Domination in 4-regular Knödel graphs

Definition 1.1

([5]). For an even integer n ≥ 2 and 1 ≤ Δ ≤ ⌊log₂n⌋, a Knödel graph W_{Δ, n}is a Δ-regular bipartite graph of even order n, with vertices (i, j), for i = 1, 2 and 0 ≤ j ≤ n/2 − 1, where for every j, 0 ≤ j ≤ n/2 − 1, there is an edge between vertex (1, j) and each vertex (2, (j+2^k − 1) mod (n/2)), for k = 0, 1, ⋯, Δ − 1.

Knödel graphs, W_{Δ, n}, are one of the three important families of graphs that have nice properties in terms of broadcasting and gossiping. There exist many important papers presenting graph-theoretic and communication properties of the Knödel graphs, see for example [4, 6, 7, 8, 9]. It is worth noting that any Knödel graph is a Cayley graph and so it is a vertex-transitive graph [10].

Xueliang et al. [11] studied the domination number in 3-regular Knödel graphs W_3,n. They obtained exact domination number for W_3,n. Mojdeh et al. [14] determined the total domination number in 3-regular Knödel graphs W_3,n. In this paper, we determine the domination number in 4-regular Knödel graphs W_4,n. The following is useful.

Theorem 1.2

([12, 13]). For any graph G of order n with maximum degreeΔ(G),n1+Δ(G)≤γ(G)≤n−Δ(G).

We need also the following simple observation from number theory.

Observation 1.3

If a, b, c, d and x are positive integers such that x^a − x^b = x^c − x^d ≠ 0, then a = c and b = d.

Definition 2.1

For any subset A = {u_i1, u_i2, ⋯, u_ik} of U with 1 ≤ i₁ < i₂ < ⋯ < i_k ≤ n2we define a sequence n₁, n₂, ⋯, n_k, called a cyclic sequence, where n_j = i_j+1 − i_j for 1 ≤ j ≤ k − 1 and n_k = n2+i₁ − i_k. For two vertices u_ij, u_ij′ ∈ A, we define index-distance of u_ij and u_ij′by id(u_ij, u_ij′) = min{|i_j − i_j′|, n2 − |i_j − i_j′|}.

Observation 2.2

Let A = {u_i1, u_i2, ⋯, u_ik} ⊆ U be a set such that 1 ≤ i₁ < i₂ < ⋯ < i_k ≤ n2and let n₁, n₂, ⋯, n_k be the corresponding cyclic-sequence of A. Then,

1. n₁+n₂+⋯+n_k = n2.

2. If u_ij, u_ij′ ∈ A, then id(u_ij, u_ij′) equals sum of some consecutive elements of the cyclic-sequence of A andn2 − id(u_ij, u_ij′) is the sum of the remaining elements of the cyclic-sequence. Furthermore, {id(u_ij, u_ij′), n2 − id(u_ij, u_ij′)} = {|i_j − i_j′|, n2 − |i_j − i_j′|}.

Proof

1. By definition of cyclic-sequence, we have n₁+n₂+ ⋯ +n_k = (i₂ − i₁)+(i₃ − i₂)+⋯+(i_k − i_{k − 1})+( n2+i₁ − i_k) = n2 as desired.

2. By vertex transitivity of Knödel graphs, without loss of generality, we assume that j′ = 1. Since 1 = j′ < j ≤ k, |i_j − i_j′| = i_j − i₁ = (i_j − i_j−1)+(i_j−1 − i_j−2)+⋯+(i₂ − i₁). So, we have |i_j − i₁| = n_j−1+n_j−2+⋯+n₁ and by (1), we have n2 − |i_j − i₁| = n_j+n_j+1+⋯+n_k. This shows that both of |i_j − i₁| and n2 − |i_j − i₁| and consequently id(u_ij, u_i1) are the sum of some consequent elements of cyclic-sequence. The proof of {id(u_ij, u_ij′), n2 − id(u_ij, u_ij′)} = {|i_j − i_j′|, n2 − |i_j − i_j′|} is straightforward. □

   Let ℳ_Δ = {2^a − 2^b : 0 ≤ b < a < Δ} for Δ ≥ 2.

Lemma 2.3

In the Knödel graph W_{Δ, n}with vertex set U ∪ V, for every i ≠ j and 1 ≤ i, j ≤ n/2, N(u_i) ∩ N(u_j) ≠ ∅ if and only if id(u_i, u_j) ∈ ℳ_Δorn2 − id(u_i, u_j) ∈ ℳ_Δ.

Proof

Without loss of generality, assume that i > j.

First, suppose that v_t ∈ N(u_i) ∩ N(u_j). Therefore, there exist two integers 0 ≤ a, b ≤ Δ − 1 such that t ≡ i+2^a − 1 ≡ j+2^b − 1 (mod n/2). Hence, (i − j) − (2^b − 2^a) ≡ 0 (mod n/2). Since 0 < i − j < n/2 and − n/2 < 2^b − 2^a < n/2, we have only two different cases: (*) (i − j) − (2^b − 2^a) = 0 and so |i − j| = 2^b − 2^a ∈ ℳ_Δ and (**) (i − j) − (2^b − 2^a) = n/2 and so n/2 − |i − j| = 2^a − 2^b ∈ ℳ_Δ. Now, by Observation 2.2(2), we have id(u_i, u_j) ∈ ℳ_Δ or n2 − id(u_i, u_j) ∈ ℳ_Δ as desired.

