Exact and approximate flexible aggregate similarity search

Abstract

Aggregate similarity search, also known as aggregate nearest-neighbor (Ann) query, finds many useful applications in spatial and multimedia databases. Given a group Q of M query objects, it retrieves from a database the objects most similar to Q, where the similarity is an aggregation (e.g., ${{\mathrm{sum}}}$, $\max$) of the distances between each retrieved object p and all the objects in Q. In this paper, we propose an added flexibility to the query definition, where the similarity is an aggregation over the distances between p and any subset of $\phi M$ objects in Q for some support $0< \phi \le 1$. We call this new definition flexible aggregate similarity search and accordingly refer to a query as a flexible aggregate nearest-neighbor ( Fann ) query. We present algorithms for answering Fann queries exactly and approximately. Our approximation algorithms are especially appealing, which are simple, highly efficient, and work well in both low and high dimensions. They also return near-optimal answers with guaranteed constant-factor approximations in any dimensions. Extensive experiments on large real and synthetic datasets from 2 to 74 dimensions have demonstrated their superior efficiency and high quality.

Appendix: Tightness of Amax

Here we show that the $(1+2\sqrt{2})$ approximation ratio of Amax is tight, by giving a concrete example. Consider the case in Fig. 25 where $\epsilon$ is an arbitrarily small positive.

Fig. 25

Amax ’s approximation bound is tight

In this case, $M=8$, $\phi =0.5$, hence $\phi M=4$ and $\phi M-1=3$. Consider $q_1$, its 3-nearest neighbors in Q are $\{q_2, q_3, q_4\}$, hence $Q_{\phi }^{q_1}=\{q_1, q_2, q_3, q_4\}$. Note that ${{\mathrm{MEB}}}(\{q_1, q_2, q_3, q_4\})$ $=\mathcal {B}(c_1, \sqrt{2} r^*)$, and ${{\mathrm{nn}}}(c_1, P)=p_2$. Now, $p_2$’s 4-nearest neighbors in Q are $\{q_4, q_3, q_2, q_1\}$. Hence, $Q_{\phi }^{p_2}=\{q_4, q_3, q_2, q_1\}$, $r_{p_2}=\max (p_2, \{q_4, q_3, q_2, q_1\})=(1+2\sqrt{2})r^*-\epsilon$.

It’s easy to verify that the results from $q_2$, $q_3$ and $q_4$ are the same as $q_1$, since $Q_{\phi }^{q_2}$, $Q_{\phi }^{q_3}$ and $Q_{\phi }^{q_4}$ are the same as $Q_{\phi }^{q_1}=\{q_1, q_2, q_3, q_4\}$. Furthermore, $q_5$, $q_6$, $q_7$ and $q_8$ are symmetric to $q_1$, $q_2$, $q_3$ and $q_4$, and $p_3$ is symmetric to $p_2$. Thus, they yield $(p_3, Q_\phi ^{p_3})$ as the answer, and $r_{p_3}=\max (p_3, \{q_5, q_6, q_7, q_8\})=(1+2\sqrt{2})r^*-\epsilon$.

As a result, Amax will return either $(p_2, Q_\phi ^{p_2})$ or $(p_3, Q_\phi ^{p_3})$ as the answer, with $r_2=r_3=(1+2\sqrt{2})r^*-\epsilon$. But in this case $p^*=p_1$, $Q_\phi ^*=\{q_1, q_2, q_3, q_4\}$, and $\max (p^*, Q_\phi ^*)=r^*$.