Game total domination for cyclic bipartite graphs

Discrete Applied Mathematics 265 (2019) 120–127

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

Game total domination for cyclic bipartite graphs✩ Yisheng Jiang, Mei Lu

∗

Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China

article

info

Article history: Received 4 October 2018 Received in revised form 26 February 2019 Accepted 14 March 2019 Available online 3 April 2019 Keywords: Domination game Total domination number Cyclic bipartite graph

a b s t r a c t Let G = (V , E) be a graph. A vertex u in G totally dominates a vertex v if u is adjacent to v in G. The total domination game played on G consists of two players, named Dominator and Staller, who alternately take turns choosing vertices of G such that each chosen vertex totally dominates at least one vertex not totally dominated by the vertices previously chosen. Dominator wishes to totally dominate the graph as fast as possible, while Staller wishes to delay the process as much as possible. The game total domination number γtg (G) (resp. the Staller-start game total domination number γtg′ (G)) of G is the number of vertices chosen when Dominator starts the game (resp. when Staller starts the game) and both players play optimally. In this paper, we determine the exact value of γtg (G) and γtg′ (G) when G is a cyclic bipartite graph. © 2019 Elsevier B.V. All rights reserved.

1. Introduction In this paper, we only consider simple graphs without isolated vertices. Let G = (V , E) be a graph. A vertex u in G totally dominates a vertex v if u is adjacent to v in G. A total dominating set of G is a vertex set S ⊆ V (G) such that every vertex of G is totally dominated by a vertex in S. The total domination game, played on a graph G consists of two players called Dominator and Staller who alternately take turns choosing a vertex from G. Each chosen vertex must totally dominate at least one vertex not totally dominated by the set of vertices previously chosen. Dominator wishes to end the game with a minimum number of vertices chosen, and Staller wishes to end the game with as many vertices chosen as possible. The game total domination number, γtg (G), is the number of vertices chosen when Dominator starts the game and both players play optimally. The Staller-start game total domination number, γtg′ (G), is the number of vertices chosen when Staller starts the game and both players play optimally. The domination game in graphs was introduced in [3] and extensively studied afterwards in [1–8,10,16,18,19] and elsewhere. The notion of total domination game was introduced recently by Henning, Klavžar and Rall in [11]. In the same paper, the authors showed that |γtg (G) − γtg′ (G)| ≤ 1 and indicated that to determine the exact value of γtg (G) and γtg′ (G) for special classes of graphs G is a challenging problem. So far, the only exact values of γtg (G) and γtg′ (G) were established in [9] for G being paths and cycles. Other results about the bounds of γtg (G) and γtg′ (G) can be found in [9–15,17]. In this paper, we determine the exact values of γtg (G) and γtg′ (G) when G is a cyclic bipartite graph. For notation and graph theory terminology, we in general follow [16]. A k-regular graph is a graph where each vertex has the same degree k. A bipartite graph is a graph whose vertices can be divided into two disjoint and independent sets U and V such that every edge connects a vertex in U and a vertex in V . If X and Y are subsets of vertices in a graph G, then the set X totally dominates the set Y in G if every vertex of Y is adjacent to at least one vertex of X . ✩ This work is partially supported by NNSFC (No. 11771247). ∗ Corresponding author. E-mail addresses: [email protected] (Y. Jiang), [email protected] (M. Lu). https://doi.org/10.1016/j.dam.2019.03.009 0166-218X/© 2019 Elsevier B.V. All rights reserved.

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Fig. 1. Illustration of cyclic bipartite graphs.

2. Main results In this section, we study the game total domination number and the Staller-start game total domination number of a cyclic bipartite graph which is defined as follows. Definition 2.1. Let n ≥ k ≥ 2 be two integers. Let Gn,k denote a cyclic bipartite graph with vertex set V (Gn,k ) = {xi |i = 1, 2, . . . , n} ∪ {yi |i = 1, 2, . . . , n} and edge set E(Gn,k ) = {(xi , yi+j )|i = 1, 2, . . . , n, j = 0, 1, . . . , k − 1}, where i + j ≡ s (mod n) with 1 ≤ s ≤ n (see Fig. 1 ). By Definition 2.1, the order of Gn,k is 2n and Gn,k is k-regular. Also, Gn,2 is a cycle of length 2n. In the following, we will denote X = {xi |i = 1, 2, . . . , n} and Y = {yi |i = 1, 2, . . . , n}. Let

{

4l

τ (n, k) = 4l + 1 4l + 2

when n = (k + 1)l, when n = (k + 1)l + j, j = 1, 2, . . . , k − 1, when n = (k + 1)l + k,

and

τ (n, k) = ′

{

4l 4l + 2

when n = (k + 1)l + j, j = 0, 1, . . . , k − 1, when n = (k + 1)l + k.