To show the if part, suppose that id(u_i, u_j) ∈ ℳ_Δ or n2 − id(u_i, u_j) ∈ ℳ_Δ. By Observation 2.2(2), we have |i − j| = 2^a − 2^b ∈ ℳ_Δ or n/2 − |i − j| = 2^a − 2^b ∈ ℳ_Δ for some integers 0 ≤ b < a ≤ Δ − 1. In the first case, i+2^b − 1 = j+2^a − 1 and so v_t ∈ N(u_i) ∩ N(u_j), where t ≡ i+2^b − 1 = j+2^a − 1 (mod n/2). In the second case, i+2^a − 1 = n/2+j+2^b − 1 and so v_t ∈ N(u_i) ∩ N(u_j), where t ≡ i+2^a − 1 = n/2+j+2^b − 1 (mod n/2). In each case we have N(u_i) ∩ N(u_j) ≠ ∅ and proof is completed. □

Lemma 3.1

For each even integer n ≥ 16, we haveγ(W4,n)=2⌊n10⌋+0n≡0 (mod 10)2n≡2,4 (mod 10).

Proof

First assume that n ≡ 0 (mod 10). Let n = 10t, where t ≥ 2. By Theorem 1.2, γ(W_4,n) ≥ n5 = 2t. On the other hand, we can see that the set D = {u₁, u₆, ⋯, u_5t−4} ∪ {v₅, v₁₀}, ⋯, v_5t} is a dominating set with 2t elements, and we have γ (W_4,n) = 2t = 2⌊ n10⌋, as desired.

Next assume that n ≡ 2 (mod 10). Let n = 10t+2, where t ≥ 2. By Theorem 1.2, we have γ (W_4,n) ≥ n5 > 2t. Suppose that γ(W_4,n) = 2t+1. Let D be a minimum dominating set of W_4,n. Then by the Pigeonhole Principal either |D ∩ U| ≤ t or |D ∩ V| ≤ t. Without loss of generality, assume that |D ∩ U| ≤ t. Let |D ∩ U| = t − a, where a ≥ 0. Then |D ∩ V| = t+1+a. Observe that D ∩ U dominates at most 4t − 4a vertices of V and therefore D dominates at most (4t − 4a) + (t+1+a) = 5t − 3a+1 vertices of V. Since D dominates all vertices of V, we have 5t − 3a+1 ≥ 5t+1 and so a = 0, |D ∩ U| = t and |D ∩ V| = t+1. Let D ∩ U = {u_i1, u_i2, ⋯, u_it} and n₁, n₂, ⋯, n_t be the cyclic-sequence of D ∩ U. By Observation 2.2, we have ∑k=1tnk=5t+1 and, therefore, there exists some k such that n_k ∈ ℳ₄ = {1, 2, 3, 4, 6, 7}. Then by Lemma 2.3, |N(u_k) ∩ N(u_k+1) ≥ 1. Hence, D ∩ U dominates at most 4t − 1 vertices of V, that is, D dominates at most (4t − 1)+(t+1) = 5t vertices of V, a contradiction. Now we deduce that γ(W_4,n) ≥ 2t+2. On the other hand the set D = {u₁, u₆, ⋯, u_5t+1} ∪ {v₅, v₁₀}, ⋯, v_5t} ∪ {v_5t+1} is a dominating set for W_4,n with 2t+2 elements. Consequently, γ(W_4,n) = 2t+2.

It remains to assume that n ≡ 4 (mod 10). Let n = 10t+4, where t ≥ 2. By Theorem 1.2, we have γ (W_4,n) ≥ n5 > 2t. Suppose that γ(W_4,n) = 2t+1. Let D be a minimum dominating set of γ (W_4,n). Then by the Pigeonhole Principal either |D ∩ U| ≤ t or |D ∩ V| ≤ t. Without loss of generality, assume that |D ∩ U| = t − a and a ≥ 0. Then |D ∩ V| = t+1+a. Observe that D ∩ U dominates at most 4(t − a) elements of V and therefore D dominates at most 4(t − a)+(t+1+a) = 5t − 3a+1 vertices of V. Since D dominates all vertices of V, we have 5t − 3a+1 ≥ |V| = 5t+2 and so −3a ≥ 1, a contradiction. Thus γ (W_4,n) > 2t+1. On the other hand, the set {u₁, u₆, ⋯, u_5t+1} ∪ {v₅, v₁₀, ⋯, v_5t} ∪ {v₃} is a dominating set with 2t+2 elements. Consequently, γ (W_4,n) = 2t+2. □

Lemma 3.2

For each even integer n ≥ 46 with n ≡6 (mod 10), we have γ (W_4,n) = 2⌊ n10⌋+3.

