CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM

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1 CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM PAWE L PRA LAT Abstract. We consier a variant of the game of Cops an Robbers, calle Containment, in which cops move from ege to ajacent ege, the robber moves from vertex to ajacent vertex but cannot move along an ege occupie by a cop). The cops win by containing the robber, that is, by occupying all eges incient with a vertex occupie by the robber. The minimum number of cops, ξg), require to contain a robber playe on a graph G is calle the containability number, a natural counterpart of the well-known cop number cg). This variant of the game was recently introuce by Komarov an Mackey, who prove that for every graph G, cg) ξg) γg) G), where γg) an G) are the omination number an the maximum egree of G, respectively. They conjecture that an upper boun can be improve an, in fact, ξg) cg) G). Observe that, trivially, cg) γg).) This seems to be the main question for this game at the moment. By investigating expansion properties, we provie asymptotically almost sure bouns on the containability number of binomial ranom graphs Gn, p) for a wie range of p = pn), showing that it forms an intriguing zigzag shape. This result also proves that the conjecture hols for some range of p or hols up to a constant or an Olog n) multiplicative factors for some other ranges).. Introuction The game of Cops an Robbers efine, along with all the stanar notation, later in this section) is usually stuie in the context of the cop number, the minimum number of cops neee to ensure a winning strategy. The cop number is often challenging to analyze; establishing upper bouns for this parameter is the focus of Meyniel s conjecture that the cop number of a connecte n-vertex graph is O n). For aitional backgroun on Cops an Robbers an Meyniel s conjecture, see the book [7]. A number of variants of Cops an Robbers have been stuie. For example, we may allow a cop to capture the robber from a istance k, where k is a non-negative integer [5], play on eges [9], allow one or both players to move with ifferent spees [, 0] or to teleport, allow the robber to capture the cops [6], make the robber invisible or runk [2, 3], or allow at most one cop to move in any given roun [6, 2, 3]. See Chapter 8 of [7] for a non-comprehensive survey of variants of Cops an Robbers. In this paper, we consier a variant of the game of Cops an Robbers, calle Containment, introuce recently by Komarov an Mackey [4]. In this version, cops move 99 Mathematics Subject Classification. 05C57, 05C80. Key wors an phrases. Containment, Cops an Robbers, vertex-pursuit games, ranom graphs. Supporte by grants from NSERC an Ryerson.

2 2 PAWE L PRA LAT from ege to ajacent ege, the robber moves as in the classic game, from vertex to ajacent vertex but cannot move along an ege occupie by a cop). Formally, the game is playe on a finite, simple, an unetecte graph. There are two players, a set of cops an a single robber. The game is playe over a sequence of iscrete time-steps or turns, with the cops going first on turn 0 an then playing on alternate time-steps. A roun of the game is a cop move together with the subsequent robber move. The cops occupy eges an the robber occupies vertices; for simplicity, we often ientify the player with the vertex/ege they occupy. When the robber is reay to move in a roun, she can move to a neighbouring vertex but cannot move along an ege occupie by a cop, cops can move to an ege that is incient to their current location. Players can always pass, that is, remain on their own vertices/eges. Observe that any subset of cops may move in a given roun. The cops win if after some finite number of rouns, all eges incient with the robber are occupie by cops. This is calle a capture. The robber wins if she can evae capture inefinitely. A winning strategy for the cops is a set of rules that if followe, result in a win for the cops. A winning strategy for the robber is efine analogously. As state earlier, the original game of Cops an Robbers is efine almost exactly as this one, with the exception that all players occupy vertices. If we place a cop at each ege, then the cops are guarantee to win. Therefore, the minimum number of cops require to win in a graph G is a well-efine positive integer, name the containability number of the graph G. Following the notation introuce in [4], we write ξg) for the containability number of a graph G an cg) for the original cop-number of G. In [4], Komarov an Mackey prove that for every graph G, cg) ξg) γg) G), where γg) an G) are the omination number an the maximum egree of G, respectively. It was conjecture that the upper boun can be strengthene an, in fact, the following hols. Conjecture. [4]). For every graph G, ξg) cg) G). Observe that, trivially, cg) γg) so this woul imply the previous result. This seems to be the main question for this variant of the game at the moment. By investigating expansion properties, we provie asymptotically almost sure bouns on the containability number of binomial ranom graphs Gn, p) for a wie range of p = pn), proving that the conjecture hols for some ranges of p or hols up to a constant or an Olog n) multiplicative factors for some other ranges of p). However, before we state the result, let us introuce the probability space we eal with an mention a few results for the classic cop-number that will be neee to examine the conjecture since the corresponing upper boun is a function of the cop number). The ranom graph Gn, p) consists of the probability space Ω, F, P), where Ω is the set of all graphs with vertex set {, 2,..., n}, F is the family of all subsets of Ω, an for every G Ω, PG) = p EG) p) n 2) EG).