The main result of this paper is the following theorem. Theorem 2.1. Let n ≥ k ≥ 2 be two integers. We have γtg (Gn,k ) = τ (n, k) and γtg′ (Gn,k ) = τ ′ (n, k). In order to prove our result, we need some lemmas. Lemma 2.1. Let n ≥ k ≥ 2 be two integers. We have

γtg (Gn,k ) ≥ τ (n, k). Proof. We consider the Dominator-start game on Gn,k . To prove a lower bound on γtg (Gn,k ), we describe a strategy for Staller that guarantees that she totally dominates exactly one new vertex on almost all of her moves. Assume, without loss of generality, that the first vertex of Dominator’s choice at the beginning is yk , which means the vertices x1 , x2 , . . . , xk have been totally dominated. Assume it is Staller’s turn to play, and denote by A the subset of vertices already totally dominated before Staller makes this move. Let AX = A ∩ X and AY = A ∩ Y . Then {x1 , x2 , . . . , xk } ⊆ AX . Also we can assume that A does not contain all vertices or the game would be finished. Let u be the vertex chosen by Dominator before Staller’s turn and v be the vertex chosen by Staller in this turn. We give Staller’s strategy as follows. (S1) If u ∈ Y , AX = X and AY = ∅, then v = x1 which implies y1 , y2 , . . . , yk are totally dominated. Note that Staller totally dominates k new vertices in this move. (S2) If u ∈ Y and AX ⊂ X or u ∈ X , AY = Y but AX ⊂ X , assume xt (1 ≤ t ≤ n) is the vertex with minimum index that is not in AX , then v = yt and therefore Staller totally dominates exactly one new vertex xt . (S3) If u ∈ X and AY ⊂ Y or u ∈ Y , AX = X but ∅ ̸ = AY ⊂ Y , then there exist a sequence of at least k consecutive vertices in Y that are already dominated, say yi , . . . , yi+l . Let yt be the first vertex in the sequence yi+l+1 , yi+l+2 , . . . , yi−1 which is not in AY . Then v = xt −k+1 is Staller’s move on which she totally dominates exactly one new vertex yt . During the game, every move of Staller therefore totally dominates exactly one new vertex except (S1). Let m be the number of moves played when the game finishes. Next, we consider the lower bound of γtg (Gn,k ) in two cases. Case 1 Dominator chooses a vertex in X first. In this case, every move of Staller totally dominates exactly one new vertex. Assume first that m is even. Let m = 2r. Then Staller made the final move and two players played r moves. In this case, Dominator totally dominates at most kr vertices, while Staller totally dominates exactly r vertices, implying that

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kr + r ≥ 2n and therefore r ≥ ⌈ k2n ⌉. Thus +1

m = 2r ≥

⎧ 4l ⎪ ⎪ ⎪ ⎪ ⎨4l + 2 ⎪ ⎪ ⎪ ⎪ ⎩4l + 4

when n = (k + 1)l, when n = (k + 1)l + j, 1 ≤ j ≤ when n = (k + 1)l + j,

⌊

k+1

⌊

k+1

⌋

2

⌋

2

,

+ 1 ≤ j ≤ k.

Hence γtg (Gn,k ) ≥ m ≥ τ (n, k) when m is even. Assume now that m is odd. Let m = 2r − 1. Then Dominator played r moves and Staller played r − 1 moves. In this case, Dominator totally dominates at most kr vertices, while Staller totally dominates exactly r − 1 vertices, implying that +1 kr + (r − 1) ≥ 2n and therefore r ≥ ⌈ 2n ⌉. Thus k+1

m = 2r − 1 ≥

⎧ ⎪ ⎪ ⎨4l + 1

when n = (k + 1)l + j, 0 ≤ j ≤

⎪ ⎪ ⎩4l + 3

when n = (k + 1)l + j,

⌊

k+1 2

⌊

⌋

k+1 2

⌋

,

+ 1 ≤ j ≤ k.

