How many zombies are needed to catch the survivor on toroidal grids?
HHOW MANY ZOMBIES ARE NEEDED TO CATCH THE SURVIVOR ONTOROIDAL GRIDS? EMBARRASSING UPPER BOUND OF O ( n ) . PAWE(cid:32)L PRA(cid:32)LAT
Abstract.
In Zombies and Survivors, a set of zombies attempts to eat a lone survivorloose on a given graph. The zombies randomly choose their initial location, and during thecourse of the game, move directly toward the survivor. At each round, they move to theneighbouring vertex that minimizes the distance to the survivor; if there is more than onesuch vertex, then they choose one uniformly at random. The survivor attempts to escapefrom the zombies by moving to a neighbouring vertex or staying on his current vertex. Thezombies win if eventually one of them eats the survivor by landing on their vertex; otherwise,the survivor wins. The zombie number of a graph is the minimum number of zombies neededto play such that the probability that they win is at least 1/2.This variant of the game was recently investigated for several graph families, such ascycles, hypercubes, incidence graphs of projective planes, and grids P n (cid:3) P n . However, un-fortunately, still very little is known for toroidal grids C n (cid:3) C n : the zombie number of C n (cid:3) C n is at least √ n/ ( ω log n ), where ω = ω ( n ) is any function going to infinity as n → ∞ , andno upper bound is known except a trivial bound of O ( n log n ). In this note, we provide anapproach that gives an embarrassing bound of O ( n ) but it is possible that (with more care-ful, deterministic, argument) it might actually give a bound of O ( n / ). On the other hand,by analyzing a specific strategy for the survivor, it seems that one could slightly improvethe lower bound to √ n/ω . In any case, we are far away from understanding this intriguingquestion. Your help is needed! Introduction
For a given connected graph G and given k ∈ N , we consider the following probabilisticvariant of Cops and Robbers, which is played over a series of discrete time-steps. In thegame of Zombies and Survivors , suppose that k zombies (akin to the cops) start the gameon random vertices of G ; each zombie, independently, selects a vertex uniformly at randomto start with. Then the survivor (akin to the robber) occupies some vertex of G . As zombieshave limited intelligence, in each round, a given zombie moves towards the survivor alonga shortest path connecting them. In particular, the zombie decreases the distance from itsvertex to the survivor’s. If there is more than one neighbour of a given zombie that is closerto the survivor than the zombie is, then they move to one of these chosen uniformly atrandom. Each zombie moves independently of all other zombies. As in Cops and Robbers,the survivor may move to another neighbouring vertex, or pass and not move. The zombieswin if one or more of them eat the survivor; that is, land on the vertex which the survivorcurrently occupies. The survivor, as survivors should do in the event of a zombie attack,attempts to survive by applying an optimal strategy; that is, a strategy that minimizes theprobability of being captured. Note that there is no strategy for the zombies; they merely Mathematics Subject Classification.
Key words and phrases.
Zombies and Survivors, Cops and Robbers.The author gratefully acknowledges support from NSERC. a r X i v : . [ m a t h . C O ] M a r PAWE(cid:32)L PRA(cid:32)LAT move on geodesics towards the survivor in each round. Note that since zombies always movetoward the survivor, he can pass at most D times, where D is a diameter of G , before beingeaten by some zombie. We note also that our probabilistic version of Zombies and Survivorswas inspired by a deterministic version of this game (with similar rules, but the zombies maychoose their initial positions, and also choose which shortest path to the survivor they willmove on) first considered in [2].Let s k ( G ) be the probability that the survivor wins the game, provided that he followsthe optimal strategy. Clearly, s k ( G ) = 1 