Cops and Robbers on Intersection Graphs^{1}^{1}1The conference versions of parts of this paper appeared in ISAAC 2013 Gavenčiak et al. (2013) and ISAAC 2015 Gavenčiak et al. (2015). For a structural dynamical diagram of the results of this paper, see http://pavel.klavik.cz/orgpad/cops_on_intersection_graphs.html (supported for Firefox and Google Chrome). The third, the fourth, and the fifth authors are supported by CEITI (P202/12/G061 of GAČR), the first, the fourth and the fifth authors are supported by Charles University as GAUK 196213.
Abstract
The cop number of a graph is the smallest such that cops win the game of cops and robber on . We investigate the maximum cop number of geometric intersection graphs, which are graphs whose vertices are represented by geometric shapes and edges by their intersections. We establish the following dichotomy for previously studied classes of intersection graphs:

The intersection graphs of arcconnected sets in the plane (called string graphs) have cop number at most 15, and more generally, the intersection graphs of arcconnected subsets of a surface have cop number at most in case of orientable surface of genus , and at most in case of nonorientable surface of Euler genus . For more restricted classes of intersection graphs, we obtain better bounds: the maximum cop number of interval filament graphs is two, and the maximum cop number of outerstring graphs is between 3 and 4.

The intersection graphs of disconnected 2dimensional sets or of 3dimensional sets have unbounded cop number even in very restricted settings. For instance, we show that the cop number is unbounded on intersection graphs of twoelement subsets of a line, as well as on intersection graphs of 3dimensional unit balls, of 3dimensional unit cubes or of 3dimensional axisaligned unit segments.
keywords:
intersection graphs, string graphs, graphs on surfaces, interval filament graphs, cop and robber, pursuit games, games on graphsproofProof \newproofclaim_proofProof of Claim
1 Introduction
The game of cops and robber on graphs has been introduced independently by Quilliot Quilliot (1978, 1983) and by Winkler and Nowakowski Nowakowski and Winkler (1983). In this paper, we investigate the game on geometric intersection graphs.
Rules of the Game. The first player, called the cops, places cops on vertices of a graph . Then the second player, called the robber, places the robber on a vertex. Then the players alternate. In the cops’ move, every cop either stays in its vertex, or moves to one of its neighbors. More cops may occupy the same vertex. In the robber’s move, the robber either stays in its vertex, or moves to a neighboring vertex. The game ends when the robber is captured which happens when a cop occupies the same vertex as the robber. The cops win if they are able to capture the robber. The robber wins if he is able to escape indefinitely.
Maximum Cop Number. For a graph , its cop number is the least number such that cops have a winning strategy on . For a class of graphs , the maximum cop number is the maximum cop number of a connected graph , possibly . The restriction to connected graphs is standard: if has connected components , then . Therefore, a graph class closed under disjoint union cannot have a bounded maximum cop number if we omit this restriction. Throughout the paper, we only work with connected graphs.
Known Results. Graphs of the cop number one were characterized already by Quilliot Quilliot (1983) and by Nowakowski and Winkler Nowakowski and Winkler (1983). These are the graphs whose vertices can be linearly ordered so that each for is a corner of , i.e., has a neighbor for some such that is adjacent to all other neighbors of . Andreae Andreae (1984) proved that regular graphs have the maximum cop number equal for all .
For part of the input, deciding whether the cop number of a graph is at most has been shown to be NPhard Fomin et al. (2010), PSPACEhard Mamino (2013) and very recently EXPTIMEcomplete Kinnersley (2015), confirming a 20 years old conjecture of Goldstein and Reingold Goldstein and Reingold (1995). In order to test whether cops suffice to capture the robber on an vertex graph, we can search the game graph which has vertices to find a winning strategy for cops. In particular, if is a fixed constant, this algorithm runs in polynomial time.
For general graphs on vertices, it is known that at least cops may be needed (e.g., for the incidence graph of a finite projective plane Prałat (2010)). Meyniel conjecture states that the cop number of a connected vertex graph is . For more details and results, see the book Bonato and Nowakowski (2011).
Geometrically Represented Graphs. We want to argue that the geometry of a graph class heavily influences the maximum cop number. For instance, the classical result of Aigner and Fromme Aigner and Fromme (1984) shows that the maximum cop number of planar graphs is 3. This result was generalized to graphs of bounded genus by Quilliot Quilliot (1985) and improved by Schroeder Schroeder (2001):
(1) 
where is the (orientable) genus of . For nonorientable surfaces, a similar result was obtained by Clarke et al. Clarke et al. (2014):
(2) 
where is the Euler genus of (also called the crosscap number of the surface is drawn on). However, the exact value of the maximum cop number is not known already for toroidal graphs ().