Hence we have that γtg (Gn,k ) ≥ m ≥ τ (n, k) when m is odd. Case 2 Staller chooses a vertex in X first. In this case, exactly k new vertices are totally dominated by Staller in this move and other moves of Staller totally dominates exactly one new vertex. Let n = (k + 1)l + j, where 0 ≤ j ≤ k. Assume that the vertex in X is first chosen by Staller at her tth move, which means X is totally dominated after Dominator’s (t − 1)th move. Note that 2t − 1 moves are needed for X to be totally dominated. Thus at most kt + (t − 1) = (k + 1)t − 1 vertices have been totally dominated after Dominator’s tth move. So we have (k + 1)t − 1 ≥ n = (k + 1)l + j which implies t ≥ l + 1. After Staller’s tth move, two players start to totally dominate vertices in Y . Dominator totally dominates at most k new vertices for each move, while Staller totally dominates exactly one new vertex for each move besides her tth move. We first claim that in total at least 2l moves (including Staller’s tth move) are needed for two players to totally dominate the vertex set Y when j ̸ = k. Note that after (2l − 1) moves, Dominator totally dominates at most k(l − 1) vertices, while Staller totally dominates exactly k+(l−1) vertices. So two players totally dominate k(l−1)+k+(l−1) = (k+1)(l−1)+k < n vertices. Therefore, we have m ≥ (2l + 1) + 2l = 4l + 1 when j ̸ = k. We next claim that in total at least (2l + 1) moves (including Staller’s tth move) are needed for two players to totally dominate the vertex set Y when j = k. After 2l moves, Dominator totally dominates at most kl vertices, while Staller totally dominates exactly k + (l − 1) vertices. So two players totally dominate kl + k + (l − 1) = (k + 1)l + k − 1 < n vertices. Therefore, we know that m ≥ (2l + 1) + (2l + 1) = 4l + 2 holds when j = k. Summarily, we know that

{ m≥

4l + 1

when n = (k + 1)l + j, 0 ≤ j ≤ k − 1,

4l + 2

when n = (k + 1)l + k.

Thus γtg (Gn,k ) ≥ m ≥ τ (n, k). □ Lemma 2.2. Let n ≥ k ≥ 2 be two integers. We have

γtg′ (Gn,k ) ≥ τ ′ (n, k). Proof. We consider the Staller-start game on Gn,k . To prove a lower bound on γtg (Gn,k ), we describe a strategy for Staller that guarantees that she totally dominates exactly one new vertex on almost all of her moves. We may assume, without loss of generality, that the first vertex of Staller’s choice is yk , which means x1 , x2 , . . . , xk have been totally dominated after Staller’s first move. In order to prove the lower bound of γtg′ (Gn,k ), we assume Staller uses the strategy almost as stated in Lemma 2.1. The only difference is that Staller’s first move is on the vertex yk . Let m be the number of moves played when the game finishes. We consider the lower bound of γtg′ (Gn,k ) in two cases. Case 1 Dominator chooses a vertex in X first. In this case, Staller guarantees that exactly one new vertex is totally dominated in each of her moves, except in her first move.

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Assume first that m is even. Let m = 2r. In this case, Dominator totally dominates at most kr vertices, while Staller totally dominates exactly k + (r − 1) vertices, implying that kr + k + (r − 1) ≥ 2n and therefore r ≥ ⌈ 2nk−+k1+1 ⌉. Then

m = 2r ≥

⎧ ⎪ ⎪ ⎨4l

when n = (k + 1)l + j, 0 ≤ j ≤

⎪ ⎪ ⎩4l + 2

when n = (k + 1)l + j,

⌊

k−1

⌊

k−1

⌋

2

⌋

2

,

+ 1 ≤ j ≤ k.

Hence γtg′ (Gn,k ) = m ≥ τ ′ (n, k) when m is even. Assume next that m is odd. Let m = 2r + 1. Then Dominator played r moves and Staller played r + 1 moves. In this case, Dominator totally dominates at most kr vertices, while Staller totally dominates exactly k + r vertices, implying that −k ⌉. Then kr + (k + r) ≥ 2n and therefore that r ≥ ⌈ 2n k+1

m = 2r + 1 ≥

⎧ ⎪ ⎪ ⎨4l + 1 ⎪ ⎪ ⎩4l + 3

when n = (k + 1)l + j, 0 ≤ j ≤ when n = (k + 1)l + j,

⌊ ⌋ k

2

⌊ ⌋ k

2

,

+ 1 ≤ j ≤ k.

