A Cop and Drunken Robber Game on n-Dimensional Infinite-Grid Graphs

A Cop and Drunken Robber (CDR) game is one variation of a famous combinatorial game, called Cops and Robbers, which has been extensively studied and applied in the area of theoretical and computer science as demonstrated by several conferences and publications. In this paper, for a natural number n, we present two strategies for a single cop to chase a drunken robber on n-dimensional infinite-grid graphs. Both strategies show that if the initial distance between the cop and the drunken robber is s, then the expected capture time is s+o(s).


Introduction
For a graph G = (V, E), we let V be the vertex set and E be the edge set. For vertices v, u ∈ V, u is a neighbor of v (vice versa) in G if u is adjacent to v. The neighbor set N(v) of v in G is the set of all neighbors of v. The distance between vertices u and v in G is the length of a shortest path from u to v. A graph G is infinite if |V| = ∞, namely G has infinitely many vertices. For a natural number n, an n-dimensional infinite-grid graph G has an n-tuple x = (x 1 , . . . , x n ) as a vertex where x i ∈ Z for all 1 ≤ i ≤ n and has an edge xy if ∑ n i=1 |x i − y i | = 1. For an n-dimensional infinite-grid graph G containing a vertex v, let N n s (v) be the set of all vertices in G with distance s from v. An infinite graph G is said to be locally finite if every vertex of G has finite degree; in particular, an infinite graph G is called r-regular if all vertices of G have degree r.
A random walk is a method that monitors every single step of (discrete) random movement of an object in a mathematical space. This process continues to play a significant role in the theory of probability. In this paper, we focus on random walks on graphs. For a graph G, we consider the directed weight graph G * to be the graph representation of the random walk on the graph G. Let V * and E * be the vertex and edge sets of G * , respectively. We obtain V * = V and if {u, v} ∈ E then (u, v) ∈ E * and (v, u) ∈ E * . The weight M uv on the edge (u, v) ∈ E * is the probability to move from u to v in a single step, i.e., ∀u ∈ V * , ∑ v∈N(u) M uv = 1. This process has been studied and applied to various fields of science and engineering such as [1][2][3][4][5].
The game of Cops and Robbers (CR) was originated in 1983 by Quilliot [6] and Nowakowski and Winkler [7] independently and has been applied to many problems in the area of theoretical computer science, such as, information seeking, robot motion planning, or network security, as demonstrated by many conferences and workshops. For example, the graph searching, theory and applications workshop (GRASTA) which has been organized every 1-2 years, has hosted many publications such as [8][9][10][11] that mainly focus on the applications of this game.
In the classical version of CR, the game is played on finite simple connected graphs. When the game starts, all cops and robbers select different vertices to be their initial positions. In each turn, all cops move simultaneously according to the team strategy from their current vertices to the neighbors. Then all robbers move from their current vertices to the neighbors to move away from the cops. This is considered one turn. A cop catches a robber, or a robber is under arrest, as soon as they move onto the same vertex, and this robber will be removed from the game. The cop team will win if they can catch all robbers within a finite number of turns, and the robber team will win if at least one of the robbers has a strategy to avoid cops in infinitely many turns. As we play this game on graphs, a graph G is said to be cop-win if only one cop is sufficient to catch all robbers in G. For example, a path or a tree are cop-win graphs. Otherwise, a graph G is said to be robber-win. The game becomes more challenging when cops chase robbers in these graphs. The cop number c(G) of a graph G is the minimum number of cops required to catch all robbers in G. Clearly, c(G) = 1 if the graph G is cop-win, and c(G) > 1 if the graph G is robber-win. For a natural number k, the k-capture time is the minimum number of turns that k cops need to catch all robbers.
There are many CR versions depending on classes of graphs that the two teams play on or depending on the rules of moves of these two teams. For example, Aigner and Fromme [12] established that if G is a planar graph, then c(G) = 3. This was a classical study which has been generalized in many topological graphs such as [13,14]. For other examples of CR studies, the readers may look at [15][16][17][18][19][20].
A very interesting CR version with different rules for moving was introduced by Kehagias and Prałin [21] when they assumed that the robber is drunk. The robber will perform a (symmetric) random walk to escape from cops while all cops play with an optimal strategy to minimize an expected value of capture time. This game is called Cops and Drunken Robbers, (CDR). Therefore, CDR is a one-player game as there is only the cop team (all cops in one team is considered one player) that needs an optimal strategy to play the game. We set notations and the game rules as follows.

