On graphs with equal dominating and c-dominating energy

Let G be the 5-vertex path P₅, with vertices v₁, v₂, v₃, v₄, v₅ and let its minimum connected dominating set be D_c = {v₂, v₃, v₄}. Then

ADc(G)=0100011100011100011100010 $$\begin{array}{} A_{D_{c}}(G) = \left( \begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \right) \end{array}$$

The characteristic polynomial of A_Dc(G) is λ⁵ – 3λ⁴ – λ³ + 5λ² + λ – 1 = 0. The minimum connected dominating eigenvalues are λ₁ = 2.618, λ₂ = 1.618, λ₃ = 0.382, λ₄ = –1.000 and λ₅ = –0.618.

Therefore, the minimum connected dominating energy is E_Dc(P₅) = 6.236.

Consider the following graph

Figure 2

Let G be a tree T as shown above and let its minimum connected dominating set be D_c = {b, d, e, f, g, h}. Then

ADc(G)=0100000000111100000001000000000101100000000111000000001110000000011101000000111000000001000000001000 $$\begin{array}{} A_{D_{c}}(G) = \left( \begin{array}{cccccccccc} 0 &1& 0& 0& 0& 0& 0& 0& 0& 0 \\ 1 &1& 1& 1& 0& 0& 0& 0& 0& 0 \\ 0 &1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0 &1& 0& 1& 1& 0& 0& 0& 0& 0 \\ 0 &0& 0& 1& 1& 1& 0& 0& 0& 0\\ 0 &0& 0& 0& 1& 1& 1& 0& 0& 0 \\ 0 &0& 0& 0& 0& 1& 1& 1& 0& 1 \\ 0 &0& 0& 0& 0& 0& 1& 1& 1& 0 \\ 0 &0& 0& 0& 0& 0& 0& 1& 0& 0 \\ 0 &0& 0& 0& 0& 0& 1& 0& 0& 0 \\ \end{array} \right) \end{array}$$

By direct calculation, we get the minimum connected dominating eigenvalues are λ₁ = 2.945, λ₂ = 2.596, λ₃ = 1.896, λ₄ = 1.183, λ₅ = –1.263, λ₆ = –1.152, λ₇ = 0.579, λ₈ = 0.000, λ₉ = –0.268 and λ₁₀ = –0.516.

Therefore, the minimum connected dominating energy is E_Dc(T) = 12.398.

The c-dominating energy of the following graphs can be calculated easily:

E_Dc(K_n) = (n – 2) + n(n−<!-- - -->2)+5 $\begin{array}{} \sqrt{n(n-2) + 5} \end{array}$, where K_n is the complete graph of order n.

In this paper, we are interested in studying the mathematical aspects of the c-dominating energy of a graph. This paper has organized as follows: The section 1, contains the basic definitions and background of the current topic. In section 2, we show the chemical applicability of c-dominating energy for molecular graphs G. The section 3, contains the mathematical properties of c-dominating energy. In the last section, we have characterized, trees, unicyclic graphs and cubic graphs and block graphs with equal minimum dominating energy and c-dominating energy. Finally, we conclude this paper by posing an open problem.

Note that the trace of A_Dc(G) = γ_c(G).

Let G = (V, E) be a graph with γ_c-set D_c. Let f_n(G, λ) = c₀λⁿ + c₁λ^n–1 + ⋯ + c_n be the characteristic polynomial of G. Then

c₀ = 1,

If λ₁, λ₂, ⋯, λ_n are the eigenvalues of A_Dc(G), then

∑<!-- ? -->i=1nλ<!-- ? -->i=γ<!-- ? -->c(G) $\begin{array}{} \displaystyle \sum\limits_{i=1}^n \lambda_i = \gamma_c(G) \end{array}$

Follows from Observation 1.

