By Ashkan Aazami, Michael D. Stilp (auth.), Moses Charikar, Klaus Jansen, Omer Reingold, José D. P. Rolim (eds.)

This quantity comprises the papers awarded on the tenth foreign Workshopon Approximation Algorithms for Combinatorial Optimization Problems(APPROX 2007) and the eleventh foreign Workshop on Randomization andComputation (RANDOM 2007), which came about at the same time at PrincetonUniversity, on August 20–22, 2007. APPROX makes a speciality of algorithmic and complexityissues surrounding the improvement of effective approximate solutionsto computationally tough difficulties.

**Read or Download Approximation, Randomization, and Combinatorial Optimization. Algorithms and Techniques: 10th International Workshop, APPROX 2007, and 11th International Workshop, RANDOM 2007, Princeton, NJ, USA, August 20-22, 2007. Proceedings PDF**

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**Example text**

Thus, the algorithm will select π(j) if π(j) ∈ Y . Finally, note that the event π(j) ∈ Y is independent of E, so Prob[π(j) ∈ S | E ∧ (a = 4)] = Prob[π(j) ∈ Y | E ∧ (a = 4)] = 12 . 3, which completes the proof of the lemma. 5 Conclusion In this paper, we have presented algorithms for a knapsack version of the secretary problem, in which an algorithm has to select, in an online fashion, a maximum-value subset from among the randomly ordered items of a knapsack problem. We gave a constant-competitive algorithm in this model, as well as a e-approximation for the k-secretary problem, in which all items have identical weights.

Finally, for a set R ⊆ U we will (x) (x) deﬁne vQ (R), wQ (R) by (x) (x) vQ (R) = v(i)yQ (i) i∈R (x) wQ (R) (x) w(i)yQ (i). 2 The Algorithm For convenience, we assume in this section that W = 1. ) Our algorithm begins by sampling a random number a ∈ {0, 1, 2, 3, 4} from the uniform distribution. The case a = 4 is a special case which will be treated in the following paragraph. If 0 ≤ a ≤ 3, then the algorithm sets k = 3a and runs the k-secretary algorithm from Section 3 (with t = n/e ) to select at most k elements.

On the other hand, Theorem 1 implies that gj ≥ v(R)/e. The lemma follows by combining these two bounds. Lemma 4. b4 ≤ 2eg4 . Proof. Assuming the algorithm chooses a = 4, recall that it splits the input into a “sample set” X = {1, 2, . . , t} and its complement Y = {t + 1, . . , n}, where t is a random sample from the binomial distribution B(n, 1/2). Recall that in the case a = 4, the algorithm aims to ﬁll the knapsack with multiple items of weight at most 1/81, and value density at least equal to the value density of the optimal solution for the sample (and a knapsack of size 1/2).