Notations and Game Rule
For a graph G = (V, E) and a non-negative integer t, we let C t ∈ V denote the position of the cop at time t and R t ∈ V denote the position of the drunken robber at time t.
We define the game rules as follows: first, the cop chooses her initial position C 0 , then the robber randomly chooses his initial position R 0 . For each turn t ∈ {1, 2, . . . }, the cop moves along one edge, {C t , C t+1 } ∈ E. Then, so does the robber, {R t , R t+1 } ∈ E. We do not allow the cop or the robber to stay at the same vertex in each turn (i.e., C t+1 = C t and R t+1 = R t ). Let T denote the capture time of this game and be defined by That means that the cop and the robber are at the same vertex. There are 2 situations; the cop arrests the robber (C t = R t−1 ) or the robber drunkenly moves to the cop (R t = C t ). In the first situation, we define R t = R t−1 since the cop already captured the robber to stay at the same vertex. If the cop cannot catch the robber in a finite number of turns, then we let T = ∞. From this assumption, the capture time T is a random variable. Then the expected capture time with given the initial positions of the cop and the robber is defined to be dct C 0 ,R 0 (G) = E(T|C 0 , R 0 ).
In their classical paper of CDR in [21], the robber chooses an initial position, uniformly random. The expected capture time of the graph G is defined by |V| .

Motivation
Interestingly, for all finite graphs, Kehagias and Prałat [21] proved that only one cop will definitely catch one drunken robber in a finite number of turns. Theorem 1 ([21]). For a finite connected graph G, the expected capture time of CDR when we play on G is finite.
However, the result of this theorem is not generally true when the graph is infinite. For example, a cop chases a drunken robber on an infinite tree whose degree of every vertex is infinity. In this graph, the drunken robber moves away from the cop with probability 1, thus making the distance between them the same in every turn. Hence, to reduce the number of robbers escaping routes in each turn, we may play CDR on locally finite graphs and the question that arises is: For a locally finite connected graph G, can a single cop catch one drunken robber by a finite number of moves (turns) when we play on G?
In this paper, we affirmatively answer this problem and further focus on CDR when the game is played on an n-dimensional infinite-grid graph. Because the number of vertices of this graph is infinity, it follows that dct(G) is undefined. Therefore, our expected capture time is dct C 0 ,R 0 (G). For more related results on CDR see [22][23][24].
This paper is organized as follows. After Section 1, we give two strategies for the cop to chase the drunken robber in an n-dimensional infinite-grid graph in Section 2. We further give all main results involving the expected capture time when the cop plays by these two strategies. In Section 3, we show all theorems, lemmas and identities that are used in establishing our main results. All the proofs of our main results are given in Section 4. In Section 5, we make a final conclusion. Some interesting conjectures are pointed out in this section too.

