Discovering recurring activity in temporal networks

Abstract

Recent advances in data-acquisition technologies have equipped team coaches and sports analysts with the capability of collecting and analyzing detailed data of team activity in the field. It is now possible to monitor a sports event and record information regarding the position of the players in the field, passing the ball, coordinated moves, and so on. In this paper we propose a new method to analyze such team activity data. Our goal is to segment the overall activity stream into a sequence of potentially recurrent modes, which reflect different strategies adopted by a team, and thus, help to analyze and understand team tactics. We model team activity data as a temporal network, that is, a sequence of time-stamped edges that capture interactions between players. We then formulate the problem of identifying a small number of team modes and segmenting the overall timespan so that each segment can be mapped to one of the team modes; hence the set of modes summarizes the overall team activity. We prove that the resulting optimization problem is $\mathrm {NP}$-hard, and we discuss its properties. We then present a number of different algorithms for solving the problem, including an approximation algorithm that is practical only for one mode, as well as heuristic methods based on iterative and greedy approaches. We benchmark the performance of our algorithms on real and synthetic datasets. Of all methods, the iterative algorithm provides the best combination of performance and running time. We demonstrate practical examples of the insights provided by our algorithms when mining real sports-activity data. In addition, we show the applicability of our algorithms on other types of data, such as social networks.

Appendix: Proof of NP-hardness

To prove the np-hardness we use the following problem.

Problem 4

(Satisfy) Assume that we are given q formulas over $\ell$ variables $\left\{ v_i\right\}$ of form $\lnot z = x \wedge y$, where x, y, and z are boolean variables or their negations. Decide whether these clauses can be simultaneously satisfied with $v_1$ being set to true.

Proposition 10

Satisfy is NP-complete.

Proof

We will prove the hardness by reduction from 3SAT. Assume an instance of 3SAT with n variables and m clauses.

For each ith clause with two literals $x \vee y$, add $\lnot c_i = \lnot x \wedge \lnot y$.

For each ith clause with three literals $x \vee y \vee z$, add two formulas $h_i = \lnot x \wedge \lnot y$ and $\lnot c_i = h_i \wedge \lnot z$.

If the ith clause contains one literal x, then refer to x as $c_i$.

Add $m - 1$ variables $v_1, \ldots , v_{m - 1}$, and formulas $v_i = v_{i + 1} \wedge c_i$, for $i = 1, \ldots , m - 2$, and $v_{m - 1} = c_{m - 1} \wedge c_m$.

It follows that ith clause can be satisfied if and only if $c_i$ can be set to true. All $c_i$s can be set to true if and only if $v_1$ can be set to true. $\square$

Proposition 11

(k, 1)-segmentation is NP-hard.

Proof

We will prove the hardness by reduction from Satisfy. Assume that we are given an instance of Satisfy with q formulas and $\ell$ variables.

We begin by specifying the vertices. The total number of vertices is $1 + 3 + 2\ell + r$, where $r = (20q + 12\ell + 2)(3 + 2\ell )$.

The first vertex is $\alpha$, and every edge will be adjacent to $\alpha$. The next three vertices are $t_1$, $t_2$, and $t_3$. Our construction will make sure that $(\alpha , t_i) \in E(G)$.

The next $2\ell$ vertices correspond to the variables and their negations, we will denote them by $v_i$ and $u_i$, for $i = 1, \ldots , \ell$. We will denote by X the set of possible edges between $\alpha$ and these vertices. Define $X' = X {\setminus } \left\{ (\alpha , u_1), (\alpha , v_1)\right\}$.

Finally, the last r vertices are auxiliary vertices that will allow us to force segmentation borders. We will denote the set of possible edges between these vertices and $\alpha$ by B.

Our interation network consists of 3 parts, which in turn consists of sections. All these sections and parts are combined consecutively.

The first part, say $P_1$, consists of $2\ell$ sections, each containing 5 time points. The first $\ell$ sections are defined as

$$\begin{aligned} \begin{array}{rllllll} (\alpha , v_i): &{} 1 &{} 1 &{} &{} &{} \\ (\alpha , u_i): &{} &{} &{} &{} 1 &{} 1 \\ (\alpha , t_1): &{} 1 &{} 1 &{} &{} 1 &{} 1 &{}\\ (\alpha , t_2): &{} 1 &{} 1 &{} &{} 1 &{} 1 &{}\\ (\alpha , t_3): &{} &{} 1 &{} 1 &{} 1 &{} &{}\\ \text {for every } e \in B: &{} 1 &{} &{} 1 &{} &{} 1 &{}\\ \end{array} \end{aligned}$$

They last $\ell$ sections are copies of the first $\ell$ sections, except that they also contain the remaining edges from X at 1st, 3rd, and 5th time point.

