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Cops and robbers on oriented toroidal grids

The game of cops and robbers is a well-known game played on graphs. In this paper we consider the straight-ahead orientations of 4-regular quadrangulations of the torus and the Klein bottle and we prove that their cop number is bounded by a constant. We also show that the cop number of every k-regularly oriented toroidal grid is at most 13.

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03/31/2021

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1 Introduction

The game of Cops and Robbers is a well-known game on graphs that was introduced by Nowakowski and Winkler nowakowski and Quilliot quilliot . There are two players, one controls the cops and the other one controls the robber. The game starts with the cops selecting some vertices as their initial positions (multiple cops can select the same vertex). Then the robber chooses his initial position. From now on, first the cops move and then the robber moves, where moving means either staying at the same position or moving to a neighbouring vertex. The game on digraphs is defined in the same way except that every move must be made along a directed edge.

The cops win the game if one of them can get to the same vertex as the robber and the robber wins if he can avoid this indefinitely. The minimum number of cops that can guarantee the robber’s capture in a graph or digraph is called the cop number of and will be denoted by .

Game of cops and robbers on graphs has received lots of attention but very little is known about the game on digraphs. Basic results have been introduced in Bonato . The paper grid shows that the game of Cops and Robbers in digraphs can be as natural and as inspiring as the game on undirected graphs. Recall that a (di)graph is a cover over a (di)graph if there is a (di)graph homomorphism (called the covering projection) which maps the edges incident with any vertex bijectively onto the edges incident with in . The following lemma from grid can be used to analyze the cops and robber game on where a good strategy of cops on is known.

Lemma 1.1.

grid Suppose that a graph or digraph is a cover over a (di)graph . If cops have winning strategy on , then they also win on , that is .

Consider a 4-regular quadrangulation of a surface. It follows by Euler’s formula that the surface is either the torus or the Klein bottle and it can be shown by using the Gauss-Bonnet Theorem that the straight-ahead walks partition the edges into cycles, all of which are noncontractible on the surface. These cycles can be split into two classes, each class consisting of pairwise disjoint cycles (we call them vertical cycles and horizontal cycles, respectively) such that each vertical and each horizontal cycle intersect (possibly more than once). By giving each of these cycles an orientation, we obtain an Eulerian digraph in which, at each vertex, the two incoming edges and two outgoing edges are consecutive in the local rotation around the vertex. The universal cover of a 4-regular quadrangulation is the 4-regular tessellation of the plane with square faces (the integer grid), and every finite quotient of the integer grid is a 4-regular quadrangulation of the torus or the Klein bottle. Four-regular quadrangulations of the torus admit a simple description. Each such quadrangulation is of the form , where are arbitrary positive integers, , and is obtained from the grid with underlying graph by identifying the “leftmost” path of length with the “rightmost” one (to obtain a cylinder) and identifying the bottom -cycle of this cylinder with the top one after rotating the top clockwise for edges. This classification can be derived by considering appropriate fundamental polygon of the universal cover (which is isomorphic to the tessellation of the plane with squares). Quadrangulations of the Klein bottle are a bit more complicated. While all toroidal quadrangulations are vertex-transitive maps, this is no longer true for the Klein bottle. For our purpose it will suffice to know that the orientable double cover of such a quadrangulation is of the form .

Note that every 4-regular quadrangulation of the torus and the Klein bottle has the Cartesian product as its covering graph. By Lemma 1.1, we will be able to restrict ourselves to deal with different orientations of . The paper grid treats the cop number of orientations of for which the corresponding digraph is vertex-transitive. In this paper we will consider more general families of orientations of the grid, which we call straight-ahead orientations, and show that their cop number is bounded by a constant.

2 Streams, confluxes, and traps

The grid we are working on is . For a vertex , the digraph induced by will be called the row of , and the subdigraph induced by will be called the column of . A line containing a vertex is either a row or a column of . We will assume that the edges of are oriented in such a way that every line is a directed -cycle in the grid. We refer to this as a straight-ahead orientation of the toroidal grid. We say that two lines in are consecutive if one of their coordinates correspond to adjacent vertices in one of the factors of . A set of consecutive lines oriented in the same direction will be called a stream and its width is the number of lines in the stream. If and are streams such that , we say that is a substream of .

