Balanced allocation on graphs

  • Authors:
  • Krishnaram Kenthapadi;Rina Panigrahy

  • Affiliations:
  • Stanford University, Stanford, CA;Stanford University, Stanford, CA

  • Venue:
  • SODA '06 Proceedings of the seventeenth annual ACM-SIAM symposium on Discrete algorithm
  • Year:
  • 2006

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Abstract

It is well known that if n balls are inserted into n bins, with high probability, the bin with maximum load contains (1 + o(1)) log n/log log n balls. Azar, Broder, Karlin, and Upfal [1] showed that instead of choosing one bin, if d ≥ 2 bins are chosen at random and the ball in serted into the least loaded of the d bins, the maximum load reduces drastically to log log n/log d + O(1). In this paper, we study the two choice balls and bins process when balls are not allowed to choose any two random bins, but only bins that are connected by an edge in an underlying graph. We show that for n balls and n bins, if the graph is almost regular with degree nε, where ε is not too small, the previous bounds on the maximum load continue to hold. Precisely, the maximum load is log log n + O(1/ε) + O(1). So even if the graph has degree nΩ(1/log log n), the maximum load is O(log log n). For general Δ-regular graphs, we show that the maximum load is log log n + O(log n/log (Δ/log4 n)) + O(1) and also provide an almost matching lower bound of log log n + log n/log (Δ log n). Further this does not hold for non-regular graphs even if the minimum degree is high.Vöcking [29] showed that the maximum bin size with d choice load balancing can be further improved to O(log log n/d) by breaking ties to the left. This requires d random bin choices. We show that such bounds can be achieved by making only two random accesses and querying d/2 contiguous bins in each access. By grouping a sequence of n bins into 2n/d groups, each of d/2 consecutive bins, if each ball chooses two groups at random and inserts the new ball into the least-loaded bin in the lesser loaded group, then the maximum load is O(log log n/d) with high probability. Furthermore, it also turns out that this partitioning into aligned groups of size d/2 is also essential in achieving this bound, that is, instead of choosing two aligned groups, if we simply choose random but possibly unaligned random sets of d/2 consecutive bins, then the maximum load jumps to Ω(log log n/log d) even if the two sets are always chosen to be disjoint.