Limited Visibility Cops and Robbers
N.E. Clarke, D. Cox, C. Duffy, D. Dyer, S. Fitzpatrick, M.E. Messinger
LLimited Visibility Cops and Robber
N.E. Clarke D. Cox C. Duffy D. Dyer S. Fitzpatrick M.E. Messinger Mathematics and Statistics, Acadia University, Canada Mathematics, Mount Saint Vincent University, Canada Mathematics and Statistics, University of Saskatchewan, Canada Mathematics and Statistics, Memorial University of Newfoundland, Canada Mathematical and Computational Sciences, University of Prince Edward Island, Canada Mathematics and Computer Science, Mount Allison University, CanadaAugust 23, 2017
Abstract
We consider a variation of the Cops and Robber game where the cops can onlysee the robber when the distance between them is at most a fixed parameter (cid:96) . Weconsider the basic consequences of this definition for some simple graph families,and show that this model is not monotonic, unlike common models where therobber is invisible. We see that cops’ strategy consists of a phase in which theyneed to “see” the robber (move within distance (cid:96) of the robber), followed by aphase in which they capture the robber. In some graphs the first phase is themost resource intensive phase (in terms of number of cops needed), while in othergraphs, it is the second phase. Finally, we characterize those trees for which k cops are sufficient to guarantee capture of the robber for all (cid:96) ≥ Cops and Robber is a well-known and well-studied pursuit-evasion game played on agraph. The classical game is played on an undirected reflexive graph by two players:the first controls a set of cops and the second controls a single robber. The cops (any orall of which may move simultaneously) move so that some cop eventually captures therobber, while the robber moves to avoid the cops. In this game, the cops and robberare always aware of the others’ positions.Cops and Robber was independently introduced by Nowakowski and Winkler [20]and Quillot [22], with both papers characterizing those graphs for which one cop wassufficient to capture the robber. Much of the work surrounding this model has been1 a r X i v : . [ c s . D M ] A ug nvolved with characterizing those graphs for which exactly k cops are necessary andsufficient to capture the robber, with even the case k = 2 being nontrivial [9]. Beyondcharacterizations, there has also been work done on minimizing the length of the game,or the period for which the robber has been uncaptured [5].Many variants of Cops and Robber have been considered. (See the book [8] for theextensive survey of Cops and Robber.) Some of these variants are due to restrictions onhow the cops can move; for example, remaining adjacent to one another [10], or alwaysmoving along a shortest path to the robber [7, 15]. Other recent work has includedlimiting the cops’ knowledge of the robber’s position. In other words, exactly whenthe cops can see the robber is limited. When the robber is completely invisible, theresulting game looks a great deal like edge-searching , another pursuit-evasion game inwhich cops chase an invisible, fast-moving robber that may stop on vertices or edges.Edge-searching and node-searching are often used to give insight into the pathwidth of agraph, though the games themselves have spawned a great deal of research in their ownrespective rights. One variation is connected edge-searching, where the area cleaned bythe searchers must induce a connected subgraph. (See, for example, [26].)Loosely, the zero-visibility cop number of a graph G is the minimum number of copsneeded to guarantee the capture of an invisible robber, played under the standard Copsand Robber movement rules (so, unlike edge-searching, the robber moves at the samespeed as the cops). This problem has been studied for a variety of graph families,and more recent work has been done on its complexity and its relation to pathwidth[23, 24, 11, 12]. A variant where the robber is invisible but capture is not guaranteed,but expected to happen eventually, has also been studied [17].Of course, this problem naturally generalizes to the (cid:96) -visibility cop number , wherea cop can see any robber that is at distance at most (cid:96) . This problem is an unusualhybrid of the classic Cops and Robber game and the zero-visibility version, as now thecops’ task may be broken down into two sub-tasks: first, see the robber, and second,capture the robber. The one-visibility Cops and Robber game was considered in [23],and versions of the one-visibility game where cops are allowed to “jump” as in edge-searching were investigated in [1, 21].The focus of this paper is the general (cid:96) -visibility Cops and Robber game. Weconsider (cid:96) ≥
1, unless explicitly stated otherwise. In Section 2.1 we formally introducethe Cops and Robber game, the zero-visibility Cops and Robber game, and the (cid:96) -visibility Cops and Robber game. We also provide definitions for a number of conceptsused throughout the paper. In Section 2.2, we observe the relationships between the (cid:96) -visibility cop number for various values of (cid:96) and we determine the (cid:96) -visibility copnumber for some graph families. Additionally, we comment on the relationship betweenthe required number of cops on a graph and the number required on any retract, and onthe role of isometric trees in computing the (cid:96) -visibility cop number of a graph. Resultsin [20, 3, 23], among others, highlight the importance of distance preserving subgraphsand retracts in both the classical game and the zero-visibility game. This is also truein the (cid:96) -visibility game, as demonstrated in Section 2.2.Many pursuit-evasion games on graphs involve the concept of monotonicity: once2 set of vertices is known to be “robber-free”, then the cops can guarantee the set canremain “robber-free” throughout the game. Interestingly, and unlike edge-searching,we show the (cid:96) -visibility Cops and Robber game is not monotonic. We demonstrate thisfact in Section 3.In Section 4, we examine the relationship between the number of cops required tosee the robber, and for which graphs this is a sufficient number to capture the robber.We show for visibility at least two, that this is the case whenever the (cid:96) -visibility copnumber of a graph and its (regular) cop number differ by at least two. We also showthis to be true for all chordal graphs for any (cid:96) ≥ (cid:96) -visibilitynumber of a tree, based on the structure of the tree. We conclude with a series of openproblems in Section 6. The classical game of Cops and Robber is played on an undirected reflexive graph withtwo players: one player controls a set of cops and the other controls a single robber.