for k < c ( G ), where c ( G ) is the cop number of G . On the other hand, s k ( G ) < k ≥ c ( G ) cops inwhich the cops always try to get closer to the robber, since with positive probability thezombies may follow such a strategy. Usually, s k ( G ) > k ≥ c ( G ); however, thereare some examples of graphs for which s k ( G ) = 0 for every k ≥ c ( G ) (consider, for example,trees). Further, note that s k ( G ) is a non-decreasing function of k (that is, for every k ≥ s k +1 ( G ) ≤ s k ( G )), and s k ( G ) → k → ∞ . The latter limit follows since the probabilitythat each vertex is initially occupied by at least one zombie tends to 1 as k → ∞ .Define the zombie number of a graph G by z ( G ) = min { k ≥ c ( G ) : s k ( G ) ≤ / } . This parameter is well defined since lim k →∞ s k ( G ) = 0. In other words, z ( G ) is the minimumnumber of zombies such that the probability that they eat the survivor is at least 1/2. Theratio Z ( G ) = z ( G ) /c ( G ) ≥ cost of being undead . Note that there are examples offamilies of graphs for which there is no cost of being undead; that is, Z ( G ) = 1 (as is thecase if G is a tree), and, as argued in [1], there are examples of graphs with Z ( G ) = Θ( n ) . This variant of the game was recently introduced in [1], where several graph familieswere investigated, such as cycles, hypercubes, incidence graphs of projective planes, andCartesian grids. In particular, the zombie number of the incidence graphs of projectiveplanes is about two times larger than the corresponding cop number; for hypercubes, thisratio is asymptotically 4 / T n = C n (cid:3) C n (so called toroidal grids ). In [1], it wasproved that the zombie number of T n is at least √ n/ ( ω log n ), where ω = ω ( n ) is anyfunction going to infinity as n → ∞ . The proof relies on the careful analysis of a strategyfor the survivor. On the other hand, trivially, z ( T n ) = O ( n log n ). Suppose that thegame is played against k = 3 n log n zombies. It is easy to see that a.a.s. every vertexis initially occupied by at least one zombie and if so the survivor is eaten immediately.Indeed, the probability that at at least one vertex is not not occupied by a zombie is at most n (1 − /n ) k ≤ n exp( − k/n ) = 1 /n = o (1).In this paper, we provide a general approach that gives immediately a tiny improvement,an upper bound of O ( n ). We investigate this technique by analyzing a few natural strategiesand show that this approach has no hope of giving an upper bound better than O ( n / ).However, it is quite possible that (with more careful, deterministic, argument) it mightactually give that bound. It is left for a further investigation. On the other hand, byanalyzing a specific strategy for the survivor, it seems that one could slightly improve thelower bound to √ n/ω . In any case, we are far away from understanding this intriguingquestion. Your help is needed! OW MANY ZOMBIES ARE NEEDED TO CATCH THE SURVIVOR ON TOROIDAL GRIDS? 3 Results
Definitions.
Let us start with a formal definition of the class of graphs we investigatein this paper. For graphs G and H , define the Cartesian product of G and H , written G (cid:3) H, to have vertices V ( G ) × V ( H ) , and vertices ( a, b ) and ( c, d ) are joined if a = c and bd ∈ E ( H )or ac ∈ E ( G ) and b = d. In this note, we consider grids formed by products of cycles. Let T n be the toroidal grid n × n , which is isomorphic to C n (cid:3) C n . For simplicity, we take the vertexset of T n to consist of Z n × Z n , where Z n denotes the ring of integers modulo n .Results in the paper are asymptotic in nature as n → ∞ . We emphasize that the notations o ( · ) and O ( · ) refer to functions of n , not necessarily positive, whose growth is bounded. Wesay that an event in a probability space holds asymptotically almost surely (or a.a.s. ) if theprobability that it holds tends to 1 as n goes to infinity. Finally, for simplicity we will write f ( n ) ∼ g ( n ) if f ( n ) /g ( n ) → n → ∞ ; that is, when f ( n ) = (1 + o (1)) g ( n ).2.2. General upper bound.
