The game of Cops and Robber on circulant graphs

Discrete Applied Mathematics 225 (2017) 64–73

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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam

The game of Cops and Robber on circulant graphs Shannon L. Fitzpatrick ∗ , John Paul Larkin University of Prince Edward Island, Canada

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Article history: Received 26 February 2016 Received in revised form 22 February 2017 Accepted 8 March 2017 Available online 15 April 2017 Keywords: Circulant graph Cops and Robber Pursuit–evasion game

abstract We examine the game of Cops and Robber on circulant graphs, and determine the copnumbers of all circulant graphs of degree at most four. We then look at wreath products, and show how they can be used to determine the copnumbers for additional classes of circulant graphs. Finally, we show how expressing a circulant graph as a wreath product relates to dismantling that graph via corners and open corners. © 2017 Elsevier B.V. All rights reserved.

1. Introduction The game of ‘Cop and Robber’ was introduced by Nowakowski and Winkler [16] and, independently, by Quilliot [15]. The game rules are as follows: given a finite, connected, reflexive graph, G, the Cop chooses a vertex of G, then the Robber chooses a vertex. Afterwards, they move alternately — each can move to an adjacent vertex or stay at his current location. If a player stays at his current location, we refer to his move as a pass. (Note that this is equivalent to the player moving along a loop.) We refer to the pair of moves by the Cop and Robber as a round in the game, with the choice of initial positions being round zero. If the Cop ever occupies the same vertex as the Robber, then we say the Robber is captured, and the game ends. We note that the game is played with perfect information; the Cop and Robber always know the other’s location. Graphs in which the Cop can always capture the Robber in a finite number of rounds are called cop-win graphs. If the Robber can avoid capture indefinitely, then the graph is robber-win. Cop-win graphs have been completely characterized [15,16] and can be recognized via a polynomial time decomposition algorithm. In [2], Aigner and Fromme posed the question: given a graph G, determine the least number of Cops required to capture a Robber on G. This least number is referred to as the copnumber of G, denoted c (G). In this version, there is a set of Cops versus a single Robber. The game begins with each Cop choosing a vertex and then the Robber choosing a vertex. They alternate moves, but in this case when it is the Cops’ turn some subset of the Cops (possibly empty) each moves to an adjacent vertex. The Cops win if at least one Cop occupies the same vertex as the Robber. We say that a graph is k-cop-win if its copnumber is at most k. The game of Cops and Robber has received much attention in recent years, due in part to its applications to network security, artificial intelligence, and robotics. Recent advances of particular note include verification by Kinnersley [11] that the problem of determining √ if k cops can capture a robber on a graph is EXPTIME-complete, and advances on Meyniel’s conjecture that c (G) = O( |V (G)|) [13,4]. For a survey of the game of Cops and Robber, the reader is directed to [5]. Cayley graphs on Zn are referred to as circulant graphs. Due to properties such as symmetry, scalability, and small average node distance, circulant graphs serve as good models for local area networks and parallel computer architectures [19]. (For

∗

Corresponding author. E-mail address: [email protected] (S.L. Fitzpatrick).

http://dx.doi.org/10.1016/j.dam.2017.03.004 0166-218X/© 2017 Elsevier B.V. All rights reserved.

S.L. Fitzpatrick, J.P. Larkin / Discrete Applied Mathematics 225 (2017) 64–73

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a survey of circulant graph see [9].) While domination number and related parameters have been extensively studied on circulant graphs ([12,18,20], for example), pursuit–evasion type games have had little consideration for this specific class of graphs. One exception is [10] in which Frankl considered the game of Cops and Robber on Cayley graphs of abelian groups. He proved that for any such a graph, G, c (G) ≤ ⌈(d + 1)/2⌉, where d is the degree of each vertex in the graph. In this paper, we determine the copnumbers for Cayley graphs on the group Zn with degree at most four. We then go on to find the copnumbers of additional Cayley graphs on Zn that can be obtained via wreath products. The wreath product, also known as the lexicographic product is of interest since the wreath product of any two circulant graphs is also a circulant graph [6]. The game of cops and robber has been previously studied on graph products, beginning with Neufeld and Nowakowski [14], and recently with Schröder [17], who determined the copnumbers of lexicographic products of graphs. 2. Definitions and preliminary results The circulant graph on n vertices with distance set S has vertex set Zn and edge set {xy|x − y ∈ S }, where S ⊆ Zn and x ∈ S implies −x ∈ S, with addition done modulo n. We denote this circulant graph Circ (n, S ). Circulant graphs are regular and vertex transitive, and are a subset of the more general family of Cayley graphs. Since S can be written as S = {s1 , −s1 , s2 , −s2 , . . . , st , −st } where 0 < s1 < s2 < · · · < st ≤ n/2, we also use the notation Circ (n; s1 , s2 , . . . , st ) to represent Circ (n, S ). The notation Circ (n; s1 , s2 , . . . , st ) is used in the majority of the paper, with the notation Circ (n, S ) mainly appearing in the final section regarding wreath products. Note that we will consider these graphs to be reflexive, although the loops are not explicitly included in the definition of circulant graphs. Also note that we limit our discussion to connected circulant graphs. Therefore, we assume gcd(n, s1 , s2 , . . . , st ) = 1, since S must generate Zn . For a graph G, V (G) and E (G) represent the vertex and edge sets, respectively. For distinct vertices u and v , we use u ∼ v to indicate that u and v are adjacent and u ̸∼ v to indicate they are nonadjacent. For v ∈ V (G) we let N (v) and N [v] denote the open and closed neighbourhoods of v , respectively. For any set S ⊆ V (G), the notation G − S denotes the graph obtained by deleting the vertices S from G, along with any edges incident to S. We let ⟨j⟩ denote the subset of Zn generated by j. That is, ⟨j⟩ = {0, j, 2j, . . .}. We let ⟨j⟩ + i represent the coset {i, j + i, 2j + i, . . .} of Zn . The subset {i, i + 1, . . . , i + j} is denoted by the interval notation [i, i + j]. In the game of Cops and Robber, we let Cop1 , . . . , Copk represent the cops. Furthermore, let Ci represent the vertex currently occupied by Copi , and R represent the vertex currently occupied by the Robber. When Cop1 moves from vertex C1 to C1 + x, we write C1 → C1 + x. We denote moves by the other Cops and the Robber similarly. We assume that, even in graphs where the Cops have a winning strategy, the Robber always moves in such a way as to maximize the number of rounds in the game. The main result of this paper is stated in Theorem 2.1, and gives the copnumbers of all circulant graphs of degree at most four. Theorem 2.1. Let G be a connected circulant graph with degree at most four. Then the following are true 1. c (G) ≤ 3 2. c (G) = 1 if and only if G is isomorphic to Circ (2; 1), Circ (3; 1), Circ (4; 1, 2), or Circ (5; 1, 2). 3. c (G) = 2 if and only if G ∼ = Circ (n; m) for any n ≥ 4, or G ∼ = Circ (n; m, k) where n ≥ 6 and n, m and k satisfy at least one of the following: (a) k = 2m (b) k = 3m (c) n = 2k (d) n = 2k + 2m (e) n = 3k, and 4m = 3k or 3m = 2k (f) n = 3m and 4m = 3k. We prove this result in stages. In this section, we prove the first two statements in the list, as well as easily verify that circulant graphs of degree two or three have copnumber at most two. In Section 3, we find the necessary conditions for a 4-regular circulant graph to have copnumber two. In Section 4, we show that these conditions are also sufficient. We begin by considering statement 2. of Theorem 2.1. This result follows immediately from the fact that circulant graphs are regular, and the only regular cop-win graphs are the complete graphs [14]. All of the complete circulant graphs of degree at most four are listed in statement 2. We now verify statement 1. of Theorem 2.1, as well as statement 3.(c), using the following upper bound due to Frankl. Theorem 2.2 ([10]). If G is a d-regular Cayley graph defined on an abelian group, c (G) ≤