(Typically, we will refer to the moves of the cops and robber, rather than of the twoplayers.) Before the game begins (considered to be on round 0), the cops choose a set ofvertices to occupy. The robber subsequently chooses an unoccupied vertex to occupy.At each round, the cops “move” and then the robber “moves”. In a “move” for the cops,each cop moves from their current vertex along an edge to a neighbouring vertex (theneighbouring vertex being the same as the initial vertex if a loop is traversed). A movefor the robber is defined similarly. Generally, we will neglect to consider loops, andinstead allow some (or all) of the cops to remain on the same vertex on an unreflexivegraph. This is referred to as a pass . The robber may also pass.The original game is played with perfect information: both the cops and robberare aware of their opponents positions in each round. The cops win if, in some round,one cop moves to the vertex occupied by the robber; in such a situation we say thecops have captured the robber. The robber wins if he avoids capture indefinitely. It isassumed that both the cops and the robber play optimally; the cops to minimize thenumber of rounds in the game, and the robber to maximize. The cop number of a graph G , denoted c ( G ), is the minimum number of cops required to capture the robber in G .The structural characterization of cop-win graphs is well known from [20, 22]: Agraph G is cop-win if and only if its vertices can be ordered v , v , . . . , v n so that for each v i , where i >
1, there exists some v j , where j < i , such that every N [ v i ] ∩ { v , . . . , v i } ⊆ N [ v j ]. This ordering is referred to as a cop-win ordering of G . Let H i be the subgraphinduced on { v , v , . . . , v i } . The vertex v i is referred to as a corner in H i and vertex v j is said to dominate v i in H i , whenever N H i [ v i ] ⊆ N H i [ v j ]. It follows that v n is a cornerin G , and H n = G . 3he rules of the (cid:96) -visibility game, where (cid:96) ≥
0, do not vary from the classical gamewhen it comes to both choice of positions in round 0 and the movement of the players.The condition for winning is also the same in both versions of the game. However, in the (cid:96) -visibility game, the cops do not have perfect information, while the robber does. Inthe (cid:96) -visibility game, we will say that the cops see the robber if any cop and the robbersimultaneously occupy some vertices x and y , respectively, such that d ( x, y ) ≤ (cid:96) . (Youcould think of the cop within distance (cid:96) seeing the robber and sharing that informationwith the other cops.) We note that for a graph G if (cid:96) ≥ diam ( G ), then (cid:96) -visibility Copsand Robber is identical to the classical game.Let c (cid:48) (cid:96) ( G ) and c (cid:96) ( G ) denote the minimum number of cops required to see and capture (respectively) the robber in the (cid:96) -visibility Cops and Robber game. It was noted in [23]that c (cid:48) ( G ) ≤ c ( G ) ≤ c ( G ) for all graphs G . Note that seeing the robber does notnecessarily imply capture of the robber: c (cid:48) ( C ) = 1 < c ( C ). However, it doesimply capture on chordal graphs (see Theorem 9).To properly describe the strategy for a finite set of cops, we require the followingnotation. Let G be a connected graph. Let L = { (cid:96) i } i = ki =1 , where (cid:96) i = (cid:96) i (0) , (cid:96) i (1) , . . . is awalk that describes the position of cop i in G , where the argument indicates the round.We call L a k -cop strategy . We say that a cop vibrates between time t and t + 2 k ifthere exists uv ∈ E ( G ) such that( (cid:96) i ( t ) , (cid:96) i ( t + 1) , . . . , (cid:96) i ( t + 2 k − , (cid:96) i ( t + 2 k )) = ( u, v, . . . , u, v, u ) . As the cops proceed through their strategies, the set of vertices that may possiblycontain the robber decrease. Borrowing from the language of edge searching, a vertexknown to not contain the robber is called clean , otherwise it is dirty . A vertex that isclean but again becomes dirty at a later time is said to be recontaminated . A set ofcops clean a subgraph by adding each vertex of the subgraph to the set of clean verticeswhile allowing no vertex of the subgraph to become recontaminated. We refer to theset of dirty vertices as the robber territory . We say that a k -cop strategy is a k -copcleaning strategy or a cleaning strategy using k cops if there is some time t so that allof the vertices are clean at time t ; that is, a time t when the robber can be guaranteedcaptured.A subgraph H of G is isometric if for any pair of vertices x, y ∈ V ( H ), d H ( x, y ) = d G ( x, y ). In the case that H is a tree, we say that H is an isometric tree in G . Recallthat a graph homomorphism is a vertex mapping such that adjacency is preserved.Given a (reflexive) graph G and a subgraph H , we say that H is a retract of G if thereis a homomorphism f : V ( G ) → V ( H ) such that for every v ∈ V ( H ), f ( v ) = v . Werefer to the mapping f as a retraction of G onto H .Throughout, we assume that graphs are finite and simple and contain a single robber.We note that there is a rich compendium of results for Cops and Robber on infinitegraphs (for example see [16]). We discuss the problem of limited visibility Cops andRobber on an infinite graph in Section 6. As we do not explore the concept of capturetime and are restricted to finite graphs, our assumption that the graph contains a single4obber is well-founded, as a strategy that is utilized to capture a single robber can berepeated until no robber remains. The notion of retract described above plays an important role in the classification ofcop-win graphs [20]. We note that if H is a retract in G , then H is also an isometricsubgraph in G . This follows from the fact that a retraction onto H is edge-preservingon G , and the identity on H . We prove that for any retract H of G , c (cid:96) ( H ) ≤ c (cid:96) ( G ).The proof of this result is similar for that of the cop number of retracts: c ( H ) ≤ c ( G )[3]. Theorem 1
For any retract H of a graph G , c (cid:96) ( H ) ≤ c (cid:96) ( G ) and c (cid:48) (cid:96) ( H ) ≤ c (cid:48) (cid:96) ( G ) . Proof : Suppose H is a retract of G with retraction f , and c (cid:96) ( G ) = k . We show c (cid:96) ( H ) ≤ k .We consider a pair of (cid:96) -visibility games, each with k cops and one robber, playedin parallel. The first game is played on G and the second game is played on H . Forthe robber, the moves in the two games will be identical. (One can think of the robberas being totally unaware of the first game, and making decisions to avoid capture aslong as possible in the second game only.) As for the cops, in the first game, each ofthe k cops will play according to a winning strategy in G . In the second game, thecops will use the moves in the first game and the retraction f to determine their moves.Specifically, for each cop C in G , there is a corresponding cop C (cid:48) in H . Whenever C moves onto a vertex v ∈ V ( G ), C (cid:48) will move onto the vertex f ( v ) in H . This is alwayspossible since f is edge-preserving.In order for the cops in H to adhere to the rules of the (cid:96) -visibility game, the copsin G cannot have information regarding the robber’s position that is unavailable to thecops in H . To verify that this is the case, consider x ∈ V ( G ) and y ∈ V ( H ) such that d G ( x, y ) ≤ (cid:96) . It follows that d H ( f ( x ) , f ( y )) ≤ d G ( x, y ) ≤ (cid:96) . Since f ( y ) = y , we have d H ( f ( x ) , y ) ≤ (cid:96) . Therefore, whenever a cop C sees the robber in G , the correspondingcop, C (cid:48) , in H also sees the robber.Since the robber is restricted to H in both games, he will be captured in the firstgame on a vertex of H . Therefore, in the second game, he will also be captured in H .Therefore, c (cid:96) ( H ) ≤ c (cid:96) ( G ). It can be similarly shown that c (cid:48) (cid:96) ( H ) ≤ c (cid:48) (cid:96) ( G ).It was shown in [19] that any isometric tree in G is also a retract in G . We, therefore,have the following corollary regarding isometric trees. Corollary 2 If T is an isometric tree in G then c (cid:96) ( T ) ≤ c (cid:96) ( G ) . For any k ≥