Strategies and Main Results
In this section, we give two strategies for the cop to chase the drunken robber. Furthermore, we give all the main results which are the expected capture time when the cop plays this game by our two strategies. First, to find the expected capture time to chase the drunken robber in any locally finite graph, we may need to make sure that the cop can catch the drunken robber within a finite number of moves. Hence, our first main theorem answers Problem 1 as detailed in the following: Theorem 2. For a locally finite connected graph G, there exists a strategy for the cop to catch the drunken robber by a finite number of moves. Now, we will give the first strategy for the cop to chase the drunken robber. The method is simple. After the cop and the robber landed their initial positions, in each turn, the cop finds a shortest path from her to the robber's position. Then she moves one step along the path. She keeps doing this every turn until she catches the robber. We call this strategy the Greedy Method. The concept of expected hitting time of Markov Chain is applied to calculate dct C 0 ,R 0 (G). Theorem 3. Let G be an n-dimensional infinite-grid graph and let the initial distance between the cop and the drunken robber be s. If the cop chases the drunken robber with the Greedy Method, then the expected capture time is if s is odd, By Theorem 3, when n = 1, the summation term is 0 which means that the expected capture time is roughly s. We further obtain the following two corollaries when n = 2 and n = 3. The proof when n = 2 is provided in Section 4.

Corollary 1.
Let G be a 2-dimensional infinite-grid graph and let the initial distance between the cop and the drunken robber be s. If the cop chases the drunken robber with the Greedy Method, then the expected capture time is Similarly, we also have the following corollary.

Corollary 2.
Let G be a 3-dimensional infinite-grid graph and let the initial distance between the cop and the drunken robber be s. If the cop chases the drunken robber with the Greedy Method, then the expected capture time is The last corollary shows that Theorem 3 yields in fact that the capture time is s + o(s), the proof of which is provided in Section 4.

Corollary 3.
Let G be an n-dimensional infinite-grid graph and let the initial distance between the cop and the drunken robber be s. If the cop chases the drunken robber with the Greedy Method, then the expected capture time is s + o(s).
In the second strategy, we apply the method from Komarov and Winkler [23] together with the Varopoulous-Carne bound for the infinite regular graph version from Tao [25] to find the expected capture time. The method consists of 4 stages. The first two stages are in the so-called Gross Progress part. To chase the robber in this part, at Stage i for i ∈ {1, 2}, the cop moves directly to the position of the robber when Stage i begins. Clearly, the cop spends s moves at Stage 1 and the number of moves for the cop at Stage 2 will be established in this section. The last two stages are in the Fine Progress part. At Stage 3, we let k be the first turn of the cop at this stage. In Turn i when i ∈ {k, k + 4, k + 8, k + 12, . . . }, the cop finds a shortest path, P i say, starting from her to the drunken robber and then, moves along P i at Turns i, i + 1, i + 2 and i + 3. She keeps doing this until the distance between her and the drunken robber at the end of Turn k + 4l is at most 3. Finally, she chases him directly in Stage 4. We call the second strategy the Gross and Fine Method. We obtain that: Theorem 4. Let G be an n-dimensional infinite-grid graph and let the initial distance between the cop and the drunken robber be s. If the cop chases the drunken robber with the Gross and Fine Method, then the expected capture time is Because n is fixed, it can be proved that

Known Results
In this section, we provide all related theorems and notations that are used in establishing our results. Let Ω be the countable state space. Suppose (X n ) n≥0 is a Markov Chain with the transition matrix P whose entries are p i,j , the probabilities that change from State i to State j for i, j ∈ N ∪ {0}. We apply the following definitions and theorems from Norris [26].

Definition 1 ([26]
). Suppose A ⊂ Ω, the hitting time of a subset A is the first time that (X n ) n≥0 reaches A when we start from the state ω. The hitting time is the random variable H A : Ω → N ∪ {∞} which is defined by Definition 2 ([26]). For a subset A ⊂ Ω, the expected hitting time of ω is the expected time of (X n ) n≥0 taken to reach A for the first time when we start from the state ω, denoted by k A ω , which is defined by

