COPS AND ROBBER WITH CONSTRAINTS

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1 COPS AND ROBBER WITH CONSTRAINTS FEDOR V. FOMIN, PETR A. GOLOVACH, AND PAWE L PRA LAT Abstract. Cops & Robber s a classcal pursut-evason game on undrected graphs, where the task s to dentfy the mnmum number of cops suffcent to catch the robber. In ths paper, we nvestgate the changes n problem s complexty and combnatoral propertes wth constranng the followng natural game parameters Fuel: The number of steps each cop can make; Cost: The total sum of steps along edges all cops can make; Tme: The number of rounds of the game. Key words. Cops and Robber games, complexty, random graphs AMS subject classfcatons. 05C80, 05C85, 68R10 1. Introducton. Cops & Robber game s a dscrete varant of the classcal Man and Lon pursut-evason problem attrbuted to Rado by Lttlewood n [22]. The game s played by two players: Cop and Robber on an undrected graph. The cop-player has a team of cops who attempt to capture the robber. At the begnnng of the game cop-player selects vertces and puts cops on these vertces. Then the robber-player puts the robber on a vertex. The players take turns startng wth the cop-player. At every move each of the cops can be ether moved to an adjacent vertex or kept on the same vertex. Smlarly, the robber-player responds by movng the robber to an adjacent vertex or by keepng hm on the same vertex. The cop-player wns f at some step of the game he succeeds to catch the robber, that s, to put one of hs cops on a vertex occuped by the robber. The game, wth one cop, was ntroduced ndependently by Wnkler and Nowakowsk [26] and by Qullot [29]. Agner and Fromme [1] ntated the study of the problem wth several cops. Dfferent combnatoral [4, 10, 30] and algorthmc [12, 13, 19] aspects of the game were studed ntensvely. We refer to surveys [3, 14, 20] and the recent monograph [8] for references on dfferent pursut-evason and search games on graphs. There are two man open problems concernng the Cops & Robber game. The frst open problem s about the upper bound on the number of cops suffcent to wn n any connected graph on n vertces. The famous conjecture, attrbuted to Meynel by Frankl [15], s that for every n-vertex connected graph O( n) cops always have wnnng strategy. It was shown by Frankl that the cop number of a graph s O(n log log n/ log n). Despte many attempts, the best known upper bound of n 2 (1 o(1))( log 2 due to Lu and Peng [23] s stll qute far from n. See also the n) works of Alon and Mehraban [2]; Freze et al. [16], and Mehraban [25] on extensons of Meynel s conjecture. The bounds on the cop number of random graphs were gven by Bollobás et al. [5] and by Luczak and Pra lat [24]. It has been shown recently by The work on ths paper was ntated at the Dagstuhl Semnar Department of Informatcs, Unversty of Bergen, N-5020 Bergen, Norway, emal: fomn@.ub.no. School of Engneerng and Computng Scences, Durham Unversty, South Road, DH1 3LE Durham, UK, emal: petr.golovach@durham.ac.uk. Supported by EPSRC under project EP/G043434/1. Department of Mathematcs, Ryerson Unversty, Toronto, ON, Canada, M5B 2K3, emal: pralat@ryerson.ca. 1

2 2 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT Pra lat and Wormald [28] that the Meynel s conjecture holds asymptotcally almost surely for random graphs. The second open problem s the computatonal complexty of the problem. Goldsten and Rengold n [19] conjectured that the game s EXPTIME-complete 1 but so far t s only known to be NP-hard [12]. The complexty of several varants of the Cops & Robber game are better understood. Goldsten and Rengold [19] proved that the verson of the Cops & Robbers game on drected graphs s EXPTIME-complete. The verson of the game on undrected graphs when the cops and the robber are gven ther ntal postons s also EXPTIME-complete [19]. The verson of the game, where each cop can make at most φ steps, the verson wth fuel constrant, s PSPACE-complete for each φ [13]. In ths paper we study the Cops & Robber game when one of the followng parameters s bounded: fuel, the number of moves φ each cop can make durng the game. In other words, each cop can pass through at most φ edges; cost, the sum σ of the number of steps that all cops can make n total. Thus the total number of edges passed by all cops s at most σ; tme, the number of rounds τ of the game. The verson of constrant tme was studed by Bonato et al. [6] and Gavenčak [18]. It was shown n [6] that for every fxed nteger τ, and nteger k beng part of the nput, decdng f at most k cops can wn wthn tme τ, s NP-complete. We show that smlar NP-completeness result holds for fxed cost constrant σ and k beng part of the nput (Theorem 3.1). For fuel constrant φ = 1 the problem s equvalent to the mnmum domnatng set problem, and thus s NP-complete. For φ 2 the problem s PSPACE-hard [13]. In ths paper, we establsh PSPACE-hardness results for the other two varants of the game, when one of the parameters, ether tme τ or cost σ s part of the nput (Theorem 3.4). Each of the varants of the Cops & Robber game s PSPACE-complete, when the correspondng parameter (τ, σ, or φ) does not exceed some polynomal of the nput length (Theorem 3.7). Ths establsh almost complete classfcaton of the complexty landscape of the game wth dfferent constrants on the resources of the players. We summarze the complexty results n the followng table. φ s σ s τ s constant s = 1, NP-compl. [13] s 2, PSPACE-compl. [13] NP-compl. NP-compl. [6] polynomal PSPACE-compl. [13] PSPACE-compl. PSPACE-compl. unbounded PSPACE-hard [13] PSPACE-hard PSPACE-hard Table 1.1 Complexty classfcaton of the Cops & Robber game wth dfferent constrants on resources. Constant means the correspondng parameter (fuel φ, tme τ, and cost σ) s not part of the nput. Polynomal means that the correspondng parameter s bounded by some polynomal of the nput length. The last lne of the table corresponds to arbtrary szes of parameters. We also present a number of results for new varants of the game played on bnomal random graphs. The varant wth Fuel Constrants and the one wth Tme Constrants behave smlarly from that perspectve. We show that the number of cops 1 Goldsten and Rengold n [19] call EXPTIME = DTIME(2 O( I ) ), where I s the nput sze; more often ths class s denoted by E or ETIME.

