We consider the game of Cops and Robber played on the Cartesian product of
two trees. Assuming the players play perfectly, it is shown that if there are
two cops in the game, then the length of the game (known as the 2-capture time
of the graph) is equal to half the diameter of the graph. In particular, the
2-capture time of the m x n grid is proved to be floor ((m+n-2)/2).Comment: 7 page

A. Mehrabian

Alspach

Bonato

Gavenčiak

Hahn

Neufeld

Nowakowski

English

arXiv

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The capture time of grids

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AbstractWe consider the game of cops and robber played on the Cartesian product of two trees. Assuming the players play perfectly, it is shown that if there are two cops in the game, then the length of the game (known as the 2-capture time of the graph) is equal to half the diameter of the graph. In particular, the 2-capture time of the m×n grid is proved to be ⌊m+n2⌋−1

Mehrabian, A.

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The capture time of grids 

Discrete Mathematics 311 (2011) 102–105
Contents lists available at ScienceDirect
Discrete Mathematics
journal homepage: www.elsevier.com/locate/disc
Note
The capture time of grids
A. Mehrabian ∗
Department of Computer Engineering, Sharif University of Technology, Azadi Avenue, Tehran, Iran
a r t i c l e i n f o
Article history:
Received 26 January 2010
Received in revised form 25 September
2010
Accepted 1 October 2010
Available online 23 October 2010
Keywords:
Cops and Robber game
Capture time
Cartesian products
Grids
a b s t r a c t
We consider the game of cops and robber played on the Cartesian product of two trees.
Assuming the players play perfectly, it is shown that if there are two cops in the game,
then the length of the game (known as the 2-capture time of the graph) is equal to half the
diameter of the graph. In particular, the 2-capture time of the m × n grid is proved to be
⌊m+n2 ⌋ − 1.
© 2010 Elsevier B.V. All rights reserved.
1. Introduction
All graphs we consider are undirected, simple, finite, and connected. Cops and robber is a vertex pursuit game played
on a graph, first introduced in [7,8]. We consider two teams of players, a set of k cops C and a single robberR. We think of
the players (cops, robber) as occupying vertices, or being at vertices; a vertex can be occupied by more than one player. The
two teams play in rounds. In round 0, first each of the cops chooses a vertex to start, and then the robber chooses a vertex.
In subsequent rounds, first each cop moves and then the robber moves. Each move takes one time unit and each player can
either stay or go to an adjacent vertex in its move. The cops win and the game ends if the robber is at a vertex occupied by
some cop; otherwise, the robber wins. Each player knows the other players’ locations during the game.
As placing a cop on each vertex guarantees that the copswin,wemay define the cop number of a graphG, denoted by c(G),
as the minimum number of cops that have a winning strategy on G. The graphs with cop number at most k are k-cop-win
graphs. For a survey of results on the cop number and related search parameters for graphs, see [1,4].
In a k-cop-win graph, how many moves does it take for the k cops to win? To be more precise, say that the length of a
game is t if the robber is captured in round t . The length is infinite if the robber can evade capture forever.We say that a play
of the gamewith k cops is optimal if its length is theminimumover all possible games, assuming the robber is trying to evade
capture for as long as possible. There may bemany optimal plays possible (for example, in a path with four vertices, one cop
may start on either vertex of the centre), but the length of an optimal game is an invariant of G. We denote this invariant by
captk(G), and call it the k-capture time of G. The capture time of a graph may be viewed as the temporal counterpart of its
cop number. The concept was first introduced in [2], and is in part motivated by the fact that in real-world networks with
limited resources, not only the number of cops but also the time it takes to capture the robber on the network is of practical
importance.
It is known that for a fixed k, one can compute captk(G) (or decide that G is not k-cop-win) in polynomial time [5].
The capture time of 1-cop-win graphs was studied in [2,3]. In this paper we study the 2-capture time of the Cartesian
∗ Corresponding address: Department of Combinatorics and Optimization, University of Waterloo, 200 University Avenue West, Waterloo, Ontario,
Canada N2L 3G1.
E-mail address: amehrabian@uwaterloo.ca.
0012-365X/$ – see front matter© 2010 Elsevier B.V. All rights reserved.
doi:10.1016/j.disc.2010.10.002
A. Mehrabian / Discrete Mathematics 311 (2011) 102–105 103
products of two trees, which are known to have cop number 2. We prove that if G is the Cartesian product of two trees,
then capt2(G) = ⌊diam(G)/2⌋. In particular, the 2-capture time of the m × n grid is ⌊m+n2 ⌋ − 1. It turns out that the
techniques used here are not strong enough to determine the exact capture time of products of more than two trees (or
higher-dimensional grids). Hence the generalization of the results to more than two trees is left open (see Section 4 for
some bounds).
In the following we will assume that after round 0, in all subsequent rounds, the robber moves before the cops. This
change clearly does not affect the cop number, and neither does it affect the capture time, because of the following
observations: first, if a game is optimal, then the robber always gets caught on the cops move. And second, since the robber
is the player who decides her starting location last, making the first move does not give her any advantage.
Basic definitions are included in Section 2, and the main result is proved in Section 3. In Section 4 some open problems
are raised.
2. Definitions
Let G be a graph.We denote the distance between two vertices u and v in G by d(u, v). The diameter of G is the maximum
distance between any two vertices of G, and is denoted by diam(G). For example, the diameter of a tree is the length of its
longest paths. We will assume that diam(T ) > 0 for all trees T in the following. The set of neighbors of a vertex u is denoted
by N(u), and N[u] is the union N(u) ∪ {u}. If c(G) ≤ k then we call an element (c1, c2, . . . , ck) ∈ (V (G))k a central k-tuple if
k cops have an optimal strategy starting from these vertices.
Theorem 1. For any tree T , we have capt1(T ) = ⌈diam(T )/2⌉.
Proof. Let d = diam(T ), and let t = capt1(T ). Let P with vertices a1, . . . , ad+1 be a longest path in T , and let v be a central
1-tuple. We must have d(v, a1) ≤ t since a robber that starts at a1 and stays there forever should be captured in t rounds,
and similarly d(v, ad+1) ≤ t . This shows d ≤ 2t thus t ≥ ⌈d/2⌉.
On the other hand, let u = a1+⌈d/2⌉; we claim that a cop starting from u can capture the robber in at most ⌈d/2⌉ rounds.
The distance from any vertex to u is at most ⌈d/2⌉. In every round, the copmoves towards the robber, and his distance from
u increases by 1. Hence after at most ⌈d/2⌉ rounds the robber is captured. 
The Cartesian product of two graphsG andH is a graphwith vertex set V (G)×V (H), andwith vertices (u1, v1) and (u2, v2)
adjacent if either u1 = u2 and v1v2 ∈ E(H), or v1 = v2 and u1u2 ∈ E(G). We denote the Cartesian product of G and H by
GH . It is not hard to check that diam(GH) = diam(G)+ diam(H).
3. The capture time of the Cartesian product of trees
The aim of this section is to prove that for trees T1 and T2,
capt2(T1T2) =