3 CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM 3 This space may be viewe as the set of outcomes of n 2) inepenent coin flips, one for each pair u, v) of vertices, where the probability of success that is, aing ege uv) is p. Note that p = pn) may an usually oes) ten to zero as n tens to infinity. All asymptotics throughout are as n we emphasize that the notations o ) an O ) refer to functions of n, not necessarily positive, whose growth is boune). We say that an event in a probability space hols asymptotically almost surely or a.a.s.) if the probability that it hols tens to as n goes to infinity. Let us now briefly escribe some known results on the classic) cop-number of Gn, p). Bonato, Wang, an the author of this paper investigate such games in Gn, p) ranom graphs an in generalizations use to moel complex networks with power-law egree istributions see [8]). From their results it follows that if 2 log n/ n p < ε for some ε > 0, then a.a.s. we have that cgn, p)) = Θlog n/p), so Meyniel s conjecture hols a.a.s. for such p. In fact, for p = n o) we have that a.a.s. cgn, p)) = + o)) log / p) n. A simple argument using ominating sets shows that Meyniel s conjecture also hols a.a.s. if p tens to as n goes to infinity see [7] for this an stronger results). Bollobás, Kun an Leaer [4] showe that if pn) 2. log n/n, then a.a.s. pn) 2 n/2 9/2 log logpn)) cgn, p)) n log n. From these results, if np 2. log n an either np = n o) or np = n /2+o), then a.a.s. cgn, p)) = n /2+o). Somewhat surprisingly, between these values it was shown by Luczak an the author of this paper [5] that the cop number has more complicate behaviour. It follows that a.a.s. log n cgn, n x )) is asymptotic to the function fx) shown in Figure enote in blue). Figure. The zigzag functions representing the orinary cop number blue) an the containability number re).

4 4 PAWE L PRA LAT Formally, the following result hols for the classic game. Theorem.2 [5, 8]). Let 0 < α < an = n) = np = n α+o). i) If < α < for some integer j, then a.a.s. 2j+ 2j ii) If 2j < α < 2j cgn, p)) = Θ j ). for some integer j 2, then a.a.s. n ) cgn, p)) = Ω, an j ) n log n cgn, p)) = O. iii) If /2 < α <, then a.a.s. ) n log n cgn, p)) = Θ. The above result shows that Meyniel s conjecture hols a.a.s. for ranom graphs except perhaps when np = n /2k)+o) for some k N, or when np = n o). The author of this paper an Wormal showe recently that the conjecture hols a.a.s. in Gn, p) [8] as well as in ranom -regular graphs [9]. Finally, we are able to state the result of this paper. Theorem.3. Let 0 < α < an = n) = np = n α+o). i) If < α < for some integer j, then a.a.s. 2j+ 2j ξgn, p)) = Θ j+ ) = ΘcGn, p)) Gn, p))). Hence, a.a.s. Conjecture. hols up to a multiplicative constant factor). ii) If < α < for some integer j 2, then a.a.s. 2j 2j n ) ξgn, p)) = Ω, an j ) n log n ξgn, p)) = O = OcGn, p)) Gn, p)) log n). j Hence, a.a.s. Conjecture. hols up to a multiplicative Olog n) factor). iii) If /2 < α <, then a.a.s. ξgn, p)) = Θn) = ΘcGn, p)) Gn, p))/ log n) cgn, p)) Gn, p). Hence, a.a.s. Conjecture. hols. It follows that a.a.s. log n ξgn, n x )) is asymptotic to the function gx) shown in Figure enote in re). The fact the conjecture hols is associate with the observation that gx) fx) = x, which is equivalent to saying that a.a.s. the ratio ξgn, p))/cgn, p)) = n o) = Gn, p)) n o). Moreover, let us mention that Theorem.3 implies that the conjecture is best possible again, up to a constant or an Olog n) multiplicative factors for corresponing ranges of p). j