3 COPS AND ROBBER WITH CONSRAINTS 3 requred to catch the robber as a functon of the average degree changes n a very ntrgung manner. We get that asymptotcally almost surely the logarthm of the cop number s asymptotc to the zg-zag functon that depends on the value of s. (Hence, n fact, we obtan an nfnte famly of zg-zags.) The thrd varant of the game, that s, the one wth Cost Constrants seems to be more challengng to nvestgate. We provde both upper and lower bounds for ths graph parameter but the shape of the zg-zag functon assocated wth ths varant stll remans an open problem. The remanng part of the paper s organzed as follows. In Secton 2 we gve basc defntons and observatons. We gve hardness proofs n Secton 3. The results about Cops & Robber on random graphs are n Secton Basc defntons and prelmnares. We consder fnte undrected graphs wthout loops or multple edges. The vertex set of a graph G s denoted by V (G) and ts edge set by E(G), or smply by V and E f ths does not create confuson. If U V (G), then the subgraph of G nduced by U s denoted by G[U]. For a vertex v, the set of vertces whch are adjacent to v s called the (open) neghborhood of v and denoted by N G (v). The closed neghborhood of v s the set N G [v] = N G (v) {v}. For Q V (G), N G [Q] = v Q N G[v] s the closed neghborhood of Q. The dstance dst G (u, v) between a par of vertces u and v n a connected graph G s the number of edges n a shortest u, v-path n G. For a postve nteger r, let N G (v, s) denote the set of vertces ( ball ) wthn dstance at most s from v; that s, N G (v, s) = {u V (G): dst G (u, v) s}. Whenever there s no ambguty, we omt the subscrpts. All logarthms wth no subscrpt are natural. The Cops & Robber game s defned as follows. Let G be a connected graph. The game s played by two players: the cop-player C and the robber player R, whch make moves alternately. The cop-player C has a team of k cops who attempt to capture the robber. At the begnnng of the game ths player selects vertces and put cops on these vertces. Then R puts the robber on a vertex. The players take turns startng wth C. At every turn each of the cops can be ether moved to an adjacent vertex or kept on the same vertex. Let us note that several cops can occupy the same vertex at some pont of the game. Smlarly, R responds by movng the robber to an adjacent vertex or keepng hm on the same vertex. It s sad that a cop catches (or captures) the robber at some round f at that round they occupy the same vertex. The copplayer wns f one of hs cops catches the robber. Player R wns f he can avod such a stuaton. For a graph G, the mnmum number k of cops suffcent for C to wn on graph G s called the cop number of G and s denoted by c(g). We consder the followng varants of the game. Let s be a postve nteger. In the frst varant of the game, each cop can make at most s moves along edges (have a bounded charge or amount of fuel ). Notce that a cop needs fuel only to move from one vertex to another, and even f a cop cannot move to adjacent vertex because he runs out of fuel, he s stll actve and the robber cannot step on a vertex occuped by such cop wthout beng caught. We denote the mnmum number of cops wth fuel at most s suffcent to wn on G by c φ s (G), and we refer to ths varant of the game as Cops & Robber wth Fuel Constrants. Another varant of constrants we study n ths paper s the case when durng the whole game the total number of moves (from a vertex to another vertex) of all the cops s at most s. In other words, f transferrng of one cop to adjacent vertex costs one unt, the total cost of all cop s movements (or the dstance traveled by the cops) s at most s. We denote the mnmum number of cops that can wn n these condtons by c σ s (G) and call ths game Cops & Robber wth Cost Constrants.

4 4 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT Fnally, we consder the game wth the number of rounds at most s, that s, the length of the game s restrcted. In other words, the cops are supposed to catch the robber wthn tme lmt s. Let c τ s (G) be the mnmum number of cops C needed to wn now and the game s called Cops & Robber wth Tme Constrants. We consder the decson verson of the problem Cops and Robber wth Fuel Constrants Input: A connected graph G and two postve ntegers k, s. Queston: Is c φ s (G) k? The decson versons of Cops and Robber wth Cost Constrants and Cops and Robber wth Tme Constrants are defned smlarly, wth the only dfference that we ask f c σ s (G) k and c τ s (G) k, correspondngly. In some of the proofs, the noton of poston wll be used. The poston of a cop at a gven moment of the game s the vertex of the graph occuped by ths cop, and we defne the poston of a team of k cops (or poston of cops) as a multset of the vertces C = (v 1, v 2,..., v k ) occuped by the cops. Notce that C s a multset, snce several cops can occupy one vertex and, as we do not dstngush the cops, t s not mportant whch cop occupes a vertex. In the case of the game wth cost or tme constrants the postons of cops s the par (C, l), where C s a multset of the vertces occuped by the cops and l s the number of moves along edges whch the cops can do or the total number of rounds left, respectvely. For the ntal poston, l = s. The poston of the robber s a vertex of the graph occuped by hm. We wll use the followng observaton. Observaton 1. For a connected graph G and a postve nteger s, c(g) c φ s (G) c τ s (G) c σ s (G). The frst two nequaltes of Observaton 1 follow drectly from the defntons of the numbers c(g), c φ s (G), c τ s (G) and c σ s (G), and the thrd one follows from the fact that t always can be assumed that at least one cop s moved at each step (otherwse the robber can ether keep hs poston or mprove t). Let the s-dstance domnaton number γ s (G) be the mnmum cardnalty of a set D V (G) wth the property that every vertex v V (G) s at dstance at most s from some vertex of D. The relaton of the game and domnaton s gven n the followng smple observaton. Observaton 2. For any connected graph G, c φ 1 (G) = c σ 1 (G) = c τ 1 (G) = γ 1 (G); γ s (G) c φ s (G) c τ s (G) c σ s (G), for every postve nteger s. 3. Complexty of Cops and Robber wth Tme and Cost Constrants. Observaton 2 ndcates that all varants of the game wth constrants are at least NPhard. In ths secton we establsh the complexty of Cops and Robber wth Tme Constrants and Cops and Robber wth Cost Constrants by showng that they are PSPACE-hard. However, we start wth the proof that when s s not a part of the nput, then Cops and Robber wth Cost Constrants s NP-complete When cost s not a part of the nput. In ths subsecton we prove that when the cost s not a part of the nput, the verson of the problem s NP-complete. Theorem 3.1. For any postve nteger s, the followng problem s NP-complete

