1 Introduction
The game of cops and robber is a twoplayer turnbased game between the first player who controls a fixed number of cops and the second player who controls a robber. The game is played on a given connected graph on which the cops pursue the robber to capture him, and the robber tries to escape. Each round of the game consists of a cops’ turn followed by a robber’s turn. In the first round, the first player locates the cops on some vertices of and then the second player locates the robber on a vertex. Afterward, the cops and the robber move consecutively; at each cops’ turn (resp. robber’s turn), every single cop (resp. the robber) can stay in his position or move to an adjacent vertex. Once a cop is in the robber’s position, the game is over, and the first player wins. The cop number of a connected graph , denoted by , is the minimum number of cops which are sufficient to capture the robber on , i.e., the minimum number such that cops have a winning strategy on . We follow [BI93, JKT08] to define the cop number of a disconnected graph as the maximum of the cop numbers of its connected components. Also, is called to be copwin, if and is called copwin, if . For instance, Petersen graph is the smallest copwin graph [BCM11] and a copwin graph has at least vertices [TY21]. For other variants of the game, see e.g. [FKL12, Luc18, Gav16, FT08].
For a copwin graph , the capture time of is the minimum number such that cops can always capture a robber on within rounds, regardless of how the robber plays. It can be easily seen that for every positive integer , the capture time of an vertex copwin graph is at most . Bonato et al. [BGHK09] showed that this bound is not tight for , proving that the capture time of all copwin graphs on vertices is at most . However, surprisingly, the upper bound happens to be asymptotically tight for all integers , i.e. there are copwin graphs with the capture time in for every integer [Kin18, BEUW20].
The problem of finding the exact value or bounds on the cop numbers of graphs has been studied extensively for decades. For a class of graphs , we say that is copbounded if there is a constant , such that for every graph , . For instance, it is easy to see that every tree is copwin, so the class of all trees is copbounded. It is a classic and wellstudied question in the literature: which classes of graphs are copbounded? One of the important classes that this question is studied is the class of graphs that exclude a (some) certain graph(s) as a subgraph, induced subgraph, or minor. Given a graph , we say that is free, subgraphfree, or minorfree if does not contain an induced subgraph, subgraph, or minor isomorphic to , respectively. Also, for a class of graphs , we say that is free if is free for every (the same definition can be stated for subgraphfree and minorfree). The first results in the context date back to 1984 when it was proved in [AF84] that all planar graphs have the cop number at most and in [And84] that the class of regular graphs are copunbounded for every integer . The latter result can be also implied from a recent result in [BHMS20] stating that the cop number of any graph with girth and minimum degree is at least . Also, there is a generalization of the former result stating that for every graph , , where is the genus of the graph [BELP21]. In addition, Andreae in [And86] proved that for every fixed graph , the class of minorfree graphs is copbounded. More precisely, if is a vertex of such that has no isolated vertex, then the cop number of every minorfree graph is at most .
Joret et al. proved in [JKT08] that for every graph , the class of free graphs is copbounded if and only if every connected component of is a path. This result implies that the class of clawfree graphs are not copbounded. In particular, if is free, then cops can capture the robber in at most moves [Siv19].
For a class of graphs , it is proved in [MS20] that if every connected component of the graphs in has a bounded diameter, then the class of free graphs is copbounded if and only if contains either a forest of paths or two graphs , , each having at least one vertex of degree three such that every connected component of is a path or a generalized claw, and every connected component of is a path or a generalized net. However, the problem of characterization of all classes for which free graphs are copbounded is still open. One interesting case is when is a class of some holes with unbounded diameters. By a hole, we mean a cycle of length at least without any chord. For instance, it is proved in [Siv20] that if is the class of all holes of length at least , , then the copnumber of every free graph is at most . It is also known that any class of graphs with bounded treewidth is copbounded. In fact, Joret et al. [JKT08] proved that for every graph , , where is the treewidth of the graph .
