Abstract

Suppose m balls are sequentially thrown into n bins where each ball goes into a random bin. It is well-known that the gap between the load of the most loaded bin and the average is i¾?mlognn, for large m. If each ball goes to the lesser loaded of two random bins, this gap dramatically reduces to i¾?loglogn independent of m. Consider a constrained setting where not all pairs of bins can be sampled. We are given a graph where each node corresponds to a bin. The process sequentially samples an edge from the graph and places a ball in the lesser loaded of its endpoints. We show the gap is at most Ologn/i¾? where i¾? is the edge expansion of the graph. Our results extend naturally to the hypergraph version of this question. Our technique involves a tight analysis of what we call the 1+β-choice process for some parameter β∈0,1: each ball goes to a random bin with probability 1-β and the lesser loaded of two random bins with probability β. For this process we show that the gap is i¾?logn/β, irrespective of m. Moreover the gap stays at i¾?logn/β in the weighted case for a large class of weight distributions. No non-trivial bounds were previously known in the weighted case, even for the 2-choice case. © 2014 Wiley Periodicals, Inc. Random Struct. Alg., 47, 760-775, 2015

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