Cops and Robbers on Geometric Graphs
aa r X i v : . [ m a t h . C O ] S e p Cops and Robbers on Geometric Graphs
Andrew Beveridge ∗ , Andrzej Dudek † , Alan Frieze ‡ , Tobias M¨uller § June 10, 2018
Abstract
Cops and robbers is a turn-based pursuit game played on a graph G . One robber ispursued by a set of cops. In each round, these agents move between vertices along theedges of the graph. The cop number c ( G ) denotes the minimum number of cops required tocatch the robber in finite time. We study the cop number of geometric graphs. For points x , . . . , x n ∈ R , and r ∈ R + , the vertex set of the geometric graph G ( x , . . . , x n ; r ) is thegraph on these n points, with x i , x j adjacent when k x i − x j k ≤ r . We prove that c ( G ) ≤ G in R and we give an example of a connected geometricgraph with c ( G ) = 3. We improve on our upper bound for random geometric graphs thatare sufficiently dense. Let G ( n, r ) denote the probability space of geometric graphs with n vertices chosen uniformly and independently from [0 , . For G ∈ G ( n, r ), we show thatwith high probability (whp), if r ≥ K (log n/n ) , then c ( G ) ≤
2, and if r ≥ K (log n/n ) ,then c ( G ) = 1 where K , K > G ( n, r ): if r ≤ K log n/ √ n then c ( G ) > K > The game of cops and robbers is a full information game played on a graph G . The game wasintroduced independently by Nowakowski and Winkler [26] and Quilliot [31]. During play, one ∗ Department of Mathematics, Statistics and Computer Science, Macalester College, Saint Paul, MN: [email protected] † Department of Mathematics, Western Michigan University, Kalamazoo, MI: [email protected] ‡ Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA: [email protected] . Supported in part by NSF Grant CCF1013110 § Centrum voor Wiskunde en Informatica, Amsterdam, the Netherlands: [email protected] . Supported in partby a VENI grant from Netherlands Organization for Scientific Research (NWO) R is pursued by a set of cops C , . . . , C ℓ . Initially, the cops choose their locations on thevertex set. Next, the robber chooses his location. The cops and the robber are aware of thelocation of all agents during play, and the cops can coordinate their motion. On the cop turn,each cop moves to an adjacent vertex, or remains stationary. This is followed by the robberturn, and he moves similarly. The game continues with the players alternating turns. The copswin if they can catch the robber in finite time, meaning that some cop is colocated with therobber. The robber wins if he can evade capture indefinitely.The original formulation [26, 31] concerned a single cop chasing the robber. These paperscharacterized the structure of cop-win graphs for which a single cop has a winning strategy.For v ∈ V ( G ), the neighborhood of v is N ( v ) = { u ∈ V ( G ) | ( u, v ) ∈ E ( G ) } and the closedneighborhood of v is N ( v ) = { v } ∪ N ( v ). When N ( u ) ⊆ N ( v ), we say that u is a pitfall . Agraph is dismantlable if we can reduce G to a single vertex by successively removing pitfalls. Theorem 1.1 ([26, 31]) G is dismantlable if and only if c ( G ) = 1 . Aigner and Fromme [1] introduced the multiple cop variant described above. For a fixedgraph G , they defined the cop number c ( G ) as the minimum number of cops for which there isa winning cop strategy on G . Among their results, they proved the following. Theorem 1.2 ([1]) If G is a connected planar graph, then c ( G ) ≤ . Various authors have studied the cop number of families of graphs [13, 12, 24, 25]. Recently,significant attention has been directed towards Meyniel’s conjecture (found in [12]) that c ( G ) = O ( √ n ) for any n vertex graph. The best current bound is c ( G ) ≤ n − (1+ o (1)) √ log n , obtainedindependently in [22, 32, 14]. The history of Meyniel’s conjecture is surveyed in [5]. For furtherresults on vertex pursuit games on graphs, see the surveys [3, 17] and the monograph [9].Herein, we study the game of cops and robbers on geometric graphs in R . Given points x , . . . , x n ∈ R and r ∈ R + , the geometric graph G = G ( x , . . . , x n ; r ) has vertices V ( G ) = { , . . . , n } and ij ∈ E ( G ) if and only if k x i − x j k ≤ r . Geometric graphs are widely used tomodel ad-hoc wireless networks [16, 34]. For convenience, we will consider V ( G ) = { x , . . . x n } ,referring to “point x i ” or “vertex x i ” when this distinction is required. Our first result gives aconstant upper bound on the cop number of 2-dimensional geometric graphs. Theorem 1.3 If G is a connected geometric graph in R , then c ( G ) ≤ . The proof of this theorem is an adaptation of the proof of Theorem 1.2. This adaptation requiresthree cops on a geometric graph to play the role of a single cop on a planar graph. We also give2n example of a geometric graph requiring 3 cops.Recent years have witnessed significant interest in the study of random graph models, moti-vated by the need to understand complex real world networks. In this setting, the game of copsand robbers is a simplified model for network security. There are many recent results on copsand robbers on random graph models, including the Erd˝os-Renyi model and random power lawgraphs [7, 23, 29, 10, 8, 30]. We add to this list of stochastic models by considering cops androbbers on random geometric graphs. A random geometric graph G on [0 , contains of n points drawn uniformly at random. Two points x, y ∈ V ( G ) are adjacent when the distancebetween them is within the connectivity radius, i.e. k x − y k ≤ r . We denote the probabilityspace of random geometric graphs by G ( n, r ). Typically, we view the radius as a function r ( n ),and then study the asymptotic properties of G ( n, r ) as n increases. We say that event A oc-curs with high probability , or whp , when P [ A ] = 1 − o (1) as n tends to infinity, or equivalently,lim n →∞ P [ A ] = 1 . For example, G ∈ G ( n, r ) is