Chasing robbers on random geometric graphs---an alternative approach
CCHASING ROBBERS ON RANDOM GEOMETRIC GRAPHS—ANALTERNATIVE APPROACH
NOGA ALON AND PAWE(cid:32)L PRA(cid:32)LAT
Abstract.
We study the vertex pursuit game of
Cops and Robbers , in which copstry to capture a robber on the vertices of the graph. The minimum number of copsrequired to win on a given graph G is called the cop number of G . We focus on G d ( n, r ),a random geometric graph in which n vertices are chosen uniformly at random andindependently from [0 , d , and two vertices are adjacent if the Euclidean distancebetween them is at most r . The main result is that if r d − > c d log nn then the copnumber is 1 with probability that tends to 1 as n tends to infinity. The case d = 2was proved earlier and independently in [4], using a different approach. Our methodprovides a tight O (1 /r ) upper bound for the number of rounds needed to catch therobber. Introduction
The game of
Cops and Robbers , introduced independently by Nowakowski and Win-kler [17] and Quilliot [22] almost thirty years ago, is played on a fixed graph G . Wewill always assume that G is undirected, simple, and finite. There are two players, aset of k cops , where k ≥ robber . The cops begin the gameby occupying any set of k vertices (in fact, for a connected G , their initial position doesnot matter). The robber then chooses a vertex, and the cops and robber move in alter-nate rounds. The players use edges to move from vertex to vertex. More than one copis allowed to occupy a vertex, and the players may remain on their current positions.The players know each others current locations. The cops win and the game ends if atleast one of the cops eventually occupies the same vertex as the robber; otherwise, thatis, if the robber can avoid this indefinitely, he wins. As placing a cop on each vertexguarantees that the cops win, we may define the cop number , written c ( G ), which isthe minimum number of cops needed to win on G . The cop number was introducedby Aigner and Fromme [1] who proved (among other things) that if G is planar, then c ( G ) ≤
3. For more results on vertex pursuit games such as
Cops and Robbers , thereader is directed to the surveys on the subject [3, 11, 13] and the monograph [7]. Themost important open problem in this area is Meyniel’s conjecture (communicated byFrankl [10]). It states that c ( n ) = O ( √ n ), where c ( n ) is the maximum of c ( G ) overall n -vertex connected graphs. If true, the estimate is best possible as one can con-struct a graph based on the finite projective plane with the cop number of order at Key words and phrases. random graphs, vertex-pursuit games, Cops and Robbers.The first author acknowledges support by an ERC Advanced grant, by a USA-Israeli BSF grant,and by the Hermann Minkowski Minerva Center for Geometry at Tel Aviv University. The secondauthor acknowledges support from NSERC and Ryerson University. a r X i v : . [ m a t h . C O ] J un NOGA ALON AND PAWE(cid:32)L PRA(cid:32)LAT least Ω( √ n ). Up until recently, the best known upper bound of O ( n log log n/ log n )was given in [10]. This was improved to c ( n ) = O ( n/ log n ) in [9]. Today we know thatthe cop number is at most n − (1+ o (1)) √ log n (which is still n − o (1) ) for any connectedgraph on n vertices (a result obtained independently by Lu and Peng [15] and Scottand Sudakov [23], see also [2, 12] for some extensions). If one looks for counterex-amples for Meyniel’s conjecture it is natural to study first the cop number of randomgraphs. Recent years have witnessed significant interest in the study of random graphsfrom that perspective [6, 8, 16, 19] confirming that, in fact, Meyniel’s conjecture holdsasymptotically almost surely for binomial random graphs [21] as well as for random d -regular graphs [20].In this note we consider a random geometric graph G d ( n, r ) which is defined as arandom graph with vertex set [ n ] = { , , . . . , n } in which n vertices are chosen uni-formly at random and independently from [0 , d , and a pair of vertices within Euclideandistance r appears as an edge—see, for example, the monograph [18].As typical in random graph theory, we shall consider only asymptotic properties of G d ( n, r ) as n → ∞ , where r = r ( n ) may and usually does depend on n . We say that anevent in a probability space holds asymptotically almost surely ( a.a.s. ) if its probabilitytends to one as n goes to infinity.2. The result and its proof
We prove the following result.
Theorem 2.1.
There exists an absolute constant c > so that if r > c nn thena.a.s. c ( G ( n, r )) = 1 . The same result was obtained earlier and independently in [4] but the proof presentedhere is quite different, provides a tight O (1 /r ) bound for the number of rounds requiredto catch the robber, and can be generalized to higher dimensions. In the proof wedescribe a strategy for the cop that is a winning one a.a.s. In [4], the known necessaryand sufficient condition for a graph to be cop-win (see [17] for more details) is used;that is, it is shown that the random geometric graph is what is called dismantlablea.a.s.Essentially the same proof we provide here gives the following. Theorem 2.2.