diam(T1T2)
2

.
Note that it is known that c(T1T2) = 2 [6].
3.1. Upper bound
Let T be a tree rooted at s. By going up or down we mean moving towards s or away from s, respectively. We say that
vertex v is a descendant of vertex u if u is contained in the unique (s, v)-path (each vertex is a descendant of itself). The
height of a vertex v, written h(v), is defined by
h(v) = max{d(v, l) : l is a leaf and a descendant of v}.
Lemma 1. Let T1, T2 be rooted trees. We denote by (u1, u2), (v1, v2), and (r1, r2) the coordinates of C1,C2 (the first and second
cops) andR (the robber) in T1T2, respectively. Suppose that the following conditions hold:
• ri is a descendant of ui and vi for i = 1, 2,
• d(v1, r1) = 1+ d(u1, r1),
• u2 = v2,
• d(u2, r2) ∈ {d(u1, r1), d(v1, r1)}.
If it isR to move, then the cops can capture her in at most m rounds, where m = ⌊(h(u1)+ h(v1)+ h(u2)+ h(v2))/2⌋.
Proof. We use induction on m. If m = 0, then h(u2) = h(v2) = 0, since u2 = v2. Thus u2 is a leaf of T2. The vertex r2 is a
descendant of u2, so r2 = u2. Hence one of d(u1, r1), d(v1, r1) is also zero, andR is already captured.
104 A. Mehrabian / Discrete Mathematics 311 (2011) 102–105
Suppose thatm > 0. The robber either moves on T1 (up or down), moves on T2 (up or down), or does not move at all. Let
(r ′1, r
′
2) be her position after the move. There are five cases to consider, in all of which we give a move for the cops, which
results in an immediate capture or a similar situation withm decreased by 1:
• The robber moves up on T1: if r1 = u1, then since R was not previously captured, d(u2, r2) > 0 = d(u1, r1). Since
d(u2, r2) ∈ {d(u1, r1), d(v1, r1)}, we must have d(u2, r2) = d(v1, r1). On the other hand, v2 = u2 and d(v1, r1) =
1+ d(u1, r1). Thus d(v2, r2) = 1. AfterR’s move we have r ′1 = v1 since v1 is the parent of u1. Now C2 can immediately
captureR. If r1 ≠ u1, then r ′1 is still a descendant of u1. The cops move down towards r2 on T2.• The robber moves down on T1: the cops move down towards r1 on T1.
• The robber moves up on T2: if r ′2 = u2, then C1 can immediately capture the robber. Otherwise, the cops move down
towards r1 on T1.
• The robber moves down on T2: the cops move down towards r2 on T2.
• The robber does not move: if d(u2, r2) = d(v1, r1), then the cops move down towards r2 on T2. Otherwise, we have
d(u2, r2) = d(u1, r1), and the cops move down towards r1 on T1. 
Remark. The point of d(u2, r2) = d(v2, r2) ∈ {d(u1, r1), d(v1, r1)} is thatR cannot pass through the cops in T1 or T2, since
she will be immediately captured if she wants to go higher than u1 (in T1) or u2 (in T2).
Lemma 2. If T1 and T2 are trees with diameters 2m+ 1 and 2n, respectively, then capt2(T1T2) ≤ m+ n.
Proof. Let P1 and P2 be longest paths in T1 and T2, respectively. Label the vertices of P1 as a1, a2, . . . , a2m+2 and the vertices of
P2 as b1, b2, . . . , b2n+1. The cops start at (u1, u2) = (am+1, bn+1), (v1, v2) = (am+2, bn+1), and suppose thatR goes to (r1, r2)
in her first move. We may assume, by symmetry, that d(r1, u1) < d(r1, v1). Select v1, v2 as roots of T1, T2. Thus h(u1) = m,
h(v1) = m + 1, and h(u2) = h(v2) = n. If d(u2, r2) ∈ {d(u1, r1), d(v1, r1)}, then we can apply Lemma 1. Otherwise, in the
first phase of their strategy, the cops try to satisfy the conditions of Lemma 1. If d(u2, r2) < d(u1, r1), then the cops move
down on T1 towards r1 to reduce d(u1, r1)−d(u2, r2) to zero. If d(u2, r2) > d(v1, r1), then the copsmove down on T2 towards