Let G be a graph of order n and size m with γ_c(G) = k. Then

EDc(G)≤<!-- = -->n(2m+k). $$\begin{array}{} \begin{split} \displaystyle E_{D_c}(G) & \le & \sqrt{n(2m+k)}. \end{split} \end{array}$$(2)

Let λ₁ ≥ λ₂ ≥ ⋯ ≥ λ_n be the eigenvalues of A_Dc(G). Bearing in mind the Cauchy-Schwarz inequality,

(∑<!-- ? -->i=1naibi)2≤<!-- = -->(∑<!-- ? -->i=1nai)2(∑<!-- ? -->i=1nbi)2 $$\begin{array}{} \displaystyle \bigg(\sum\limits_{i=1}^n a_i b_i \bigg)^2 \le \bigg(\sum\limits_{i=1}^n a_i \bigg)^2 \bigg(\sum\limits_{i=1}^n b_i \bigg)^2 \end{array}$$

we choose a_i = 1 and b_i = |λ_i|, which by Theorem 3 implies

EDc2=(∑<!-- ? -->i=1n|λ<!-- ? -->i|)2≤<!-- = -->n(∑<!-- ? -->i=1n|λ<!-- ? -->i|2)=n∑<!-- ? -->i=1nλ<!-- ? -->i2=2(2m+k). $$\begin{array}{} \begin{split} \displaystyle E_{D_c}^2 &=& \bigg(\sum\limits_{i=1}^n|\lambda_i| \bigg)^2 \\ & \leq & n\bigg(\sum\limits_{i=1}^n|\lambda_i|^2 \bigg) \\ &=& n \sum\limits_{i=1}^n \lambda^2_i \\ &=& 2(2m + k). \end{split} \end{array}$$

□

Let G be a graph of order n and size m with γ_c(G) = k. Let λ₁ ≥ λ₂ ≥ ⋯ ≥ λ_n be a non-increasing arrangement of eigenvalues of A_Dc(G). Then

EDc(G)≥<!-- = -->2mn+nk−<!-- - -->α<!-- a -->(n)(|λ<!-- ? -->1|−<!-- - -->|λ<!-- ? -->n|)2 $$\begin{array}{} \begin{split} \displaystyle E_{D_c}(G) & \ge & \sqrt{2mn+nk-\alpha(n)(|\lambda_1|-|\lambda_n|)^2} \end{split} \end{array}$$

where α<!-- a -->(n)=n[n2](1−<!-- - -->1n[n2]) $\begin{array}{} \alpha(n) = n [\frac{n}{2}](1-\frac{1}{n}[\frac{n}{2}]) \end{array}$, where [x] denotes the integer part of a real number k.

Let a₁, a₂, ⋯, a_n and b₁, b₂, ⋯, b_n be real numbers for which there exist real constants a, b, A and B, so that for each i, i = 1, 2, ⋯, n, a ≤ a_i ≤ A and b ≤ b_i ≤ B. Then the following inequality is valid (see [6]).

∣<!-- | -->n∑<!-- ? -->i=1naibi−<!-- - -->∑<!-- ? -->i=1nai∑<!-- ? -->i=1nbi∣<!-- | -->≤<!-- = -->α<!-- a -->(n)(A−<!-- - -->a)(B−<!-- - -->b), $$\begin{array}{} \begin{split} \displaystyle \mid n\sum\limits_{i =1}^n a_ib_i-\sum\limits_{i =1}^n a_i \sum\limits_{i=1}^n b_i \mid & \le & \alpha(n)(A-a)(B-b), \end{split} \end{array}$$(4)

where α<!-- a -->(n)=n[n2](1−<!-- - -->1n[n2]) $\begin{array}{} \alpha(n) = n [\frac{n}{2}](1-\frac{1}{n}[\frac{n}{2}]) \end{array}$. Equality holds if and only if a₁ = a₂ = ⋯ = a_n and b₁ = b₂ = ⋯ = b_n.

We choose a_i := |λ_i|, b_i := |λ_i|, a = b := |λ_n| and A = B := |λ₁|, i = 1, 2, ⋯, n, inequality (4) becomes

|n∑<!-- ? -->i=1n|λ<!-- ? -->i|2−<!-- - -->(∑<!-- ? -->i=1n|λ<!-- ? -->i|)2|≤<!-- = -->α<!-- a -->(n)(|λ<!-- ? -->1|−<!-- - -->|λ<!-- ? -->n|)2. $$\begin{array}{} \begin{split} \displaystyle |n \sum\limits_{i=1}^{n}|\lambda_{i}|^2-\bigg(\sum\limits_{i=1}^n|\lambda_i| \bigg)^2| & \le & \alpha(n)(|\lambda_1|-|\lambda_n|)^2. \end{split} \end{array}$$(5)