5 COPS AND ROBBER WITH CONSRAINTS 5 Input: A connected graph G and a postve nteger k. Queston: Is c σ s (G) k? Proof. To prove hardness, we reduce the well-known NP-complete Domnatng Set problem [17]. Gven a graph G, we construct the graph H as follows: for each vertex v V (G), we add a path P v of length s 1 wth one endpont n v. All vertces of P v except v are new. Observe that for s = 1, H = G. We prove that G has a domnatng set of sze k f and only f c σ s (H) k. Let S V (G) be a domnatng set of sze at most k n G. We put cops on the vertces of S n H. There s a vertex v V (G) such that the robber occupes a vertex of the path P v. Snce S s a domnatng set n G, we have that ether v S or v s adjacent to a vertex u S. In the second case the cop from u s moved to v at the frst step. Now n both cases the robber occupes a vertex of P v v, the vertex v s occuped by a cop, and the cops can make at least s 1 moves along edges. It remans to observe that cops can capture the robber by movng the cop from v toward the other end of P v. Hence, k cops have a wnnng strategy on H and c σ s (H) k. Consder now a wnnng strategy of k cops on H. Suppose that (C, s) s the ntal poston of the cops n t. Let S = {v V (G) C V (P v ) }. Clearly, C k. We prove that S s a domnatng set n G. Assume that, contrary to ths clam, there s a vertex u V (G) such that N G [u] S =. Let w be the endpont of P u dfferent from u. Snce vertces of P u are not occuped by the cops and there are no cops n the neghborhood of u n G, all cops are at dstance at least s+1 from w. Therefore, the robber can occupy ths vertex at the begnnng of the game and stay safely there untl the end of the game. Ths contradcts the exstence of a wnnng strategy for the cops from the consdered ntal poston. Hence S s a domnatng set of sze at most k n G. To complete the proof, we have to show that for fxed σ, Cops and Robber wth Cost Constrants s n NP. In order to do t, we observe that a wnnng strategy of the cops on G can be descrbed as a drected rooted tree of all possble moves of the robber (where the frst move s a choce of the ntal poston) and correspondng moves of the cops. Each node of the tree correspond to the current postons of the cops and robber (the root corresponds only to the ntal poston of the cops). The sze of ths rooted tree s at most O( V (G) s ) because out-degrees of ts nodes are at most V (G) and the heght of the tree s at most s. It remans to observe that such a tree certfes ncluson of our problem n NP, because t may be checked n polynomal tme whether the strategy of cops s wnnng. For Cops and Robber wth Tme Constrants, NP-completeness was proved n [6], but the reducton descrbed above works for t and Cops and Robber wth Fuel Constrants as well. Usng t and the fact that Domnatng Set s a W[2]- complete problem (see the book of Downey and Fellows [11] for an ntroducton to parameterzed complexty), we derve the followng corollary. Corollary 3.2. For any fxed postve nteger s, the Cops and Robber wth Cost Constrants (Tme Constrants, Fuel Constrants, respectvely) problem s W[2]-hard parameterzed by the number of cops. Combned wth the non-approxmablty for the domnatng set problem [31], the same reducton mples the followng. Corollary 3.3. For any fxed postve nteger s, there s a constant c > 0 such that there s no polynomal tme algorthm to approxmate c σ s (G) (c τ s (G), c φ s (G), respectvely) wthn a multplcatve factor of c log n where n = V (G), unless P = NP.

6 6 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT 3.2. PSPACE-hardness of Cops and Robber wth Tme and Cost Constrants. When σ and τ are a part of the nput then Cops and Robber wth Cost Constrants and Cops and Robber wth Tme Constrants become PSPACE-hard. The followng theorem s the man result of ths subsecton. Theorem 3.4. Cops and Robber wth Cost Constrants and Cops and Robber wth Tme Constrants are PSPACE-hard. Proof. We reduce from the PSPACE-complete Quantfed Boolean Formula n Conjunctve Normal Form (QBF) problem. For a set of Boolean varables x 1, x 2,..., x n and a Boolean formula F = C 1 C 2 C m, where C j s a clause, the QBF problem asks whether the expresson Φ = Q 1 x 1 Q 2 x 2 Q n x n F s true, where for every, Q s ether or. We assume that n 2 s even, Q = for odd {1,..., n} and Q = whenever s even. It s known that QBF remans PSPACE-complete wth these restrctons [17]. Gven a quantfed Boolean formula φ, we construct an nstance (G, n, s) of our problem such that Φ s true f and only f the cop-player can wn on G wth n cops for s = n + 1 n both games. Constructng G.. For every Q x we ntroduce a gadget graph G : Construct vertces x, x. If Q =, then we construct vertces y, y and edges x y, x y. If Q =, then we construct a vertex y and jon t wth x, x by edges. Construct a path of length s wth endponts z, w and jon x, x wth z by edges. Construct two paths of length s 1 wth endponts u (1), v (1) and u (2), v (2) respectvely, jon u (1), u (2) wth z by edges. Introduce two vertces r and r and jon them wth x and x by paths of length s 1 respectvely. Construct two paths of length s 3 wth endponts p (1), q (1) and p (2), q (2) respectvely, jon p (1), p (2) wth x, x by edges. For Q = and Q =, we denote the obtaned graphs as G ( ) and G ( ) respectvely (see Fg. 3.1). Let X = {x, x }, U = {u (1), u (2) }, P = {p (1), p (2) }, Y = {y, y } for G ( ) and Y = {y } for G ( ). Usng these gadgets we construct G as follows Construct gadget graphs G for {1,..., n}. For each {2,..., n}, jon the vertces from the set Y 1 wth all vertces from Y. For each {2,..., n}, jon the vertces of U wth all vertces from the sets Y j for j {1,..., 1}. For each {1,..., n 1}, jon the vertces of P wth all vertces from the sets Y j for j { + 1,..., n}. For each even {2,..., n}, jon the vertces from the set Y 1 wth all vertces from X. For each odd {3,..., n 1}, jon the vertces from the set Y 1 wth the vertex z. Jon x n and x n by an edge. Add vertces C 1, C 2,..., C m correspondng to the clauses and jon them wth the unque vertex of Y n (recall that Q n = ) by edges.

7 COPS AND ROBBER WITH CONSRAINTS 7 v (1) v (2) w v (1) v (2) w u (1) u (2) q (1) q (2) u (1) u (2) q (1) q (2) z p (1) p (2) z p (1) p (2) r r r r x x x x y y y G ( ) G ( ) Fg Graphs G ( ) and G ( ) for s = 3. For {1,..., n} and j {1,..., m}, the vertex x s joned wth C j by an edge f C j contans the lteral x, and x s joned wth C j f C j contans the lteral x. Connectons between gadgets G 1 and G s shown n Fg. 3.2 and 3.3, and constructon of G s shown n Fg. 3.4 u (1) 1 u (2) 1 u (1) u (2) z z p (1) 1 p (1) p (2) 1 p (2) 1 x 1 x 1 x x y 1 y 1 y Fg Connectons of G 1 ( ) and G ( ). Now we prove the followng two lemmata. Lemma 3.5. If Φ = true, then n cops have a wnnng strategy on G n Cops & Robber wth Cost Constrants and Cops & Robber wth Tme Constrants. Proof. By Observaton 1, t s suffcent for the proof to descrbe a wnnng strategy for the cop-player for Cops & Robber wth Cost Constrants. The cops start by occupyng vertces z 1,..., z n. If the robber occupes a vertex of some (z, w )-path of G, then the cop from the vertex z moves toward the robber n ths path. Snce the path has length s, the robber wll be captured by ths cop. If the robber occupes )-path or (x, r )-path or (x, r )-path of G, then the cop from the vertex z moves to u (j) or x or x respectvely and then moves toward the robber n ths path. Snce these paths have length s 1, the robber wll be captured. If the robber occupes a vertex of some (p (j), q (j) )-path of G for < n, then the cop a vertex of some (u (j), v (j)