If and are edge-disjoint streams and the set is not empty, we will call a conflux (see Figure 1). The vertices in a conflux with the minimum number of neighbours in are called corners. Notice that the set of corners of a conflux is never empty, and if , then it can have four vertices (if ), two vertices ( and with ) or one vertex ().

If has four corners, then a corner is main

if it has an odd number of outneighbours in

and secondary otherwise (see Figure 1). However, if has one or two corners, they will all be referred to as main.

Figure 1: The black vertices form a (maximal) conflux. Corners and of the conflux are its main corners, while and are secondary corners.

We will always assume that the robber is forced to move from its current position. We can make sure this happens by chasing him with a cop. For Lemma 2.2 and Lemma 2.3, we will assume that one cop is chasing the robber so the robber is forced to move. We will use to denote the current position of the robber, and for the position of the cop .

Lemma 2.2.

Let be a conflux with one cop on each main corner. If and , then the robber will be captured or his movements will be restricted to a stream containing .

Proof.

Notice that when , there is a main corner cop in the same row (column) as . Let us call this cop and the other one . We may assume the incoming edge to comes from above due to the symmetry of the argument. In order to leave , the robber must step on the column where is or leave through the bottom.

The strategy for and will be the following: If the robber moves towards ’s column, then stays where he is and copies the robber’s move. If the robber moves down, then copies the robber’s move, and stays in the same place if it is in the same column but different row as the robber, copies the robber’s move if he is in the same row, and moves towards the robber’s column otherwise.

By following this strategy, the robber and are always on the same row, so the robber cannot leave crossing ’s column or he will be captured. This means that the robber can move horizontally at most times, where is the stream formed by the columns containing vertices of . Therefore, the robber’s movements are restricted to . ∎

Notice that in the case where the streams that form have the same width, two cops guarantee the capture of the robber. However, once the robber’s movements have been restricted to a stream, one extra cop will guarantee the capture. Note that for Lemma 2.2 to work, we need to set up the trap before the robber enters it. It is possible to set a slightly different trap that works regardless of where the robber’s in-neighbors are, but we need one more cop to do this.

Lemma 2.3.

Let be a conflux with one cop on each main corner and one cop in the secondary corner of without out-neighbors in if such corner exists. If the robber is in , then he will be captured or his movements will be restricted to a stream containing .

Proof.

Let and be the streams such that . If we have that , then all the corners of are main and by Lemma 2.2 we are done. If and is the secondary corner of with a cop, take the substreams and of and ,respectively, that contain just the row (column) through the vertex . After at most moves, the robber will be on a vertex of or . Since the main corners of both and are covered by cops, then an application of Lemma 2.2 gives us the desired result. ∎

Given a grid , we can define the conflux digraph of , which we will denote by , as the digraph whose vertex consists of all maximal confluxes of , and where is an edge of whenever there exist vertices and such that . Note that is isomorphic to (with straight-ahead orientation) where is the number of maximal column streams and is the number of maximal row streams.

There is a natural correspondence between and the elements of

. This correspondence allows us to represent each move of the robber or a cop by the addition of a vector in

to its current position.

Given a vertex with , we can define the sets

Observation 2.4.

For any vertex and any line in , and .

The sets and will be called the secondary diagonal and main diagonal of , respectively. Geometrically speaking, if we think of the arcs of the digraph as vectors, is the set of all the vertices of in the diagonal line through defined by the sum of the arcs leaving , and is the set of vertices in the line orthogonal to that one. Notice that for every vertex with , we have that is an element of . This value will be called the type of the vertex, , and two vertices will be of opposite types if their types are additive inverses in . Elements of will be refered to as types. Notice that all the vertices in a conflux have the same type, so we can define where .

Let be a vertex in and the maximal conflux containing . We define the horizontal escape distance of , as the length of the shortest directed path starting at and ending at a vertex outside of using only horizontal arcs (adding ). Analogously, we define the vertical escape distance of and denote it with . The escape distance of is .

Lemma 2.5.

Let and be confluxes of such that . If there are cops in the main corners of and , and the robber is in , then the robber will be captured or its movements will be restricted to a stream.

Proof.

