Can one compute the PageRank score of a single, arbitrary node in a graph, exploring only a vanishing fraction of the graph? We provide a positive answer to this extensively researched open question. We develop the first algorithm that, for any n -node graph, returns a multiplicative $(1\pmε)$-approximation of the score of any given node with probability $(1-δ)$, using at most $O\big(n^2/3 łn(n)^1/3 łn(1/δ)^2/3 ε^-2/3 \big) = \tildeO (n^2/3 )$ queries which return either a node chosen uniformly at random, or the list of neighbours of a given node. Alternatively, we show that the same guarantees can be attained by fetching at most $O\big( E^4/5 d^-3/5 łn(n)^1/5 łn(1/δ)^3/5 ε^-6/5 \big) = \tildeO (E^4/5 )$ arcs, where E is the total number of arcs in the graph and d is its average degree.
Brief announcement: On approximating pageRank locally with sublinear query complexity
Peserico, Enoch;Pretto, Luca
2018
Abstract
Can one compute the PageRank score of a single, arbitrary node in a graph, exploring only a vanishing fraction of the graph? We provide a positive answer to this extensively researched open question. We develop the first algorithm that, for any n -node graph, returns a multiplicative $(1\pmε)$-approximation of the score of any given node with probability $(1-δ)$, using at most $O\big(n^2/3 łn(n)^1/3 łn(1/δ)^2/3 ε^-2/3 \big) = \tildeO (n^2/3 )$ queries which return either a node chosen uniformly at random, or the list of neighbours of a given node. Alternatively, we show that the same guarantees can be attained by fetching at most $O\big( E^4/5 d^-3/5 łn(n)^1/5 łn(1/δ)^3/5 ε^-6/5 \big) = \tildeO (E^4/5 )$ arcs, where E is the total number of arcs in the graph and d is its average degree.Pubblicazioni consigliate
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