connected whp if r = q log n + ω ( n ) n . (Here and inthe remainder of this paper, ω ( n ) denotes an arbitrarily slowly growing function.) For this andfurther results on G ( n, r ), see the monograph [28].We improve on the bound of Theorem 1.3 when our random geometric graph is sufficientlydense. Essentially, we determine thresholds for which we can successfully adapt known pursuitevasion strategies to the geometric graph setting. Typical analysis of G ( n, r ) focusses on thehomogeneous aspects of the resulting graph, resulting from tight concentration around theexpected structural properties. Our cop strategies rely on these homogeneous aspects.When studying G ∈ G ( n, r ), it is often productive to tile [0 , into small squares, chosenso that whp, there is a vertex in each square, and vertices in neighboring squares are adjacentin G . We then use the induced grid on these vertices to analyze properties of G , cf. [4, 11]. Itis easy to show that the 2-dimensional grid has cop number 2. When our random geometricgraph is dense enough, we can adapt a winning two cop strategy on the grid to obtain a winningstrategy on G ( n, r ). Theorem 1.4
There is a constant K > such that the following holds. If G ∈ G ( n, r ) on [0 , with r ≥ K (log n/n ) then c ( G ) ≤ whp. A further increase in the connectivity radius leads to an even denser geometric graph, so thateventually the cops and robbers game on G ( n, r ) becomes quite similar to a turn-based pursuitevasion game on [0 , . Such pursuit evasion games on R d and in polygonal environments havebeen well studied, using winning criteria such as capture [33, 20, 6] and line-of-sight visibility[21, 15, 18]. It is known [33, 20] that pursuers can win the capture game in R d if and only if3he evader starts in the interior of the convex hull of the initial pursuer locations. Furthermore,a single pursuer can always catch the quarry in a bounded region, such as [0 , . We use thedismantlable criterion of Theorem 1.1 to prove that a sufficiently dense G ( n, r ) also requires asingle pursuer. Theorem 1.5
There is a constant K > such that the following holds. If G ∈ G ( n, r ) on [0 , with r ≥ K (log n/n ) , then c ( G ) = 1 whp. We note that Theorem 1.5 was proven independently by Alon and Pra lat [2] using a graphpursuit algorithm in the spirit of [33, 20].Finally we also give a lower bound of the cop number of G ( n, r ) proving that some randomgeometric graphs beyond the connectivity threshold require at least two cops. This answers aquestion of Alon [2]. Theorem 1.6
There is a constant K > such that the following holds. If G ∈ G ( n, r ) on [0 , with r ≤ K log n/ √ n , then c ( G ) > whp. We do not know whether any of our multiple cop bounds are tight. We are particularly hopefulthat the bound for arbitrary geometric graphs can be improved.
We begin by setting some notation. For x ∈ R and r ∈ R , define the ball B ( x, r ) = { y ∈ R : k x − y k ≤ r } .In the standard formulation of cops and robbers, the cops are first to act in each round. Incontinuous pursuit evasion games, the evader is usually first to act. The distinction is merelynotational, and we choose to view the robber as the first to act in each round. This leads toa more intuitive notation for the game state in our proofs below. Indeed, our cops are alwaysreacting to the robber’s previous move (which was made according to some unknown strategy),so it is useful to group these two moves together in a single round.We formally describe the game of cops and robbers using this notational convention. Beforethe game begins, the ℓ cops place themselves on the graph at vertices C , . . . , C ℓ . Then thegame begins. In the first round, the robber chooses his location R . Next the cops beginthe chase, moving to vertices C , . . . , C ℓ where C j ∈ N ( C j ). For i ≥
2, the i th round startsin configuration ( R i − , C i − , . . . , C i − ℓ ). The robber is first to act, leading to configuration4 R i , C i − , . . . , C i − ℓ ) where R i ∈ N ( R i − ) at the start of the i th cop turn. Next, the cops movesimultaneously to yield configuration ( R i , C i , . . . , C iℓ ) at the end of the i th round. The copswin if C ik = R i for some finite i, k . Otherwise the robber wins.Finally, we note that the winning cop criteria has an equivalent formulation. Namely, thecops win if there are finite i, k such that R i ∈ N ( C i − k ). Indeed, C k would subsequently capturethe evader on his i th move, achieving C ik = R i . Of course, if R i / ∈ N ( C i − k ) for all k , then therobber cannot be caught in the current round, and his evasion continues. In this section, we prove Theorem 1.3. Let G = G ( x , . . . x n ; r ) be a fixed geometric graph. Wesay that a cop C controls a path P if whenever the robber steps onto P , then he steps onto C or is caught by C on his responding move. Let diam( G ) denote the diameter of the graph.Aigner and Fromme [1] prove the following. Lemma 3.1 ([1])
Let G be any graph, u, v ∈ V ( G ) , u = v and P = { u = v , v , . . . v s = v } ashortest path between u and v . A single cop C can control P after at most diam( G ) + s moves. It takes C at most diam( G ) moves to reach P , and then at most s moves to take control of P .We have the following simple corollary which will be useful for geometric graphs. Corollary 3.2