We study intersection representations in which a graph is represented by a map for some ground set such that the edges of are described by the intersections: . The ground set and the images of are usually somehow restricted to get particular classes of intersection graphs. For example, the wellknown interval graphs have and every is a closed interval.
All these graph classes admit large cliques, so their genus is unbounded and the bound (1) of the maximum cop number does not apply. On the other hand, existence of large cliques does not imply big maximum cop number since only one cop is enough to guard a maximal clique. For instance, chordal graphs, which are intersection graphs of subtrees of a tree, may have arbitrary large cliques but their maximum cop number is 1.
String Graphs. The class of string graphs (STRING) is the class of intersection graphs of strings: and every is a bounded curve, i.e., a continuous image of the interval in . It is known that every intersection graph of arcconnected sets in the plane is a string graph. For instance, boxicity graphs (BOX), which are intersection graphs of dimensional intervals in , are string graphs when .
The class of outerstring graphs (OUTERSTRING) consists of all string graphs having string representations with each string in the upper halfplane, intersecting the axis in exactly one point, which is an endpoint of this string.
The class of interval filament graphs (INTERVAL FILAMENT), introduced by Gavril Gavril (2000), consists of intersection graphs of interval filaments, where an interval filament defined on an interval is the graph of a continuous function such that and for all . In our description, we identify the filament with the function , i.e., we use instead of . It holds that . (The first inequality is strict from Theorem 1.1(i) and (ii), the second follows from Rok and Walczak (2014).)
Let be an arbitrary orientable surface of genus . We consider a generalization of string graphs for and every is a bounded curve in , and we denote this class by GENUS STRING. It can be seen that any intersection graph of arcconnected subsets of belongs to GENUS STRING. It is known that every graph embeddable to a surface of genus can be represented by a contact representation of disks on a suitable Riemann surface of genus ; so . (It is strict since arbitrarily large complete graphs belong to GENUS STRING, but not to GENUS .) Similarly. let EULERGENUS STRING denote intersection graphs of bounded curves on a (possibly nonorientable) surface of Euler genus (equal to the crosscap number of the surface).
Note that while we could work only with Euler genus for both orientable and nonorientable surfaces, the bounds obtained for orientable genus are significantly lower that for Euler genus (since is the best possible general bound).
Intersection Graphs of Disconnected and Higher Dimensional Sets. As stated, all intersection classes of graphs of arcconnected sets in the plane are subclasses of string graphs. Other classes of intersection graphs are obtained either for disconnected sets, or for sets in dimensions higher than two.
For a graph , it linegraph, denoted by , is the intersection graph of the edges of . Let LINE denote the class of all linegraphs. Observe that each linegraph can be represented as an intersection graph of twoelement subsets of a line. Thus, linegraphs provide a simple example of intersection graphs of disconnected sets. As shown by Dudek et al. Dudek et al. (2014), the cop number of is related to the cop number of via the inequalities
In particular, the cop number of linegraphs is unbounded.
Many other geometric intersection classes can be seen as generalizations of linegraphs. Among the most studied are the interval graphs (INTERVAL) where and every is a union of closed intervals. And for its subclass unit interval graphs (UNIT INTERVAL), all intervals of have the length one. It follows that the cop number is unbounded on all these classes.
For sets in higher dimensions, notice that every graph has a representation by 3dimensional strings. Therefore, to get interesting classes of graphs, we have to further restrict geometry of the sets. Aside already described BOX, we consider the following classes for : the intersection graphs of axis parallel segments GRID, the intersection graphs of dimensional unit cubes UNIT CUBE, the intersection graphs of dimensional balls BALL, and the intersection graphs of dimensional unit balls UNIT BALL.
Our Results. It has been asked at several occasions, last during the Banff Workshop on Graph Searching in October 2012, whether intersectiondefined graph classes (other than interval graphs) have bounded maximum cop numbers. The classes in question have included circle graphs, intersection graphs of disks in the plane, graphs of boxicity 2, and others. A recent paper Beveridge et al. (2012) shows that the maximum cop number of intersection graphs of unit disks (UNIT BALL) is between 3 and 9. We solve this question in a general way by proving a dichotomy for previously studied classes of geometric intersection graphs in Theorems 1.1 and 1.3. For an overview of the results presented in this paper, see Fig. 1.
Theorem 1.1
The following bounds for the maximum cop number hold:

.

.

.

.

.