We are going to play the game on the toroidal grid T n . Letus consider the family F of strategies of the survivor for the first M = M ( n ) := (cid:98) n/ (cid:99) moves of the game. Here, a strategy is simply a sequence of moves of the survivor whichis not affected neither by initial distribution of zombies nor by their behaviour during thegame. Moreover, we may assume that zombies do not eat the survivor immediately whenthey catch him; instead, they walk with him to the end of this sequence of T moves andthen do their job. Hence, since there are n vertices to choose from for the starting positionof the survivor and 5 options in each round (“go west”, “go east”, “go north”, “go south”,and “stay put”), the number of strategies in F is n M = n (cid:98) n/ (cid:99) . Finally, let F ⊆ F be asubfamily of 5 (cid:98) n/ (cid:99) strategies of the survivor that finish his walk at vertex (0 ,
0) of T n .Let us concentrate on a given strategy S ∈ F finishing at vertex ( a, b ) of T n . For any x, y ∈ Z such that | x | + | y | ≤ M , let p S ( x, y ) be the probability that a zombie starting atvertex ( a + x, b + y ) eats the survivor using strategy S for the first M moves. Clearly, ifa zombie starts at vertex at distance larger than M from ( a, b ), then it is impossible forher to catch the survivor (even intelligent player would not be able to reach ( a, b ) in M moves!); hence, in this situation p S ( x, y ) = 0. Note also that restricting staring positions forzombies to the subgraph around ( a, b ) guarantees that the survivor and zombies starting atthis subgraph do not leave it during the fist M rounds. As a result, the game during these M rounds is played as if it was played on the square grid ( P n (cid:3) P n ) centred at ( a, b ), not thetoroidal one. Moreover, since T n is a vertex transitive graph, without loss of generality, wemay assume that the survivor finishes his walk at vertex (0 , t ( S ) = M (cid:88) x = − M M (cid:88) y = − M p S ( x, y ) . As mentioned earlier, due to the fact that T n is vertex transitive, we have t n := min S∈F t ( S ) = min S∈F t ( S ) . Now, we are ready to state our first observation.
Theorem 2.1.
A.a.s. k = n /t n zombies catch the survivor on T n . Hence, z ( T n ) = O ( n /t n ) .Proof. Let X , X , . . . be a sequence of independent random variables, each of them being theBernoulli random variable with parameter p = 1 / X i = 1) = Pr( X i = 0) = 1 / PAWE(cid:32)L PRA(cid:32)LAT for each i ∈ N ). This (random) sequence will completely determine the behaviour of all thezombies. Formally, let us first fix a permutation π of k zombies. Then, in each round, weconsider all zombies, one by one, using permutation π . If there is precisely one shortest pathbetween a given zombie and the survivor, the next move is determined; otherwise, the nextrandom variable X i from the sequence guides it (for example, if X i = 0, the zombie moveshorizontally; otherwise, she moves vertically).Our goal is to show that a.a.s. the survivor is eaten during the fist M = (cid:98) n/ (cid:99) rounds,regardless of the strategy he uses. In order to show it, let us pretend that the game is playedon a real board but, at the same time, there are n M auxiliary boards where the game isplayed against all strategies from F . An important assumption is that the same sequence X , X , . . . is used for all the boards.Since zombies select initial vertices uniformly at random, the probability that a givenzombie wins against a given strategy S ∈ F is at least M (cid:88) x = − M M (cid:88) y = − M p S ( x, y ) n ≥ t n n . Since zombies play independently, the probability the survivor using strategy S is not eatenduring the first M rounds is at most (cid:18) − t n n (cid:19) k ≤ exp (cid:18) − t n kn (cid:19) = exp( − n ) = o (cid:18) n n/ (cid:19) . (Note that, in particular, if two zombies start at the same vertex ( x, y ), each of them catchesthe survivor with probability p S ( x, y ) and the corresponding events are independent.) Itfollows that the expected number of auxiliary games where the survivor is not eaten is o (1)and so a.a.s. he loses on all auxiliary boards. As a result, a.a.s. zombies win the real gametoo, regardless of the strategy used by the survivor. Clearly, the survivor should makedecisions based on the behaviour of the zombies (who make decisions at random and so veryoften behave in a sub-optimal way); however, if the survivor wins the real game, at leastone auxiliary game is also won by some specific strategy which we showed cannot happena.a.s. That is why we needed the trick with the sequence X , X , . . . guiding all the gamesconsidered. (cid:3) Deriving p S ( x, y ) . In this subsection, we first derive a recursive formula for calculating p S ( x, y ) for a given strategy S ∈ F . Consider any strategy
S ∈ F : at the beginning of round i (1 ≤ i ≤ M = (cid:98) n/ (cid:99) ), the survivor occupies vertex ( x i , y i ), zombies make their movesand then the survivor moves to ( x i +1 , y i +1 ); round i is finished and round i + 1 starts. Asbefore, since T n is vertex transitive, without loss of generality we may restrict ourselves toa subfamily F of strategies for the first M moves that bring the survivor to vertex (0 , M rounds; that is, ( x M , y M ) = (0 , p i S ( x, y ), theprobability that a zombie occupying ( x, y ) at the beginning of round i catches the survivorusing strategy S by the end of round M . Note that p S ( x, y ) = p S ( x, y ). Clearly, p M S (0 ,
0) = p M S (1 ,
0) = p M S ( − ,
0) = p M S (0 ,
1) = p M S (0 , −
1) = 1 , and p M S ( x, y ) = 0 if | x | + | y | > M ). Moreover, for any round 1 ≤ i < M we have the following OW MANY ZOMBIES ARE NEEDED TO CATCH THE SURVIVOR ON TOROIDAL GRIDS? 5 (x ,y ) i-1 i-1
Figure 1.
