Connected domination game played on Cartesian products

Lemma 1.1

(Chooser Lemma). Consider the connected domination game with Chooser on a graph G. Suppose that in the game Chooser picks k vertices, and that both Dominator and Staller play optimally. Then at the end of the game the number of played vertices is at most γ_cg(G) + k and at least γ_cg(G) – k.

Theorem 2.1

[13,Theorem 1] If G is a connected graph, then γ_cg(G) ≤ 2γ_c(G) – 1.

Proof

Let ℓ = γ_c(G) and let D = {x₁, …, x_ℓ} be a connected dominating set in G, where for every i ∈ [ℓ], the set {x₁, …, x_i} induces a connected subgraph of G. As D is a connected dominating set, such an ordering of the vertices of D is clearly possible.

The strategy of Dominator is to consecutively play the vertices x₁, …, x_ℓ, in particular, his first move is on the vertex x₁. Suppose now that x_i, i ≤ 2, is the next vertex to be played by Dominator according to his strategy and that he is not able to play it.

It is possible that Dominator cannot play x_i because x_i was already played by Staller in the course of the game. If this is the case, Dominator simply plays x_i+1, provided this is a legal move. If it is not and also x_i+1 was already played by Staller, we continue this process until we reach a vertex x_j, j ≥ i, that was neither played earlier by Staller nor it is playable by Dominator. Because of the latter fact, all the vertices from N[x_j] are already dominated at this stage of the game. If each of the vertices x_j+1, …, x_ℓ was either already played by Staller or is unplayable by Dominator, then the set of vertices selected so far constitutes a connected dominating set of G. Indeed, all the vertices from each of N[x₁], …, N[x_ℓ] are dominated, hence G is dominated because {x₁, …, x_ℓ} is a dominating set. Moreover, it is a connected dominating set because the strategy of Dominator preserves the connectedness (and Staller must follow it anyway).

Assume therefore that there exits a vertex x_k, j + 1 ≤ k ≤ ℓ, that is undominated and let k be the smallest possible index of such a vertex. If x_k is adjacent to some vertex already played, then Dominator plays x_k. Otherwise, consider an arbitrary neighbor x_s of x_k from the sequence x₁, …, x_k–1. If x_s was played in the game played so far, then Dominator can extend the game by playing x_k. So suppose that x_s was not played before. This means that x_s was (and is) an unplayable vertex. In particular, its neighbor x_k must have been dominated by some vertex already played. But this means again that Dominator can play x_k because it is adjacent to an already played vertex.

By the above strategy of Dominator, the game is finished no later than at the moment when Dominator plays x_ℓ. If this happens and if Dominator played all the vertices x₁, …, x_ℓ, then the number of vertices played is exactly 2ℓ – 1 = 2γ_c(G) – 1. If, however, some of the vertices x_i were played by Staller or became unplayable during the game, the number of vertices played is even smaller.□

Theorem 2.2

If G and H are connected graphs, then

Proof

Let X_G be a connected dominating set of G with |X_G| = γ_c(G). Since X_G induces a connected subgraph of G, the set of vertices X_G × V(H) induces a connected subgraph of G ◻ H and hence it is a connected dominating set of G ◻ H. Therefore, γ_c(G ◻ H) ≤ 2γ_c(G)n(H) and thus γ_cg(G ◻ H) ≤ 2γ_c(G)n(H) – 1 by Theorem 2.1. By a parallel argument (or by referring to the commutativity of the Cartesian product) we also have γ_cg(G ◻ H) ≤ 2γ_c(H)n(G) – 1.□

Corollary 2.3

[13, Theorem 5] If G is a connected graph, then γ_cg(G ◻ K_m) ≤ min {2mγ_c(G)) – 1, 2n(G) – 1}.

Theorem 3.1

If n ≥ 3 and m ≥ 1, then

Proof

The statement is clear for m = 1 and easily verified for m = 2. For m = 3 it holds γ_cg(K_1,n ◻ P₃) = 5. Indeed, the upper bound follows from Theorem 2.2. The lower bound can be shown with a simple case analysis. In the remaining part of the proof, we consider m ≥ 4.