We note that the strategies of cops in all upper bounds are geometric and their description is constructive, using an intersection representation of . If only the graph is given, we cannot generally construct these representations efficiently since recognition is NPcomplete for string graphs Kratochvíl (1991) and interval filament graphs Pergel (2007), and open for the other classes. Nevertheless, since the state space of the game has states and the number of cops is bounded by a constant, we can use the standard exhaustive game space searching algorithm to obtain the following:
Corollary 1.2
There are polynomialtime algorithms computing the cop number and an optimal strategy for the cops for any interval filament graph in time , outerstring graph in time , string graph in time and a string graph on a surface of a fixed genus (resp. Euler genus ) in time (resp. ), even when representations are not given.∎
Furthermore, our results can be used as a polynomialtime heuristic to prove that a given graph is not, say, a string graph, by showing that . For instance, a graph of girth 5 and the minimum degree at least 16 is not a string graph since : in any position of 15 cops with the robber on , at least one neighbor of is nonadjacent to the cops.
On the other hand, when sets are not arcconnected, we prove that even in very restricted geometric settings that cop numbers are unbounded. The main lemma states that when we subdivide all edges of a graph by a same number of vertices, increases by at most one. Since all these classes contain certain subdivisions of all graphs or all cubic graphs, we get the following:
Theorem 1.3
The classes LINE, INTERVAL, UNIT INTERVAL, GRID, BOX, UNIT CUBE, BALL, and UNIT BALL have the maximum cop number equal .
Outline. In Section 2, we show that is 2. In Section 3, we show that is between 3 and 4. Aigner and Fromme Aigner and Fromme (1984) show the classical result that one cop can guard a shortest path in any graph. In Section 4, we extend this result to show that five cops can guard a shortest path together with its neighborhood. This is used in Section 5 to show that is at most 15. In Section 6, we combine the previous result with the approach of Quilliot Quilliot (1985) to simultaneously show the bounds for boundedgenus orientable surfaces and boundedEulergenus nonorientable surfaces. In Section 7, we prove that cop numbers are unbounded for intersection graphs of disconnected or 3dimensional sets.
Preliminaries. Let be a graph. For , we let denote the subgraph of induced by , and . For a vertex , we use the open neighborhood and the closed neighborhood . Similarly for , we put and .
Let be a string representation of . Without loss of generality, we may assume that there is only a finite number of string intersections in the representation, that strings never only touch without either also crossing each other, or at least one of them ending, that no three or more strings meet at the same point, and that no string selfintersects. This follows from the fact that strings can be replaced by piecewise linear curves with finite numbers of linear segments without affecting their intersection graph. For more details see Kratochvíl et al. (1986). We always assume and maintain these properties.
Suppose that we have some strategy for the cops. For a vertex , the robber cannot safely move to if the strategy ensures that he is immediately captured after moving to . Let . We say that the strategy guards if it ensures that the robber cannot safely move to any vertex in . We say that the robber is confined to , if the strategy ensures that the robber is immediately captured by moving to any vertex in . Notice that the robber is confined to if and only if he stands in and is guarded.
2 Capturing Robber in Interval Filament Graphs
In this section, we show that the maximum cop number of interval filament graphs is equal to two, thus establishing Theorem 1.1(i).
If a filament is defined on , we call the left endpoint and the right endpoint of . We assume that the filaments have pairwise distinct endpoints and the defining intervals are always nontrivial. In the description, we move the cops on the representation , and we say that a cop takes a filament if it is placed on the vertex which this filament represents. We shall assume that the robber never moves into the neighborhood of a vertex taken by a cop, and a cop capture the robber immediately if he stands on a neighboring vertex.
Filaments and Regions. It is important that each filament splits the halfplane into two regions: the unbounded top region and the bottom region. A filament is nested in a filament if is contained in the bottom region of . We say that the robber is/stays in a region if he is/stays on filaments entirely contained in this region. The robber is confined by if a cop is placed in and the robber is in the bottom region of .
Lemma 2.1
Suppose that the robber is confined by . Then he stays in the bottom region of as long as there is a cop on .
To move from one region to another, the robber has to use a filament which crosses . But then is a neighbor of , and the cop captures the robber in the next turn.∎
A filament is called top in if it maximizes the value over all filaments defined for . Suppose that is the leftmost and is the rightmost endpoint of the representation. We have a sequence of top filaments as we traverse from to . We note that one filament can appear several times in this sequence. See Fig. 2 for an example.
Let be top in . Each filament together with the upward ray starting at separates the halfplane into three regions: the left region, the bottom region and the right region. The key property is that there is no filament intersecting the left and right regions and avoiding . Also note that the division of filaments into the regions is the same for all in the same top part of .
Lemma 2.2
Suppose that a cop stands on and the robber is in the right region of . If the cop moves to a neighboring filament with , the robber cannot move to the left region of .
