By Alexander A. Ageev, Alexander V. Kononov (auth.), Thomas Erlebach, Christos Kaklamanis (eds.)

This e-book constitutes the completely refereed post-proceedings of the 4th foreign Workshop on Approximation and on-line Algorithms, WAOA 2006, held in Zurich, Switzerland in September 2006 as a part of the ALGO 2006 convention event.

The 26 revised complete papers awarded have been conscientiously reviewed and chosen from sixty two submissions. subject matters addressed via the workshop are algorithmic online game thought, approximation sessions, coloring and partitioning, aggressive research, computational finance, cuts and connectivity, geometric difficulties, inapproximability effects, mechanism layout, community layout, packing and protecting, paradigms, randomization ideas, real-world purposes, and scheduling problems.

**Read or Download Approximation and Online Algorithms: 4th International Workshop, WAOA 2006, Zurich, Switzerland, September 14-15, 2006. Revised Papers PDF**

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**Extra resources for Approximation and Online Algorithms: 4th International Workshop, WAOA 2006, Zurich, Switzerland, September 14-15, 2006. Revised Papers**

**Example text**

Erlebach and C. ): WAOA 2006, LNCS 4368, pp. 29–42, 2006. c Springer-Verlag Berlin Heidelberg 2006 30 D. Amzallag, J. Naor, and D. Raz and p(i, i , j) = 0 for every j ∈ / Si )1 . This means that the interference caused as a result of a coverage of a client by more than one base station depends on the geographical position of the related “client”. Followed by the above setting, we denote by Q(i, j) the net contribution of base station i to client j, for every j ∈ J, i ∈ I, after incorporating the interference.

First, we show that this problem remains hard. of 2e−1 Theorem 2. The k4k-budgeted cell planning problem is NP-hard. Proof. Via a reduction from the budgeted maximum coverage problem. Consider an instance of the budgeted maximum coverage problem, that is, a collection of subsets S = {S1 , . . , Sm } with associated costs {ci }m i=1 over a domain of elements X = {x1 , . . , xn }, and a budget L. We can construct an instance of k4k-BCPP such that an optimal solution to this problem gives an optimal solution to the budgeted maximum coverage problem.

Note also that the minimum-length chain for Θ−x ends at slot x, and so if y is present in this chain, it cannot follow a downward link; otherwise the chain would contradict Lemma 1, since x is above y. Thus we may conclude that y ends up in position y ≤ y. Since slot y is in range for bidder x by assumption, we also have that y is in range for bidder x; thus we can construct a candidate solution for Θ−y by replacing (in Θ−x ) bidder y with bidder x. We may conclude that Θ−y ≥ Θ−x + cy (vx − vy ) ≥ Θ−x + cy (vx − vy ).