aa r X i v : . [ m a t h . C O ] J un JÉRÉMIE TURCOTTE AND SAMUEL YVON
Abstract.
We show that the cop number of any graph on 18 or fewer vertices is at most 3. Thisanswers a specific case of a question posed by Baird et al. on the minimum order of 4-cop-wingraphs, first appearing in 2011. We also find all 3-cop-win graphs on 11 vertices, narrow down thepossible 4-cop-win graphs on 19 vertices and get some progress on finding the minimum order of3-cop-win planar graphs. Introduction
The game of cops and robbers was first defined by Quilliot in [26] and Nowakowski and Winkler in[24]. Playing on a connected, undirected and finite graph, the cops try to catch a robber. The copsand the robber alternate turns. On the first turn, each cop selects a starting vertex followed by therobber. At each subsequent turn, each player may either stay put or move to an adjacent vertex. Ifat any point the robber and one of the cops share a vertex, the cops win. The robber wins if it hasa strategy ensuring it is never caught by the cops. At all times, the positions of the cops and therobber are known by all. Furthermore, the cops may coordinate their strategies, and they are allowedto share vertices.For a connected graph G , we denote by c ( G ) the minimal number of cops which can always catchthe robber on G . Introduced by Aigner and Fromme in [2], c ( G ) is called the cop number of G . If c ( G ) = k , we say G is k -cop-win.The cop-number is the most frequently studied parameter related to the game of cops and robbers,but other parameters, such as the capture time, are also studied. See [11] for a quick overview ofthis field or [14] for a more in-depth introduction. A multitude of variants of this game have beenconsidered in recent years, but in this paper we only study the classical version.While there has been a good deal of research on the cop number of graphs, often on specific classesof graphs, there are still surprisingly many elementary open questions. We consider here the questionof finding the minimum order of k -cop-win graphs, more specifically for k = 4 . This question wasposed by Baird et al. in [5], although the concept appeared in two preprints by two subsets of theauthors in 2011 [6, 9]. In [5], the analogous question for 3-cop-win graphs is solved: the Petersengraph (see Figure 1) is the smallest 3-cop-win graph.We denote by V ( G ) and E ( G ) , respectively, the set of vertices and of edges of G . We denote by M k the minimum order of k -cop-win graphs; formally, M k = min {| V ( G ) | : G connected graph, c ( G ) = k } .Interestingly, Hosseini proved in [20] that M k < M k +1 , confirming the intuition that if one scans allgraphs by increasing order, one cannot find a ( k + 1) -cop-win graph before finding a k -cop-win graph.The problem of finding the minimum order of 4-cop-win graphs has received some interest since itwas first raised. Hosseini proved in [19] that M ≥ , and that such a minimal graph is 3-connected,provided it does not contain a vertex of degree 2. This problem is also referenced in [11].It is suggested in [5] that the value of M might be 19. Indeed, the smallest known 4-cop-win graphis the Robertson graph (see Figure 1). This graph was first discovered by Robertson in [28] as thesmallest 4-regular graph with girth 5. A ( d, g ) -cage is a smallest regular graph of degree d and girth Mathematics Subject Classification.
Primary 05C57; Secondary 05C35, 05C85, 90C35, 68V05.
Key words and phrases.
Cops and robbers, Cop number, Extremal problems, Graph construction, Computer-assistedproof. g . For instance, the Petersen graph is the unique (3 , -cage and the Robertson graph is the unique (4 , -cage.It has been proved in [2] that graphs with girth at least 5 have a cop number of a least theirminimum degree. This result has since been generalized in [17], and recently in [15]. One easilydeduces that the cop number of the Robertson graph is therefore at least 4. It is easily seen in thefigure that placing a cop on each of the three exterior vertices only leaves 4 unprotected vertices,which form independent edges. A last cop may then easily capture the robber, thus the Robertsongraph is 4-cop-win (this argument appears in [8]). It is suggested in [5] that the smallest d -cop-wingraphs might be the ( d, -cages. α α α α α β β β β β (a) The Petersen graph (b)
The Robertson graph Figure 1.
Some small ( d, -cage graphsThe main result of this article is to confirm that M = 19 . Although we are not able to prove acomplete uniqueness result, we narrow down the possible 4-cop-win graphs on 19 vertices.Although our proof is not directly based on those in [5] and [19], there are certainly some commonelements. In particular, we also break down the problem by maximum degree and find properties ofpotential 4-cop-win graphs by constructing explicit strategies.While we are able to get many interesting results formally, this article makes extensive use ofcomputational methods to verify the remaining cases. All of the code and data produced in the writingof this article are available online at [30]. This includes not only the final results, but the graphs wegenerate in the intermediate algorithms, precise counts of the number of graphs we generate at everystep in these algorithms, and the time required for almost all computations. All of the computationsare separated in small parts to facilitate verification. At various points in this article, we will alsodiscuss possible improvements and alternative computational approaches.2. Notation and previous results
In this section, we introduce most of the notation used in the article. We also cite previously knownresults that will be useful.When considering a graph G , we will respectively denote by n ( G ) , d G ( u ) δ ( G ) and ∆( G ) thenumber of vertices of G , the degree of a vertex u in G , the minimum degree of G and the maximumdegree of G , although we will usually omit the G when the choice of graph is easily deduced. If u is a Computer-generated drawing [21]. -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 3 vertex of G , N ( u ) will denote the (open) neighbourhood of u and N [ u ] = N ( u ) ∪ { u } will denote theclosed neighbourhood of u .If S ⊆ V ( G ) , S c will denote the complement (in V ( G ) ) of S , h S i will denote the subgraph of G induced by S , and G − S will denote h S c i . If S = { x } , we will use the notation G − x instead of G − S .Similarly, if H is a subgraph of G , then G − H will denote G − V ( H ) .Denote by P the Petersen graph, as seen in Figure 1. As P is -regular with girth 5, we getthat c ( P ) ≥ , and as it contains a dominating set of size , we know that c ( G ) = 3 . The followingtheorem was proved by Baird et al., first by computer verification and then formally. Theorem 2.1. [5]
Let G be a connected graph.(1) If n ≤ , then c ( G ) ≤ .(2) If n = 10 , then c ( G ) ≤ , unless G is the Petersen graph.In particular, M = 10 . The proof of the previous theorem makes use of the following lemma, which will also be useful tous.
Lemma 2.2. [5]
Let G be a connected graph. If ∆ ≥ n − , then c ( G ) ≤ . A simple, visual proof of this lemma is available in [8]. We now define a useful concept, which hasbeen used many times to study the game of cops and robbers.
Definition 2.3.
Let G be a graph. If H is an induced subgraph of G , we say H is a retract of G ifthere exists a mapping f : V ( G ) → V ( H ) such that:(1) If xy ∈ E ( G ) , then f ( x ) f ( y ) ∈ E ( H ) or f ( x ) = f ( y ) .(2) f | V ( H ) : V ( H ) → V ( H ) is the identity mapping.Such a mapping f is called a retraction. This definition formalizes the intuitive idea that G can be "folded" onto H , where each edge musteither be sent onto an edge or onto a vertex. Those familiar with graph homomorphisms will noticethat condition (1) states that f is a homomorphism from G to H , if we consider H to be reflexive(that is, if we add a loop at each vertex of H ). This requirement for reflexivity is a consequence ofallowing the cops and the robber to stay on a vertex during their turn, implying a loop on each vertex.The concept of retracts has been central in the study of the game of cops and robbers.If G is disconnected, denote G , . . . , G t the connected components of G . By extension, we maydefine the cop number of a disconnected graph by c ( G ) = max ≤ i ≤ t c ( G i ) .These definitions allow us to state the following result of Berarduci and Intriglia, which we will usemany times to reduce the number of cases we need to consider. Theorem 2.4. [7] If G is a connected graph and H is a retract of G , then c ( H ) ≤ c ( G ) ≤ max { c ( H ) , c ( G − H ) + 1 } . A specific case of this theorem is the following, which will often be easier to use.
Corollary 2.5. If G is a connected graph, u is a vertex of G and K is a union of some connectedcomponents of G − N [ u ] , then c ( G − K ) ≤ c ( G ) ≤ max { c ( G − K ) , c ( K ) + 1 } . In particular, if c ( K ) ≤ k − , then c ( G ) ≤ k if and only if c ( G − K ) ≤ k .Proof. It is easy to verify that f : V ( G ) → V ( G − K ) defined by f ( x ) = ( u if x ∈ V ( K ) x otherwiseis a retraction. It is only left to apply Theorem 2.4 to H = G − K . (cid:3) JÉRÉMIE TURCOTTE AND SAMUEL YVON
One trivial consequence of this corollary is that if the cop number of every component of G − N [ u ] is at most k − , then c ( G ) ≤ k . One can also see this directly by leaving a fixed cop on u and playingwith k − cops on the connected component of G − N [ u ] in which the robber is located.We then easily get the following result, which is implicit in [19]. Corollary 2.6. If G is a connected graph and ∆ > n − , then c ( G ) ≤ .Proof. If ∆ > n − and u is a vertex of maximum degree, then | V ( G − N [ u ]) | < . By Theorem2.1, every connected component of G − N [ u ] has cop number at most 2. The last remark yields theresult. (cid:3) Finally, we recall a well known concept in the study of the game of cops and robbers.
Definition 2.7.