Getting p i − S ( x, y ) from p i S ( x, y ). → → Table 1.
The survivor does not move: p M S → p M − S → p M − S recursive formula. If x (cid:54) = x i and y (cid:54) = y i , then p i S ( x, y ) = p i +1 S ( x ± , y )2 + p i +1 S ( x, y ± , where “ ± ” is “+” or “ − ” so that | x i − ( x ± | < | x i − x | and | y i − ( y ± | < | y i − y | (zombiesmove towards the survivor, either horizontally or diagonally; the decision is made uniformlyat random). If x = x i and y (cid:54) = y i , then p i S ( x, y ) = p i +1 S ( x, y ± x (cid:54) = x i and y = y i , then p i S ( x, y ) = p i +1 S ( x ± , y )(zombies move horizontally towards the survivor).Reversing the direction, we get the following recursive relationship between p i S ( x, y ) and p i − S ( x, y ). (See Figure 1 for an illustration.) If x (cid:54) = x i − and y (cid:54) = y i − , then half of the weightof p i S ( x, y ) is moved horizontally and half of it is moved vertically to the two neighbours of( x, y ) that are further away from ( x i − , y i − ). If x (cid:54) = x i − but y = y i − , then half of theweight of p i S ( x, y ) is moved vertically to both “vertical” neighbours of ( x, y ); in additionto this, p i S ( x, y ) is added to the “horizontal” neighbour of ( x, y ) that is further away from( x i − , y i − ). We proceed in a symmetric way if x = x i − and y (cid:54) = y i − . Finally, we put p i − S ( x i − , y i − ) = p i S ( x i − + 1 , y i − ) = p i S ( x i − − , y i − )= p i S ( x i − , y i − + 1) = p i S ( x i − , y i − −
1) = 1 . PAWE(cid:32)L PRA(cid:32)LAT → → Table 2.
The survivor goes straight up: p M S → p M − S → p M − S Table 1 presents the first two iterations for the strategy where the survivor does not moveat all. Of course, for this simple strategy there is no need to use recursive formula. Clearly, p S ( x, y ) = 1 if the distance from ( x, y ) to (0 ,
0) is at most M (that is, | x | + | y | ≤ M ) and p S ( x, y ) = 0 otherwise. The next example, presented in Table 2 is more interesting; thistime the survivor goes straight up. In both cases, vertex ( x i − , y i − ) that is used to get p i − S from p i S is coloured red.2.4. Upper bound: z ( T n ) = O ( n ) . Since not only all the weight contributing to (cid:80) x,y p i S ( x, y )is preserved in (cid:80) x,y p i − S ( x, y ) but each time the total weight increases by at least 4, it triv-ially follows that t n ≥ M = 4 (cid:98) n/ (cid:99) ∼ n . As a result, from Theorem 2.1 we get the followingupper bound for z ( T n ). Corollary 2.2. z ( T n ) = O ( n ) . This improves the trivial upper bound of O ( n log n ) but it is embarrassing that this isthe best we can do. Is there an upper bound of n − ε for some ε >
0? This remains an openquestion.2.5.
Examples of 8 strategies.
