1 Introduction
Pursuitevasion games are games played between two teams of players, who take turns moving within the confines of some abstract arena. Typically, one team – the pursuers – are tasked with catching the members of the other team – the evaders – whose task it is to evade capture indefinitely. The study of such games has led to their application in a number of realworld scenarios, one widelystudied example of which would be their application to the problem of guiding robots through realworld environments [8]. From a theoretical standpoint, other variants of the game have been studied for their intrinsic links to important graph parameters; for example, in one particular variant in which each pursuer can, in a single turn, move to an arbitrary vertex of the given graph , it is well known that establishing the number of pursuers it takes to catch one evader also establishes the treewidth of [20].
The variant most closely resembled by the one considered in this paper was first studied separately by Quilliot [18], and by Nowakowski and Winkler [15], as the discrete Cops and Robbers game. In essence, the games these authors considered were the same: one cop (pursuer) and one robber (evader) take turns moving across the edges (or remaining at their current vertex) of a given graph , with the cop aiming to catch the robber, and the robber attempting to avoid capture. (By ‘catching the robber’ we mean that the cop is able to occupy the same vertex as the robber within .) In this paper, we consider a variant of the Cops and Robbers game with an almost identical set of rules to the one considered in [18, 15], but broaden the class of viable game arenas to include the edgeperiodic graphs [6]. As such, we call the game edgeperiodic Cops and Robbers, or EPCR for short. Informally, such graphs can be thought of as traditional static graphs equipped with an additional function, mapping each edge to a pattern of length that dictates in which time steps is present within each consecutive period of steps. (Formal definitions of EPCR as well as the class of edgeperiodic graphs are given in Section 2.) The class of edgeperiodic graphs could also be considered a subclass of socalled temporal graphs [14].
The blanket term ‘temporal graphs’ covers a broad, and relatively new area of interest to mathematicians and computer scientists, which looks to examine the inherent properties of graphs that have had incorporated into their combinatorial structure some element of timevariance. As far as we are aware, cops and robbers type games (in fact, pursuitevasion games in general) have not yet been studied in the context of temporal graphs.
The contribution of this paper is twofold: in Section 3, we consider the problem of deciding, given an edgeperiodic graph , whether a game of edgeperiodic cops and robbers played on is won by the cop, or won by the robber. We exploit a connection (that was previously noted in [11]) between the game of cops and robbers (the one considered in [18, 15]) and reachability games, which are a well studied class of 2player games with strong connections to formal verification. Utilising reachability games as a framework, we extend existing techniques in order to solve the one cop, one robber variant of cops and robbers on a wider collection of graphs: specifically, what we show is that, given an edgeperiodic graph whose edges are each labelled with patterns of length , it is possible to decide in time whether cop or robber wins on (where denotes the least common multiple of the set of all edgepattern lengths in ). We also show that by applying the general form of this result to some subclasses of the class of edgeperiodic graphs, we are able to obtain polynomial upper bounds on the amount of time taken to determine the winner. Further, we show that it is possible to construct a strategy for the winning player using an algorithm with the same time bound.
In Section 4, we consider a subclass of the edgeperiodic graphs, the members of which have underlying cycle graphs. We provide proof of an upper bound of on the length required by any such cycle in order to guarantee that it is robberwin, where if , and otherwise. Also provided within this section are lower bound constructions showing that there exist copwin edgeperiodic cycles of length and in the and cases, respectively.
1.1 Related Work
The introduction of pursuitevasion type combinatorial games is most often attributed to Torrence D. Parsons, who studied a problem in which a team of rescuers search for a lost spelunker in a circular cave system [16]. By representing the cave as a cycle graph, he showed that one rescuer is not enough to guarantee that the spelunker is found, but that two are. In a similar vein to the work of Parsons, the Cop and Robber problem, in which one cop attempts to catch a robber in a given graph , was introduced independently by both Quilliot [18], and by Nowakowski and Winkler [15]. Their papers characterise precisely those graphs for which one cop is enough to guarantee that the robber is caught. Aigner and Fromme [1] considered a generalised variant of the game, in which cops attempt to catch a single robber; their paper introduced the notion of the copnumber of a graph, i.e., the minimum number of cops required to guarantee that the robber is caught.