Suppose that there are three cops C − , C, C + chasing robber R on G . Considera shortest ( u, v ) -path P = { u = v , v , . . . , v s = v } . After k ≤ diam( G ) + 2 s moves, the cop C controls P , and ( C k − , C k , C k + ) = ( v i − , v i , v i +1 ) , where we set v − = u and v s +1 = v .Proof. Start with the three cops colocated on any vertex of P . The cops attain this controllingconfiguration in two phases. In phase one, cops move as one until they control the path, as inLemma 3.1. In phase two, C remains in control of the path while C − , C + obtain their properpositions within s moves. Assume that until round j ≥ C + is colocated with C .If C stays put on v i in round j , then C + moves to v i +1 . If C moves from v i to v i − then C + stays put on v i . Otherwise, both C and C + move to v i +1 . After at most s rounds, C must eitherstay put or move left, and C + attains his proper position. Similarly, C − attains his positionwithin s rounds.Geometric graphs are frequently non-planar. Because of crossing edges, simply keeping R from stepping onto P does not necessarily prevent him from moving from one side of P to5he other. We say that R crosses P at time t if the closed segment R t − R t has nonemptyintersection with the closed segments corresponding to the edges of P . The additional guardsflanking C ensure that once the three cops are positioned as in Corollary 3.2, R cannot cross P . On a geometric graph, we say that a set of cops patrols a path P if they control P andwhenever R crosses P , he is caught in the subsequent cop move. Lemma 3.3
Let P = { v , . . . , v t } be a shortest path on a geometric graph G ( x , . . . , x n ; r ) .Suppose that the cops C − , C, C + are located on v i − , v i , v i +1 respectively, and that cop C controls P . Then these three cops patrol P .Proof. If the robber steps onto P then C will capture him. Suppose that the robber cancross P without losing the game, and does so from position R t to R t +1 . We characterize someconstraints on the location of R t . Consider the configuration ( R t , C t − − , C t − , C t − ) prior torobber’s crossing. This occurs in round t , after the robber move but before the cop moves. Atthis point, the cops are positioned on three successive vertices of P . We claim that R t / ∈ B ( C t , r ).Indeed, if C t − = C t (so that the cops are stationary in round t ), then C can actually catch R at time t , a contradiction. Otherwise C t ∈ { C t − − , C t − } , so one of these flanking cops cancatch R at time t −
1, also a contradiction.Next, we observe that the robber cannot be far from the cops. Let ( R t , C t − − , C t − , C t − ) =( R t , v i − , v i , v i +1 ) . First of all, R t / ∈ B ( v i − , r ) ∪ B ( v i +2 , r ). Indeed, if R t is close to eitherof v i − , v i +2 then R could step onto that vertex in round t + 1 without being caught by C ,contradicting the fact that C controls P . Secondly, R t cannot be within 2 r of any path vertex v j where | i − j | > P between v i − and v i +2 . The region forbidden to R t along this subpath is shown in Figure 3.1(a).Without loss of generality, assume that R crosses P so that R t R t +1 intersects v i v i +1 or v i +1 v i +2 . Now R t +1 / ∈ B ( v i , r ) ∪ B ( v i +1 , r ); otherwise either C or C + can immediately catchhim. Suppose that R t R t +1 crosses v i v i +1 where R t / ∈ B ( v i , r ) ∪ B ( v i +2 , r ) and R t +1 / ∈ B ( v i , r ) ∪ B ( v i +1 , r ), as shown in Figure 3.1(b). We have k v i − v i +1 k ≤ r and k R t − v i k > r . This meansthat the angle ∡ v i R t v i +1 < π/
2; otherwise in the triangle v i v i +1 R t , this obtuse angle forces r ≥ k v i − v i +1 k > k v i − R t k > r , a contradiction. Likewise, since k R t +1 − v i +1 k > r , we musthave ∡ v i R t +1 v i +1 < π/
2. Therefore max { ∡ R t v i R t +1 , ∡ R t v i +1 R t +1 } > π/
2, and the resultingobtuse triangle forces k R t − R t +1 k > r , a contradiction. Therefore R cannot cross P by crossing v i v i +1 . An identical argument, replacing v i with v i +2 , shows that R cannot cross v i +1 v i +2 .Therefore, R cannot cross P .We now prove that if G is a connected geometric graph in R , then c ( G ) ≤ . t − C t C t + v i − v i − v i v i +1 v i +2 C t + C t v i v i +1 v i +2 R t R t +1 (a) (b)Figure 3.1: (a) The robber must cross between v i − and v i +2 , but R t cannot lie in the grayregion B ( v i − , r ) ∪ B ( v i , r ) ∪ B ( v i +2 , r ). (b) The geometry of the quadrilateral v i R t v i +1 R t +1 shows that the robber cannot cross P at edge v i v i +1 without ending in B ( C t , r ) ∪ B ( C t + , r ). Proof of Theorem 1.3 . The proof is a direct adaptation of the Aigner and Fromme [1] proofof Theorem 1.2. In our proof, we need 3 cops to patrol a shortest path of a geometric graph,instead of the single cop required to control a shortest path of a planar graph. The idea of theproof of Aigner and Fromme is divide the pursuit into stages. In stage i , we assign to R a certainsubgraph H i , the robber territory , which contains all vertices which R may still safely enter,and to show that, after a finite number of cop-moves, H i is reduced to H i +1 ( H i . Eventually,there is no safe vertex left for the robber. In each iteration, at most two shortest paths in H i must be controlled. For a planar graph, this requires one cop per path, and the third cop movesto control another shortest path in H i . For geometric graphs, Lemma 3.3 shows that 3 copscan patrol any shortest path of a geometric graph. Using that lemma in place of Lemma 3.1,the proof of Aigner and Fromme for planar graphs with 3 cops becomes a proof for geometricgraphs with 9 cops. See [1] for the proof details.It is an open question whether this upper bound on the cop number can be improved forthe class of geometric graphs. Here we construct a geometric graph that requires 3 cops, whichleaves a considerable gap to our upper bound. Aigner and Fromme [1] proved that any graphwith minimum degree δ ( G ) ≥ g ( G ) ≥ c ( G ) ≥ δ ( G ). We describe a geometricgraph G on 1440 vertices with unit connectivity radius which has girth 5 and minimum degree3, so that c ( G ) ≥
3. A representative subgraph of G appears in Figure 3.2. Start with anannulus having inner radius 55 and outer radius 57. Within the annulus, we create an innerand outer strip of pentagons. Each pentagon corresponds to a one degree angle (or π/ r : θ ) where θ is in degrees. For integral θ , 1 ≤ θ ≤ θ )7nd at (57 : θ + 1 / / θ ), (56 .