1, a k -dominating set of G is a subset S of V ( G ) such that every vertexin V ( G ) \ S is at most distance k from some vertex of S . The k -domination numberof G , denoted γ k ( G ), is the minimum cardinality of a k -dominating set of G . When k = 1, this is simply referred to as the domination number of G . We begin by statingsome obvious, but useful inequalities. 5 roposition 3 For any connected graph G ,1. c (cid:48) (cid:96) ( G ) ≤ c (cid:96) ( G ) ;2. c (cid:48) (cid:96) ( G ) ≤ γ (cid:96) ( G ) ;3. c ( G ) ≥ c ( G ) ≥ c ( G ) ≥ · · · ≥ c diam ( G ) ( G ) = c ( G ) ;4. c (cid:48) ( G ) ≥ c (cid:48) ( G ) ≥ c (cid:48) ( G ) ≥ · · · ≥ c (cid:48) rad ( G ) ( G ) = 1 . We next state some easy results for paths, cycles, complete bipartite graphs andcomplete graphs. On a cycle C n with n ≥ (cid:96) + 2, an (cid:96) -visibility cop chooses a vertexto initially occupy and can see a total of 2 (cid:96) + 1 vertices (including the vertex occupiedby the cop). The cop first moves to an adjacent vertex and can now see a vertex thatthe cop could not previously see. As there is a delay between the cops move and therobber’s, the cop can effectively see 2 (cid:96) + 2 vertices of the graph during any given round. Proposition 4
For (cid:96) ≥ ,1. c (cid:96) ( P n ) = c (cid:48) (cid:96) ( P n ) = 1 for any n ≥ ;2. c (cid:96) ( C n ) = 2 for any n ≥ ;3. c (cid:48) (cid:96) ( C n ) = (cid:40) if n ≥ (cid:96) + 3 , otherwise;4. c (cid:96) ( K m,n ) = 2 for any ≤ m ≤ n ;5. c (cid:48) (cid:96) ( K m,n ) = 1 for any ≤ m ≤ n ;6. c (cid:96) ( K n ) = c (cid:48) (cid:96) ( K n ) = 1 for any n ≥ . There are two cases of particular note in Proposition 4. While it has already been notedthat seeing is not the same as capturing (as in the graph C ), part 3 of Proposition 4goes a step further. Consider the cycle C in the 1-visibility game. The cop chooses avertex to occupy and can “see” a total of 3 vertices (including the vertex he occupies).In his first move, he moves to an adjacent vertex. He can now see a vertex he could notpreviously see. If the robber is not there, then the cop does not see the robber, but nowthe set of vertices that can possibly contain the robber, that is, the robber territory , isonly a single vertex. That is, the cop now knows exactly where the robber is located!In this case, “locating” is stronger than seeing.It is also interesting to note that c (cid:96) ( G ) = 1 for any graph G with a universalvertex. The graph K n is merely a classic example, and particularly interesting because c ( K n ) = (cid:100) n/ (cid:101) .In Section 4, we investigate the relationship between c (cid:48) (cid:96) ( G ) and c (cid:96) ( G ), and show thatfor cop-win graphs c (cid:96) ( G ) − c (cid:48) (cid:96) ( G ) is at most one. In Section 5 we see that the difference6etween c (cid:96) ( G ) and c ( G ) can be arbitrarily large. We now show that difference betweenboth γ ( G ) − c (cid:48) ( G ) and γ ( G ) − c ( G ) can be arbitrarily large.In Figure 1, we have a graph G such that c ( G ) = 1, as a cop-win ordering is givenby the order of the vertices. By placing a single cop on v , the cop can see every vertexexcept v and v . If the robber has not already been seen, the cop’s first move is to v , and the robber is seen. Thus, c (cid:48) ( G ) = 1. Since there is no dominating vertex, it isstraightforward to see that γ ( G ) = 2. However, the next result shows that c ( G ) = 2,and therefore, seeing does not imply capture even on a cop-win graph. v v v v v v v v Figure 1: A graph G in which c ( G ) = 1, c (cid:48) ( G ) = 1, and c ( G ) = 2. Observation 5
For the graph G given in Figure 1, c ( G ) = 2 . Proof : Let G be the graph shown in Figure 1 and for a contradiction, suppose c ( G ) = 1. Consider the final move for the cop: the robber must have occupied a vertex x and the cop occupied a vertex y such that N [ x ] ⊆ N [ y ] (i.e. x was a corner). Thenany vertex to which the robber can move, the cop can also move. (If N [ x ] (cid:54)⊆ N [ y ],then the robber could have moved to a vertex z / ∈ N [ y ] and avoided capture in thesubsequent round.) Note that G has two corners, v and v . Without loss of generality,assume the robber occupied v immediately prior to capture and the cop occupied v .In the previous round, the cop must have occupied v or v before his penultimate moveto v (if the cop had occupied v or v , then he could have moved to v instead of v andcaptured the robber earlier). Suppose the cop occupied v before moving to v . Afterthe cop moved to v , the robber moved to v . In the previous round, the robber musthave occupied a vertex of N [ v ]. If the robber had occupied v or v then he could havemoved to either v or v and the cop would not know to which vertex the robber hadmoved. Consequently, the cop would not know whether to move to v or to v in hispenultimate move, which yields a contradiction. If the robber occupied v or v , thenhe could have moved to v instead of v . In this case, even assuming the cop could seethat the robber moved to v , the cop cannot capture the robber in two moves, which isa contradiction. Hence, c ( G ) >
1. 7ince γ ( G ) = 2, two 1-visibility cops can occupy dominating vertices of G andcapture the robber immediately.We next provide a family of graphs that shows that the difference between γ and c can be arbitrarily large. We begin with k copies of the graph G given in Figure 1, forfinite k . Let G k be the graph obtained by merging vertices of degree 5 from k copies of G as shown in Figure 2. … v v v v v v v v v Figure 2: A graph G k for which γ ( G k ) = k + 1, but c (cid:48) ( G k ) = 1 and c ( G k ) = 2.Clearly γ ( G k ) = k + 1. However, we can easily observe that c (cid:48) ( G k ) = 1. Initiallyplace a 1-visibility cop at an endpoint of the path indicated with dotted edges (labeledas v in Figure 2). If the cop does not see the robber, then the cop moves along thepath indicated by dotted edges. If the robber was initially located at v or v , thenthe cop will see the robber. Otherwise, the robber was not initially in the leftmost“copy” of G . A similar argument shows that if the cop continues to move along thepath indicated by dotted edges, he will eventually see the robber. With the addition ofa second 1-visibility cop, this strategy can easily capture the robber: a 1-visibility copoccupies v while the other occupies v . If the robber is located in the leftmost “copy”of G , he is captured immediately. Otherwise, one cop moves from v to v and thento v . After both cops occupy v , one cop moves to v (cid:48) (labeled in Figure 2) and thecops follow the same strategy to clean the next “copy” of G . The graph G is a retractof G k , since every unlabelled can be mapped to v , while every labelled vertex can bemapped to itself. By Theorem 1, we know c ( G k ) ≥ c ( G ) = 2. Combined with theabove argument, this yields c ( G k ) = 2.Since c ( K n ) = 1 and there exist trees with arbitrarily large 1-visibility cop number,the 1-visibility cop number is not closed under minors in general. However graph G k does highlight the existence of a minor relationship, based on cut-vertices, cut-edges,and isometric subgraphs. Observation 6