For each fixed d > there exists a constant c d > so that if r d − >c d log nn then a.a.s. c ( G d ( n, r )) = 1 . In all dimensions the proof gives that a.a.s. the cop can win in O (1 /r ) steps and, aswe mention below, this is tight; namely, a.a.s. the robber can ensure not to be caughtin less steps. Therefore, the capture time for this range of parameters is Θ(1 /r ) a.a.s.We make no attempt to optimize the absolute constants in all arguments below, aimingto propose an argument which is as simple as possible.In order to prove Theorem 2.1 it is convenient to describe first a cleaner proof ofthe corresponding result for the continuous (infinite) graph G ( r ) whose vertices are all HASING ROBBERS ON RANDOM GEOMETRIC GRAPHS—AN ALTERNATIVE APPROACH 3 of the points of [0 , , where two of them are adjacent if and only if their distance isat most r . This is a natural variant of the well-known problem of the Lion and theChristian in which (perhaps surprisingly) the Christian (counterpart of the robber inour game) has a winning strategy; see, for example, [5] for more details. In our game,the cop (counterpart of the lion) has a winning strategy. This is a essentially a knownresult [24, 14], but the proof described here differs from the known ones and, cruciallyfor us, can be easily modified to yield a proof of Theorem 2.1 and Theorem 2.2. Theorem 2.3. c ( G ( r )) = 1 for any r > .Proof. We show that c ( G ( r )) = 1 for any r >
0, by describing a winning strategy forthe cop. In the first step, the cop places himself at the center O of [0 , . After eachmove of the robber, when he is located at a point R , the cop catches him if he can (thatis, if the distance between him and the robber is at most r ); otherwise, he moves to apoint C that lies on the segment OR , making sure his distance from the robber is atleast, say, r / O increasesby at least r /
5. As this square distance cannot be more than 1 /
2, this implies thatthe cop catches the robber in at most O (1 /r ) steps. RR R'Z'ZC C'O l Figure 1.
Catching the bad guy on G ( r ).Here is the proof showing that the cop can indeed achieve the above in each stepof the game. Suppose that the cop is located at C and the robber at R , where C lieson OR (and the distance between C and R is at least r / R to R (cid:48) followed by a move of the cop from C to C (cid:48) . Let Z denote the midpoint of CR and let (cid:96) be the line through Z perpendicular to OZ —seeFigure 1. Without loss of generality, choose a coordinate system so that OR is a vertical NOGA ALON AND PAWE(cid:32)L PRA(cid:32)LAT line (and hence (cid:96) is a horizontal one), and assume R (as well as C and Z ) are below O .Note that R (cid:48) , the new location of the robber, may be assumed to be below the line (cid:96) ,since otherwise the distance between C and R (cid:48) is at most the distance between R and R (cid:48) , meaning that the cop can catch the robber, winning the game. Suppose then that R (cid:48) is below (cid:96) and let C (cid:48) be the intersection point of the horizontal line through C withthe line OR (cid:48) . Let also Z (cid:48) denote the intersection point of the horizontal line through R (cid:48) with the line containing OR —see Figure 1 one more time. Now it is easy to seethat RR (cid:48) ≥ R (cid:48) Z (cid:48) > CC (cid:48) , as the triangle RR (cid:48) Z (cid:48) is a right-angle triangle and the twotriangles OCC (cid:48) and OZ (cid:48) R (cid:48) are similar. Hence, the cop may move to C (cid:48) if he decides todo so, as CC (cid:48) < RR (cid:48) ≤ r . Consider the following two possible cases. Case 1: CC (cid:48) > r/
2. In this case if the cop moves to C (cid:48) then its square distance to O increases by CC (cid:48) > r /
4. If C (cid:48) is too close to R (cid:48) , we shift him towards O slightly (thatis, by less than r / C (cid:48) and R (cid:48) is at least r / O by less than 2 r /
100 = r / r / − r / > r /
5. Thus, in thiscase the cop can make a step as required.
Case 2: | CC (cid:48) | ≤ r/
2. In this case the cop can move to C (cid:48) and then walk along the line OR (cid:48) at least distance r/ R (cid:48) (without passing it, since otherwise the game endsand the cop wins). As clearly OC (cid:48) ≥ OC , in this case the cop increases its distancefrom O by more than r/ r /
4. As before,it may be the case that he gets too close to R (cid:48) and then he backups slightly by lessthan r / G ( r ) the cop can indeed increase its square distance from O byat least r / O (1 /r ) steps. (cid:3) Modifying the above argument to get a winning strategy for G ( n, r ) (a.a.s.) is nottoo difficult. The cop will follow essentially the same strategy, but will always placehimself at a vertex of the graph which is sufficiently close to where he wants to be inthe continuous game.More precisely, for each point X of the unit square whose distance from the center O is at least r/ T ( x ) defined below will indeed bewell defined; in our argument this will always be the case) and whose distance fromthe boundary is at least r / (again, in our argument this will always be the case),we define an isosceles triangle T ( X ) as follows. One vertex is X , and the segment oflength r /
100 on the line OX starting at X (and going towards O ) is the height of T ( X ). The base is orthogonal to it and of length r / . Despite the fact that thereare infinitely many triangles, it is not difficult to show that if the area of such a triangleis large enough, then a.a.s. G ( n, r ) contains a vertex inside each such triangle. Lemma 2.4.