Theorem 5 ([26]
). The expected hitting time is the minimal non-negative solution to the system of linear equations Definition 3. Let A ⊂ Ω, A is called the set of all absorbing states when We call the expected hitting time the expected absorbing time if A ⊂ Ω is the set of all absorbing states. If |A| = 1, then we write k A ω as k ω and H A (ω) as H(ω).
For a result in random walks in graphs, we use the Varopoulous-Carne bound for infinite regular graphs version from Tao [25]. Theorem 6 ( [25]). Let G be an infinite regular graph containing a vertex y. For an integer t ≥ 1, let x, x 1 , x 2 , . . . , x t be a t step random walk on G. Then the probability that x t = y, the random walk terminates at the vertex y, is at most 2 exp(−d(x, y) 2 /2t).
We finish this section with useful identities. The first one is called the Pascal Identity, which can be found in Mazur [27]. For natural numbers n and d, we have The next one is an extension of (1) which is called the Hockey Stick Identity. This identity can be found in Jones [28] as follows The following identity can be found in Gould [29] at Equation (5.17). For arbitrary real numbers x and y, we obtain that The last identity is called Jensen's inequality of concave function, which was established by Jensen in 1906 and can be found in Durrett [30]. For any concave function f , we obtain that

Proofs
Throughout this section, when no ambiguity occurs, we call the drunken robber simply the robber.

Proof of Theorem 2
Recall that each vertex in G is represented by an n-tuple (x 1 , x 2 , . . . , x n ). So C t = (c t 1 , c t 2 , . . . , c t n ) and R t = (r t 1 , r t 2 , . . . , r t n ) are the positions of the cop and the robber at time t, respectively. Let d(C t , R t * ) be the distance between the cop at time t and the robber at time t * ∈ {t, t − 1} defined by We need the following observation in the proofs of Theorems 2 and 3.

Observation 1.
For d(C t , R t ) ≥ 2, the distance between the cop and the robber on the next move will be either the same or 2 steps closer.
Proof. Suppose the distance between the cop and the robber is s i.e., d(C t , R t ) = s. Then, after the cop moves, the distance will be d(C t+1 , R t ) = s − 1. Since the robber moves randomly, it follows that the distance will be either d( Now, we are ready to establish the proof of Theorem 2. Proof of Theorem 2. Let G be a locally finite graph with maximum degree r. It suffices to prove that dct C 0 ,R 0 (G) < ∞. Suppose the initial positions of the cop and the robber are C 0 , R 0 , respectively, and the initial distance between the cop and the robber is s. By Observation 1, the probability that the robber uses the shortest path back to the cop is at least ( 1 r ) s 2 . Regardless of the current positions of the cop and the robber at time t, the probability that the cop catches the robber after at most s 2 rounds is at least = ( 1 r ) s 2 . Let the initial positions of the cop and the robber be the random variable Y. Then we obtain the expected capture time to be Since t ≥ t s s, we obtain From the floor function · , we obtain t s to be an integer with t s = i for every is ≤ t < (i + 1)s. Therefore, Since the corresponding events for times t + is is mutually independent for all i ∈ N, it follows that P(T > is|Y = y) ≤ P(T > s|Y = y) i . Now, we can see that dct C 0 ,R 0 (G) is finite by

Proof of Theorem 3: Greedy Method and Expected Hitting Time
Although our first strategy, the Greedy Method, seems simple by letting the cop chase the robber directly after every turn, interestingly, this simple strategy is an effective move, as it was shown in Section 2 that the capture time by this strategy is just s + o(s) which is the same as that of the Gross and Fine Method, the second strategy. Recall that d(C t , R t * ) is the distance between the cop at time t and the robber at time t * ∈ {t, t − 1} which is defined by Then the capture time can be defined by From the definition of capture time, we let X t be the random variable representing the set of all possible distances between the cop and the robber. X t is defined by X t = d(C t , R t ).
All possible values of X t , the state space, are every non-negative integer. We obtain that the state 0 becomes the absorbing state. Hence, when d(C 0 , R 0 ) = s.