Hence γtg′ (Gn,k ) = m ≥ τ ′ (n, k) holds when m is odd. Case 2 Staller chooses a vertex in X first. In this case, Staller guarantees that exactly k vertices are totally dominated after her first move and after her first move in X , and exactly one vertex is totally dominated after her other moves during the game. Let n = (k + 1)l + j, where 0 ≤ j ≤ k. Assume that Staller first chooses a vertex in X at her tth move, which means X is totally dominated after Dominator’s (t − 1)th move. Note that at most k(t − 1) + k + (t − 2) = (k + 1)t − 2 vertices have been totally dominated after Dominator’s (t − 1)th move. So we have (k + 1)t − 2 ≥ n = (k + 1)l + j which implies

{ t≥

when n = (k + 1)l + j, 0 ≤ j ≤ k − 1, when n = (k + 1)l + k.

l+1 l+2

Therefore, in order to totally dominate the vertex set X , the number of moves is no less than

{ 2t − 2 ≥

2l 2l + 2

when n = (k + 1)l + j, 0 ≤ j ≤ k − 1, when n = (k + 1)l + k.

After Staller’s tth move, two players start to totally dominate vertices in Y . Dominator totally dominates at most k vertices for each move, while Staller totally dominates exactly one vertex for each move besides her tth move. We first claim that in total at least 2l moves are needed for both players to totally dominate the vertex set Y when j ̸ = k. After (2l − 1) moves, Dominator totally dominates at most k(l − 1) vertices, while Staller totally dominates exactly k + (l − 1) vertices. So two players totally dominate k(l − 1) + k + (l − 1) = (k + 1)(l − 1) + k < (k + 1)l + j = n. Therefore, we have m ≥ 2l + 2l = 4l when j ̸ = k. We next claim that in total at least (2l + 1) moves for both players to totally dominate the vertex set Y when j = k. After 2l moves, Dominator totally dominates at most kl vertices, while Staller totally dominates exactly k + (l − 1) vertices. So two players totally dominate kl + k + (l − 1) = (k + 1)l + k − 1 < (k + 1)l + k = n. Therefore, we have m ≥ (2l + 2) + (2l + 1) = 4l + 3 when j = k. Now we have that

{ m≥

4l 4l + 3

when n = (k + 1)l + j, 0 ≤ j ≤ k − 1, when n = (k + 1)l + k.

Thus γtg′ (Gn,k ) ≥ m ≥ τ ′ (n, k). □ We have already established lower bounds on γtg (Gn,k ) and γtg′ (Gn,k ), respectively. Upper bounds on γtg (Gn,k ) and γtg (Gn,k ) can be obtained from the following two lemmas. ′

Lemma 2.3. Let n ≥ k ≥ 2 be two integers. We have

γtg (Gn,k ) ≤ τ (n, k). Proof. We consider the Dominator-start game on Gn,k . To prove an upper bound on γtg (Gn,k ), we describe a strategy for Dominator. We denote the vertex in X which is first chosen by Dominator or Staller when the game is played by xa−k+1 , which means ya−k+1 , ya−k+2 , . . . , ya are the k first vertices totally dominated in Y . Let X = X1 ∪ X2 ∪ · · · ∪ X⌈ n ⌉ k+1 and Y = Y1 ∪ Y2 ∪ · · · ∪ Y⌈ n ⌉ , where Xi = {xi(k+1)−k , xi(k+1)−k+1 , . . . , xi(k+1) } for i = 1, 2, . . . , ⌈ k+n 1 ⌉ − 1, Yi = k+1

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Y. Jiang and M. Lu / Discrete Applied Mathematics 265 (2019) 120–127 ⌈

n

{ya+i(k+1)−k , ya+i(k+1)−k+1 , . . . , ya+i(k+1) } for i = 1, 2, . . . , ⌈ k+n 1 ⌉ − 1, X⌈ k+n 1 ⌉ = X \ ∪j=k+1 1

⌉−1

⌈

n

⌉−1

1 Xj , and Y⌈ n ⌉ = Y \ ∪j=k+ Yj . 1 k+1 By Definition 2.1, all neighbours of yi(k+1) (resp. xa+i(k+1)−k+1 ) are contained in Xi (resp. Yi ) for 1 ≤ i ≤ ⌈ k+n 1 ⌉ − 1. Note that the partition of Y will be obtained after the vertex xa−k+1 is chosen. The subset Xi or Yi which contains (k + 1) elements is called a full tuple; otherwise it is called a remainder tuple. Note that there is at most one remainder tuple either in the partition of the sets X or Y . If all the vertices of one full tuple (resp. remainder tuple) are totally dominated, then we say this full tuple (resp. remainder tuple) is totally dominated. Now we give Dominator’s strategy as follows. Dominator chooses the vertex yn as his first move so that if X⌈ n ⌉ is a remainder tuple, it is totally dominated; and if it k+1