In [10, 4, 2], the authors develop reductions from the standard game of cops and robbers to a directed game graph, and specify algorithms that can decide, for a given graph, whether cop or robber wins. In [11], Kehagias and Konstantinidis note a connection between the formulations of [10, 4, 2] and reachability games. Reachability games are a wellstudied class of 2player tokenpushing games, in which two players push a token along the edges of a directed graph in turn – one with the aim to push the token to some vertex belonging to a prespecified subset of the graph’s vertex set, and the other with the aim to ensure the token never reaches such a vertex [9]. It is well known that the winner of a reachability game played on a given directed graph can be established in polynomial time [3, 9]. For more information regarding cops and robbers/pursuitevasion games, as well as their connection to reachability games, we refer the reader to [9, 3, 11, 12, 5, 17, 8].
In this paper, we consider the game of cops and robbers within the context of a temporal graph model. Temporal graphs are a relatively new object of interest, and incorporate an aspect of timevariance into the combinatorial structure of traditional static graphs [14]. One previously considered way of viewing a temporal graph is as a sequence of subgraphs of a given underlying graph (where is the lifetime of the graph) [13], with each subgraph indexed by the time steps . For problems within this model, it is often natural to assume that each subgraph in all time steps is connected [13]. The edgeperiodic graphs considered in this paper differ in that this connectivity assumption is dropped – similar graphs were introduced in [6]. For a detailed account of the theory of temporal graphs thus far, as well as the various models/problems that have been studied, we refer the reader to, for example, [7, 13, 14].
2 Graph Model and Game Rules
For any positive integer we write for the set . We begin with our definition of edgeperiodic graphs:
Definition 1 (Edgeperiodic graph ).
We define an edgeperiodic graph to be a temporal graph with underlying graph and infinite lifetime, and an additional relation , which maps each edge to a pattern of length . Each consists of boolean values, such that is present in a time step if and only if ; otherwise, is not present in time step . Additionally, assume that for any edge , for at least one ; this implies that every edge is present at least once in any period of time steps.
(We note that any whose edges each have patterns consisting of all ’s can be treated, under our model, as equivalent to the traditional static graph.) In this paper, we consider a game of cop(s) and robber(s) identical in its rule set to the one introduced in [18] and [15] (in particular, the variant with 1 cop and 1 robber), but broaden the class of viable gamearenas to include all possible edgeperiodic graphs . We call the resulting game edgeperiodic cop(s) and robber(s), or EPCR for short – the following paragraph specifies its rule set, as well as the separate winning conditions for C and R:
Rules of EPCR. Initially, the two players (cop C and robber R) each select a start vertex on a given edgeperiodic graph . C chooses first, followed by R, whose choice is made in full knowledge of C’s choice. After the start vertices have been chosen, in each subsequent time step , players take alternating turns moving over an edge in the graph that is incident to their current vertex (or indeed choosing to remain at their current vertex), following the convention that in any particular time step, C moves first, in full knowledge of R’s position, followed by R; again, R’s move is made with full knowledge of the move that C just made. In all subsequent turns, each player makes their moves in full knowledge of the other player’s previous move. In line with Definition 1, whenever C or R are situated at a vertex during some time step and it is their turn to make a move, they may only traverse those edges such that . The game terminates only when, at the end of either player’s move, C and R are situated at the same vertex in . If there exists a strategy for C which ensures that the game terminates, we say that is copwin. Otherwise, there must exist a strategy for R which enables infinite evasion of C; in this case we call robberwin.
3 Determining the Winner of a Game of Epcr
Within this section, our aim is to prove the following theorem (throughout the following, we use to denote the lowest common multiple of a set of integers ):
Theorem 2.
Let be an edgeperiodic graph of order , and let . Then, it can be decided in time whether or not is copwin.
The proof relies primarily on a transformation from a given edgeperiodic graph to a finite directed game graph . The transformation is such that the playing of an instance of EPCR on is, in some sense, equivalent to the playing of a 2player tokenpushing game (specifically a reachability game, which will be properly defined in due course) played on . To establish this equivalence, we need a way of translating a particular state of an instance of EPCR played on to a corresponding state in the reachability game played on . To this end, the following definition properly introduces the notion of a position in a game of EPCR, played on an edgeperiodic graph .
