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The Fault-Tolerant Group Steiner Problem Rohit Khandekar IBM Watson Joint work with G. Kortsarz and Z. Nutov.

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Presentation on theme: "The Fault-Tolerant Group Steiner Problem Rohit Khandekar IBM Watson Joint work with G. Kortsarz and Z. Nutov."— Presentation transcript:

1 The Fault-Tolerant Group Steiner Problem Rohit Khandekar IBM Watson Joint work with G. Kortsarz and Z. Nutov

2 Fault-tolerant group Steiner problem Given: A weighted graph G(V,E), a collection of subsets (groups) g i  V and a root r. The goal: Find a minimum weight subgraph in which for each g i, at least 2 vertices have edge (or vertex) disjoint paths to r.

3 An example g1g2g3 g1g2g3 r

4 Previous work on fault-tolerant problems: Steiner networks Steiner Network: Instance: A complete graph with edge (or vertex) costs, connectivity requirements r(u,v) Objective: Min-cost subgraph with r(u,v) edge (or vertex) disjoint uv-paths for all u,v in V k-edge-Connected Subgraph: r(u,v) = k for all u,v, edge-disjointness k-vertex-Connected Subgraph: r(u,v) = k for all u,v, vertex-disjointness

5 Previous work on Steiner Network Edge case: A sequence of papers reaching a 2-approximation [Jain 98] Vertex case: Labelcover hard [Kortsarz, Krauthgamer, Lee 04] k ε approximation is unlikely for some universal ε>0 [Chakraborty, Chuzhoy, Khanna 08] Undirected and directed problems are equivalent for k>n/2 [Lando, Nutov 08] O(log n)-approximation for metric cost [Cheriyan, Vetta 05] O(k 3 log n)-approximation [Chuzhoy, Khanna 09]

6 2-connectivity problems (like ours) 2-edge-connected subgraph spanning k vertices: O(log n log k) [Lau, Naor, Salavatipour, Singh 09] (fault tolerant version of k-MST) Same problem with 2-vertex-connectivity: O(log n log k) [Chekuri, Korula 08] Finding buy at bulk trees with 2-vertex-disjoint paths from the terminals to the root: O(log 3 n) [Antonakopoulos, Chekuri, Shepherd, Zhang 07]

7 Our results ProblemEdge caseVertex case FTGS-2 (each group has 2 vertices) 3.55 VC hard O(log 2 n) FTGS-k (each group has k vertices) O(k log 2 n) FTGS (with disjoint groups) O(  n log n) GS hard O(  n log n) GS hard FTGS directedLabel Cover Hard FTGS = Fault tolerant group Steiner, GS = Group Steiner, VC = Vertex cover

8 Our results ProblemEdge caseVertex case FTGS-2 (each group has 2 vertices) 3.55 VC hard O(log 2 n) FTGS-k (each group has k vertices) O(k log 2 n) FTGS (with disjoint groups) O(  n log n) GS hard O(  n log n) GS hard FTGS directedLabel Cover Hard FTGS = Fault tolerant group Steiner, GS = Group Steiner, VC = Vertex cover

9 Why is our problem difficult? Known algorithms for Group Steiner tree are based on approximating the given metric by tree metrics [Bartal 98], [Fakcharoenphol, Rao, Talwar 03] and solving the problem on trees. This reduction does not preserve the connectivity information and hence cannot be used here. An intriguing question: Can we approximate Group Steiner problem without first transforming the graph into a tree?

10 Algorithm for FTGS-2 (edge case) As |g i | = 2, all terminals must be connected to the root in any feasible solution. Therefore we first find a STEINER TREE T connecting all terminals to the root (1.55-approximation). Then we augment T to a feasible FTGS-2 solution.

11 Violated sets Say that X  V is violated if there is only one edge leaving X, but there should be two edges leaving X (i.e., X does not contain r but contains a group). Claim: If X and Y are violated, either X U Y and X ∩ Y are both violated, or X-Y and Y-X are both violated. Such a family of violated sets is called “uncrossable”.

12 For any violated set X, the set X ∩ T must be a sub-tree of T containing an entire group. Subtrees are laminar! (i.e., either two subtrees are disjoint or one is contained in the other.) Why are violated sets uncrossable? gigi X ∩ T

13 The two cases g1g1 g2g2 X ∩ T Y ∩ T X-Y=X and Y-X=Y g1g1 g2g2 X ∩ T Y ∩ T X∩Y=Y and XUY=X

14 Consequence The problem of finding a minimum cost cover of an uncrossable family admits 2 approximation (Primal-Dual) [Goemans, Goldberg, Plotkin, Shmoys, Tardos 94]. Therefore, overall we get 1.55 + 2 = 3.55 approximation. It is also easy to see that the problem is Vertex Cover hard.

15 Algorithm for FTGS-2 (vertex case) First step: Steiner tree (same) Second step: Augmentation problem is now different u1u1 u2u2 u g

16 The augmentation problem Theorem: The group g is satisfied iff either u 1 or u 2 is 2-vertex-connected with r. u1u1 u2u2 u g

17 The augmentation problem u1u1 u2u2 u g r

18 Profit(v) = number of groups g for which v serves the role of either u 1 or u 2 u1u1 u2u2 u g

19 Density version of 2-vertex-connected graph problem Given a graph with profits on vertices, find a subgraph H that minimizes the ratio of cost(H) to the profit of vertices that are 2-vertex-connected to r in H. O(log n)-approximation [Chekuri, Korula 08] This combined with the set-cover analysis gives O(log 2 n)-approximation.

20 FTGS-k (|g| ≤ k for all groups g) A similar argument with a careful counting gives O(k log 2 n) approximation if groups are assumed to be disjoint.

21 Thanks!

22 How many groups can u1 or u2 cover? g3g1 g4g3 P=2 P=3 r g4 g1 g3 g2 P=4 P=2 P=1 g2


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