In this subsection, we present a few natural strategiesfrom family F : the survivor . . . (a) . . . does not move.(b) . . . performs a random walk.(c) . . . goes straight down.(d) . . . moves along the diagonal (that is, moves down and then immediately left in eachpair of the two consecutive rounds).(e–h) . . . goes along edges of a square ( k times): k = 1 , , n to estimate t ( S ). Asdiscussed earlier, p S ( x, y ) = 0 if | x | + | y | > M = (cid:98) n/ (cid:99) so it makes sense to use the followingscaling: t ( S ) / ( n /
8) (note that the number of vertices at distance at most M is asymptoticto 4( M / ∼ n / t ( S ) / ( n / ∼
1. However, for strategies (d) and (h), it seemsthat different scaling is appropriate, namely, t ( S ) /n / . Strategy (d) is analyzed in one ofthe following subsections and it will become clear why this scaling is more appropriate. Wepresent our results in Table 3. Finally, functions p S ( x, y ) are presented visually on Figures 2and 3 (dark colours correspond to values of probabilities that are close to 1, light ones tovalues close to 0).2.6. Reduction to one dimensional problem.
For a while, our conjecture was that t n = Ω( n ) and, as a result, z ( T n ) = O ( n ). In order to simplify the analysis, we may project OW MANY ZOMBIES ARE NEEDED TO CATCH THE SURVIVOR ON TOROIDAL GRIDS? 7 (a) stay put (b) move randomly(c) go down (d) go diagonally
Figure 2.
Examples of 4 strategies for the first M moves n (a) (b) (c) (d) (d)’ (e) (h) (h)’1,000 1 0.699986 0.506009 0.124296 0.491322 0.244716 0.281038 1.1108992,000 1 0.694902 0.503004 0.089770 0.501829 0.236672 0.172459 0.9640764,000 1 0.692208 0.501504 0.064498 0.509900 0.229853 0.122445 0.9680138,000 1 0.690806 0.500758 0.046374 0.518481 0.225787 0.089200 0.997286 Table 3. t ( S ) / ( n /
8) and t ( S ) /n / ((d)’ and (h)’) for various strategies p S ( x, y ) onto x ; that is, we may concentrate on q i S ( x ) = M (cid:88) y = − M p i S ( x, y ) . PAWE(cid:32)L PRA(cid:32)LAT (e) one square (f) two squares(g) three squares (h) eight squares
Figure 3.
Examples of 4 more strategies for the first M movesClearly, no recursive formula for getting q i − S ( x ) from q i S ( x ) exists but, by setting up asimple coupling, we can get the following lower bound: q i S ( x ) ≥ w i ( x ), where w M (0) = 3, w M ( −
1) = w M (1) = 1, and for each 2 ≤ i ≤ M and 1 ≤ j ≤ Mw i − ( x i − ) = w i ( x i − ) + 2 w i − ( x i − ± j ) = w i ( x i − ± ( j − w i ( x i − ± j )2 + δ j =1 , where δ j =1 = 1 if j = 1 and 0 otherwise. (Revisiting Figure 1 might be helpful to see thiscoupling.) Alternatively, one can apply a slightly weaker coupling to get: q i S ( x ) ≥ z i ( x ),where z M (0) = 3, z M ( −
1) = z M (1) = 1, and for each 2 ≤ i ≤ M and 1 ≤ j ≤ Mz i − ( x i − ) = z i ( x i − ) + 2 z i − ( x i − ± j ) = z i ( x i − ± ( j − z i ( x i − ± j )2 . OW MANY ZOMBIES ARE NEEDED TO CATCH THE SURVIVOR ON TOROIDAL GRIDS? 9
It is easy to show (by induction) that the difference between two consecutive terms, that is | z i ( x ) − z i ( x + 1) | , is always at most 4. Indeed, the property is clearly satisfied for i = M ;for 2 ≤ i ≤ M and 1 ≤ j ≤ M we get | z i − ( x i − ) − z i − ( x i − ± | = (cid:12)(cid:12)(cid:12)(cid:12) z i ( x i − ) + 2 − z i ( x i − ) + z i ( x i − ± (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z i ( x i − ) − z i ( x i − ± | ≤ , and | z i − ( x i − ± j ) − z i − ( x i − ± ( j + 1)) | = (cid:12)(cid:12)(cid:12)(cid:12) z i ( x i − ± ( j − z i ( x i − ± j )2 − z i ( x i − ± j ) + z i ( x i − ± ( j + 1))2 (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) z i ( x i − ± ( j − − z i ( x i − ± j )2 + z i ( x i − ± j ) − z i ( x i − ± ( j + 1))2 (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z i ( x i − ± ( j − − z i ( x i − ± j ) | | z i ( x i − ± j ) − z i ( x i − ± ( j + 1)) | ≤ . In particular, it follows that the sequence ( z