5 CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM 5 Note that in the above result we skip the case when np = n /k+o) for some positive integer k or np = n o). It is one for a technical reason: an argument for the lower boun for ξgn, p)) uses a technical lemma from [5] that, in turn, uses Corollary 2.6 from [20] which is state only for np = n α+o), where α /k for any positive integer k. Clearly, one can repeat the argument given in [20], which is a very nice but slightly technical application of the polynomial concentration metho inequality by Kim an Vu. However, in orer to make the paper easier an more compact, a reay-to-use lemma from [5] is use an we concentrate on the linear parts of the graph of the zigzag function. Nonetheless, similarly to the corresponing result for cgn, p)), one can expect that, up to a factor of log O) n, the result extens naturally also to the case np = n /k+o) as well. On the other han, there is no problem with the upper boun so the case when np = n /k+o) for some positive integer k is also investigate see below for a precise statement). Moreover, some expansion properties that were use to prove that Meyniel s conjecture hols for Gn, p) [8] are incorporate here to investigate sparser graphs. The rest of the paper is evote to prove Theorem Proof of Theorem Typical properties of Gn, p) an useful inequalities. Let us start by listing some typical properties of Gn, p). These observations are part of folklore an can be foun in many places, so we will usually skip proofs, pointing to corresponing results in existing literature. Let N i v) enote the set of vertices at istance i from v, an let N i [v] enote the set of vertices within istance i of v, that is, N i [v] = 0 j i N jv). For simplicity, we use N[v] to enote N [v], an Nv) to enote N v). Since cops occupy eges but the robber occupies vertices, we will nee to investigate the set of eges at istance i from a given vertex v that we enote by E i v). Formally, E i v) consists of eges between N i v) an N i v), an within N i v). In particular, E v) is the set of eges incient to v. Finally, let P i v, w) enote the number of paths of length i joining v an w. Let us start with the following lemma. Lemma 2.. Let = n) = pn ) log 3 n. Then, there exists a positive constant c such that a.a.s. the following properties hol in Gn, p) = V, E). i) Let S V be any set of s = S vertices, an let r N. Then N r [v] c min{sr, n}. v S Moreover, if s an r are such that s r < n/ log n, then N r [v] = + o))sr. ii) Gn, p) is connecte. v S

6 6 PAWE L PRA LAT iii) Let r = rn) be the largest integer such that r n log n. Then, for every vertex v V an w N r+ v), the number of eges from w to N r v) is at most b, where { 250 if n 0.49 b = 3 log n if n 0.49 < n. log log n Proof. The proof of part i) can be foun in [8]. The fact that Gn, p) is connecte is well known see, for example, []). In fact, the sharp) threshol for connectivity is p = log n/n so this property hols for even sparser graphs. For part iii), let us first expose the rth neighbourhoo of v. By part i), we may assume that N r [v] = + o)) r < 2 r. For any w V \ N r [v], the probability that there are at least b eges joining w to N r v) is at most If n 0.49, then 2 r q := b ) p b 2e r b ) b n ) b ) 2e r+ b =. bn ) b 2e n log n q n 0.005b = on 2 ), bn provie that b is large enough say, b = 250). For n 0.49 < n an so r = ), we observe that ) b 2e q = exp + o))b log b) = on 2 ), b provie b = 3 log n/ log log n. The claim follows by the union boun over all pairs v, w. The proof of the lemma is finishe. The next lemma can be foun in [5]. See also [3] for its extension.) Lemma 2.2. Let ε an α be constants such that 0 < ε < 0., ε < α < ε, an let = n) = pn ) = n α+o). Let l N be the largest integer such that l < /α. Then, a.a.s. for every vertex v of Gn, p) the following properties hol. i) If w N i [v] for some i with 2 i l, then P i v, w) 3 ii) If w N l+ [v] an l+ 7n log n, then P l+ v, w) 6 lα iii) If w N l+ [v] an l+ < 7n log n, then P l+ v, w) 42 lα. iα l+. n log n. Moreover, a.a.s. iv) Every ege of Gn, p) is containe in at most ε cycles of length at most l + 2. We will also use the following variant of Chernoff s boun see, for example, []): Lemma 2.3 Chernoff Boun). If X is a binomial ranom variable with expectation µ, an 0 < δ <, then ) Pr[X < δ)µ] exp δ2 µ 2

