F,F^{\prime})$ is an $x$-exchangeable pair and $c(F)-c(F^{\prime})\leq 0$}, then $x_{0}=(x\backslash F_{0})\cup F^{\prime}_{0}$ is a (j+1)th $c$-best solution of $\mathcal{X}$.

Proof.

By induction on $j\geq 1$.

By using the condition (3) of the definition of $\alpha$-bases systems, any $x^{\prime}\in\mathcal{X}\backslash\{x\}$ can be expressed as $x^{\prime}=x\backslash(\bigcup_{i=1}^{t}F_{i})\cup(\bigcup_{i=1}^{t}F^{\prime}_{i})$ for some $x$-exchangeable pairs $(F_{i},F^{\prime}_{i})\subseteq(x\backslash x^{\prime})\times(x^{\prime}\backslash x),i=1,...,t$. Since $x$ is a $c$-best solution then $\mathcal{F}_{x}=\{(F,F^{\prime})$ $x$-exchangeable pairs such that $c(F)-c(F^{\prime})>0$}=$\O$. Thus $c(x_{0})=c(x)-(c(F_{0})-c(F^{\prime}_{0}))\leq c(x)-\sum_{i=1}^{t}(c(F_{i})-c(F^{\prime}_{i}))=c(x^{\prime})$. So $x_{0}$ is the 2nd $c$-best solution.

Suppose now that $j\geq 2$ and let $x_{i}$ be the ith $c$-best solution for $i=1,2,...,j$.

For any subset $X\subseteq\mathcal{F}_{x}$ we can get a $x_{i}=x\backslash(\bigcup_{(F,F^{\prime})\in X}F)\cup(\bigcup_{(F,F^{\prime})\in X}F^{\prime})$ and $c(x_{i})=c(x)-\sum_{(F,F^{\prime})\in X}(c(F)-c(F^{\prime}))\leq c(x)$ ($X=\O$ gives $x=x_{j}$ itself and $X=\mathcal{F}_{x}$ gives the $c$-best solution). It follows that $x_{j+1}=x_{0}=x\backslash F_{0}\cup F^{\prime}_{0}$ because of a similar argument as for $j=1$. ∎

This proof gives an algorithm for finding K best solutions in $\alpha$-bases systems. The algorithm consists of finding the best solution first ($O(C(m))$) and then the 2nd best by adding a subset to the (best) solution ($O(|E|-r)$), finding the matched subsets of our (best) solution forming an $x$-echangeable pair ($O(\theta)$) and choosing the best subset of this solution forming an exchangeable pair ($O(r)$). By repeating this procedure $K$ times, the running time complexity of this algorithm is $O(C(m)+Kr(|E|-r)\theta)$ where $\theta$ is the running time complexity of the oracle used to find exchangeable pairs.

3. KthTSP is pseudopolynomial when TSP is polynomial

First we need to prove that Hamilton cycles of a complete graph verify the properties (1)-(3) of $\alpha$-bases systems.

Theorem 3.1**.**

Hamilton cycles of a complete graph form a 2-bases system.

Proof.

For Hamilton cycles, E is the set of edges of a given complete graph $K_{n}$.

Property (1): It is clear that $\alpha=2$.

Property (2): It is clear that $r=n$.

Property (3): Let $H$ and $H^{\prime}$ two distinct Hamilton cycles and $d(H,H^{\prime})=|H\backslash H^{\prime}|$. We will prove this property by induction on $d(H,H^{\prime})$.

If $d(H,H^{\prime})=2$ (respectively 3) then let $F=H\backslash H^{\prime}$ and $F^{\prime}=H^{\prime}\backslash H$. It is not difficult to see that $H^{\prime}=(H\backslash F)\cup F^{\prime}$, $|F|=|F^{\prime}|=2=\alpha$ (respectively $=3=\alpha+1$) and $(F,F^{\prime})$ is an $H$-exchangeable pair.

If $d(H,H^{\prime})=2p$ (respectively $2p+1$), with $p\geq 2$, then there exists a circuit $C=\{e,e^{\prime},f,f^{\prime}\}$ (of cardinality 4) such that $\{e,f\}\subseteq H\backslash H^{\prime}$, $e^{\prime}\in H^{\prime}\backslash H$, $f^{\prime}\notin H$ and $H^{\prime\prime}=H\Delta C=H\backslash\{e,f\}\cup\{e^{\prime},f^{\prime}\}$ is a Hamilton cycle. It is clear that $d(H^{\prime\prime},H^{\prime})\leq d(H,H^{\prime})-1$. By induction, $H^{\prime}$ can be expressed in means of $H^{\prime\prime}$ and $H^{\prime\prime}$-exchangeable pairs. If $f^{\prime}\in H^{\prime}$ then we are done. Else, one of the removed $H^{\prime\prime}$-exchangeable pairs should contain $f^{\prime}$ and by subsituting $H^{\prime\prime}$, we will get an $H$-exchangeable pair with components of cardinality 3. ∎

Since finding exchangeable pairs corresponds to choose 2 nonadjacent edges (respectively 3 edges) from a Hamilton cycle and to find 2 nonadjacent edges (respectively 3 edges) such that exchanging between them gives a new Hamilton cycle then $O(\theta)=O(1)$. It follows that the running time complexity of our algorithm for KthTSP is $O(C(m)+Knm)$. Then we can state our main result.

Corollary 3.2**.**

KthTSP is pseudopolynomial when TSP is polynomial.

Proof.

If TSP is polynomial for (special instances of) complete graphs then $C(m)$ is polynomial and we are done.

If TSP is polynomial for special classes of graphs, then we can put an infinity weight to removed edges from the corresponding complete graph and we get the same result. ∎

Note that, with a natural modification, our algorithm works for arbitrary weights and for Max KthTSP.

4. Conclusion

We have generalized an algorithm described in [11] for a generalization of bases of a matroid. By applying this algorithm to Hamilton cycles, we have proved that KthTSP is psudopolynomial when TSP is polynomial. Future investigations can be applying this algorithm for appropriate combinatorial objects.