 d+1  2

.

We note that Circ (n; m) is 2-regular for n ≥ 3, and Circ (n; m, k) is 3-regular when n = 2k. When n ̸= 2k, then Circ (n; m, k) is 4-regular. Therefore, by Theorem 2.2, and statement 2. of Theorem 2.1, we have the following result.

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Corollary 2.3. Let G be a connected circulant graph 1. If G ∼ = Circ (n; m) and n ≥ 4, then c (G) = 2. 2. If G ∼ = Circ (n; m, k), then c (G) ≤ 3. Furthermore, if n = 2k and n ≥ 6, then c (G) = 2. We have, therefore, determined the copnumbers of each of the circulant graphs of maximum degree at most three. 3. Necessary conditions for c (Circ (n; m, k )) = 2 We now determine the necessary conditions for 4-regular circulant graphs to be 2-cop-win. Since we know which of these graphs are cop-win, we therefore determine the necessary condition for 4-regular circulant graphs to have copnumber two. In Theorem 3.1, we prove a weaker version of the conditions in the third statement of Theorem 2.1. In particular, additional conditions for n = 3k and n = 3m have been excluded. These conditions are addressed later, in Lemmas 3.2 and 3.3. Theorem 3.1. Let G be a 4-regular circulant graph. If c (G) = 2, then G ∼ = Circ (n; m, k) where n, m and k satisfy at least one of the following 1. k = 2m 2. k = 3m 3. n = 2k + 2m 4. n = 3k 5. n = 3m. Proof. Let G = Circ (n; m, k), where gcd(n, m, k) = 1. Assume c (G) = 2. Recall that 1 ≤ m < k ≤ ⌊ 2n ⌋. However, since n = 2k implies that G is 3-regular, it must be the case that n > 2k. Consider the game of Cops and Robber played with two Cops. Since G is 2-cop-win, in any optimal (cop-winning) game one Cop moves onto the Robber in the final round. Therefore, as the Cops and Robber enter the final round, it must be the case that N [R] ⊆ N [C1 ] ∪ N [C2 ], where R ̸∈ {C1 , C2 }. Therefore, there exist vertices x, y and z in V (G) such that x ̸∈ {y, z } and N [x] ⊆ N [y] ∪ N [z ]. We now show that such vertices x, y and z exist only if n, m and k satisfy at least one of the five conditions listed in the theorem. Since G is vertex transitive, we may assume x = 0. It follows that x ∼ y or x ∼ z. Without loss of generality, assume that x ∼ y. Then y ∈ {m, k, n − m, n − k}. However, of the four possible values of y, we only need to consider two cases: y = m or y = k. The other two cases follow by symmetry. We begin by showing that, in each case, either k ∈ {2m, 3m} or n ∈ A where A = {2k + m, 2m + k, 2k + 2m, 3k, 3k + m, 3k − m, 3m, 3m + k}. Case 1: y = m. It follows that N [x] = {0, m, k, n − k, n − m} and N [y] = {0, m, 2m, m + k, n + m − k}. If k ∈ N [y], then k = 2m, since m < k < m + k. If {n − k, n − m} ∩ N [y] ̸= ∅, then n ∈ {2m + k, m + 2k, 3m}, since n > 2m. Therefore, {k, n − k, n − m} ∩ N [y] ̸= ∅ implies k ∈ {2m, 3m} or n ∈ A. Now assume that {k, n − k, n − m} ∩ N [y] = ∅. It follows that {k, n − k, n − m} ⊆ N [z ], and z ∈ (N [k] ∩ N [n − k] ∩ N [n − m]) − {0}. Note that N [k] − {0} = {k − m, k, k + m, 2k}, N [n − k] − {0} = {n − 2k, n − k − m, n − k, n − k + m} and N [n − m] − {0} = {k − m, n − k − m, n − 2m, n − m}. Let Nk , Nn−k and Nn−m represent each of the three sets, respectively. It follows that z is one of the four values listed in Nk . We now examine these four possibilities. Case 1A: z = k − m. It follows that k − m ∈ Nn−k . Since n > 2k, it follows that n = 3k − m and n ∈ A. Case 1B: z = k. It follows that k ∈ Nn−k . Since k < n − k + m, it follows that k ∈ {n − 2k, n − m − k, n − k}. Since n > 2k, this implies either n = 3k or n = 2k + m. Therefore, n ∈ A. Case 1C: z = k + m. It follows that k + m ∈ Nn−m , and either n = 2k + 2m, n = k + 3m or n = k + 2m. Therefore, n ∈ A. Case 1D: z = 2k. It follows that 2k ∈ Nn−m , and either n = 3k + m, n = 2k + 2m or n = 2k + m. Therefore, n ∈ A. Therefore, in Case 1, if the required vertices x, y and z exist, then k ∈ {2m, 3m} or n ∈ A. Case 2: y = k. It follows that N [x] = {0, m, k, n − k, n − m} and N [y] = {0, k − m, k, k + m, 2k}. If m ∈ N [y], then k = 2m, which is a contradiction. If {n − k, n − m} ∩ N [y] ̸= ∅, then either n = 2k, n = 2k + m, n = 3k, or n = k + 2m. Since n > 2k, it follows that n ∈ A. We, therefore, may assume that {m, n − k, n − m} ∩ N [y] = ∅, and {m, n − k, n − m} ⊆ N [z ]. Then z ∈ (N [m] ∩ N [n − k] ∩ N [n − m]) − {0}. Let Nm = N [m] − {0}. Define Nn−k and Nn−m similarly. We now consider the possible values of z based on the fact that z ∈ Nm . Recall that Nm = {m, 2m, m + k, n + m − k}. If z = m, then we have the previous case. So we look at the three other possible values of z. Case 2A: z = 2m. Since z ∈ Nn−k , 2m ∈ {n−2k, n−k−m, n−k, n−k+m}. It follows that n = 2k+2m, n = 3m+k, n = 2m+k or n = k + m. Since n > 2k > k + m, it follows that n ̸= k + m and n ∈ A. Case 2B: z = m + k. Since z ∈ Nn−m , m + k ∈= {k − m, n − m − k, n − 2m, n − m}, and n = 2k + 2m, n = 3m + k or n = 2m + k. Therefore, n ∈ A. Case 2C: z = n − k + m. Since n − k + m ∈ Nn−m , either k = 3m or k = 2m. We have shown that if c (G) = 2, then either k ∈ {2m, 3m} or n ∈ A. However, if n = 2k + m or n = 2m + k, then G ∼ = Circ (n; 1, 2). This follows from the fact that kS = S or mS = S, respectively, where S = {−2, −1, 1, 2}. Similarly, if n = 3k + m, n = 3k − m, or n = 3m + k, then G ∼ = Circ (n; 1, 3). Therefore, if c (G) = 2, then G ∼ = Circ (n; m, k) where k ∈ {2m, 3m} or n ∈ {2k + 2m, 3k, 3m}. 