7 CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM 7 an if δ > 0, ) Pr[X > + δ)µ] exp δ2 µ. 2 + δ 2.2. Upper boun. First, let us eal with ense graphs that correspon to part iii) of Theorem.3. In fact, we are going to make a simple observation that the containability number is linear if G has a perfect or a near-perfect matching. The result will follow since it is well-known that for p = pn) such that pn log n, Gn, p) has a perfect or a near-perfect) matching a.a.s. As usual, see [], for more etails.) Lemma 2.4. Suppose that G on n vertices has a perfect matching n is even) or a near-perfect matching n is o). Then, ξg) n. Proof. Suppose first that n is even. The cops start on the eges of a perfect matching; two cops occupy any ege of the matching for a total of n cops. All vertices of G can be associate with unique cops. The robber starts on some vertex v. One ege incient to v the ege vv that belongs to the perfect matching use) is alreay occupie by a cop in fact, by two cops, associate with v an v ). Moreover, the remaining cops can move so that all eges incient to v are protecte an the game ens. Inee, for each ege vu, the cop associate with u moves to vu. The case when n is o is also very easy. Two cops start on each ege of a nearperfect matching which matches all vertices but u. If u is isolate, we may simply remove it from G an arrive back to the case when n is even. Recall that the cops win if all eges incient with the robber are occupie by cops. As this property is vacuously true when the robber starts on an isolate vertex, we may assume that she oes not start on u.) Hence, we may assume that u is not isolate. We introuce one more cop on some ege incient to u. The total number of cops is at most 2 n + = n; again, 2 each vertex of G can be associate with a unique cop an the proof goes as before. Now, let us move to the following lemma that yiels part i) of Theorem.3. We combine an ajust ieas from both [5] an [8] in orer to inclue much sparser graphs. Cases when α = /k for some positive integer k are also covere. Lemma 2.5. Let = n) = pn ) log 3 n. Suppose that there exists a positive integer r = rn) such that Then, a.a.s. n log n) 2r+ n log n) 2r. ξgn, p)) = O r+ ). Proof. Since our aim is to prove that the esire boun hols a.a.s. for Gn, p), we may assume, without loss of generality, that a graph G the players play on satisfies the properties state in Lemma 2.. A team of cops is etermine by inepenently choosing each ege of e EG) to be occupie by a cop with probability C r /n, where C is a large) constant to be etermine soon. It follows from Lemma 2.i) that G has + o))n/2 eges. Hence, the expecte number of cops is equal to + o)) n 2 Cr n = + o))cr+. 2

