A 2-Approximation for the k-Prize-Collecting Steiner Tree Problem

Abstract

We consider the $k$-Prize-Collecting Steiner Tree Problem. An instance is composed of an integer k and a graph G with costs on edges and penalties on vertices. The objective is to find a tree spanning at least k vertices which minimizes the cost of the edges in the tree plus the penalties of vertices not in the tree. This is one of the most fundamental network design problems and is a common generalization of the Prize-Collecting Steiner Tree Problem and the $k$-Minimum Spanning Tree Problem. Our main result is a 2-approximation algorithm, which improves on the currently best known approximation factor of 3.96 and has a faster running time. The algorithm builds on a modification of the primal-dual framework of Goemans and Williamson, and reveals interesting properties that can be applied to other similar problems.

Proof of Lemma 7

In this appendix, we prove Lemma 7. We show that, if $\mathtt {GW}(0,\emptyset )$ returns a tree T with at least k vertices, then T is a 2-approximation. The proof is adapted from [9].

Let $(T,{\mathcal {B}})$ be the tuple returned by $\mathtt {GP}(0,\emptyset )$, and remember that the growth-phase computes a corresponding laminar collection ${\mathcal {L}}$ associated with vector y. Also, let be the output of $\mathtt {GW}(0,\emptyset )$, and observe that is the tree returned by $\mathtt {PP}(T,{\mathcal {B}})$.

Recall that ${\mathcal {L}}(S)$ denotes the collection of subsets in ${\mathcal {L}}$ which contain some but not all vertices of a subset S, and that ${{\mathcal {L}}_v}$ denote the collection of subsets in ${\mathcal {L}}$ which contain a vertex v. Thus, ${\mathcal {L}}_r(S)$ is the collection of subsets in ${\mathcal {L}}(S)$ which contain the root r.

We bound the cost of into two steps. Lemma 1 bounds the cost of edges of , and Lemma 2 bounds the penalty of vertices not in .

Lemma 1

Let be the tree returned by $\mathtt {GW}(0,\emptyset )$, then

Proof

Since every edge in is tight for (y, 0), we have

We prove by induction that, at the beginning of each iteration of $\mathtt {GP}(0,\emptyset )$,

At the beginning of the growth-phase, $y = 0$, then the inequality holds. Thus, assume that the inequality holds at the beginning of an iteration. Consider the laminar collection ${\mathcal {L}}$ at the beginning of this iteration, let ${\mathcal {A}}$ be the collection of active maximal subsets in ${\mathcal {L}}^*$ which contain some vertex of , and let ${\mathcal {I}}$ be the collection of non-active maximal subsets in ${\mathcal {L}}^*$ which contain some vertex of .

If some maximal subset in ${\mathcal {L}}^*$ contains , then neither side of the inequality changes, thus assume that no maximal subset contains . Suppose that the variable $y_L$ of each active subset L is increased by $\varDelta$ in this iteration. Hence, the left side of the inequality increases by , and the right side increases by $2 |{\mathcal {A}}\setminus {\mathcal {A}}_r|\varDelta$. We claim that

and thus the inequality is maintained at the end of the iteration.

Note that ${\mathcal {A}}\cup {\mathcal {I}}$ partition . Thus, we can create a graph $T'$ from  by contracting each subset in ${\mathcal {A}}\cup {\mathcal {I}}$. As is a tree and a subgraph of T, it follows that the graph $T'$ is a tree because T is ${\mathcal {L}}$-connected by invariant (gp3). Let S be a non-active subset in ${\mathcal {I}}$, and observe that S is in ${\mathcal {B}}$. Since is pruned with ${\mathcal {B}}$, Corollary 2 implies that the degree of S on is not one. Therefore, the degree of each vertex of $T'$ corresponding a subset in ${\mathcal {I}}$ is at least 2. It follows that

Since the subset containing the root r is always active, exactly one subset in ${\mathcal {A}}$ contains r, hence ${|{\mathcal {A}}\setminus {\mathcal {A}}_r| = |{\mathcal {A}}|-1}$, and thus , which shows the claim. $\square$

Lemma 2

Let be the tree returned by $\mathtt {GW}(0,\emptyset )$, then

Proof

By Corollary 1, every vertex not spanned by T is contained in a subset in ${\mathcal {B}}$, and since $\mathtt {PP}(T,{\mathcal {B}})$ only deletes from T subsets in ${\mathcal {B}}$, we have that every vertex not spanned by is contained in a subset in ${\mathcal {B}}$. Since ${\mathcal {B}}$ is laminar, there is a collection $\{B_1,\cdots ,B_m\}$ of subsets in ${\mathcal {B}}$ which partitions the vertices in . Because every subset in ${\mathcal {B}}$ is tight for (y, 0),

$\square$

Now, we can prove Lemma 7.

Lemma 7 Let be the tree returned by $\mathtt {GW}(0,\emptyset )$, and $T^*$ be an optimal solution. Then,

Proof

Let $L^*$ be the minimal subset in ${\mathcal {L}}$ containing all the vertices of $T^*$. Also, let ${\mathcal {P}}$ be the collection of subsets in ${\mathcal {L}}$ which contain all the vertices of $L^*$. Observe that $\{{\mathcal {L}}(L^*),{\mathcal {L}}[V \setminus L^*],{\mathcal {P}}\}$ partitions ${\mathcal {L}}$. Also, since $L^*$ contains r, it follows that ${{\mathcal {P}}\subseteq {\mathcal {L}}_r}$.

Combining Lemmas 1 and 2 we have

Observe that the subsets considered in the terms are disjoint. Thus, we can simplify the indices of the summation as

Now, since y respects $c$ and $\pi ^0$, using Lemma 1 for ${\lambda }= 0$,

$\square$