It is easy to see that if the robber is in and is forced to move, then he will enter . Since the main corners of and are covered, the result follows from Lemma 2.2. ∎

3 The -regularly oriented grid

We say that a grid is -regularly oriented if for every maximal stream in . The cases where have been covered in grid , so in this section we will assume that is a -regularly oriented grid with . Let and be vertices in . We say is a main shadow of if:

  1. ,

  2. , and

  3. .

We say that is a secondary shadow of if:

  1. ,

  2. , and

  3. .

Notice that we get equivalent definitions by changing condition iii) for

  • .

In the case when , we will call a main (or secondary) shadow of the robber. We say a vertex is a diagonal shadow of a vertex if is a secondary shadow of or a main shadow of . Again, if we will use the term diagonal shadow of the robber. The following result states that if a cop is in a diagonal shadow of the robber and the robber moves, there is a move that the cop can make that keeps him in a diagonal shadow of the robber. Notice that if the type of the vertex the robber is in changes when he moves, the diagonal that the cop must be in will change from secondary to main, or vice versa.

Lemma 3.6.

Let be vertices such that and take .

  • If is a main shadow of , then is a diagonal shadow of and .

  • If is a secondary shadow of , then is a diagonal shadow of and .

The proof is left to the reader. Figure 2 is added for the guidance.

Figure 2: Proof of Lemma 3.6.

The next lemma will give us another way of restricting the robber’s moves in the case of a -regularly oriented grid.

Lemma 3.7.

Let be vertices such that , is a diagonal shadow of , , and is a diagonal shadow of , is the intersection of the row of and the column of , and is the intersection of the row of and the column of . If is orthogonal to , then there exists an integer such that .

Proof.

Let and . If , then , and therefore and . If , then . In this case, we can see that . Since , we get that there is an integer such that . ∎

Figure 3: (a) The robber (white vertex ) is not able to cross the mirror line (dashed line) or he will be caught by the cop (black vertex ). (b) One of the possible cases for Lemma 3.7.

The geometric interpretation of Lemma 3.7 is key (see Figure 3b): If a cop moves in such a way that he remains in a diagonal shadow of the robber, then there exists a set of vertices that touches every line in (Observation 2.4) that the robber cannot step on (and therefore cannot cross it) without being captured. This set corresponds to the vertices in the orthogonal bisector of the “line segment” from to shown as the dotted line in the figure. We will refer to this line as the mirror of the corresponding shadow of the robber. Two mirrors and are parallel if the two types of their vertices are the same.

Recall that we are working on a -regularly oriented grid. Given two parallel mirror lines and , and a type orthogonal to the type of a vertex in , the distance between and , denoted by , is the minimum positive integer such that or . Notice that for any two different parallel mirrors, and .

Let be a vertex of , and a type orthogonal to . We define

Because of Observation 2.4, we know that every vertex is in for some . Given two vertices of opposite types, and , the diagonal distance, which we will denote by , between and is the minimum such that (or equivalently, ). Notice that if and only if . We define

where has been chosen so that . Notice that if and are confluxes of such that , then and are defined in .

Observation 3.8.

If and are maximal confluxes of a -regularly oriented grid , and , then we have and .

For the rest of the section, all the confluxes will be assumed to be maximal. For the Lemmas 3.9, 3.10, 3.11 and 3.12, and will denote confluxes, whose main corners are covered by cops.

Lemma 3.9.

Suppose and that the robber is in a conflux in whose type is orthogonal to . If the robber enters , then the cops can capture a main shadow of the robber.

Proof.

If the robber is in , in order to enter the robber must enter a conflux . In either case, a main shadow of the robber will enter . Since the main corners of both and are covered by cops, we capture a main shadow of the robber by applying Lemma 2.2 to the corresponding shadow. ∎

Lemma 3.10.

Let be a type orthogonal to , and suppose that satisfies and . If the robber is in , then we can force him to move to a conflux outside of .

Proof.

We can assume the robber is at a vertex of type or . If the robber is in , then by forcing him to move he must enter a conflux whose type is parallel to . If , then , and so by forcing him to move he will exit . If , then we have , so he will exit if we force him to move. ∎

Lemma 3.11.

Let and be the same as in the hypothesis of Lemma 3.10 and take such that and in for some . If the robber is in , then we can force him to move to a vertex out of or we capture his main shadow.

Proof.

