Indexable sub-trajectory matching using multi-segment approximation: a partition-and-stitch framework

Abstract

With advances in base technologies for moving objects, many studies have been conducted on the construction of databases of the trajectories of moving objects, including the diverse applications related to the trajectories. Most previous studies deal with whole trajectory matching, which finds the trajectories T in the database similar to a given query trajectory Q ‘as a whole.’ However, we often want to find those T containing the sub-trajectories $T_{\mathrm{sub}} \, (\subseteq T)$ that are similar to Q. This problem is known as sub-trajectory matching  and is more complicated than whole trajectory matching since the query trajectory Q can be of any length and the matching sub-trajectories $T_{\mathrm{sub}}$ can be at any position in the data trajectories T. In this paper, we present a novel indexing-based sub-trajectory matching algorithm using multi-segment approximation. Our algorithm partitions a data trajectory into multiple component segments and then stores the individual segments in an index. The query trajectory is also partitioned into its component segments, and the search for similar segments for each query segment is efficiently performed using the index. The sub-trajectories similar to the query trajectory are reconstructed by our ‘stitching’ algorithm using the individual segments retrieved from the index. Our stitching algorithm is novel and innovative in the sense that it facilitates segment-wise partitioning and indexing of data trajectories. Without stitching, only trajectory-wise operations would be affordable, which causes severe storage space overhead and degradation in search performance. Our study is the first that uses indexing in sub-trajectory matching. We define a (multi-segment) trajectory similarity measure that extends a widely used single-segment similarity measure proposed by Lee et al. (in: Proceedings of ACM SIGMOD international conference on management of data (SIGMOD), 2007; in: Proceedings of IEEE international conference on data engineering (ICDE), 2008; Proc VLDB Endow (PVLDB) 1(1):1081–1094, 2008) by using the Hausdorff distance. We perform extensive experiments to compare our method with EDS (Xie, in: Proceedings of ACM SIGMOD international conference on management of data (SIGMOD), 2014), which has been proved to outperform all representative point-based measures in terms of accuracy and performance. The accuracy of our similarity measure is better than EDS by up to 52.0%, and our algorithm significantly outperforms that using EDS by up to 22,543 times. The performance of our algorithm is linearly scalable in the size of the database, which is an essential property for handling large-scale databases.

Appendix

Lemma 5

For any two segments$L_i$and$L_j$, the following always holds:

$$\begin{aligned} {\mathrm{dist}}(L_i, L_j) \ge d_{{\mathrm{seg}},0}(L_i, L_j), \end{aligned}$$

(11)

where${\mathrm{dist}}(L_i, L_j) = w_\perp \cdot d_\perp (L_i, L_j) + w_\parallel \cdot d_\parallel (L_i, L_j) + w_\theta \cdot d_\theta (L_i, L_j)$and$w_\perp = w_\parallel = w_\theta = 1$ [16].

Fig. 17

Similarity measure between two segments $L_i$ and $L_j$

Proof

Let ${\mathcal {L}}_i$ and ${\mathcal {L}}_j$ be two lines containing two segments $L_i$ and $L_j$, respectively. Let $p_s$ and $p_e$ be the projection points of two end points $s_j$ and $e_j$ of $L_j$ onto ${\mathcal {L}}_i$, respectively. Without loss of generality, we assume $d(s_j, p_s) \le d(e_j, p_e)$. We also assume that $L_i$ is longer than $L_j$ as in [16]. We prove for the following three cases:

Case 1: $p_s$ is located on $L_i$.

As shown in Fig. 17a, $d_{{\mathrm{seg}},0}(L_i, L_j) = l_{\perp 1}$. Thus,

$$\begin{aligned} {\mathrm{dist}}(L_i, L_j)&\ge {} d_\perp (L_i, L_j) = \frac{l_{\perp 1}^2 + l_{\perp 2}^2}{ l_{\perp 1} + l_{\perp 2}} \\&\ge {} \frac{l_{\perp 1}^2 + l_{\perp 1} l_{\perp 2}}{ l_{\perp 1} + l_{\perp 2}} = l_{\perp 1} \\&= {} d_{{\mathrm{seg}},0}(L_i, L_j). \end{aligned}$$

Case 2: $p_s$ is located behind $e_i$.

As shown in Fig. 17b, $d_{{\mathrm{seg}},0}(L_i, L_j) = d(e_i, s_j)$. Thus,

$$\begin{aligned} {\mathrm{dist}}(L_i, L_j)&\ge {} d_{\perp }(L_i, L_j) + d_\parallel (L_i, L_j) \\&\ge {} l_{\perp 1} + l_{\parallel 2} \ge d(e_i, s_j) \\&= {} d_{{\mathrm{seg}},0}(L_i, L_j). \end{aligned}$$

Case 3: $p_s$ is located in front of $s_i$.

Let $p_j$ be the projection point of $s_i$ in $L_i$ onto ${\mathcal {L}}_j$, then $d_{{\mathrm{seg}},0}(L_i, L_j) \le d(s_i, p_j)$. If it holds that $l_{\parallel 1}' \le l_{\parallel 1}''$, then $d_{\parallel }(L_i, L_j) = l_{\parallel 1}'$. Thus,

$$\begin{aligned} {\mathrm{dist}}(L_i, L_j)&\ge {} d_\perp (L_i, L_j) + d_\parallel (L_i, L_j) \ge l_{\perp 1} + l_{\parallel 1}' \\&\ge {} d(s_i, s_j) \ge d(s_i, p_j) \\&\ge {} d_{{\mathrm{seg}},0}(L_i, L_j). \end{aligned}$$

If it holds that $l_{\parallel 1}' \ge l_{\parallel 1}''$, then $d_\parallel (L_i, L_j) = l_{\parallel 1}''$. Thus,

$$\begin{aligned} {\mathrm{dist}}(L_i, L_j)&= {} d_\perp (L_i, L_j) + d_\parallel (L_i, L_j) + d_\theta (L_i, L_j) \\&\ge {} l_{\perp 1} + l_{\parallel 1}'' + d_\theta \ge d(s_i, e_j) \ge d(s_i, p_j) \\&\ge {} d_{{\mathrm{seg}},0}(L_i, L_j). \end{aligned}$$

Therefore, combining these three cases, it always holds that ${\mathrm{dist}}(L_i, L_j) \ge d_{{\mathrm{seg}},0}(L_i, L_j)$. $\square$