Let v be a cut-vertex or an endpoint of a cut-edge in graph G . Forany (cid:96) ≥ , we have c (cid:96) ( G ) ≤ ≤ i ≤ α { c (cid:96) ( H i ) } , where H , H , . . . , H α are the subgraphsinduced by the deletion of N (cid:96) [ v ] from G . v is a cut-vertex (or endpoint of a cut-edge) in a graph G , then a 1-visibilitycop can occupy v . The robber is then restricted to some subgraph H i . The remainingmax i ∈ [ α ] { c ( H i ) } cops then search each subgraph H , H , . . . , H α one by one until therobber is caught. Many cops and robber games involve the idea of monotonicity. Essentially a strategyis monotonic if, once a portion of the graph is known to be free of the robber, thenthe cops have a strategy to guarantee that the robber cannot re-enter such a portion.Edge-searching requires no extra cops to guarantee a monotonic strategy; alternatively,we say that edge-searching is monotonic [18, 4]. However, connected edge-searchingrequires strictly more cops to maintain monotonicity; that is, connected edge-searchingis not monotonic [26].In the classical Cops and Robber game, the robber is visible, and so the idea ofmonotonicity has not been of interest to researchers. However, in the zero-visibility Copsand Robber game, which conceptually is quite similar to edge-searching, monotonicitybecomes a natural concern. It was shown in [11] that the zero-visibility cop and robbergame is clearly not strictly monotonic (due to the usefulness of vibrating, as describedin Section 1). It is also not weakly monotonic – even allowing the recontaminationassociated with vibrations, more cops are required to show that the size of the robberterritory (considered immediately after the cops’ move) is always non-increasing overtime.Certainly, there are similar considerations in (cid:96) -visibility Cops and Robber. Let S k denote the robber territory immediately after the cops have taken their k th move, butbefore the robber moves. We define the (weakly) monotonic (cid:96) -visibility cop number ofa graph G , mc (cid:96) ( G ), to be the minimum number of cops required to guarantee captureof the robber in G with the restriction that S k +1 ⊆ S k for all moves after the initialplacement.Consider a perfect binary tree T of depth 3. Let every edge of T be subdivided2 (cid:96) + 1 times; call the resulting tree T (cid:96) . Let the root of T (cid:96) be r , the “left” vertex ofdepth 2 (cid:96) + 2 be a , the right descendent of a that is 2 (cid:96) + 2 levels below be b , and thevertices on the path from a to b be labelled v , v , . . . , v (cid:96) +1 , with v adjacent to a and v (cid:96) +1 adjacent to b , as shown in Figure 3. Theorem 7
For (cid:96) ≥ , c (cid:96) ( T (cid:96) ) = 2 , and mc (cid:96) ( T (cid:96) ) = 3 . Proof : One (cid:96) -visibility cop is not sufficient to guarantee capture of the robber (seeSection 5 for further detail). We present a (non-monotonic) strategy for two cops, C and C , to capture the robber.First, place one cop on each of the leaves to the left of vertex a . They may bothmove up the tree, eventually both ending at a . Then, cop C moves from a to v (cid:96) ,and in subsequent moves will vibrate between v (cid:96) and v (cid:96) +1 until all vertices below b are9 rb v v v v v Figure 3: The tree T (cid:96) when (cid:96) = 2.robber-free. Thus, C will guard the branch of T (cid:96) to the left of a , as in every secondmove, a will be surveilled; that is, C will see the robber and be able to guaranteecapture, by Theorem 9. Then cop C may move to the left descendant of b that is (cid:96) levels above a leaf, and proceed to the corresponding right descendant of b . This cleansall the descendants of a , at which point both C and C may proceed to r , and cleanthe right half of T (cid:96) in the reverse of this strategy. Thus, c (cid:96) ( T (cid:96) ) = 2.However, this is not a monotonic strategy; the vertex b is known to be robber-free( C even occupies it), but after that, b is unseen for three successive moves, allowingrecontamination. This may be repaired by introducing a third cop, C . After cleaningthe left branch of a , while C vibrates between v (cid:96) and v (cid:96) +1 , C may move to vertex b , while C proceeds as in the previous strategy; again after cleaning both branchesof a , all cops may move to r and repeat the reverse of the strategy. This shows that mc (cid:96) ( T (cid:96) ) ≤
3, and it is straightforward to show that there is no 2-cop strategy thatpreserves monotonicity.
We now attempt to answer the question “For which graphs does seeing imply capture?”We have seen in Proposition 4 that seeing does not imply capture for complete bipartitegraphs and some cycles. However, in these examples, the cops could see almost theentire graph from fixed starting positions. In general, this will not be the case, andthe searching phase of the cops’ strategy may be the phase that is the most resourceintensive. 10irst, we investigate this problem for the (cid:96) -visibility game when (cid:96) ≥
2. We thenexamine the problem on chordal and cop-win graphs.
Theorem 8
For any graph G and (cid:96) ≥ , either c (cid:48) (cid:96) ( G ) = c (cid:96) ( G ) or c ( G ) ≤ c (cid:96) ( G ) ≤ c ( G ) + 1 . Proof : Consider a graph G and a set of cops of size m where m = max { c (cid:48) (cid:96) ( G ) , c ( G )+1 } and (cid:96) ≥
