Processing Reverse Nearest Neighbor Queries Based on Unbalanced Multiway Region Tree Index

Abstract

In many applications and scenarios, there are opportunities for processing reverse nearest neighbor (RNN) queries, which are derived from and more complex than nearest neighbor (NN) queries. Generally, processing NN queries involves sophisticated data structures and methods, and has been very well addressed for low-dimensional data (usually less than 10); while efficiently processing exact NN or RNN queries for high dimensional data remains a challenging problem. This paper proposes an algorithm of evaluating RNN queries in higher dimensional l_p spaces. The main idea of our algorithm is that an RNN query can be processed efficiently based on relevant information easily available and retrievable from memory. The data space containing a finite dataset is divided into multiple small regions forming an unbalanced multiway region tree, then an index containing important information is created by using the tree and the sorted lists of tuples in the dataset. The algorithm consists of two pruning approaches and a verification method based on the index and the characteristics of l_p spaces. Extensive experiments are conducted to demonstrate the excellent performance of our algorithm over various datasets and to show that it outperforms existing state-of-the-art methods CSD, VR-RNN, SFT and TPL.

Appendix: Proofs for Lemmas

Lemma 1.

Consider the dataset R and its UR-tree Index with the region set {P_i: i = 0, 1, ⋅⋅⋅, s}. For a query point Q, a region P_i and R_i = R ∩ P_i (0 ≤ i ≤ s), if d(c, Q) > r_max + d_max, then ∀t ∈ R_i, t ∉ RNN(Q), where c is the center-point of P_i, r_max = d(c, y), d_max = max{d(t, t_NN) | ∀t ∈ R_i}, and y = (y₁, ⋅⋅⋅, y_n) is the max-point of P_i.

Proof of Lemma 1:

For a query point Q, a region P_i and R_i = R ∩ P_i, as is shown in Fig.A, d(c, Q) > r_max + d_max = d(c, y) + max{d(t, t_NN) | ∀t ∈ R_i}, c is the center-point of P_i, and y = (y₁, ⋅⋅⋅, y_n) is the max-point of P_i.

Fig. A.

Illustration of the proof of Lemma 1.

For arbitrary t ∈ R_i, according to the triangle inequality of distance function, we have

$$ \begin{array}{*{20}c} {d\left( {{\varvec{c}},{\varvec{t}}} \right) + d\left( {{\varvec{t}},Q} \right) \, \ge d\left( {{\varvec{c}},Q} \right)} \\ {d\left( {{\varvec{t}},Q} \right) \, \ge d\left( {{\varvec{c}},Q} \right) - d\left( {{\varvec{c}},{\varvec{t}}} \right)} \\ \end{array} $$

Since d(c, Q) > r_max + d_max, we have.

$$ \begin{array}{*{20}c} {d\left( {{\varvec{t}},Q} \right)\, > \,r_{max} \, + \,d_{max} \, - \,d({\varvec{c}},{\varvec{t}})} \\ {d\left( {{\varvec{t}},Q} \right)\, - \,d_{max} \, > \,r_{max} \, - \,d({\varvec{c}},{\varvec{t}})} \\ \end{array} $$

Moreover, d(c, t) ≤ d(c, y) = r_max as y is the max-point of P_i. Therefore,

$$ \begin{array}{*{20}c} {d\left( {{\varvec{t}},Q} \right)\, - \,d_{max} \, > \,0} \\ {d\left( {{\varvec{t}},Q} \right)\, > \,d_{max} \, = \,max\{ d\left( {{\varvec{t}},{\varvec{t}}_{NN} } \right) \, |\forall {\varvec{t}}\, \in \,{\varvec{R}}_{i} \} \, \ge \,d({\varvec{t}},{\varvec{t}}_{NN} )} \\ \end{array} $$

Thus, query point Q is not the nearest neighbor of tuple t, that is, t ∉ RNN(Q).

We restate Lemma 2 that summarizes some characteristics of l_p spaces, and then we prove it.

Lemma 2.

Let two arbitrary points x = (x₁, ⋅⋅⋅, x_n), y = (y₁, ⋅⋅⋅, y_n) ∈ ℜⁿ, and a constant σ > 0. Then.

(1°) ||x||_p ≤ n^(1/p−1/q) ||x||_q, for 1 ≤ p < q.

(2°) ||x||_∞ ≤ ||x||_p ≤ n^1/p||x||_∞, for 1 ≤ p < ∞.

