One-pass trajectory simplification using the synchronous Euclidean distance

Abstract

Various mobile devices have been used to collect, store and transmit tremendous trajectory data, and it is known that raw trajectory data seriously wastes the storage, network bandwidth and computing resource. To attack this issue, one-pass line simplification ($\textsf {LS}$) algorithms have been developed, by compressing data points in a trajectory to a set of continuous line segments. However, these algorithms adopt the perpendicular Euclidean distance, and none of them uses the synchronous Euclidean distance ($\textsf {SED}$), and cannot support spatiotemporal queries. To do this, we develop two one-pass error bounded trajectory simplification algorithms ($\textsf {CISED}$-$\textsf {S}$ and $\textsf {CISED}$-$\textsf {W}$) using $\textsf {SED}$, based on a novel spatiotemporal cone intersection technique. Using four real-life trajectory datasets, we experimentally show that our approaches are both efficient and effective. In terms of running time, algorithms $\textsf {CISED}$-$\textsf {S}$ and $\textsf {CISED}$-$\textsf {W}$ are on average 3 times faster than $\textsf {SQUISH}$-$\textsf {E}$ (the fastest existing $\textsf {LS}$ algorithm using $\textsf {SED}$). In terms of compression ratios, $\textsf {CISED}$-$\textsf {S}$ is close to and $\textsf {CISED}$-$\textsf {W}$ is on average $19.6\%$ better than $\textsf {DPSED}$ (the existing sub-optimal $\textsf {LS}$ algorithm using $\textsf {SED}$ and having the best compression ratios), and they are $21.1\%$ and $42.4\%$ better than $\textsf {SQUISH}$-$\textsf {E}$ on average, respectively.

Appendix: Proofs

Proof of Proposition 1

Let $P'_{s+i}$ ($i\in [1, k]$) be the intersection point of line segment $\overrightarrow{{ P_sQ}}$ and the plane $P.t - P_{s+i}.t$ = 0. It suffices to show that the intersection point $P'_{s+i}$ satisfies that (1) $P'_{s+i}.t = P_{s+i}.t$, (2) $P'_{s+i}.x$ = $P_s.x + w\cdot (Q.x - P_s.x)$, and (3) $P'_{s+i}.y$ = $P_s.y + w\cdot (Q.y - P_s.y)$, where $w= \frac{P'_{s+i}.t - P_{s}.t}{Q.t - P_{s}.t}= \frac{P_{s+i}.t-P_s.t}{Q.t-P_s.t}$. Hence, $P'_{s+i}$ is indeed a synchronized point of $P_{s+i}$w.r.t.$\overrightarrow{{ P_sQ}}$, and the distance $|\overrightarrow{{ P_{s+i}P'_{s+i}}}|$ from $P_{s+i}$ to $P'_{s+i}$ is the synchronous distance of $P_{s+i}$ to $\overrightarrow{{ P_sQ}}$.

We assume first that $\sqcap _{i=1}^{k}$$\mathcal {C}{(P_s, {\mathcal {O}}(P_{s+i}, \epsilon ))}$$\ne \{P_s\}$. Then, there must exist a point Q in the area of the synchronous circle $\mathcal {O}{(P_{s+k}, \epsilon )}$ such that $\overrightarrow{{ P_sQ}}$ passes through all the cones $\mathcal {C}{(P_s, {\mathcal {O}}(P_{s+i}, \epsilon ))}$$i\in [1, k]$. Hence, $Q.t = P_{s+k}.t$. We also have $sed(P_{s+i}, \overrightarrow{{ P_sQ}}) = |\overrightarrow{{ P'_{s+i}P_{s+i}}}| \le \epsilon$ for each $i \in [1, k]$ since $P'_{s+i}$ is in the area of circle $\mathcal {O}{(P_{s+i}, \epsilon )}$.

Conversely, assume that there exists a point Q such that $Q.t = P_{s+k}.t$ and $sed(P_{s+i}, \overrightarrow{{ P_sQ}})\le \epsilon$ for all $P_{s+i}$ ($i \in [1,k]$). Then, $|\overrightarrow{{ P'_{s+i}P_{s+i}}}| \le \epsilon$ for all $i \in [1, k]$. Hence, we have $\sqcap _{i=1}^{k}$$\mathcal {C}{(P_s, {\mathcal {O}}(P_{s+i}, \epsilon ))}$$\ne \{P_s\}$. $\square$

Proof of Proposition 2

By Proposition 1, it suffices to show that $\sqcap _{i=1}^{k}$$\mathcal {O}^c{(P^c_{s+i}, r^c_{s+i})}$$\ne \emptyset$ if and only if $\sqcap _{i=1}^{k}$$\mathcal {C}{(P_s, {\mathcal {O}}(P_{s+i}, \epsilon ))}$$\ne \{P_s\}$, which is obvious. Hence, we have the conclusion. $\square$

