Network Design with Coverage Costs

We study network design with a cost structure motivated by redundancy in data traffic. We are given a graph, g groups of terminals, and a universe of data packets. Each group of terminals desires a subset of the packets from its respective source. The cost of routing traffic on any edge in the network is proportional to the total size of the distinct packets that the edge carries. Our goal is to find a minimum cost routing. We focus on two settings. In the first, the collection of packet sets desired by source-sink pairs is laminar. For this setting, we present a primal-dual based 2-approximation, improving upon a logarithmic approximation due to Barman and Chawla (2012). In the second setting, packet sets can have non-trivial intersection. We focus on the case where each packet is desired by either a single terminal group or by all of the groups, and the graph is unweighted. For this setting we present an O(log g)-approximation. Our approximation for the second setting is based on a novel spanner-type construction in unweighted graphs that, given a collection of g vertex subsets, finds a subgraph of cost only a constant factor more than the minimum spanning tree of the graph, such that every subset in the collection has a Steiner tree in the subgraph of cost at most O(log g) that of its minimum Steiner tree in the original graph. We call such a subgraph a group spanner.Comment: Updated version with additional result

Approximating the minimum directed tree cover

Given a directed graph $G$ with non negative cost on the arcs, a directed tree cover of $G$ is a rooted directed tree such that either head or tail (or both of them) of every arc in $G$ is touched by $T$. The minimum directed tree cover problem (DTCP) is to find a directed tree cover of minimum cost. The problem is known to be $NP$-hard. In this paper, we show that the weighted Set Cover Problem (SCP) is a special case of DTCP. Hence, one can expect at best to approximate DTCP with the same ratio as for SCP. We show that this expectation can be satisfied in some way by designing a purely combinatorial approximation algorithm for the DTCP and proving that the approximation ratio of the algorithm is $\max\{2, \ln(D^+)\}$ with $D^+$ is the maximum outgoing degree of the nodes in $G$.Comment: 13 page

Hitting Diamonds and Growing Cacti

We consider the following NP-hard problem: in a weighted graph, find a minimum cost set of vertices whose removal leaves a graph in which no two cycles share an edge. We obtain a constant-factor approximation algorithm, based on the primal-dual method. Moreover, we show that the integrality gap of the natural LP relaxation of the problem is \Theta(\log n), where n denotes the number of vertices in the graph.Comment: v2: several minor changes

Approximating k-Forest with Resource Augmentation: A Primal-Dual Approach

In this paper, we study the $k$-forest problem in the model of resource augmentation. In the $k$-forest problem, given an edge-weighted graph $G(V,E)$, a parameter $k$, and a set of $m$ demand pairs $\subseteq V \times V$, the objective is to construct a minimum-cost subgraph that connects at least $k$ demands. The problem is hard to approximate---the best-known approximation ratio is $O(\min\{\sqrt{n}, \sqrt{k}\})$. Furthermore, $k$-forest is as hard to approximate as the notoriously-hard densest $k$-subgraph problem. While the $k$-forest problem is hard to approximate in the worst-case, we show that with the use of resource augmentation, we can efficiently approximate it up to a constant factor. First, we restate the problem in terms of the number of demands that are {\em not} connected. In particular, the objective of the $k$-forest problem can be viewed as to remove at most $m-k$ demands and find a minimum-cost subgraph that connects the remaining demands. We use this perspective of the problem to explain the performance of our algorithm (in terms of the augmentation) in a more intuitive way. Specifically, we present a polynomial-time algorithm for the $k$-forest problem that, for every $\epsilon>0$, removes at most $m-k$ demands and has cost no more than $O(1/\epsilon^{2})$ times the cost of an optimal algorithm that removes at most $(1-\epsilon)(m-k)$ demands