Let be on and on and we have . Suppose that the cop moves from to and the robber stands on a filament defined on . We know that , so does not intersect the left region of . And since is top, there is no path going to the left region which avoids . So the robber cannot move there.∎
Proof of Theorem 1.1(i). We are ready to prove that the maximum cop number of interval filament graphs is equal to two.
[Theorem 1.1(i)] Since interval filament graphs contain all cycles , one cop has no winning strategy. (Note that is a circle, circulararc and function graph.) Therefore two cops are necessary.
We describe a strategy how to capture a robber with two cops. We call one cop the guard, and the other one the hunter. The strategy proceeds in phases. Every phase starts with both cops on a filament such that the robber is confined by it. The guard stays on till the robber is either captured, or confined by the hunter in some filament nested in . By Lemma 2.1, the robber can only move in the bottom region of . If the confinement by happens, then the guard moves to the filament taken by the hunter, ending the phase. In the next phase the hunter proceeds with capture the robber inside the bottom region of .
For the initial phase, we imagine that the guard takes some imaginary filament above all filaments of so the robber is confined to its bottom region, i.e., to the entire graph . We can choose both cops to start the first phase at a filament with leftmost left endpoint and therefore top in .
Suppose that we are in a phase where the guard is placed on . Let be the subgraph of induced by the vertices whose filaments are nested in , and let be the connected component of containing the vertex occupied by the robber. Since the guard stays at till the robber is confined in some nested , the strategy ensures that the robber must remain in throughout the phase, because any vertex in is adjacent to the vertex guarded by the guard.
Let be the sequence of top intervals in the restriction of to the vertices of . The hunter first goes to . When he arrives to , the robber cannot be in the left region of since there is no filament of contained there. Now suppose that the hunter is in and assume the induction hypothesis that the robber is not in the left region of . If the robber is confined in , the phase ends with the guard moving towards . If the robber is in the right region of , the hunter moves to the neighbor with maximal index . By Lemma 2.2 the robber cannot move to the left region of so he is either in the bottom, or the right region. The robber cannot stay in the right regions forever since has no filament of contained in the right region, so eventually the robber is confined in or captured directly.
Since there are only finitely many filaments nested in each other, the strategy proceeds in finitely many phases and the robber is captured.∎
With a small modification, we can prove that this strategy captures the robber in turns. Suppose that initially both cops are placed in the filament with the leftmost endpoint and there are phases. Let be the graph the robber is confined to by the guard on in the phase , so and let . Let , and note that contains all top filaments of .
During the th phase the hunter moves to any top filament of in at most 2 moves (note that there must be a filament in which simultaneously intersects and a top filament of ), then to the leftmost top filament of in at most moves using a shortest path in , then takes at most steps over the top filaments of . Finally, when the hunter confines the robber in , it takes the guard at most steps to get to by a similar argument. Since and the number of phases is also bounded by , we have used turns.
3 Capturing Robber in OuterString Graphs
In this section, we prove that the maximum cop number of outerstring graphs is between 3 and 4, thus establishing Theorem 1.1(ii). Our strategy is similar to the one described in Section 2.
String Pairs and Regions. For a given outerstring representation of , let be the ordering of the vertices of by the coordinates of the unique intersection of with the axis. We say that is on the left of and is on the right of if .
Every pair of intersecting outerstrings divides the halfplane into at least two regions: the unbounded top region, the bottom region incident with an interval of the axis, and possibly several middle regions. The middle regions do not play any role in our strategy since no string is entirely contained in them. The strings entirely contained in the bottom region are surrounded by and , a robber on a vertex surrounded by and , each occupied by a cop, is confined by and .
The following lemma can be proved the same way as Lemma 2.1:
Lemma 3.1
Suppose that the robber is confined by and . Then he stays in the bottom region of as long as there are cops on and .∎
The strings partition the upper halfplane into several regions, of which exactly one is unbounded. We say that a string is external, if it has at least one point on the boundary of the unbounded region. We have a sequence of external strings sorted by their appearance on the boundary of the unbounded region, each external string may appear several times in the sequence. See Fig. 3 for an example.
Lemma 3.2
Suppose that a cop is placed in an external string and the robber is in some nonintersecting string on the right of . If the cop stays in , the robber cannot move to a string on the left of .
Observe that separates nonintersecting strings on the left of it from those on the right. Thus, to get to a string on the left of , the robber has to move to and the cop captures him.∎
Proof of Theorem 1.1(ii). We are ready to prove that the maximum cop number of outerstring graphs is equal to three or four.