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We now examine the circulant graph Circ (3k; m, k) in more detail. Lemma 3.2. If connected circulant graph Circ (3k; m, k) is 2-cop-win then either k = 2m, k = 3m, 4m = 3k or 3m = 2k. Proof. Suppose G = Circ (3k; m, k) where gcd(3k, m, k) = 1. Assume that k ̸∈ {2m, 3m}. As in the previous proof, if G is 2-cop-win, then there exist vertices x, y, and z in V (G) such that x ̸∈ {y, z } and N [x] ⊆ N [y]∪ N [z ]. Without loss of generality, let x = 0, and assume y ∈ {m, k}. Case 1: y = m. It follows that N [x] = {0, m, k, 2k, 3k − m} and N [y] = {0, m, 2m, m + k, 2k + m}. We note that the elements of each set are between 0 and n − 1, inclusive, and are written in increasing order. Since k ̸= m and k ̸= 2m, it follows that {k, 2k, 3k − m} ∩ N [y] = ∅, and {k, 2k, 3k − m} ⊆ N [z ]. If z = k, then N [z ] ⊆ [0, 2k], and 3k − m ̸∈ N [z ]. If z = 2k, then 2k + m = 3k − m and k = 2m, which is a contradiction. Therefore, z ∈ [1, k − 1] ∪ [k + 1, 2k − 1] ∪ [2k + 1, 3k − 1]. If z ∈ [1, k − 1], then 2k ̸∼ z since |2k − z | > k. If z ∈ [2k + 1, 3k − 1], then k ̸∼ z since |z − k| > k. Therefore, z ∈ [k + 1, 2k − 1]. However, since |z − k| < k, k ∈ N [z ] implies z − k = m. Similarly, 2k ∈ N [z ] implies 2k − z = m. It follows that k = 2m, which is a contradiction. Therefore, the required vertex, z, does not exist. Case 2: y = k. It follows that N [x] = {0, m, k, 2k, 3k − m} and N [y] = {0, k − m, k, k + m, 2k}. Since m and 3k − m are not in N [y], then {m, 3k − m} ⊆ N [z ]. It follows that z ∈ (N [m] ∩ N [3k − m]) − {0}. Therefore, z ∈ {m, 2m, m + k, 2k + m} ∩ {k − m, 2k − m, 3k − 2m, 3k − m}. Since z = m is covered by the previous case, we may assume z ̸= m. Furthermore, by symmetry, we may assume that z ̸= 3k − m. Then z ∈ {2m, m + k, 2k + m} ∩ {k − m, 2k − m, 3k − 2m}. We consider three possible values of z, based on its membership in {k − m, 2k − m, 3k − 2m}. Case 2A: z = k − m. Since z is also in {2m, m + k, 2k + m} and k − m < k + m, it follows that k − m = 2m. However, this implies k = 3m, which is a contradiction. Case 2B: z = 2k − m. Since z is also in {2m, m + k, 2k + m}, it follows that either 2k − m = 2m or 2k − m = m + k. The latter implies that k = 2m, which is a contradiction. Hence, 2k − m = 2m and 2k = 3m. Case 2C: z = 3k − 2m. It follows that z is also in {2m, m + k, 2k + m}. Since 3k − 2m = 2k + m implies k = 3m, it must be the case that 3k − 2m = 2m or 3k − 2m = m + k. Therefore, 3k = 4m or 2k = 3m. Hence, if k ̸∈ {2m, 3m}, then either 4m = 3k or 3m = 2k.  For our final result in this section, we examine the circulant graph Circ (3m; m, k) in more detail. Lemma 3.3. If connected circulant graph Circ (3m; m, k) is 2-cop-win, then 3m = 2k or 4m = 3k. Proof. Let G be the connected circulant graph Circ (3m; m, k) where n = 3m and 3m ̸= 2k. Therefore, n > 2k. This implies < 2m. As in the proof of Theorem 3.1, if G is 2-cop-win then there are vertices x, y and z, such that x ̸∈ {y, z } and k < 3m 2 N [x] ⊆ N [y] ∪ N [z ]. Without loss of generality, assume x = 0 and y ∈ {m, k}. Case 1: y = m. It follows that N [x] = {0, m, k, 3m − k, 2m} and N [y] = {0, m, 2m, m + k, 4m − k}. Since m < k < 2m and m < 3m − k < 2m, neither k nor 3m − k is in N [y]. Therefore, both elements are in N [z ]. Therefore, z ∈ Nk ∩ N3m−k , where Nk = N [k] − {0} = {k − m, k, k + m, 2k} and N3m−k = N [3m − k] − {0} = {3m − 2k, 2m − k, 3m − k, 4m − k}. Note that the elements of each of these two sets have been written in increasing order. We now consider each possible value of z, based on its membership in N3m−k . Case 1A: z = 3m − 2k. Since 3m − 2k < k, 3m − 2k ∈ Nk implies 3m − 2k = k − m, and 3k = 4m. Case 1B: z = 2m − k. Since k − m < 2m − k < k, 2m − k ̸∈ Nk . Case 1C: z = 3m − k. Since k < 3m − k < k + m, 3m − k ̸∈ Nk . Case 1D: z = 4m − k. Since 3m > 2k, 4m − k > m + 2k − k = m + k. Therefore, 4m − k ∈ Nk implies 4m − k = 2k and 4m = 3k. Therefore, there exists a vertex z such that N [x] ⊆ N [y] ∪ N [z ] only if 4m = 3k. Case 2: y = k. It follows that N [x] = {0, m, k, 3m − k, 2m} and N [y] = {0, k − m, k, k + m, 2k}. Note that the elements of N [x] and N [y] are written in increasing order. Since m < k and 2k < 3m, it follows that k − m < m < k and k < 2m < k + m. Therefore, {m, 2m} ∩ N [y] = ∅, and {m, 2m} ⊆ N [z ]. We may assume that z ̸= m, since this would be covered by the previous case. By symmetry, we can also assume that z ̸= 2m, since 2m = n − m. Hence, z ∈ [1, m − 1] ∪ [m + 1, 2m − 1] ∪ [2m + 1, 3m − 1]. If z ∈ [1, m − 1], then |z − m| < m and z ̸∼ m. If z ∈ [m + 1, 2m − 1], then |z − m| < m and z ̸∼ m. Finally, if z ∈ [2m + 1, 3m − 1], then |z − 2m| < m and z ̸∼ 2m. We, therefore, have a contradiction, and no such vertex z exists. Therefore, assuming n > 2k, there exist vertices x, y, and z such that N [x] ⊆ N [y] ∪ N [z ] and x ̸∈ {y, z } only if 4m = 3k.  We have verified that if G ∼ = Circ (n; m, k) and c (G) = 2, then G satisfies at least one of conditions 3.(a)–(f) in Theorem 2.1.