8 8 PAWE L PRA LAT It follows from Chernoff s boun that the total number of cops is Θ r+ ) a.a.s. The robber appears at some vertex v V G). Let X EG) be the set of eges between N r v) an N r+ v). It follows from Lemma 2.i) that X + o)) N r v) 2 r+. Our goal is to show that with probability on ) it is possible to assign istinct cops to all eges e in X such that a cop assigne to e is within istance r + ) of e. Note that here, the probability refers to the ranomness in istributing the cops; the graph G is fixe.) If this can be one, then after the robber appears these cops can begin moving straight to their assigne estinations in X. Since the first move belongs to the cops, they have r + ) steps, after which the robber must still be insie N r [v], which is fully occupie by cops. She is trappe insie N r [v], so we can sen an auxiliary team of, say, 2 r+ cops to go to every ege in the graph inuce by N r [v], an the game ens. Hence, the cops will win with probability on ), for each possible starting vertex v V G). It will follow that the strategy gives a win for the cops a.a.s. Let Y be the ranom) set of eges occupie by cops. Instea of showing that the esire assignment between X an Y exists, we will show that it is possible to assign bu) istinct cops to all vertices u of N r+ v), where bu) is the number of neighbours of u that are in N r v) that is, the number of eges of X incient to u) an such that each cop assigne to u is within istance r + ) from u. Note that this time istance is measure between vertex u an eges which is non-stanar. In this paper, we efine it as follows: ege e is at istance at most r + ) from u if e is at istance at most r from some ege ajacent to u.) Inee, if this can be one, assigne cops run to u, after r rouns they are incient to u, an then sprea to eges between u an N r v); the entire X is occupie by cops. In orer to show that the require assignment between N r+ v) an Y exists with probability on ), we show that with this probability, N r+ v) satisfies Hall s conition for matchings in bipartite graphs. Suppose first that n 0.49 an fix b = 250. It follows from Lemma 2.iii) that bu) b for every u N r+ v). Set k 0 = max{k : k r < n}. Let K N r+ v) with K = k k 0. We may apply Lemma 2.i) to boun the size of u K N r[u] an the number of eges incient to each vertex. It follows that the number of eges of Y that are incient to some vertex in u K N r[u] can be stochastically boune from below by the binomial ranom variable Bin ck r /3), C r /n), whose expecte value is asymptotic to Cc/3)k 2r+ /n Cc/3)k log n. Using Chernoff s boun we get that the probability that there are fewer than bk eges of Y incient to this set of vertices is less than exp 4k log n) when C is a sufficiently large constant. Hence, the probability that the sufficient conition in the statement of Hall s theorem fails for at least one set K with K k 0 is at most k 0 ) Nr+ v) k 0 exp 4k log n) n k exp 4k log n) = on ). k k= Now consier any set K N r+ v) with k 0 < K = k N r+ v) 2 r+ if such a set exists). Lemma 2.i) implies that the size of u K N r[u] is at least cn, so we k=

9 CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM 9 expect at least cn /3) C r /n = Cc/3) r+ eges of Y incient to this set. Again using Chernoff s boun, we euce that the number of eges of Y incient to this set is at least 2b r+ b N r+ v) bk with probability at least exp 4 r+ ), by taking the constant C to be large enough. Since N r+ v) k=k 0 + Nr+ v) k ) exp 4 r+ ) 2 2r+ exp 4 r+ ) = on ), the necessary conition in Hall s theorem hols with probability on ). Finally, suppose that > n Since Lemma 2.4 implies that the result hols for > n, we may assume that n. In fact, for > n we get a better boun of n rather than O 2 ) that we aim for.) This time, set b = 3 log log n/ log n to make sure bu) b for all u N r+ v). The proof is almost the same as before. For small sets of size at most k 0 = Θn/), we expect Cc/3)k 3 /n Cc/3)kn 0.47 eges, much more than we actually nee, namely, bk. For large sets of size more than k 0, we moify the argument slightly an instea of assigning b cops to each vertex of N r+ v), we notice that the number of cops neee to assign is equal to u K bu) X 2r+. There might be some vertices of N r+ v) that are incient to b eges of X but the total number of incient eges to K is clearly at most X.) The rest is not affecte an the proof is finishe. The next lemma takes care of part ii) of Theorem.3. Lemma 2.6. Let = n) = pn ) log 3 n. Suppose that there exists an integer r = rn) 2 such that Then, a.a.s. n log n) 2r n log n) 2r. ) n log n ξgn, p)) = O. Proof. We mimic the proof of the previous lemma so we skip etails focusing only on ifferences. A team of cops is etermine by inepenently choosing each ege of e EG) to be occupie by a cop with probability C log n/ r, for the total number of cops Θn log n/ r ) a.a.s. The robber appears at some vertex v V G). This time, X EG) is the set of eges between N r v) an N r v) an X 2 r. We show that it is possible to assign b = 250 istinct cops to all vertices u of N r v) such that a cop assigne to u is within istance r from u. The efinition of k 0 has to be ajuste. Set r k 0 = max{k : k r < n}. Let K N r v) with K = k k 0. The expecte number of eges of Y that are incient to some vertex in u K N r [u] is at least ck r )/3)C log n/ r ) = Cc/3)k log n, an the rest of the argument is not affecte. Now consier any set K N r v) with