Observation 2.
For an n-dimensional infinite-grid graph, let d(C 0 , R 0 ) = s be the initial distance between the cop and the robber. To end the game, the cop captures the robber by her last move when s is odd and the robber drunkenly moves to the cop with probability 1/2n when s is even.
Proof. If s is odd, before the game ends, the distance will be 1 from Observation 1, in the next turn the cop will guarantee to catch the robber. If s is even, the distance between the cop and the robber is 2 from Observation 1 before the game ends. In the last turn, the cop will move to a vertex adjacent to the robber. Since the robber performs a symmetric random walk with deg(R t ) = 2n, the probability that the robber will move to the cop is 1/2n.
The following theorem is needed to calculate probabilities in the transition matrix. Recall that N n s (v) is the set of all vertices at distance s from v.

Theorem 7.
Let G be an n-dimensional infinite-grid graph containing a vertex v. Then the number of vertices at distance s from v is Proof. We may let w be the vertex at distance s from v. Let the integer d, 1 ≤ d ≤ n, be the number of dimensions that |v k − w k | is nonzero. There are ( n d ) ways to choose d. For each nonzero |v k − w k |, there are 2 possibilities of v k − w k which are ±|v k − w k |. Let j 1 , j 2 , . . . , j d be dimensions that and (6). Therefore, the number of vertices at distance s from v is Let P be the transition matrix defined by [p i,j ] i,j∈N∪{0} such that each entry p i,j is defined to be the probability of changing the distance between the cop and the robber from i to j. We can see that if the distance between the cop and the robber is zero that means the cop has already caught the robber. Therefore, this continues forever. We say that the distance 0 is an absorbing state, i.e., p 0,0 = 1 and p 0,j = 0 for all j ∈ N. If the distance s is 1, then the cop will catch the robber in the next move, i.e., p 1,0 = 1 and p 1,j = 0 for all j ∈ N. From Observation 1, the probability p i,j is nonzero if i = j and j = i − 2. For other integers j, the probability is zero. When s = 2, the cop moves to the robber first and the distance between them is just 1. From Observation 2, the probability that the robber drunkenly moves to the cop is 1/2n, i.e., p 2,0 = 1/2n, p 2,2 = (2n − 1)/2n. We obtain the transition matrix P as follows: By the Law of Total Probability for a conditional probability, let B d be the set of states which C t = (c t 1 , . . . , c t n ), R t = (r t 1 , . . . , r t n ) has d terms of |c t 1 − r t 1 |, . . . , |c t n − r t n | that are nonzero. We see that the set {B d } d∈N partitions sample space into n parts. Thus, . Therefore, the probability p s,s when s > 2 will be Now, we are ready to prove Theorem 3.

Proof of Theorem 3.
For the case when s = 1, it is trivial since the cop moves first. Therefore, she will capture the robber in the first turn. For the case when s = 2, we need to wait for the robber to drunkenly move to the cop, which takes 2n turns on average. For the case when s ≥ 3, we obtain by Theorem 5 that k s = 1 + p s,s k s + p s,s−2 k s−2 which can be solved as After continuing this procedure repeatedly, we obtain k s = 1 p s,s−2 + 1 p s−2,s−4 + · · · + k s * where s * = s(mod2). By (7) and k 0 = 0, k 1 = 1 we obtain if s is odd, if s is even.
Next, we consider the case when s is odd. We may use (1) to simplify each term of the summation as follows:

by shifting index d at the denominator)
The case when s is even can be simplified similarly. Hence, the expected hitting time can be deduced to if s is odd, if s is even.
Since we have by (5) that dct C 0 ,R 0 (G) = k s when d(C 0 , R 0 ) = s, the proof is complete.

Proof of Corollary 1
Proof of Corollary 1. We only need to show when s is odd as the other case is the same. Let n = 2; we obtain From the property of Euler-Mascheroni constant, we obtain So k s ≤ s − 2 + 1 2 (γ + ln( s−1 2 )). The proof is complete.
As we mentioned earlier that Corollary 2 can be proved similarly, we may omit the proof and present one for Corollary 3 in the following subsection.