[Theorem 1.1(ii)] Figure 4 shows a connected outerstring graph with the cop number 3. It remains to show that four cops are always sufficient.
Among the four cops, there are two guards and two hunters. The strategy is divided into phases. During each phase, the two guards stand on a pair of intersecting strings and confining the robber. For the initial phase, we imagine that the guards take some imaginary strings around the entire representation, so the robber is confined to their bottom region, i.e., to the entire graph . We can choose all cops to start the first phase in the leftmost string which is external.
In each phase, let and be the pair of adjacent strings occupied by the guards, confining the robber. Let be the subgraph induced by the strings entirely contained in the bottom region, and let be the connected component of containing the vertex with the robber. By Lemma 3.1, the robber is confined to . The hunters move in and either capture the robber, or make the robber confined to a smaller subgraph by taking two intersecting strings and such that the robber is placed in their bottom region. Then the guards move to and , and the next phase begins.
Let be the sequence of external strings in the restriction of to the vertices of . The hunters start by taking and its rightmost neighbor . Suppose that after several moves the hunters take intersecting external strings and . We want the strategy to preserve the following property: either the robber occupies a string between and , or he occupies a string on the right of . It is certainly satisfied in the beginning since there are no strings on the left of .
Suppose that the property holds when the hunters occupy and , where . The hunter taking stays there, while the other hunter moves from to the rightmost external neighbor of . By Lemma 3.2, the robber cannot move from the right of to the left of it. Therefore, the robber either appears between and , or he is on the right of . Since the sequence of external strings is finite, the hunters either capture the robber, or he is confined by some and , so the phase ends after finitely many steps.
Since there are only finitely pairs of intersecting strings nested in each other, the strategy proceeds in finitely many phases and the robber is captured.∎
Similarly as in Section 2, we can show that the strategy requires at most a linear number of moves.
4 Guarding Shortest Paths and Curves in String Graphs
In this section, we build a crucial tool for designing our strategy to capture the robber using 15 cops in any string graph. The main result shows that 5 cops are able to guard a shortest curve in a string representation together with the strings intersecting it.
Guarding Shortest Paths. We recall a classical lemma of Aigner and Fromme Aigner and Fromme (1984):
Lemma 4.1 (Aigner and Fromme (1984))
Let be a shortest path between a pair of vertices of a graph . Then a single cop has a strategy to guard , after a finite number of initial moves.
In Aigner and Fromme (1984), this result is essential to prove that the maximum cop number of planar graphs is three. The idea is that one can cut the planar graph by protecting several shortest paths. Consider a planar embedding. The strategy is to protect two shortest paths and from to such that the robber is confined to the subgraph between and . A third shortest path in is chosen and guarded by the third cop. The robber has to choose one of the smaller subgraphs of to which he is confined. It is shown that can be guarded by just two paths, so one of the cops can be freed and the strategy can be iterated.
Guarding Retracts. There is the following simple generalization of Lemma 4.1. It was stated in Berarducci and Intrigila (1993) (in a different form) and we believe that this statement should be more known. A retract from to an induced subgraph is a map such that for all , and for every either , or .
Lemma 4.2
Let be a retract of . Then cops have a strategy in to position one of them, in finite number of steps. After the positioning, the cop can guard while the remaining ones are free.
The strategy for cops plays on as if a robber standing on is placed on . By the definition of the retract, moves by the distance at most 1 with each turn of the game. Therefore, cops have a strategy to “capture” in finitely many turns. The cop standing on can then follow the robber, to be always on for the current robber’s position . If the robber steps on , he is immediately captured by the cop. Note that the remaining cops are no longer required.∎
This implies Lemma 4.1 since a shortest path is a retract and paths have the cop number equal to 1.
Guarding Neighborhoods of Shortest Paths. We want to apply a similar idea to string graphs. Unfortunately, guarding a shorting path is not sufficient to prevent the robber to move from one side of to the other one. We need a stronger tool to geometrically restrict the robber. We show that five cops are sufficient to guard which prevents to the robber to use any string crossing the protected path; see Fig. 5.
Before stating the lemma, we add another definition. Suppose that the robber is confined by the strategy to . We say that a path in is shortest relative to , if it is shortest in . The path does not have to be shortest in , it just have to be shortest with respect to the robber’s confinement to :
Lemma 4.3
Let be a shortest path relative to and let the robber be confined to . Then five cops have a strategy to guard , after a finite number of initial moves.
It follows from Lemma 4.1 applied to that one cop, called the sheriff, has a strategy to guard since the robber can only move in . The four additional cops, called the deputies, follow the sheriff and stand at neighboring vertices of . More precisely, suppose that the path consists of the vertices . When the sheriff stands at , the deputies stand at , , and (with the convention that and here refer to the vertex , and and refer to ). As the sheriff moves along the path according to the strategy, the deputies follow him. The initial setup procedure is analogous to the one in Lemma 4.1.