Definition 3 (Position in ).
The current state of a game of EPCR played on an edgeperiodic graph is determined by 4 individual pieces of information: (1) the vertex currently occupied by C; (2) the vertex currently occupied by R; (3) the player whose turn it is to move; and (4) the current time step t. To this end, we can define a position in to be a tuple,
where is C’s current vertex, is R’s current vertex, is the player whose turn it is to move next, and is the current time step.
We call any position such that a terminating position, since this indicates that both players are situated on the same vertex and hence C has won. We now formally introduce the notion of reachability games [9]:
Definition 4 (Reachability game ).
A reachability game is a directed graph , given as a 3tuple:
where is a partition of the state set ; is a set of directed edges; and is a set of final states.
The game is played by two opposing players, Player 0 and Player 1; and are the (disjoint) sets of Player 0/Player 1 owned nodes, respectively. One can imagine a token being placed at some initial vertex (call it ) at the start of the game. Depending on whether or , we can then imagine the corresponding player selecting one of the outgoing edges of , and pushing the token along that edge. When the token arrives at the next vertex, the corresponding player then selects an outgoing edge and pushes the token along it. This process then continues – such a sequence of moves constitutes a play of the reachability game on . Formally, a play is a (possibly infinite) sequence of vertices in , such that for all . We say that a play is won by Player 0 if there exists some such that . Otherwise, is of infinite length and for no is ; in this latter case, we say that is won by Player 1.
3.1 Transformation
We now detail our transformation from a given edgeperiodic graph to a reachability game : let be a transformation function that takes as argument a given edgeperiodic graph , so that the notation denotes the game graph on which each play of a reachability game corresponds to a sequence of moves performed by C and R in a game of EPCR on . As before, we denote by the least common multiple of the set of integers . Further, let . We let , and go on to define its individual components below:
State set . We define the state set (i.e., vertex set) of our directed game graph to be a set of 4tuples, each corresponding to a position in the game of EPCR on as follows:
Keeping in line with Definition 4, we also let be the sets of Player 0 (or Cop) owned nodes, and Player 1 (or Robber) owned nodes, respectively. In order to justify this formulation (in particular, to justify the range of variable – the current time step component of any state ), recall from Definition 3 that a position of a game of EPCR consists of a cop vertex , a robber vertex , a variable which represents the player whose turn it is to move next, and a current time step . We wish to capture with our finite directed game graph all possible positions of EPCR on . Since, by Definition 1, the lifetime of is infinite, we cannot simply create a state corresponding to each possible position ; the infinite number of time steps would result in an infinite game graph . It is not hard to see that in the th step, all edge patterns will finish (since divides for all ), and so in the following step, all patterns will restart. As such, we can view the temporal structure of our edge set as an infinitely repeating pattern, and by letting range over the integers in we are able to properly capture this structure using only a finite number of states.
Edge set . In order to construct the edge set , we consider all pairs of states and such that and . We can then see as the set of edges such that if and only if the states and satisfy all of the conditions below:

,

,

,

if and , then ,

if and , then .
Condition (1) ensures that any sequence of moves constituting a play in alternate between C and R, which keeps in line with the rules of EPCR. Condition (2) ensures that any state , reachable in one move from a Cowned state , is such that is adjacent to in (or is in fact , indicating that C has waited at the current vertex); that the robber’s new vertex does not change; and, that , keeping in line with the rule stating that C moves first in any given time step, followed by R who must also make a move in step . On the other hand, Condition (3) ensures that, once R pushes the token from some Rowned state , R’s new vertex is adjacent to in (or equal to ); that C’s vertex remains the same, and that the state to which the token is pushed onto is a state in which the current time step is advanced by one. Conditions (4)(5) ensure that both players can only make moves across edges that are incident to their current vertex, as well as present in the current time step; on the other hand, they also ensure that either player always has the ability to remain at their current vertex in any step if they should choose to do so.
Set of final states . Let , so that the set of final states consists of all those states that correspond to a position in such that C is positioned on the same vertex as R. This adheres to the rules of EPCR, which state that the game terminates only when this condition is met by the current position.
3.2 Proof of Theorem 2
We first introduce the elements of the theory of reachability games that are required for the proof of Theorem 2, starting with the definition of the attractor set:
Definition 5 (Attractor set [3]).