2. Since m ≥ c (cid:48) (cid:96) ( G ), there is a strategy on G for the m cops to eventually seethe robber. Therefore, at some point in the game, a cop C is within distance (cid:96) of therobber. Suppose this occurs in round t (cid:48) .In any given round, if the robber occupies vertex r at the end of the round, we let r (cid:48) denote the vertex occupied by the robber at the end of the previous round. If weassociate the robber with vertex r , then we refer to r (cid:48) as the robber’s shadow. Startingin round t (cid:48) + 1, we assume that C moves on a geodesic between himself and the robber’sshadow until indicated otherwise. As a result, C maintains a distance of at most (cid:96) fromthe robber’s shadow at all times. Now, the other m − m − ≥ c ( G ), these m − C (cid:48) adjacent to the robber. Assumethe latter. Now, C (cid:48) will continue to move onto the robber’s shadow in each round, and C will abandon his previous strategy and join the other m − C (cid:48) is always withing distance two of the robber, and therefore, the other m − m − ≥ c ( G ), these m − c (cid:96) ( G ) ≤ m .If m = c ( G ) + 1, then c ( G ) ≤ c (cid:96) ( G ) ≤ c ( G ) + 1. If m = c (cid:96) ( G ), then c (cid:48) (cid:96) ( G ) ≤ c (cid:96) ( G ) ≤ c (cid:48) (cid:96) ( G ). The result follows.It is well-known that chordal graphs provide a large family of graphs G for which c ( G ) = 1 [20]. However it is easy to construct examples of chordal graphs that require atleast two cops to capture the robber with limited visibility (see Theorem 17). However,in this section we show that for (cid:96) -visibility Cops and Robber, chordal graphs do havethe property that if at any point there is a cop at distance no more than (cid:96) from therobber, then that cop can eventually capture the robber. We begin by stating severaluseful definitions from [25]. A vertex of graph G is simplicial if its neighbourhood in G induces a clique. A simplicial elimination ordering is an ordering v n , . . . , v for deletionof vertices so that each vertex v i is a simplicial vertex of the remaining graph inducedby { v , . . . , v i } . It is well-known that G has a simplicial elimination ordering if and onlyif G is chordal.In the following proof, using zero-visibility as a base case, we show that if a cop seesthe robber, then in a finite number of subsequent rounds the distance between that copand the robber decreases. Theorem 9 If G is a chordal graph and (cid:96) ≥ , then c (cid:48) (cid:96) ( G ) = c (cid:96) ( G ) . Furthermore, oncethe robber has been seen by a cop, that cop can capture the robber. roof : Let G be a chordal graph and fix a simplicial elimination ordering of thevertices: v n , v n − , . . . , v . We proceed by inducting on (cid:96) . As the case where (cid:96) = 0 istrivial, we suppose c (cid:96) − ( G ) = c (cid:48) (cid:96) − ( G ) for some (cid:96) ≥
1. Furthermore, we suppose oncethe robber has been seen by an ( (cid:96) − c (cid:48) (cid:96) ( G )-many (cid:96) -visibility cops follow a strategy that allows an (cid:96) -visibility cop C to see the robber. Thus, at some point, C occupies a vertex that is distance (cid:96) fromthe vertex occupied by the robber, R ; suppose they are located on vertices v i and v x ,respectively. We assume it is the robber’s turn to move, as otherwise that cop will moveto a vertex that is distance (cid:96) − (cid:96) = 1 then the cop occupies thesame vertex as the robber) and by the induction hypothesis, C can capture the robber.Observe that the robber cannot always increase (or decrease) his index (accordingto the simplicial ordering). Thus, there is some step where the robber moves from v x to v y to v z , where y > x and y > z . Suppose C occupies v i and R occupies v x where d ( v i , v x ) = (cid:96) . We will now show that C will eventually capture the robber.After R moves to v y , C moves to any neighbour of v i that is distance (cid:96) − v x ;call such a neighbour v j . If (cid:96) = 1 then v x = v j and the robber is captured, otherwise, R moves to v z . Since the vertices are indexed by a simplicial ordering, v y ’s lower-indexedneighbours form a clique. Thus v x is adjacent to v z and d ( v j , v z ) = (cid:96) −
1. Since C nowoccupies a vertex that is distance (cid:96) − R , by the inductivehypothesis C can now capture the robber using an ( (cid:96) − (cid:96) is the diameter of the graph. Then the cop can see theentire graph and will capture the robber, thus the above result provides an alternateproof that chordal graphs are cop-win.Theorem 9 raises the question: if G is chordal, what is c (cid:48) (cid:96) ( G )? In the followingsection we fully answer this case for trees, a subfamily of chordal graphs. We providea complete structural characterization for trees and show that for each k ∈ N , thereexists a chordal graph G such that c (cid:48) (cid:96) ( G ) ≥ k .Another question naturally arises, given that chordal graphs are cop-win: if G iscop-win, does c (cid:48) (cid:96) ( G ) = c (cid:96) ( G )? In general, the answer is no. In Figure 1, we have acop-win graph G such that c (cid:48) ( G ) = 1, but c ( G ) = 2. Furthermore, when (cid:96) ≥ G is cop-win and c (cid:48) (cid:96) ( G ) (cid:54) = c (cid:96) ( G ), then 1 ≤ c (cid:96) ( G ) ≤ (cid:96) ≥ G , c (cid:96) ( G ) − c (cid:48) (cid:96) ( G ) ≤ Recall that for a tree T , the height of the tree h ( T ) is given by min v ∈ V ( T ) { ec ( v ) } , where ec ( v ) is the eccentricity of v . In [14], the author gives an upper bound for the 1-visibilitycop number of a trees as a function of the height of the tree. The result is restated inTheorem 10, and the proof is provided for completeness. Theorem 10 [14] Given a tree T , c ( T ) ≤ (cid:108) h ( T )3 (cid:109) . roof : We note that on a tree, once a cop sees the robber, the robber will be capturedin a finite number of rounds (the cop simply moves onto the robber’s previous positionwhich eventually forces the robber onto a leaf). Therefore, in the 1-visibility game, itsuffices to show that at some point, a cop occupies a vertex adjacent to the robber. Webegin by verifying that for any tree with height at most three, a single cop can capturethe robber.First, suppose T is a tree with height at most two. Assume T is rooted at its centre(or an endpoint of its centre), w . If h ( T ) ≤
1, then one cop initially positioned at w willcapture the robber in the next round. Suppose h ( T ) = 2 and N ( w ) = { w , w , . . . , w m } for some m ≥
2. The cop then performs the walk w, w , w, w , w, . . . , w, w m , alternatingbetween a neighbour of w and w itself. The robber is prevented from moving onto avertex in N [ w ], since the cop will either see him immediately or in the next round.Therefore, to avoid capture, the robber must occupy some vertex x such that d ( w, x ) = 2and pass in every round. Eventually the robber will be seen by the cop.Now, suppose T is a tree rooted at its centre, r , such that h ( T ) = 3. For each w ∈ N ( r ), the cop, in turn, performs the walk described in the previous paragraph. Ifthe robber is in the subtree rooted at w , he will be seen by the cop. Furthermore, ifthe robber moves onto r , he will either be seen by the cop immediately, or he will beseen by the cop in the next round when the cop moves onto some w ∈ N ( r ). Therefore,if the robber is in a subtree of T rooted at some w ∈ N ( r ) in the initial round of thegame, he cannot move out of that subtree without being seen by the cop. It followsthat the robber will be seen by the cop after a finite number of rounds.Now assume that for any rooted tree T (cid:48) of height at most 3 k for some k ≥ c ( T (cid:48) ) ≤ k . Consider a tree T rooted at its centre r , such that h ( T ) ≤ k + 3. Let N ( r ) = { x , x , . . . , x m } . For each x ∈ V ( T ), let T x be the subtree of T rooted at x . Itfollows that if d ( r, x ) = 3, then h ( T x ) ≤ k and c ( T x ) ≤ k .For each i = 1 , . . . , m we let X i = { x | d ( r, x ) = 3 and d ( x i , x ) = 2 } . It follows thatthe vertices in X i are descendants of both r and x i . Without loss of generality, assumethat for some m (cid:48) ≥ X , . . . X m (cid:48) are non-empty, and either m (cid:48) = m or X j = ∅ for j = m (cid:48) + 1 , . . . , m .The cops’ strategy is as follows: a cop, C , initially occupies x , while a set of k cops, C , occupy vertices in the subtree T x for some x ∈ X . The cop C vibrates between x and the parent of x in T . Meanwhile, the set C of k cops clean T x . Once T x iscleaned, assuming the robber has not been seen, we mark vertex x . (All vertices in X are initially unmarked.) We then iteratively choose an unmarked vertex y in X andmove the cops in C to y . Only then does C change its vibrating pattern and begin tovibrate between x and the parent of y . (Note that x and y may have the same parent.)The cops in C then clean T y . This is repeated until all vertices in X are marked orthe robber is seen. Since the robber will be seen if he moves onto any neighbour of x ,once T y is cleaned for some y ∈ X , it cannot be recontaminated during this phase.As a result, the subtree rooted at x is cleaned. Furthermore, if the robber ever movesonto the root r of T , he will either be seen immediately by C or will been seen the nextround when C moves onto x . 13 T T ... r ... ... ... ...... r r r x x x z y y y z z each rx i -path is of length ` each z i x i -path is of length ` Figure 4: An element of T k,(cid:96) For each i = 1 , , . . . m −
1, once the subtree T x i is cleaned, C moves on the path x i rx i +1 and the cops in C move onto the subtree T x i +1 . Together they then clean T x i +1 .(If i + 1 ≤ m (cid:48) , they use the strategy in the previous paragraph. Otherwise, the subtree T x i +1 has height at most one and is cleaned by C .) We note that since C is adjacent to r at some point in every round, the robber cannot move onto r without being captured.As as result once T x i is cleaned, it cannot be recontaminated. It follows that the robberwill eventually be seen.By noticing that in a tree of height h ≤ (cid:96) + 1 the robber may never pass throughthe root unobserved we arrive at the following. Observation 11 If T is a tree with height h ≤ (cid:96) + 1 , (cid:96) ≥ with respect to root r ,then c (cid:96) ( T ) = 1 . In [12], the authors provide a full characterization for the zero-visibility cop numberof trees. Here we extend this work for (cid:96) > T be a tree with a root r . Let x, y be vertices of T where x is a successor of y with respect to r . The subtree rooted at x is the component of T − xy that contains x .We call a vertex at distance q from r a q -descendent of r . Definition 12
Let T k,(cid:96) be the family of trees defined in the following way: • T ,(cid:96) = { K }• T k,(cid:96) is the set of trees, T , that can be formed as follows: let T , T , T ∈ T k − ,(cid:96) . Let r , r , r be vertices of T , T , T respectively. Then T is formed from the disjointunion of T , T , T , together with paths of length (cid:96) + 2 from each of r , r , r , toa common endpoint, q . r , r , r are any vertices of T , T , T , respectively. They need not be thevertices q identified in the construction of elements of T k − ,(cid:96) . Lemma 13 If T ∈ T k,(cid:96) then c (cid:96) ( T ) = k . Proof : Let T ∈ T k,(cid:96) with vertices labelled as in Figure 4, with paths of length (cid:96) from r to x i and from z i to r i . Observe that the subtrees T , T , and T are in T k − ,(cid:96) .Inductively, c (cid:96) ( T i ) = k − ≤ i ≤
3. We first assume that k − T .Let T (cid:48) i be the subtree rooted at z i . We now show that for all i there exists a roundduring which all k − T (cid:48) i . Assume, for some cleaning strategy C for T using k − k − T (cid:48) i . Let w x ( t )be the vertex occupied by cop C x at the end of round t in C for 1 ≤ x ≤ k −
1. Define w (cid:48) x ( t ) as follows. w (cid:48) x ( t ) = w x ( t ) if w x ( t ) ∈ V ( T i ), r i if w x ( t ) ∈ V ( T (cid:48) i ) \ V ( T i ) y i otherwise.In this strategy, k − T i , a contradiction. For each i thereexists a round so that all k − T (cid:48) i .Let t be the earliest round for which exactly one leaf is cleaned at the end of round t and this leaf remains clean during each subsequent round. Let v be this leaf andwithout loss of generality, let it be in T .By the above argument, there exists a round during which all k − T (cid:48) and there is a different round during which all k − T (cid:48) . We note that thesetimes must occur after t , otherwise by the argument above, we have a contradiction,as k − T or T . Of T (cid:48) and T (cid:48) , assume without loss of generalitythat T (cid:48) is the first to contain all k − t > t .Let t be the round after t and before t that all cops are on the path from r to r or are in T . Assume, without loss of generality, that from round t to round t that therobber maintains a distance of (cid:96) + 2 from the nearest cop. Then, at the end of round t , all cops are in T (cid:48) and the robber may be at r . At the end of each subsequent round,the robber maintains (at least) distance (cid:96) + 2 from the nearest cop and will eventuallyrecontaminate v ∈ T , which yields a contradiction. Therefore k − T .We now construct a k -cop cleaning strategy for T k . Place all of the cops on x . Onecop, C , will vibrate between x and y . The remaining k − k − T . Note that r is not necessarily theroot of T , but as the ( k )th cop prevents the robber from leaving T unobserved, theremaining k − C is on y in everysecond time-step, this ensures that once T is cleaned using the k − x and repeat this process with T . Since C is on x in every second time step while T is being cleaned, the robber cannot pass through r T . Hence we have cleaned T using k cops. Corollary 14
For all T ∈ T k,(cid:96) there exists a cleaning strategy for T using k cops inwhich the root of T is never unseen for two consecutive timesteps. Lemma 15
Let T be a tree with root r such that each of the subtrees rooted at an (cid:0)(cid:6) i (cid:7)(cid:1) -descendant have (cid:96) -visibility cop number at most k − , for some i ≤ (cid:96) + 2 . Thereexists a cleaning strategy for T using k − cops such that r is occupied every secondtime-step. Proof : Let x be a neighbour of r . Let S x be the set of (cid:0)(cid:6) i (cid:7)(cid:1) -descendant of r that isalso a descendants of x . If a single cop C vibrates between x and r , and if S x (cid:54) = ∅ , thenthe remaining k − S x . If S x = ∅ , thenmoving C to x suffices to clean the subtree rooted at x . Repeating this process for allneighbours of r gives the required cleaning strategy. Lemma 16