8 8 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT u (1) 1 u (2) 1 u (1) u (2) z 1 p (1) 1 p (2) 1 z p (1) p (2) x 1 x 1 x x y 1 y y Fg Connectons of G 1 ( ) and G ( ). u (1) 1 u (2) 1 u (1) 2 u (2) 2 z p (1) z 2 p (1) 1 p (2) 1 p (2) x 2 x 2 C 1 C 2 x 1 x 1 y 1 y 1 y 2 Fg Constructon of G for Φ = x 1 x 2 (x 1 x 2 ) (x 1 x 2 ), n = 2, and s = 3. from the vertex z +1 moves to x +1, and then he moves to y +1. If the robber tres to move to x or x, then he s captured by the cop from z n one step. Hence, the robber has to stay on the path. Now the cop from y +1 moves to p (j) and after that he moves toward the robber n ths path. Snce the path has length s 3, we have that the robber wll be captured. If = n, then the cop from the vertex z moves to x. If the robber moves to x, then he s captured by the cop from x n one step. So, the robber stays on the path. Now the cop from x moves to p (j) and after that he moves toward the robber n ths path. Thus we assume that the robber s on a vertex of some set Y or n {C 1,..., C m }. We consder two cases. Case 1. The robber occupes a vertex of some set Y. For each j {1,..., 1}, the cop from z j moves ether to x j or to x j. We assume that the choce of x j corresponds to the value true of the varable x j and the choce of x j corresponds to the value false of x j. Snce φ = true, the varables x 1,..., x 1 can be assgned values such that Q x... Q n x n F = true. The cop from the vertex z j moves accordng to the value of x j. If x j = true, then he moves to the vertex x j, otherwse he moves to x j. Now nductvely for j {,..., n}, we assume that the robber occupes a vertex of Y j and we move the cop from z j accordng to the followng subcases. a) Q j =. If the robber occupes the vertex y j, then the cop from z j moves to

9 COPS AND ROBBER WITH CONSRAINTS 9 x j, and f the robber s on y j, then the cop moves to x j. We agan suppose that a placement of a cop on x j corresponds to the value true of the varable x j, and a movng a cop to x j corresponds to the value false. Notce that now the robber choses the value of the varable x j. Now the robber should make hs move: If the robber stays n hs old poston, then he wll be captured n one step by the cop whch s ether on x j or x j. If the robber moves to a vertex of X j+1, then he wll be captured n one step by the cop from z j+1 n G j+1 ( ). If the robber moves to a vertex of Y j 1, then agan he wll be captured n one step by the cop whch s ether on x j 1 or x j 1 n G j 1 ( ). If the robber moves to a vertex u (1) t or u (2) t for t > j, then he wll be captured n one step by the cop from z t. If the robber moves to a vertex p (1) t or p (2) t for t < j, then he wll be captured n one step by the cop from x t or x t. Hence he should move ether to a vertex of Y j+1 or to one of the vertces C 1,..., C m f j = n to avod the capture. b) Q j =. Then the robber occupes y j. The cop from z j moves ether to x j or to x j. The vertex s chosen n such a way that t corresponds to the value of the varable x j for whch (and for already assgned values of the varables x 1,..., x j 1 ) Q j+1 x j+1... Q n x n F = true. Then smlarly to the Subcase a), If the robber stays n hs old poston, then he wll be captured n one step by the cop whch s ether on x j or x j. If the robber moves to a vertex of X j unoccuped by the cops, then he wll be captured n one step ether by the cop from z j+1 n G j+1 ( ) f j < n or by the cop from X j f j = n. If the robber moves to a vertex of Y j 1, then agan he wll be captured n one step by the cop whch s ether on x j or x j n G j ( ). If the robber moves to a vertex u (1) t or u (2) t for t > j, then he wll be captured n one step by the cop from z t. If the robber moves to a vertex p (1) t or p (2) t for t < j, then he wll be captured n one step by the cop from x t or x t. Hence, the robber s ether captured by the next step or moves to a vertex of Y j+1 or to one of the vertces C 1,..., C m f j = n. Fnally, the robber s ether captured or occupes some vertex C t. Notce that f the robber s not captured yet, then the cops made exactly n = s 1 moves along edges. Observe also, that the cops have chosen the vertces of the sets X 1,..., X n such that F = true for the correspondng values of boolean varables. Hence there s a cop on a vertex adjacent wth C t and he captures the robber by the next move. Case 2. The robber occupes some vertex C j n the begnnng of the game. For each {1,..., n} such that C j contans ether x or x, the cop from z moves ether to x or to x accordng to the ncluson of the lteral. If x n, x n are not ncluded n C j, then addtonally the cop from z n moves to x n. Clearly, the robber should ether stay on C j or move to y n. In the frst case he s captured by one of the cops on x, x, and he s captured by the cop from x n or x n n the second case. To complete the proof of PSPACE-hardness, t remans to prove the followng lemma.

10 10 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT Lemma 3.6. If Φ = false, then the robber has a wnnng strategy aganst n cops on G n Cops & Robber wth Cost Constrants and Cops & Robber wth Tme Constrants. Proof. By Observaton 1, now t s suffcent for the proof, to descrbe a wnnng strategy for the robber-player for Cops & Robber wth Tme Constrants. Assume that the cops have chosen ther ntal postons. If there s a path (z, w )-path n some G such that all vertces of the path are unoccuped by the cops, then we place the robber on w. Snce there are no cops at dstance at least s + 1 from w, the wnnng strategy for the robber s trval he should stay on w. Suppose now that for each (z, w )-path, there s a cop on one of the vertces of the path. We have n cops. Hence exactly one cop occupes one vertex of each path. Moreover, ths cop should occupy z f the cops do not want the robber to wn by a trval strategy, snce otherwse the vertces v (1), v (2), r, r have no cops at dstance at least s + 1, and the robber can safely stay on one of these vertces. Denote ths cop by P. The robber s placed on a vertex of Y 1. The choce of the vertex and further moves of the robber are descrbed nductvely for {1,..., n}. Suppose that for each j n, the cop P j s on the vertex z j, and each 1 j <, P j s ether on x or x. Assume also that values of the varables x 1,..., x 1 are already defned and Q x... Q n x n = false for ths assgnment. a) Q =. Snce Q x... Q n x n = false, there s a value of x for whch Q +1 x Q n x n = false. If ths value s true, then the robber s placed on y and he s placed on y otherwse. Observe, that the value of x s chosen by the robber-player. In the followng cases the robber has a straghtforward wnnng strategy. One of the cops P j for 1 j < leaves the vertex of X j. If j 1 or j = 1 and the cop from z does not move smultaneously to X 1, then the robber chooses one of the vertces p (t) j not occuped by P j, moves to p (t) j, and then moves by hs subsequent moves to q(t) j. Observe that n ths case the cops already made at least two moves and any cop have to make at least two further moves to reach the vertex p (t) j. Snce ths (p (t) j, q(t) j )-path s not occuped by the cops when P j leaves X j, ths s a wnnng strategy for the robber. One of the cops P j for < j n leaves the vertex z j. Then the robber chooses one of the vertces u (t) j then moves by hs subsequent moves to v (t) j not occuped by P j and moves to u (t) j, and. Snce ths (u (t) j, v(t) j )-path s not occuped by the cops when P j leaves z j, ths s a wnnng strategy for the robber. The cop P moves to the vertex of X not adjacent wth the robber s poston. Then the robber moves to the vertex of X adjacent wth hs current poston, and then moves by hs subsequent moves to ether r or r and wns. The cop P moves from z to a vertex of X 1 but the cop from X 1 does not go to z by the same step. Then the robber agan moves to the vertex of X adjacent wth hs current poston, and then moves by hs subsequent moves to ether r or r and wns. The cop-player has the followng remanng possbltes. Exactly two cops P 1 and P move: the cop P 1 moves from hs current poston n X 1 to z and the cop P moves to X 1. We call a move

