The K Group Nearest-Neighbor Query on Non-indexed RAM-Resident Data

Abstract

Data sets that are used for answering a single query only once (or just a few times) before they are replaced by new data sets appear frequently in practical applications. The cost of buiding indexes to accelerate query processing would not be repaid for such data sets. We consider an extension of the popular (K) Nearest-Neighbor Query, called the (K) Group Nearest Neighbor Query (GNNQ). This query discovers the (K) nearest neighbor(s) to a group of query points (considering the sum of distances to all the members of the query group) and has been studied during recent years, considering data sets indexed by efficient spatial data structures. We study (K) GNNQs, considering non-indexed RAM-resident data sets and present an existing algorithm adapted to such data sets and two Plane-Sweep algorithms, that apply optimizations emerging from the geometric properties of the problem. By extensive experimentation, using real and synthetic data sets, we highlight the most efficient algorithm.

G. Roumelis, M. Vassilakopoulos, A. Corral and Y. Manolopoulos—Work funded by the GENCENG project (SYNERGASIA 2011 action, supported by the European Regional Development Fund and Greek National Funds); project number 11SYN_ 8_1213.

A. Corral—Supported by the MINECO research project [TIN2013-41576-R].

Appendix

Lemma: The sum of dx-distances between one given point $p(x,y) \in P$ and all points of the query set Q (sumdx(p, Q)):

A :

Is minimized at the median point q[m] (where q[m] is the array notation of $q_m$),

B :

For all $p.x \ge q[m].x$, sumdx is constant or increasing with the increment of x, and

C :

For all $p.x < q[m].x$, sumdx is increasing while x decreases.

Fig. 5.

The point p has K query points on the left and the point $p'$ $(p'.x > p.x)$ has $K'$ query points on the left.

Proof: Property A has been proved in [18]. To prove property B, for every point $p \in P$ and $q \in Q$, we use

If the point p has K query points on the left ($p.x < q[K-1].x$) and $M-K$ query points on the right (Fig. 5), then:

For another point $p' \in P$ with $p'.x > p.x$ which has $K'$ query points on the left (Fig. 5) and $M-K'$ query points on the right, it is:

The difference between dx-distances of the points $p'$ and p is: . If the set of the query points Q has cardinality M and this is an even number then there are two medians q[m1] and q[m2], while if M is odd then there is only one median point q[m].

B.1 M is even and $q[m1].x \le p.x < p'.x$ then $M \le 2K \le 2K'$ so $(2K-M)\ge 0$, $(p'.x-p.x) \ge 0$ and $(K'-K)p'.x - \displaystyle \sum _{i=K}^{K'-1} q[i].x \ge 0$ because $p'.x \ge q[i].x$, whereas $K \le i \le K'$

B.2 All of the above apply to M if it is odd and it is only one median point $q[m].x \le p.x < p'.x$. It is proven that for all points p on the right of the median query point the sum of dx-distances is increasing.

C For both types of cardinality of the query set Q and for the case $p.x < p'.x < q[m].x$ it is: $\varDelta sumdx = (2K-M)(p'.x-p.x) + 2(K'-K)p'.x \ - 2 \displaystyle \sum _{i=K}^{K'-1} q[i].x \ \le (2K-M)(p'.x-p.x) + 2(K'-K)p'.x \ - 2(K'-K)p.x \ = 2(K-M)(p'.x-p.x) \ + 2(K'-K)(p'.x-p.x) \ = (2K-M+2K'-2K)(p'.x-p.x) \ = (2K'-M)(p'.x-p.x) < 0$. It is proven that for all points p on the left of the median query point the sum of dx-distances is strictly decreasing.    $\square$