Since EGc(G)=∑<!-- ? -->i=1n|λ<!-- ? -->i|,∑<!-- ? -->i=1n|λ<!-- ? -->i|2=∑<!-- ? -->i=1n|λ<!-- ? -->i|2=2m+k $\begin{array}{} \displaystyle E_{G_c}(G) = \sum\limits_{i=1}^n|\lambda_i|,~ \sum\limits_{i=1}^n |\lambda_i|^2 = \sum\limits_{i=1}^n |\lambda_{i}|^2 = 2m+k \end{array}$ and EDc(G)≤<!-- = -->n(2m+k), $\begin{array}{} \displaystyle E_{D_c}(G) \le \sqrt{n(2m+k)}, \end{array}$ the inequality (5) becomes

n(2m+k)−<!-- - -->(EDc)2≤<!-- = -->α<!-- a -->(n)(|λ<!-- ? -->1|−<!-- - -->|λ<!-- ? -->n|)2(EDc)2≥<!-- = -->2mn+nk−<!-- - -->α<!-- a -->(n)(|λ<!-- ? -->1|−<!-- - -->|λ<!-- ? -->n|)2. $$\begin{array}{} \begin{split} \displaystyle n(2m+k)-(E_{D_c})^2 & \le & \alpha(n)(|\lambda_1|-|\lambda_n|)^2 \\ (E_{D_c})^2 & \ge & 2mn+nk-\alpha(n)(|\lambda_1|-|\lambda_n|)^2. \end{split} \end{array}$$

Hence equality holds if and only if λ₁ = λ₂ = ⋯ = λ_n.□

Let G be a graph of order n and size m with γ_c(G) = k. Let λ₁ ≥ λ₂ ≥ ⋯ ≥ λ_n be a non-increasing arrangement of eigenvalues of A_Dc(G). Then

EDc(G)≥<!-- = -->2mn+nk−<!-- - -->n24(|λ<!-- ? -->1|−<!-- - -->|λ<!-- ? -->n|)2. $$\begin{array}{} \begin{split} \displaystyle E_{D_c}(G) & \geq & \sqrt{2mn+nk-\frac{n^2}{4}(|\lambda_1|-|\lambda_n|)^2}. \end{split} \end{array}$$

Since α<!-- a -->(n)=n[n2](1−<!-- - -->1n[n2])≤<!-- = -->n24 $\begin{array}{} \alpha(n) = n [\frac{n}{2}](1-\frac{1}{n}[\frac{n}{2}]) \le \frac{n^2}{4} \end{array}$, therefore by (3), result follows.□

Let G be a graph of order n and size m with γ_c(G) = k. Let λ₁ ≥ λ₂ ≥ ⋯ ≥ λ_n be a non-increasing arrangement of eigenvalues of A_Dc(G). Then

EGc(G)≥<!-- = -->|λ<!-- ? -->1||λ<!-- ? -->2|n+2m+k|λ<!-- ? -->1|+|λ<!-- ? -->n|. $$\begin{array}{} \begin{split} \displaystyle E_{G_c}(G) & \geq & \frac{|\lambda_1||\lambda_2|n+2m+k}{|\lambda_1|+|\lambda_n|}. \end{split} \end{array}$$(7)

Let a₁, a₂, ⋯, a_n and b₁, b₂, ⋯, b_n be real numbers for which there exist real constants r and R so that for each i, i = 1, 2, ⋯, n holds ra_i ≤ b_i ≤ Ra_i. Then the following inequality is valid (see [11]).

∑<!-- ? -->i=1nbi2+rR∑<!-- ? -->i=1nai2≤<!-- = -->(r+R)∑<!-- ? -->i=1naibi. $$\begin{array}{} \begin{split} \displaystyle \sum\limits_{i=1}^n b^2_i + rR\sum\limits_{i=1}^n a^2_i & \le & (r+R)\sum\limits_{i=1}^n a_i b_i. \end{split} \end{array}$$(8)

Equality of (8) holds if and only if, for at least one i, 1 ≤ i ≤ n holds ra_i = b_i = Ra_i.