Proof

Let n = 10t+6 and t ≥ 4. By Theorem 1.2, we have γ (W_4,n) ≥ n5 > 2t+1. The set D = {u₁, u₆, ⋯, u_5t+1} ∪ {v₅, v₁₀, ⋯, v_5t} ∪ {v₂, v₃} is a dominating set with 2t+3 elements. Thus, 2t+2 ≤ γ (W_4,n) ≤ 2t+3. We show that γ(W_4,n) = 2t+3. Suppose to the contrary that γ(W_4,n) = 2t+2. Let D be a minimum dominating set of W_4,n. Then by the Pigeonhole Principal either |D ∩ U| ≤ t+1 or |D ∩ V| ≤ t+1. Without loss of generality, assume that |D ∩ U| = t+1 − a, where a ≥ 0. Then |D ∩ V| = t+1+a. Note that D ∩ U dominates at most 4t+4 − 4a vertices and therefore D dominates at most (4t+4 − 4a)+(t+1+a) = 5t − 3a+5 vertices of V. Since D dominates all vertices of V, we have 5t − 3a+5 ≥ 5t+3 and so a = 0 and |D ∩ U| = |D ∩ V| = t+1. Also we have |V − D| = 4t+2 ≤ |N(D ∩ U)| ≤ 4t+4, and similarly, |D − U| = 4t+2 ≤ |N(D ∩ V)| ≤ 4t+4.

Let D ∩ U = {u_i1, u_i2, ⋯, u_it+1} and n₁, n₂, ⋯, n_t+1 be the cyclic-sequence of D ∩ U. By Observation 2.2, we have ∑k=1t+1nk = 5t+3 and, therefore, there exists k′ such that n_k′ < 5. Then n_k′ ∈ ℳ₄ and by Lemma 2.3, |N(u_ik′) ∩ N(u_ik′+1)| ≥ 1. Hence, D ∩ U dominates at most 4t+3 vertices from V and therefore 4t+2 ≤ N(D ∩ U) ≤ 4t+3.

If |N(D ∩ U)| = 4t+3, then for each k ≠ k′ we have n_k ∉ ℳ₄. If there exists k″ ≠ k′ such that n_k″ ≥ 8, then 5t+3 = Σk=1t+1nk ≥ 1+8+5(t − 1) = 5t+4, a contradiction. By symmetry we have n₁ = n₂ = ⋯ = n_t = 5, n_t+1 = 3 and D ∩ U = {u₁, u₆, ⋯, u_5t+1}. Observe that D ∩ U doesn’t dominate vertices v₃, v₅, v₁₀, ⋯, v_5t and so {v₃, v₅, v₁₀, ⋯, v_5t} ⊆ D. Thus D = {u₁, u₆, ⋯, u_5t+1, v₃, v₅, v₁₀, ⋯, v_5t}. But the vertices u₄, u₅, u_5t−2, u_5t+2 are not dominated by D, a contradiction.

Thus, |N(D ∩ U)| = 4t + 2. Then there exists precisely two pairs of vertices in D ∩ U with index-distances belonging to ℳ₄. If there exists an integer 1 ≤ i′ ≤ t+1 such that n_i′+n_i′+1 ∈ ℳ₄, then min{n_i′, n_i′+1} ≤ 3. Then min{n_i′, n_i′+1} ∈ ℳ₄ and \max{n_i′, n_i′+1} ∉ ℳ₄. Now we have max{n_i′, n_i′+1} = 5, min{n_i′, n_i′+1} ∈ {1, 2}, and n_i ∉ ℳ₄, for each i∉{i′, i′+1}. Now a simple calculation shows that the equality 5t+3 = ∑k=1t+1nk does not hold. (Note that if each n_i is less than 8, then we have ∑k=1t+1nk ≤ 2+5+5(t − 1) = 5t+2; otherwise we have ∑k=1t+1nk ≥ 1+5+8+5(t − 2) = 5t+4.) Thus there exist exactly two indices j and k such that n_j, n_k ∈ ℳ₄ and {n₁+n₂, n₂+n₃, ⋯, n_t+n_t+1, n_t+1+n₁} ∩ ℳ₄ = ∅. By this hypothesis, the only possible cases for the cyclic-sequence of D ∩ U are those demonstrated in Table 1.

Note that each column of Table 1 shows the cyclic sequence of D ∩ U. We show that each case is impossible. For this purpose, we show that the cyclic-sequence of D ∩ U posed in the column i, for i ≥ 1 is impossible.

i = 1). If n₁ = 8, n₂ = 1, n₃ = 4, n₄ = ⋯ = n_t+1 = 5 and D ∩ U = {u₁, u₉, u₁₀, u₁₄, u₁₉, ⋯, u_{5t − 1}}, then D ∩ U does not dominate the vertices v₅, v₆, v₇, v₁₈, v₂₃, v₂₈, …, v_5t+3.