Consider a path P_m = u₁ … u_m (m ≥ 4) and a star K_1,n on the vertex set v₁, …, v_n+1 (n ≥ 3) where v_n+1 is the center. In K_1,n ◻ P_m, the layer K1,nuj will be denoted simply by K1,nj . In a connected domination game on K_1,n ◻ P_m, let A_i (resp. Ai′ ) denote the set of integers j ∈ [m] for which at least one vertex from the layer K1,nj was played within the moves d₁, s₁, …, d_i (resp. d₁, s₁, …, d_i, s_i). By connectivity, both A_i and Ai′ are sets of consecutive integers. We will use the notation A_i = [a_i, b_i] and Ai′=[ai′,bi′]. By Theorem 2.2, we have γ_cg(K_1,n ◻ P_m) ≤ 2m – 1. To prove the lower bound we describe a strategy of Staller that maintains the following property for every i.

1. After the move s_i of Staller, either at least two vertices are played from every layer K1,nj with j∈[ai′,bi′]∖ {1, m}, or there is only one layer K1,nj∗ (j^* = ai′ > 1 or j^* = bi′ < m) among them which is played only once. In the latter case, the vertex played in the layer K1,nj∗ is a leaf of the layer, and there are at least two leaves played from the neighboring layer.

Property (⋆) and the strategy of Staller will ensure that from every K1,nj , with 2 ≤ j ≤ m – 1, at least two vertices will be played during the game. The below rule (S1) in Staller’s strategy clearly establishes (⋆) after s₁. Then, we will suppose that it is true before Dominator’s move d_i (i ≥ 2). Note that, under this condition, we may have at most two layers K1,nj with 2 ≤ j ≤ m – 1 which are played exactly once after the move d_i.

1. If Dominator plays the center of a layer K1,nj as his first move d₁, Staller replies by playing the leaf (v₁, u_j) of the same star; if Dominator plays a leaf from K1,nj , Staller picks the center (v_n+1, u_j) as s₁.

Later, if Dominator plays the first vertex from K1,nbi as d_i, that is, if b_i = bi−1′ , and b_i < m, then Staller replies according to the following rules:

1. If at least two vertices from the neighboring layer K1,nbi−1 have been played in the game so far, Staller chooses a vertex from K1,nbi . This clearly can be done as at most one vertex from K1,nbi+1 has been dominated so far.

2. If only one vertex from K1,nbi−1 has been played so far, Staller chooses a leaf from K1,nbi−1 . Such a legal move exists as (⋆) was true after the move s_i–1.

The rules are analogous (and we do not repeat them) for the case when Dominator plays the first vertex from K1,nai as d_i (and a_i > 1). If Dominator plays a vertex from K1,nm or K1,n1 as d_i, or he plays a vertex from a layer K1,nj which has been played earlier (i.e.,j∈[ai−1′,bi−1′]), Staller replies as follows:

1. If b_i < m and K1,nbi is played only once, Staller chooses a vertex from K1,nbi . This clearly can be done as at most one vertex from K1,nbi+1 has been dominated so far. The rule is similar for the case when a_i > 1 and K1,nai is played only once.

2. If both K1,nbi and K1,nai are played at least twice, Staller plays a vertex from K1,nbi or K1,nai , if such a legal move exists.

3. If there are no playable vertices in K1,nbi (and b_i < m), then all vertices of the layer have been played. If the situation is similar for K1,nai , Staller plays the leaf (v₁, u_bi+1) if b_i < m – 1, and she plays (v₁, u_ai–1) if b_i ≥ m – 1 and a_i > 2.

What remains is only the case when all the layers K1,nj with 2 ≤ j ≤ m – 1 have been played at least twice.

1. Staller plays a vertex from K1,nm−1 or from K1,n2 . Note that it is always possible if the game is not yet over.

One can check that property (⋆) is preserved if Staller applies the above strategy. Remark that, by (S5) and (S6), Staller plays a first vertex from a K1,nj only if all vertices of the neighboring layer have already been played. Moreover, if Staller plays the first vertex in a layer, then this vertex corresponds to a leaf of the star. Consequently, if we have two neighboring layers which are played only once, then these played vertices must be leaves. In particular, we may have the following two situations before the end of the game:

1. If Staller finishes the game and in the last turn a vertex from K1,nm becomes dominated, then the center of this star is not played and, therefore, at least n + 1 vertices were played from these two stars. Moreover, every layer K1,nj , 2 ≤ j ≤ m – 2, is played at least twice and the domination of K1,n1 needs at least one further vertex. We may infer that at least n + 1 + 2(m – 3) + 1 ≥ 2m – 1 vertices were played during the game. (The argumentation is similar if a vertex from K1,n1 was dominated in the last turn.)