Proof of Corollary 3
Proof of Corollary 3. From Theorem 3, it suffices to show that the summation term is o(s). We will only show when s is even since the case when s is odd can be proved similarly.
For a fixed j, we have by the inequality ( n It remains to show that lim s→∞ f (s) s = 0.

Proof of Theorem 4: Gross and Fine Method
In this section, we find the expected capture time when the cop chases the robber by the Gross and Fine Method introduced in Section 2. First, we may establish the upper bound of the number of vertices at a given length of distance. Observation 3. Let G be an n-dimensional infinite-grid graph containing a vertex v. Then, for positive integers q and s such that q < s, the number of vertices at distances from q to s of v, In particular, | ∪ s i=q N n i (v)| < 2e(s+n) n n .

Proof.
We obtain by Theorem 7 that When q ≥ d, we obtain that In both cases, by (2), we obtain that Thus, by (8), we obtain . By Identity (3) with x = n, y = s and k = d, we obtain that as ( n k ) < (ne/k) k and this completes the proof. Throughout this section, we may let D i be the distance between the cop and the robber when Stage i terminates and T i be the number of turns that the cop spends until Stage i terminates for 1 ≤ i ≤ 4. By the rule of Stage 1, we obtain that E(T 1 ) = s because the cop moves directly to the robber's initial position. We establish the following lemma to find E(D 1 ). Lemma 1. Let x 0 , x 1 , . . . , x t be a t-step random walk on an n-dimensional infinite-grid graph.
Proof. For a vertex x and a t-step random walk x 0 , x 1 , . . . , x t starting at x 0 and terminating at x t , we may let p t (x 0 , x) be the probability that x t = x. Additionally, we may select c = 2 ln 2t 2e(t+n) n n . Thus, By Observation 3 and Theorem 6, we obtain that Clearly, Hence, Let R s be the position of the robber after he moves randomly s steps. The first stage will end when the cop reaches R 0 . Hence, E(D 1 ) = E(d(R 0 , R s )). By Lemma 1, E(D 1 ) ≤ 1 + 2s ln 2s 2e(s+n) n n . By the strategy of Stage 2, the cop moves approximately the distance between her and the robber at the end of Stage 1. That is E(T 2 ) = E(D 1 ) ≤ 1 + 2s ln 2s 2e(s+n) n n . Now we are finding E(D 2 ). Recall that the robber stopped at R s from Stage 1. By (4) and Lemma 1, we obtain that 2e(s + n) n n ≤ 1 + 1 + 2s ln 2s 2e(s + n) n n 2 ln 2s 2e(s + n) n n = 1 + 2 ln 2s 2e(s + n) n n 1 + 2s ln 2s 2e(s + n) n n ≤ 1 + 2(ln 2s 2e(s + n) n The following observation shows that for any 4-step random walk x 0 , . . . , x 4 , we have d(x 0 , x 4 ) < 4 with probability 40n 2 −74n+41 Hereafter, we are in the Fine Progress. At the beginning of Stage 3, we let k be the first turn of the cop at this stage. That means the cop is at C k . As we mentioned earlier, for each Turn i where i ∈ {k, k + 4, k + 8, k + 12, . . . }, the cop finds a shortest path, P i say, from her to the robber and then she will move on P i for Turns i, i + 1, i + 2 and i + 3. She keeps doing this until the distance between her and the robber at the end of Turn k + 4l is at most 3. Clearly the robber walks randomly 4 steps every time that the cop starts finding a new path P i . Thus, the distance between the cop and the robber decreases in each re-targeting with probability 40n 2 −74n+41 8n 3 by Observation 4. By the above discussion, for j = 1 + i−k 4 , we let W j be a {0, 1}-random variable so that W j = 1 when the distance between the cop and the robber decreases during Turns i to i + 3.
Hence, P(W j = 1) = 40n 2 −74n+41 8n 3 for all j ∈ N. Furthermore, we let S j = W 1 + · · · + W j and let {W l } l∈N be the random process whose terminating time is τ where S τ = D 2 − 3. Specifically, the process will be finished when the distance between them has decreased D 2 − 3 which implies that two of them are at most 3 steps apart after the robber has moved. Then, we may apply Wald's identity [31] to show that . As there are 4 turns for each W j , we have This is the end of Stage 3. At Stage 4, the cop starts chasing the robber directly while the distance between them is at most 3. The first move of the cop at this stage is one step forward to the robber. Once the robber moves backward to the cop, he will be caught by the next move of the cop. Thus, we may let P(c ← r) be the probability that the robber moves backward one step to the cop and P(cr →) be the probability that the robber moves away from the cop.
This competes the proof of Theorem 4.