The sequence is recursively defined as follows:
We can see that the sets , as defined above, are a sequence of subsets of that are monotone with respect to setinclusion. We then let
Since is finite, we are able to view the set as the least fixed point of the sequence .
From Definition 5 it follows by induction that, from those states such that and for any , Player 0 is able to force the sequence of play into some state within moves, by selecting for each such a successor state such that and for some . On the other hand we have that, from any state (again, let and for any ), Player 1 can not avoid forcing the sequence of play into a state (for some ); from the definition of (), it again follows by induction that play will be forced into some state in at most steps. This brings us to the following wellknown result from the reachability games literature, which will be useful in proving Theorem 2:
Theorem 6 (Berwanger [3]).
In a given reachability game , Player 0 has a winning strategy from any state , and Player 1 has a winning strategy from any state .
Recall now that the transformation produces, from a given edgeperiodic graph , a directed game graph such that there is a correspondence between every possible position in the game of EPCR on with some state in , and vice versa. Using the notation to refer to the state in that corresponds to the position in the game of EPCR on , we can compute the set for our game graph and thus, on invocation of Theorem 6, state the following lemma:
Lemma 7.
Cop can force a win from a position if and only if the state satisfies .
Note that one consequence of Lemma 7 is the following: In a game of EPCR on starting from a position such that , the robber can force the sequence of moves to never reach any state , and, as such, the game can be won by R.
Lemma 8.
An edgeperiodic graph is copwin if and only if there exists a vertex such that for all .
Proof.
() Assume not, so that is copwin but there exists no vertex such that for all . Then for every , there exists at least one vertex such that the state . Let C choose its start vertex as (i.e., sets ), and let R set . Since R chooses in full knowledge of C’s choice of , it follows that R can force the equivalent reachability game on to begin from a state , hence winning the reachability game regardless of C’s choice of . Notice that this implies that there exists a winning strategy for R in the game of EPCR on ; this is a contradiction since, by assumption, is copwin.
() Assume C chooses as its start vertex, i.e., sets . By doing so, the equivalent reachability game on can be forced to start at some state regardless of R’s choice of , since for all . Hence, regardless of R’s choice of , C wins the reachability game on , and, as a result, can win the game of EPCR on by picking its start vertex as ; the lemma follows. ∎
The proof of the main theorem will also make use of a further known result from the reachability games literature; for the following, let be a given directed game graph.
Theorem 9 (Grädel et al.[9]).
There exists an algorithm which computes the set in time .
Given the above, all is in place for the proof of Theorem 2:
Proof of Theorem 2.
Since , produces, given an edgeperiodic graph , a directed game graph , such that . To see this, observe first that for an arbitrary position in a game of EPCR on , there are ways to choose , ways to choose , and a further 2 ways to choose . By definition of the transformation function, has states for time steps only, and so in total we have that , as claimed. Next, note that each state has at most edges leading away from it to other states. This is because in the corresponding position in the game of EPCR on , the player whose turn it currently is has at most choices of moves across edges – at most edges leading to other vertices plus the choice of remaining at the current node. Since there are states , it follows that .
Combining the above with the result of Theorem 9, we can conclude that the attractor set (that is, the set of all states from which Player 0, i.e., C, has a winning strategy) of any graph can be computed in time . By Lemma 8, we can then verify whether or not is copwin by checking if there exists at least one vertex such that , for all ; if such a exists, the algorithm will return YES, otherwise the algorithm will return NO. Carrying out this check can clearly take at most time, and the theorem follows. ∎
We also note, as a direct consequence of Theorem 2, that as long as is polynomial in and , then the winner of a given graph can be decided in polynomial time. Furthermore, if the labels are bounded by some constant for all , then the winner can be decided in time.
As well as being able to decide whether or not a given edgeperiodic graph is copwin or not, we would like to be able to compute a strategy for the winning player of the game of EPCR on a given graph . One common way to view a strategy for Player i (), in a general infinite game played on a game graph (where ), is as a partial function . Here, can be seen as the set of all prefixes (of any play in ) that end in a state , with dictating to Player i the appropriate move to play, based on the history of these prefixes.