Let T be a tree with root r such that each of the subtrees rooted at an i -descendant of r have (cid:96) -visibility cop number at most k − for some i < (cid:96) + 2 .There exists a cleaning strategy for T using k − cops such that r is seen every secondtime-step. Proof : Let S r be the set of 1-descendants of r . For x ∈ S r let S x be the (cid:0)(cid:4) i (cid:5)(cid:1) -descendants of r that are descendants of x . By Lemma 15, a subtree rooted at anelement of S x can be cleaned using k − r is seen in every secondtime-step. As such we can clean the subtree rooted at x using k − S x in left to right order such that r is seenin every second time-step. Observe that any vertex that is a descendant of x that isnot contained in a subtree rooted at an element of S x is cleaned during this process.Repeating this process for each of the elements of S r gives a cleaning strategy for T using k − r is seen in every second time-step. Note that since r is seenin every second time-step, once a subtree rooted at an element of S r is clean it cannotbe recontaminated. Theorem 17 If T is a tree, then c (cid:96) ( T ) = k , where k the greatest integer such that T contains a subgraph from T k,(cid:96) . Proof : Let T be a tree and k the greatest integer such that T contains a subgraphfrom T k,(cid:96) . Clearly T needs at least k cops by Lemma 13 and Corollary 2. It suffices toshow that if c (cid:96) ( T ) = k , then T must contain some T k ∈ T k,(cid:96) .Now, let k be the least integer such that there exists a tree T such that c (cid:96) ( T ) ≥ k and does not contain some element of T k,(cid:96) . Since k is the least such integer, T containsan element of T k − ,(cid:96) . We proceed in two cases.16 ase I: T contains no pair of vertex disjoint elements of T k − ,(cid:96) .Let H be any minimal subtree of T such that c (cid:96) ( H ) = k − T \ H is connected.By minimality of k , H has as a subgraph an element of T k − ,(cid:96) . By construction of H ,there is a single vertex x of such that x has a single neighbour not in H . Let x bethis neighbour. Since H is minimum, each of the subtrees rooted at a 1-descendant of x that are contained in H require at most k − T \ H containsno element of T k − ,(cid:96) . Therefore, by the minimality of k , c (cid:96) ( T \ H ) < k −
1. Thus T can be cleaned using k − x and cleaning all of thesubtrees of T \ { x } one at a time, each using k − Case II: T contains at least two elements of T k − ,(cid:96) that share no vertices in T . Let H and H (cid:48) be minimal vertex disjoint subtrees of T such that c (cid:96) ( H ) = c (cid:96) ( H (cid:48) ) = k − T \ H and T \ H (cid:48) is connected. We choose H and H (cid:48) to be at maximumdistance among all possible pairs of minimal vertex disjoint subtrees of T that satisfythese properties.Let P = x , x , . . . , x d be the path from H to H (cid:48) such that only x and x d arecontained in H and H (cid:48) , respectively. (See Figure 5.) We note that we need not have x or x d be the vertex identified as q in the construction of elements of T k − ,(cid:96) . Let r be a centre vertex this path. We consider T to be rooted at r . For 2 ≤ i ≤ d −
1, let X i be the subgraph containing x i induced by the deletion of E ( P ) from T . We show c (cid:96) ( T ) = k − H , each of the subtrees of H \ { x } , say H , H , . . . H t , requireat most k − H can be cleaned using k − x and using the remaining k − H a (1 ≤ a ≤ t ). Similarly, if every vertex excluding those in H (cid:48) are clean and all ofthe cops are at x d , we can clean the vertices of H (cid:48) so that no vertex of T \ H (cid:48) becomesrecontaminated to complete the cleaning of T with k − d ≥
3. To complete the proof it suffices to show that X i can be cleaned using at most k − x i is seen by a cop in at most every second time-step for 1 < i < d .By Lemma 16 each of X , . . . X t can be cleaned so that x i (1 ≤ i ≤ t ) is seen inevery second time-step for t < min { (cid:96) + 2 , (cid:98) d (cid:99)} .If d ≤ (cid:96) + 2, then observe that for d ≤ i ≤ (cid:96) + 2 a subtree rooted at a i -descendentof x i cannot require k − T (cid:48)(cid:48) would containan element of T k − ,(cid:96) with q = x i , T = H , T = H (cid:48) , T = T (cid:48)(cid:48) . Thus, by Lemma 16, each X i can be cleaned so that x i is seen in every second time-step for i ≤ (cid:98) d (cid:99) .By symmetry, each of the trees X (cid:98) d (cid:99) , . . . X d − can be cleaned by applying Lemma16. This completes the proof. Corollary 18
Given a tree T , c (cid:96) ( T ) ≤ (cid:108) h ( T )2 (cid:96) +2 (cid:109) . Proof : Observe that every tree contained in T k,(cid:96) has height at least k · (2 (cid:96) + 2).Therefore a tree T of height h ( T ) contains no element of T(cid:100) h ( T )2 (cid:96) +2 (cid:101) +1 ,(cid:96) . By Theorem 1717 . . x i x i +1 x i . . . ... x i x i ( i x i x r . . . x H x d H Yx x x x d . . . H X X X X i x i . . .x ... H H H t ... x d X d Figure 5: T in Case II of the proof of Theorem 17.18e have c (cid:96) ( T ) ≤ (cid:108) h ( T )2 (cid:96) +2 (cid:109) . In the original game of Cops and Robber, cops can see the location of the robberthroughout the game. In the other extreme, in zero-visibility Cops and Robber, copscannot see the location of the robber unless a cop occupies the same vertex as therobber. The more general (cid:96) -visibility game covers the spectrum in between the originalCops and Robber and zero-visibility Cops and Robber and we have seen this problemto be distinct from the two extremes. For example, Theorem 17 showed that for anychoice of (cid:96) ≥ k ≥
1, there exists a tree such that c ( T ) = 1 and c (cid:96) ( T ) = k .Naturally then, we ask for a graph G with inequality amongst the cop numbers acrossthe visibility spectrum. We note that when (cid:96) is the radius of G , c (cid:96) ( G ) ≤ c ( G ) + 1,since one cop sitting on a centre vertex can maintain sight of the robber while the c ( G )other cops capture the robber. Therefore, increasing (cid:96) past the radius of G will lead to c (cid:96) ( G ) decreasing in value. We therefore, pose the following problem regarding visibilityacross the spectrum. Open Problem 1
Provide a graph G for which c ( G ) > c ( G ) > · · · > c rad ( G ) ( G ) >c ( G ) . In Section 4, we show that for (cid:96) ≥
1, if an (cid:96) -visibility cop sees the location of arobber on a chordal graph, then that cop can capture the robber. Thus, although c (cid:48) (cid:96) ( G )cops are needed to see the location of the robber; after this point, only one (cid:96) -visibilitycop is necessarily in order to eventually capture the robber. Which cop-win graphs,beyond chordal graphs (including trees), have this property? Open Problem 2
Given a graph G , suppose that if an (cid:96) -visibility cop, C , is withindistance (cid:96) of the robber in some round t , then C can then capture the robber in round t (cid:48) for some t (cid:48) > t , where the play of the other cops in rounds t + 1 through t (cid:48) is irrelevant.Characterize such graphs. From Theorem 8, we have established a relationship between c (cid:96) ( G ) − c (cid:48) (cid:96) ( G ) and c (cid:96) ( G ) − c ( G ), whenever (cid:96) ≥