10 0 PAWE L PRA LAT k 0 < K = k N r v) 2 r if such a set exists). The size of u K N r [u] is at least cn, so we expect at least ) ) C log n cn) = Ccr n log n Ccr 3 r 3 2r 3 eges of Y incient to this set. Hence, the number of eges of Y incient to this set is at least 2b r b N r v) bk with probability at least exp 4 r ), by taking the constant C to be large enough. The argument we ha before works again, an the proof is finishe Lower boun. The proof of the lower boun is an aaptation of the proof use for the classic cop number in [5]. The two bouns, corresponing to parts i) an ii) in Theorem.3, are prove inepenently in the following two lemmas. Lemma 2.7. Let < α < for some integer j, c = cj, α) = 3, an 2j+ 2j 2jα = n) = np = n α+o). Then, a.a.s. ξgn, p)) > K := 30c2j + ) ) j+. Proof. Since our aim is to prove that the esire boun hols a.a.s. for Gn, p), we may assume, without loss of generality, that a graph G the players play on satisfies the properties state in Lemmas 2. an 2.2. Suppose that the robber is chase by K cops. Our goal is to provie a winning strategy for the robber on G. For vertices x, x 2,..., x s, let C x,x 2,...,x s i v) enote the number of cops in E i v) that is, at istance i from v) in the graph G \ {x, x 2,..., x s }. Right before the robber makes her move, we say that the vertex v occupie by the robber is safe, if for some neighbour x of v we have C x v), an 30c2j+) ) i+ C x 2iv), C x 2i+v) 30c2j + ) for i =, 2,..., j such a vertex x will be calle a ealy neighbour of v). The reason for introucing ealy neighbours is to eal with a situation that many cops apply a greey strategy an always ecrease the istance between them an the robber. As a result, there might be many cops right behin the robber but they are not so angerous unless she makes a step backwars by moving to a vertex she came from in the previous roun, a ealy neighbour! Moreover, note that a vertex is calle safe for a reason: if the robber occupies a safe vertex, then the game is efinitely not over since the conition for C x v) guarantees that at most a small fraction of incient eges are occupie by cops. Since a.a.s. G is connecte see Lemma 2.ii)), without loss of generality we may assume that at the beginning of the game all cops begin at the same ege, e. Subsequently, the robber may choose a vertex v so that e is at istance 2j + 2 from v see Lemma 2.i) applie with r = 2j + to see that almost all vertices are at istance 2j + from both enpoints of e). Hence, even if all cops will move from e to E 2j+ v) after this move, v will remain safe as no boun is require for C x 2j+v). Of course,

