Sublinear time approximation of the cost of a metric k-nearest neighbor graph

被引:0
|
作者
Czumaj, Artur [1 ]
Sohler, Christian [2 ,3 ]
机构
[1] Univ Warwick, Coventry, W Midlands, England
[2] Univ Cologne, Cologne, Germany
[3] Google Res, Zurich, Switzerland
来源
PROCEEDINGS OF THE THIRTY-FIRST ANNUAL ACM-SIAM SYMPOSIUM ON DISCRETE ALGORITHMS (SODA'20) | 2020年
基金
英国工程与自然科学研究理事会;
关键词
MINIMUM SPANNING TREE; WEIGHT; ALGORITHMS;
D O I
暂无
中图分类号
TP301 [理论、方法];
学科分类号
081202 ;
摘要
Let (X, d) be an n-point metric space. We assume that (X, d) is given in the distance oracle model, that is, X = {1, : : :, n} and for every pair of points x, y from X we can query their distance d(x, y) in constant time. A k-nearest neighbor (k-NN) graph for (X, d) is a directed graph G = (V,E) that has an edge to each of v's k nearest neighbors. We use cost(G) to denote the sum of edge weights of G. In this paper, we study the problem of approximating cost(G) in sublinear time, when we are given oracle access to the metric space (X, d) that defines G. Our goal is to develop an algorithm that solves this problem faster than the time required to compute G. We first present an algorithm that in (O) over tilde epsilon(n(2)/k) time with probability at least 2/3 approximates cost(G) to within a factor of 1 + epsilon. Next, we present a more elaborate sublinear algorithm that in time (O) over tilde epsilon(min{nk(3/2), n(2)/kg) computes an estimate (cost) over bar of cost(G) that satisfies with probability at least 2/3 vertical bar cost(G) - (cost) over bar vertical bar <= epsilon . (cost(G) + mst(X)), where mst(X) denotes the cost of the minimum spanning tree of (X, d). Further, we complement these results with near matching lower bounds. We show that any algorithm that for a given metric space (X, d) of size n, with probability at least 2/3 estimates cost(G) to within a 1 + epsilon factor requires Omega(n(2)/k) time. Similarly, any algorithm that with probability at least 2/3 estimates cost(G) to within an additive error term epsilon . (mst(X)+ cost(X)) requires Omega(epsilon) (min{nk(3/2), n(2)/k}) time.
引用
收藏
页码:2973 / 2992
页数:20
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