i ( x i )) i = M goes up by 2 if the survivor “pauses”(with respect to this projection; that is, when x i = x i +1 ) and goes down by at most 2 ifhe “moves” (that is, when x i (cid:54) = x i +1 ). Hence, if a strategy S has the property that atsome point t ≤ M/ s more rounds when the survivor pauses than moves, then t ( S ) = Ω( s ). Since, without loss of generality, we may assume that the survivor pauses atfor at least M/ y instead of x if needed), this “almost” implysome nontrivial bound for t n . For a while, we were hoping to be able to show that thereexists an ε > z ( x ) ≥ εn which would imply t n = Θ( n ) and this in turn wouldimply z ( T n ) = O ( n ). Unfortunately, it is not true. Based on simulations, we were ableto identify that strategy (d) might create a problem and serve as a counterexample whichturned out to be the case. We will show in the next section that the conjecture was toooptimistic and, in fact, t n = O ( n / ). But there is still a possibility that t n = Θ( n / ) andso perhaps z ( T n ) = O ( n / ).2.7. t ( S ) = O ( n / ) for strategy (d). Consider strategy (d) discussed above; that is,suppose that the survivor starts at vertex ( (cid:98) M/ (cid:99) , (cid:100) M/ (cid:101) ), where M = (cid:98) n/ (cid:99) , and goes to(0 ,
0) along the diagonal (that is, moves “South” and then immediately “West” in each pairof the two consecutive rounds). We will show the following result:
Theorem 2.3. t ( S ) = O ( n / ) for strategy (d).Proof. In order to estimate t ( S ), we need to estimate p S ( x, y ) for each x, y ∈ Z such that | x | + | y | ≤ M . Due to the symmetry, we may assume that x ≥ y . Suppose that a givenzombie starts the game at vertex ( x, y ). We will partition the part of the grid we investigateinto 4 regions and deal with each of them separately. See Figure 4. Let c ∈ R + be somefixed, large enough, constant.Region R1 : Suppose that x ≥ M/ c √ n . The zombie moves randomly “North” or“West” until her position and a position of the survivor match horizontally or vertically. Ifthey match horizontally, she will never be able to catch the survivor—the distance will bepreserved till the end of the game. On the other hand, if coordinates are matched vertically, Figure 4.
Strategy (d) under microscopethen there is a chance. The probability of this event can be estimated (see discussion forRegion R3 below) but we do not need it here (we may use a trivial upper bound of 1 forthis conditional probability). It is enough to notice that if the vertical match occurs, thenat some point of the game a zombie moved x − (cid:98) M/ (cid:99) more often “West” than she moved“North”. Hence, p S ( x, y ) ≤ P (cid:16) S t ≥ t/ x − (cid:98) M/ (cid:99) ) for some t ≤ M (cid:17) , where X , X , . . . is a sequence of independent random variables, each of them being theBernoulli random variable with parameter p = 1 /
2, and S t = (cid:80) ti =1 X t . It follows that p S ( x, y ) ≤ exp (cid:18) − Ω (cid:18) ( x − (cid:98) M/ (cid:99) ) M (cid:19)(cid:19) . (For example by converting S t into a martingale and using Hoeffding-Azuma inequality.)The contribution to t ( S ) from Region R1 is then at most (cid:88) x ≥ M/ c √ n O ( n ) exp (cid:18) − Ω (cid:18) ( x − (cid:98) M/ (cid:99) ) M (cid:19)(cid:19) = O ( n ) (cid:88) x ≥ c √ n exp (cid:18) − Ω (cid:18) x n (cid:19)(cid:19) = O ( n / ) , provided that c is large enough.Region R2 : Suppose now that y ≤ − M/ − c √ n . The zombie moves “North” or “East”,and it is expected that players match horizontally after r rounds at which point the zombieoccupies vertex (ˆ x, ˆ y ), whereˆ x = M/ x O (1) = M x O (1) r = 2(ˆ x − x ) + O (1) = 2 (cid:18) M − x (cid:19) + O (1)ˆ y = y + r y + (cid:18) M − x (cid:19) + O (1) = M − x y + O (1) . OW MANY ZOMBIES ARE NEEDED TO CATCH THE SURVIVOR ON TOROIDAL GRIDS? 11
The distance from (ˆ x, ˆ y ) to (0 ,
