“At least one” caching

We consider a variant of the caching problem, where each request is a set of pages of a fixed size, instead of a single page. In order to serve such a request, we require at least one of those pages to be present in the cache. Each page is assumed to have unit size and unit cost for getting loaded into the cache. We prove lower bounds on the competitive ratio for this problem in both the deterministic and the randomized settings. We also give online algorithms for both settings and analyze them for competitive ratio. 1 The problem Let [n] be the set of pages, each of unit size and unit loading cost, and let k be the size of the cache. Suppose that for some l, at each time instant t, we are given a request set At ∈ ([n] l ) . We need to ensure that the cache contains at least one page from At. For this, we may evict some pages from the cache and load an equal number of pages; each at unit cost. The goal is, as usual, to minimize the cost of serving all the requests. For a sequence ρ = A1, . . . , Am of request sets, let OPT(ρ) denote the minimum cost required to serve ρ. For an algorithm A, let A(ρ) denote the cost incurred by algorithm A(ρ). Note that the problem is interesting only when n ≥ k + l and hence we make this assumption throughout this document. We consider the offline problem first. It can be easily proved that this problem is NP-hard, even for l = 2 and arbitrary k. To prove this, we reduce from the Vertex Cover problem. Let G = (V,E), k be an instance of Vertex Cover. We treat the vertices as pages. The cache size is k and it contains k pages other than the vertices initially. The edges are requested in arbitrary order. It is easy to see that it is possible to serve the requests with a cost at most k, iff G has a vertex cover of size k. In the subsequent sections we analyze the online problem in determinisitc and randomized settings. 2 Lower Bound for Determinisitic Online Algorithms We prove a lower bound of ( k+l k ) − 1 on the competitive ratio of any deterministic algorithm. In other words, we prove