On the other hand, a memoryless strategy can be viewed more simply – as a partial function . Such a strategy can be employed in games where a correct move for a player depends not on the entire statehistory of some play (or a prefix of) , but only on the current state. It is wellknown that reachability games fall into this category [9]; since EPCR reduces to a reachability game, we are thus able to make use of the following result from the literature:
Theorem 10 (Berwanger [3]).
Given a reachability game , one can compute in time a memoryless winningstrategy for Player 0 from any state , and a memoryless winningstrategy for Player 1 from any state .
As such, given a directed game graph (with ), Theorem 10 tells us that it suffices to compute, for the winning player, a memoryless winning strategy , with the value of depending on the winner of the reachability game on . The following theorem shows that it is possible to interpret any such as a strategy for the winning player in the corresponding game of EPCR on :
Theorem 11.
Let be an arbitrary edgeperiodic graph and . Then, depending on whether is copwin or not, one can compute in time either a memoryless winning strategy enabling C to capture R, or a memoryless winning strategy enabling R to evade capture infinitely.
Proof.
Let be the return value of the algorithm from Theorem 2 when provided as input. First, we construct a strategy for the winning player of the equivalent reachability game , and go on to show how such a strategy can then be interpreted as a strategy for the corresponding game of EPCR on .
First, consider the case in which . Then we know is copwin and, by Lemma 8, we know that there exists some vertex such that for all . As such, the initial stage of our strategy for C should consist of computing such a vertex , and setting the cop start vertex in to . Now, using Theorem 10, we can compute a memoryless winningstrategy . The initial stage of identifying some vertex such that for all takes time, and the algorithm of Theorem 10 takes time at most ; it follows that the overall construction of a Cstrategy for the reachability game takes time. Such a strategy for can then be interpreted as strategy for C in the game of EPCR on by first selecting start vertex . From then onward, whenever it is turn, in order to establish the appropriate move to play given a current position , C constructs from it a state , and checks the component of the state . Finally, it is guaranteed that is adjacent to (or possibly ) during , due to the way the transformation from to has been defined.
In the situation in which , then we know that is robberwin, and thus know by Lemma 8 that for every , there exists at least one vertex such that the state . Thus, the initial stage of our strategy for R involves the construction of a mapping from each possible that C might choose as its start vertex, to a vertex satisfying the aforementioned nonmembership condition. Application of Theorem 10 then allows us to construct a memoryless winning strategy, , from all states . The construction of involves checking, for each of possible start vertices for C, at most vertices in order to identify which combination of and satisfies . Hence, this initial phase can take at most time. Similar to before, the algorithm of Theorem 10 can take at most time, and hence we have that the overall construction of a strategy for R can take at most time, as claimed. Finally, interpreting as a strategy for R in the game of EPCR on is the same as for C; the only difference is the way in which the start vertex is selected – R waits until C has selected a start vertex , and then chooses its own start vertex as . ∎
We remark that Theorems 2 and 11 can be generalised to a setting with cops at the expense of increasing the algorithm’s running time to . The idea is to fix an arbitrary ordering of the cops, and create layers of states during every time step (one for each of the cops’ moves, followed finally by the robber’s move). By allowing in each time step for the players to play their moves in this serialised fashion the resulting game graph would require layers with states in each, with at most edges leading from every state to states in the following layer.
4 An Upper Bound on the Length Required to Ensure an EdgePeriodic Cycle is RobberWin
In this section, we consider a restricted subclass of the edgeperiodic graphs – in particular, we consider the subclass of edgeperiodic cycles . We provide an upper bound on the length required of any edgeperiodic cycle , to ensure that it is robberwin. First, we show that any edgeperiodic infinite path whose edgepattern lengths originate from a set of integers with finite size is robberwin, and second show how the strategy for such infinite paths can be adapted to the cycle case. Throughout the following, we refer to cop as C and robber as R. We always assume that we are given an edge periodic cycle , and that the set of integers contains the lengths of every bit pattern that maps at least one edge to. Additionally, we write LCM to denote the least common multiple of the numbers in the set .
We first present a lemma for infinite paths, which will also allow us to handle the case in which the cop chases the robber around the cycle in a fixed direction.
Lemma 12.
Let be an infinite edgeperiodic path, and assume that is finite. Then, starting from any time step , if then there exists a winning strategy for R from any vertex with distance at least from C’s start vertex, and at least LCM otherwise.
