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The Communication Complexity of Set Intersection and Multiple Equality Testing

Published 30 Aug 2019 in cs.IT, cs.CC, and math.IT | (1908.11825v2)

Abstract: In this paper we explore fundamental problems in randomized communication complexity such as computing Set Intersection on sets of size kk and Equality Testing between vectors of length kk. Sa\u{g}lam and Tardos and Brody et al. showed that for these types of problems, one can achieve optimal communication volume of O(k)O(k) bits, with a randomized protocol that takes O(log<sup></sup>k)O(\log<sup>*</sup> k) rounds. Aside from rounds and communication volume, there is a \emph{third} parameter of interest, namely the \emph{error probability} perrp_{\mathrm{err}}. It is straightforward to show that protocols for Set Intersection or Equality Testing need to send Ω(k+logperr<sup>1)\Omega(k + \log p_{\mathrm{err}}<sup>{-1}) bits. Is it possible to simultaneously achieve optimality in all three parameters, namely O(k+logperr<sup>1)O(k + \log p_{\mathrm{err}}<sup>{-1}) communication and O(log<sup></sup>k)O(\log<sup>*</sup> k) rounds? In this paper we prove that there is no universally optimal algorithm, and complement the existing round-communication tradeoffs with a new tradeoff between rounds, communication, and probability of error. In particular: 1. Any protocol for solving Multiple Equality Testing in rr rounds with failure probability 2<sup>E2<sup>{-E} has communication volume Ω(Ek<sup>1/r)\Omega(Ek<sup>{1/r}). 2. There exists a protocol for solving Multiple Equality Testing in r+log<sup>(k/E)r + \log<sup>*(k/E) rounds with O(k+rEk<sup>1/r)O(k + rEk<sup>{1/r}) communication, thereby essentially matching our lower bound and that of Sa\u{g}lam and Tardos. Our original motivation for considering perrp_{\mathrm{err}} as an independent parameter came from the problem of enumerating triangles in distributed (CONGEST\textsf{CONGEST}) networks having maximum degree Δ\Delta. We prove that this problem can be solved in O(Δ/logn+loglogΔ)O(\Delta/\log n + \log\log \Delta) time with high probability 11/poly(n)1-1/\operatorname{poly}(n).

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