Meanwhile, the structure of oddholefree and evenholefree graphs have been studied widely in the literature, mostly motivated by the study and generalization of perfect graphs (for instance, see
[SS16, CCV04, CRST10, dV13]). This gives rise to the following natural question that whether the classes of oddholefree and evenholefree graphs are copbounded. At first glance, one may easily check that oddholefree graphs are copunbounded since subdividing all edges of a graph does not decrease its cop number [JKT08, BI93]. How about when we also exclude claws? One may see that (claw, oddhole)free graphs are also copunbounded. To see this, we need a notion called clique substitution defined in [JKT08]. Given a graph , its clique substitution is the graph obtained from by replacing each vertex of by a clique of the size such that, if is adjacent to in , then a vertex in is adjacent to a vertex in in a way that every vertex in a clique has exactly one neighbor outside . Joret et al. [JKT08] proved that if is the clique substitution of , then . Also, it is clear that whatever is, is always oddholefree and clawfree as well. Hence, if is a copunbounded class, then the class of clique substitutions of the graphs in is also copunbounded, and they are evidently (claw, oddhole)free. So, we haveCorollary 1.
The class of all (claw, oddhole)free graphs is copunbounded.
It is not so straightforward to answer the same question for evenholefree graphs. It is proved that the treewidth of (triangle, evenhole)free graphs is at most [CdHV18], so the cop number of (triangle, evenhole)free graphs is at most . On the other hand, (claw, evenhole)free graphs contain all cliques, so the class has unbounded treewidth. In this paper, we prove, in contrast, that the class of (claw, evenhole)free graphs is copbounded.
Theorem 2.
If is a (claw, evenhole)free graph with vertices, then and the capture time of is at most .
Our proof also gives an insight into the structure of such graphs. It remains open to answer the following question in general.
Question 3.
Is it true that the class of evenholefree graphs is copbounded?
Recently, there have been many pieces of work on the study of evenholefree graphs. Here, we survey some most important results in this area. A decomposition theorem for evenholefree graphs is provided in [dV13] strengthening a previously known decomposition result in [CCKV02]. For a survey on the structural results, see [Vuš10]. Chudnovsky and Seymour in [CS19] settled a conjecture by Reed and proved that every evenholefree graph has a bisimplicial vertex. A bisimplicial vertex is a vertex whose neighborhood can be partitioned into two cliques. Also, Chudnovsky et al. [CSSS21] motivated by an application in condensed matter physics and quantum information theory, proved that every nonnull (claw, evenhole)free graph has a simplicial clique. By a simplicial clique, we mean a clique such that for every vertex in , the neighbors of outside form a clique.
The treewidth of evenholefree graphs has also been studied extensively. In particular, Cameron et al. [CdHV18] providing a structure theorem for (cap, evenhole)free graphs proved that they are bounded, i.e., their chromatic number is bounded by their clique number. This raises the question of whether the treewidth of an evenholefree graph can be bounded by its clique number? This question was answered negatively by Sintiari and Trotignon [ST19] by constructing a family of (, evenhole)free called layered wheels, whose treewidth is unbounded. Moreover, it is proved in [AAK20] that every class of evenholefree graphs that excluding a fixed graph as a minor has bounded treewidth and Abrishami et al. [ACV20], resolving a conjecture given in [AAK20], proved that evenholefree graphs of bounded degree have bounded treewidth.
The structure of forthcoming sections is as follows. In Section 2, we investigate the structure of (claw, evenhole)free graphs inside and between its distance levels. In Section 3, we study the structure of holes in (claw, evenhole)free graphs and their setting with respect to the distance levels. In Section 4, we give an algorithm for the game of cops and robber on (claw, evenhole)free graphs which guarantees that the cop number of these graphs is at most three. Finally, in Section 5, we improve the algorithm to prove Theorem 2.
1.1 Notations and Conventions
Given a graph and a vertex , the set of neighbors of in is denoted by and, whenever there is no ambiguity, we drop the subscript. For a subset of vertices , the set of neighbors of in is denoted by , i.e. . The induced subgraph of on is denoted by which is a graph with vertex set and all connections as in the graph . The graph obtained from by removing all vertices in is denoted by . Also, for a pair of nonadjacent vertices , the graph is obtained from by adding the edge to . For two disjoint subsets of vertices , we say that is complete (resp. incomplete) to if every vertex in is adjacent (resp. nonadjacent) to every vertex in . Also, we say that is connected to if is not incomplete to .
A sequence of distinct vertices is called a path of length if for every , is adjacent to . Sometimes we consider a direction on and say that is a path from to . So, the vertices and are defined as the last vertex before and the first vertex after , respectively. Also, for some indices , , the subpath is denoted by .