35 : 2 θ + 0 . .
85 : 2 θ + 1) and (56 : 2 θ + 1 .
5) for integral θ , 1 ≤ θ ≤ .
995 and 0 . G on 1440 vertices with c ( G ) = 3. The eightcircles show the connectivity neighborhood for each type of vertex.We must have c ( G ) = 3 since G is planar. Indeed, there is a simple winning strategy forthree cops. Have cop C remain stationary on any interior vertex. Place cops C , C on verticeson the inner and outer boundaries, separated by half a degree. In each step, one of the boundarycops can take a clockwise step along his boundary while preventing the robber from crossingthe shortest path between C , C . Eventually the robber cannot move counterclockwise becauseof C , C , and cannot move clockwise because of C . G ( n, r ) In this section, we prove Theorem 1.4. Our winning two cop strategy is similar to a winningstrategy on the grid P n (cid:3) P m . One cop catches the robber’s “shadow” in a copy of P n , whilethe other catches the robber’s shadow in a copy of P m . On subsequent moves, either therobber moves towards the boundary, or at least one cop decreases his distance from the robber.8ventually, the robber hits the boundary, and the cops close in for the win. Our cop strategybelow follows along similar lines, but accommodates the full range of robber movement.It is convenient to split the proof of Theorem 1.4 into two parts, a probabilistic part and adeterministic part. Let V = { x , . . . , x n } ⊂ [0 , and let r ≥ s >
0. Let us say that the tuple( x , . . . , x n ; r, s ) satisfies condition (M) when the following holds: (M) For every x ∈ [0 , and every y ∈ B ( x, r ) ∩ [0 , , we have V ∩ B ( x, r ) ∩ B ( y, s ) = ∅ .All the probability theory needed in the proof of Theorem 1.4 is contained in the followinglemma. Lemma 4.1
Let us set s := 5 p log n/n . Let x , . . . , x n ∈ [0 , be chosen i.i.d. uniformly atrandom, and let r ≥ s be arbitrary. Then ( x , . . . , x n ; r, s ) satisfies condition (M) whp. Proof:
Let us set t := 1 / lp n/ n m . Then t = (1 + o (1)) p n/n and it is of the form t = 1 /k with k ∈ N an integer. We can thus tile [0 , into 1 /t squares of dimension t × t . Let Z denote the number of these squares that do not contain any point of x , . . . , x n . Then E [ Z ] = (1 /t ) · (1 − t ) n ≤ (1 /t ) e − nt = (1 + o (1)) n n e − (1+ o (1))2 log n = o (1) . Thus, whp each square contains at least one x i .Now let us assume that each square of our dissection indeed contains a point of x , . . . , x n and pick an arbitrary x ∈ [0 , and y ∈ B ( x ; r ) ∩ [0 , . If k x − y k < r − t √ y is completely contained in B ( x ; r ) (because the diameterof a t × t square is t √ x i that lies inside this square will clearly do as k y − x i k ≤ t √ < s . Let us thus assume r − t √ ≤ k x − y k ≤ r , and let z ∈ [ x, y ] be chosen onthe segment between x and y in such a way that k z − x k = r − t √
2. Then the square of ourdissection that contains z is contained in B ( x ; r ) and the point x i inside this square satisfies k y − x i k ≤ k y − z k + k z − x i k ≤ t √ ≤ s . (cid:4) Lemma 4.2
Suppose that ( x , . . . , x n ; r, s ) with x , . . . , x n ∈ [0 , and < s < r / satisfycondition (M) . Then c ( G ( x , . . . , x n ; r )) ≤ . Proof:
We can assume without loss of generality that r ≤ √ G is a cliqueand a single cop will be able to catch the robber in a single move. We start by describing thestrategy of the cops. The two cops act independently (i.e. the action of C does not depend on9he position or movement of C and vice versa). First, we describe only the movements of C .Cop C will follow a similar strategy, described below.We introduce notation for a series of lines and points. Suppose the robber is at point R t .Let L t be the vertical line through R t . Let P t denote the point on L exactly r/ R t provided this point is above the x -axis. Otherwise P t is the point on the x -axis exactly below R t . Similarly, we define the horizontal line L t and the point P t to the left of R t on L . Forsimplicity, we occasionally refer to L , L , P , P (without the superscript) to refer to these linesand points with respect to the current position of R .At time t = 0, C starts at a vertex C := x j that is within s of the origin (0 , t ; such an x j exists because of (M) . In each round, the robber will first choose his new location R t +1 . Thecop then chooses a point y ∈ B ( C t , r ) ∩ [0 , and finds an x i ∈ B ( C t , r ) ∩ B ( y, s ) (such an x i exists because of property (M) ) and chooses as his new location C t +11 := x i . The strategy of C has three phases:S1: Cop C moves right until he reaches a point within s of L and within r/ of the x -axis.S2: While staying within r/ of L , cop C moves to within s of the point P .S3: Cop C tries to stay as close to P as he can. Stage S1:
During stage S1, cop C moves as follows. Let y be the point of B ( C t , r ) closestto L t +11 . Then C moves to a point x i ∈ B ( C t , r ) ∩ B ( y, s ). If y ∈ L then stage S1 ends.Otherwise, the cop travels a horizontal distance of at least r − s . Thus, stage S1 lasts no morethan ⌈ / ( r − s ) ⌉ < /r rounds, since he can keep jumping right by at least r − s and he willreach L before he reaches the right boundary of the unit square (note the cop either starts tothe left of L or within s of L ). Observe that, by the end of stage S1, the y -coordinate of C is at most s · /r < r/ (as s < r / ). Stage S2:
In this stage, the cop will always stay as close to L as he can, and will movecloser to his target point P if he can. The round starts with C t within s of L t and within r/ of the x -axis. If R t has y -coordinate smaller than r/ R t is above C t .If P t +11 ∈ B ( C t , r ) then we can pick an x i ∈ B ( C t , r ) ∩ B ( P t +11 , s ) and set C t +11 := x i , therebyending stage S2. Otherwise, the cop’s move depends on how the robber moves. We classify thepossible robber moves into four (non-exclusive) types, depending on where the robber jumps,as shown in Figure 4.1. Writing this displacement in polar coordinates ( d : θ ), the four typesare 101: d ≤ r/ r/ < d ≤ r and 7 π/ ≤ θ ≤ π/ r/ < d ≤ r and 2 π/ ≤ θ ≤ π/ r/ < d ≤ r and − π/ ≤ θ ≤ π/ R does a T1 move, then we compute J t +1 := R t +1 − R t . We can write J t +1 = ( ℓ cos α, ℓ sin α )with ℓ ≤ r/
2. Assuming C t is within r/ of L t , we can move at most r (cos( α ) / / ) to theleft or right to reach L t +11 . Thus y := (cid:0) R t +1 x , C t + r (cid:0) − cos α − / (cid:1)(cid:1) T ∈ L t +11 ∩ B ( C t , r ),where R t +1 x is the x -coordinate of R t +1 . We pick x i ∈ B ( C t +11 , r ) ∩ B ( y, s ) and set C t +11 := x i .Observe x i is within s of L t +11 and that the distance between C and R has decreased by atleast r (cid:0) − sin α − cos α − / (cid:1) − s ≥ r (cid:0) − √ − / − / (cid:1) > r/ R does a T2 move, then L moves left or right by at most r cos( π/
6) = √ r/ R moves down by at least r sin( π/
6) = r/
2. Assuming that C t is within r/ of L t , we canthus move sideways by at most ( √ / / ) r and reach L t +11 . We can therefore pick a point y ∈ L t +11 ∩ B ( C t , r ) that is at least ( − √ / − / ) r − s > r/ R t +1 than C t is to R t . Again we pick x i ∈ B ( C t +11 , r ) ∩ B ( y, s ) and set C t +11 := x i .If R does a T3 or T4 move then we compute y := R t +1 − R t + C t , (if y [0 , then wetake the point y ′ on ∂ [0 , with minimum distance to y ) we pick x i ∈ B ( C t , r ) ∩ B ( y, s ) andwe set C t +11 := x i . Note that this way the distance of C to P cannot increase by more than s . Stage S3:
At present it is not yet clear whether stage S2 will ever finish (and also wemay not be able to stay within r/ of L indefinitely). If however we do get to stage S3, weobserve that R cannot make a T1 or T2 move without getting caught by the cop immediately(see Figure 4.2). Therefore, during stage S3, we act exactly as in the case of stage S2 where R does a T3 or T4 move. This concludes the description of the first cop’s movements.Suppose that during the first T = 1000 /r moves of the game the robber does not get caught.Stage S1 will have finished after at most 10 /r moves. Since s · T < r/ , we will be able to11 C Figure 4.2: If C is within r/ of the point r/ R , then R can no longer make T1 orT2 moves.stay within r/ of L for the remaining moves until T , and assuming we reach stage S3 atsome time t < T we will be able to stay within r/ of P for the remaining moves until T .Thus stage S2 will have finished as soon as we have done at most 14 /r moves of type T1 or T2(the first 10 /r may occur during stage S1 and after that we move closer to P by at least r/ T moves, at most 14 /r robber moves are of typeT1 or T2.Completely analogously we can define a strategy for the second cop C that will ensure thatin the first T moves no more than 14 /r robber moves are of type T1 or T3. Cop C tries toattain position on the horizontal line L through R . The stages of his strategy are:S ′
1: Cop C moves up until he reaches a point within s of L and within r/ of the y -axis.S ′
2: While staying within r/ of L , cop C moves to within s of the point P .S ′
3: Cop C tries to stay as close to P as he can.Observe that whenever R does a T4 move, then the sum of his coordinates increases by atleast min − π/ ≤ θ ≤ π (sin θ + cos θ ) r √ − ! r. Meanwhile, if the robber makes a T T T r √ θ = 5 π/ T moves, then the sum of robbers coordinates at time T is at least R Tx + R Ty ≥ ( T − /r ) · (cid:16) √ − (cid:17) r − (28 /r ) · r √
2= 972 (cid:16) √ − (cid:17) − √ > . But this is impossible, since the robber stays inside the unit square. It follows that R getscaught by the cops within the first T moves. (cid:4) roof of Theorem 1.4: Follows from Lemmas 4.1 and 4.2 by taking K = 3 · . (cid:4) G ( n, r ) In this section, we prove Theorem 1.5 by showing that when r ≥ K (log n/n ) / the randomgeometric graph is dismantlable. We begin by setting some notation. Let c := ( , ) denotethe center of the unit square [0 , . Let us write N c ( i ) := { ≤ j ≤ n : k x i − x j k ≤ r, and k x j − c k < k x i − c k} . In other words, N c ( i ) is the set of (indices) of vertices adjacent to x i and closer to the center c than x i . We will prove the following lemma. Lemma 5.1
There is a constant K > such that the following holds. Suppose r ≥ K (log n/n ) / .Whp the following holds for all ≤ i ≤ n : either k x i − c k < r/ , or there is a j ∈ N c ( i ) suchthat N c ( i ) ⊆ N c ( j ) . Assuming that Lemma 5.1 holds, the proof of Theorem 1.5 is straightforward dismantling ofthe random geometric graph.
