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Massively Parallel Algorithms for Finding Well-Connected Components in Sparse Graphs

8 May 2018
Sepehr Assadi
Xiaorui Sun
Omri Weinstein
ArXiv (abs)PDFHTML
Abstract

A fundamental question that shrouds the emergence of massively parallel computing (MPC) platforms is how can the additional power of the MPC paradigm be leveraged to achieve faster algorithms compared to classical parallel models such as PRAM? Previous research has identified the sparse graph connectivity problem as a major obstacle to such improvement: While classical logarithmic-round PRAM algorithms for finding connected components in any nnn-vertex graph have been known for more than three decades, no o(log⁡n)o(\log{n})o(logn)-round MPC algorithms are known for this task with truly sublinear in nnn memory per machine. This problem arises when processing massive yet sparse graphs with O(n)O(n)O(n) edges, for which the interesting setting of parameters is n1−Ω(1)n^{1-\Omega(1)}n1−Ω(1) memory per machine. It is conjectured that achieving an o(log⁡n)o(\log{n})o(logn)-round algorithm for connectivity on general sparse graphs with n1−Ω(1)n^{1-\Omega(1)}n1−Ω(1) per-machine memory may not be possible, and this conjecture also forms the basis for multiple conditional hardness results on the round complexity of other problems in the MPC model. We take an opportunistic approach towards the sparse graph connectivity problem, by designing an algorithm with improved performance guarantees in terms of the connectivity structure of the input graph. Formally, we design an algorithm that finds all connected components with spectral gap at least λ\lambdaλ in a graph in O(log⁡log⁡n+log⁡(1/λ))O(\log\log{n} + \log{(1/\lambda)})O(loglogn+log(1/λ)) MPC rounds and nΩ(1)n^{\Omega(1)}nΩ(1) memory per machine. As such, this algorithm achieves an exponential round reduction on sparse "well-connected" components (i.e., λ≥1/polylog(n)\lambda \geq 1/\text{polylog}{(n)}λ≥1/polylog(n)) using only nΩ(1)n^{\Omega(1)}nΩ(1) memory per machine and O~(n)\widetilde{O}(n)O(n) total memory, and still operates in o(log⁡n)o(\log n)o(logn) rounds even when λ=1/no(1)\lambda = 1/n^{o(1)}λ=1/no(1).

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