Assume that the robber moves to a vertex , and suppose that is adjacent to a vertex , which is adjacent to . Then the strategy necessarily moves the sheriff to one of the vertices , since otherwise the robber could step on in two moves without being immediately captured by the sheriff, contradicting the properties of the strategy. Therefore, after the cops’ move, there is at least one cop on , and so if the robber moves to , he is captured immediately.∎
To guard with the cops moving only on , five cops are necessary as shown in Fig. 5b. When we say that five cops start guarding a path, we do not explicitly mention the initial time required to position them onto the path and assume that the strategy waits for enough turns.
Unfortunately, the result of Lemma 4.3 cannot be straightforwardly extended to retracts. Even if the retract has bounded degree (so the number of deputies required to guard the vertices in distance at most 2 is bounded), it is not possible to move the deputies together with the sheriff in the required way.
Guarding Shortest Curves. Our strategy for string graphs is geometric, based on string representations. To simplify its description, we introduce the concept of shortest curves as particular curves through the string representation of some shortest path.
Let be a string graph together with a fixed string representation . Suppose that the robber is confined to and let be a shortest path relative to from to . Suppose that we choose and fix two points and . Let be a curve from to such that for every , the curve has a connected intersection with , and these intersections are ordered on in the same order as the vertices of . We call a shortest curve of , relative to with endpoints and . A curve is called a shortest curve relative to if it is a shortest curve of some shortest path relative to . We may omit when it is clear from the context.
The shortest path corresponding to a shortest curve is uniquely defined by the sequence of strings whose intersection with has nonzero length. By guarding a shortest curve , we mean guarding of the corresponding shortest path . The length of is the number of its strings; we note that its Euclidean length plays no role.
Corollary 4.4
Let the robber be confined to and let be a shortest curve relative to in a string representation. Then five cops can prevent the robber from entering any string intersecting , after a finite number of initial moves.
Let be the shortest path defining . By guarding , five cops prevent the robber from entering strings intersecting . See Fig.5a for illustration.∎
Observation 4.5
Any subcurve of a shortest curve relative to is a shortest curve relative to .∎
5 Capturing Robber in String Graphs
In this section, we show that the maximum cop number of string graphs is at most 15. Our strategy is inspired by the strategy for 3 cops in planar graphs Aigner and Fromme (1984). The key difference is that we use Lemma 4.3 instead of Lemma 4.1, so we require 5 cops for each shortest path instead of 1. Therefore, our strategy requires cops.
Segments, faces and regions. Consider a set of curves/strings in . The topological arcconnected components of are called faces and their topological closures are closed faces; every face is an open set. We assume that the number of intersections of is finite, so the number of faces is also finite.
A segment of a curve is a maximal arcconnected subset of not containing any intersection with another curve in . The number of segments is also finite. A region is a closed subset of obtained as a closure of a union of some of the faces.
Consider a string representation . For , we denote the topological closure of by , the topological interior by , and the boundary . We say that a vertex is contained in if . We denote the subgraph of induced by the vertices contained in by . Two curves sharing only their endpoints are said to to be internally disjoint.
Lemma 5.1
Let and be two internally disjoint curves with endpoints and , let be the closed face of containing the string with the robber, and let be the connected component of containing the robber. If and are shortest curves with respect to , each guarded by five cops, then the robber is confined to .
It follows from Corollary 4.4 applied to and .∎
Additionally, below we use the following topological lemma.
Lemma 5.2
Let and be two internally disjoint simple curves from to , where . Let be a closed face of . Let be a simple curve from to , going through at least one inner point of . Then every face of is bounded by two simple internally disjoint curves and , where for some and .
Without loss of generality, we may assume that is the inner face of , otherwise we can apply the circular inversion. We know that is an open arcconnected set by definition, so is a simple closed Jordan curve.
We first establish that for some . Observe that would imply that is not a simple curve. There is a point . The reason is that otherwise we would have , so , which contradicts that intersects .
We argue that it is not possible that there exist both and . If both would exists, there would be a curve from to separating from in , not intersecting . By Jordan curve theorem, this contradicts that is a curve from to through ; see Fig. 6a. However, we necessarily have one such , for . Without loss of generality, we assume that exists and no exists, so .