Thus we have D ∩ V = {v₅, v₆, v₇, v₁₈, v₂₃, v₂₈, …, v_5t+3}. But D does not dominate three vertices u₈, u₁₂, u₁₃, a contradiction.

i = 2). If n₁ = 4, n₂ = 1, n₃ = 8, n₄ = ⋯ = n_t+1 = 5 and D ∩ U = {u₁, u₅, u₆, u₁₄, u₁₉, ⋯, u_{5t − 1}}. Then D ∩ U does not dominate the vertices v₁₀, v₁₁, v₁₆, v₁₈, v₂₃, v₂₈, …, v_5t+3. Thus we have D ∩ V = {v₁₀, v₁₁, v₁₆, v₁₈, v₂₃, v₂₈, …, v_5t+3}. But D does not dominate three vertices u₂, u₁₂, u_5t+1, a contradiction.

i ∈ {3, 4}). As before, we obtain that u₈ ∉ D. But N(u₈) = {v₈, v₉, v₁₁, v₁₅} ⊆ N(D ∩ U) and therefore N(u₈) ∩ D = ∅. Hence, N[u₈] ∩ D = ∅ and D does not dominate u₈, a contradiction.

i ≥ 5). As before, we obtain that u₃∉ D. But N(u₃) = {v₃, v₄, v₆, v₁₀} ⊆ N(D ∩ U) and therefore N(u₃) ∩ D = ∅. Hence, N[u₃] ∩ D = ∅ and D does not dominate u₃, a contradiction.

Consequently, γ(W_4,n) = 2t+3, as desired. □

Lemma 3.2 determines the domination number of W_4,n when n ≡ 6 (mod 10) and n ≥ 46. The only values of n for n ≡ 6 (mod 10) are thus 16, 26 and 36. We study these cases in the following lemma.

Lemma 3.3

For n ∈ {16, 26, 36}, we have:

Proof

For n = 16, by Theorem 1.2 we have γ (W_4,16) ≥ 165 > 3. On the other hand, the set D = {u₁, u₂, v₆, v₇} is a dominating set for W_4,16, and therefore γ (W_4,16) = 4.

For n = 26, by Theorem 1.2 we have γ (W_4,26) ≥ 265 > 5. On the other hand, the set D = {u₁, u₄, u₉, u₁₀, v₁, v₂, v₆} is a dominating set for W_4,26 and therefore 6 ≤ γ (W_4,26) ≤ 7. We show that γ (W_4,26) = 7. Suppose, on the contrary, that γ (W_4,26) = 6. Let D be a minimum dominating set for W_4,26. Then by the Pigeonhole Principal either |D ∩ U| ≤ 3 or |D ∩ V| ≤ 3. If |D ∩ U| = 3 − a, where a ≥ 0, then |D ∩ V| = 3+a. Now, the elements of D ∩ U dominate at most 4(3 − a) elements of V and D dominates at most 4(3 − a) + (3+a) = 15 − 3a vertices of V. Thus 15 − 3a ≥ |V| = 13, which implies a = 0 and |D ∩ U| = |D ∩ V| = 3. Let |D ∩ U| = {u₁, u_i, u_j}, where 1 < i < j ≤ 13 and n₁ = i − 1, n₂ = j − i, n₃ = 13+1 − j. Since n₁+n₂+n₃ = 13, we have {n₁, n₂, n₃} ∩ ℳ₄ ≠ ∅.

If ℳ₄ includes at least three numbers of n₁, n₂, n₃, n₁+n₂, n₂+n₃, n₃+n₁, then by Lemma 2.3, D ∩ U dominates at most 4 × 3 − 3 = 9 vertices of V and |D ∩ V| ≥ 13 − 9 = 4, a contradiction.

If ℳ₄ includes exactly one number of n₁, n₂, n₃, then we have, by symmetry, n₁ = 3, n₂ = 5, n₃ = 5 and D ∩ U = {u₁, u₄, u₉}, N(D ∩ U) = {v₁, v₂, v₃, v₄, v₅, v₇, v₈, v₉, v₁₀, v₁₁, v₁₂}. Then {v₆, v₁₃} ⊆ D. But {u₁, u₄, u₉, v₆, v₁₃} does not dominate the vertices {u₂, u₇, u₈, u₁₁} and we need at least 2 other vertices to dominate this four vertices, and hence |D| ≥ 7, a contradiction.

Thus, we assume that ℳ₄ includes two numbers of n₁, n₂, n₃ and {n₁+n₂, n₂+n₃, n₃+n₁} ∩ ℳ₄ = ∅. We thus have five possibilities for the cyclic-sequence of D ∩ U that are demonstrated in Table 3. Note that each column of Table 3 shows the cyclic sequence of D ∩ U. We show that each case is impossible. For this purpose, we show that the cyclic sequence of D ∩ U posed in the column i, for i ≥ 1 is impossible.

i = 1) If n₁ = 1, n₂ = 4, n₃ = 8, then D ∩ U = {u₁, u₂, u₆} and N(D ∩ U) = {v₁, v₂, ⋯, v₉, v₁₃}. Thus {v₁₀, v₁₁, v₁₂} ⊆ D and D = {u₁, u₂, u₆, v₁₀, v₁₁, v₁₂}. But D does not dominate the vertex u₁₃, a contradiction.

i = 2) If n₁ = 1, n₂ = 8, n₃ = 4, then D ∩ U = {u₁, u₂, u₁₀} and N(D ∩ U) = {v₁, v₂, ⋯, v₁₁, v₁₃}. Thus {v₆, v₇, v₁₂} ⊆ D and D = {u₁, u₂, u₁₀, v₆, v₇, v₁₂}. But D does not dominate the vertex u₈, a contradiction.