Proof.
First, notice that since we assume that is finite, so must be LCM. Let C pick its initial vertex . Let R’s initial vertex be denoted by , and assume without loss of generality that will be some vertex in that lies to the right of . As such, we will from here onward denote by the path starting at and extending infinitely to the right.
Consider the set and its constituent elements. There are two cases – either (1) there exists such that is not a multiple of – then, , since it cannot be the case that for any ; or (2) for every , for some integer ; then, . With this in mind, define if (1) holds and if (2) holds. Now, let us define the strips () to be finite subpaths of , such that for all edges , can first be traversed by C in some time step . Note that and hence each must contain at least two edges. By convention, we call the leftmost and rightmost edges (vertices) of any its first and last edges (vertices), respectively. Note also that the last vertex of and the first vertex of are one and the same, for all .
Assume from now on that C moves right whenever possible. It is safe to do so since, otherwise, C may only be positioned at the same vertex or further left than when following this strategy. The strategy for R is as follows: pick to be the first vertex of and move right (i.e., away from C) whenever possible.
We now demonstrate that R’s strategy is a winning one. Let and denote the first time step that player is able to traverse the first/last edge of , respectively. Note that and that . Combining the two gives that , which implies that C can never catch R in any step in which the edge leading to both player’s right belongs to .
We next show that R cannot be caught when the edge leading to C’s right belongs to and the edge leading to R’s right belongs to . Let and recall that . Since the strips are defined to consist of all edges crossed in the period , and since , it follows that at time , there is at least one more edge of that remains to be crossed by C. This gives that and yields the claim. Combining this with the earlier observation that , it follows that there exists a strategy for R starting from the first vertex of .
Finally, recall that when , we have that each consists of at most edges, otherwise it consists of at most LCM. Hence, there exists a winning strategy for R starting from some vertex with distance at most or LCM from , depending on the condition satisfied by LCM. The lemma follows by noticing that the above strategy also works when R is initially positioned at any vertex further to the right than the first vertex of . ∎
Theorem 13.
Let be an edgeperiodic cycle on vertices and . Then, if , is robberwin (where if , and otherwise).
Proof.
We let and denote the vertex at which C and R are positioned at the start of time step , respectively. Consider now some edge
and classify its vertices as a ‘left’ and ‘right’ vertex arbitrarily; let the left vertex of each edge be the right vertex of the following edge in the cycle. We proceed by specifying a strategy for
R. Initially, let C choose ; R should choose to be the vertex antipodal to in . (Ifis odd then
R should select to be either of the two vertices that are furthest away from ; we will refer to both these vertices as antipodal to , and treat vertices in all steps in the same way.) We now distinguish between two modes of play, Hide and Escape, and specify R’s strategy in each of them.Hide mode: A Hide period begins in step and in any step such that and are antipodal, but and were not. As such, any game in which R follows our strategy begins in a Hide period. The Hide period beginning at step consists of the steps such that and are antipodal, but and are not. Any Hide period is followed directly by an escape period, which will start in step .
R’s Hide strategy: If the game is in a Hide period during step , R should observe C’s choice of , and always try to move to a vertex antipodal to it. We claim that R cannot be caught in any step belonging to a Hide period. To see this, observe that regardless of whether or , we have that . As a result, antipodal vertices in are at least distance 2 apart from one another, and the claim follows.
Escape mode: An Escape period always begins in a step such that step was the last step of some Hide period. As such, an Escape period consists of steps , such that each and are not antipodal, but and are. The last step of the Escape period is then , and the first step of the next Hide period is .
R’s Escape strategy: Assume that some Escape period starts in step . Then, at the start of step , and were antipodal to one another, and during step , we had a situation in which C was able to move towards R in some direction, but the edge incident to leading in the same direction was not present. Now, recall that if , so that , then ; and if so that , then . Therefore, since and are antipodal in , if holds we have that the distance between them is at least and if holds, the distance between them is at least LCM. Observe now that we are able to view any edgeperiodic cycle of finite length as an infinite path whose edge patterns repeat infinitely often. Combining these two facts, it then follows from Lemma 12 that when the Escape period starts in step , there exists a strategy for R (which started in the previous step from vertex ) which will enable it to stay alive until the Hide period ends.