Let and be two arbitrary adjacent vertices in . Also, let . Now, we define to be the connected component of containing . For each positive integer , define the level as the set of vertices of distance from in the graph , i.e. . For simplicity, we use the notations and . For every vertex , the set of neighbors of in is denoted by and the set of neighbors of in is denoted by . Every connected component of the induced subgraph of on , i.e. , is called a level component in .
2 The Structure of Level Components
In this section, we study the structure of a (claw, evenhole)free graph. Let be a (claw, evenhole)free graph and and the levels ’s be defined as in Subsection 1.1. Here, we investigate the form of the level components in each level and how they are connected to other levels of the graph .
Since is clawfree, it is an easy observation that, for every vertex , , the set of neighbors of in the next level , , is a clique. It is because has a neighbor in and if it has two nonadjacent neighbors in , then there is a claw with the center . Thus,
Lemma 4.
For every vertex , is a clique.
Now, we define a kind of path which is forbidden in .
Definition 5.
Let be a path of odd length such that , for some , and , and all other vertices are in . We say that is a forbidden path, if either is an induced path or is a hole.
In the following lemma, we prove that there is no forbidden path in .
Lemma 6.
There is no forbidden path in the graph .
Proof.
Suppose that is a forbidden path such that , for some , and , and all other vertices are in .
Let be two shortest paths from to and from to , respectively. Then, there is an edge with one endpoint in and another in , because if the vertex is the first common vertex of and , then, by Lemma 4, the neighbors of in form a clique and thus, the neighbors of in and are adjacent. Now, let be the first edge between and in the sense that is the first vertex in (by passing from to ) which has a neighbor in and is the closest neighbor of to . Both vertices and are located in the same level, since otherwise, by Lemma 4, there is another edge between and which is a contradiction with the choice of and . Note that if is adjacent to , then and .
But now, is an even hole. To see this, suppose that be adjacent to a vertex . Then, and, by Lemma 4, is adjacent to , contradicting the fact that is an induced path. ∎
Corollary 7.
For every edge in , either or .
Proof.
Suppose that there are two vertices such that and . Now, is a forbidden path, contradicting Lemma 6. ∎
In the next result, we determine the structure of inside each level . First, we need the following lemma from [Wol65]. Let be a vertex in a graph and be the connected component of containing . The vertex is called a king vertex in , if is complete to .
Lemma 8.
[Wol65] Every connected free graph has a king vertex.
By a clique, we mean a connected graph whose vertex set is partitioned into three nonempty sets such that and are cliques and is incomplete to .
In the following lemma, we determine the structure of in each level.
Lemma 9.
For every integer , every level component in is either a clique or a clique.
Proof.
First, we claim that the graph is free. To see this, suppose that there is an induced path in . Since is nonadjacent to , by Lemma 4, . Now, since is adjacent to both and , by Corollary 7, . But since is nonadjacent to , , which means , contradicting Corollary 7. Therefore, is free, and thus, by Lemma 8, each of its level components has a king vertex.
Let be a level component of and be the set of king vertices of . If , then is a clique, and we are done. So, suppose that , which means there are at least two nonadjacent vertices that are adjacent to a vertex .
By Corollary 7, there is a vertex . By Lemma 4, is nonadjacent to . Define and . Since is complete to , by Lemma 4, is a clique. Also, is a clique, since otherwise, there must be two nonadjacent vertices and would induce a claw, a contradiction.
Also, is incomplete to . To see this, suppose that is adjacent to . Since , there are vertices and such that is nonadjacent to and is nonadjacent to . Then induces a or a on which is in contradiction with the fact that is free. This proves that is a clique. ∎
The next lemma determines the connections between two consecutive levels and .
Lemma 10.
Let be a level component in . Also, let be the set of all vertices in which has a neighbor in .

If is a clique, then for every , we have either , or .

If is a clique, then there is an ordering on the vertices in , say , such that
Proof.
First, note that, by Lemma 4, every has no neighbor in other level components in , i.e. .

Suppose that is a clique. Note that has a neighbor in , since if is incomplete to , then has a neighbor , and for every and , induces a claw on . So, without loss of generality, suppose that is adjacent to some vertex . Also, let be a vertex in . Now, we prove that . To see this, suppose that is nonadjacent to a vertex . Then, by Corollary 7, and has a common neighbor in . Since is nonadjacent to , by Lemma 4, is also nonadjacent to . Thus, would be a forbidden path, contradicting Lemma 6. This proves that is complete to .