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4. Sufficient conditions for c (Circ (n; m, k )) = 2 In this section, we verify that Circ (n; m, k) is 2-cop-win for each of the conditions 3.(a)–(f) in Theorem 2.1. In many of the proofs, we adapt a strategy used in [10]. This is first done in Lemma 4.1, where we consider a version of Cops and Robber where the robber is active. We refer to a Robber as active if he cannot pass in any round, and we let c ′ (G) denote the minimum number of cops required to capture an active Robber. Lemma 4.1. Suppose G = Circ (n; m, k) is a connected circulant graph. If either gcd(n, k + m) = 1 or gcd(n, k − m) = 1, then c ′ (G) ≤ 2. Proof. Let G = Circ (n; m, k) where gcd(n, m, k) = 1 and gcd(n, k + m) = 1. Let the Cops, C1 and C2 , both begin on vertex 0 and let the Robber begin on some vertex r. The Cops’ strategy relies on them reacting to the Robber’s move, so the Cops begin by passing in round one. In the table below, the four possible moves available to the Robber are listed in the second column. For each type of move, the Cops’ response in the next round is provided. Type

Robber Move

1 2 3 4

R R R R

→R+m →R+k →R−m →R−k

Cop1 Move C1 C1 C1 C1

→ C1 − k → C1 − m → C1 − m → C1 − k

Cop2 Move C2 C2 C2 C2

→ C2 + m → C2 + k → C2 + k → C2 + m

Since gcd(n, k + m) = 1, there is some −j ∈ Zn such that −j(k + m) = r. Suppose we look at the minimum round of completed play such that the Robber has either made j total moves of type 1 or 2, or n − j total moves of type 3 or 4. Now suppose the Cops move in response, and we examine all of the players’ positions immediately afterwards (midway through the round). Let ti represent the number of moves of type i that have been made by the Robber at this point in the game. Since the Cops began with a pass, each of the Cops has also made ti moves of type i, as defined for each of them. Recall that the Cops started at vertex 0, and the Robber started at vertex r. Therefore, at this point in the game, R = r + t1 m + t2 k − t3 m − t4 k, C1 = −t1 k − t2 m − t3 m − t4 k, and C2 = t1 m + t2 k + t3 k + t4 m. If t1 + t2 = j, then R = r + t1 m + t2 k − t3 m − t4 k

= −j(k + m) + (j − t2 )m + (j − t1 )k − t3 m − t4 k = −t2 m − t1 k − t3 m − t4 k = C1 . If t3 + t4 = n − j, then R = r + t1 m + t2 k − t3 m − t4 k

= −j(k + m) + t1 m + t2 k + (t4 + j)m + (t3 + j)k = t1 m + t2 k + t4 m + t3 k = C2 . Therefore, the Cops have a winning strategy, and c ′ (G) ≤ 2. The case where gcd(n, k − m) = 1 can be proved similarly by simply replacing m with −m in all instances.



In verifying the conditions in Statement 2. of Theorem 2.1, we will also repeatedly make use of the following result due to Ádám [1]. Lemma 4.2 ([1]). For any circulant graph Circ (n, S ) and integer µ such that gcd(n, µ) = 1, Circ (n, S ) ∼ = Circ (n, µS ), where µS = {µs|s ∈ S }. For Circ (n; m, k) such that k = 2m, we can modify the strategy in Lemma 4.1 to obtain the following result. Lemma 4.3. The connected circulant graph Circ (n; m, 2m) is 2-cop-win. Proof. Let G be the connected circulant graph Circ (n; m, 2m). Since gcd(n, m) = 1, it follows from Lemma 4.2 that G∼ = Circ (n; 1, 2). We now adapt the methods used in Lemma 4.1 to obtain a winning strategy. Specifically, we add a type 0 move, which accounts for the Robber’s option to pass. We provide the following table, which gives the Cops’ move in response to every possible move by the Robber. As in Lemma 4.1, the Cops begin by choosing vertex 0, but in this instance, we write the Robber’s initial choice of vertex as −r. The Cops pass in the first round.