0) is d = ˆ x − ˆ y = x − y + O (1) and so r + d = M/ − y + O (1) ≥ M + c √ n + O (1). It follows that in order for the zombie to have a chance to win, horizontalmatch has to occur much later than expected. Arguing as before, we can estimate theprobability of this event and show that the contribution to t ( S ) from Region R2 is O ( n / ),provided that c is large enough.Region R3 : Suppose now that x ≥ y + c √ n , y ≥ − M/ c √ n , and x ≤ M/ − c √ n .This case seems to be the most interesting. As for the previous region, it is expected thatplayers match horizontally when the zombie occupies vertex (ˆ x, ˆ y ) and the distance betweenplayers is k = (cid:18) M − y (cid:19) − (cid:18) M − x (cid:19) = x − y ≥ c √ n. Arguing as before, we can show that with probability 1 − exp( − Ω( − k /M )) not only thishappens but at that time the distance between players, Y , is at least k/ Y = y ≥ k/
2, we aim now to estimate the probability that the survivoris eaten. Assume that y is even; the odd case can be dealt similarly. Consider a sequence oftwo consecutive rounds. At the beginning of each pair of rounds, before the survivor goes“South”, we measure the absolute difference between the corresponding x -coordinates of theplayers, to get a sequence Z , Z , . . . , Z y/ of random variables. Clearly, Z = 0 and thesurvivor is eaten if and only if Z y/ = 0. Indeed, if Z y/ >
0, then the zombie ends up linedup horizontally before getting close to the survivor and from that point on she will continuekeeping the distance. If Z t >
0, then, Z t +1 = Z t + 1 with probability 1 / Z t − / Z t with probability 1 / . On the other hand, if Z t = 0, then the first move of the zombie is forced (she goes “North”)and so Z t +1 = (cid:40) / / . We can couple this process with a lazy random walk and one can show that P ( Z y/ = 0) =Θ(1 / √ y ) = Θ(1 / √ k ). The contribution to t ( S ) from Region R3 is O ( n ) O ( n ) (cid:88) k = c √ n (cid:18) exp (cid:18) − Ω (cid:18) k n (cid:19)(cid:19) + 1 √ k (cid:19) = O ( n / ) + O ( n ) (cid:90) O ( n ) √ n dxx = O ( n / ) , provided that c is large enough.The number of vertices not included in the three Regions we considered is O ( n / ) and sothe total contribution is O ( n / ) and the proof is finished. (cid:3) Potential improvement on the lower bound: z ( T n ) = √ n/ω , where ω = ω ( n ) isany function going to infinity as n → ∞ . Let ω = ω ( n ) be any function going to infinityas n → ∞ . Suppose that the survivor uses strategy (d) (regardless what zombies are doing).This time, he continues moving “diagonally” forever (of course, unless he is eaten earlier).In the previous section, in order to avoid problems with a boundary effect, we restrictedourselves to sub-graph around (0 , and show that if the survivor is not eaten for long enough, all the zombies will stay behindhim keeping their distances forever. We do not do it here as the improvement would beminor anyway. One can show that, a.a.s. k = √ n/ω zombies cannot catch the survivor on T n and so z ( T n ) > √ n/ω . Indeed, after extending the argument, the probability that nozombie catches the survivor would be (cid:18) − t ( S ) n (cid:19) k = exp (cid:18) − O (cid:18) k √ n (cid:19)(cid:19) = exp( − o (1)) ∼ . This would only be a small improvement comparing to the lower bound in [1] and so we arenot formalize the argument here. Is there a lower bound of n / ε for some ε >
0? Thisremains an open question.
References [1] A. Bonato, D. Mitsche, X. Per´ez-Gimen´ez, and P. Pra(cid:32)lat, A probabilistic version of the game of Zombiesand Survivors on graphs,
Theoretical Computer Science (2016), 2–14.[2] S.L. Fitzpatrick, J. Howell, M.E. Messinger, D.A. Pike, A deterministic version of the game of zombiesand survivors on graphs, Preprint 2015.
Department of Mathematics, Ryerson University, Toronto, ON, Canada
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