Now, we prove that is complete to and incomplete to . To see this, note that since is incomplete to and is adjacent to , by Lemma 4, is incomplete to . On the other hand, if is nonadjacent to some vertex , then would be a claw. This completes the proof of (i).

Suppose that is a clique. We prove that for every , we have either , or . For the contrary, assume that there are two vertices such that and . Therefore, would be a forbidden path, contradicting Lemma 6. This proves that the inclusion induces a total ordering on .
Now, let be the vertex with the largest number of neighbors in among all vertices in . If , then is nonadjacent to a vertex . The vertex has a neighbor , so , which is in contradiction with the choice of .
∎
3 Holes in (Claw, Evenhole)free Graphs
In this section, we study the structure of holes in the graph . Let be a hole in . The first and last levels and which intersect is called the first and last level of , i.e. and . Also, all levels , , are called inner levels of . For every level , , a level component in that contains some vertices of is called a component of in . Also, define
Note that, in the definition of , we have excluded the components of in its first level. In the next two lemmas, we prove that every hole in has a specific structure with respect to the levels ’s (see Figure 1).
Lemma 11.
Let be a hole in the graph . The hole ,

has no edge with both endpoints in a level , for some , and

has exactly one edge with both endpoints in its first level.
Proof.

Let be the largest integer such that contains an edge of , say , and suppose that is not the first level of . Orient the cycle in a direction such that the edge is directed from to . Let and consider the induced subgraph of on . Now, define as the connected component of containing the edge (note that is a subpath of which lies in ).
First, suppose that there is no edge in adjacent to with both endpoints in . Now, if does not start from , then let be the last vertex of in that is before and let be an arbitrary neighbor of in and if starts from , then define and be the neighbor of in . Similarly define as the first vertex of in after and as an arbitrary neighbor of in (if ends in , then define and as the neighbor of in ). Now, is a forbidden path, contradicting Lemma 6. To see this, note that is incomplete to , because if , then and if and is adjacent to a vertex in , then is adjacent to which is in contradiction with the fact that is a hole. With a similar discussion, is incomplete to .
Finally, suppose that there is another edge of adjacent to the edge in and, without loss of generality, name this edge as . Define and exactly as above and let and be a common neighbor of and in which exists by Corollary 7. Hence, again is a forbidden path, a contradiction.

By (i), has an odd number of edges in its first level . Suppose that has more than one edge in its first level. Therefore, has at least edges in . So, has a subpath of the form
where and .
The vertices and have a common neighbor in and, the vertices and have a common neighbor in Now, the path is a forbidden path, because is a subpath of and so is induced. Also, due to Lemma 4, and are both nonadjacent to inner vertices of .
∎
Lemma 12.
The hole has exactly one vertex in its last level and exactly two vertices in its other levels.
Proof.
First, we prove a claim.
Claim 12.1.
Let be a subpath of the hole such that , for some , and . Then, if is not the first level of , then and otherwise .
To prove the claim, note that if , then by Lemma 4, and are adjacent, which is a contradiction with the fact that is a hole.
Also, if is not the first level, cannot be in due to Lemma 11(i).
This proves the claim.
Now, we prove the lemma. Let be a vertex of in its last level . Then, by Lemma 11(i), the two neighbors of in are in . Now, by Claim 12.1 and Lemma 11(ii), the neighbors of and in are in . Finally, we have two distinct vertices of in its first level . We claim that and are adjacent, which completes the proof. If and are not adjacent, then and have neighbors , respectively, in (that might be the same). By Claim 12.1, and are in first level of . So, contains at least two edges and in its first level which is a contradiction with Lemma 11(ii). This completes the proof. ∎
Lemma 12 reveals the structure of the hole with respect to the levels ’s. So, has a configuration as Figure 1.
A hole is called dominated if there is a vertex in such that is complete to . It is clear that in every clawfree graph, the only dominated hole is a cycle which is the induced subgraph of a wheel . The next lemma states that every nondominated hole has exactly one component in its first and last level and exactly two components in other levels. So, if is not dominated, then , where is the length of , and if is dominated, then .
Lemma 13.
Let be a hole in . If there is a component in an inner level such that , then is a which is dominated by a king vertex of .