Let x, u be distinct vertices of G . If N ( x ) ⊆ N [ u ] , we say x is cornered by u orthat x is a corner . We note that this is a slight variation on the classical notion of a corner (or irreducible vertex), asit normally requires ux to be an edge, see [24]. We may now get a further simplification of Corollary2.5. Corollary 2.8. [7]
Let G be a connected graph and x be a corner of G . If c ( G − x ) ≥ , then c ( G ) = c ( G − x ) . If c ( G − x ) = 1 , then c ( G ) ∈ { , } .Proof. If x is cornered by u , notice that x is isolated in G − N [ u ] . (cid:3) Computational results for 3-cop-win graphs
In this section, we find some 3-cop-win graphs on at most vertices respecting some degreeconditions. These results will be useful in the following sections.We will do this by computing the cop number of every graph with the desired order and degreeconditions. Graph generation in this section and in Section 5 is done using the geng function providedwith the nauty/Traces package (version 26r12) [23].The algorithm to compute the cop number is similar to that proposed, in particular, in [12, 16, 27],which we have implemented in the Julia language [10, 29]. For a given k (which will be between and in our case), the algorithm determines whether c ( G ) ≤ k or c ( G ) > k .To test the validity of our implementation, we have compared the results for the cop number ofconnected graphs up to 10 vertices to those in [5]. Due to a small discrepancy between the counts, ourtallies of cop-win graphs has also been verified by implementing a dismantling algorithm [24] and bycomparing with the implementation at [1]. To test our code for higher cop numbers, we have also runit on some cage graphs on which we know 3 cops lose. The results of these tests make us confident inthe correctness of our implementation.We first define a variant of the Petersen graph. Definition 3.1.
We say a connected graph G is a cornered Petersen graph if G contains a corner x such that G − x ≃ P . There are 6 such graphs up to isomorphism. We denote them P i , i = 1 , . . . , ,as seen in Figure 2. We now solve a question raised in [5], classifying the 3-cop-win graphs on 11 vertices, even if it iscomputationally.
Proposition 3.2. If G is a connected graph such that n = 11 , then c ( G ) = 3 if and only if G ≃ P i for some ≤ i ≤ . Otherwise, c ( G ) ≤ .Proof. Firstly, it is clear by Theorem 2.1 and Corollary 2.8 that the cornered Petersen graphs are3-cop-win.We would like to show that these graphs are the only graphs on 11 vertices with cop number 3,and that all other graphs have cop number at most 2.By Lemma 2.2, we may only consider graphs such that ∆ ≤ n − . We generate all graphson 11 vertices such that ∆ ≤ and classify each graph according to its cop number. The results are -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 5 m ′ m P m ′ m P m ′ m P m ′ m P m ′ m P m ′ m P Figure 2.
The cornered Petersen graphspresented in Table 1 (the counts are up to isomorphism). The 5 graphs found are the graphs P i for i = 2 , . . . , , which concludes the proof. (cid:3) This is an interesting phenomenon: the -cop-win graphs on 11 vertices are all retracts of theunique 3-cop-win graph on 10 vertices. This behaviour does not occur for the 2-cop-win graphs: theminimum 2-cop-win graph is the -cycle, on which the 5-cycle does not retract. Although we willnot have any answer for this question in this article, it would be interesting to know whether ingeneral (even for 4-cop-win graphs only), the k -cop-win graphs on M k + 1 vertices can be retractedon k -cop-win graph(s) on M k vertices.In Section 6, we will also need the following lemma. Lemma 3.3.
There exist • connected graphs G on vertices with ∆ ≤ , • connected graphs G on vertices with ∆ ≤ , • connected graphs G on vertices with ∆ ≤ , and • connected graphs G on vertices with ∆ ≤ such that c ( G ) = 3 . All other connected graphs considered with these orders and maximum degreesare such that c ( G ) ≤ .Proof. Firstly, all graphs on at most 14 vertices have cop number at most 3. This is a direct conse-quence of knowing that M ≥ [19]. For cases where ∆ ≥ , this is a direct consequence of Corollary2.6. For ∆ = 2 , the graph is either a path or a cycle. For ∆ = 3 , see the results of Table 4.We generate, up to isomorphism, all connected graphs on vertices such that ∆ ≤ and on vertices such that ∆ ≤ . We classify these graphs according to their cop number. Afterwards, wealso count which of the graphs on 12 vertices with ∆ ≤ and c ( G ) ≥ are such that ∆ ≤ . Theresults are in Table 1. (cid:3) While the counts are presented to summarize the results, the precise 3-cop-win graphs are the focusof our attention as we use them in the following sections.To achieve these results, we exhaustively computed the cop number of every connected graph thatsatisfied our maximum degree constraints. Since we proceeded by exhaustion, the run time of thesecomputations is somewhat long due to the high number of graphs, especially in the case of ∆ = 14 .We note that a more clever approach might yield faster calculation time.
JÉRÉMIE TURCOTTE AND SAMUEL YVON
Cop number n Degree bounds Number of graphs ≥
311 ∆ ≤
12 ∆ ≤ - - - 80
12 ∆ ≤
13 ∆ ≤
14 ∆ ≤ Table 1.
Cop number breakdown for connected graphs on 11-14 vertices with somedegree restrictionsThe first and most obvious improvement is to only look at graphs with a minimum degree of atleast 2, which can reduce the number of graphs to consider by up to around 50%. If we already knowthe 3-cop-win graphs on one fewer vertices, we can then just consider all possible ways to attach anextra vertex of degree 1 to those graphs. Using this method, we can get all connected 3-cop-wingraphs of a given order.However, this method is still an exhaustive search. Again, a more clever approach would be toconsider every 2-cop-win graph G ′ on n − ∆ − vertices, add a vertex u with ∆ neighbours andconsider each way of adding edges between N ( u ) and G ′ (up to isomorphism), then checking whichof these graphs are 3-cop-win. We would recommend the interested reader to try this approach.An even more efficient approach would be to use the algorithm of Section 6 to build candidate3-cop-win graphs, by merging 2-cop-win graphs on fewer vertices. As we will see later, although thismethod can reduce significantly the calculation time, in practice it requires some effort to make sureall the possible cases are considered. For the size of graphs we are considering, this may not benecessary. 4. Graphs with high maximum degree
In this section, we consider the cop number of graphs G such that ∆ = n − or ∆ = n − . Westart by investigating some properties of the game of cops and robbers on the Petersen graph and it’svariants, the cornered Petersen graphs. Many of the arguments in this section are extremely simpleonce visualized. For this reason, we have provided many figures representing visualizations of thesituation in some of the proofs. Of course, we cannot provide figures for every case, so we encouragethe reader to draw out the graphs while reading the proof, especially regarding player movements.By considering the Petersen graph as the Kneser graph KG , [4], one easily gets the followingwell-known result (although maybe not with this precise formulation), an illustration of the fact thatthe Petersen graph is highly transitive. Definition 4.1.
We say a set of 3 vertices { x, y, z } is a strong stable set if it is a stable set and if N ( x ) ∩ N ( y ) ∩ N ( z ) = ∅ . Lemma 4.2. (a) If { x, y, z } and { x ′ , y ′ , z ′ } are strong stable sets of P , then there exists an automorphism φ of P such that φ ( x ) = x ′ , φ ( y ) = y ′ and φ ( z ) = z ′ .(b) If ab and a ′ b ′ are two edges of P , then there exists an automorphism φ of P such that φ ( a ) = a ′ , φ ( b ) = b ′ . This property is known as being arc-transitive. We use the labels m , m ′ on the graphs P i for i = 1 , . . . , , as shown in Figure 2. In particular, foreach of these graphs, P i − m ≃ P . We see that m ′ always corners m , which will be very useful. Wealso note that as m, m ′ / ∈ V ( P ) , we can say that P − m = P − m ′ = P .As stated in Theorem 2.1, we know that c ( P ) = 3 . In the next two lemmas, we show that althoughtwo cops do not have a winning strategy, they have a lot of power as to which positions can be reached.These lemmas would be very easy to establish computationally, but we consider that formalizing thestrategies is worthwhile. -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 7 Lemma 4.3. If { x, y, z } is a strong stable set of P i − m , then there exists a strategy for 2 cops on P i to reach the following situation.(1) The robber is on x , except possibly in the case x = m ′ and i ∈ { , } , where the robber iseither on m ′ or m .(2) The cops are on y and z .(3) It is the cops’ turn.Proof. Let us first consider the case of P . Consider the labelling of P with α i and β i as shown inFigure 1. Observe that if two cops are on β j , β j +1 (working in modulo 5), they can directly move toany pair β j ′ , β j ′ +1 .Without loss of generality, we may consider that x = α , y = β , z = β , as for all other strongstable sets we can apply the automorphism of Lemma 4.2(a). For some k , we start the game withtwo cops on β k and β k +1 (modulo 5). Notice that if the robber is on α j , moving the cops to β j and β j +1 forces the robber to move to α j − . By repeating this strategy, the cops can essentially make therobber turn in circles on the outer 5-cycle of P . At the end of every cops’ turn (except the first), therobber is on α j and the cops are on β j and β j +1 , for some j . The cops repeat until the robber is on α : it is now the cops’ turn and the game is in the desired situation. Observe that this strategy worksfor any initial choice of k . This will be useful later, as for any vertex w ∈ P , we may choose an initialposition such that one of the cops is in N [ w ] . We call this the chasing strategy for the Petersen graph.An example is illustrated in Figure 3. Even though this might be a very simple idea, this strategy iscritical for the rest of this section as it enables more complicated strategies. r c c (a) Initial position r c c (b)
After 1 cop turn r c c (c)
After 1 robber turn r cc (d)
After 2 cop turns rcc (e)
Desired position
Figure 3.