For b_i := |λ_i|, a_i := 1 r := |λ_n| and R := |λ₁|, i = 1, 2, ⋯, n inequality (8) becomes

∑<!-- ? -->i=nn|λ<!-- ? -->i|2+|λ<!-- ? -->1||λ<!-- ? -->n|∑<!-- ? -->i=1n1≤<!-- = -->(|λ<!-- ? -->1|+|λ<!-- ? -->n|)∑<!-- ? -->i=1n|λ<!-- ? -->i|. $$\begin{array}{} \begin{split} \displaystyle \sum\limits_{i=n}^{n}|\lambda_i|^2+|\lambda_1||\lambda_n|\sum\limits_{i=1}^{n}1 \le (|\lambda_1|+|\lambda_n|)\sum\limits_{i=1}^n|\lambda_i|. \end{split} \end{array}$$(9)

Since ∑<!-- ? -->i=1n|λ<!-- ? -->i|2=∑<!-- ? -->i=1nλ<!-- ? -->i2=2m+k,∑<!-- ? -->i=1n|λ<!-- ? -->i|=EDc(G), $\begin{array}{} \displaystyle \sum\limits_{i=1}^{n}|\lambda_i|^2 = \sum\limits_{i=1}^{n}\lambda_i^2 = 2m+k,~ \sum\limits_{i=1}^n|\lambda_i| = E_{D_c}(G), \end{array}$ from inequality (9),

2m+k+|λ<!-- ? -->1||λ<!-- ? -->n|n≤<!-- = -->(λ<!-- ? -->1+λ<!-- ? -->n)EDc(G) $$\begin{array}{} \begin{split} \displaystyle 2m+k+|\lambda_1||\lambda_n|n & \le & (\lambda_1+\lambda_n)E_{D_c}(G) \end{split} \end{array}$$

Hence the result.□

Let G be a graph of order n and size m with γ_c(G) = k. If ξ = |detA_Dc(G)|, then

EDc(G)≥<!-- = -->2m+k+n(n−<!-- - -->1)ξ<!-- ? -->2n. $$\begin{array}{} \begin{split} \displaystyle E_{D_c}(G) & \ge & \sqrt{2m+k+n(n-1)\xi^\frac{2}{n}}. \end{split} \end{array}$$(10)

(EDc(G))2=(∑<!-- ? -->i=1n|λ<!-- ? -->i|)2=∑<!-- ? -->i=1n|λ<!-- ? -->i|2+∑<!-- ? -->i≠<!-- ? -->j|λ<!-- ? -->i||λ<!-- ? -->j|. $$\begin{array}{} \begin{split} \displaystyle (E_{D_c}(G))^2 &=& \bigg(\sum\limits_{i=1}^n |\lambda_i| \bigg)^2 \\ &=& \sum\limits_{i=1}^n |\lambda_i|^2+\sum\limits_{i \ne j} |\lambda_i||\lambda_j|. \end{split} \end{array}$$

Employing the inequality between the arithmetic and geometric means, we obtain

1n(n−<!-- - -->1)∑<!-- ? -->i≠<!-- ? -->j|λ<!-- ? -->i||λ<!-- ? -->j|≥<!-- = -->(∏<!-- ? -->i≠<!-- ? -->j|λ<!-- ? -->i||λ<!-- ? -->j|)1n(n−<!-- - -->1). $$\begin{array}{} \begin{split} \displaystyle \frac{1}{n(n-1)} \sum\limits_{i \ne j} |\lambda_i||\lambda_j| & \ge & \bigg(\prod\limits_{i \ne j} |\lambda_i||\lambda_j| \bigg)^{\frac{1}{n(n-1)}}. \end{split} \end{array}$$

Thus,

(EDG)2≥<!-- = -->∑<!-- ? -->i=1n|λ<!-- ? -->i|2+n(n−<!-- - -->1)(∏<!-- ? -->i≠<!-- ? -->j|λ<!-- ? -->i||λ<!-- ? -->j|)1n(n−<!-- - -->1)≥<!-- = -->∑<!-- ? -->i=1n|λ<!-- ? -->i|2+n(n−<!-- - -->1)(∏<!-- ? -->i≠<!-- ? -->j|λ<!-- ? -->i|2(n−<!-- - -->1))1n(n−<!-- - -->1)=2m+k+n(n−<!-- - -->1)ξ<!-- ? -->2n. $$\begin{array}{} \begin{split} \displaystyle (E_{D_G})^2 & \ge & \sum\limits_{i=1}^n |\lambda_i|^{2}+n(n-1)\bigg(\prod\limits_{i \ne j} |\lambda_i||\lambda_j| \bigg)^{\frac{1}{n(n-1)}} \\ & \ge & \sum\limits_{i=1}^n|\lambda_i|^{2}+n(n-1)\bigg(\prod\limits_{i \ne j} |\lambda_i|^{2(n-1)} \bigg)^{\frac{1}{n(n-1)}} \\ &=& 2m+k+n(n-1)\xi^{\frac{2}{n}}. \end{split} \end{array}$$