11 COPS AND ROBBER WITH CONSRAINTS 11 of ths type swtchng, and we assume that these cops exchange ther names P 1 and P after ths move. In ths case the robber stays n hs current poston. Exactly one cop P moves to the vertex of X adjacent wth the robber s poston. Then the robber moves to a vertex of Y +1 or to some vertex of {C 1,..., C m } f = n. b) Q j =. The robber s placed on y. In the followng cases the robber wns drectly. One of the cops P j for 1 j < leaves the vertex of X j. Then the robber chooses one of the vertces p (t) j then moves by hs subsequent moves to q (t) j not occuped by P j, moves to p (t) j, and. Snce ths (p (t) j, q(t) j )-path s not occuped by the cops when P j leaves X j, ths s a wnnng strategy for the robber. One of the cops P j for < j n leaves the vertex z j. Then the robber chooses one of the vertces u (t) j not occuped by P j, moves to u (t) j, and. Snce ths (u (t) j, v(t) j )-path then moves by hs subsequent moves to v (t) j s not occuped by the cops when P j leaves z j, ths s a wnnng strategy for the robber. The cop P moves from z to ether u (1) or u (2). Then the robber moves to the vertex of X adjacent to hs current poston, and then moves by hs subsequent moves to ether r or r. Only one case remans: exactly one cop P moves to the vertex of X. If he moves to x, then we let x = true and x = false otherwse. Observe that now the cop-player chooses the value of x. Then the robber moves to a vertex of Y +1 or to some vertex of {C 1,..., C m } f = n. It remans to defne the strategy for the case when the robber moves to a vertex of {C 1,..., C m }. Now we can assume that the varables x 1,..., x n have values for whch F = false. Hence, there s a clause C j = false. Observe that the total number of steps s at least n = s 1. Moreover, f the cops made at least one swtchng step, then the total number of steps s s and the game s over. It follows that by each step exactly one cop was moved and they occupy vertces x, x accordng to the values of the correspondng varables. Therefore, the cops cannot move by the last step to C j. The robber moves to ths vertex and stays there. Now the proof of the theorem follows from Lemmata 3.5 and Incluson n PSPACE. Theorem 3.4 establshes only PSPACE-hardness of Cops and Robber wth Cost Constrants and Cops and Robber wth Tme Constrants. In what follows, we prove that these problems are n PSPACE when the parameter s bounded by some polynomal of the nput sze or f t s assumed that these ntegers are encoded n unary. Theorem 3.7. For every ntegers s, k 1 and an n-vertex graph G, t s possble to decde whether c σ s (G) k (c τ s (G) k respectvely) by makng use of space O(s n O(1) ). Proof. The proof s constructve. We descrbe a recursve algorthm whch solves Cops and Robber wth Cost Constrants. Note that we can consder only strateges of the cop-player such that at least one cop s moved to an adjacent vertex. Otherwse, f all cops are stayng n old postons, the robber can only mprove hs poston. Our algorthm uses a recursve procedure W (P, u), whch for a poston of the

12 12 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT cops P = (C, l), C = (v 1,..., v k ) such that l s, and a vertex u V (G), returns true f k cops can wn startng from the poston P aganst the robber whch starts from the vertex u, and the procedure returns f alse otherwse. Clearly, k cops can capture the robber on G f and only f there s an ntal poston P 0 such that for any u V (G), W (P 0, u) = true for l = s. If l = 0, then W (P, u) = true f and only f u = v for some 1 k. Suppose that l > 0. Then W (P, u) = true n the followng cases: u = v for some 1 k, u N G (v ) for some 1 k, there s a poston P = (C, l ), where C = (v 1,..., v k ) and l < l, such that the cops can go from P to P n one step n such a way that exactly l l cops move to ther new postons along edges, and for any u N G [u], W (P, u ) = true. Snce all postons can be lsted (wthout storng them) by usng polynomal space, the number of possble moves of the robber s at most n, and the depth of the recurson s at most s, the algorthm uses space O(s n O(1) ). The algorthm for Cops and Robber wth Tme Constrants s almost the same. The only dfference s that for l > 0, W (P, u) = true f and only f u = v for some 1 k, or u N G (v ), or there s a poston P = (C, l 1), where C = (v 1,..., v k ), such that the cops can go from P to P n one step, and for any u N G [u], W (P, u ) = true. 4. Random Graphs. In ths secton, we present asymptotc results for the game of Cops & Robber played on a bnomal random graph G(n, p). The secton s organzed as follows. In Subsecton 4.1 we defne the probablty space we are nterested n, and brefly descrbe results for the orgnal game played on a random graphs. In Subsecton 4.2 we descrbe our man results for the the varant wth Fuel Constrants as well as the one wth Tme Constrants. Upper and lower bounds are nvestgated separately n Subsectons 4.3 and 4.4, respectvely. The proofs are based on technques and deas from [24, 28]. Fnally, n Subsecton 4.5 we dscuss the thrd varant of the game, namely the one wth Cost Constrants, n the context of random graphs. A non-trval upper bound s provded for ths case but the behavor of ths varant of the game remans to be nvestgated Defntons and the orgnal game. Let us recall the classc model of random graphs that we study n ths paper. The bnomal random graph G(n, p) s defned as a random graph wth vertex set [n] = {1, 2,..., n} n whch a par of vertces appears as an edge wth probablty p, ndependently for each such a par. As typcal n random graph theory, we shall consder only asymptotc propertes of G(n, p) as n, where p = p(n) may and usually does depend on n. We say that an event n a probablty space holds asymptotcally almost surely (a.a.s.) f ts probablty tends to one as n goes to nfnty. Let us frst brefly descrbe some known results on the cop number of G(n, p). Bonato, Hahn, and Wang [7] started nvestgatng such games n G(n, p) random graphs. Bonato, Wang, and the thrd author of ths paper [9], generalzed these result to sparser random graphs and ther generalzatons used to model complex networks wth a power-law degree dstrbuton. From ther results t follows that f