Typical application of the chasing strategy on the Petersen graph.We now consider the cases of P and P . In both cases, observe that m and m ′ are completelyindistinguishable: N ( m ) = N ( m ′ ) . It is then easily seen that the strategy for 2 cops on P or P will be the same as the strategy developed above for P , except that the robber may choose to go toeither m or m ′ (we apply the strategy for P by considering the robber to be on m ′ whenever it isactually on m ). This is essentially a watered-down version of the well-known argument used to prove,in particular, Theorem 2.4.Finally, we consider the cases P i , i ∈ { , . . . , } . Our goal is to apply the strategy of P developedabove, with only slight modifications. Using that strategy, we choose initial positions for the cops in JÉRÉMIE TURCOTTE AND SAMUEL YVON P i − m such that one of the cops is in N [ m ′ ] (it is described above why this is possible). If the robberchooses m as an initial position, this cop may then move to m ′ . As m ′ corners m , the robber cannotmove without being captured. The other cop may then, within a few turns, capture the robber. Thus,the robber will choose an initial vertex in P i − m . Now, as long as the robber is not on m , copy thestrategy for P . Suppose that, at some point, the robber moves to m . In the strategy above, therobber is adjacent to a cop before every of its turns. Thus, this cop can move to a vertex adjacentto m . One easily verifies that in all graphs P i for i ∈ { , . . . , } , if one cop is adjacent to m , there isat most one other escape route for the robber. As P i − m ≃ P has diameter 2, the other cop canmove to block this escape route. Thus, while applying this strategy, the robber will never move to m .Hence, the strategy copied from P yields the desired final position. (cid:3) By weakening the condition that it is the cops’ turn at the end of the strategy, we can get morefreedom as to where we can place the cops, enabling more strategies.
Lemma 4.4. If x, y, z are any three distinct vertices of P i − m , then there exists a strategy for 2 copson P i to reach the following situation.(1) The robber is on x , except possibly in the case x = m ′ and i ∈ { , } , where the robber iseither on m ′ or m .(2) The cops are on y and z .(3) It is the robber’s turn.Proof. Without loss of generality, we show the statement for P . For this lemma, generalizing to thecornered Petersen graphs is immediate.We first consider the case where xz ∈ E ( P ) . We will enumerate the main cases and conclude bysymmetry for the others. We may assume that x = α and z = β (using the labelling from Figure1), all other possibilities can be solved using the automorphisms of Lemma 4.2 (b).We apply Lemma 4.3 to place the robber on vertex α and the cops on the vertices specified inTable 2 (always forming a strong stable set), and then specify the additional move required to placethe cops in the desired final position.Final position for cops Position after applying Lemma 4.3 Movements α , β β , β β → α , β → β β , β β , β β → β , β → β β , β β , β β → β , β → β α , β α , β α → α , β → β Table 2.
Strategy on P to bring the robber to α with the cops in the desired finalposition, where at least one cop will be adjacent to robber.It is easily seen that all other choices of y are analogous by reflection of the graph relative to thevertical axis.We use a similar approach for the case where xz / ∈ E . We may suppose without loss of generalitythat x = α and z = β : it is easily verified that any two non-adjacent vertices can be expanded intoa strong stable set, then apply Lemma 4.2 (a). To further reduce the number of cases, we can alsoassume that xy / ∈ E (if xy ∈ E , switching the roles of y and z brings us back to the previous case), aswe can see in Table 3. (cid:3) In the next lemmas, we will consider consider graphs with the following properties, with the goalof eventually showing that these do not exist. We state these properties now to avoid repetition.
Hypothesis 4.5.
Let G be a connected graph and u be a vertex of G such that c ( G ) > and G − N [ u ] ≃ P i , for some ≤ i ≤ . -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 9 Final position for cops Position after applying Lemma 4.3 Movements α , β α , β α → α , β → β α , β α , β α → α , β → β β , β β , β β → β , β → β β , β β , β β → β , β → β β , β β , β β → β , β → β Table 3.
Strategy on P to bring the robber to α with the cops in the desired finalposition, where neither cop will be adjacent to robber.In the cases of ≤ i ≤ , we may in particular consider that m, m ′ ∈ V ( G ) by fixing the isomor-phism. In the cases of i = 5 , , as the labels m and m ′ can be switched, we will always suppose m ′ to be the vertex of the two which has the greatest degree in G (if both have the same degree, thenwe choose arbitrarily). To simplify notation, we will denote B u = V ( G − N [ u ] − m ) . It is easily seenthat in all cases h B u i ≃ P .The approach will be to build up a number of structural properties of G by showing that otherwisethere exists a winning strategy for cops, yielding a contradiction. We start by proving that allvertices in B u have a neighbour in N ( u ) . Lemma 4.6.
Consider Hypothesis 4.5. For all x ∈ B u , | N ( x ) ∩ N ( u ) | ≥ .Proof. Let x ∈ B u . Suppose that | N ( x ) ∩ N ( u ) | = 0 and that there exists a neighbour v of x in B u such that | N ( v ) ∩ N ( u ) | ≥ . If x is adjacent to m but x = m ′ , we also suppose v = m ′ (this additionalhypothesis will be useful later). We note that in the case with x = m ′ , then by our choice of m ′ weknow that m also has no neighbours in N ( u ) .We show this situation yields a winning strategy for 3 cops.Let y, z be the other neighbours of x in B u and let w ∈ N ( v ) ∩ N ( u ) . The situation is portrayedin Figure 4. We start by placing a cop on u , which will only move if the neighbour enters N [ u ] . Thisis commonly referred to as a stationary cop. As long as this cop stays on u , the robber is stuck in G − N [ u ] ≃ P i . Thus, the two other cops may apply the strategy from Lemma 4.4 on G − N [ u ] toplace the robber on x , a cop on y and a cop on z . In the special case where i ∈ { , } and x = m ′ ,the robber might actually be on m . z v xy m uwN ( u ) Figure 4.
Example situation during the proof of Lemma 4.6. Unused or unknownvertices and edges are omitted.
During the last turn of this strategy, the cop on u moves to w . It is now the robber’s turn. Inall cases, all of the robber’s neighbours in B u are protected by cops, there is a cop adjacent to therobber, and the robber has no neighbour in N ( u ) .The robber is caught unless it can move to an unprotected vertex outside B u (necessarily m ), whichonly happens if the robber is on x (we cannot be in the special case, as in this case m is adjacent to v ), if x is adjacent to m , and if m is adjacent to neither y or z . We may suppose this is the case.If x = m ′ , the cop on w moves back to u and the cop on y moves to m ′ : as m ′ covers all neighboursof m in B u and u covers all neighbours in N ( u ) , the robber is trapped and will be caught one turnlater.Now, suppose that x = m ′ , but that x is adjacent to m . In particular, x is also adjacent to m ′ .Recall that, in this case, we supposed that v = m ′ . Thus, m ′ is either y or z . The cop on m ′ staysand the cop on w moves back to u , trapping the robber on m . The last cop may then capture therobber within a few turns.In all cases, a contradiction is reached with the hypothesis that c ( ˆ G ) > .In other words, as soon as we know that a vertex of B u (other than m ′ ) has a neighbour in N ( u ) ,we can say the same for its 3 neighbours in B u . Thus, in order to prove the lemma, it suffices to showthat there exists at least one vertex of B u \ { m ′ } that has a neighbour in N ( u ) (because h B u i − m ′ isnecessarily connected).Suppose the contrary: no vertex of B u \ { m ′ } has a neighbour in N ( u ) . If we are in the case of G − N [ u ] ≃ P , then G would be disconnected (as B u \ { m ′ } = B u ), which is a contradiction. Inthe remaining cases, we explicit a winning strategy for 3 cops. Place a stationary cop on a vertex t ∈ N [ m ] ∩ N [ m ′ ] (one easily verifies that in all cases this set is non-empty), and place another on u (the third cop can be placed anywhere initially). The robber must choose an initial position in B u .As any exit from B u will go through m or m ′ (by our hypothesis), the stationary cop guarantees thatthe robber will never leave B u . The two other cops then have a winning strategy on B u \ N [ t ] , whichcontains at most 7 vertices. (cid:3) We now characterize the intersection of neighbourhoods of vertices in N ( u ) with B u . Lemma 4.7.
Consider Hypothesis 4.5. If w ∈ N ( u ) , then N ( w ) ∩ B u does not contain a subset { a, b, c } of distinct vertices such that :(1) ab / ∈ E ( G ) ;(2) c / ∈ N ( a ) ∩ N ( b ) ( c is not the common neighbour of a and b in h B u i );(3) c / ∈ N ( x ) for x ∈ N ( a ) ∩ N ( b ) ∩ B u ( c is not adjacent to the common neighbour of a and b in B u ).Proof. Suppose that N ( w ) ∩ B u does contains a subset { a, b, c } respecting these conditions. We explicita winning strategy for 3 cops on G , which will lead to a contradiction. We denote by x the commonneighbour of a and b in B u and by d the neighbour of x in B u that is neither a or b .Let z be a vertex of B u such that { x, c, z } is a strong stable set of h B u i (it is easily seen that anystable set of size 2 in the Petersen graph can be expanded into a strong stable set). The situation isportrayed in Figure 5.We place a cop on u at the start of the game, and then use Lemma 4.3 to place the other cops on c and z , and the robber on x . In the case where x = m ′ and i ∈ { , } and the robber is actually on m , switch the labels for m and m ′ . Notice that all three properties still hold after this switch.It is now the cops’ turn. The cop on c moves to w , the cop on z moves to either d or a neighbourof d (this is possible because the Petersen graph has diameter 2), and the cop on u stays still. Allneighbours of x in N ( u ) are covered by the cop on u , a and b covered by the cop on w , and d iscovered by the 3rd cop, which is either on d or on a neighbour of d . Thus, the robber either stays puton x or moves to m (if m exists).If the robber stayed on x and cannot be immediately captured, the cop which is adjacent to d movesto d . Now, the robber cannot stay put without being captured: we may then assume the robber movesto m . -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 11 dz a xc b m uwN ( u ) Figure 5.
Example situation during the proof of Lemma 4.7. Unused or unknownvertices and edges are omitted.If m ′ is a or b , the cop on w moves to m ′ . If m ′ = d , the cop that is on d or adjacent to d moves to(or stays on) d . In both cases, there is now a cop on m ′ , which, together with the cop on u , guaranteesthat the robber is now stuck on m . The third cop may capture the robber within a few turns.If m ′ = x , then, by definition, N ( m ) ∩ B u ⊆ { a, b, d, x } . As previously, move a cop to d (if it isnot already there). The pair of cops on d and w cover this set, hence the robber cannot move. At thenext cops’ turn, the cop on d moves to m ′ , and at the following turn capture the robber.In all cases, there is a contradiction as c ( G ) > . (cid:3) Lemma 4.8.