Proof.
Let be a hole in with a component in an inner level such that . Let be vertices of such that , and such that and are subpaths of . Note that if is the last level of , then . Since is a clique, by Lemma 10(i), there is a king vertex such that is adjacent to all vertices . Also, let be the induced subpath of in .
Now if , then to prevent the hole to be an even hole, must be adjacent to only one of or . Without loss of generality, suppose that is adjacent to . Now forms a claw, a contradiction. So, and is the last level of . Also, to prevent to be an even hole, is adjacent to .
Now, if and are nonadjacent, then forms a claw, which is a contradiction. Hence, is adjacent to and thus, is the first level of and is a dominated . This completes the proof. ∎
In the following lemma, we prove that every nondominated hole in passes through king vertices of the components of .
Lemma 14.
Let be a hole of . If is not a dominated , then every vertex of in a component is a king vertex of . Also, if is a dominated , then the vertex in the last level of is a king vertex.
Proof.
Let be an arbitrary hole in . Suppose that is a nonking vertex of a component . Thus, is a clique. First, suppose that is the last level of (Here, could be either dominated or not). So, there are nonadjacent vertices which are both adjacent to . By Lemma 10, there is a king vertex of such that is adjacent to and a nonking vertex of such that is adjacent to but nonadjacent to . Now, is a claw, a contradiction.
Now, suppose that is not the last level of . Also, suppose that is not a dominated . Since is not a king vertex of , there is a king vertex of and a nonking vertex of such that is nonadjacent to . Also, let be a common neighbor of and in . Let be the subpath of in such that , , and (if is the last level of , then ).
First, suppose that is nonadjacent to both and . In this case, we claim that is a forbidden path. To see this, first note that since is not a dominated , by Lemma 13, and are in two distinct components of in and thus, is nonadjacent to . Therefore, by Lemma 4, is nonadjacent to and is nonadjacent to . This proves that is a forbidden path, a contradiction with Lemma 6.
Now, suppose that is adjacent to either or . Without loss of generality, suppose that is adjacent to . Since does not induce a claw, we have is adjacent to . Now, if , then would induce a claw, and if , then would induce a claw, where is the first vertex in after . This completes the proof. ∎
The following theorem, which is the main result of this section, illustrates the connection of two holes in .
Theorem 15.
Let and be two holes in . Then either , or . In the latter case, and share the same component in their first level.
Proof.
Let be the common component of in the highest possible level . So, the components of and in are disjoint. Let and be the subpaths of and , respectively, such that and (they exist since is not the first level of and ). Also, for , let be the unique neighbor of in . One of the following two cases may occur.
Case 1.
The vertices and are distinct and nonadjacent.
Then, and are not king vertices of . Therefore, by Lemma 14, and are both dominated and is not the last level of both . Let , , be the unique vertex in . Since and are nonadjacent vertices in the unique component of and in , the set is incomplete to the set and induces on both sets a . Also, since the components of and in are disjoint, is nonadjacent to . So, by Lemma 4, is incomplete to and is incomplete to . This implies that would be a hole , a contradiction.
Case 2.
The vertex is adjacent to , or .
Subcase 2.1.
.
First, note that if , then by Lemma 4, is adjacent to , which is a contradiction with the choice of . So, assume that and are distinct and adjacent.
For , let be the unique vertices in , , and , respectively. Also, let be the subpath of from to in . Note that, it is possible that whenever is the last level of (see Figure 1(a)). We show that the path
is a forbidden path. To prove this, we need the following claim.
Claim 15.1.
The subgraph is an induced subgraph of .
Note that and are induced paths since they are subpaths of and , respectively.
To prove the claim, suppose that there are adjacent vertices and .
Without loss of generality, suppose that , for some , and .
If , then it contradicts the fact that is the last level that and meet in a common component.
So, .
If is not the last level of , then has a neighbor and by Lemma 4, and are adjacent which is, again, contradicts how is chosen.
So, is the last level of .
Let be the two neighbors of in .
Therefore, and thus, induces a claw.
This proves the claim.
Note that every has only one neighbor in , since otherwise, by Lemma 4, is not an induced subgraph, contradicting Claim 15.1. Also, is nonadjacent to , since otherwise, is an even hole. Similarly, is nonadjacent to . Hence, is a forbidden path, contradicting Lemma 6.