13 COPS AND ROBBER WITH CONSRAINTS 13 2 log n/ n p < 1 ε for some ε > 0, then a.a.s. c(g(n, p)) = Θ(log n/p). A smple argument usng domnatng sets shows that c(g(n, p)) = o(log n) a.a.s. f p tends to 1 as n goes to nfnty (see [27] for ths and stronger results). Recently, Bollobás, Kun and Leader [5] showed that for p(n) 2.1 log n/n, we get that a.a.s. 1 (np) 2 n(log log(np) 9)/(2 log log(np)) c(g(n, p)) n log n. From these results, f np 2.1 log n and ether np = n o(1) or np = n 1/2+o(1), then a.a.s. c(g(n, p)) = n 1/2+o(1). Somewhat surprsngly, between these values t was shown by Luczak and the thrd author of ths paper [24] that the cop number has more complcated behavor. It follows that a.a.s. log n c(g(n, n x 1 )) s asymptotc to the functon g(x) shown n Fgure 4.1(a). (a) g(x) (b) h 3(x) (b) h 5(x) Fg The zgzag functons. Below we precsely state the result. Theorem 4.1 ([24]). Let 0 < α < 1, let j 1 be nteger, and let d = d(n) = (n 1)p = n α+o(1) If 2j+1 < α < 1 2j, then a.a.s. c(g(n, p)) = Θ(d j ) If 2j < α < 1 2j 1, then a.a.s. n d j = O( c(g(n, p)) ) ( n ) = O d j log n. 3. If α = 1/(2j) or α = 1/(2j + 1), then a.a.s. c(g(n, p)) = d j+o(1). In partcular, t follows from [5, 24] that the cop number s a.a.s. of order O( n log n), provded that d 2.1 log n. Recently, Wormald and the thrd author of ths paper removed unnecessary log n factor and extended the result to sparser graphs and random d-regular graphs, provng Meynel s conjecture for random graphs [28] Man results for the two varants of the game. In ths subsecton, we descrbe the behavor of the cop number wth two constrants we consder n ths paper: the varant wth Fuel Constrants as well as the one wth Tme Constrants. From the pont of vew of random graphs, asymptotc behavors of c φ s (G(n, p)) and

14 14 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT c τ s (G(n, p)) are essentally the same. All our proofs work for both varants of the game, and we state result for the Fuel Constrants varant only. The varant wth Cost Constrants s dscussed n Subsecton 4.5. Ths varant s stll not fully nvestgated. It follows from Observaton 1 that for any s N and any graph G we have that c τ s (G) c φ s (G) c(g). However, t turns out that ntroducng constrants n the random graph case does not affect substantally the cop number, provded that the graph s dense enough; that s, the average degree s at least n 1/(2s). Wth a log n tmes more cops (or sometmes just O(1) tmes more) we can catch the robber n a few steps only, whch provdes an upper bound of c τ s (G). On the other hand, for sparse random graphs the dstance s-domnaton number γ s (G(n, p)) s a.a.s. large whch mples that the constraned cop number has to go up. We wll show that a.a.s. c τ s (G) (and thus c φ s (G) as well) has the same order as ths obvous lower bound. In partcular for s = 3, 5, t follows that a.a.s. log n c τ s (G(n, n x 1 )) (and log n c φ s (G(n, n x 1 )) as well) s asymptotc to the functon h s (x) shown n Fgure 4.1(b) and (c), respectvely. Here s a precse result for any s N. Theorem 4.2. Let 0 < α < 1, let s 1 be nteger, and let d = d(n) = (n 1)p = n α+o(1). () If α < 1 2s 1, then a.a.sċφ s(g(n, ( n ) p)) = Θ d s log n. () If 1 2j+1 < α < 1 2j () If 1 2j < α < 1 2j 1 for some nteger j < s, then a.a.s. c φ s (G(n, p)) = Θ(d j ). for some nteger j < s, then a.a.s. n d j = O( c φ s (G(n, p)) ) ( n ) = O d j log n. (v) If α = 1/(2j) or α = 1/(2j + 1) for some nteger j < s, then a.a.s. c φ s (G(n, p)) = d j+o(1). Exactly the same statement holds for c τ s (G(n, p)) Upper bound. In ths secton, we provde an upper bound for the man theorem of ths secton, Theorem 4.2. The result s actually slghtly stronger than the one stated n the man theorem. Frst of all, we prove t for all random graphs but qute sparse ones (the average degree at least log 3 n) whereas the man theorem focuses on dense graphs (the average degree n α+o(1) for some α (0, 1)). Second of all, we allow s = s(n) to be a functon of n whereas the man theorem assumes ths to be arbtrarly large but fxed constant. We wll use the followng expanson-type property of random graphs that s proved n [28]. Lemma 4.3 ([28]). Suppose that d = p(n 1) log 3 n. Let G = (V, E) G(n, p). Then the followng property holds a.a.s. Let Q V be any set of q = Q vertces, and let s = s(n) N. Then N(v, s) 1 3 mn{qds, n}. v Q Wth ths tool n hand we are ready to prove the followng theorem. Theorem 4.4. Let s = s(n) 1 be nteger and let d = d(n) = (n 1)p.