r2 to reduce d(u2, r2) − d(v1, r1). In both cases they will eventually succeed to satisfy d(u2, r2) ∈ {d(u1, r1), d(v1, r1)}. As
soon as they reach this situation, the first phase finishes. In the second phase, they will use the strategy given in Lemma 1
to captureR. We will now analyze the capture time of this strategy.
Initially h(u1)+ h(v1)+ h(u2)+ h(v2) = 2m+ 2n+ 1. In each round of the first phase, the cops decrease the left side by
2. If the first phase takes t rounds, then by Lemma 1, the second phase will take at most ⌊(2m+ 2n+ 1− 2t) /2⌋ rounds.
Consequently,
capt2(T1T2) ≤ t + ⌊(2m+ 2n+ 1− 2t)/2⌋ = m+ n. 
Corollary 1. When T1 and T2 are trees, capt2(T1T2) ≤ ⌊diam(T1T2)/2⌋.
Proof. Clearly diam(T1T2) = diam(T1) + diam(T2) and the corollary is proved by conditioning on the parity of diam(T1)
and diam(T2), and noticing that adding a leaf to either of T1, T2 does not decrease the capture time of T1T2. 
3.2. Lower bound
A vertex u is a corner if there exists another vertex v with N[u] ⊆ N[v].
Lemma 3. Let G be a 2-cop-win graph, u be a vertex of G contained in an induced cycle C of length 4 with the property that for
every vertex v ∈ V (G), |N(v) ∩ V (C)| ≤ 2 holds. If (c1, c2) is a central 2-tuple, then
d(u, c1)+ d(u, c2) ≤ 2 capt2(G)+ 1.
Proof. Let t = capt2(G). It can be seen that none of the vertices of C is a corner, hence one cop alone cannot captureR even
ifR is restricted to move on C . Let d1 = d(c1, u) and d2 = d(c2, u), and assume, by symmetry, that d1 ≤ d2. SinceR will be
captured in at most t rounds if she starts at u and remains there, we have d1 ≤ t . There are two cases to consider:
d1 = t: Suppose thatR starts at u and does not move in the first t − 1 rounds. Then after t − 1 rounds, C1 should be in a
vertex c1 adjacent to u. Since u is not a corner, it has a neighbor u′ with u′ ∉ N[c1]. Thus ifRmoves to u′ in round
t , then C2 must immediately capture her. Therefore d2 ≤ t + 1. This gives d1 + d2 ≤ 2t + 1.
d1 < t: Since C1 alone cannot capture R even if R is restricted to move on C , C2 should be able to reach C in at most t
rounds. Therefore, d2 ≤ t + 2 and d1 + d2 ≤ t − 1+ t + 2 = 2t + 1. 
Lemma 4. When T1 and T2 are trees, capt2(T1T2) ≥ ⌊diam(T1T2)/2⌋.
Proof. Let G = T1T2, m = diam(T1), and n = diam(T2). Let P1, P2 be longest paths in T1 and T2, respectively. Label the
vertices of P1 as a1, a2, . . . , am+1, and vertices of P2 as b1, b2, . . . , bn+1. Let u = (a1, b1) and v = (am+1, bn+1). The vertex
A. Mehrabian / Discrete Mathematics 311 (2011) 102–105 105
u is contained in the cycle formed by the four vertices (a1, b1), (a1, b2), (a2, b2), and (a2, b1), and satisfies the conditions of
Lemma 3. The vertex v satisfies the conditions of Lemma 3, too. Let (c1, c2) be a central 2-tuple. By triangle inequality and
Lemma 3,
2 diam(G) = 2d(u, v) ≤ d(c1, u)+ d(c1, v)+ d(c2, u)+ d(c2, v) ≤ 4 capt2(G)+ 2.
We are done after dividing by 2 and noticing that capt2(G) is an integer. 
Theorem 2. For every two trees T1, T2 we have
capt2(T1T2) = ⌊diam(T1T2)/2⌋.
Corollary 2. For every two trees T1, T2 we have
capt1(T1)+ capt1(T2)− 1 ≤ capt2(T1T2) ≤ capt1(T1)+ capt1(T2).
Corollary 3. The 2-capture time of the m× n grid is equal to ⌊(m+ n)/2⌋ − 1.
Proof. The m × n grid is the Cartesian product of a path with m vertices and a path with n vertices, and has diameter
m+ n− 2. 
4. Open problems
In the following we write capt(G) instead of captc(G) G. The cop number of the Cartesian product of n trees is known to
be ⌈(n+ 1)/2⌉; see [6].
We have seen that if T1, T2 are trees, then
capt(T1T2) =

