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Spatial inverse query processing

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Abstract

Traditional spatial queries return, for a given query object q, all database objects that satisfy a given predicate, such as epsilon range and k-nearest neighbors. This paper defines and studies inverse spatial queries, which, given a subset of database objects Q and a query predicate, return all objects which, if used as query objects with the predicate, contain Q in their result. We first show a straightforward solution for answering inverse spatial queries for any query predicate. Then, we propose a filter-and-refinement framework that can be used to improve efficiency. We show how to apply this framework on a variety of inverse queries, using appropriate space pruning strategies. In particular, we propose solutions for inverse epsilon range queries, inverse k-nearest neighbor queries, and inverse skyline queries. Furthermore, we show how to relax the definition of inverse queries in order to ensure non-empty result sets. Our experiments show that our framework is significantly more efficient than naive approaches.

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Notes

  1. We note that the proposed techniques can be easily adapted to use any L p -Norm as well as weighted euclidean distance.

  2. The later observation follows by the definition of domination, which allows to prune an object o if there is another object o which is at least as close to o as q in all dimension, and close in at least one dimension. In the case o = q, the case applies where distances in all dimensions are equal, so nothing can be pruned.

  3. Only objects within the same partition are considered for the domination relation.

  4. Obtained and modified from http://www.cs.fsu.edu/~lifeifei/SpatialDataset.htm. The original source is the Digital Chart of the World Server (http://www.maproom.psu.edu/dcw/).

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Acknowledgements

This work was supported by a grant from the Germany/Hong Kong Joint Research Scheme sponsored by the Research Grants Council of Hong Kong (Reference No. G HK030/09) and the German Academic Exchange Service (Proj. ID 50149322).

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Correspondence to Andreas Züfle.

Appendix: Proof of Lemma 3

Appendix: Proof of Lemma 3

We first require the following corollary:

Corollary 2

Let Q ∈ ℝd be a set of points and \(C\subseteq Q\) be the vertices of the convex hull of Q in d . Then, for each point p ∈ ℝd , the farthest point in Q to p must be in C as well.

Proof

Consider any point p ∈ ℝd and its farthest point q ∈ Q. Then all points in Q must be located in the hyper-sphere centered at o with radius d(o,q). Now, we can proof the above lemma by contradiction assuming that q is not a convex-hull vertex. If q is assumed to be within the convex-hull (not lying on the margin of the convex hull), then the hyper-sphere splits the convex-hull into points that are inside the sphere and points that are out-side of the sphere as shown for q 2 in Fig. 20. Consequently, the convex hull contains points that are farther from o than q which contradicts the assumption. Now, we assume that q lies on the margin (but not on a vertex) of the convex hull which corresponds to a region of a hyper-plane like q 3 in our example. If we move along this hyper-plane starting from q, we are still within the convex-hull but leave the hyper-sphere of o. Consequently, again, the convex hull contains points that are farther from o than q which again contradicts the assumption.□

Fig. 20
figure 20

Illustration of Corollary 2

Now we can use Corollary 2 to prove Lemma 3:

Proof

By definition of q ref it holds that

$$ q_{\rm ref}=\mathrm{argmax}_{q\in Q}(dist(o,q)) $$

Since the vertices C of the convex hull of Q consists only of points in Q, Corollary 2 leads to

$$ q_{\rm ref}=\mathrm{argmax}_{c\in C}(dist(o,c)) $$

Thus,

$$ \forall c\in C:dist(o,q_{\rm ref})\geq dist(o,c) $$

and since \(\mathcal{H}\subseteq C\):

$$ \forall p\in \mathcal{H}:dist(o,q_{\rm ref})\geq dist(o,p). $$

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Bernecker, T., Emrich, T., Kriegel, HP. et al. Spatial inverse query processing. Geoinformatica 17, 449–487 (2013). https://doi.org/10.1007/s10707-012-0162-y

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