15 COPS AND ROBBER WITH CONSRAINTS 15 () If n 1/(2j+1) d (n log n) 1/(2j) for some nteger j < s, then a.a.s. c φ s (G(n, p)) = O(d j β), where β = max{n log n/d 2j+1, 1} = O(log n). () If (n log n) 1/(2j+2) d n 1/(2j+1) for some nteger j < s, then a.a.s. ( n ) c φ s (G(n, p)) = O d j+1 log n. () If d (n log n) 1/(2s), then a.a.s. ( n ) c φ s (G(n, p)) = O d s log n. Before we start the proof, let us menton that () actually works for any value of d. Ths wll provde a non-trval upper bound for the thrd varant of the game we dscuss n Subsecton 4.5. Here, we put restrcton on d to focus on the range that gves the best upper bound. Note that when d = n 1/(2j+1) for some j < s, then both () and () gve the same bound of O(n j/(2j+1) log n). Smlarly, f d = (n log n) 1/(2j) for some j < s, then both () and () gve a bound of O( n log n). Proof. We show that the upper bounds hold for an arbtrary graph G possessng the propertes stated n Lemma 4.3; then that lemma mples the theorem. We start wth (). The team of cops s determned by ndependently choosng each vertex of v V to be occuped by a cop wth probablty Cd j β/n, where C s a constant to be determned soon. The total number of cops s (1 + o(1))cd j β a.a.s. The robber appears at some vertex v V. We wll show below that a.a.s. for each vertex v V t s possble to assgn dstnct cops to all vertces u n N(v, j) such that a cop assgned to u s wthn dstance (j + 1) of u. (Note that here, a.a.s. refers to the randomness n dstrbutng the cops; the graph G s fxed.) If ths can be done, then after the robber appears these cops can begn movng straght to ther assgned destnatons n N(v, j). Snce the frst move belongs to the cops, they have j + 1 s steps, after whch the robber must stll be n N(v, j), whch s fully occuped by cops. He s caught and the game ends after at most s rounds. In order to show that the assgnments we requre exst a.a.s. for every vertex v, we show that the random placement of cops a.a.s. satsfes the Hall condton for matchngs n bpartte graphs. Set q 0 = max{q : qd j+1 < n}. Fx any vertex v V and let Q N(v, j) wth Q = q q 0. It follows by applyng the condton n Lemma 4.3 to bound the sze of u Q N(u, j + 1), that the number of cops occupyng ths set of vertces can be stochastcally bounded from below by the bnomal random varable Bn( qd j+1 /3, Cd j β/n) wth expected value asymptotc to C 3 qd2j+1 β/n C 3 q log n. We next use a consequence of Chernoff s bound (see e.g. [21, p. 27 Cor. 2.3]), that ( ) P( X EX εex)) 2 exp ε2 EX 3 for 0 < ε < 3/2. Ths mples that the probablty that there are fewer than q cops n ths set of vertces s less than exp( 4q log n) when C s a suffcently large constant.

16 16 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT Hence, the probablty that the necessary condton n the statement of Hall s theorem fals for at least one set Q wth Q q 0 s at most q 0 ( ) N(v, j) q 0 exp( 4q log n) n q exp( 4q log n) = o(n 1 ). q q=1 Let Q N(v, j) wth q 0 < Q = q N(v, j) 2d j. (The last nequalty s a rough estmaton. In fact, t follows from (4.1) below that N(v, j) = (1+o(1))d j.) It follows from Lemma 4.3 that the sze of u Q N(u, j +1) s at least n/3, so we expect at least C 3 dj cops n ths set. Usng Chernoff s bound agan, we get that the number of cops s at least 2d j N(v, j) Q wth probablty at least 1 exp( 4d j ), by takng the constant C to be large enough. Snce N(v,j) q=q 0+1 ( N(v, j) q q=1 ) exp( 4d j ) 2d j 2 2dj exp( 4d j ) = o(n 1 ), the necessary condton n Hall s theorem holds wth probablty 1 o(n 1 ). Hence, the assgnment can be done wth at least ths probablty, for each possble startng vertex v V. It follows that the strategy s a wnnng one for the cops a.a.s. Ths fnshes the proof of (). One can adjust the proof of () to get (). Note that n ths case for any set Q N(v, j), t follows from Lemma 4.3 that the sze of u Q N(u, j + 1) s at least Q d j+1 /6, snce Q 2d j and so Q d j+1 2n. Therefore, the bottleneck when checkng the Hall s condton s not for sets of large cardnalty anymore, ths tme we need to adjust the number of cops so that the condton holds for small sets. The team of cops s determned by ndependently choosng each vertex of v V to be occuped by a cop wth probablty C log n/d j+1, where C s a large enough constant. The expected number of cops n u Q N(u, j + 1) s at least C Q log n/6 and we are fne. Fnally, let us nvestgate (). The game has to end after s steps (at least n the case of the game wth tme constrants). Hence, we need to restrct ourselves and search for cops n a ball of radus s. Smlarly as before, the team of cops s determned by ndependently choosng each vertex of v V to be occuped by a cop wth probablty C log n/d s, where, as usual, C s a large enough constant. For any set Q N(v, s 1), we expect Q log n/3 cops n u Q N(u, s) and the proof works exactly as before Lower bound. In ths secton, we provde a lower bound of the cop number for the man theorem of ths secton, Theorem 4.2. In order to do t we need another expanson-type property of random graphs. Lemma 4.3 provdes a lower bound for the sze of a unon of neghbourhoods; ths tme we are nterested n fndng an upper bound for t. In order to prove the followng lemma, an adjustment of the proof from [28] s needed. Lemma 4.5. Suppose that d = p(n 1) log 3 n. Let G = (V, E) G(n, p). Then the followng property holds a.a.s. Let Q V be any set of q = Q vertces, and let s = s(n) N be such that qd s 1 2n log n. Then N(v, s) < n. v Q

17 COPS AND ROBBER WITH CONSRAINTS 17 In partcular, a.a.s. γ s (G(n, p)) > n log n 2d. s Proof. Let Q V, q = Q, and consder the random varable X = X(Q) = N[Q]. We wll bound X from below n a stochastc sense. There are two thngs that need to be estmated: the expected value of X, and the concentraton of X around ts expectaton. It s clear that ( EX = n 1 d ) q (n q) n 1 ( = n exp dq ) (1 + O(d/n)) (n q) n = dq(1 + O(log 1 n)) provded dq n/ log n. It follows from Chernoff s bound that the expected number of sets Q that have N[Q] d Q > εd Q and Q n/(d log n) s, for ε = 2/log n, at most ( n q exp ε2 q log 3 ) n = exp (q log n 43 ) 3 q log n = o(1). q 1 q 1 So a.a.s. f Q n/(d log n) then N[Q] = d Q (1 + O(1/ log n)) where the bound n O() s unform. We may assume ths statement holds. Gven ths assumpton, we have good bounds on the ratos of the cardnaltes of N[Q], N[N[Q]] = v Q N(v, 2), and so on. We consder ths up to the s th terated neghborhood provded qd s n/ log n and thus s = O(log n/ log log n). Then the cumulatve multplcatve error term s (1 + O(log 1 n)) s = (1 + o(1)), that s, N(v, s) = (1 + o(1))qds (4.1) v Q for all q and s such that qd s n/ log n. Ths establshes the property n ths case. Suppose now that qd s = cn log n wth 1/ log 2 n < c = c(n) 1/2. Usng (4.1), we have that U = v Q N(v, s 1) has cardnalty (1+o(1))qds 1 = O(n/ log 2 n) = o(n). Now V \ N[U] has expected sze e c log n n(1 + o(1)) n(1 + o(1)), snce c 1/2. Chernoff s bound can be used agan n the same way as before to show that wth hgh probablty V \ N[U] s concentrated near ts expected value, and hence that a.a.s. V \ N[U] > 2 1 n for all q and s n ths case. Thus the statement holds also n ths case. Lemma 4.5 mples easly the followng general lower bound that s tght for sparse graphs wth the expected degree d n 1/(2s 1). Corollary 4.6. Suppose that d = p(n 1) log 3 n. Let G = (V, E) G(n, p) and s = s(n) N. Then a.a.s. ( ) n log n c φ s (G) = Ω. d s