Again, we can assume that the robber is in a vertex of type or . By forcing him to move he will enter a conflux of type parallel to . If , the robber is forced to move to . Otherwise, an Lemma 3.9 guarantees the capture of a main shadow of the robber. ∎

Lemma 3.12.

Let be such that . If the robber is not in , then we can capture his main shadow.

Proof.

This follows directly from the fact that by forcing the robber to move, he must enter a conflux whose type is parallel to , all of which are contained in . ∎

All the previous results in this section either assume that we already captured a diagonal shadow of the robber or include some assumption about the current position of the robber in their statements. The following is the first result that makes no such assumptions.

Lemma 3.13.

Seven cops can capture a diagonal shadow of the robber in .

Proof.

First, one of the cops will be chasing the robber in order to force him to move, so we only need to show that six cops can capture the robber if he is forced to move. Let and be confluxes such that and . We move 4 cops to their main corners. By Lemma 3.10 we can make sure that the robber is in (i.e. out of ). By applying Lemma 3.9 we can see that, once the robber is in , he cannot enter without having his shadow captured. We will show that we can either capture a diagonal shadow of the robber or force the conditions of Lemma 3.12.

Let be a conflux such that and cover its main corners with two cops. Notice that Lemma 3.11 guarantees that either we capture a diagonal shadow of the robber (in which case we are done) or that the robber is in . In the latter case, the cops in can be released and we can rename as . This can be done until the diagonal shadow of the robber is caught or . In the later situation, since the robber is , by applying Lemma 3.12 we can capture the robber’s diagonal shadow.

The basic idea of the proof of Theorem 3.15 is to successively capture diagonal shadows of the robber such that their mirrors get closer until the distance between them is two, and then use the remaining cops to capture the robber between those mirrors. However, it is clear that in order to effectively restrict the robber’s movements we need two different mirrors. However, there is no way to guarantee that if we have a cop in a diagonal shadow of the robber and we use Lemma 3.13 again we won’t capture the shadow where we already have a cop. A simple way around this problem is to use Lemma 3.13 twice at the same time. This is what the following result deals with.

Lemma 3.14.

Thirteen cops can capture two main diagonal shadows of the robber simultaneously. Moreover, we can actually guarantee that the distance between the mirrors of the diagonal shadows is two.

Proof.

Like in the proof of Lemma 3.13, a cop will force the robber to move, so we only need to show that twelve cops can achieve the desired result if the robber is forced to move at such step. For each cop used in the strategy of Lemma 3.13, we will use one more cop in the following way: If , we will choose , where . Notice that , so we can move in such a way that he stays in the shadow of . In this way, by using the strategy of Lemma 3.13 with the first set of six cops and maintaining the copies of the cops in their shadows, we will capture two main shadows of the robber simultaneously. Let and be the mirrors of these shadows, and and the cops moving in these shadows.

If we are done, so we can assume it is greater than two. Suppose the type of the confluxes that we used in the application of Lemma 3.13 is . Notice that whenever , there exist maximal confluxes and whose types are parallel to such that and are mutually disjoint.

Since we are using one cop to guard each mirror, we have ten free cops. By using six of those ten cops to repeat the strategy moving positioning the cops in confluxes in of type we will capture a new diagonal shadow. If is the mirror corresponding to this new shadow and is the cop guarding it, notice that and the robber is either between and or between and , so we can release either or . Since the robber’s movements are restricted to a strictly smaller set, induction over the distance between the mirrors gives us that the distance between mirrors is two. ∎

With this we are ready to prove the main theorem of this section:

Theorem 3.15.

For every , if is a -regularly oriented grid, then .

Proof.

Since we have cops, we can use one to chase the robber and force him to move. That means we have twelve free cops. By Lemma 3.14, we can assume 10 of those cops are free and that the robber is restricted to the vertices between two mirrors at distance two. If we manage to capture a main diagonal shadow of the robber between the mirrors whose type is parallel to the mirrors, then we capture the robber.

Let and be confluxes such that , , and and and guard the main diagonals of each of these four confluxes with two cops. We can assume the robber is in a vertex of type orthogonal to . Notice that . An application of Lemma 3.12 to and guarantees that the robber is in . By Lemma 3.10 applied to and , and to and , we can guarantee that the robber is in .