S.L. Fitzpatrick, J.P. Larkin / Discrete Applied Mathematics 225 (2017) 64–73

Type

Robber’s Move

0 1 2 3 4

R R R R R

→R →R−1 →R+2 →R+1 →R−2

Cop1 ’s Move C1 C1 C1 C1 C1

→ C1 − 1 → C1 − 2 → C1 + 1 → C1 + 1 → C1 − 2

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Cop2 ’s Move C2 C2 C2 C2 C2

→ C2 → C2 − 1 → C2 + 2 → C2 + 2 → C2 − 1

Suppose we look at the minimum round of completed play such that either r total moves of type 0, 1 or 2 have been made by the Robber, or n − r total moves of type 3 or 4 have been made by the Robber. Now suppose the Cops move as indicated in response to each of the Robber’s moves. Define ti as in the proof of Lemma 4.1. It follows that, at this point in the game, R = r − t1 + 2t2 + t3 − 2t4 , C1 = −t0 − 2t1 + t2 + t3 − 2t4 and C2 = −t1 + 2t2 + 2t3 − t4 . If t0 + t1 + t2 = r, then R = −r − t1 + 2t2 + t3 − 2t4

= −(t0 + t1 + t2 ) − t1 + 2t2 + t3 − 2t4 = −t0 − 2t1 + t2 + t3 − 2t4 = C1 . If t3 + t4 = n − r, then R = −r − t1 + 2t2 + t3 − 2t4

= (t3 + t4 ) − t1 + 2t2 + t3 − 2t4 = −t1 + 2t2 + 2t3 − t4 = C2 . Hence, the Cops have a winning strategy, and the connected circulant graph Circ (n; m, 2m) is 2-cop-win.



To show that Circ (n; m, 3m) is 2-cop-win, we describe a strategy for the Cops that has two parts. First, the Cops move so that, after a finite number of moves, |C − R| = 2 for some C ∈ {C1 , C2 }. Once this is accomplished, that same Cop mimics the Robber’s movements. That is, if the Robber moves R → R + j, then the Cop immediately responds with C → C + j. This will restrict the Robber’s movements, and allow the second Cop to capture him. Lemma 4.4. The connected circulant graph Circ (n; m, 3m) is 2-cop-win. Proof. Suppose G = Circ (n; m, 3m), with gcd(n, m, 3m) = 1. It follows that G ∼ = Circ (n; 1, 3). We begin by proving the following claim, which, in turn, describes the first phase of the Cops’ strategy. Claim. After a finite number of rounds, the Cops can move so that immediately following their move either C = R or |C − R| = 2, for some C ∈ {C1 , C2 }. Proof of Claim. We begin the game with Cop1 on vertex 1, Cop2 on vertex n − 1, and the Robber on vertex r. If r = 0, then the Robber can be captured on the Cops’ next move. We may, therefore, assume 1 ≤ r ≤ n − 1. Note that every move is of the form x → x + j where j ∈ {3, 1, −1, −3}. We now proceed by induction on the round number. Suppose at the beginning of round t, C1 = t , C2 = n − t and t ≤ R ≤ n − t. If R = t or R = n − t, then the Robber has been captured. If R = t + 1 or R = n − t − 1, then one of the Cops can move to capture the Robber. If R = t + 2 or R = n − t − 2, then |C1 − R| = 2 or |C2 − R| = 2, respectively. The Cops now pass, and the statement of the Claim is satisfied. If R = t + 3 or R = n − t − 3, then either moving C1 → C1 + 1 or C2 → C2 − 1 would put one of the Cops into the position required for the Claim. We, therefore, may assume that t + 4 ≤ R ≤ n − t − 4. Suppose the Cops move to t + 1 and n − t − 1, respectively, and the active Robber moves to vertex s. Since the Robber was previously on a vertex in [t + 4, n − t − 4], it follows that s ∈ [t + 1, n − t − 1]. Therefore, at the beginning of Round t + 1, C1 = t + 1, C2 = n − t − 1 and t + 1 ≤ R ≤ n − t − 1. Therefore, by induction, we see that after round t the claim is either satisfied, or R ∈ [t + 1, n − t − 1]. However, after a finite number of rounds [t + 1, n − t − 1] = ∅. Therefore, at some point in the game, immediately after the Cops’ move either C = R or |C − R| = 2 for some C ∈ {C1 , C2 }. Assume that |C − R| = 2 for some C ∈ {C1 , C2 }, and it is the Robber’s move. Without loss of generality, let C1 = n − 2 and R = 0. If the Robber moves R → R + 1, R → R − 1 or R → R − 3, Cop1 would capture the Robber in the next round. Therefore, the Robber must move R → R + 3 or pass. In the next round, Cop1 mimics the Robber’s move. Therefore, to avoid capture by Cop1 , the Robber must again move R → R + 3 or pass. We may, therefore, assume that for the remainder of the game, the Robber continues to move R → R + 3 or pass, while Cop1 mimics his move. The Robber is now confined to the subgraph, H, where V (H ) = ⟨3⟩ and E (H ) = {ij|j = i + 3}. It follows that H is a cycle. Cop2 now moves onto a vertex of H, if she is not already on such a vertex. Once Cop2 is on H, she only makes moves of the type C2 → C2 − 3 for the remainder of the game. We, therefore, have R and C2 playing on a cycle, H, where at each round Cop2 is moving in one direction around the cycle, and the Robber is either passing or moving the other direction around the cycle. Therefore, after a finite number of moves, Cop2 captures the Robber. 

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The next circulant graphs we verify to be 2-cop-win are Circ (3k; m, k) where 4m = 3k or 3m = 2k, and Circ (3m; m, k) where 4m = 3k. This is done over two lemmas. For each of Circ (3k; m, k) where 3m = 2k, and Circ (3m; m, k) where 4m = 3k, we see that the class of graphs actually consists of a single graph isomorphic to Circ (9; 1, 3). For the remaining class, Circ (3k; m, k) where 4m = 3k, we use a modified version of the strategy described in Lemma 4.1 that allows Cop1 to control the movements of the Robber. As in the previous proof, this is followed by Cop1 mimicking the movement of the Robber, which allows Cop2 to capture the Robber. Lemma 4.5. If 4m = 3k or 3m = 2k, then the connected circulant graph Circ (3k; m, k) is 2-cop-win. Proof. Suppose G = Circ (3k; m, k), where 4m = 3k or 3m = 2k. If 3m = 2k, then 3|k and G ∼ = Circ (9j; 2j, 3j) for some integer j. Since G is connected, it follows that j = 1 and G ∼ = Circ (9; 2, 3). By Lemma 4.2, G ∼ = Circ (9; 1, 3). If 4m = 3k, it follows by the similar argument that G ∼ = Circ (12; 3, 4). By Lemma 4.4, Circ (9; 1, 3) is 2-cop-win. It now suffices to show that Circ (12; 3, 4) is 2-cop-win. Playing the game on Circ (12; 3, 4), assume two Cops start on 0, and the Robber starts on r. We now consider a variation the Cops’ strategy in the proof of Lemma 4.1. In this case, the Robber can pass, so we have added a type 0 move to address this. Type