Consider Hypothesis 4.5. If w ∈ N ( u ) , then there exists a vertex of B u dominating N ( w ) ∩ B u .Proof. If | N ( w ) ∩ B u | ≤ , the result is trivial, as diam( h B u i ) = 2 .If | N ( w ) ∩ B u | ≥ , suppose the statement is false. As h B u i does not contain a triangle, not allvertices of N ( w ) ∩ B u can be pairwise adjacent: we can choose a, b ∈ N ( w ) ∩ B u such that a, b arenot adjacent. Denote x the common neighbour of a, b in B u . By our previous supposition, x does notdominate N ( w ) ∩ B u , thus we can choose c ∈ N ( w ) ∩ B u not adjacent to x . The subset { a, b, c } thencontradicts Lemma 4.7. (cid:3) In particular, every vertex of N ( u ) can have at most 4 neighbours in B u because h B u i is 3-regular.We also note that in some of the cases there is a unique choice for this dominating vertex, in particularwhen N ( w ) ∩ B u has 3 or 4 vertices. We are now ready to strengthen Lemma 4.6. Lemma 4.9.
Consider Hypothesis 4.5. For all x ∈ B u , the following holds.(1) If x / ∈ N [ m ′ ] , then | N ( x ) ∩ N ( u ) | ≥ .(2) If x ∈ N [ m ′ ] , then | N ( x ) ∩ N ( u ) | ≥ .Proof. (1) Suppose the contrary: there exists x ∈ B u \ N [ m ′ ] such that | N ( x ) ∩ N ( u ) | ∈ { , } (by Lemma 4.6, | N ( x ) ∩ N ( u ) | ≥ ). We explicit a winning strategy for 3 cops on G . Denoteby w , w the neighbours of x in N ( u ) (if there is only one neighbour, set w = w ) and by y , y , y the neighbours of x in B u .By Lemma 4.8, there exists a vertex of B u dominating the neighbourhood of w in B u . As x is in this neighbourhood, we know this dominating vertex (there might be more than onepossible choice) is in { y , y , y , x } . This is also true for w . Thus, we can pick at most 2elements of { y , y , y , x } that dominate all neighbours of w , w in B u . Without loss of generality (by symmetry of y , y , y in the Petersen graph), we assume the2 elements can be picked in { y , y , x } . By Lemma 4.6, y must have a neighbour t in N ( u ) .The situation is portrayed in Figure 6. y y xy m uw w tN ( u ) Figure 6.
Example situation during the proof of Lemma 4.9 (1). Unused or unknownvertices and edges are omitted.We place one cop on u . We use Lemma 4.4 to place the robber on x and the two othercops on y , y . During the last move of this strategy, the cop on u moves to t . It is now therobber’s turn.The robber on x cannot move to a neighbour inside of B u (there are cops on y and y ,and y is covered by the cop on t ) and there are cops adjacent to the robber. As x / ∈ N [ m ′ ] ,we know that m / ∈ N ( x ) . Thus, the robber has no choice but to move to either w or w .Without loss of generality, let us say the robber moves to w .Denote by a the vertex dominating the neighbours of w in B u . We recall that a is either y , y or x . We now move the cop on t back to u . Of the two cops on y and y , one must beable to move to a , and does so. If m ′ ∈ B u (that is, if we are not in the case of P ), the thirdcop moves to either m ′ or a neighbour of m ′ . After this move, all escapes in N ( u ) are coveredby the cop on u , all escapes in B u are covered by the cop on a , and the robber cannot staystill as u is adjacent to w . Thus, the robber is caught one move later unless m the robber canmove to m . In this case, the third cop can now move to m ′ and trap the robber. Leaving thecops on u and m ′ fixed, the cop on a can then go capture the robber. This is a contradictionas c ( G ) > .(2) As the proof will be very similar to the previous case, we outline the main differences.If x = m ′ and i ∈ { , } , suppose that m ′ and m each have at most one neighbour each in N ( u ) . Then, consider w and w these vertices and apply the same strategy as above. Eventhough the robber can choose to go to either m ′ or m , both cases will be covered. If therobber moves to m , then one of the cops moves to x = m ′ to trap it. Thus, either m or m ′ must have 2 or more neighbours in N ( u ) . As we have selected m ′ to have the greatest degreeof the two, the statement follows for this case.Consider that x ∈ N [ m ′ ] but x ′ = m ′ or i / ∈ { , } (as we covered that case above). Ourgoal is to prove that x cannot have a unique neighbour in N ( u ) . Suppose the contrary, wedenote by w this neighbour. As in the above strategy, one cop’s role will be to cover thevertex dominating the neighbourhood of w in B u , or if this x , then to be on an adjacentvertex. For the second cop, we will want it to be on m ′ , or on a neighbour of m ′ if x = m ′ . Inthe notation of the original case, this could be seen as setting w = m . If the robber moves -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 13 to m , then one of the cops will move to m ′ . Recalling that another robber will return to u ,the last cop will be able to go capture the robber. (cid:3) We are now ready to prove the desired results.
Proposition 4.10. If G is a connected graph such that ∆ ∈ { n − , n − } and n ≤ , then c ( G ) ≤ .Proof. (1) We consider ∆ = n − . Let u be a vertex of maximum degree. We know that | V ( G − N [ u ]) | =10 .If G − N [ u ] is disconnected, every of it’s connected components has cop number at most2, as no connected component can contain at least 10 vertices. Applying Corollary 2.5 yieldsthe desired result. Otherwise, G − N [ u ] must be connected. Suppose that c ( G ) > . Then, c ( G − N [ u ]) > , and by Theorem 2.1, G − N [ u ] must be isomorphic to P .Then, G and u satisfies the conditions of Hypothesis 4.5.By Lemma 4.9, every vertex of B u has at least 3 neighbours in N ( u ) . As B u has 10 vertices,there are at least 30 edges between N ( u ) and B u .As n ≤ , we have that ∆ ≤ . By Lemma 4.8, every vertex of N ( u ) has at most neighbours in B u . Thus, there are at most ≤ edges between N ( u ) and B u .This is a contradiction, as we have claimed there are at least 30 but at most 28 edgesbetween N ( u ) and B u . Thus, c ( G ) ≤ .(2) We consider ∆ = n − . Let u be a vertex of maximum degree. We know that | V ( G − N [ u ]) | =11 .Let us first consider the case where G − N [ u ] is disconnected. If every connected componenthas cop number at most 2, then, as in the previous case, we are done. By Theorem 2.1, theonly other case is if one component is isomorphic to P and the other is an isolated vertex x . By applying Corollary 2.5, c ( G ) ≤ if and only if c ( G − x ) ≤ . As G − x satisfies theconditions of the previous case of this proposition, we conclude that c ( G − x ) ≤ .We may now consider that G − N [ u ] is connected. By Proposition 3.2, G − N [ u ] ≃ P i ,for some ≤ i ≤ . Suppose that that c ( G ) > . Then, G and u satisfies the condition ofHypothesis 4.5.By Lemma 4.9, every vertex of B u has at least 2 neighbours in N ( u ) , and each vertex notin N [ m ′ ] (of which there are at least 6) has at least 3 neighbours in N ( u ) . In total, there areat least 26 edges between N ( u ) and B u .By Lemma 4.8, each vertex of N ( u ) has at most neighbours in B u . As n ≤ , wehave that ∆ ≤ . Thus, there are at most ≤ edges between N ( u ) and B u . This is acontradiction, as we have claimed there are at least 26 but at most 24 edges between N ( u ) and B u . (cid:3) These results will be used to prove that M = 19 , but we would also like to reduce the number ofpossible 4-cop-win graphs on 19 vertices. This will be possible with more work, but we first need thefollowing definition. Definition 4.11.
Consider Hypothesis 4.5. Let w ∈ N ( u ) such that | B u ∩ N ( w ) | = 4 . The vertex x of B u dominating B u ∩ N ( w ) will be called the projection of w . If x is the projection of k vertices of N ( u ) , we will call p ( x ) = k the projection multiplicity of x . By Lemma 4.8, this is well defined and the projection of a vertex is unique.
Observation 4.12. If x ∈ B u , | N ( x ) ∩ N ( u ) | ≥ P y ∈ N [ x ] ∩ B ( u ) P ( y ) . In particular, if y ∈ B u hasprojection multiplicity k , then each vertex in N ( y ) ∩ B u has degree at least k . Proof.
We recall that when a vertex y has projection multiplicity k , this means that k vertices of N ( u ) have for neighbours in B u exactly N [ y ] ∩ B u , giving each vertex in this set at least k neighboursin N ( u ) .Noting that the projection of a vertex is unique, we see that the neighbours x inherits from eachprojection on it or it’s neighbours in B u are pairwise distinct. The lower bound follows immediatelyby summing the projective multiplicity for each vertex in N [ x ] . (cid:3) We now see an interesting property of projections.
Lemma 4.13.
Consider Hypothesis 4.5. Let x ∈ B u \ N [ m ′ ] .(1) If | N ( x ) ∩ N ( u ) | = 3 , then p ( x ) = 0 .(2) More generally, p ( x ) ≤ | N ( x ) ∩ N ( u ) | − .Proof. (1) Suppose the contrary, we explicit a winning strategy for 3 cops.Suppose that x is the projection of a vertex w of N ( u ) : w is adjacent to x and to eachneighbour of x in B u .As | N ( x ) ∩ N ( u ) | = 3 , x has two other neighbours in N ( u ) , which we will denote by t and t . If t has a neighbour in B u other than x , choose one and denote it r . If not, thenchoose r to be any neighbour of x in B u . We choose r similarly. The situation is portrayedin Figure 7. r r x ut wt N ( u ) Figure 7.