Finally, since every step belongs to either a Hide period or an Escape period, we have shown that C can never catch R, and the proof is complete. ∎
We now give lower bounds on the length required of a strictly edgeperiodic cycle to ensure that it is robberwin.
Theorem 14.
There exists an edgeperiodic cycle of length with edge pattern lengths in the set that is both copwin and satisfies .
Proof.
Let be an integer and consider an edgeperiodic cycle of length with edge pattern lengths in . Let consecutive edges have patterns of length , and all remaining edges have pattern . We refer to the subpath of consisting of the two edges with period as the path, and the subpath with edges labelled with as the path.
We now specify a strategy for C and show that it is in fact a winning strategy: Let C position itself initially at either of the two vertices belonging to the path that are distance from one extreme point of the path, and from the opposite extreme point (where distance is taken to mean the length of the path to that extreme point that avoids the edges of the path). Call that chosen vertex , and notice that it splits the path into two subpaths that intersect only in – one of length which we will call , the other of length that we call . If R chooses its initial position to be some vertex lying on , then C can move along all edges of in the first steps. Since the only way for R to leave without running into C is via the path, C will catch R in these steps, since no edge of the path is present until step . If R chooses its initial position as some vertex lying on , then in the first steps C can traverse all edges of . Again, the only way for R to leave without encountering C is via the path. This time R will be able to traverse one edge of the path, but will be stuck at the middle vertex of the path until time step . Since, in this step (which is the th step), C will be positioned at the vertex that lies on both the path and , C will move first and catch R. It remains to be shown that R will be caught if it chooses the middle vertex of the path as its start vertex: here, R will not be able to move until step , so C should traverse all edges of in the first steps. Then, in step , C will be at one endpoint of the path, with R at the middle vertex of the path – in this step, C will move first and catch R. Since we have shown that C wins in all cases, and since , the theorem follows. ∎
A small amount of modification to the construction in the proof of Theorem 14 yields an additional lower bound for the case:
Theorem 15.
There exists an edgeperiodic cycle of length with edge periods in the set that is both copwin and satisfies .
Proof.
Perform the construction from the proof of Theorem 14, taking to be odd. Select one of the vertices that has distance and from opposite ends of the path, calling that vertex . Consider the strategy from the proof of Theorem 14, and notice that there are two edges that C may cross in the second step. Select either one of these edges and replace its current pattern of with a pattern of the form (with length ). C should now select its initial vertex as and follow the strategy in the proof of Theorem 14 – this works since the edge with pattern has been selected so that it is present whenever C’s strategy crosses that edge. Now, notice that is odd, and so we have that . Since the constructed cycle has length , the theorem follows. ∎
5 Conclusion
We considered the problem of deciding whether a single cop can catch a single robber on a given graph, but extended the class of viable game arenas to include the edgeperiodic graphs. For a given vertex edgeperiodic graph whose edges are labelled with patterns of lengths in the set , we showed, amongst other things, that there exists an algorithm with running time that decides whether the cop or robber wins, and computes a strategy for the winning player. One natural open question that we find particularly interesting is the following: what is the complexity of deciding whether cop or robber wins when the least common multiple of the label lengths, , is not bounded by a polynomial? We note that , where is Chebyshev’s function [19]. This shows that, given a graph with at least edges and at least one edge labelled by each of the integers , the transformation function described in Section 3.1 would produce a directed graph of order , i.e., exponential in , and as such the algorithm of Theorem 2 would have exponential running time. It would be interesting to establish whether there exists a better algorithm, or whether the problem is hard for this case. More generally, one could also examine the cops and robbers problem within the context of other temporal graph models.
In the second part of the paper, we obtained a upper bound on the length required of any edgeperiodic cycle to ensure that it is robberwin, where if , and otherwise. We also provided lower bound constructions of copwin cycles that have lengths and in the and cases, respectively. As an example of a potentially interesting further direction, one could attempt to tighten the gap between upper and lower bounds on the minimum length of edgeperiodic cycles to be guaranteed to be robberwin in both of these cases.
Acknowledgements
The authors would like to thank an anonymous reviewer for a suggestion leading to the runningtime for the variant with cops mentioned at the end of Section 3.
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