2. If Dominator finishes the game, at least two vertices are chosen from every K1,nj , if 2 ≤ j ≤ m – 1. The domination of K1,n1 and K1,nm needs at least two further vertices. Hence, at least 2m – 2 vertices were played. On the other hand, the number of chosen vertices must be odd as, by our assumption, Dominator finishes the game. This proves γ_cg(K_1,n ◻ P_m) ≥ 2m – 1.

This finishes the proof of the equality.□

Theorem 3.2

If n ≥ 3 and m ≥ 3, then

Proof

For m = 3, it can easily be checked that γ_cg(K_1,n ◻ C₃) = 5. Thus we only consider m ≥ 4.

By Theorem 2.2, γ_cg(K_1,n ◻ C_m) ≤ 2m – 1. To prove the lower bound, we use a similar strategy of Staller as in the proof of Theorem 3.1. We also use the same notation, but now the addition is done modulo m (with values in [m]), so the sets A_i and Ai′ are sets of consecutive integers modulo m. The main difference is only that in the strategy on K_1,n ◻ P_m layers K1,n1 and K1,nm are studied differently, but on the K_1,n ◻ C_m the special K_1,n-layers are the ones which are played last or not played at all. Similarly, all the conditions like 1 < ai′ and bi′ < m must be replaced with |bi′−ai′|≤m−3. Apart from that, Staller’s strategy and the final counting of the moves remains the same.□

Theorem 4.1

If m ≥ 4, then

Proof

Let P₃ = v₁v₂v₃ and P_m = u₁ … u_m. We will also use the notations introduced in the proof of Theorem 3.1.

To prove that γ_cg(P₃ ◻ P_m) ≤ 2m – 2, we consider the following strategy of Dominator in a connected domination game with Chooser and apply the Chooser Lemma 1.1. Let d₁ = (v₂, u₂). Then, Staller has four different legal replies.

1. If Staller responds with the move s₁ = (v₁, u₂), then Chooser picks the vertices (v₂, u₃), …, (v₂, u_m–1). Dominator’s strategy is to play d₂ = (v₂, u_m). Then, only the vertex (v₃, u₁) remains undominated and the game finishes with a set of m + 1 played vertices. By Chooser Lemma, we have t ≤ (m + 1) + (m – 3) = 2m – 2, where t is the length of the connected domination game under the described strategies of the players.

2. The reply s₁ = (v₃, u₂) is treated analogously as the move s₁ = (v₁, u₂).

3. If s₁ = (v₂, u₃), Chooser picks (v₂, u₄), …, (v₂, u_m) and Dominator plays d₂ = (v₂, u₁). By Chooser Lemma, we have t ≤ m + (m – 3) = 2m – 3.

4. If s₁ = (v₂, u₁), Chooser picks (v₂, u₃), …, (v₂, u_m–1) and Dominator plays d₂ = (v₂, u_m). By Chooser Lemma, we have t ≤ m + (m – 3) = 2m – 3, again.

This proves that Dominator can ensure t ≤ 2m – 2 under any strategy of Staller, that is, γ_cg(P₃ ◻ P_m) ≤ 2m – 2.

To prove the other direction, that is, γ_cg(P₃ ◻ P_m) ≥ 2m – 2, we consider the strategy (S1)–(S7) of Staller and property (⋆) as these are given in the proof of Theorem 3.1. In fact, as P₃ = K_1,2, we have the same line of the proof except the final counting which is the following for the case of P₃ ◻ P_m. By Staller’s strategy, at least two vertices were played from every P3j with 2 ≤ j ≤ m – 1. Moreover, the domination of V(P31)∪V(P32) and V(P3m−1)∪V(P3m) needs at least two further vertices. Hence, we get γ_cg(P₃ ◻ P_m) ≥ 2(m – 2) + 2 = 2m – 2 and the theorem follows.□

Theorem 4.2

If m ≥ 4, then

Proof

Let m ≥ 4 and let V(C_m) = [m]. All the exponents of the P₃-layers will be taken modulo m.