is a full tuple, all but one vertex is totally dominated in X⌈ n ⌉ . All other Dominator’s moves depend on Staller’s previous k+1 move. Let u be a new vertex totally dominated by Staller’s previous move. (If there is more than one new vertex totally dominated by Staller’s previous move, then we choose any one of them as u.) (D1) u ∈ Xt (1 ≤ t ≤ ⌈ k+n 1 ⌉ − 1) and Xt is not totally dominated after Staller’s previous move. Assume u = x(k+1)t −k+s , 0 ≤ s ≤ k. In this case, at least one of the following statements is true: (a) {x(k+1)t −k , x(k+1)t −k+1 , . . . , x(k+1)t −k+s } are totally dominated; (b) {x(k+1)t −k+s , x(k+1)t −s+1 , . . . , x(k+1)t } are totally dominated. If (a) holds, then Dominator chooses the vertex y(k+1)t +s as his next move so that Xt is totally dominated. If (b) holds, then Dominator chooses the vertex y(k+1)t −k+s−1 as his next move so that Xt is totally dominated. (D2) u ∈ Xt (1 ≤ t ≤ ⌈ k+n 1 ⌉) and Xt is totally dominated after Staller’s move. If there is a full tuple Xi (1 ≤ i ≤ ⌈ k+n 1 ⌉ − 1) such that all the vertices in Xi are not totally dominated, then Dominator chooses the vertex yi(k+1) so that all vertices in Xi \ {xi(k+1)−k } are totally dominated. If such Xi does not exist but there is a full tuple Yi (1 ≤ i ≤ ⌈ k+n 1 ⌉ − 1) such that all the vertices in Yi are not totally dominated, then Dominator chooses the vertex xa+i(k+1)−k+1 so that all vertices in Yi \ {ya+i(k+1) } are totally dominated. (D3) u ∈ Y⌈ n ⌉ and n ̸ ≡ 0 (mod k + 1). k+1

In this case, the vertex xa−k+1 is chosen by Staller and so the remainder tuple in Y is totally dominated. Then Dominator chooses vertex xa−2k+1 as his next move so that {ya−2k+1 , ya−2k+2 , . . . , ya−k } are totally dominated. (D4) u ∈ Yt (1 ≤ t ≤ ⌈ k+n 1 ⌉ − 1) and Yt is not totally dominated after Staller’s previous move. In this case, the remainder tuple in Y is already totally dominated. Now Dominator can use a similar strategy as (D1) to ensure that Yt is totally dominated after his next move. (D5) u ∈ Yt (1 ≤ t ≤ ⌈ k+n 1 ⌉ − 1) or u ∈ Y⌈ n ⌉ but n ≡ 0 (mod k + 1) and Yt is totally dominated after Staller’s previous k+1 move. In this case, Dominator can use a similar strategy as (D2). In brief, the Dominator’s strategy guarantees that there is at least one new full tuple totally dominated after two players move once. Now, we divide the game into two phases. Phase 1: Two players only choose vertices in Y , which means no vertex in Y is totally dominated. Phase 2: Some vertices in Y are totally dominated. Therefore, the remainder tuple in Y (if it exists) is totally dominated. In order to obtain the upper bound of γtg (Gn,k ), we will consider the following three cases. Case 1 n = (k + 1)l. In this case, there is no remainder tuple. We first have the following two claims. Claim 1 holds obviously by Dominator’s strategy. Claim 1. In Phase 1 of the game after Dominator’s tth move, there are at least (t − 1) full tuples which are totally dominated and one extra full tuple which contains at most one vertex not yet totally dominated, where t ≥ 1. Claim 2. In Phase 2 of the game after Dominator’s tth move, there are at least (t − 1) full tuples which are totally dominated and one extra full tuple which contains at most one vertex not yet totally dominated. Proof of Claim 2. By Dominator’s strategy, we just need to show that the result holds when Staller makes the first move to totally dominate vertices in Y in her (t − 1)th move or Dominator makes the first move to totally dominate vertices in Y in his tth move. Assume that Staller makes the first move that totally dominates vertices in Y and this is her (t − 1)th move. By Dominator’s strategy, the full tuple Yl = {ya−k , ya−k+1 , ya−k+2 , . . . , ya } is totally dominated after Dominator’s tth move. By Claim 1, there are at least (t − 2) totally dominated tuples and one extra tuple contains at most one vertex not yet totally dominated after Dominator’s (t − 1)th move. Thus the result holds. Assume that Dominator makes the first move that totally dominates vertices in Y and this is his tth move. Then the full tuple Yl = {ya−k , ya−k+1 , ya−k+2 , . . . , ya } contains at most one vertex which is not yet totally dominated. In this case, we know that X is already totally dominated after Staller’s (t − 1)th move. By Claim 1, there are at least (t − 2) full tuples are totally dominated and one extra full tuple contains at most one vertex not yet totally dominated after Dominator’s (t − 1)th move. Since X is not totally dominated after Dominator’s (t − 1)th move, t − 2 < l, that is l ≥ t − 1. Note that X has l full tuples. So there are l ≥ t − 1 full tuples in X are totally dominated and one extra full tuple in Y which contains at most one vertex not yet totally dominated after Dominator’s tth move. Thus the result holds. ■