i = 3) If n₁ = 2, n₂ = 3, n₃ = 8, then D ∩ U = {u₁, u₃, u₆}, N(D ∩ U) = {v₁, v₂, ⋯, v₁₀, v₁₃}.Thus {v₅, v₁₁, v₁₂} ⊆ D and D = {u₁, u₃, u₆, v₅, v₁₁, v₁₂}. But D does not dominate the vertices u₇ and u₁₃.

i = 4) If n₁ = 2, n₂ = 8, n₃ = 3, then D ∩ U = {u₁, u₃, u₁₁} and N(D ∩ U) = {v₁, v₂, ⋯, v₆, v₈, v₁₀, v₁₁, v₁₂}. Thus {v₇, v₉, v₁₃} ⊆ D and D = {u₁, u₃, u₁₁, v₇, v₉, v₁₃}. But D does not dominate the vertex u₅, a contradiction.

i = 5) If n₁ = 4, n₂ = 4, n₃ = 5, then D ∩ U = {u₁, u₅, u₉} and N(D ∩ U) = {v₁, v₂, ⋯, v₁₀, v₁₂}. Thus {v₇, v₁₁, v₁₃} ⊆ D and D = {u₁, u₃, u₁₁, v₇, v₉, v₁₃}. But D does not dominate the vertices u₂ and u₃.

Consequently, γ(W_4,26) = 7.

We now consider the case n = 36. By Theorem 1.2, γ (W_4,36) ≥ 365 > 7. On the other hand, the set D = {u₁, u₂, u₁₀, u₁₁, v₆, v₇, v₁₅, v₁₆} is a dominating set for the graph W_4,36 and, therefore, γ (W_4,36) = 8. □

We now consider the case n ≡ 8 (mod 10). For n = 18, 28 and 38 we have the following lemma.

Lemma 3.4

For n ∈ {18, 28, 38}, we have:

Proof

For n = 18, by Theorem 1.2 we have γ (W_4,18) ≥ 185 > 3. But the set D = {u₁, u₂, v₆, v₇} is a dominating set for W_4,18 and, therefore, γ (W_4,18) = 4.

For n = 28, by Theorem 1.2 we have γ (W_4,28) ≥ 285 > 5 and the set D = {u₁, u₆, u₁₁, u₁₃, v₃, v₅, v₉} is a dominating set for W_4,28 and, therefore, 6 ≤ γ (W_4,28) ≤ 7. Suppose on the contradiction that γ (W_4,28) = 6 and D is a minimum dominating set for γ (W_4,28). By the Pigeonhole Principal, either |D ∩ U| ≤ 3 or |D ∩ V| ≤ 3. If |D ∩ U| = 3 − a and a ≥ 0, then |D ∩ V| = 3+a. Now, the elements of D ∩ U dominate at most 4(3 − a) elements of V and D dominates at most 4(3 − a)+(3+a) = 15 − 3a vertices of V. Thus 15 − 3a ≥ |V| = 14. Therefore, a = 0 and so |D ∩ U| = |D ∩ V| = 3. Let |D ∩ U| = {u₁, u_i, u_j} and 1 < i < j ≤ 14 and n₁ = i − 1, n₂ = j − i, n₃ = 14 + 1 − j. Since n₁+n₂+n₃ = 14, we have {n₁, n₂, n₃} ∩ ℳ₄ ≠ ∅. If ℳ₄ includes at least two numbers in {n₁, n₂, n₃, n₁+n₂, n₂+n₃, n₃+n₁}, then D ∩ U dominates at most 4 × 3 − 2 = 10 vertices of V and |D ∩ V| ≥ 14 − 10 = 4, a contradiction.

The only remaining case is that ℳ₄ includes exactly one of the three numbers n₁, n₂ and n₃ and also {n₁+n₂, n₂+n₃, n₃+n₁} ∩ ℳ₄ = ∅. By symmetry we have n₁ = 4, n₂ = 5, n₃ = 5 and D ∩ U = {u₁, u₅, u₁₀}, N(D ∩ U) = {v₁, v₂, v₃, v₄, v₅, v₆, v₈, v₁₀, v₁₁, v₁₂, v₁₃} thus {v₇, v₉, v₁₄} ⊆ D and D = {u₁, u₅, u₁₀, v₇, v₉, v₁₄}. But D does not dominate the vertex u₃. That is a contradiction and therefore γ (W_4,28) = 7. For n = 38, by Theorem 1.2 we have γ (W_4,38) ≥ 385 > 7 and the set D = {u₁, u₆, u₁₁, u₁₆, u₁₈, v₃, v₅, v₁₀, v₁₃, v₁₅} is a dominating set for W_4,38 and therefore 8 ≤ γ (W_4,38) ≤ 10. Let γ (W_4,38) < 10 and D is a dominating set for γ (W_4,38) with |D| = 9. Then by the Pigeonhole Principal either |D ∩ U| ≤ 4 or |D ∩ V| ≤ 4. If |D ∩ U| = 4 − a and a ≥ 0, then |D ∩ V| = 5+a. Now, the elements of D ∩ U dominate at most 4(4 − a) elements of V and D dominates at most 4(4 − a)+(5+a) = 21 − 3a vertices of V. Thus 21 − 3a ≥ |V| = 19 that results a = 0 and we have |D ∩ U| = 4 and |D ∩ V| = 5. Let |D ∩ U| = {u₁, u_i, u_j, u_k}, where 1 < i < j < k ≤ 19 and n₁, n₂, n₃, n₄ be the cyclic-sequence of D ∩ U. Since n₁+n₂+n₃+n₄ = 19, we have {n₁, n₂, n₃, n₄} ∩ ℳ₄ ≠ ∅. If ℳ₄ includes at least three numbers of {n₁, n₂, n₃, n₄, n₁+n₂, n₂+n₃, n₃+n₄, n₄+n₁}, then D ∩ U dominates at most 4 × 4 − 3 = 13 vertices of V and |D ∩ V| ≥ 19 − 13 = 6, a contradiction.