Proof of Theorem 1.5:
We can induce a strict ordering of the vertices according to theirdistance from the center c , in descending order. Indeed, for any vertices x, y , P ( k x − c k = k y − c k ) = 0. By Lemma 5.1, the outermost vertex is a pitfall, and can be removed. Wecontinue to remove vertices until the remaining vertices lie in B ( c, r/ c ( G ) = 1. (cid:4) The remainder of this section is devoted to proving Lemma 5.1, which requires a series ofintermediate geometric lemmas. For x, y ∈ R , let us write W ( x, y ; r ) := { z ∈ R : B ( z, r ) ⊇ B ( x, r ) ∩ B ( y, k x − y k ) } . (1)Let [ x, y ] denote the line segment between these two points. Note thatif z ∈ [ x, y ] then W ( x, y ; r ) ⊆ W ( x, z ; r ) . (2)Indeed, we have B ( x, r ) ∩ B ( z, k x − z k ) ⊆ B ( x, r ) ∩ B ( y, k x − y k ) so that W ( x, y ; r ) ⊇ W ( x, z ; r ).Observe that area( W ( x, y ; r )) does not depend on the exact locations of x, y , but only on k x − y k r . We can thus denote A ( d, r ) := area( W ( x, y ; r )) for an arbitrary pair x, y with k x − y k = d .By observation (2), the area A ( d, r ) is nonincreasing in d for a fixed r .We give a simpler geometric characterization of W ( x, y ; r ) when k x − y k = d > r . Let p , p denote the two intersection points of ∂B ( x, r ) and ∂B ( y, d ). Denote W ′ ( x, y ; r ) := B ( p , r ) ∩ B ( p , r ) , as shown in Figure 5.1(a). y xp p W ′ D ′ D (a) (b)Figure 5.1: (a) The set W ′ = W ′ ( x, y ; r ). (b) When closed discs intersect, the smaller disc D contains the shortest arc on the bigger disc D ′ between the intersection points of the boundaries. Lemma 5.2 If k x − y k = d > r then W ′ ( x, y ; r ) = W ( x, y ; r ) . Proof:
Pick any z ∈ W ( x, y ; r ). We must have p , p ∈ B ( z, r ), which means that z ∈ B ( p , r ) ∩ B ( p , r ) Therefore W ( x, y ; r ) ⊆ W ′ ( x, y ; r ).Picking any z ∈ W ′ ( x, y ; r ), we have p , p ∈ B ( z, r ). Observe that if a closed disc D intersects a disc D ′ of the same or larger radius then D contains the shortest circular arc along ∂D ′ between the two intersection points of ∂D and ∂D ′ , see Figure 5.1(b). So B ( z, r ) containsthe part of ∂B ( x, r ) between p and p that lies inside B ( y, d ). Using that d > r , B ( z, r ) alsocontains the part of ∂B ( y, d ) between p and p that falls inside B ( x, r ). Thus B ( z, r ) contains ∂ ( B ( x, r ) ∩ B ( y, k x − y k )). Because both B ( z, r ) and B ( x, r ) ∩ B ( y, k x − y k ) are convex, it nowalso follows that B ( x, r ) ∩ B ( y, d ) ⊆ B ( z, r ). This shows that W ′ ( x, y ; r ) ⊆ W ( x, y ; r ). (cid:4) We now compute a lower bound for A ( d, r ) for distant vertices x, y .14 y xp p d h rW s Figure 5.2: Determining the area of W = W ( x, y ; r ). Lemma 5.3 If d = K · max (cid:0) r, / √ (cid:1) where K > is a sufficiently large constant, then A ( d, r ) = Ω( r ) . Proof:
Choose x, y ∈ R with k x − y k = d . The geometry of W = W ( x, y, r ) is shown inFigure 5.2. We have area( W ) = 4 (cid:16) πr (cid:16) α π (cid:17) − r cos( α ) sin( α ) (cid:17) = r · (2 α − sin(2 α )) . (3)Indeed, the expression πr (cid:0) α π (cid:1) equals the area of a slice of opening angle α out of a disc ofradius r , and the term r cos( α ) sin( α ) equals the area of a triangle with sides h = r cos( α )and s = r sin( α ). Also note that d = h + ( d − s ) and r = h + s , giving s = r / d = min (cid:18) r K √ , r K (cid:19) = Ω( r ) . Thus, sin( α ) = s/r = Ω( r ), and because sin( x ) = x + o ( x ), this also gives α = Ω( r ). Theapproximation x − sin( x ) = x / o ( x ), together with (3), proves the lemma. (cid:4) Our next lemma places a lower bound on area( W ( x, c ; r )) where c = ( , ) is the center ofthe unit square. Lemma 5.4
For all x ∈ [0 , with k x − c k ≥ r/ , we have area (cid:0) W ( x, c ; r ) ∩ [0 , ∩ B ( c, k x − c k ) (cid:1) = Ω( r ) . Proof:
Pick the point ˜ c on the line L containing x and c , so that c ∈ [˜ c, x ] and k x − ˜ c k = d = K · max( r, / √ W ( x, ˜ c ; r ) ⊆ W ( x, c ; r ). Provided15 c ˜ c Ld (0 ,
0) (1 , c such that c ∈ [˜ c, x ].that K is sufficiently large, we have diam( W ( x, ˜ c ; r )) < r/ . Furthermore, both the anglebetween ∂B ( p , r ) and the line L at their intersection points, and the angle between ∂B ( p , r )and the line L at their intersection points will be less than 1 degree. It follows directly that W ( x, ˜ c ; r ) ⊆ [0 , ∩ B ( c, k x − c k ) for every x ∈ [0 , \ B ( c, r/ (cid:4) We conclude this section with the proof of our main lemma: that for every vertex x i suchthat k x i − c k > r/