Let . The proof will be by induction on . If , Lemma 3.12 guarantees the capture of the robber, so we can assume . Suppose . Since the robber is in , we can release the cops in and and move them to the main corners of the confluxes and . Again, an application of Lemma 3.10 with and guarantees that the robber is in , and using Lemma 3.12 with and gives that the robber is in . This now gives us that the robber is in , so if we rename as we get that , so the result follows by induction. ∎

It is important to mention that the only part of the proof where we use cops is during the application of Lemma 3.14. The rest of the proof only uses cops, so finding a more efficient way of capturing two diagonal shadows simultaneously would improve the bound for the cop number of .

4 Paddles

Now we return to the general case of straight-ahead oriented toridal grid . We begin by establishing more general conditions which guarantee that the robber is confined to a stream, i.e., the robber will be forced to stay in the subgraph of the stream as he will be caught if trying to leave it. As before, we assume that the robber is forced to move.

Lemma 4.16.

Let be a stream and let and be the lines which form the boundary of . Suppose that the robber is in , at distance from and at distance from . Let be the closest vertices (using distances in the undirected grid) of to , respectively. Suppose further that there are distinct cops and , such that can move to in moves and can move to in moves. Then by using and we can ensure that the robber will be caught or confined to .

Proof.

If or then the robber is already caught. Otherwise, whatever the robber’s move, we update , , and accordingly. Now for each , if is at then he remains in place; otherwise, he moves towards . We will show that the conditions of the lemma are maintained. If the robber moved in the direction of the stream, then and each move in the direction of the stream and and are unchanged. In this case the robber’s move increases and by at most , and the cops’ moves immediately decrease and by , to a minimum of ; thus and . If the robber moved towards , then decreases by and increases by . The cops’ moves now decrease and by , to a minimum of 0, and again and . The case in which the robber moved towards is similar. ∎

Given a conflux , we refer to the secondary corner with no outneighbours in as the terminal corner of . Let be the main corner of with a vertical (respectively, horizontal) edge leaving (but no other edge leaving , unless has only one vertex); then we refer to the vertical (respectively, horizontal) outneighbour of as the vertical guard post (respectively, horizontal guard post) of . If is maximal, then we refer to the vertex outside with the same outneighbours as the terminal corner as the terminal guard post of .

Lemma 4.17.

Let be a vertical stream, be a horizontal stream and let be the conflux . Suppose that the robber is in , for each , let and be the distance from to the boundary of in the direction of and in the opposite direction, respectively. Suppose that there are distinct cops , and such that can reach the vertical guard post of in moves, can reach the horizontal guard post of in at most moves, and can reach either the terminal corner or, if is maximal, the terminal guard post of in moves. Then by committing , and we can ensure that the robber will be caught or confined to either or .

Proof.

If the robber does not leave on his move, then with the cops’ moves we will decrease , and by , to a minimum of zero; then it is clear that the conditions of the lemma still hold. Since the robber can make only finitely many such moves, we may assume that the robber leaves on his move. Suppose without loss of generality that he leaves and remains in . In this case, before the robber’s move we must have had , and hence and . Let and be the boundary lines of , such that there is a directed path in from to . Let and be the closest vertices of and , respectively, to . Now observe that can move to in at most moves (by first moving to the terminal corner or terminal guard post of ), while can move to in at most moves (since is the vertical guard post of ). We move each cop one step along the appropriate directed path; now Lemma 4.16 implies that we can ensure the robber will be caught or confined to . ∎

Our strategy to catch the robber will be based on blocking streams of maximum width and thus confining the robber to be between two vertical or horizontal streams. The basic tool is the following. Let us denote by the largest width of a stream in . To avoid technicalities in the forthcoming proofs, we will assume that . This assumption is justified by the covering Lemma 1.1.

Lemma 4.18.

Let be a stream of maximum width, and let . Let be formed by the lines of that are at distance at least from the boundary lines of . Then by committing cops, and temporarily using additional cops, we can ensure that after a finite number of steps, if the robber enters then he will be caught or confined to a stream.

Proof.