Conclusions and Discussion
To sum up, we extend the study on CDR from Kehagias and Prałat [21] by playing this game on infinite graphs. Although Kehagias and Prałat proved in Theorem 1 that only one cop can catch the drunken robber when we play on finite connected graphs, we discover that this is not true for infinite connected graphs. An infinite tree whose degree of every vertex is infinity is a counterexample. Thus, we prove in Theorem 2 that there exists a strategy for the cop to catch the drunken robber by a finite number of turns when the game is played on a locally finite connected graph. In particular, when the graph is an ndimensional infinite-grid, we present two strategies which are the Greedy Method and the Gross and Fine Method for the cop to chase the drunken robber. We prove in Theorems 3 and 4 that if d(C 0 , R 0 ) = s (the initial distance between the cop and the drunken robber is s), then the expected capture time from both methods is s + o(s).
We believe that in any other strategy, the cop will catch the drunken robber with at least s + o(s) moves. Thus, we conjecture that: Conjecture 1. In any strategy, the expected capture time in n-dimensional infinite-grid graphs with d(C 0 , R 0 ) = s is at least s + o(s).
Next, we may let G ∞ r be the class of infinite connected r-regular (un-weighted) graphs. By Theorems 3 and 4, we can see that there exists an n-dimensional infinite-grid graph in the class G ∞ 2n whose expected capture time is at most s + o(s) when the cop plays with an optimal strategy. Let T r be the class of infinite trees whose all vertices have degree r. Clearly T r ⊆ G ∞ r . We have the expected capture time of T r as follows: Theorem 8. Let T r ∈ T r . If d(C 0 , R 0 ) = s, then there exists a strategy for the cop whose expected capture time is dct C 0 ,R 0 (T r ) = if s is even.
Proof. We may prove this theorem when the cop plays with the Greedy Method. Then the expected capture time is calculated by expected hitting time in Theorem 3. For the r-regular tree, the movement of the robber is equivalent to the un-symmetric random walk with the probability as follows: for i ≥ 2. Further p 0,0 is an absorbing state and p 1,0 = 1. By Theorem 5, we obtain k s = 1 + p s,s k s + p s,s−2 k s−2 = r + k s−2 Using recurrence relation with k 1 = 1 and k 2 = r, we obtain if s is even.
Thus, by Theorem 8, when the initial distance between the cop and the drunken robber is s, there exist infinite r-regular graphs T r which the expected capture time in T r is rs 2 + o(s). We also believe that the expected capture time between the cop and the drunken robber on T r is maximum among all graphs in G ∞ r when the cop plays with an optimal strategy. Finally, as we mentioned in the proof of Theorem 2 that the cop needs at least s/2 turns to catch the drunken robber, we conjecture that Conjecture 2. For all r ∈ N and c ∈ [1, r], there exists a graph in G ∞ r with d(C 0 , R 0 ) = s such that the expected capture time is cs 2 + o(s) when the cop plays with an optimal strategy (the strategy that the cop tries her best to minimize the expected capture time).