By Alexander A. Ageev, Alexander V. Kononov (auth.), Thomas Erlebach, Christos Kaklamanis (eds.)
This ebook constitutes the completely refereed post-proceedings of the 4th overseas 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 provided have been rigorously reviewed and chosen from sixty two submissions. themes addressed via the workshop are algorithmic online game idea, approximation periods, coloring and partitioning, aggressive research, computational finance, cuts and connectivity, geometric difficulties, inapproximability effects, mechanism layout, community layout, packing and protecting, paradigms, randomization options, real-world purposes, and scheduling problems.
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Extra info for Approximation and Online Algorithms: 4th International Workshop, WAOA 2006, Zurich, Switzerland, September 14-15, 2006. Revised Papers
We say that this is a downward link if i > i; otherwise it is an upward link. Lemma 1. The minimum length chain for Θ−i does not contain a downward link followed by an upward link. Proof. Suppose it does contain such a sequence. Then, some bidder i1 moved from slot i1 to slot i2 > i1 , and bidder i2 moved from slot i2 to a slot i3 < i2 . An alternate solution, and thus a candidate solution for Θ−i is to have bidder i1 move from slot i1 to slot i3 , have bidder i2 remain in slot i2 , and keep everything else the same.
Naor. The budgeted maximum coverage problem. Information Processing Letters, 70:39–45, 1999. 11. N. Linial and N. Nisan. Approximate inclusion-exclusion. Combinatorica, 10:349– 365, 1990. 12. M. Sviridenko. A note on maximizing a submodular set function subject to knapsack constraint. Operations Research Letters, 32:41–43, 2004. Online Dynamic Programming Speedups Amotz Bar-Noy1, Mordecai J. hk Abstract. Consider the Dynamic Program h(n) = min1≤j≤n a(n, j) for n = 1, 2, . . , N . For arbitrary values of a(n, j), calculating all the h(n) requires Θ(N 2 ) time.
We must show that x does not envy y, and that y does not envy x. If y is out of range of bidder x, then certainly x does not envy y. If x is in range of bidder y, then by Lemma 2(i), we get Θ−y ≥ Θ−x + cy (vx − vy ). Substituting for Θ−y and Θ−x using the deﬁnitions of py and px , we get Θ + py − cy vy ≥ Θ + px − cx vx + cy (vx − vy ) ⇐⇒ cy vx − py ≤ cx vx − px = ux . Thus x does not envy y. Similarly, Lemma 2(ii) shows that y does not envy x. Now consider some bidder z not assigned in Θ. We must show that bidder z does not envy any bidder that is assigned a slot in the desired range of z.
Approximation and Online Algorithms: 4th International Workshop, WAOA 2006, Zurich, Switzerland, September 14-15, 2006. Revised Papers by Alexander A. Ageev, Alexander V. Kononov (auth.), Thomas Erlebach, Christos Kaklamanis (eds.)