Without loss of generality we may assume that is a vertical stream, and that the edges of its lines are all directed upwards. Let and be the boundary lines of , and let and be the lines outside adjacent to and , respectively. We use two formations of cops, which we call paddles: an inner paddle (respectively, outer paddle) is a formation of 32 cops evenly spaced at distance along each of and (respectively, and ), where each horizontal line has either two or zero cops in each paddle. By our assumption that , every paddle has a row containing two cops such that rows below it contain no cops of the paddle. Let us consider the union of this row together with rows above it that contain cops of our paddle. The union of these rows is the domain of the paddle.

Claim 4.19.

If the robber enters the intersection of the domain of a paddle with along a horizontal edge, then he will be caught or confined to a stream.

To prove the claim, we first observe that there is a pair of cops at the same row or below at a vertical distance of at most . If the paddle is an inner paddle, Lemma 4.16 implies that the robber will be caught or confined to . Hence we may assume that the paddle is an outer paddle. If , then and . Then the robber must have been at the same vertex as a cop on the previous move, which is a contradiction; hence, . Let be the maximal conflux containing the robber; then has height at most . Suppose without loss of generality that the horizontal edges in are directed from to . Then the robber is at distance from and at distance from . There is a cop on above the top row of , at a distance of at most ; this cop can reach the terminal guard post of in at most moves. Further, there is a pair of cops above or level with at a vertical distance at most . Let be the cop in this pair which is on , and let be the cop in this pair which is on . We let if ; otherwise, we let be the closest cop above on . Let and be the distances from to the top and bottom rows of , respectively. Then can reach the vertical guard post of in at most moves, while can reach the horizontal guard post of in at most moves, where the inequality follows from the definition of . Now , and each make their first moves along their respective paths, and the claim follows by Lemma 4.17. Note that , and are cops in the paddle that might be above the domain of the paddle. This is the reason that we use 30 instead of 26 cops in the paddles.

Claim 4.20.

The cops forming an inner paddle can reform to form an outer paddle with the same domain in at most moves, and vice versa.

To prove the claim, we observe that for any vertex of there is a directed path of length at most to the horizontally adjacent vertex of : move up times until there is an edge to , move to , and then move down times. Similarly there is a directed path of length at most from any vertex of , or to the horizontally adjacent vertex of , or respectively; the claim now follows immediately.

When we will use Claim 4.20 and have a paddle change from inner to outer, or vice versa, we will term this as reforming of the paddle. Note that the domain of the paddle does not change during the reforming process. Now suppose we have formed the cops into two paddles, each consisting of cops. Observe that the domains and of these paddles each contain rows. Since having a larger domain only helps us, we may assume that and contain rows each.

We first show that once we have set up the appropriate circumstances, we can make sure that the robber remains within either or indefinitely. To achieve this we define five states, and show that regardless of the robber’s move we can either stay in the same state or go to the next state (where State 5 is followed by State 1). We say that a domain is moving up (respectively moving down) if the corresponding paddle is an inner (respectively outer) paddle. We say that a domain has steps to start moving up or down if it is in the process of reforming and will complete this process in moves, and that it is active otherwise. We say that we switch the domain when we reform the corresponding paddle to move in the opposite direction. Note that the States 1-5 assume that our domain is moving up. The classification of states will be equally true if we reverse the vertical directions or relabel the paddles, so that we may do that at any time.

State 1: is moving up, is moving down, and occupy the same rows and the robber is on one of these rows.

In this state the robber can only force us out of State 1 by leaving the occupied rows. If he does so and moves above the top row of , then we move up, switch and enter State 2. If he leaves the rows of and below, then we move down, switch and enter State 2 (with the role of and exchanged). Otherwise we remain in State 1.

State 2: is moving up, is reforming and has steps to start moving up, is rows above and the robber is rows below the top row of , where .

When we come from State 1 to State 2 we have , and . Later in this state if the robber moves above the top row of then we move up; then increases by and . Otherwise we keep stationary; then increases by at most and remains the same. Hence increases by at most ; since decreases by , the inequality still holds and we stay in State 2 until , when we enter State 3.

State 3: and are both moving up, is rows above and the robber is rows below the top row of , where .

In this state if the robber moves above the top row of then we move both and up; then remains the same and . Otherwise we move only up; then decreases by while increases by at most . So the inequality still holds and we stay in State 3 unless , when we enter State 4.