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By Claim 2, we know that there are at least (2l − 1) full tuples are totally dominated and one extra full tuple contains at most one vertex not yet totally dominated after Dominator’s (2l)th move which implies there are at least (k + 1)(2l − 1) + k = 2n − 1 vertices are totally dominated after Dominator has played his (2l)th move. There is therefore at most one remaining vertex not yet totally dominated. Hence, all vertices are totally dominated after 4l moves. That is

γtg (G(k+1)l,k ) ≤ 4l = τ ((k + 1)l, k). Case 2 n = (k + 1)l + j, 1 ≤ j ≤ k − 1. In this case, there are two remainder tuples containing j < k vertices. We first have the following two claims. Claim 3 holds obviously by Dominator’s strategy. Claim 3. In Phase 1 of the game after Dominator’s tth move, the remainder tuple in X is totally dominated and there are at least (t − 1) full tuples which are totally dominated, where t ≥ 1. Claim 4. In Phase 2 of the game after Dominator’s tth move, the remainder tuples in X and Y are totally dominated, there are at least (t − 1) full tuples which are totally dominated and one extra full tuple which contains at least (k − 1 − j) totally dominated vertices. Proof of Claim 4. By Dominator’s strategy, we just need to show that the result holds when Staller makes the first move to totally dominate vertices in Y in her (t − 1)th move or Dominator makes the first move to totally dominate vertices in Y in his tth move. Assume Staller makes the first move to totally dominate vertices in Y in her (t − 1)th move. By Dominator’s strategy, the 2k vertices {ya−2k+1 , ya−2k+2 , . . . , ya } are totally dominated after Dominator’s tth move. These 2k vertices contain the (k + 1) vertices in the full tuple Yl = {ya−j−k , ya−j−k+1 , . . . , ya−j }, j vertices in the remainder tuple {ya−j+1 , ya−j+2 , . . . , ya } and (k − 1 − j) vertices in the full tuple Yl−1 . By Claim 3, the result holds. Assume that Dominator makes the first move to totally dominate vertices in Y in his tth move. In this case, we know that the vertices in X are already totally dominated. By Claim 3, there are at least (t − 2) full tuples which are totally dominated after Dominator’s (t − 1)th move. Since X is not totally dominated after Dominator’s (t − 1)th move, t − 2 < l, that is l ≥ t − 1. So at least l ≥ t − 1 full tuples in X are totally dominated, the remainder tuples in X and Y are totally dominated and one extra full tuple in Y contains at least (k − j) totally dominated vertices after Dominator’s tth move. Thus the result holds. ■ By Claim 4, we know that the remainder tuples in X and Y are totally dominated, at least (2l) full tuples are totally dominated and one extra full tuple contains at least (k − 1 − j) totally dominated vertices after Dominator’s (2l + 1)th move. Hence there are at least (k + 1)2l + 2j + k − j − 1 ≥ 2(k + 1)l + 2j = 2n vertices which are totally dominated after Dominator’s (2l + 1)th move. Therefore,