There exists an absolute constant c > so that a.a.s. every triangle T ( X ) contains a vertex of G ( n, r ) , provided r > c log nn .Proof. Let us start with a fixed collection F of O ((1 /r ) ) rectangles, each of area Ω( r ),so that every triangle T ( X ) fully contains at least one of these rectangles. To do so, for HASING ROBBERS ON RANDOM GEOMETRIC GRAPHS—AN ALTERNATIVE APPROACH 5 each point Y in an 10 r by 10 r grid in the unit square take the rectangle of width r / and height r / in which Y is the midpoint of the edge of length 10 r andthe other edge is in direction Y O . It is clear that every T ( X ) under consideration ( X not too close to O nor to the boundary) fully contains at least one such a rectangle.In order to complete the proof it is enough to show that a.a.s. each rectangle in F contains at least one vertex of G ( n, r ). The area of each such rectangle is r / , andhence the probability it contains no vertex is (cid:18) − r (cid:19) n ≤ e − c log n/ . Since there are O ((1 /r ) ) = O ( n ) rectangles, the desired result follows by the unionbound for, say, c = 10 , as needed. (cid:3) Now, let us come back to the main result of this section, since we have all necessaryingredients.
Proof of Theorem 2.1.
Since we aim for a statement that holds a.a.s., it follows fromLemma 2.4 that we may assume that every triangle T ( X ) contains at least one vertex.As we already mentioned, the cop plays the continuous strategy, but whenever he wantsto place himself at a point X , he chooses an arbitrary vertex x ∈ V of T ( X ) to go to.The line R (cid:48) x is now not necessarily identical to the line R (cid:48) O , but the angle betweenthem is sufficiently small to ensure that in the computations above for the continuouscase we do not lose much. That was the reason we ensured that R (cid:48) and X are never tooclose in the continuous algorithm, and as the triangle T ( X ) is thin, the angle betweenthese two lines is smaller than r/ . This completes the proof. (cid:3) As we already mentioned, essentially the same proof works for general dimension.The continuous game in dimension d is nearly identical to the one in dimension 2. Inthe first step, the cop places himself at the center O of [0 , d . After each move of therobber, when he is located at a point R , the cop catches him if he can, otherwise, hemoves to a point C that lies on the segment OR , making sure his distance from therobber is at least, say, r / O increases by atleast r /
5. Indeed, since in each round the center O , the location of the cop C , the oldlocation of the robber R and his new location R (cid:48) lie in a two dimensional plane (since O , C and R lie on a line) the analysis is identical to the planar case. As the squaredistance of the cop from the center cannot exceed d/
4, this implies that the cop catchesthe robber in at most O ( d/r ) steps.In the discrete game we let T ( X ) be a cone with height r /
100 on the line connectingthe center O to X , and basis of radius r / centered at X . The probabilistic estimategiven in the proof of Lemma 2.4 shows that a.a.s. every such cone T ( X ) contains avertex of our graph, provided r d − > c d log nn . We can thus repeat the arguments in theproof of the planar case to show that the assertion of Theorem 2.2 holds.Finally, note that the robber can keep escaping for Ω(1 /r ) steps a.a.s. For simplicity,we describe the strategy for the robber for the plane but this also holds for any dimen-sion, for the same reason. As before, we start with the continuous variant of the game. NOGA ALON AND PAWE(cid:32)L PRA(cid:32)LAT
Initially, the robber places himself at distance bigger than r from the cop ensuring he isnot too far from the center O of the square. At each step, when the robber located at R has to move, he moves distance r exactly in the direction perpendicular to RC , wherethe choice of the direction (among the two options), is such that its square distancefrom O increases by at most r (that is, the angle ORR (cid:48) is at most π/ r aftereach such step. In the discrete case, the robber simply chooses a nearby point, makingsure his distance from the cop is at least what it would have been in the continuouscase. References [1] M. Aigner, M. Fromme, A game of cops and robbers,
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Sackler School of Mathematics and Blavatnik School of Computer Science, TelAviv University, Tel Aviv 69978, Israel and Institute for Advanced Study, Princeton,New Jersey, 08540, USA
E-mail address : [email protected] Department of Mathematics, Ryerson University, Toronto, ON, Canada, M5B 2K3
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