State 4: and are both moving up and occupy the same rows, and the robber is rows below the top row of .

If and the robber moves down then we switch and enter State 5. Otherwise we move and only if the robber goes above their top row, and remain in State 4.

State 5: is moving up, is reforming and has steps to start moving down, and occupy the same rows and the robber is rows below the top row of , where and .

In this case we keep both and stationary and stay in State 5 until , when we enter State 1.

We next show that we can reach one of these states. We form our cops into four paddles of cops each, with domains , , and . Initially all of these domains occupy the same rows. Then we begin with and moving up and and moving down. At some step, the robber will occupy either the top row of and or the bottom row of and . Without loss of generality he occupies the top row of and . At this point we enter State 3, and release the cops from and .

Now if at any point the robber is outside , Claim 4.19 implies that he cannot re-enter without being caught or confined to a stream. To force the robber to leave , we choose an arbitrary row and place two cops at either end of the intersection of this row with . One cop on each end remains stationary, while the remaining two cops move in the direction of . If the robber does not leave , then after some finite time he will be on the same row as one of the pairs of cops, at which point he is confined to a stream. ∎

In the sequel we will use Lemma 4.18 on streams that may not be maximum width. The only feature where we needed to use that has large width was that any crossing stream in the domain of the paddle is not wider than . In order to have this property, some confluxes of the considered stream will be guarded (by involving 3 cops for each such conflux by using Lemma 4.17).

Theorem 4.21.

If is any straight-ahead orientation of a toroidal grid, then .

Proof.

For each stream mentioned below, is formed by the lines of at distance at least from the boundary lines of (or all the lines of , if ), where . For any subgraph of , we define to be a stream of maximum width intersecting .

Throughout the proof, let be the territory of the robber, i.e., vertices of the graph that the robber can reach without getting caught. Also let be where is the union of guarded streams.

Let and be three widest streams in . We will have two cases.

Case 1. and are (without loss of generality) vertical streams. Then Lemma 4.18 shows that we can commit cops (and temporarily using another 60 cops) to guard and . The robber is confined between two of these streams, say and . Now we can release the cops used to guard . Now, redefine to be equal to (i.e., the widest stream intersecting the territory of the robber). We can continue this approach to shrink the territory of the robber until is a horizontal stream, which brings us to Case 2.

Case 2. and are vertical streams but is a horizontal one. Then we can commit cops to guard and . Let and be the intersection of and with , respectively. Using Lemma 4.17, we can commit cops to guard and and then using 60 cops we can guard . Note that there is no stream wider than that intersects in . Also note that since the horizontal movement of the robber is bounded between and , we do not need (even temporarily) an extra 60 cops to guard . Now, let be the next widest stream intersecting .

Subcase 2a. If is a vertical stream, then use another 3 cops to guard and use 60 cops to guard . Now, based on the position of the robber, redefine and , release the third set of 60 cops, shrink the territory of the robber and repeat Subcase 2a.

Subcase 2b. If is a horizontal stream, then commit cops to protect and and use 60 cops to guard . Now the robber is confined between and .

Let be the next widest stream intersecting . Without loss of generality we can assume that it is a vertical one. Commit cops to protect and and use 60 cops to guard . Now, based on the position of the robber, redefine and , release the fifth set of 60 cops, shrink the territory of the robber and repeat Subcase 2b until we catch the robber.

Note that we have used at most 5 sets of 60 cops to guard () and at most cops to guard the intersections of these streams. Also, to avoid complication, we use one cop to force the robber to move. Therefore by using at most 319 cops we can capture the robber. ∎

References

  • (1) Anthony Bonato and Richard J. Nowakowski. The game of cops and robbers on graphs. volume 61 of Student Mathematical Library. American Mathematical Society, Providence, RI, 2011.
  • (2) Seyyed Aliasghar Hosseini and Bojan Mohar. Game of cops and robbers in oriented quotients of the integer grid. Discrete Mathematics, 341(2):439–450, 2018.
  • (3) Richard Nowakowski and Peter Winkler. Vertex-to-vertex pursuit in a graph. Discrete Mathematics, 43(2):235–239, 1983.
  • (4) A Quilliot. Jeux et pointes fixes sur les graphes. Thèse de 3ème cycle, Université de Paris, VI:131–145, 1978.