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Reverse k nearest neighbors queries and spatial reverse top-k queries

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Abstract

Given a set of facilities and a set of users, a reverse k nearest neighbors (RkNN) query q returns every user for which the query facility is one of the k closest facilities. Almost all of the existing techniques to answer RkNN queries adopt a pruning-and-verification framework. Regions-based pruning and half-space pruning are the two most notable pruning strategies. The half-space-based approach prunes a larger area and is generally believed to be superior. Influenced by this perception, almost all existing RkNN algorithms utilize and improve the half-space pruning strategy. We observe the weaknesses and strengths of both strategies and discover that the regions-based pruning has certain strengths that have not been exploited in the past. Motivated by this, we present a new regions-based pruning algorithm called Slice that utilizes the strength of regions-based pruning and overcomes its limitations. We also study spatial reverse top-k (SRTk) queries that return every user u for which the query facility is one of the top-k facilities according to a given linear scoring function. We first extend half-space-based pruning to answer SRTk queries. Then, we propose a novel regions-based pruning algorithm following Slice framework to solve the problem. Our extensive experimental study on synthetic and real data sets demonstrates that Slice is significantly more efficient than all existing RkNN and SRTk algorithms.

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Notes

  1. Although the partitions with different sizes can be used, in this paper, we use equally sized partitions so that the partition that contains a point p can be identified in O(1).

  2. Lemma 20 in “Appendix” section shows that dist(u, f) can never be equal to \(dist(f,p) + dist(u,p)\)) even when f lies on the ray \(L^\theta \).

  3. The observation 1 in [5] can be summarized as follows. For a circle C centered at q, let x be the point such that \(dist(f,x)= mindist(f,C)\) where mindist(f, C) is the minimum distance between f and any point of the circle C. Let u be a user located on the perimeter of this circle. Then, dist(f, u) monotonically increases if u moves along the circle in clockwise or counterclockwise direction from x. Since f lies outside the partition, x also lies outside the partition, and this implies that dist(u, f) is minimum when u is at one of the end points of the arc.

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Acknowledgments

Muhammad Aamir Cheema is supported by ARC DE130101002 and DP130103405. Xuemin Lin is supported by NSFC61232006, NSFC61021004, ARC DP120 104168 and DP110102937. The research of Ying Zhang is supported by ARC DP130103245 and DP110104880. Wenjie Zhang is supported by ARC DP150103071 and DP150102 728.

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Authors and Affiliations

  1. School of Computer Science and Engineering, The University of New South Wales, Sydney, Australia

    Shiyu Yang, Xuemin Lin & Wenjie Zhang

  2. Faculty of Information Technology, Monash University, Melbourne, Australia

    Muhammad Aamir Cheema

  3. University of Technology, Sydney, Australia

    Ying Zhang

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  1. Shiyu Yang
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  4. Ying Zhang
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Correspondence to Muhammad Aamir Cheema.

Appendix

Appendix

Lemma 20

Consider a facility f, a ray \(L^\theta \), a critical point p on \(L^\theta \) and a user u that lies on \(L^\theta \) and \(dist(u,q) > d^\theta \). Then, \(dist(u,f)\ne dist(f,p) + dist(p,u)\).

Proof

Note that \(dist(u,f)\le dist(f,p) + dist(u,p)\) due to the triangular inequality. Since p and u lie on \(L^\theta \), \(dist(u,f) = dist(f,p) + dist(p,u)\) if and only if f also lies on \(L^\theta \) (i.e., \(\theta =0^\circ \)) and p lies between u and f (see Fig. 30). We prove by contradiction that p cannot lie between u and f.

Assume that p lies between u and f as shown in Fig. 30. Since \(\theta =0^\circ \), \(dist(p,q)=\frac{dist(q,f)^2 - \Delta ^2}{2(\Delta + dist(q,f)\cos {0})}=\frac{dist(q,f) - \Delta }{2}\). As \(dist(q,f) > |\Delta |\), we have \(dist(p,q) < \frac{dist(q,f) + dist(q,f)}{2}\). In other words, \(dist(p,q) < dist(q,f)\). Furthermore, since \(dist(u,q) > d^\theta \) (i.e., \(dist(u,q) > dist(p,q)\)), this implies that p cannot be between f and u which contradicts the assumption. \(\square \)

Fig. 30
figure 30

Lemma 20

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Yang, S., Cheema, M.A., Lin, X. et al. Reverse k nearest neighbors queries and spatial reverse top-k queries. The VLDB Journal 26, 151–176 (2017). https://doi.org/10.1007/s00778-016-0445-2

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