(3°) d_p(x, y) > σ if |x_i − y_i|>σ for some 1 ≤ i ≤ n, where the distance function d_p(⋅,⋅) is induced by ||⋅||_p.

We will present the proof of (1°) by using Hölder’s inequality.

Hölder’s Inequality:

Assume that r and s are in the open interval (1, ∞) with 1/r + 1/s = 1. Then, for arbitrary a = (a₁, a₂, ⋅⋅⋅, a_n), b = (b₁, b₂, ⋅⋅⋅, b_n) ∈ ℜⁿ,

$$ \left| {\left| {{\varvec{ab}}} \right|} \right|_{1} \, \le \,\left| {\varvec{a}} \right|_{r} \left| {\left| {\varvec{b}} \right|} \right|_{s} $$

that is,

$$ \sum\nolimits_{i = 1}^{n} {\left| {a_{i} b_{i} } \right| \le \left( {\sum\nolimits_{i = 1}^{n} {\left| {a_{i} } \right|^{r} } } \right)}^{1/r} \left( {\sum\nolimits_{i = 1}^{n} {\left| {b_{i} } \right|^{s} } } \right)^{1/s} $$

Proof of Lemma 2:

(1°) Let s = q/p and r = q/(q − p). Then s > 1, r > 1 and 1/r + 1/s = 1 since 1 ≤ p < q. Suppose that a_i = 1 and b_i = |x_i|^p, i = 1, ⋅⋅⋅, n. By Hölder’s inequality, we have.

$$ \begin{gathered} \sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{p} = \sum\nolimits_{i = 1}^{n} {(1 \cdot \left| {x_{i} } \right|^{p} )\, = \sum\nolimits_{i = 1}^{n} {\left| {a_{i} b_{i} } \right| \le \left( {\sum\nolimits_{i = 1}^{n} {\left| {a_{i} } \right|^{r} } } \right)} } }^{1/r} \left( {\sum\nolimits_{i = 1}^{n} {\left| {b_{i} } \right|^{s} } } \right)^{1/s} \hfill \\ = \,n^{1/r} \left( {\sum\nolimits_{i = 1}^{n} {\left| {b_{i} } \right|^{s} } } \right)^{1/s} = n^{(q - p)/q} \left( {\sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{ps} } } \right)^{1/s} = n^{(q - p)/q} \left( {\sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{q} } } \right)^{p/q} \hfill \\ \end{gathered} $$

Then,

$$ \begin{gathered} \left( {\sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{p} } } \right)^{1/p} \le \left( {n^{(q - p)/q} \left( {\sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{q} } } \right)^{p/q} } \right)^{1/p} = \,n^{(q - p)/qp} \left( {\sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{q} } } \right)^{1/q} \hfill \\ = \,n^{(1/p - 1/q)} \left( {\sum\nolimits_{i = 1}^{n} {\left| {x_{i} } \right|^{q} } } \right)^{1/q} = \,n^{(1/p - 1/q)} \left| {\left| {\varvec{x}} \right|} \right|_{q} \hfill \\ \end{gathered} $$

That is, ||x||_p ≤ n^(1/p−1/q)||x||_q by the definition ||x||_p = ($\sum^{n}_{i=1}$|x_i|^p)^1/p.

(2°) By the definition of l_p-norm ||⋅||_p, (1°) and ||x||_q → ||x||_∞ when q → ∞, we have.

$$ \left| {\left| {\varvec{x}} \right|} \right|_{\infty } \, \le \,\left| {\left| {\varvec{x}} \right|} \right|_{p} and \, \left| {\left| {\varvec{x}} \right|} \right|_{p} \, \le \,n^{1/p} \left| {\left| {\varvec{x}} \right|} \right|_{\infty } $$

That is,

$$ \left| {\left| {\varvec{x}} \right|} \right|_{\infty } \, \le \,\left| {\left| {\varvec{x}} \right|} \right|_{p} \, \le \,n^{1/p} \left| {\left| {\varvec{x}} \right|} \right|_{\infty } {\text{fo}}r \, 1\, \le \,p\, < \,\infty $$

(3°) If |x_i − y_i| > σ for some 1 ≤ i ≤ n, then ||x − y||_∞ ≥ |x_i − y_i| > σ. Thus, d_p(x, y) = ||x − y||_p ≥ ||x − y||_∞ > σ by using (2°).