If we wish that ℳ₄ includes exactly one number out of n₁, n₂, n₃, n₄, we have three cases:

i = 1) If n₁ = 4, n₂ = 5, n₃ = 5, n₄ = 5 and D ∩ U = {u₁, u₅, u₁₀, u₁₅}, then {v₇, v₉, v₁₄, v₁₉} ⊆ D but {u₁, u₅, u₁₀, u₁₅, v₇, v₉, v₁₄, v₁₉} does not dominate the vertices u₃ and u₁₇. For dominating u₃ and u₁₇, we need two vertices and therefore |D| ≥ 10, a contradiction.

i = 2) If n₁ = 1, n₂ = 8, n₃ = 5, n₄ = 5 and D ∩ U = {u₁, u₂, u₁₀, u₁₅}, then {v₆, v₇, v₁₂, v₁₄, v₁₉} ⊆ D and D = {u₁, u₂, u₁₀, u₁₅, v₆, v₇, v₁₂, v₁₄, v₁₉} but D does not dominate the vertices u₈ and u₁₇.

i = 3) If n₁ = 8, n₂ = 1, n₃ = 5, n₄ = 5 and D ∩ U = {u₁, u₉, u₁₀, u₁₅}, then {v₅, v₆, v₇, v₁₄, v₁₉} ⊆ D and D = {u₁, u₉, u₁₀, u₁₅, v₅, v₆, v₇, v₁₄, v₁₉} but D does not dominate the vertex u₈.

Now we consider the cases that ℳ₄ includes exactly two numbers of the cyclic-sequence n₁, n₂, n₃, n₄ and {n₁+n₂, n₂+n₃, n₃+n₄, n₄+n₁} ∩ ℳ₄ = ∅. By symmetry we have ten cases:

i = 1) If n₁ = 1, n₂ = 9, n₃ = 1, n₄ = 8 and D ∩ U = {u₁, u₂, u₁₁, u₁₂}, then {v₆, v₇, v₁₀, v₁₆, v₁₇} ⊆ D and D = {u₁, u₂, u₁₁, u₁₂, v₆, v₇, v₁₀, v₁₆, v₁₇} but D does not dominate the vertex u₈.

i = 2) If n₁ = 2, n₂ = 8, n₃ = 1, n₄ = 8 and D ∩ U = {u₁, u₃, u₁₁, u₁₂}, then {v₅, v₇, v₉, v₁₆, v₁₇} ⊆ D and D = {u₁, u₃, u₁₁, u₁₂, v₅, v₇, v₉, v₁₆, v₁₇} but D does not dominate the vertex u₁₈.

i = 3) If n₁ = 4, n₂ = 1, n₃ = 9, n₄ = 5 and D ∩ U = {u₁, u₅, u₆, u₁₅}, then {v₁₀, v₁₁, v₁₄, v₁₇, v₁₉} ⊆ D and D = {u₁, u₅, u₆, u₁₅, v₁₀, v₁₁, v₁₄, v₁₇, v₁₉} but D does not dominate the vertex u₂.

i = 4) If n₁ = 9, n₂ = 1, n₃ = 4, n₄ = 5 and D ∩ U = {u₁, u₁₀, u₁₁, u₁₅}, then {v₅, v₆, v₇, v₉, v₁₉} ⊆ D and D = {u₁, u₁₀, u₁₁, u₁₅, v₅, v₆, v₇, v₉, v₁₉} but D does not dominate the vertices u₁₃ and u₁₄.

i = 5) If n₁ = 3, n₂ = 2, n₃ = 9, n₄ = 5 and D ∩ U = {u₁, u₄, u₆, u₁₅}, then {v₁₀, v₁₂, v₁₄, v₁₇, v₁₉} ⊆ D and D = {u₁, u₄, u₆, u₁₅, v₁₀, v₁₂, v₁₄, v₁₇, v₁₉} but D does not dominate the vertices u₂ and u₈.

i = 6) If n₁ = 9, n₂ = 2, n₃ = 3, n₄ = 5 and D ∩ U = {u₁, u₁₀, u₁₂, u₁₅}, then {v₅, v₆, v₇, v₉, v₁₄} ⊆ D and D = {u₁, u₁₀, u₁₂, u₁₅, v₅, v₆, v₇, v₉, v₁₄} but D does not dominate the vertex u₁₆.