1
2
2−
i=1
diam(Ti)

.
It would be interesting to extend this result for more trees. Unfortunately, there is no easy generalization of the efficient
capturing strategy given in Lemmas 1, 2. Nevertheless, the author was able to prove the following bounds.
If T1, T2, T3 are trees, then
1
2
3−
i=1
diam(Ti)

≤ capt(T1T2T3) ≤ 1+
3−
i=1
diam(Ti).
The lower bound has a proof similar to Lemma 4, and the proof for the upper bound uses some ideas from [6] and some new
ideas.
Weaker bounds can be proved when the number of trees becomes larger than three: if T1, T2, . . . , Tn are trees, then
capt

n

i=1
Ti

≤
n−
i=1
(2⌈
i
2 ⌉ − 1) diam(Ti).
Finding better bounds for higher-dimensional grids (which are Cartesian products of more than two paths) would also be
interesting.
References
[1] B. Alspach, Searching and sweeping graphs: a brief survey, Matematiche (Catania) 59 (2006) 5–37.
[2] A. Bonato, P. Golovach, G. Hahn, J. Kratochvíl, The capture time of a graph, Discrete Math. 309 (2009) 5588–5595.
[3] T. Gavenčiak, Cop-win graphs with maximal capture-time, Discrete Math. 310 (2010) 1557–1563.
[4] G. Hahn, Cops, robbers and graphs, Tatra Mt. Math. Publ. 36 (2007) 1–14.
[5] G. Hahn, G. MacGillivray, A note on k-cop, l-robber games on graphs, Discrete Math. 306 (2006) 2492–2497.
[6] S. Neufeld, R. Nowakowski, A game of cops and robbers played on products of graphs, Discrete Math. 186 (1998) 253–268.
[7] R. Nowakowski, P. Winkler, Vertex to vertex pursuit in a graph, Discrete Math. 43 (1983) 235–239.
[8] A. Quilliot, Jeux et pointes fixes sur les graphes, Ph.D. Dissertation, Université de Paris VI, 1978.


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