Robber Move

0 1 2 3 4

R R R R R

→R →R+3 →R+4 →R−3 →R−4

Cop 1 Move C1 C1 C1 C1 C1

→ C1 − 3 → C1 − 4 → C1 − 3 → C1 − 3 → C1 − 4

Cop 2 Move C2 C2 C2 C2 C2

→ C2 → C2 + 3 → C2 + 4 → C2 + 4 → C2 + 3

Since gcd(12, 7) = 1, we have j such that r = −7j. We define ti as in Lemma 4.3, and examine the game at the point where either t0 + t1 + t2 = j or t3 + t4 = 12 − j. It follows that R = r + 3t1 + 4t2 − 3t3 − 4t4 , C1 = −3t0 − 4t1 − 3t2 − 3t3 − 4t4 and C2 = 3t1 + 4t2 + 4t3 + 3t4 . Note the value of C1 differs by 4t0 from value of C1 described in the proof of Lemma 4.1, while R and C2 have the same values. Therefore, we either have C1 − R = 4t0 or C2 = R. Assume the former. Since addition is done modulo 12, it follows that either C1 = R or |C1 − R| = 4. Therefore, we can assume the latter, and relabel the graph so that R = 0 and C1 = 4. Assume it is the Robber’s turn to move. (If not the Cops can pass.) The Robber will be captured either immediately, or in the next round, if he passes or moves to either 4 or −4. Therefore, to avoid capture, the Robber must move to either 3 or −3. Cop1 now mimics the Robbers’ move to ensure that he continues to make moves of type 1 or type 3. It follows that the Robber is confined to the cycle induced on {0, 3, 6, 9}. While Cop1 has the Robber confined to this cycle, Cop2 moves to vertex 0. Since the Robber cannot pass without capture by Cop1 , the Robber will eventually be captured by Cop2 .  Lemma 4.6. If 4m = 3k, then the connected circulant graph Circ (3m; m, k) is 2-cop-win. Proof. Suppose G = Circ (3m; m, k) where gcd(3k, m, k) = 1 and 4m = 3k. As in the proof of Lemma 4.5 it is straightforward to show that G ∼ = Circ (9; 3, 4) ∼ = Circ (9, 1, 3). By Lemma 4.4, Circ (3m; m, k) is 2-cop-win.  There is one final class of graphs that we verify to be 2-cop-win — the connected circulant graphs Circ (2k + 2m; m, k). For these graphs, we use a different technique than previously used. For such a graph G, we show that a cycle can be obtained by successively removing open corners. A vertex x is an open corner in G if there is a second vertex y such that N (x) ⊆ N (y). This is to be distinguished from a corner in G which is a vertex x such that N [x] ⊆ N [y] for some other vertex y. We then use the following result, due to Clarke and Nowakowski [8], to prove that G is 2-cop-win. Lemma 4.7 ([8]). Suppose G has an open corner, x. If G − x is 2-cop-win, then G is 2-cop-win. We are now prepared to prove the final result in this section. Lemma 4.8. The connected circulant graph Circ (2m + 2k; m, k) is 2-cop-win. Proof. Let G = Circ (2m + 2k; m, k) where gcd(2m + 2k, m, k) = 1. It follows that gcd(m, k) = 1, and that at least one of m or k is odd. We show that in either case, G ∼ = Circ (2m + 2k; 1, k + m − 1). Suppose m is odd. Then gcd(2m + 2k, m) = 1. Otherwise, there is some prime p > 2 such that p divides m and p divides 2m + 2k. This implies p divides 2k, and since p > 2, it follows that p divides k. This contradicts the fact that gcd(2m + 2k, m, k) = 1. Note that in the following, addition is done modulo 2m + 2k. Since gcd(2m + 2k, m) = 1, it follows that ⟨m⟩ = V (G), and G ∼ = Circ (2m + 2k; 1, t ), with isomorphism f (im) = i. Furthermore, since (k + m − 1)m = (m − 1)(k + m) + k and m − 1 is even, it follows that (k + m − 1)m = k. Therefore, t = k + m − 1, and G ∼ = Circ (2m + 2k; 1, k + m − 1). If k is odd, it can be similarly show that Circ (2m + 2k; m, k) ∼ = Circ (2m + 2k; 1, k + m − 1) by just interchanging k and m in the previous paragraph.

