Finding the longest common sub-pattern in sequences of temporal intervals

Abstract

We study the problem of finding the longest common sub-pattern (LCSP) shared by two sequences of temporal intervals. In particular we are interested in finding the LCSP of the corresponding arrangements. Arrangements of temporal intervals are a powerful way to encode multiple concurrent labeled events that have a time duration. Discovering commonalities among such arrangements is useful for a wide range of scientific fields and applications, as it can be seen by the number and diversity of the datasets we use in our experiments. In this paper, we define the problem of LCSP and prove that it is NP-complete by demonstrating a connection between graphs and arrangements of temporal intervals. This connection leads to a series of interesting open problems. In addition, we provide an exact algorithm to solve the LCSP problem, and also propose and experiment with three polynomial time and space under-approximation techniques. Finally, we introduce two upper bounds for LCSP and study their suitability for speeding up 1-NN search. Experiments are performed on seven datasets taken from a wide range of real application domains, plus two synthetic datasets. Lastly, we describe several application cases that demonstrate the need and suitability of LCSP.

Appendix: Proof of the properties described in Sect. 4.3.2

• 1. Supposing that $LCS(i,j)$ was composed of more than one interval, then there must exist a pair of intervals with the same label in $\{{S_A}_1,\ldots ,{S_A}_{i-1}\}$ and $\{{S_B}_1,\ldots ,{S_B}_{i-1}\}$. That is a contradiction since it would imply that not all previous sub-problems yield $\emptyset$ as their solution.

• 2. By applying the operation $LCS(p,q)\otimes (i,j)$ or, equivalently selecting from $LCS(p,q)$ only the intervals that induce similar relations to the corresponding interval of $i$ and $j$, we make sure that the interval corresponding to $i$ and $j$ has the same relations to the previous intervals in the produced arrangement. Conversely, the existing intervals have the same relations to the correspondent of $i$ and $j$. Additionally, pairs of existing intervals of $LCS(p,q)$ have identical relations with their correspondents in $\mathcal {A}$ and $\mathcal {B}$; this was examined when each interval was added to the solution of the previous sub-problems.

• 3. In other words, the $\otimes$ operator does not discard extra intervals. Suppose that the maximal CSPs are correctly retrieved for all previous sub-problems $LCS(p,q)$, but not for $LCS(i,j)$. This would imply that an interval belonging to a maximal CSP of $\{ {{E_S}_A}_1, \ldots , {{E_S}_A}_i\}$ and $\{ {{E_S}_B}_1, \ldots , {{E_S}_B}_j\}$ (where $A_i$ is matched to $B_j$) exists but was not selected for $LCS(i,j)$. But since the not-selected interval belongs to a maximal CSP then it has the same relation to ${S_A}_i$ and ${S_B}_j$. So, since the relations are the same, the interval would have been selected for $LCS(i,j)$, which contradicts to the previous. Thus, the algorithm at point $(i,j)$ returns maximal CSPs of $\{ {{E_S}_A}_1, \ldots , {{E_S}_A}_i\}$ and $\{ {{E_S}_B}_1, \ldots , {{E_S}_B}_j\}$ that matches ${{E_S}_A}_i$ to ${{E_S}_B}_j$.

• 4. Suppose there exists a maximal CSP that matches $A_i$ to $B_j$ but was not discovered. This would imply that by removing the interval corresponding to $A_i$ and $B_j$, one is left with common a sub-pattern $s$. Then, either $s\subseteq r,r\in \mathcal {M}_{i-1,j-1}$ or not. In the first case, $s$ must have been retrieved when performing $r\otimes (i,j)$, so this cannot be. So, it can only be that $s$ is maximal but then it must hold that $s\in \mathcal {M}_{i-1,j-1}$. Contradiction.

  An alternative approach is that in the Cartesian graph $G_{AB}$ (see proof of Theorem 2 for exact definition), this corresponds to finding all maximal cliques containing the vertex $u$ labeled $(i,j)$ by checking all previously found maximal cliques and for each one returning its intersection with the neighbors of $u$.