Let . Let be the first point of going along from to and last such point. If was not connected, let be an endpoint of one segment of ; necessarily, since . However, it is not possible since would have to contain a point of , as shown in Fig. 6b, contradicting the definitions of and . Therefore is connected and we can take , getting a connected curve .∎
Restricted Graphs and Strategies. Given a closed region , let restricted to , denoted , be the intersection graph of the strings of . This restriction may remove vertices (represented by the strings outside ), may remove edges (intersections outside ) and may split vertices whose strings leave and reenter at least once; every arcconnected part of forms a new vertex . The newly obtained vertices are called the splits of . The graph is again a string graph with its representation, denoted by , directly derived from . Note that this operation preserves the faces and strings in and all representation properties assumed above, namely the vertex set of is finite. Also, the number of segments does not increase.
Lemma 5.3
Let be a region. If is a shortest curve relative to , and is a subcurve of , then is a shortest curve relative to in and .
Observe that the underlying path of is preserved, and if any is split in , we use intersecting . The rest follows from Observation 4.5, and the fact that no path is shortened in .∎
We now show that certain strategies for a restricted graph can be used in the original graph.
Lemma 5.4
Let be a region such that the robber is confined to . Suppose that there exists a strategy capturing the robber in , confining him for the entire strategy to . Then there exists a strategy for the same number of cops capturing the robber on , if the robber is initially confined to .
The strategy proceeds as , with the following exception. When moves a cop to a split of , the strategy move this cop to ; note that this move is always possible. It is key that the robbers choices in are not extended, so he is confined to it by and captured. ∎
Proof of Theorem 1.1(iii). We are ready to prove that the maximum cop number of string graphs is at least 3 and at most 15.
[Theorem 1.1(iii)] The lower bound of 3 cops follows from the graph in Fig. 4. It remains to argue that there exists a strategy using 15 cops. Our strategy proceeds in phases, monotonously shrinking the confinement of the robber. In the beginning of each phase, the robber is confined to either (A) by a single cop guarding a cutvertex separating from the rest of the graph, or (B) by ten cops guarding two shortest curves forming a simple (nonselfintersecting) cycle surrounding . In each phase, we decrease the number of vertices in or , so the robber is caught after finitely phases.
Let be the union of the currently guarded paths and vertices; by Lemma 4.3, if the robber moves to , he is captured. Let be the component of containing the vertex with the robber, and let . Since our strategy confines the robber to for the rest of the game, we can leave out the remaining vertices and assume that . Let be the number of segments of .
Claim 5.5
Let , the robber stands on , and one of the following holds:

and 1 cop guards a vertex .

, 10 cops guard two shortest curves and relative to between points to such that forms a simple cycle, and additionally where is the closed face of containing .
Then 15 cops have a strategy to capture the robber.
[Claim] We prove this claim by induction on and . The claim obviously true when and . The strategy proceeds differently according to which of (A) and (B) is satisfied.
Case (A). If , then move the cop guarding to start guarding . Let , we further leave out the irrelevant vertices, so as above. The rest follows from the induction hypothesis, with the assumption (A), applied to with and .
Let for . Let be a point of . We choose be a shortest curve in between some and , and be the subcurve of between and . Without loss of generality, forms a simple cycle; if not, we can shorten it by choosing different points and .
We start guarding and with 10 cops. Let be the closed face of containing a string with the robber. We denote , we leave out the irrelevant vertices, so as above. We use the induction hypothesis, with the assumption (B), applied to for and . By Lemma 5.4, the strategy on from the induction hypothesis implies a strategy on .
Case (B). First suppose that there exists no shortest curve in between and intersecting . By Menger theorem, there must be a cutvertex separating from . Our strategy guards with one cop, and then stops guarding . Let , leaving out the irrelevant vertices, so as above. The rest follows from the induction hypothesis, with the assumption (A), applied to with and .
Otherwise, let be a shortest curve relative to in from to intersecting . The strategy starts guarding with the five free cops. Then, let be the closed face of containing the string on which the robber stands. By Lemma 5.2, we have that where is a subcurve of , is a subcurve of , and form a simple cycle. We free the five cops stop guarding , where , and we restrict the guarding of and to and , which are shortest curves by Observation 4.5.
Let , leave out the irrelevant vertices, so as above. The rest follows from the induction hypothesis, with the assumption (B), applied to with and . By Lemma 5.4, the strategy on from the induction hypothesis implies a strategy on .
The theorem follows by guarding an arbitrary vertex with one cop, so . We leave out the irrelevant vertices, so . We use Claim 5.5 with the assumption (A) for .∎
6 Capturing Robber in String Graphs on Bounded Genus Surfaces
In this section, we generalize the results of the previous section to graphs having a string representation on a fixed surface, and we prove Theorem 1.1(iv).