i = 7) If n₁ = 8, n₂ = 3, n₃ = 5, n₄ = 3 and D ∩ U = {u₁, u₉, u₁₂, u₁₇}, then {v₃, v₆, v₇, v₁₁, v₁₄} ⊆ D and D = {u₁, u₉, u₁₂, u₁₇, v₃, v₆, v₇, v₁₁, v₁₄} but D does not dominate the vertex u₁₆.

i = 8) If n₁ = 3, n₂ = 6, n₃ = 5, n₄ = 5 and D ∩ U = {u₁, u₄, u₁₀, u₁₅}, then {v₆, v₉, v₁₂, v₁₄, v₁₉} ⊆ D and D = {u₁, u₄, u₁₀, u₁₅, v₆, v₉, v₁₂, v₁₄, v₁₉} but D does not dominate the vertex u₁₇.

i = 9) If n₁ = 3, n₂ = 5, n₃ = 6, n₄ = 5 and D ∩ U = {u₁, u₄, u₉, u₁₅}, then {v₆, v₁₃, v₁₄, v₁₇, v₁₉} ⊆ D and D = {u₁, u₄, u₉, u₁₅, v₆, v₁₃, v₁₄, v₁₇, v₁₉} but D does not dominate the vertices u₂ and u₈.

i = 10) If n₁ = 3, n₂ = 5, n₃ = 5, n₄ = 6 and D ∩ U = {u₁, u₄, u₉, u₁₄}, then {v₃, v₆, v₁₃, v₁₈, v₁₉} ⊆ D and D = {u₁, u₄, u₉, u₁₄, v₃, v₆, v₁₃, v₁₈, v₁₉} but D does not dominate the vertices u₇ and u₈.

Hence, W_4,38 has not any dominating set with 9 vertices and W_4,38 = 10 as desired. □

Now for n ≥ 48 with n ≡ 8 (mod 10) we determine the domination number of W_4,n as follows.

Lemma 3.5

For each even integer n ≥ 48, n ≡ 8 (mod 10), we have γ (W_4,n) = 2⌊ n10⌋+4.

Proof

Let n = 10t+8, where t ≥ 4. By Theorem 1.2, we have γ (W_4,n) ≥ n5 > 2t+1. The set D = {u₁, u₆, ⋯, u_5t+1} ∪ {v₅, v₁₀, ⋯, v_5t} ∪ {v₃, v_{5t − 2}, v_5t+3} is a dominating set with 2t+4 elements, and so, 2t+2 ≤ γ (W_4,n) ≤ 2t+4. We show that γ (W_4,n) = 2t+4.

First, assume that γ(W_4,n) = 2t+2. Let D be a minimum dominating set of W_4,n. Then by the Pigeonhole Principal either |D ∩ U| ≤ t+1 or |D ∩ V| ≤ t+1. Without loss of generality assume that |D ∩ U| = t+1 − a, where a ≥ 0. Then |D ∩ V| = t+1+a. Observe that D ∩ U dominates at most 4t+4 − 4a vertices of V, and therefore, D dominates at most (4t+4 − 4a)+(t+1+a) = 5t − 3a+5 vertices of V. Since D dominates all vertices in V, we have 5t − 3a+5 ≥ 5t+4 and so a = 0. Then |D ∩ U| = |D ∩ V| = t+1. Also we have |V − D| = 4t+3 ≤ |N(D ∩ U)| ≤ 4t+4 and |U − D| = 4t+3 ≤ |N(D ∩ V)| ≤ 4t+4. Let D ∩ U = {u_i1, u_i2, ⋯, u_it+1} and n₁, n₂, ⋯, n_t+1. By Observation 2.2, we have Σk=1t+1nk = 5t+4 and therefore there exist k′ such that n_k′ ∈ ℳ₄. By Lemma 2.3, D ∩ U dominates at most 4t+3 vertices from V. Then |N(D ∩ U)| = |N(D ∩ V)| = 4t+3 and k′ is unique. If there exists 1 ≤ k″ ≤ t+1 such that n_k″ ≥ 8, then 5t+4 = Σk=1t+1nk ≥ n_k′+n_k″+5(t − 1) ≥ 1+8+5(t − 1) = 5t+4 which implies n_k′ = 1, n_k″ = 8 and for each k∉{k′, k″} we have n_k = 5. Now in each arrangement of the cyclic sequence of D ∩ U, we have one adjacency between 1 and 5. Then we have two vertices in D ∩ U with index-distance equal to 6, a contradiction. Thus for k ≠ k′ we have n_k = 5 and n_k′ = 4. We have (by symmetry) n₁ = n₂ = ⋯ = n_t = 5 and n_t+1 = 4 and D ∩ U = {u₁, u₆, ⋯, u_5t+1}. Now D ∩ U doesn’t dominate the vertices v₃, v₅, v₁₀, ⋯, v_5t and so {v₃, v₅, v₁₀, ⋯, v_5t} ⊆ D. Thus D = {u₁, u₆, ⋯, u_5t+1, v₃, v₅, v₁₀, ⋯, v_5t}, but D does not dominate two vertices u_5t−2 and u_5t+3, a contradiction.