First, we show that γ_cg(P₃ ◻ C_m) ≤ 2m – 2. The basic strategy of Dominator is to play a legal move in the center of some P₃-layer. Clearly this is possible in the D-game. Inductively, let P3i,…,P3j,i≤j, be the P₃-layers in which at least one vertex has been played so far. We may assume without loss of generality that Staller’s last move was in P3j . If this move was in the center of P3j , then Dominator replies with the move on the center of P3j+1 . Otherwise, Dominator selects the center of P3j . After m – 1 moves of Dominator we have two cases. If all the P₃-layers were played, then by the strategy of Dominator, the whole graph is dominated, hence at most 2m – 3 moves were played. Otherwise, one P₃-layer has not been played, say P3j . In this case Dominator has played all the centers of the remaining m – 1 P₃-layers. But then Staller has played at least one non-center vertex from the layers P3j−1 and P3j+1 . Consequently, at most one vertex from P3j is not yet dominated after the (m – 1)st move of Dominator, which means that in the next move Staller must finish the game. Hence the game finishes in no more than 2m – 2 moves.

Proving that γ_cg(P₃ ◻ C_m) ≥ 2m – 2 goes along the same lines as in the proof for P₃ ◻ P_m in Theorem 4.1, where the strategy of Staller is modified as in the proof of Theorem 3.2.□

Theorem 5.1

If G is a connected graph and k ≥ min {4Δ(G) + α(G), 2n(G) – 1}, then

Proof

Let G be a connected graph on n vertices with maximum degree Δ and with independence number α. During the game, a K_k-layer will be called played if at least one vertex has been already played from it. A K_k-layer will be called dominated if all of its vertices have been dominated, otherwise, it is undominated. Clearly, if a K_k-layer is played, then it is dominated. Observe that, by Theorem 2.2, we have γ_cg(K_k ◻ G) ≤ 2n – 1.

First suppose that k ≥ 2n – 1 and consider the following simple strategy of Staller: she always chooses a (legal) vertex from a dominated K_k-layer. While the game is not over, Staller can follow this strategy. Indeed, by the connectivity of G, if there is an undominated K_k-layer in K_k ◻ G, then we have two neighboring layers Kkg and Kkg′ (i.e., gg′ ∈ E(G)) such that Kkg is dominated but a vertex (h, g′) ∈ V( Kkg′ ) is undominated. Then, Staller may play (h, g). Hence, after the (n – 1)st move of Staller, at most n – 1 K_k-layers are played. Each remaining unplayed layer is certainly not dominated since it would need at least k ≥ 2n – 1 vertices played in the neighboring layers. This proves γ_cg(K_k ◻ G) ≥ 2n – 1 under the condition k ≥ 2n – 1.

Now, suppose that k ≥ 4Δ + α and consider a connected domination game on K_k ◻ G. We assign blue, green or red color to those vertices which were played in the game. If Dominator plays a vertex d_i = v from a K_k-layer such that the layer was undominated before his move, the vertex d_i is colored blue. If d_i is blue and Staller plays s_i from the same K_k-layer as her next move, then s_i is blue, otherwise s_i is green. If Dominator plays d_i from a dominated layer, then both d_i and s_i are red. We say that a K_k-layer is a blue layer if it contains a blue vertex.^[1] A K_k-layer is white if it remains unplayed at the end of the game and also if it becomes dominated before it is played. When the game finishes, each K_k-layer is either blue or white and, denoting the number of blue or white layers by ℓ_b or ℓ_w respectively, we have ℓ_b + ℓ_w = n.

Consider the following strategy of Staller: After Dominator’s move d_i, Staller plays a legal vertex from the same K_k-layer if possible; otherwise, Staller plays a legal vertex from any dominated layer. As we have seen earlier, the latter choice is always possible (if the game is not over) as there must be an undominated layer which has a neighboring dominated layer.

Now, we prove the following claims.

Claim 1

At the end of the game, the number of green vertices is at most α.