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Without loss of generality, let G = Circ (2m + 2k; 1, k + m − 1). For any i ∈ V (G), N (i) = {i + 1, i − 1, i + k + m − 1, i − k − m + 1} = {i + 1, i − 1, i + k + m − 1, i + k + m + 1} and N (i + k + m) = {i + k + m + 1, i + k + m − 1, i + 2k + 2m − 1, i + 1} = {i + k + m + 1, i + k + m − 1, i − 1, i + 1}. Therefore, for each i ∈ V (G), N (i) = N (i + k + m) and every vertex in V (G) is an open corner. Let G0 = G and, for each i ∈ [0, k + m − 1], let Gi+1 = Gi −{i}. Note that Gi is the subgraph of G induced on [i, 2k + 2m − 1]. For each i ∈ [0, . . . , k + m − 1], i is a corner in Gi , since i and i + k + m have identical open neighbourhoods in Gi . By Lemma 4.7, if Gi+1 is 2-cop-win, then Gi is 2-cop-win. Therefore, if Gk+m is 2-cop-win, then G is 2-cop-win. Hence, it only remains to verify that Gk+m is 2-cop-win. Then V (Gk+m ) = {k + m, k + m + 1, . . . , 2k + 2m − 1}. Since vertices i and j in Gk+m are adjacent if and only if |i − j| ∈ {1, k + m − 1} it is straightforward to see that Gk+m is a cycle, and, therefore, is 2-cop-win. Hence, G is 2-copwin.  With this final case, we have now proved all parts of Theorem 2.1, and determined the copnumbers of all circulant graphs of degree at most four. 5. Wreath products of circulant graphs The wreath product of graphs G and H is denoted G w r H (adopting the notation used in [12]), where V (G w r H ) = V (G) × V (H ), and E (G w r H ) = {(x1 , y1 )(x2 , y2 )|x1 x2 ∈ E (G), or x1 = x2 and y1 y2 ∈ E (H )}. In other words, G w r H is obtained by (1) replacing each vertex of G with a copy of H, and (2) adding all possible edges between two copies of H, whenever they have replaced adjacent vertices of G. We will say that two copies of H are adjacent if the corresponding vertices in G are adjacent. The wreath product is of interest since the wreath product of any two circulant graphs is also a circulant graph [6]. We begin this section with some general results regarding wreath products and copnumber. We can find an upper bound on the copnumber of wreath products by considering retracts. We say that a subgraph F of G is a retract of G if there is function f : V (G) → V (F ) such that for any edge xy in G, f (x)f (y) is an edge in F . Furthermore, f (x) = x for every vertex in V (F ). In other words, f is a graph homomorphism such that f is the identity on V (F ). It was show in [15] that the retract of any cop-win graph is also cop-win. Retracts were also used extensively in [2] to describe winning strategies for three cops in planar graphs. The following theorem regarding the relationship of the copnumber of G and a retract F of G was given in [3]. Theorem 5.1 ([3]). If F is a retract of G, then c (F ) ≤ c (G). A retract can be found in the wreath product G w r H by taking a single vertex from each copy of H. Call the subgraph induced on this subset of vertices F . It follows that F is retract of G w r H, where the required mapping associates every vertex in a given copy of H with a single vertex in that copy. Since F ∼ = G, we have the following lower bound on c (G wr H ). Corollary 5.2. For graphs G and H , c (G w r H ) ≥ c (G). We can now find the copnumber of G w r H, given the copnumbers of G and H. The following result appears in [17], using the terminology of lexicographic product. We include an alternative proof to that in [17]. Lemma 5.3. For any connected graphs G and H, with G  K1 , 1. if c (G) ≥ 2, then c (G w r H ) = c (G). 2. c (G w r H ) = 1 if and only if c (G) = c (H ) = 1. 3. if c (G) = 1 and c (H ) ≥ 2, then c (G w r H ) = 2. Proof. Suppose k Cops and a Robber play on G w r H, where k = c (G). The Cops can take their winning strategy in G, and apply it on G w r H, by moving between copies of H exactly as they would move between corresponding vertices of G. Therefore, in some round, following the Cops’ move, at least one Cop, say Cop1 , is on the same copy of H as the Robber. Refer to this copy of H as H1 . If the Robber moves to an adjacent copy of H, say H2 , he will be captured in the next round. This is due to the fact that each vertex in H1 is adjacent to every vertex in H2 . Therefore, we may assume that the Robber and Cop1 stay in H1 for the remainder of the game. We now consider three cases, based on the copnumbers of G and H. Case 1: k ≥ 2. There is at least one other Cop besides Cop1 . Therefore, Cop2 can move onto a copy of H adjacent to H1 . Once this is achieved, the Robber will be captured by Cop2 in the next round. Therefore, G w r H is k-cop-win, and c (G w r H ) ≤ c (G). By Corollary 5.2, c (G w r H ) = c (G), and statement 1. of the theorem has been proved. Case 2: k = 1 and c (H ) = 1. We know that Cop1 can keep the Robber confined to H1 . Since Cop1 has a winning strategy in H1 , Cop1 will capture the Robber. Hence, c (G w r H ) = 1. Case 3: k = 1 and c (H ) ≥ 2. In this case, we play the game with two Cops. We know that Cop1 can confine the Robber to H1 . Therefore, Cop2 can move to a copy of H adjacent to H1 , and then capture the Robber. Hence, c (G w r H ) ≤ 2. We now show that c (G w r H ) = 1 implies c (G) = c (H ) = 1. By Corollary 5.2, c (G w r H ) = 1 implies that c (G) = 1. Suppose c (H ) > 1 and we play the game on G w r H with a single Cop. First Cop1 chooses a vertex, v . Let H1 denote the