2, there is a j ∈ N c ( i ) such that N c ( i ) ⊆ N c ( j ). Proof of Lemma 5.1:
We can assume without loss of generality that r ≤ √ k x i − c k < r/ i ). Let Z denote the number of indices i such that k x i − c k ≥ r/ j ∈ N c ( i ) such that N c ( j ) ⊇ N c ( i ). Then E Z can be boundedabove by: E [ Z ] ≤ n Z [0 , \ B ( c,r/ (cid:0) − area( W ( x, c ; r ) ∩ [0 , ) (cid:1) n − d x ≤ n (cid:0) − Ω( r ) (cid:1) n − ≤ n exp (cid:2) − Ω( nr ) (cid:3) Thus, if we chose K sufficiently large we have E Z ≤ exp[log n − Ω( nr )] = exp[ − Ω(log n )] = o (1). So the assertion of the lemma holds whp. (cid:4) G ( n, r ) near the connectivity threshold is not cop-win In this section, we prove that some random geometric graphs require at least two cops. Inparticular, when we are near the connectivity threshold, the graph is not dismantlable whp.
Proof of Theorem 1.6:
Without loss of generality we can assume r ≥ p log n/n , becauseby a result of Penrose [27] our graph is disconnected whp for smaller choices of r (obviously a16isconnected graph is not cop-win). We will show that there is a small constant K > r ≤ K log n/ √ n then whp the graph is not dismantlable.Intuitively, we are hunting for a subset of [0 , as shown in Figure 6.1. Start with an N -gonwith side length ρ , slightly smaller than r . Draw a small disc B ( c i , ρ ) around each corner,where ρ + 2 ρ = r. We want each disc B ( c i , ρ ) to contain exactly one vertex of G , say x i .Next, we consider the sets B ( x i − , r ) ∩ B ( x i +1 , r ). We want this intersection to contain no othervertices besides x i . If we can find such a structure, it creates a cycle { x , . . . x N } in G such that x i the only vertex in G that is adjacent to both x i − , x i +1 (addition modulo N ). Therefore G is not dismantlable because none of the x i will ever become pitfalls. B ( c i − , ρ ) B ( c i , ρ ) B ( c i +1 , ρ ) B ( x i − , r ) ∩ B ( x i +1 , r ) x i − x i x i +1 ρ r r Figure 6.1: For an N -gon with side length ρ , we want each B ( c i , ρ ) to contain a single vertex,and we want each B ( x i − , r ) ∩ B ( x i +1 , r ) to contain no additional vertices.We now prove the existence of such a structure. Let N denote the number of vertices ofthe cycle; we will specify this value later. Set ρ = r − r/N and ρ = r/ N . Consider aregular N -gon Γ ⊆ [0 , , whose edges each have length ρ . (Once we fix our choice of N , weshall see later that Γ fits easily inside the unit square [0 , .) Let us label the corners of Γas c , . . . , c N − for convenience, where of course c i is next to c i − and c i +1 (addition of indicesmodulo N ). We will insist that, for each 0 ≤ i ≤ N − x j i ∈ B ( c i , ρ ) with { x , . . . , x n } ∩ B ( c i , ρ ) = { x j i } , (4)and the point x j i is also the unique common neighbor of the two points x j i − and x j i +1 , i.e. { x , . . . , x n } ∩ B ( x j i − , r ) ∩ B ( x j i +1 , r ) = { x j i } . (5)17bserve that k c i +1 − c i − k = 2 ρ sin (cid:18) π ( N − N (cid:19) = 2 ρ cos (cid:16) πN (cid:17) = 2 r (cid:0) − /N (cid:1) (cid:0) − O (cid:0) /N (cid:1)(cid:1) = 2 r − O (cid:0) r/N (cid:1) using the Taylor approximation cos( x ) = 1 − x + O ( x ). Hence for any x ∈ B ( c i +1 , ρ ) and y ∈ B ( c i − , ρ ) we also have k x − y k = 2 r − O ( r/N ). Let us write W ( x, y ) := B ( x, r ) ∩ B ( y, r ).By the same computation as equation (3),area( W ( x, y )) = r (2 β − sin(2 β )) = O ( r β ) , where β is a small angle with cos β = k x − y k /r = 1 − O (1 /N ), so that β = O (1 /N ) (againusing the Taylor expansion of cosine), Hencearea( W ( x, y )) = O ( r /N ) . (6)Rather than computing directly in the standard random geometric graph, it helps to considera “Poissonized” version. Consider an infinite sequence x , x , . . . of random points, i.i.d. uni-formly at random on the unit square. The ordinary random geometric graph, which we willdenote by G O for the rest of the proof, is just G ( x , . . . , x n ; r ). Now let Z = d Po( n ) be a Poissonrandom variable of mean n , independent of the points x , x , . . . and consider the random ge-ometric graph G ( x , . . . , x Z ; r ) on the points x , . . . , x Z which we will denote by G P . Observethat the points