Now, assume that γ(W_4,n) = 2t+3. Let D be a minimum dominating set of W_4,n. Then by the Pigeonhole Principal either |D ∩ U| ≤ t+1 or |D ∩ V| ≤ t+1. Without loss of generality, suppose |D ∩ U| = t+1 − a, where a ≥ 0. Then |D ∩ V| = t+2+a. Observe that D ∩ U dominates at most 4t+4 − 4a vertices of V and, therefore, D dominates at most (4t+4 − 4a)+(t+2+a) = 5t − 3a+6 vertices of V. Since D dominates all vertices in V, we have 5t − 3a+6 ≥ 5t+4 and a = 0, |D ∩ U| = t+1 and |D ∩ V| = t+2. Also we have 4t+2 ≤ |N(D ∩ U)| ≤ 4t+4. Since 4t+2 ≤ |N(D ∩ U)| ≤ 4t+4, at most two elements of n₁, n₂, ⋯, n_t+1 can be in ℳ₄. If x is the number of 5′s in the cyclic-sequence of D ∩ U, then by Observation 2.2, we have Σk=1t+1nk = 5t+4 ≥ 1+1+8(t − x − 1)+5x and, therefore, 3x ≥ 3t − 10 which implies x ≥ t − 3. Thus t − 3 elements of the cyclic sequence are equal to 5. The sum of the remaining four values of the cyclic sequence is 19, and at most two of them are in ℳ₄. In the last case of Lemma 3.3, for n = 38, we identified all such cyclic sequences and placed them in two tables, Table 5 and Table 6. We now continue according to Table 5 and Table 6.

In the case (i = 1) in Table 5 we have n₁ = 4, n₂ = ⋯ = n_t+1 = 5 and D ∩ U = {u₁, u₅, u₁₀, ⋯, u_5t}. Thus {v₇, v₉, v₁₄, ⋯, v_5t+4} ⊆ D. But {u₁, u₅, u₁₀, ⋯, u_5t, v₇, v₉, v₁₄, ⋯, v_5t+4} does not dominate the vertices u₃ and u_5t+2. For dominating u₃ and u_5t+2, we need two vertices and therefore |D| ≥ 2t+4, a contradiction. In the cases (i ∈{2, 3}) in Table 5 we have to add 5′s to the end of the cyclic sequence and construct the corresponding set D with 2t+3 elements. In both cases we obtain that N[u₈] ∩ D = ∅. Then D is not a dominating set, a contradiction.

In the case (i = 1) in Table 6 we cannot add a 5 to the cyclic sequence, since by adding a 5 to the cyclic-sequence we obtain two consecutive values of the cyclic sequence which one is 5 and the other is 1 and their sum is 6 which belongs to ℳ₄, a contradiction.

In the case (i = 2) in Table 6 we cannot add a 5 to the cyclic sequence, since by adding a 5 to the cyclic sequence we obtain two consecutive values of the cyclic sequence which one is 5 and the other is 1 or 2, and their sum is 6 or 7, which belongs to ℳ₄, a contradiction.

In the cases (i ∈ {3, 4, 5, 6}) in Table 6 we have to add 5′s to the end of the cyclic sequence and construct the corresponding set D with 2t+3 elements. In (i = 3), we obtain that N[u₂] ∩ D = ∅, in (i = 4), we obtain that N[{u₁₃, u₁₄}] ∩ D = ∅, in (i = 5), we obtain that N[{u₂, u₈}] ∩ D = ∅, and in (i = 6), we obtain that N[u₁₆] ∩ D = ∅. In all four cases, D is not a dominating set, a contradiction.

In the case (i = 7) in Table 6, by adding 5′s to the cyclic-sequence, we obtain some different new cyclic-sequences. We divide them into three categories.

1. n₁ = 8, n₂ = 3 and n₃ = 5. In this category, the constructed set, D, does not dominate u₁₆, a contradiction.

2. n₁ = 8, n₂ = 5, n₃ = 3 and n₄ = 5. In this category, the constructed set, D, does not dominate u₁₅ a contradiction.

3. n₁ = 8, n₂ = n₃ = 5 and if n_i = n_j = 3, then |i − j| ≥ 2. In this category, the constructed set, D, does not dominate u_5i+1, a contradiction. (Notice that this category does not appear for n ≤ 5.) In the cases (i ∈{8, 9, 10}) in Table 6, by adding 5’s to the cyclic-sequences, we obtain some different new cyclic-sequences. we divide them into two categories.

4. n₁ = 3 and n_t+1 = 5. In this category, the constructed set, D, does not dominate u₂, a contradiction.

5. n₁ = 3 and n_t+1 = 6. In this category, the constructed set, D, does not dominate u₇ and u₈. Hence, D is not a dominating set, a contradiction.

Hence, γ (W_4,n) = 2t+4 = 2⌊ n10⌋+4 □

Now a consequence of Lemmas 3.1, 3.2, 3.33, 3.4 and 3.5 implies the following theorem which is the main result of this section.

Theorem 3.6

For each integer n ≥ 16, we have:

Problem 1

Obtain the domination number of k-regular Knödel graphs for k ≥ 5.

Problem 2

Obtain the total domination number of k-regular Knödel graphs for k ≥ 4.

Problem 3

Obtain the connected domination number of k-regular Knödel graphs for k ≥ 3.

Problem 4

Obtain the independent domination number of k-regular Knödel graphs for k ≥ 3.