11 CONTAINMENT GAME PLAYED ON RANDOM GRAPHS: ANOTHER ZIG-ZAG THEOREM again, without loss of generality we may assume that all cops pass for the next roun an stay at e before starting applying their best strategy against the robber.) Hence, in orer to prove the lemma, it is enough to show that if the robber s current vertex v is safe, then she can move along an unoccupie ege to a neighbour y so that no matter how the cops move in the next roun, y remains safe. For 0 r 2j, we say that a neighbour y of v is r-angerous if i) an ege vy is occupie by a cop for r = 0), or i ii) C v,x r y) 3 30c2j+)) for r = 2i or r = 2i, where i =, 2,..., j), where x is a ealy neighbour of v. We will check that for every r {0,,..., 2j}, the number of r-angerous neighbours of v, which we enote by angr), is smaller than. Clearly, since v is safe, 22j+) ang0) C x v) 30c2j + ) 22j + ). Suppose then that r = 2i or r = 2i for some i {, 2,..., j}. Every r-angerous i neighbour of v has at least 3 30c2j+)) cops occupying E r+) v). On the other han, every ege from E r+) v) is incient to at most 2 vertices at istance at most r from v. Moreover, Lemma 2.2i) implies that there are at most c paths between v an any w N r v). Finally, by the assumption that v is safe, we have C x 2iv), C x 2i+v) ) i+, 30c2j+) provie that i j ; the corresponing conitions for C x 2jv) an C x 2j+v) are trivially true, since both can be boune from above by K, the total number of cops. Combining all of these yiels ) i angr) 2c C x 3 30c2j + ) r+)v) 2c 2 + o))c x r+v) ) i+ 5c, 30c2j + ) an consequently angr), as require. Thus, there at most /2 of neighbours 22j+) of v are r-angerous for some r {0,,..., 2j}. Since we have + o)) neighbours to choose from see Lemma 2.i)), there are plenty of neighbours of v which are not r-angerous for any r = 0,,..., 2j an the robber might want to move to one of them. However, there is one small issue we have to eal with. In the efinition of being angerous, we consier the graph G \ {v, x} whereas in the efinition of being safe we want to use G \ {v} instea. Fortunately, Lemma 2.2iv) implies that we can fin a neighbour y of v that is not only not angerous but also x oes not belong to the 2j-neighbourhoo of y in G \ {v}. It follows that vy is not occupie by a cop an C v ry) < 3 i =, 2,..., j. We move the robber to y. Now, it is time for the cops to make their move. Because of our choice of the vertex y, we can assure that the esire upper boun for C v ry) require for y to be safe will hol for r {, 2,..., 2j }. Inee, the best that the cops can o to try to fail i 30c2j+)) for r = 2i or r = 2i, where

12 2 PAWE L PRA LAT the conition for C v ry) is to move all cops at istance r an r + from y to r- neighbourhoo of y, an to make cops at istance r stay put, but this woul not be enough. Thus, regarless of the strategy use by the cops, y is safe an the proof is finishe. Lemma 2.8. Let < α < for some integer j, c = cα) = 6 2j 2j 2j )α = n) = np = n α+o). Then, a.a.s. ξgn, p)) K := 30 c2j + ) ) j+ n 2j. Proof. The proof is very similar to that of Lemma 2.7. The only ifference is that checking the esire bouns for ang2j ) an ang2j) is slightly more complicate. As before, we o not control the number of cops in E 2j v) an E 2j+ v), clearly C x 2jv) an C x 2j+v) are boune from above by K, the total number of cops. We get ) j ang2j ) 2 c C x 3 30 c2j + ) 2j)v) 2 c 2 + o)) K 5 c 30 c2j + ) ) j+, an consequently ang2j ), as require. Note that we have room to spare 22j+) here but we cannot take avantage of it so we o not moify the efinition of being 2j )-angerous.) Let us now notice that a cop at istance 2j + from v can contribute to the angerousness of more than c neighbours of v. However, the number of paths of length 2j joining v an w is boune from above by c 2j /n see Lemma 2.2ii) an note that 2j = n 2jα+o) 7n log n, since 2jα > ). Hence, ) j ang2j) 2 c2j 3 30 c2j + ) an, as esire, ang2j) basically the same. 22j+) n Cx 2j+)v) 2 c2j n ) j+ 5 c, 30 c2j + ) 2 + o)) K an. Besies this moification the argument remains References [] N. Alon, A. Mehrabian, Chasing a Fast Robber on Planar Graphs an Ranom Graphs, Journal of Graph Theory 782) 205), [2] D. Bal, A. Bonato, W. Kinnersley, P. Pra lat, Lazy Cops an Robbers on hypercubes, Combinatorics, Probability an Computing, to appear. [3] D. Bal, A. Bonato, W. Kinnersley, P. Pra lat, Lazy Cops an Robbers playe on ranom graphs an graphs on surfaces, preprint 204. [4] B. Bollobás, G. Kun, I. Leaer, Cops an robbers in a ranom graph, Journal of Combinatorial Theory Series B ), [5] A. Bonato, E. Chiniforooshan, P. Pra lat, Cops an Robbers from a istance, Theoretical Computer Science 4 200),

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