The second part, say $P_2$, consists of 2q sections, each containing 7 time points. Let $c_i = (\lnot z = x \wedge y)$ be the ith formula. By using the same letters to represent the corresponding vertices, taking account negations, we define the ith section, where $i = 1, \ldots , k$, as

$$\begin{aligned} \begin{array}{rllllllll} (\alpha , x): &{} 1 &{} 1 &{} &{} &{} 1 \\ (\alpha , y): &{} 1 &{} &{} &{} 1 &{} 1 \\ (\alpha , z): &{} &{} 1 &{} 1 &{} 1 &{} &{} &{} 1\\ (\alpha , t_1): &{} 1 &{} 1 &{} &{} 1 &{} 1 &{} 1 &{} 1\\ (\alpha , t_2), (\alpha , t_3): &{} 1 &{} 1 &{} 1 &{} 1 &{} 1 &{} 1 &{} 1\\ \text {for every } e \in B: &{} 1 &{} &{} 1 &{} &{} 1 &{} 1 &{} 1 \\ \end{array} \end{aligned}$$

The $(q + i)$th section is a copy of ith segment, except that they also contain the remaining edges from X at 1st, 3rd, and 5th–7th time points.

The last part, say $P_3$, consists of $10q + 6\ell + 2$ sections, each consisting of 1 single time point. Each section contains B, $(\alpha , t_i)$, and $(\alpha , u_1)$. Moreover, every even section contains edges in $X'$.

We set $k = 20q + 12\ell + 2$. We claim that Satisfy is true if and only if the optimal segmentation has a score of

$$\begin{aligned} \begin{aligned} \sigma =&{\left| P_1\right| }/2 \times (3(2\ell - 2) + 2) + {\left| P_2\right| }/2 \times (5(2\ell - 3) + 12) + {\left| P_3\right| } / 2 \times (2\ell - 2)\quad . \end{aligned} \end{aligned}$$

We will prove this in several steps.

Step (i): Every $e \in B$ is contained in every segment exactly once. First, note that this segmentation is possible since B occurs at k different time points, the optimal cost of any such segmentation is bounded by r / 2, the number of possible edges times half the number of segments. Note that each $e \in B$ occurs at the exact same time point. Thus there is an optimal solution with every $e \in B$ either present or absent from the core. Assume that there is a segment that disagrees with the core. Then the cost is at least r. Consequently, every segment must contain every $e \in B$. Since B occurs at k different time points, each segment can contain only one instance of each $e \in B$.

Step (ii): It follows immediately, that the borders of the sections are included in the borders of the optimal segmentation. Moreover, each section in $P_1$ part is divided into 3 segments, each section in $P_2$ is divided into 5 segments, each section in $P_3$ corresponds to exactly 1 segment.

Step (iii): $(\alpha , t_i) \in E(G)$, $u_1 \in E(G)$ and $v_1 \notin E(G)$. This follows immediately from the fact that each section in $P_3$ corresponds to one segment, and ${\left| P_3\right| } > k / 2$, that is, $P_3$ contains the majority of the segments.

Step (iv): The cost of i th and $(i + 1)$ th section in $P_1$ is at least $3(2\ell - 2) + 2$. This bound is reached if and only if G contains either $u_i$ or $v_i$, but not both. First note, that the middle segment in both sectons contains the 3rd time point. This means that the remaining edges in X will occur exactly 3 times in 6 segments. Thus, they induce a cost of $3(2\ell - 2)$. A brute-force enumeration now implies that the involved edges induce a cost of at least 1, and this is possible if and only if G contains either $u_i$ or $v_i$, but not both.

Step (v): The cost of i th and $(i + 1)$ th section in $P_2$ is at least $5(2\ell - 3) + 12$. This bound is reached if and only if $(\alpha , z) \notin E(G) \Leftrightarrow (\alpha , x) \in E(G)$ and $(\alpha , y) \in E(G)$. First note, that the 2nd segment in both sectons contains the 3rd time point, and the 4th and 5th segments consists of exactly one time point. This implies that the remaining edges in X will occur exactly 5 times in 10 segments. Thus, they induce a cost of $5(2\ell - 3)$. A brute-force enumeration now implies that the involved edges induce a cost of at least 6, and this is possible if and only if $(\alpha , z) \notin E(G) \Leftrightarrow (\alpha , x) \in E(G)$ and $(\alpha , y) \in E(G)$.

Step (vi): The cost of an odd and even section in $P_3$ is equal to $2\ell - 2$. This follows from the fact that the edges in $X'$ occur exactly once in these two sections.

Step (vii): Steps (iv)–(vi) imply that $\sigma$ is a lower bound for the optimal score. This bound is reached if and only if, the lower bounds of each section is reached. This can happen if and only if each sentence in Satisfy can be satisfied. $\square$