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Extra resources for Algorithms - Sequential, Parallel - A Unified Appr.

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Doing this for each edge set Ti yields a feasible solution to the original instance using exactly m i=1 m+ 2 R 2 e∈Ti \Pi R d(e) buses. This is at most m d(e) = m + 2 i=1 e∈Ti \Pi e∈T \Q R d(e) ≤ 3OP T by Proposition 2 and Proposition 3, where OP T is the optimal number of buses required in any feasible solution. Together with Lemma 1, this proves Theorem 1. One may notice that the bounds given in Propositions 2 and 3 are necessarily also respected by fractional solutions to the LP relaxation of (IP ).

3 log n Note that 2−i · ki ≥ k. Consider an instance of classical survivable i=0 network design problem over terminals in Ti ∪ {r} with connectivity requirement 2 from every node in Ti to root. In the following lemma we show that we can compute a 2-edge-connected subgraph Hi over Ti ∪ {r} of cost at most O(2i · opt∗ ). This describes how to perform Step 7. 2 in [14]. Lemma 8. In Step 7, For each 0 ≤ i ≤ 3 log n , we can find a 2-edge-connected subgraph Hi of cost at most 2i+3 · opt∗ containing terminals Ti ∪ {r}.

At each iteration, we guess the starting point v ∗ (by trying all |V |−1 possibilities). Using Proposition 1, the resulting problem we are left with is to find a v∗ − s walk in G of length at most d(v ∗ , s) + R visiting the maximum number of uncovered nodes in W . Such a problem is well known in the literature as the Orienteering Problem, and can be approximated within a constant [3, 4]. , see [13]), we may then obtain an O(log C)-approximation algorithm for the SBP and show that the integrality gap of (IP) is at most O(log C) (we refer to the full version of this paper for a rigorous argument).

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