Proof

Suppose that s_i, which is the ith move of Staller, is a green vertex. Then d_i, the previous move of Dominator, was the first vertex played from the undominated layer Kkg and no further vertex could be played from Kkg after his move. The latter property means that every neighboring layer was already dominated (either played or not) right after the move d_i. Consequently, no blue vertex can be played from the neighboring layers in the game after d_i. Therefore, if every green vertex s_j of the product is associated with the vertex f(s_j) from V(G) such that d_j ∈ V( Kkf(sj) ) holds for the previous move of Dominator, the set F = {f(s_j) : s_j is green} is an independent vertex set in G and |F| ≤ α follows.(◻)

Claim 2

If Kkg is a white layer, then at least 2Δ red vertices were played from the neighboring layers.

Proof

Since Kkg is white, its k vertices are dominated by k different (blue, green, red) vertices of the neighboring layers. (This remains true, even if later some vertices from Kkg are played.) By definition, every neighboring K_k-layer may contain at most two blue vertices and there are at most Δ neighboring layers. By Claim 1, there exist at most α green vertices. Hence, the number of red vertices in the neighboring layers is at least k – 2Δ – α. Since k ≥ 4Δ + α, the number of such red vertices is at least 2Δ.(◻)

Now, denote by r and p the number of red vertices and played vertices, respectively, at the end of the game. The number of edges between the red vertices and the white layers is at least 2Δ ⋅ ℓ_w by Claim 2. On the other hand, since Δ is the maximum vertex degree, the number of these edges is at most Δ ⋅ r. This yields 2ℓ_w ≤ r. Moreover, p – r, which is the total number of blue and green vertices, equals either 2ℓ_b or 2ℓ_b – 1. (The latter occurs if Dominator finishes the game by playing a blue vertex.) Hence, 2ℓ_b – 1 ≤ p – r. We have the following inequalities:

and, since ℓ_b + ℓ_w = n, we may conclude 2n – 1 ≤ p. This finishes the proof of γ_cg(K_k ◻ G) = 2n(G) – 1 under the condition k ≥ 4Δ + α.□

Theorem 5.2

If n ≥ 1, then

Proof

We prove the theorem using imagination strategy. Two games will be played in the proof, one on K_n+1 ◻ G and one on K_n ◻ G. In order to avoid confusion between vertices of those two graphs, we introduce the notation x_[n] = {x₁, …, x_n}.

The real game is played on K_n+1 ◻ G with vertex set v_[n+1] × V(G), while Staller imagines a game on K_n ◻ G with vertex set u_[n] × V(G). Define a function f : v_[n] × V(G) → u_[n] × V(G) by f : (v_i, g) ↦ (u_i, g) for every i ∈ [n] and g ∈ V(G), with the inverse f^–1 : (u_i, g) ↦ (v_i, g). Even though f(v_n+1, g) is not formally defined, we will simply say that it is not a legal move in the imagined game on K_n ◻ G. Additionally, in the real game the label v_n+1 is given to a vertex of K_n+1 such that the underlying G-layer was played last among all G-layers (or not played at all).

Dominator plays optimally in the real game and Staller plays optimally in the imagined game. Let D_R and D_I denote the set of already played vertices in the real and in the imagined game, respectively. Staller’s strategy is to ensure that until the imagined games ends it holds (i) |D_R| = |D_I| and (ii) f(N[D_R] ∩ (v_[n] × V(G))) ⊆ N[D_I]. If this is true, then the number of moves needed to finish the real game is at least the number of moves needed in the imagined game. As Staller plays optimally in the imagined game and Dominator plays optimally in the real game, we have γ_cg(K_n ◻ G) ≤ γ_cg(K_n+1 ◻ G).

Consider the strategy of Staller. She always replies optimally in the imagined game and tries to copy her move to the real game. By the property (ii), this move surely dominates new vertices in the real game, but might not be connected to a previous move in the real game. Let x = s_i be the optimal move of Staller in the imagined game. If f^–1(x) is a legal move in the real game, then Staller copies it to the real game (clearly preserving properties (i) and (ii)). But if f^–1(x) is not a legal move in the real game, then let y be the move in the imagined game that first dominated x. Now set x = y and iteratively repeat the above procedure until Staller reaches a vertex x such that f^–1(x) is a legal move in the real game. Clearly, the procedure stops (as the graph is finite) and properties (i) and (ii) are preserved.

Let d_i = (v_k, g), k ∈ [n + 1], g ∈ V(G), be an (optimal) move of Dominator in the real game. Staller copies his move to the imagined game by the following rules.