Example situation during the proof of Lemma 4.13 (1). Unused or un-known vertices and edges are omitted.We start by placing one cop on u . Using Lemma 4.4 (recall that x = m ′ , which avoids theexceptional case), we place the robber on x and the two other cops on r and r . During thelast move of that strategy, move the cop from u to w . It is now the robber’s turn.As there is a cop on w , the robber cannot stay in B u . As x / ∈ N [ m ′ ] , x is not adjacentto m . If t had a neighbour in B u other than x , then the cop on r blocks the robber frommoving to t , and similarly for t .Thus, the only scenario in which the robber does not get captured immediately after movingis if (without loss of generality), t only has one neighbour in B u , and the robber moves to t . In this case, the cop on r is adjacent to x . The cop on w moves back to to u , and thecop on r moves to x . The third cop (on r ) moves to m ′ or a neighbour of m ′ . The robber iscaught one turn later, unless it can go to m . In this case, the third cop can move to m ′ andtrap the robber. The cop on x can capture the robber within a few turns. This contradictsthat c ( G ) > . -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 15 (2) The strategy is similar to the previous case. Suppose to the contrary that x has projectionmultiplicity at least | N ( x ) ∩ N ( u ) | − . Then, there is at most 1 neighbour of x in N ( u ) which does not project onto x . Choose t to be this vertex (if there is any) and select thecorresponding r as above. Choose r to be any other neighbour of x in B u : t covers allvertices projecting onto x . The rest of the strategy is identical. (cid:3) We are now ready for the desired result.
Proposition 4.14. If G is a connected graph such that n = 19 and ∆ ∈ { , } , then c ( G ) ≤ .Proof. Suppose c ( G ) > . Let u be a vertex of maximal degree in G .(1) We consider ∆ = 8 . Recall the arguments of the proof of Proposition 4.10. In particular, wecan consider that G − N [ u ] ≃ P .There are at most
4∆ = 32 edges between N ( u ) and B u , by Lemma 4.8. By Lemma 4.9,each vertex in B u has at least 3 neighbours in N ( u ) : there are at least 30 edges between B u and N ( u ) . Thus, there are at most 2 extra edges. By extra edges, we mean that there areedges which, if removed, would leave each vertex in B u with exactly the lower bound numberof neighbours in B u , as specified in Lemma 4.9. Then, there are at least 8 vertices in B u incident to exactly 3 such edges.Furthermore, if there are fewer than vertices of N ( u ) that each have exactly 4 neighboursin B u , then there cannot be at least 30 edges between N ( u ) and B u . Thus, P x ∈ B u p ( x ) ≥ .Recall that Lemma 4.13 states that no vertex in N ( u ) with 3 neighbours in B u = B u \ N [ m ′ ] can be a projection. If all vertices of B u have exactly 3 neighbours in N ( u ) , this is a directcontradiction.Otherwise, there are at most 2 vertices which can have non-zero projective multiplicity.Denote them by a , a (if there is only one vertex, a = a ). Then, p ( a ) + p ( a ) ≥ . Let x ∈ N [ a ] ∩ N [ a ] ∩ B u (which exists as P has diameter 2). As x is adjacent to all projections, x must be adjacent at least 6 vertices of N ( u ) (observation 4.12). As x also has 3 neighboursin B u , the degree of x is at least , which is a contradiction.(2) We consider ∆ = 7 . Recall the arguments of the proof of Proposition 4.10. In particular, wecan say that G − N [ u ] ≃ P i , for some ≤ i ≤ .There are at most
4∆ = 28 edges between N ( u ) and B u , by Lemma 4.8. By Lemma 4.9,each vertex in B u \ N [ m ′ ] has at least 3 neighbours in N ( u ) and each vertex in B u ∩ N [ m ′ ] has at least 2 neighbours in N ( u ) : in total, there are at least 26 edges between B u and N ( u ) .Using the same argument as above, depending on the number of edges between B u and N ( u ) , we can find between 5 and 7 vertices in N ( u ) which have 4 neighbours each in B u , andthus the total projection multiplicity of B u is as follows.(a) 26 edges: P x ∈ B u p ( x ) ≥ (b) 27 edges: P x ∈ B u p ( x ) ≥ (c) 28 edges: P x ∈ B u p ( x ) ≥ Recall that Lemma 4.13 states that no vertex in B u \ N [ m ′ ] with 3 neighbours in N ( u ) canbe a projection. Also, if x ∈ B u \ N [ m ′ ] , p ( x ) ≤ | N ( x ) ∩ N ( u ) | − : if x has 4 neighbours in N ( u ) it can be the projection of at most 2 vertices.If all vertices in B u \ N [ m ′ ] have exactly 3 neighbours in N ( u ) , then this implies all projec-tions will be vertices in N [ m ′ ] : at least 5 vertices project on m ′ or on a neighbour. Thus, m ′ will have at least 5 neighbours in N ( u ) . As m ′ also has at least 3 neighbours in B u , d ( m ′ ) ≥ ,which is impossible as ∆ = 7 . This situation includes the case in which there are exactly 26edges between B u and N ( u ) .Suppose there is exactly one vertex x of B u \ N [ m ′ ] with exactly 4 neighbours in N ( u ) ,with all others having exactly 3. This vertex will have projection multiplicity at most 2, sothe total projection multiplicity of vertices of N [ m ′ ] is at least 4. Thus, m ′ will have at least N ( u ) , which is impossible as this would imply there are 29 edges between B u and N ( u ) (the initial 26, 1 more from x and 2 more from m ′ ).Suppose now there are 2 vertices x , x of B u \ N [ m ′ ] with 4 neighbours in N ( u ) . Thesetwo additional edges bring the total to 28. Thus, the total projection multiplicity is at least7. There are at most 2 vertices projecting onto x and 2 vertices projecting on x . Thus, atleast 3 vertices project onto vertices in N [ m ′ ] . This a contradiction, as m ′ must have exactly2 neighbours in N ( u ) , otherwise there would be more than 28 edges between B u and N ( u ) .Considering that with ∆ = 7 , no vertex of B u can have 5 or more neighbours in N ( u ) , thereare no cases left.In all possible cases, a contradiction was found: we therefore conclude that G − N [ u ] P i , ≤ i ≤ ,and thus c ( G ) ≤ . (cid:3) Graphs with maximum degree 3
In this section, we consider the cop number of graphs with maximum degree 3. We start with themain result of this section.
Proposition 5.1. If G is a connected graph such that ∆ ≤ and n ≤ , then c ( G ) ≤ .Proof. We first prove the statement for δ ≥ . For ≤ n ≤ , we generate all graphs such that δ ≥ and ∆ ≤ . We then classify each graph according to its cop number. We present the results inTable 4, which shows that no such graph with cop number at least 4 exists. We have also extractedthe 3-cop-win graphs. Cop number n G : δ ≥ , ∆ ≤ ≥
458 7 450 1 0 Table 4.
Cop number breakdown for connected subcubic graphs.We now considers graphs which contain vertices of degree 1. We know that removing a vertex ofdegree 1 from a graph does not change the cop number (as the vertex of degree 1 is cornered by itsneighbour). We successively remove vertices of degree 1 from the graph. We eventually either getto a graph such that δ ≥ and n ≥ (in which case the above results can now be applied) or weeventually get to a graph of order at most 9 (in which case we apply Theorem 2.1). (cid:3) Notwithstanding the slight improvement of considering δ ≥ , the approach here is clearly far fromoptimal. The algorithm described in the following section is an example of a possibly better strategy.However, as we will see, this algorithm would not be the most efficient for maximum degree 3 : tocompute potential 4-cop-win graphs on 19 vertices, one would still need to compute subcubic 3-cop-win graphs on 15 vertices. A potentially more interesting algorithm for building possible 4-cop-winsubcubic graphs would consist in building graphs around long shortest paths, see Lemma 4 in [2],which describes how a cop can protect a shortest path. Nonetheless, our exhaustive testing approach -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 17 is not without it’s advantages, as we can use it to gain further knowledge on the cop number of smallgraphs.In fact, Hosseini, Mohar and Gonzalez Hermosillo de la Maza [3] have recently showed that studyingthe cop number for graphs with ∆ ≤ is of interest for the study of the cop number at large. Inthis regard, we consider that getting a distribution of the cop-number of small subcubic graphs mightbe interesting, even if it is somewhat skewed by adding the condition δ ≥ . Our calculations showthat not only there are no 4-cop-win subcubic graphs on at most 20 vertices, but that that subcubic3-cop-win graphs are overwhelmingly rare for these orders.The exhaustive search approach also gives us progress on a related problem. Arguably, the mostwell-known result on the game of cops and robbers is Aigner and Fromme’s proof that the cop numberof any planar graph is at most 3, see [2]. This yields the analogous question of finding the minimumorder of 3-cop-win planar graphs, and an enumeration of such graphs. The smallest known planar3-cop-win graph is the dodecahedral graph, see Figure 8. It is easy to see that this graph requires 3cops, as it has girth 5 and is 3-regular. It has been conjectured that the dodecahedral graph is theunique smallest 3-cop-win planar graph (for instance in [13], although the article claims the conjectureto have been originally formulated in [5], a claim which we were not able to verify). Figure 8.
The dodecahedral graph There are some partial results for this problem. In [19], Hosseini proved that a minimal 3-cop-winplanar graph must be 2-connected. Furthermore, Pisantechakool and Tan have shown in [25] that anyplanar graph on 19 or fewer vertices must contain a winning position for 2 cops, although it has notbeen proved that the cops can bring the game to this winning state. Using the computations in theproof of Proposition 5.1, we are able to get more evidence supporting the conjecture.
Corollary 5.2. If G is a connected planar graph such that ∆ ≤ and n ≤ , then c ( G ) ≤ , unless G is the dodecahedral graph.Proof. We simply test the 3-cop-win graphs found in the proof of Proposition 5.1 for planarity [21].The only such graph which was planar was the dodecahedral graph. (cid:3) Remaining cases
In this section, we consider the few remaining cases needed to prove that M = 19 , and also worktowards reducing the possible 4-cop-win graphs on 19 vertices. More precisely, we consider graphssuch that n = 17 with ∆ = 4 , n = 18 with ∆ = 4 , , and n = 19 with ∆ = 4 . Computer-generated drawing [21].
