On Approximating Degree-Bounded Network Design Problems

Abstract

Directed Steiner Tree (DST) is a central problem in combinatorial optimization and theoretical computer science: Given a directed graph $G=(V, E)$ with edge costs $c \in {\mathbb {R}}_{\ge 0}^E$, a root $r \in V$ and k terminals $K\subseteq V$, we need to output the minimum-cost arborescence in G that contains an $r \rightarrow t$ path for every $t \in K$. Recently, Grandoni, Laekhanukit and Li, and independently Ghuge and Nagarajan, gave quasi-polynomial time $O(\log ^2k/\log \log k)$-approximation Algorithms for the problem, which are tight under popular complexity assumptions. In this paper, we consider the more general Degree-Bounded Directed Steiner Tree (DB-DST) problem, where we are additionally given a degree bound $d_v$ on each vertex $v \in V$, and we require that every vertex v in the output tree has at most $d_v$ children. We give a quasi-polynomial time $(O(\log n \log k), O(\log ^2 n))$-bicriteria approximation: The Algorithm produces a solution with cost at most $O(\log n\log k)$ times the cost of the optimum solution that violates the degree constraints by at most a factor of $O(\log ^2n)$. This is the first non-trivial result for the problem. While our cost-guarantee is nearly optimal, the degree violation factor of $O(\log ^2n)$ is an $O(\log n)$-factor away from the approximation lower bound of $\Omega (\log n)$ from the set-cover hardness. The hardness result holds even on the special case of the Degree-Bounded Group Steiner Tree problem on trees (DB-GST-T). With the hope of closing the gap, we study the question of whether the degree violation factor can be made tight for this special case. We answer the question in the affirmative by giving an $(O(\log n\log k), O(\log n))$-bicriteria approximation Algorithm for DB-GST-T.

Appendix: Omitted Proofs

Proof of Lemma 2.1

We assume $n \ge 4$; otherwise, if $n = 3$, then we have $2n/3 + 1 = 3$, and $\mathrm {root}(T)$ satisfies the condition. Our goal is to find a vertex u with $n/3 < |\Lambda ^*(u)| \le 2n/3+1$. Start from $u = \mathrm {root}(T)$ in the tree, and thus, we have $\Lambda ^*(u) > 2n/3 + 1$. Let v be the child of u with the biggest $|\Lambda ^*(v)|$. So, $|\Lambda ^*(v)| \ge (|\Lambda ^*(u)|-1)/2> n/3$. We then replace u with v. So $|\lambda ^*(u)|$ has decreased but the condition $|\Lambda ^*(u)| > n/3$ is maintained. Thus, if we repeat the process, we will eventually find a u with $n/3< |\Lambda ^*(u)| \le 2n/3+ 1$. $\square$