γtg (G(k+1)l+j,k ) ≤ 4l + 1 = τ ((k + 1)l + j, k), 1 ≤ j ≤ k − 1. Case 3 n = (k + 1)l + k. In this case, there are two remainder tuples containing k vertices. We first have the following two claims. Claim 5 holds obviously by Dominator’s strategy. Claim 5. In Phase 1 of the game after Dominator’s tth move, the remainder tuple in X is totally dominated and at least (t − 1) full tuples are totally dominated, where t ≥ 1. Claim 6. In Phase 2 of the game after Dominator’s tth move, the remainder tuples in X and Y are totally dominated, at least (t − 2) full tuples are totally dominated and one extra full tuple contains at most one vertex not yet totally dominated. Proof of Claim 6. By Dominator’s strategy, we just need to show that the result holds when Staller makes the first move to totally dominate vertices in Y in her (t − 1)th move or Dominator makes the first move to totally dominate vertices in Y in his tth move. Assume that Staller makes the first move to totally dominate vertices in Y in her (t − 1)th move. By Dominator’s strategy, the 2k vertices {ya−2k+1 , ya−2k+2 , . . . , ya } are totally dominated after Dominator’s tth move. These 2k vertices contain the k vertices in the remainder tuple {ya−k+1 , ya−k+2 , . . . , ya } and k vertices in the full tuple Yl . By Claim 5, the result holds. Assume that Dominator makes the first move to totally dominate vertices in Y in his tth move. By Dominator’s strategy, we know that X is already totally dominated. By Claim 5, we know that at least (t − 2) full tuples are totally dominated after Dominator’s (t − 1)th move. Since X is not totally dominated after Dominator’s (t − 1)th move, t − 2 < l, that is l ≥ t − 1. So at least l ≥ t − 1 full tuples in X are totally dominated and the remainder tuples in X and Y are totally dominated after Dominator’s tth move. Thus the result holds. ■

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By Claim 6, we know that the remainder tuples in X and Y are totally dominated, at least (2l − 1) full tuples are totally dominated and one extra full tuple contains at most one vertex not yet totally dominated after Dominator’s (2l + 1)th move. Hence there are at least (k + 1)(2l − 1) + 2k + k = (k + 1)2l + 2k − 1 = 2n − 1 totally dominated vertices after Dominator’s (2l + 1)th move. There is therefore at most one remaining vertex not yet totally dominated. Hence, on Staller’s (2l + 1)th move, all vertices are totally dominated. That is

γtg (G(k+1)l+k,k ) ≤ 4l + 2 = τ ((k + 1)l + k, k). □ Lemma 2.4. For all n ≥ k ≥ 2, n, k ∈ Z+ , we have

γtg′ (Gn,k ) ≤ τ ′ (n, k).

Proof. We consider the Staller-start game on Gn,k . We may assume, without loss of generality, that the first vertex of Staller’s choice is yn , which means the vertices xn−k+1 , xn−k+2 , . . . , xn are totally dominated after Staller’s first move. Moreover, the vertices ya−k+1 , ya−k+2 , . . . , ya will be defined as the k first vertices totally dominated in Y when the game is played. We partition the sets X and Y into subsets in the same way as in the proof of Lemma 2.3. We also define the full tuple and the remainder tuple in the same way as before. If all the vertices of one full tuple (resp. remainder tuple) are totally dominated, then we say this full tuple (resp. remainder tuple) is totally dominated. In order to prove the upper bound of γtg′ (Gn,k ), we assume Dominator uses the strategy almost as stated in the proof of Lemma 2.3 except for two differences. One difference is that Staller plays the first move instead of Dominator. Another difference is that Dominator chooses yn−k as his first move if n > k and chooses x1 as his first move if n = k. We also divide the game into two phases in the same way as in the proof of Lemma 2.3. In order to prove the upper bound of γtg′ (Gn,k ), we consider the following two cases. Case 1 n = (k + 1)l + j, 0 ≤ j ≤ k − 1. In this case, there are two remainder tuples containing j < k vertices if j ̸ = 0. We first have the following claims. Claim 7 is obvious by Dominator’s strategy. Claim 7. In Phase 1 of the game after Dominator’s tth move, the remainder tuple in X for 1 ≤ j ≤ k − 1 is totally dominated and at least t full tuples are totally dominated, where t ≥ 1. Claim 8. In Phase 2 of the game after Dominator’s tth move, the remainder tuples in X and Y (if they exist) are totally dominated, at least t full tuples are totally dominated and one extra full tuple contains at least (k − j − 1) totally dominated vertices. Proof of Claim 8. By Dominator’s strategy, we just need to show that the result holds when Staller makes the first move to totally dominate vertices in Y in her tth move or Dominator makes the first move to totally dominate vertices in Y in his tth move. Assume that Staller makes the first move to totally dominate vertices in Y in her tth move. Then all vertices in X are totally dominated after Dominator’s (t −1)th move. By Dominator’s strategy, there are 2k vertices {ya−2k+1 , ya−2k+2 , . . . , ya } in Y which are totally dominated after Dominator’s tth move. These 2k vertices contain the full tuple Yl = {ya−j−k , ya−j−k+1 , . . . , ya−j }, the remainder tuple {ya−j+1 , ya−j+2 , . . . , ya } if 1 ≤ j ≤ k − 1 and (k − 1 − j) vertices of Yl−1 . By Claim 7, the result holds. Assume that Dominator makes the first move to totally dominate vertices in Y in his tth move. In this case, X is already totally dominated. By Claim 7, we know that at least (t − 1) full tuples are totally dominated after Dominator’s (t − 1)th move. Since X is not totally dominated after Dominator’s (t − 1)th move, t − 1 < l, that is l ≥ t. So at least l ≥ t full tuples in X , the remainder tuple {ya−j+1 , ya−j+2 , . . . , ya } if 1 ≤ j ≤ k − 1 and (k − j) vertices of Yl are totally dominated after Dominator’s tth move. Thus the result holds. ■ By Claim 8, we know that the remainder tuples in X and Y (if they exist) are totally dominated, at least 2l full tuples are totally dominated and one extra full tuple contains at least (k − 1 − j) totally dominated vertices after Dominator’s (2l)th move. Hence at least (k + 1)2l + 2j + k − j − 1 ≥ 2(k + 1)l + 2j = 2n vertices are totally dominated after Dominator has played his (2l)th move. Therefore,