Definitions. We assume familiarity with basic topological concepts related to curves on surfaces, such as genus, Euler genus, noncontractible closed curves, the fundamental group of surfaces and graph embedding properties. A suitable treatment of these notions can be found in Prasolov (); Mohar and Thomassen (2001).
A walk in a graph is a sequence of vertices where and adjacent; repetitions of vertices and edges are allowed. A walk is called closed if . Let denote the length of . For walks and with , we denote the concatenation by . Let be the reversal of and let .
A curve is a continuous function from the interval to the surface, and it is closed if . The concatenation of curves is defined naturally whenever , and similarly is the reversal and . We use the following topological lemma, following from the properties of the fundamental group; see Prasolov ().
Lemma 6.1 (Prasolov ())
Let , and be three curves on a surface from to . If the closed curve is noncontractible, then at least one of and is noncontractible.
Consider a string representation of on a surface . We represent the combinatorial structure of by an auxiliary multigraph embedded on defined as follows. The vertices of are the endpoints of the strings of and the intersection points of pairs of strings of . The edges of correspond to segments of strings of , i.e., to subcurves connecting pairs of vertices appearing consecutively on a string of . By representing by , we can use the welldeveloped theory of graph embeddings on surfaces.
Walks Imitating Noncontractible Curves. We introduce a relation between a walk in and a curve on , allowing us to easily transition between the two. We say that a walk in imitates a curve on the surface if can be partitioned into a sequence of consecutive subcurves of positive length such that and . A closed walk imitates a noncontractible curve if there is a noncontractible curve imitated by .
Lemma 6.2
Let be a string representation of on an orientable (resp. nonorientable) surface of genus (resp. Euler genus ) and let be a closed walk in imitating a noncontractible curve. Then every connected component of has a string representation on a surface of genus at most (resp. Euler genus at most ).
Note that the proof and te arguments are the same for orientable genus and Euler genus.
If has an embedding on a surface of genus (resp. Euper genus ), then has a string representation on this surface and we are done. Suppose then that this is not the case, i.e., is a graph of genus (resp Euler genus ). Therefore its embedding on is a 2cell embedding, i.e., every face of is homeomorphic to a disk.
Let be the noncontractible curve imitated by . The curve traces a closed walk in . Since is noncontractible, contains a noncontractible simple cycle of . By standard results on 2cell embeddings (see (Mohar and Thomassen, 2001, Chapter 4.2)), the genus (resp. the Euler genus) of every connected component of is strictly smaller than the genus of .
Let be the string representation . The auxiliary multigraph is a subgraph of , and hence each of its connected components has an embedding on a surface of genus (resp. Euper genus ). This embedding corresponds to a string representation of a connected component of on a surface of genus (resp. Euper genus ).∎
Lemma 6.3
If has no string representation in the plane, then for every string representation of on a surface there is a closed walk in imitating a noncontractible curve.
Since is not planar, the embedding of contains a noncontractible cycle (see (Mohar and Thomassen, 2001, Chapter 4.2)), which corresponds to a noncontractible curve on . This curve is imitated by a closed walk of .∎
Lemma 6.4
Let be a string representation of on a surface , let be two vertices, and let , , be three walks from to . If imitates a noncontractible closed curve, then at least one of and imitates a noncontractible closed curve.
Let be a noncontractible closed curve imitated by . Looking at the consecutive subcurves of corresponding to vertices of , we have that with imitated by and imitated by . Let be the first point of and be the last point of .
Now let be any  curve imitated by and observe that is imitated by and is imitated by . By Lemma 6.1, at least one of and is noncontractible and the lemma follows.∎
Lemma 6.5
On a graph with a string representation on a surface and a shortest closed walk imitating a noncontractible curve, 10 cops have a strategy to guard after a finite number of initial moves.
If , the cops may occupy every vertex of for the rest of the game and we are done. Otherwise we divide into two almostequally long walks , , where is from to , where , are edges. We have , such that ; see Fig. 7. Note that .
We claim that both and are shortest paths in . If is not a shortest path, let be a shortest path from to , so . Then both closed walks and would be shorter than :
By Lemma 6.4, at least one of them is noncontractible which contradicts the assumption. Similarly, there exists no path from to with .
Therefore we may use Lemma 4.3 (with ) to guard and with ten cops.∎
Proof of Theorem 1.1(iv,v). We are ready to prove that the maximum copnumber of GENUS STRING graphs is at least and at most , and of EULERGENUS STRING graphs is at least and at most .
[Theorem 1.1(iv,v)] For the lower bound, consider the incidence graph of a projective plane of order . It has at most than edges, so its genus is at most . It is regular graph with girth 6, so