A note on the generalized min-sum set cover problem

In this note, we consider the generalized min-sum set cover problem. In this problem we are given as input a universe U of n elements, a collection S = {S 1 , . . . , S m } of subsets S i of U , and a covering requirement K(S) for each S ∈ S, where K(S) ∈ {1, 2, . . . , |S|}. The output of any algorithm for the problem is an ordering of the n elements. Let C S be the position of the K(S)th element of S in the ordering. The goal is to find an ordering that minimizes S∈S C S . This problem is a generalization of the min-sum set cover problem (in which K(S) = 1 for all S ∈ S), introduced by Feige, Lovász, and Tetali [3], and the min-latency set cover problem (in which K(S) = |S| for all S ∈ S), introduced by Hassin and Levin [4]. This generalization was introduced by Azar, Gamzu, and Yin [1] in the context of a ranking problem.

Because the problem is NP-hard, Azar, Gamzu, and Yin give an O(log r)-approximation algorithm for the problem, where r = max S∈S |S|. This was improved to a constant factor randomized approximation algorithm by Bansal, Gupta, and Krishnaswamy [2]. They introduce a new linear programming relaxation for the problem and show how to use randomized rounding to achieve a performance guarantee of 485. 1 In this paper, we show that by altering their algorithm using some concepts from α-point scheduling (see Skutella [6] for a survey), we are able to improve their algorithm and obtain a performance guarantee of about 28, which is an order of magnitude better. 2 We now briefly review their algorithm and analysis, and then state the ideas we introduce to obtain an improvement. Their algorithm begins with solving the following linear programming relaxation of the problem, where the variable y S,t for t ∈ [n] (here and in the following the set {1, . . . , n} is denoted by [n]) and set S ∈ S indicates whether C S < t or not, and x e,t for e ∈ U and t ∈ [n] indicates whether element e is assigned to the tth Date: October 15, 2018. 1 They observe in their paper that they did not try to optimize the constants in their analysis. 2 Here we would like to point out that 28 ∈ O( √ 485). While we still use several ideas from their algorithm and analysis, we modify it in several key ways. In particular, we discard the idea of stages, and we use the idea of a random α-point for each element e; in particular, after modifying the solution x * to a solution x in a way similar to theirs, we then randomly choose a value α e ∈ [0, 1] for each e ∈ U . Let t e,αe be the first time t for which t t ′ =1 x e,t ′ ≥ α e . We then schedule elements e in the order of nondecreasing t e,αe . The improvements in analysis come from scrapping the stages (so we don’t need to account for the possibility of O i being too large) and using α-point scheduling; in particular, we introduce a parameter α and look for the last point in time t S,α in which y * S,t < α (the Bansal et al. paper uses α = 1/2). Choosing α randomly gives our ultimate result. We turn to the full analysis in the next section.

Let x * , y * be an optimum solution to the linear program above. Let Q > 0 be a constant determined later. Construct a new solution x from x * as follows: Initialize x := Q • x * ; for t = 1 to ⌊n/2⌋ set x e,2t := x e,2t + x e,t .

Moreover, for each e ∈ U and t ≤ ⌊n/2⌋

x * e,t ′ , (2)

x * e,t ′ .

Proof. We start by giving an alternative view on the definition of x above. Notice that

That is, I(t ′ ) is precisely the subset of indices t ′′ such that x * e,t ′′ contributes to x e,t ′ . For a fixed t ∈ [n] and t ′′ ≤ t, let J(t ′′ ) be the subset of all indices t ′ ≤ t such that x * e,t ′′ contributes to x e,t ′ , i. e., J(t ′′ ) = {t ′ ≤ t : t ′′ ∈ I(t ′ )}. By definition of I(t ′ ) and J(t ′′ )

This concludes the proof of (1).

In order to prove (2), simply notice that for each t ′′ ∈ {1, . . . , t} there is t ′ ∈ {t + 1, . . . , 2t} such that t ′′ ∈ I(t ′ ); then (2) follows from (4). Finally, (3) also follows from (4) since t ′ ∈ I(t ′ ) for all t ′ . Algorithm: As discussed above, for each e ∈ U we independently choose α e ∈ [0, 1] randomly and uniformly. Let t e,αe denote the first point in time t when t t ′ =1 x e,t ′ ≥ α e . In our final solution, we sequence the elements e ∈ U in order of nondecreasing t e,αe ; ties are broken arbitrarily.

For S ∈ S and some fixed α ∈ (0, 1), let t S,α be the last point in time t for which y * S,t < α. We observe that the contribution of set S to the objective function of the linear program is

since in time step t it holds that t S,α ≥ t for all α ∈ [0, 1] such that α > y * S,t , or for (1 -y * S,t ) of the possible α. We now bound the probability that we have fewer than K(S) elements from S with t e,αe ≤ t S,α in terms of Q and α.

Then for integer i ≥ 0, Pr {e ∈ S : t e,αe ≤ 2 i • t S,α } < K(S) ≤ p i+1 .

A := e ∈ S :

Then observe that for any e ∈ A it holds that Pr[t e,αe ≤ 2 i • t S,α ] = 1. By the properties of the linear program,

so that by ( 3)

More generally, it follows from induc

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