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Algorithms and Uncertainty Summer Term 2021

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Thomas Kesselheim April 15, 2021

Alexander Braun Due: April 21, 2021 at noon

Algorithms and Uncertainty Summer Term 2021

Exercise Set 1

If you want to hand in your solutions for this problem set, please send them via email to alexander.braun@uni-bonn.de - make sure to send a pdf-file which contains your name and your email address. Of course, submitting solutions in groups is also possible. We will discuss this sheet in the tutorials on April 22.

Exercise 1: (1+3 Points)

Consider the following algorithm for the ski rental problem: Buy a pair of ski on your first skiing day for a price of B.

(a) Give a sequence σ of skiing/non-skiing days such that c(ALG(σ)) = c(OPT(σ)).

(b) Is there anα >0 such that this algorithm is strictlyα-competitive? If yes, give a proof, otherwise a counterexample.

Exercise 2: (3 Points)

The Online Bipartite Vertex Cover problem is defined as follows: We are given a bipartite graph with vertices V = L∪R. The nodes in L are offline which means they are present initially. Nodes in R are online and revealed one at a time together with its incident edges.

Each vertex v ∈V has a cost cv. We need to maintain a feasible vertex cover in every step with the goal of minimizing the overall incurred costs.

Consider the ski rental problem in a simplified version, i.e. we assume that every day is a skiing day but we do not know the number of days in advance. Show in a constructive way that the ski rental problem is a special case of the Online Bipartite Vertex Cover problem.

Exercise 3: (4 Points)

We want to show that the assumptions (completeness and triangular inequality) in the lecture concerning the online Steiner tree problem are indeed without loss of generality. Therefore, consider an α-competitive online algorithm for the online Steiner tree problem on complete graphs satisfying the triangular inequality (a.k.a. metric Steiner tree problem). Show how to convert this algorithm into anα-competitive one for the general online Steiner tree problem.

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