As in Section 4, our main tool will be knowing that if a graph G is 4-cop-win, then for each vertex u , c ( G − N [ u ]) ≥ . We know there are relatively few such graphs. Since we will be attempting toconstruct minimal 4-cop-win graphs, we know that c ( G − N [ u ]) < and so c ( G − N [ u ]) = 3 . In thecases of ∆ = n − or ∆ = n − , these graphs were only the Petersen and cornered Petersen graphs.As these were very few and very similar, we were able to build structural properties that allowedus to show that the graphs were not 4-cop-win. As they had somewhat a large maximum degree, acomputational approach would have been difficult due to the fact that there are too many possibleedges we need to consider.For the cases we will now consider, a computational approach is possible. On the other hand, aformal approach would be difficult, although certainly not impossible given a sufficient amount of time.Most graphs found in Lemma 3 contain the Petersen graph as an induced subgraph, so modifyingthe strategy to take these vertices into account would most likely be possible. But, just as we saw,adding even a single vertex yields significant complications for the proof. This would only becomemore complex with multiple additional vertices. Furthermore, some of the graphs do not containthe Petersen graph as an induced subgraph, and would need to be considered separately. As a finalblow, this proof method would not scale very well, as the more vertices we add the further away fromPetersen graphs we stray. For these reasons, we have mostly investigated the computational approach.Our goal is to build graphs which are possibly 4-cop-win: graphs G for which we cannot say that c ( G ) ≤ simply by looking at G − N [ u ] for the vertices u of maximum degree. Throughout, we willcall these graphs candidate 4-cop-win graphs.The simplest idea, which we have briefly discussed in Section 3, would be simply to consider a3-cop-win graph G ′ on 12 or 13 vertices, add a vertex u of chosen maximum degree and its neighbour-hood, and then look at every possible ways of joining N ( u ) to G ′ by respecting the maximum degreecondition. Even by reducing the number of cases by isomorphism, the number of graphs to consider ismassive, especially in the case ∆ = 5 : we must be a tad smarter. We present the Merging Algorithmas a way to generate potential 4-cop-win graphs, which we then test using a standard cop-numberalgorithm.We briefly introduce some notation. In general, when considering a graph G and a vertex u , thedegree of u will always refer to the degree of G in u . If we want to discuss the degree of u in someinduced subgraph H , we will refer to it as the H -degree of u . In general, if we say there exists a vertexof H -degree r , we are also implicitly stating that this vertex is in H .6.1. Presentation of the Merging Algorithm.
Quick Overview.
Our approach to build candidate 4-cop-win graphs will be the following. Let v and v be non-adjacent vertices, which we will in general choose to be a pair with the highestpossible degree.Then, knowing the computational results of Section 3, we are able to determine every possibleoption for G = G − N [ v ] and G = G − N [ v ] . We denote by L and L the sets of 3-cop-win graphsin which G and G are respectively chosen from.We want to determine every possible graph G , with maximum degree ∆ , which can be formed withthis structure. We will call the process the Merging Algorithm, which we will now describe.6.1.2. Input of the Algorithm.
Integers n, D , D = ∆ and sets of isomorphism classes of graphs L and L , such that(1) the graphs in L and L are 3-cop-win and have maximum degree at most ∆ ,(2) the graphs in L have n − D − vertices and the graphs in L have n − D − vertices.6.1.3. Output of the Algorithm.
The algorithm returns all graphs G on n vertices and maximum degreeexactly ∆ which contain a pair of non-adjacent vertices v and v , with the following 4 properties.Denote G = G − N [ v ] and G = G − N [ v ] . Then,(1) v and v have degree respectively D and D ,(2) G ∈ L and G ∈ L , and -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 19 (3) for all other vertex u of degree ∆ , G − N [ u ] ∈ L , and(4) if D < D , then the set of vertices of G of maximum degree forms a clique and v and v have at least 1 common neighbour.Isomorphic graphs may be omitted from the results, as we are not interested in the precise labellingsof the graphs.6.1.4. Phase 1 of the Algorithm.
We first choose some G and G from L and L respectively, wewill repeat the rest of the algorithm for each possible choice of G and G . We also choose strictlypositive integers d and d such that D − d = D − d , d ≤ D and d ≤ D , we will also considerevery possible choice.We then consider every possible choice of v ∈ V ( G ) and v ∈ V ( G ) such that v has G -degree d and v has G -degree d (we can of course only choose v and v up to automorphism in G and in G ). For each choice of vertices, consider every possible way of identifying G − N [ v ] and G − N [ v ] , by computing every isomorphism between these graphs. If there are none, this branch ofthe algorithm simply doesn’t yield a graph. Using this identification, we may then merge the graphsby union, keeping the closed neighbourhoods of v and v distinct.If this process has created vertices of degree greater than ∆ , we throw out the graph, as the restof the algorithm can only raise the degree again, yielding graphs we do not want to consider.We now add vertices which are not in V ( G ) ∪ V ( G ) . The only vertices which are neither in V ( G ) and V ( G ) are those which are to be adjacent to both v and v . Thus, we add D − d = D − d common neighbours to v and v , which ensures that the degrees D and D are respected. All verticesof G are now in the graph, but there possibly exists some missing edges, which we will add in thesecond phase. The result of the current phase is called a "base graph". Illustrated in Figure 9 is sucha base graph.It is easily seen in Figure 9 that the construction implicitly partitions the vertices into six sets : { v } , N ( v ) \ N ( v ) , V ( G ) \ ( N [ v ] ∪ N [ v ]) , N ( v ) ∩ N ( v ) , N ( v ) \ N ( v ) , { v } . If two graphs G, G ′ aregenerated with the same properties (same choices of G , G , v , v but by choosing a different identi-fication), we may be able to reduce the number of cases to consider: if φ is an isomorphism between G and G ′ such that φ ( S ) = S for each S being one of these 6 (we call this a strong isomorphism),we can consider to these graphs to be duplicates: each base graph, once the algorithm is over, will betransformed into the same candidate 4-cop-win graphs (again, up to isomorphism).6.1.5. Phase 2 of the Merging Algorithm.
The goal of this phase is to complete the 4-cop-win candi-dates graphs. As the base graph contain all required vertices, we now need to add the missing edges.We know that we do not want to add any edge such that both ends are in G or both ends in G as these are chosen to be induced subgraphs of G . Furthermore, we have already created all incidentedges to either v or v , by giving them the desired number of neighbours. Thus, we only need toconsider adding edges which are either(1) between N ( v ) ∩ N ( v ) and { v , v } c , including edges with both ends in N ( v ) ∩ N ( v ) , or(2) between N ( v ) \ N ( v ) and N ( v ) \ N ( v ) .We proceed by considering every vertex (first those in N ( v ) ∩ N ( v ) , then those in N ( v ) \ N ( v ) )and creating a new graph for every possible subset of new edges. Of course, we only consider subsetssuch that the degree will be at most ∆ . We repeat this step on the new graphs for the next vertex.We are also able to reduce some cases by isomorphism in this case. As above, we consider two graphsto be equivalent if at any step in the process the two graphs can be related by some isomorphism whichhas the property that the final graphs they will generate will be identical, up to isomorphism. Theadditional consideration here is that we must distinguish vertices for which we have already consideredadding extra and those for which this remains to be done. This additional piece of information iscrucial to ensure we are indeed considering every possible set of additional edges. Thus, the conditionwill be that the isomorphism φ not only preserves the six sets as above, but also identifies the verticesin N ( v ) for which we have not yet run the second part of the algorithm with other vertices with the v v G G N ( v ) \ N ( v ) N ( v ) \ N ( v ) N ( v ) ∩ N ( v ) ( N [ v ] ∪ N [ v ]) c Figure 9.
Example of the first part of the Merging Algorithm. Here, the base graphwas generated using parameters n = 18 , D = D = ∆ = 5 and d = d = 3 .same property. As this procedure is often lengthy, we only apply this improvement on lists of graphsof reasonable length.After considering every possible way of adding edges, we can throw out all graphs G such that G − N [ u ] is not a 3-cop-win graph for every vertex u of maximum degree (by construction, we do notneed to verify this for v and v ). We can also also remove isomorphic graphs. We note that as wehave split up the computations in many pieces, we only compare graphs which were generated withthe same choice of G .6.1.6. Specific cases.
As one can deduce from the parameters and output section, the algorithm canbe divided in two main cases.In the first case, D = D . In other words, the resulting graphs contain non-adjacent vertices v and v of maximum degree. In general, we apply the merging for every possible way (up to automorphism)of choosing vertices v in G and v in G . Applying this naively may yield multiple isomorphic graphs, -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 21 as different choices of v can yield the same graph. This happens when the resulting graph containsa vertex of maximum degree which has multiple non-adjacent vertices of maximum degree. We cantweak the algorithm to partially avoid this problem.For each graph G in L , we first define a total ordering on it’s vertices as follows. We list thevertices by decreasing degree, where the choice of order on vertices of same degree is arbitrary exceptthat vertices which are equivalent by automorphism are consecutive in this order (or we could definethe ordering on the classes of vertices up to automorphism). Then, when considering some choiceof v , we will add in the remaining of the algorithm the restriction that vertices greater than v inthis order do not have maximum degree in G . We will do the computations by decreasing v . Inessence, the graphs G where the vertices which are greater in this order have maximum degree in G will already have been considered in the algorithm.A particular case of the above is when G already contains multiple vertices of degree ∆ . Whenchoosing any vertex v other than the vertex u which is maximal in the chosen order, no graph willbe generated: u would have degree ∆ in G , which we have excluded. For this reason, we not even tryand simply do the merging algorithm for one choice of v when G contains a vertex of degree ∆ .It is not directly obvious that this simplification is compatible with the one we described earlier,which was considering strongly isomorphic partially-constructed graphs equivalent. It suffices to seethat automorphically equivalent vertices of G have the same restriction on their degree: either bothare allowed to have degree ∆ in G or neither is. Indeed, this property is preserved when consideringthe isomorphism φ between two of the partially constructed graph : as φ preserves in particular { v } , N ( v ) \ N ( v ) and V ( G ) \ ( N [ v ] ∪ N [ v ]) , we know that φ restricted to the vertices of G is anisomorphism of G .In the second case of the algorithm, D < D . We no longer apply the improvements to the MergingAlgorithm we described in the previous case, but we still apply some minor modifications to the basealgorithm to fulfill the condition (4) of Section 6.1.3. Observe that at any point in the algorithm, ifthe graph contains non-adjacent vertices both of degree ∆ , we can throw out this graph, since thevertices will also be non-adjacent in the final graphs. We also only test for choices of d such that d < D (and thus d < D ) to only build graphs where v and v have common neighbours.6.1.7. Validity of the Algorithm.