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copy of H in G containing v . The Robber then chooses a vertex in H1 according to his winning strategy in H. Now, as long as Cop1 stays in H1 , the Robber can avoid capture. Therefore, Cop1 must move to an adjacent copy of H, say H2 . The Robber responds by moving to a vertex of H2 . Again, since he has a winning strategy in H, he can avoid capture in H2 . Therefore, if the Robber always moves so that at the end of each round he is in the same copy of H as Cop1 , then the Robber can avoid capture forever. This is a contradiction, so c (H ) = 1. This, together with Cases 2 and 3, proves statements 2. and 3. of the theorem.  We now provide the following proposition due to Broere and Hattingh [6] that will allow us to find additional classes of circulant graphs with copnumber two. For convenience, we let Circ (n, S ) represent the circulant graph Circ (n, s1 , . . . , sk ) where S = {s1 , . . . , sk , −sk , . . . , −s1 }, as described in Section 2. Furthermore, for an integer n and a set X , we let nX = {nx|x ∈ X }. Therefore, nZm + s = ⟨n⟩ + s = {s, n + s, 2n + s, . . . , n(m − 1) + s}, with addition done modulo m. Proposition 5.4 ([6]). If G = Circ (n, S ) and H = Circ (m, T ), then G w r H ∼ = Circ (nm, nT ∪ (∪s∈S (nZm + s))). Further, the subgraph of F induced by choosing one element from each coset of ⟨n⟩ is isomorphic to G, and the subgraph of F induced by any coset of ⟨n⟩ is isomorphic to H. For example, let G = Circ (mn, {1, n − 1, n + 1, 2n − 1, 2n + 1, . . . , (n − 1)m − 1, (n − 1)m + 1, nm − 1}) be a connected circulant graph. It can be verified using Proposition 5.4, that G is isomorphic to the wreath product of a cycle and an empty graph. Specifically, G ∼ = Cn wr Km , since Cn = Circ (n, {1, n − 1}) and Km = Circ (m, ∅). Therefore, by Lemma 5.3, c (G) = 2. Therefore, Circ (2m + 2k; m, k) can also be shown to have copnumber two by noting that Circ (2m + 2k; m, k) ∼ = Cm+k w r K2 . We also note that for any connected circulant graph G, G w r Kn has the property that every vertex is an open corner. In fact, any two vertices in the same copy of Kn in G w r Kn have exactly the same open neighbourhoods. Therefore, we can successively remove open corners of G w r Kn and obtain the graph G. We can make a similar statement regarding G w r Kn and corners. A vertex x is a corner in G if there is a second vertex y such that N [x] ⊆ N [y]. Therefore, for G w r Kn , we can successively remove corners and obtain the graph G. Identifying corners and open corners plays an important role in determining copnumber. In characterizing cop-win graph, it was shown that every cop-win graph has a corner [15]. Furthermore, the proof of this can be easily modified to show that if G has a corner, x, then c (G − {x}) = c (G). This, together with a slight modification to the proof of Lemma 4.7 in [8], tells us that for any vertex x in G such that x is either a corner or open corner, if G − x is k-cop-win, then G is k-cop-win, where k ≥ 2. It turns out that for a non-complete connected circulant graph G, there are open corners or corners in G exactly when there is a graph H such that G ∼ = H wr Km , respectively, where m ≥ 2. This will be proved in Theorem 5.7. = H w r Km or G ∼ Furthermore, given a circulant graph Circ (n, S ), we can determine whether it can be expressed as one of these two wreath products by examining the distance set S. This is due to the following result due to Kumar and MacGillivray [12], and a similar result for Km that is proved immediately afterwards. Lemma 5.5 ([12]). The graph G = Circ (mn, S ) is isomorphic to the wreath product of a circulant graph on n vertices and Km if and only if S ∪ {0} is a union of cosets of ⟨n⟩. Lemma 5.6. The graph G = (mn, S ) is isomorphic to the wreath product of a circulant graph on n vertices and Km if and only if S is the union of cosets of ⟨n⟩. Proof. Let H = (n, T ) and suppose G ∼ = H w r Km for some m > 1. Then, by Proposition 5.4, G is isomorphic to Circ (mn, S ) where S = n∅ ∪ (∪t ∈T (nZm + t )) = ∪t ∈T (nZm + t ). Therefore, S is the union of cosets of nZm = ⟨n⟩. Suppose G is a circulant graph such that G = (mn, S ), and S = (⟨n⟩ + s1 ) ∪ (⟨n⟩ + s2 ) ∪ · · · ∪ (⟨n⟩ + sk ), where 0 < si < n for each i ∈ [1, k]. Now consider the coset ⟨n⟩ + si . For any x and y in this coset, |x − y| = jn for some j. Since there is no integer j such that jn ∈ S, it follows that the coset ⟨n⟩ + sj is an independent set in G. Furthermore, since each si ∈ S has an inverse modulo mn in S, it follows that si has an inverse, modulo n, in {s1 , s2 , . . . , sk }. It follows from Proposition 5.4 that G∼ = H w r Km where H ∼ = Circ (n, T ) and T = {s1 , s2 , . . . , sk }.  We say that a circulant graph G is reducible if there is some graph H such that G ∼ = H w r Km or G ∼ = H wr Km for some m ≥ 2. Otherwise G is irreducible. It follows that each reducible graph, G has an irreducible subgraph, Q . Furthermore, G can be expressed as a wreath product in terms of Q , as described in the next paragraph. Let K be the set of graphs such that (1) K contains all complete graphs on at least two vertices, (2) K contains all empty graphs on at least two vertices, and (3) if F1 ∈ K and F2 ∈ K , then F1 w r F2 ∈ K . It follows that if G is irreducible, then there is some irreducible graph H, and some graph K ∈ K such that G ∼ = H wr K . Theorem 5.7. The connected circulant graph G is reducible if and only if G has either a corner or an open corner. Proof. Suppose G is the connected circulant graph Circ (n, S ), where S = {s1 , s2 , . . . , sr }. We note that, since G is regular, N (x) ⊆ N (y) implies N (x) = N (y), and N [x] ⊆ N [y] implies N [x] = N [y]. Suppose G has a corner. Without loss of generality, it follows that for some j ∈ [1, n − 1], N [j] = N [0]. Suppose we have chosen the smallest positive j possible. Then, {j, s1 + j, s2 + j, . . . , sr + j} = {0, s1 , s2 , . . . , sr }, and j = st for some t ∈ [1, r ].

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Furthermore, assuming s1 < s2 < · · · < sr < n, si + j = si+t , where s0 = 0, and the subscripts of s are recorded modulo r + 1. It follows that si+wt = si + w j for all w ≥ 0. It follows that N [0] is the union of cosets of ⟨j⟩. Since S ∪ {0} = N [0], by Lemma 5.5, G is isomorphic to the wreath product of a circulant graph and Km for some m. Therefore, G is reducible. Suppose G has an open corner. Without loss of generality, assume that vertex 0 is an open corner. It follows that for some j ∈ [1, n − 1], N (j) = N (0). Then, {s1 + j, s2 + j, . . . , sr + j} = {s1 , s2 , . . . , sr }, and s1 + j = st +1 for some t. Furthermore, si + j = si+t for all i ≥ 1, where the subscripts of s are recorded modulo r + 1. It follows that N (0) is the union of cosets of ⟨j⟩. Since S = N (0), by Lemma 5.6, G is isomorphic to the wreath product of a circulant graph and Km for some m ≥ 2. Therefore, G is reducible. Now suppose G is reducible. If G ∼ = H wr Km for some H and m ≥ 2, then it is straightforward to see that for any pair vertices x and y in the same copy of Km in G (that is, the same coset of S), N [x] = N [y]. If G ∼ = H wr Km , it is also straightforward to see that N (x) = N (y) whenever x and y are in the same copy of Km in G (same coset of S).  Having established the relationship between irreducible circulant graphs and wreath products, we give one final result regarding tandem-win graphs. In [7], a variation of the game of Cops and Robber was introduced in which two Cops and a Robber play the game, and the two Cops must be within distance one of each other at the end of each round. This is referred to as tandem play, and graphs in which two Cops can always win, given this restriction, are referred to as tandem-win graphs. Obviously, tandem-win graphs are a subset of the set of 2-cop-win graphs. In [8], tandem win graphs were characterized as follows. Theorem 5.8 ([8]). Suppose x is a corner or an open corner in graph G. The graph G is tandem-win if and only if G − {v} is tandem-win. Corollary 5.9. A reducible circulant graph G is tandem-win if and only if there is a tandem-win irreducible circulant graph H such that G ∼ = H wr K for some K ∈ K . Acknowledgement First author was partially supported by a grant from Natural Sciences and Engineering Research Council of Canada (Grant Number 238874). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]

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