1. If f(d_i) is a legal move in the imagined game, then Staller imagines Dominator played f(d_i) = (u_k, g) in the imagined game.

2. If f(d_i) is not a legal move, but there exists a playable vertex p in Kng in the imagined game, then Staller imagines Dominator played p.

3. If neither (D1) nor (D2) can be applied, then Staller imagines Dominator played any legal move in the imagined game.

All the rules clearly preserve (i) and the rule (D1) also preserves (ii).

If k ∈ [n], but f(d_i) is not a legal move, then either f(d_i) dominates no new vertex in the imagined game or it is not connected to a previously played vertex. In the first case, the property (ii) still holds after Staller imagines the ith move of Dominator by the rule (D2) or (D3). We now explain that the second case is not possible. Let x be the already played neighbor of d_i in the real game. By the property (ii), N[x] and f(N[x] ∩ (v_[n] × V(G))) are both dominated, in particular, d_i and f(d_i) are both dominated. Thus a neighbor of f(d_i) was already played in the imagined game.

Next, consider the case k = n + 1. By the definition of the vertex v_n+1, the move d_i cannot be the first move of the game. By induction on the number of moves on ^v_n+1G, we will prove that the property (ii) is preserved and also that if Staller cannot apply (D2) at Dominator’s move d_i, then the whole set u_[n] × N[g] is already dominated in the imagined game.

Firstly, consider the first move d_i0 = (v_n+1, g₀) played on ^v_n+1G in the real game. Property (ii) is preserved no matter where Staller imagines his move. Suppose that Staller cannot apply (D2) after the move d_i0. If a vertex was already played in Kng0 in the imagined game, then Staller cannot apply (D2) only if u_[n] × N[g₀] is already dominated (otherwise she could play another move on Kng0 ). So from now on, suppose also that no vertex was already played in Kng0 in the imagined game. As d_i0 is a legal move of Dominator in the real game, it is adjacent to an already played vertex x = (v_k, g₀), k ∈ [n], in the real game. Then x was a move of Dominator in the real game, N[(u_k, g₀)] is already dominated in the imagined game, in particular Kng0 is already dominated. Thus all vertices in Kng0 are adjacent to an already played vertex in the imagined game. As none of them is a legal move, it means that u_[n] × N[g₀] is already dominated.

Next, consider the case when d_i is some later move on ^v_n+1G. If Staller can apply (D2), then the property (ii) is clearly satisfied. Suppose now that Staller cannot apply (D2). If d_i is adjacent to an already played vertex in v_[n] × {g} in the real game, then the proof goes along the same lines as above for d_i0. So the only remaining case is if d_i is not adjacent to an already played vertex in v_[n] × {g} in the real game. This means it is adjacent to an already played vertex d′ = (v_n+1, g′), g ≠ g′ ∈ V(G), in the real game and no vertex was played so far in v_[n] × {g} in the real game. We distinguish two cases.

The first case is that d_i dominates no new vertex in v_[n] × {g}. Then this set is already dominated in the real game and u_[n] × {g} is dominated in the imagined game, thus the connectedness condition is satisfied for every vertex in u_[n] × {g}. As Staller cannot apply (D2), u_[n] × N[g] is already dominated and property (ii) holds.

The second case is that d_i dominates a new vertex in v_[n] × {g}. If u_[n] × {g} is already dominated in the imagined game, then Staller could apply rule (D2), unless u_[n] × N[g] is already dominated. Otherwise, there is an undominated vertex in u_[n] × {g} in the imagined game. This means that a move (u_l, g′) was played in the imagined game on u_[n] × {g′} (otherwise u_[n] × {g} ⊆ u_[n] × N[g′] would be already dominated). But then Staller can imagine Dominator plays (u_l, g) in the imagined game to dominate the undominated vertex in u_[n] × {g}, hence applying (D2).□

Theorem 6.1

If G and H are connected graphs on at least two vertices, then

Proof

The vertices of V(G ◻ H) will be denoted as {(g, h); g ∈ G, h ∈ H}. Let A_i (resp. Ai′ ) denote the set of all g ∈ G such that at least one vertex in the layer ^gH has been played within moves d₁, s₁, …, d_i (resp. d₁, s₁, …, d_i, s_i). Let B_i ⊆ A_i be the smallest connected subset of A_i that dominates N[A_i]. We similarly define Bi′ ⊆ Ai′ . Note that at each step (and at the end of the game) it holds |A_i| ≥ |B_i| and | Ai′ | ≥ | Bi′ |.