18 18 F. V. FOMIN, P. A. GOLOVACH AND P. PRA LAT Proof. Suppose that q = n log n 2d cops try to catch the robber. It follows from s Lemma 4.5 that a.a.s. for every startng confguraton for cops, there s a vertex that s at the dstance at least s + 1 from any of the cops. The robber starts the game on ths vertex and remans safe to the end of the game. In short, we use the fact that c φ s (G) γ s (G). For denser graphs (d > n 1/(2s 1) ), a lower bound for the classc game s useful, snce c φ s (G) c(g). It turns out that these bounds are tght. In other words, ntroducng more constrants on cops s not affectng substantally the cop number. The followng results have been shown n [24] for the classc cop number, whch provde a lower bounds for our varants of the game. 1 Theorem 4.7 ([24]). Let 2j+1 < α < 1 2j for some natural j 1, c = c(j, α) = 3 1 2jα and d = d(n) = (n 1)p = nα+o(1). Let s = s(n) N. Then a.a.s. c φ s (G(n, p)) c(g(n, p)) [ d ] j. 3cj 1 Theorem 4.8 ([24]). Let 2j < α < 1 2j 1 for some natural number j 2, 3 c = c(α) = 1 (2j 1)α and d = d(n) = (n 1)p = nα+o(1). Let s = s(n) N. Then a.a.s. [ d ] j n c φ s (G(n, p)) c(g(n, p)) 3 cj cd 2j. Note that n the above result the case when np = n 1/k+o(1), for some natural k s skpped. It was done for techncal reasons. However, t was verfed that, up to a factor of log O(1) n, the result extends to the case np = n 1/k+o(1) as well. Fnally, as we already mentoned, very dense random graphs were nvestgated n [9, 27]. We know that f np = n α+o(1) wth α [1/2, 1], then a.a.s. c(g(n, p)) = n 1 α+o(1). Specfcally, to get an upper bound t was proved that a.a.s. there s a domnatng set of cardnalty n 1 α+o(1). If ths s the case, then one move s enough to catch the robber. We have that a.a.s. c σ s (G(n, n α 1 )) = n 1 α+o(1) for any s 1 and 1/2 α 1. Ths mples that the same statement holds for c φ s (G(n, n α 1 )) and cτ s(g(n, n α 1 )), whch gves us the whole pcture for a wde range of d log 3 n Cops & Robber wth Cost Constrants. In ths secton, we dscuss the last varant of the game, the one wth Cost Constrants. From the perspectve of random graphs, ths varant seems to be the most challengng one. When the total dstance s bounded by n η+o(1) for some η < 1/2, our upper bound sometmes does not math the lower bound of the orgnal cop number. We stll do not know the shape of the functon assocated wth ths varant of the game. Let us note that the proof of Theorem 4.4() provdes an upper bound for the game wth cost constrants as well. We restrcted the range of d before for the best outcome but, as we already mentoned, the statement holds for any value of d. Let j N. For d (n log n) 1/(2j+1), the strategy requrng O(d j ) cops was presented; each cop made at most (j + 1) moves for a total dstance traveled by cops of (1 + o(1))(j + 1)d j. For

19 COPS AND ROBBER WITH CONSRAINTS 19 d (n log n) 1/(2j+1), more cops were needed (O ( n d log n ) ) but only a small fracton j+1 of them were actually movng ((1 + o(1))d j ). Agan, each actve cop moved exactly (j + 1) tmes for a total dstance of (1 + o(1))(j + 1)d j. We get the followng corollary. Corollary 4.9. Let j 1 be nteger, let d = d(n) = (n 1)p, and consder a game played on G(n, p). Then, a.a.s. cops have a wnnng strategy requrng: () O(d j β) cops, where β = max{n log n/d 2j+1, 1}, () (1 + o(1))(j + 1)d j moves, n total. In partcular, f s (1 + ε)(j + 1)d j for some j N and ε > 0, then a.a.s. c σ s (G(n, p)) = O(d j β), where β = max{n log n/d 2j+1, 1}. Upper bounds correspondng to j = 1, 2, 3 as well as the one correspondng to the strategy based on the domnaton number that works for any s 1 are presented n Fgure 4.2(a). (Although t has nothng to do wth j from the prevous corollary, let us say that ths unversal bound corresponds to j = 0.) (a) strateges for j {0, 1, 2, 3} (b) l 2 5 (x) (c) l 2 5 (x) and the lower bound Fg More zgzag functons. Note that f the total dstance s bounded by n η+o(1) for some η 1/2, the addtonal restrcton does not prevent us from choosng the optmal value of j and we get exactly the same zgzag as before (see Fgure 4.1(a)). The upper bound matches the lower bound for the orgnal game. Unfortunately, t s not the case when η < 1/2. Ths tme we need to choose the largest value of j such that s (1+ε)(j+1)d j for some ε > 0. If no such j exsts, we need to use the unversal strategy (j = 0). The functon assocated wth an upper bound s not contnuous anymore and bounds do not match for many denstes. However, t would not be surprsng f ths s the behavor of the cop number for ths varant of the game. Ths remans an open problem. In order to llustrate the behavor of the upper bound, we consder the followng example: suppose that s = n η+o(1) wth η = 2 5 and the graph has average degree of d = n x+o(1) for some x (0, 1). To get the best outcome, we choose the largest value of j such that j < η x ; that s j = η x 1. In partcular, for x 2 5 we take j = 0 to get a bound of n 1 x+o(1). For 1 5 x < 2 5, we select j = 1 but a bound changes at x = 1 3 : we have nx+o(1) for 1 3 x 2 5 and n1 2x+o(1) for 1 5 x < 1 3. Ths s the only 2 lower peak we get for ths value of η. If 5(k+1) x < 2 5k, then we take j = k and we get a bound of n 1 (k+1)x+o(1). The functon l 2 (x) assocated wth the case of η = s presented n Fgure 4.2(b). The functon s repeated n Fgure 4.2(c) together wth the lower bound. For η (0, 1 2 ), the shape of l η(x) s smlar. If η > 2 5 we wll see more lower peaks (but stll a fnte number); however, there wll be none of them for η 1 3. For a sparse graphs (x small enough), l η (x) = 1 η x x and so lm x 0 l η (x) = 1 η.