x , . . . , x Z constitute a Poisson process of intensity n on the unit square, whichhas the convenient properties that for every A ⊆ [0 , the number of points that fall in A is a Poisson random variable with mean n · area( A ), and that for any two disjoint sets A, B the number of points in A is independent of the number of points in B (cf. [19]). This makes G P slightly easier to handle than G O . We shall first do our probabilistic computations for thePoissonized version G P and then we’ll derive the results for the original model G O from thosefor the Poissonized one.Let us say the polygon Γ is good if it satisfies the demands of equations (4) and (5) with Z swapped for n . Employing the useful independence properties of the Poisson process we nowsee that P [Γ is good] = (cid:0) P [Po(( nπr / N ) = 1] (cid:1) N · P [Po( n · O ( r /N )) = 0]= (cid:0) ( nπr / N ) exp( − nπr / N ) (cid:1) N · exp( − O ( nr /N ))= exp (cid:0) N log( nπr / − N log N − O ( nr /N ) (cid:1) . Considering the right hand side of the first inequality, the first term is the probability thatthe N discs B ( c i , ρ ) contain exactly one random point, and the second term is the probability18hat the N sets ( B ( x i − , r ) ∩ B ( x i +1 , r )) \ B ( c i , ρ ) contain no random points. We now choose N = ⌈ (cid:0) nπr (cid:1) / ⌉ and choose K > P (Γ is good) ≥ exp (cid:16) − O (cid:16) √ nr (cid:17)(cid:17) ≥ exp (cid:18) −
12 log n (cid:19) = n − because r ≤ K log n/ √ n by assumption. Also note that as promised before, the polygon Γ fitseasily inside the unit square as it has diameter O ( rN ) = O ( r ( nr ) / ) = o (1).Let us now place shifted copies Γ , . . . , Γ M of Γ inside the unit square in such a way thatthey are contained in [0 , and their centers are separated by at least 10 diam(Γ) = Θ( rN ) =Θ( n / r / ) = n − / o (1) . (Recall we assumed without loss of generality that r = Ω( p log n/n ).)Then we can place M = Ω((1 /rN ) ) = n − o (1) such shifted copies, with their centers forminga lattice in [0 , . Let X denote the number of Γ i s that are good. Now notice that the eventsthat the Γ i are good are independent of each other as they concern disjoint areas of the plane.Hence X is distributed like a binomial with parameters M = n − o (1) and p ≥ n − . Thus: P [ X = 0] = (1 − p ) M ≤ e − Mp ≤ e − n / − o (1) = o (1) . So X > G O again. Let X P denote the number of goodΓ i s under the Poisson model, and let X O denote the number of good Γ i s under the originalmodel. We have, with K > P [ X O = 0 | X P >
0] = P ∞ z =0 P [ X O = 0 | X P > , Z = z ] P [ X P > | Z = z ] P [ Z = z ] ≤ P ∞ z =0 P [ X O = 0 | X P > , Z = z ] P [ Z = z ] ≤ P n + K √ nz = n − K √ n P [ X O = 0 | X P > , Z = z ] P [ Z = z ]+ P [ | Z − n | > K √ n ] . (7)By Chebyschev’s inequality we have P (cid:2) | Z − n | > K √ n (cid:3) ≤ Var( Z ) / ( K √ n ) = 1 /K . Now consider the term P [ X O = 0 | X P > , Z = z ]. If z = n then it clearly equals 0. Let us take n − K √ n ≤ z < n . If we condition on the event that X P > , Z = z , then we can fix a goodΓ i , say with “corners” ( x i , . . . , x i N ). If X O = 0 then the set A := S Nj =1 W ( x i j − , x i j +1 ) mustcontain one of the points x z +1 , . . . , x n . By equation (6), area( A ) = N · O ( r /N ) = O ( r /N ).Thus, for n − K √ n < z < n we have P [ X O = 0 | X P > , Z = z ] ≤ ( n − z ) · O ( r /N ) ≤ K √ n · O ( r/ √ n )= o (1) , N = ⌈ (cid:0) πnr (cid:1) / ⌉ . Observe that the o (1) bound is uniform over all n − K √ n < z < n .Similarly, if we condition on the event that X P > , Z = z with n < z ≤ n + K √ n , wecan pick an N -tuple ( x i , . . . , x i N ) uniformly at random from all N -tuples that are “corners”of a good Γ i . The indices i , . . . , i N are a uniformly random sample (without replacement)from { , . . . , z } . Now, if X O = 0, it must hold that one of i , . . . i N is larger than n . Note P ( i j > n ) = ( z − n ) /z for j = 1 , . . . , N , and so P [ X O = 0 | X P > , Z = z ] ≤ N (cid:18) z − nz (cid:19) ≤ ( πnr ) (cid:18) K √ nn (cid:19) = Kπ n − r = o (1) . Observe that again the o (1) bound is uniform over all z considered. Combining these boundswith (7) we get P [ X O = 0 | X P > ≤ /K + P n + K √ nz = n − K √ n o (1) · P [ Z = z ]= 1 /K + o (1) . By sending K → ∞ , we see that P [ X O = 0 | X P >
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