Considering the algorithm itself is relatively straightforward, we donot present a complete proof of the validity of the algorithm. We however present a few key pointstowards a formal proof.Consider a graph G respecting the conditions described in the Section 6.1.3. Choose v to be anyvertex of degree D in G such that G − N [ v ] contains at least one vertex respecting the conditions for v in Section 6.1.3. Denote S this non-empty set of possible choices for v . Once this choice is made,we set G = G − N [ v ] , G = G − N [ v ] , D = d ( v ) , D = d ( v ) , d = d G ( v ) and d = d G ( v ) .In the case D < D , we choose v to be any vertex of S . If D = D , then choose v to be maximalin S relative to the order on the vertices of G − N [ v ] as described in the previous section.It is easy to verify that d − d = d G ( v ) − d G ( v )= | N G − N [ v ] ( v ) | − | N G − N [ v ] ( v ) | = | N ( v ) \ ( N [ v ] ∩ N ( v )) | − | N ( v ) \ ( N ( v ) ∩ N [ v ]) | = | N ( v ) \ ( N ( v ) ∩ N ( v )) | − | N ( v ) \ ( N ( v ) ∩ N ( v )) | = ( | N ( v ) | − | N ( v ) ∩ N ( v ) | ) − ( | N ( v ) | − | N ( v ) ∩ N ( v ) | )= | N ( v ) | − | N ( v ) | = D − D . Thus, the pair of degrees d and d is indeed considered in the Merging Algorithm. It is then easyto see that the first part of the algorithm has considered this case. In particular, for the case D = D ,choosing v as maximal in S implies that all vertices of G − N [ v ] which are greater in this order donot have maximal degree in G , which is consistent with the simplification that we implemented. Then, in the second part of the algorithm we consider adding every possible edge not totallycontained in G or G (or at least, up to isomorphism), while still respecting some degree conditions(which we have just seen to be consistent). We thus see that G has indeed been constructed by theMerging Algorithm.6.2. Results.
We will now use this algorithm to build all possible 4-cop-win graphs. We will use someadditional heuristics in some cases to reduce the number of cases to consider, which we will explain indetail in the proof of the following proposition. Our implementation of the algorithm is done in theWolfram language [21].
Proposition 6.1.
Let G be a connected graph such that either(1) n = 17 and ∆ = 4 ,(2) n = 18 and ∆ ∈ { , } , or(3) n = 19 and ∆ = 4 .If every proper induced connected subgraph H of G respects c ( H ) ≤ , then c ( G ) ≤ , unless G is theRobertson graph.Proof. Let u be any vertex of G . We know that G − N [ u ] has at most n − ≤ vertices. If G − N [ u ] is disconnected, it must contain at least one component K which has at most 8 vertices.Then, Theorem 2.1 implies that c ( K ) ≤ . Furthermore, our hypothesis implies that c ( G − K ) ≤ ,as G − K is necessarily a connected induced subgraph of G . By Corollary 2.5, we get that c ( G ) ≤ max { c ( G − K ) , c ( K ) + 1 } ≤ . Thus, for the remainder of the proof, we assume that for every vertex u , G − N [ u ] is connected.Likewise, we can assume that G does not contain a corner x . Indeed, G − x is necessarily connectedand has cop number at most 3, therefore Corollary 2.8 would then imply that G also has cop numberat most 3.We may also assume that c ( G − N [ u ]) = 3 : if c ( G − N [ u ]) ≤ , placing a stationary cop on u implies that c ( G ) ≤ .Before going further, we define a property P with the usual definition: a property P is a functionfrom a set to a Boolean value. For instance, if C is the property of being 3-cop-win, then C ( P ) iswhether the Petersen graph is 3-cop-win (which is true).With this language, we can bring together the last assumptions. We define property M as follows : G is a graph respecting the hypotheses of the proposition such that G − N [ u ] is a connected 3-cop-wingraph for every vertex u of G and such that G does not contain a corner. By the previous discussion,it suffices to show the proposition for graphs respecting M .We now define property P . A graph G is said to have property P if G contains two non-adjacentvertices of degree ∆ . We use the Merging Algorithm to generate all graphs G such that M ( G ) that respect property P , and then compute their cop numbers. More precisely, we choose n and ∆ according to the case we are consider, D = D , and L = L to be the set of 3-cop-win graphson n − ∆ − vertices with maximum degree at most ∆ , as computed in Lemma 3.3. We note thatthe Merging Algorithm computes a somewhat larger class of graphs than we want. In particular,the Merging Algorithm does not exclude graphs which contain corners and in it’s last step only tests G − N [ u ] for vertices of maximum degree.The summary results are presented in Table 5. For more detail, we also split up the graphs relativeto the various possible maximum degrees of G , although we of course always merge with all of thepossible graphs G , not only the G with the same maximum degree. We note that there are no3-cop-win graphs with maximum degree 3 on 13 vertices (which can also be seen in Table 4), and thatthe 3-cop-win graphs with maximum degree 3 on 11 and 13 vertices are 3-regular (and thus the onlypossible value of d is 3).We note that the 4-cop-win graphs found on 19 vertices are actually all copies of the Robertsongraph, which can be see in Figure 1. In fact, the 3 copies correspond to 3 different choices of G whichcan yield the Robertson graph. -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 23 Cop number ∆ n ∆ G d Base graphs Final graphs ≥
123 0 0 0 0 0
10 0 0 0 0 018
911 0 0 0 0 05 18
534 18645 0 3455 15190 0
111 24238 0 1494 22744 0
88 698809 0 82882 615927 0
22 12778 0 4960 7818 0
Table 5.
Results of the first wave of computations using the Merging Algorithm.It presents the counts for the graphs built with the property that they contain 2non-adjacent vertices of maximum degree. In particular, d = d and ∆ = D = D .Furthermore, G is chosen with maximum degree ∆ .It is also interesting to note that for all cases with ∆ = 4 , not only is the Robertson graph theonly 4-cop-win graph, but there are no other candidate 4-cop-win graphs. It would appear that whenmerging, too many vertices of high degree are created: either a vertex of degree 5 or more is created(in which case the graph is immediately thrown out) or there are "too many" vertices of degree 4,such that there is always some u of maximum degree and G − N [ u ] not 3-cop-win.With these results, we will then only consider graphs which do not have property P . In otherwords, the graphs left to consider are those such that the set of vertices of maximum degree of G forms a clique. This is a very restrictive property, and will be very useful.Note that graphs G such that M ( G ) and ∆ = 4 respect property P : let u be a vertex of maximumdegree in G . Consider G ′ = G − N [ u ] . If G ′ contains a vertex of degree 4, P ( G ) is satisfied. Otherwise,we must have ∆( G ) = 3 . If G ′ is not 3-regular, it is at most 2 cop-win (by the results mentionedabove) and therefore G is not a 4-cop-win candidate. Therefore, any vertex in G ′ that was adjacentto a vertex of N ( u ) is also of degree 4 in G and not adjacent to u . Thus, P ( G ) is verified. We cantherefore suppose ∆( G ) = 5 . Furthermore, since P ( G ) is false, we can assume that if there existstwo vertices of maximum degree, they must be adjacent (as they must form a clique: otherwise, P issatisfied).We now define property P . We say a graph G has property P if G contains two non-adjacentvertices v and v such that v has degree either 3 or 4, v has degree , v and v have a commonneighbour, and G − N [ v ] has maximum degree at most . Then, we compute the graphs G such that M ( G ) and P ( G ) , but not P ( G ) . Precisely, we set n = 18 , D = ∆ = 5 , D to either or , L to bethe 3-cop-win graphs on vertices with maximum degree at most (if we choose G with maximumdegree 5, then the generated graphs automatically respect property P ), and L to be the 3-cop-wingraphs on respectively either or vertices with maximum degree at most . We have computedthese lists L and L in Lemma 3.3. The results of this computation are presented in Table 6.We note that as the number of possible vertices of maximum degree is generally smaller than before,there are fewer graphs thrown out because G − N [ u ] not being a 3-cop-win graph. Furthermore, asthe graphs on 14 vertices with maximum degree 3 are 3-regular, when choosing any of these graphs itis impossible for d to be anything other than .We see that none of the graphs are 4-cop-win. We claim that all graphs M ( G ) implies that either P ( G ) or P ( G ) .Let G be as graph such that M ( G ) . If P ( G ) ; we are done. Let us then consider that P ( G ) is falseand show that P ( G ) must be true. As discussed earlier, we may only consider the case where ∆ = 5 . Cop number D G ∆ G d Base graphs Final graphs ≥ Table 6.