Staller’s strategy is to play s_i on a vertex above B_i, meaning that s_i ∈ B_i × V(H). If this is not possible, she plays any legal move. We prove that this strategy assures that after each move of Staller it holds i ≥ | Bi′ |.

As n(H) ≥ 2, Staller can make her first move in the same H-layer where Dominator played, thus 1 ≥ | B1′ |. Suppose the property holds for the first i – 1 moves of Staller and consider the situation after Dominator’s ith move, say d_i = (g, h). We now distinguish two cases.

1. The vertex d_i newly dominates only vertices above A_i (i.e. d_i ∈ A_i × V(H)).

   In this case, g is not added to B_i, so B_i = Bi−1′ . No matter where Staller replies, we have |B_i| ≤ | Bi′ | ≤ |B_i| + 1 = | Bi−1′ | + 1. But then i – 1 ≥ | Bi−1′ | ≥ | Bi′ | – 1, which implies i ≥ | Bi′ |.

2. The vertex d_i dominates at least one vertex not above A_i, say a vertex (g′, h).

   1. Bi−1′ dominates N[A_i].

      In this case B_i = Bi−1′ and i ≥ | Bi′ | as in the previous case.

   2. Bi−1′ does not dominate N[A_i].

      This means that at least the vertex g′ ∈ N[A_i] is not dominated by Bi−1′ , so no vertex in ^g′H, except for (g′, h), is dominated after Dominator’s ith move. We have B_i = Bi−1′ ∪ {g} and d_i is the only already played vertex in ^gH. As n(H) ≥ 2, Staller can reply in the same H-layer to newly dominate a vertex in ^g′H. Thus | Bi′ | = |B_i| = | Bi−1′ | + 1 ≤ (i – 1) + 1 = i.

If Dominator is the one who finishes the game in his jth move, then his last move is not as in the case 2.(b) above (otherwise Staller has a legal reply and the game is not over). Thus j – 1 ≥ | Bj−1′ | = |B_j| ≥ γ_c(G), which means j ≥ γ_c(G) + 1. Hence the number of moves is at least j + (j – 1) ≥ 2 γ_c(G) + 1.

On the other hand, if Staller ends the game with her jth move, then her strategy assures that j ≥ | Bj′ | ≥ γ_c(G), thus the number of moves is at least 2 γ_c(G). This completes the proof in the case n(H) = 2.

If n(H) ≥ 3, consider again the situation where Staller ended the game. {We distinguish two cases. If the last vertex was added to the set Bj′ not in the last two moves of the game but earlier, then clearly the number of moves is greater than 2 γ_c(G) + 1. Otherwise, in the last two steps, one of the players added the last vertex to the set Bj′ . If the move d_j added the last vertex to Bj′ = B_j, then this situation is as in case 2.(b). Thus after the moves d_j, s_j, at most 2 < n(H) vertices in the layer ^g′H are dominated. But then the game is not yet finished and at least one more move is needed. On the other hand, if the move s_j added a vertex to Bj′ ≠ B_j (and it follows from the above cases that this can only occur if Dominator did not add a vertex to B_j), then in the newly dominated H-layer only 1 < n(H) vertex is dominated, hence at least one more move is needed. It follows that if n(H) ≥ 3, the game ends after at least 2 γ_c(G) + 1 moves.□

Problem 7.1

Find additional infinite families of graphs which attain the upper bound of Theorem 2.1.

Problem 7.2

Determine γ_cg(P_n ◻ P_m), γ_cg(P_n ◻ C_m), and γ_cg(C_n ◻ C_m).

Problem 7.3

Does there exist a constant c ∈ ℝ⁺, such that Theorem 5.1 remains valid provided that 4Δ + α is replaced with cΔ?

Problem 7.4

Given a graph class 𝓖 that does not contain disconnected graphs, determine the constant

Problem 7.5

Do there exist families of graphs for which the lower bound of Theorem 6.1 is sharp and none of the factors is K₂ or K₃?