γtg′ (G(k+1)l+j,k ) ≤ 4l = τ ′ ((k + 1)l + j, k), 0 ≤ j ≤ k − 1. Case 2 n = (k + 1)l + k. In this case, there are two remainder tuples containing k vertices. We first have the following claims. Claim 9 is obvious by Dominator’s strategy. Claim 9. In Phase 1 of the game after Dominator’s tth move, the remainder tuple in X is totally dominated, at least (t − 1) full tuples are totally dominated and one extra full tuple contains at most one vertex not yet totally dominated, where t ≥ 1.

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Claim 10. In Phase 2 of the game after Dominator’s tth move, the remainder tuples in X and Y are totally dominated and at least (t − 1) full tuples are totally dominated. Proof of Claim 10. By Dominator’s strategy, we just need to show that the result holds when Staller makes the first move to totally dominate vertices in Y in her tth move or Dominator makes the first move to totally dominate vertices in Y in his tth move. Assume that Staller makes the first move to totally dominate vertices in Y in her tth move. By Claim 9, the remainder tuple in X is totally dominated, at least (t − 2) full tuples are totally dominated and one extra full tuple contains at most one vertex not yet totally dominated after Dominator’s (t − 1)th move. By Dominator’s strategy, Dominator makes his tth move to totally dominate the full tuple in X which contains at most one vertex not yet totally dominated. Remark that Staller totally dominate the remainder tuple in Y on her tth move. Therefore, the result holds. Assume that Dominator makes the first move to totally dominate vertices in Y in his tth move. By Dominator’s strategy, we know that X is already totally dominated. By Claim 9, we know that at least (t − 2) full tuples are totally dominated after Dominator’s (t − 1)th move. Since X is not totally dominated after Dominator’s (t − 1)th move, t − 2 < l, that is l ≥ t − 1. So at least l ≥ t − 1 full tuples in X , the remainder tuples in X and Y are totally dominated after Dominator’s tth move. Thus the result holds by induction. ■ By Claim 10, we know that the remainder tuples in X and Y are totally dominated and at least (2l) full tuples are totally dominated after Dominator’s (2l + 1)th move. Hence at least (k + 1)(2l) + 2k = 2n vertices are totally dominated after Dominator has played his (2l + 1)th move. That is

γtg′ (G(k+1)l+k,k ) ≤ 4l + 2 = τ ′ ((k + 1)l + k, k). □ By Lemmas 2.1 to 2.4, Theorem 2.1 holds immediately. Let k = 2. By Theorem 2.1, we get the game total domination number and the Staller-start game total domination number of even cycles which were given in [9]. Corollary 2.1 ([9]). For all positive even numbers n ≥ 4, we have 4l

{

γtg (Cn ) = γtg (G 2n ,2 ) = 4l + 1 4l + 2

when n = 6l, when n = 6l + 2, when n = 6l + 4,

and

γtg′ (Cn ) = γtg′ (G 2n ,2 ) =

{

4l 4l + 2

when n = 6l, 6l + 2, when n = 6l + 4.

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