Results of the second wave of computations with the Merging Algorithm.It presents the counts for the graphs G built with the property that G − N [ v ] hasmaximum degree 4, v and v always have a common neighbour (in particular d This is only one of many possible computational approaches to solvingthe problem. We now discuss a few improvements and alternatives that the interested reader maywant to apply.Our approach was based on merging 3-cop-win graphs relative to non-adjacent vertices v , v . Itis easy to see that one could instead choose v and v to be adjacent. Even if the construction wouldbe somewhat different, the ideas are similar. In particular, after proving that G does not containnon-adjacent vertices of maximum degree, we could have proved that G does not contain any adjacentvertices of maximum degree, instead of considering v of smaller degree. This would then leave onlythe case with a single vertex of maximum degree to be treated. With some additional heuristics orwith a simplification of the methods we used, this could be dealt with more specificity.We decided against this approach for few reasons. Although our approach required us to computemore 3-cop-win graphs than otherwise, it allowed us to implement only one Merging Algorithm.Furthermore, computing the 3-cop-win graphs on 14 vertices with maximum degree (at most) 4 allowedus to simultaneously handle on the case on n = 18 with d ( v ) = 3 , and build the candidate 4-cop-wingraphs on 19 vertices with maximum degree 4.Another method would be to not only merge graphs relative to pairs of vertices, but varying sizesof subsets. This approach would certainly reduce the number of intermediate graphs generated bythe algorithm: instead of pruning out graphs after adding edges, we could build up a larger partof the graph. The difficulty lies in implementing this approach. When merging only 2 graphs, it iseasy to determine between which vertices we possibly want to add edges. When merging more thantwo graphs, relative to either adjacent or non-adjacent vertices, we must keep track of which pairs ofvertices do not have an edge because it is not an edge in one of the G i , or because the pair has notbeen considered yet.Although at the expense of some computation time, we have chosen not to implement these im-provements in order to keep the code as simple as possible. Indeed, the simplicity of the code reducesthe chances of it being erroneous, as well as making it easier to verify. As the proof is completelydependent on the results of the algorithm, we felt this compromise was justified.A last idea essentially combines the processes of generating the graphs and testing their cop number.Let G be a connected graph and e be some edge of G . In general, it is unclear whether removing e will help the robber or help the cops, as this depends on many other factors. If we consider aslightly modified ruleset so that the robber can use the edge e but not the cops, we might achievesome results. Denote c ′ the cop number of this modified game. With these rules, c ( G ) ≤ c ′ ( G ) , asthe new edge can only benefit the robber. Furthermore, c ′ ( G − e ) ≤ c ′ ( G ) because, removing e canonly help the cops, as they were not allowed to use it anyways. Thus, both c ( G ) and c ( G − e ) arebounded above by c ′ ( G ) . If we modify the algorithm which calculates the cop number to take intoaccount the robber-only edge e (a fairly easy modification), we could then determine whether both G and G − e have cop number at most 3 simultaneously. This generalizes to larger subsets of edges.Afterwards, in theory, we reduce by a significant amount the number of cases to consider. It is notclear how many such "special edges" we can assign in G without the cop number increasing. We leaveimplementing and studying this approach as a problem. Modifying slightly the rules of the game tostudy the cop number has been done many times before. For instance, cop-only edges are studied in[18] and allowing the cops to teleport in [22]. Main results We are now ready to prove the desired results. Theorem 7.1. If G is a connected graph such that n ≤ , then c ( G ) ≤ .Proof. This is a direct consequence of Corollary 2.6 and Propositions 4.10, 5.1 and 6.1. (cid:3) Considering there exists a known 4-cop-win graph on 19 vertices, the Robertson graph, we get thefollowing corollary. Corollary 7.2. M = 19 We also want to narrow down the possible 4-cop-win graphs on 19 vertices. Theorem 7.3. Let G be a connected graph such that n = 19 . If ∆ ≤ or ∆ ≥ , then c ( G ) ≤ . If ∆ = 4 , then c ( G ) ≤ , unless G is the Robertson graph.Proof. This is a direct consequence of Corollary 2.6 and Propositions 4.14, 5.1 and 6.1. (cid:3) We leave filling the missing cases in this theorem as a conjecture. Conjecture 7.4. There does not exist a connected graph G such that n = 19 , ∆ ∈ { , } and c ( G ) = 4 . This would then show that the Robertson graph is the unique 4-cop-win graph on 19 vertices. Witha better implementation of the algorithm, in some low overhead programming language such as C,and with a few good ideas, this problem seems within reach. On the other hand, finding M with themethods used in this article is clearly unfeasible.It is asked in [5] whether the minimum d -cop-win graphs are ( d, -cage graphs for every d . Althoughwe now have further evidence pointing towards this conjecture, any general proof of this statement isstill beyond our grasp. Acknowledgments We would like to thank Seyyed Aliasghar Hosseini for introducing us to this problem and for manyhelpful discussions. We would also like to thank Ben Seamone and Geňa Hahn for their support overthe course of this project.We also acknowledge that most computations were done on the Université de Montréal - Départe-ment de mathématiques et de statistique computer network.The authors are partially supported by the Natural Sciences and Engineering Research Council ofCanada (NSERC) and the Fonds de Recherche du Québec - Nature et technologies (FRQNT). References [1] A. Afanassiev. Cop-Number. GitHub, Available at https://github.com/Jabbath/Cop-Number , 2017-2020.[2] M. Aigner and M. Fromme. A game of cops and robbers. Discrete Applied Mathematics , 8(1):1 – 12, 1984.[3] S. Aliasghar Hosseini, B. Mohar, and S. Gonzalez Hermosillo de la Maza. Meyniel’s conjecture on graphs of boundeddegree. arXiv e-prints , page arXiv:1912.06957, Dec 2019.[4] J. Baez. Petersen graph. Available at https://blogs.ams.org/visualinsight/2015/07/01/petersen-graph/ ,Jul 2015.[5] W. Baird, A. Beveridge, A. Bonato, P. Codenotti, A. Maurer, J. McCauley, and S. Valeva. On the minimum orderof k-cop-win graphs. Contributions to Discrete Mathematics , 9:70–84, 2014.[6] W. Baird and A. Bonato. Meyniel’s conjecture on the cop number: a survey. Preprint, Available at https://math.ryerson.ca/~abonato/papers/meyniel_0311.pdf , 2011.[7] A. Berarducci and B. Intrigila. On the cop number of a graph. Advances in Applied Mathematics , 14(4):389 – 403,1993.[8] A. Beveridge. The petersen graph is the smallest 3-cop-win graph. GRAScan Workshop, Available at https://math.ryerson.ca/~abonato/GRASCan/AndrewBeveridge.pdf , 26 May 2012.[9] A. Beveridge, P. Codenotti, A. Maurer, J. McCauley, and S. Valeva. The petersen graph is the smallest 3-cop-wingraph. Preprint, Available at https://arxiv.org/pdf/1110.0768v1.pdf , 2011.[10] J. Bezanson, A. Edelman, S. Karpinski, and V. B. Shah. Julia: A fresh approach to numerical computing. SIAMreview , 59(1):65–98, 2017. -COP-WIN GRAPHS HAVE AT LEAST 19 VERTICES 27 [11] A. Bonato. Conjectures on Cops and Robbers , pages 31–42. Springer International Publishing, Cham, 2016.[12] A. Bonato, E. Chiniforooshan, and P. Prałat. Cops and robbers from a distance. Theoretical Computer Science ,411(43):3834 – 3844, 2010.[13] A. Bonato and B. Mohar. Topological directions in cops and robbers. Journal of Combinatorics , 11, 09 2017.[14] A. Bonato and R. Nowakowski. The Game of Cops and Robbers on Graphs . 09 2011.[15] P. Bradshaw, S. A. Hosseini, B. Mohar, and L. Stacho. On the cop number of graphs of high girth, 2020.[16] N. E. Clarke and G. MacGillivray. Characterizations of k-copwin graphs. Discrete Mathematics , 312(8):1421 –1425, 2012.[17] P. Frankl. Cops and robbers in graphs with large girth and cayley graphs. Discrete Applied Mathematics , 17(3):301– 305, 1987.[18] P. Frankl. On a pursuit game on cayley graphs. Combinatorica , 7(1):67–70, Jan. 1987.[19] S. A. Hosseini. Game of Cops and Robbers on Eulerian Digraphs . PhD thesis, Simon Fraser University, 2018.[20] S. A. Hosseini. A note on k-cop-win graphs. Discrete Mathematics , 341(4):1136 – 1137, 2018.[21] W. R. Inc. Mathematica, Version 12.1. Champaign, IL, 2020.[22] F. Lehner. On the cop number of toroidal graphs, 2019.[23] B. D. McKay and A. Piperno. Practical graph isomorphism, ii. Journal of Symbolic Computation , 60:94 – 112,2014.[24] R. Nowakowski and P. Winkler. Vertex-to-vertex pursuit in a graph. Discrete Mathematics , 43(2):235 – 239, 1983.[25] P. Pisantechakool and X. Tan. On the conjecture of the smallest 3-cop-win planar graph. In T. Gopal, G. Jäger,and S. Steila, editors, Theory and Applications of Models of Computation , pages 499–514, Cham, 2017. SpringerInternational Publishing.[26] A. Quilliot. Problèmes de jeux, de point fixe, de connectivité et de représentation sur des graphes, des ensemblesordonnés et des hypergraphes . PhD thesis, Université de Paris VI, 1978.[27] J. Rickert. Cops and Robbers, Bachelor Thesis, 2017.[28] N. Robertson. The smallest graph of girth 5 and valency 4. Bull. Amer. Math. Soc. , 70(6):824–825, 11 1964.[29] J. F. Seth Bromberger and other contributors. JuliaGraphs/LightGraphs.jl: an optimized graphs package for theJulia programming language, 2017.[30] J. Turcotte and S. Yvon. Code and data complement - 4-cop-win graphs have at least 19 vertices. Available at , 2020. Départment de mathématiques et de statistique, Université de Montréal, Montréal, Canada E-mail address : [email protected] URL : Département d’informatique et de recherche opérationnelle, Université de Montréal, Montréal,Canada E-mail address ::