2: if c (cid:96) ( G ) − c (cid:48) (cid:96) ( G ) >
0, then c (cid:96) ( G ) − c ( G ) ≤
1. We havealso established that when G is chordal c (cid:96) ( G ) − c (cid:48) (cid:96) ( G ) = 0 for any (cid:96) ≥
0. However, ingeneral, we do not currently have bounds on these differences.
Open Problem 3
For each (cid:96) ≥ , are the following differences bounded?1. c (cid:96) ( G ) − c (cid:48) (cid:96) ( G ) c (cid:96) ( G ) − c ( G ) 19ecause “seeing implies capture” on chordal graphs, we are interested in the numberof (cid:96) -visibility cops required to see the location of the robber, c (cid:48) (cid:96) . However, c ( G ) is not alower bound for this parameter, as evidenced by Proposition 4: for example c (cid:48) ( C ) = 1and yet c ( C ) = c ( C ) = 2. Open Problem 4
For (cid:96) ≥ , find a lower bound for c (cid:48) (cid:96) ( G ) . Open Problem 5
Suppose there is a cost associated with increasing the visibility ofthe cops. We then want to consider the ratio c (cid:96) ( G ) c (cid:96) +1 ( G ) . What is the closure of c (cid:96) ( G ) c (cid:96) +1 ( G ) ? In [12], the authors provide a non-trivial family of graphs G n and show that given G ∈ G n and an integer k >
0, the problem of deciding whether c ( G ) ≤ k is NP-complete. To do this, the authors relate c ( G ) to the problem of computing pathwidthof G , which is known to be NP-complete [2]. By restriction this gives directly that givena graph G , and integers (cid:96) ≥ k > c (cid:96) ( G ) ≤ k isNP-complete. However, as pathwidth can be computed in polynomial time for severallarge classes of graphs, it gives hope that there are families of graphs for which the (cid:96) -visibility cop number may be efficiently computed.We conclude with a variant of Cops and Robbers. In Section 1, we observe thatonce a 1-visibility cop, C has seen the robber (i.e. has occupied a vertex adjacent tothe vertex occupied by the robber), C can “tail” the robber, by following the path ofvertices previously occupied by the robber. Then C can see the location of the robber ateach subsequent round – but of course this strategy of illuminating previous locations ofthe robber may not lead to capture. It does however, lead us to a time-delayed variantof Cops and Robber. During round 0, cops choose a set of vertices to occupy and thenthe invisible robber chooses a vertex to occupy. During each round t >
0, the copsmove and then the robber moves and then the vertex occupied by the robber at the endof round t − (cid:96) -visibility cop occupies a vertex distance (cid:96) from thevertex occupied by the robber and follows the “tailing” strategy outlined above. Denoteby c t ( G ), the minimum number of cops required to capture the robber on a graph G .Obviously c ( G ) ≤ c t ( G ) ≤ γ ( G ) for all graphs G , so the parameter is well-defined, andit is easy to find graphs for which strict inequality holds.Throughout we have assumed that all graphs are finite, simple and contain a singlerobber. However, following work on Cops and Robber for infinite graphs [16] it ispossible to consider (cid:96) -visibility Cops and Robber on infinite graphs. The result ofTheorem 9 does not extend to infinite chordal graphs with two-way infinite paths. In[6] the authors show that the cop-number of the Rado graph is ℵ . By [13], the sameis true for almost all countably infinite graphs. However, there do exist infinite graphsfor which the cop-number is finite. A first question in the study of (cid:96) -visibility Copsand Robber on infinite graphs is to classify those infinite graphs G and those integers (cid:96) such that c (cid:96) ( G ) = N , but c (cid:96) +1 ( G ) is finite.20 Acknowledgements
N.E. Clarke acknowledges research support from NSERC (2015-06258). D. Cox ac-knowledges research support from NSERC and Mount Saint Vincent University. C. Duffyacknowledges research support from AARMS. D. Dyer acknowledges research supportfrom NSERC. M.E. Messinger acknowledges research support from NSERC (grant ap-plication 356119-2011) and Mount Allison University.
References [1] T.V. Abramovskaya, F.V. Fomin, P.A. Golovach, and M. Pilipczuk, How to huntan invisible rabbit on a graph.
Eur. J. Comb. (2016)12–26.[2] S. Arnborg, D.G. Corneil, and A. Proskurowski, Complexity of finding embeddingsin a k -tree. SIAM J. on Algebraic and Discrete Methods (2) (1987) 277–284.[3] A. Berarducci and B. Intrigila, On the cop number of a graph. Adv. in Appl. Math. (1993) 389–403.[4] D. Bienstock and P. Seymour, Monotonicity in graph searching. J. Algorithms (1991) 239–245.[5] A. Bonato, P. Golovach, G. Hahn, and J. Kratochv´ıl, The capture time of a graph. Discrete Math. (18) (2009) 5588–5595.[6] A. Bonato, G. Hahn, and C. Wang, The cop density of a graph.
Contrib. DiscreteMath. Theoretical Computer Science (2016) 2–14.[8] A. Bonato and R.J. Nowakowski,
The Game of Cops and Robbers on Graphs ,American Mathematical Society, Providence, Rhode Island, 2011.[9] N.E. Clarke and G. MacGillivray, Characterizations of k -cop-win graphs. DiscreteMath. (8) (2012) 1421–1425.[10] N.E. Clarke and R.J. Nowakowski, Tandem-win graphs.
Discrete Math. (2005)56–64.[11] D. Dereniowski, D. Dyer, R.M. Tifenbach, and B. Yang, Zero-visibility cops androbber and the pathwidth of a graph.
J. Comb. Optim. (3) (2015) 541–564.[12] D. Dereniowski, D. Dyer, R.M. Tifenbach, and B. Yang, The complexity of zero-visibility cops and robber. Theoret. Comput. Sci. (2015) 135–148.2113] P. Erd¨os and A. R`enyi, Asymmetric graphs.
Acta Math. Acad. Sci. Hungar (1963) 295–315.[14] F. Yang, 1-visibility Cops and Robber Problem, Honour’s thesis, University ofPrince Edward Island, 2012.[15] S.L. Fitzpatrick, J. Howell, M.E. Messinger, and D.A. Pike, A deterministic versionof the game of zombies and survivors on graphs. Discrete Appl. Math. (2016)1-12.[16] G. Hahn, F. Laviolette, N. Sauer, and R.E. Woodrow, On cop-win graphs.
DiscreteMath. , , 27-41.[17] A. Kehagias, D. Mitsche, and P. Pra(cid:32)lat, Cops and invisible robbers: The cost ofdrunkenness. Theoret. Comp. Sci. (2013) 100–120.[18] A.S. LaPaugh, Recontamination does not help to search a graph.
J. Assoc. Comput.Mach. (1993) 224–245.[19] R. Nowakowski and I. Rival, On a class of isometric subgraphs of a graph, Combi-natorica (1) (1982) 79–90.[20] R.J. Nowakowski and P. Winkler, Vertex-to-vertex pursuit in a graph, DiscreteMath. (1983) 235–239.[21] P. Gordinowicz, Those Magnificent Blind Cops in Their Flying Machines withSonars , at GRASTA 2017.[22] A. Quillot,
Th`ese d’Etat , Universit´e de Paris VI, 1983.[23] A. Tang, Cops and robber with bounded visibility, Master’s thesis, Dalhousie Uni-versity, 2004.[24] R. To˘si´c, Vertex-to-vertex search in a graph, Graph Theory (Dubrovnik 1985)(1985) 233–237.[25] D.B. West,
Introduction to Graph Theory , 2nd Ed., Prentice-Hall, New Jersey,2001.[26] B. Yang, D. Dyer, and B. Alspach, Sweeping graphs with large clique number.
Discrete Math.309