Proof of Proposition 3

We shall prove this by contradiction. Assume first that ${\mathcal {R}}^*_{l} \sqcap {\mathcal {R}}_{s+l+1}$ has more than m edges. Then, it must have two distinct edges $\overrightarrow{{ A_i}}$ and $\overrightarrow{{ A_{i'}}}$ with the same label j$(1\le j \le m)$, originally from ${\mathcal {R}}_{s+i}$ and ${\mathcal {R}}_{s+i'}$ ($1\le i< i' \le l+1$). Note that here ${\mathcal {R}}_{s+i} \sqcap {\mathcal {R}}_{s+i'} \ne \emptyset$ since ${\mathcal {R}}^*_l \sqcap {\mathcal {R}}_{s+l+1} \ne \emptyset$. However, when ${\mathcal {R}}_{s+i} \sqcap {\mathcal {R}}_{s+i'} \ne \emptyset$, the intersection ${\mathcal {R}}_{s+i} \sqcap {\mathcal {R}}_{s+i'}$ cannot have both edge $\overrightarrow{{ A_i}}$ and edge $\overrightarrow{{ A_{i'}}}$, as all edges with the same label are in parallel (or overlapping) with each other by the above definition of inscribed regular polygons. This contradicts with the assumption. $\square$

Proof of Proposition 4

The inscribed regular polygon ${\mathcal {R}}_{s+l+1}$ has m edges, and intersection polygon ${\mathcal {R}}^*_l$ has at most m edges by Proposition 3. As the intersection of two m-edges convex polygons can be computed in O(m) time [25], the intersection of polygons ${\mathcal {R}}^*_l$ and ${\mathcal {R}}_{s+l+1}$ can be done in O(1) time for a fixed m. $\square$

Proof of Proposition 5

If ($\overrightarrow{{ A}} \sqcap \overrightarrow{{ B}} \ne \emptyset$and$\overrightarrow{{ A}} \times \overrightarrow{{ B}} < 0$and$P_{e_A} \not \in {\mathcal {H}}(\overrightarrow{{ B}})$) or ($\overrightarrow{{ A}} \sqcap \overrightarrow{{ B}} \ne \emptyset$and$\overrightarrow{{ A}} \times \overrightarrow{{ B}} \ge 0$and$P_{e_B} \in {\mathcal {H}}(\overrightarrow{{ A}})$), then as all edges in the same edge group $E^j$ ($1\le j\le m$) are in parallel with each other and by the geometric properties of regular polygon ${\mathcal {R}}_{s+k+1}$, it is easy to find that, for each position of $\overrightarrow{{ A}}$ between its original to its opposite positions, we have (1) $\overrightarrow{{ A}} \sqcap \overrightarrow{{ B}} = \emptyset$, and (2) either $P_{e_A} \not \in {\mathcal {H}}(\overrightarrow{{ B}})$ or $P_{e_B} \in {\mathcal {H}}(\overrightarrow{{ A}})$. Hence, by the advance rule (1) of algorithm $\textsf {CPolyInter}$ in Sect. 2.4, edge $\overrightarrow{{ A}}$ is always moved forward until it reaches the opposite position of its original one. From this, we have the conclusion. $\square$

Proof of Proposition 6

If ($\overrightarrow{{ A}} \sqcap \overrightarrow{{ B}} \ne \emptyset$and$\overrightarrow{{ A}} \times \overrightarrow{{ B}} \ge 0$and$P_{e_B} \not \in {\mathcal {H}}(\overrightarrow{{ A}})$) or ($\overrightarrow{{ A}} \sqcap \overrightarrow{{ B}} \ne \emptyset$and$\overrightarrow{{ A}} \times \overrightarrow{{ B}} < 0$and$P_{e_A} \in {\mathcal {H}}(\overrightarrow{{ B}})$), then it is also easy to find that, for each position of $\overrightarrow{{ B}}$ between its original to its target positions (i.e., the edge after the one having the same edge group as $\overrightarrow{{ A}}$), we have (1) $\overrightarrow{{ A}} \sqcap \overrightarrow{{ B}} = \emptyset$, and (2) either $P_{e_B} \not \in {\mathcal {H}}(\overrightarrow{{ A}})$ or $P_{e_A} \in {\mathcal {H}}(\overrightarrow{{ B}})$. Hence, by the advance rule (2) of algorithm $\textsf {CPolyInter}$ in Sect. 2.4, edge $\overrightarrow{{ B}}$ is always moved forward until it reaches the target position. From this, we have the conclusion. $\square$

Proof of Proposition 8

If $\sqcap _{i=s+1}^{e}{{\mathcal {C}}(P_s, P_{s+i}, \epsilon /2)} \ne \{P_s\}$, then by Proposition 1, there exists a point Q, $Q.t = P_{s+k}.t$, such that $sed(P_{s+i}, \overrightarrow{{ P_sQ}})\le \epsilon /2$ for all $i \in [1,k]$. By the triangle inequality essentially, $sed(P_{s+i}, \overrightarrow{{ P_sP_{s+k}}})\le sed(P_{s+i}, \overrightarrow{{ P_sQ}}) + |\overrightarrow{{ QP_{s+k}}}| \le \epsilon /2+\epsilon /2 = \epsilon$. $\square$

Proof of Proposition 9

By Proposition 2 and the nature of inscribed regular polygon, it is easy to find that for any point $Q \in {\mathcal {R}}^*_k$w.r.t. plane $t_c=P_{s+k}.t$, there is $sed(P_{s+i}, \overrightarrow{{ P_sQ}})\le \epsilon$ for all points $P_{s+i}$ ($i \in [1,k]$). From this, we have the conclusion. $\square$