□

If λ₁(G) is the largest minimum connected dominating eigenvalue of A_Dc(G), then λ<!-- ? -->1≥<!-- = -->2m+γ<!-- ? -->c(G)n. $\begin{array}{} \lambda_1 \ge \frac{2m+\gamma_c(G)}{n}. \end{array}$

Let X be any non-zero vector. Then we have λ<!-- ? -->1(A)=maxX≠<!-- ? -->0{X′AXX′X}, $\begin{array}{} \lambda_1(A) = max_{_{X \ne 0}} \{\frac{X' A X}{X' X}\}, \end{array}$ see [16]. Therefore, λ₁(A_Dc(G)) ≥ J′AJJ′J=2m+γ<!-- ? -->c(G)n. $\begin{array}{} \frac{J'AJ}{J'J} = \frac{2m+\gamma_c(G)}{n}. \end{array}$□

If G is a graph of order n and size m and 2m + γ_C(G) ≥ n, then

EDc(G)≤<!-- = -->2m+γ<!-- ? -->c(G)n+(n−<!-- - -->1)[(2m+γ<!-- ? -->c(G))−<!-- - -->(2m+γ<!-- ? -->c(G)n)2]. $$\begin{array}{} \begin{split} \displaystyle E_{D_c}(G) & \le & \frac{2m+\gamma_c(G)}{n}+\sqrt{(n-1)\bigg[(2m+\gamma_c(G))-\bigg(\frac{2m+\gamma_c(G)}{n}\bigg)^{2} \bigg]}. \end{split} \end{array}$$(11)

Bearing in mind the Cauchy-Schwarz inequality,

(∑<!-- ? -->i=1naibi)2≤<!-- = -->(∑<!-- ? -->i=1nai)2(∑<!-- ? -->i=1nbi)2. $$\begin{array}{} \displaystyle \bigg(\sum\limits_{i=1}^n a_i b_i \bigg)^2 \le \bigg(\sum\limits_{i=1}^{n}a_i \bigg)^2 \bigg(\sum\limits_{i=1}^n b_i \bigg)^2. \end{array}$$

Put a_i = 1 and b_i = |λ_i| then

(∑<!-- ? -->i=2naibi)2≤<!-- = -->(n−<!-- - -->1)(∑<!-- ? -->i=2nbi)2(EDc(G)−<!-- - -->λ<!-- ? -->1)2≤<!-- = -->(n−<!-- - -->1)(2m+γ<!-- ? -->c(G)−<!-- - -->λ<!-- ? -->12)EDc(G)≤<!-- = -->λ<!-- ? -->1+(n−<!-- - -->1)(2m+γ<!-- ? -->c(G)−<!-- - -->λ<!-- ? -->12). $$\begin{array}{} \begin{split} \displaystyle \bigg(\sum\limits_{i=2}^n a_i b_i \bigg)^2 & \le & (n-1)\bigg(\sum\limits_{i=2}^n b_i \bigg)^2 \\ (E_{D_c}(G)-\lambda_1)^2 & \le &(n-1)(2m+\gamma_c(G)-\lambda^2_1) \\ E_{D_c}(G) & \le & \lambda_1+\sqrt{(n-1)(2m+\gamma_c(G)-\lambda^2_1)}. \end{split} \end{array}$$

Let

f(x)=x+(n−<!-- - -->1)(2m+γ<!-- ? -->c(G)−<!-- - -->x2). $$\begin{array}{} \begin{split} \displaystyle f(x) &=& x+\sqrt{(n-1)(2m+\gamma_c(G)-x^2)}. \end{split} \end{array}$$(12)

For decreasing function

f′(x)≤<!-- = -->0⇒<!-- ? -->1−<!-- - -->x(n−<!-- - -->1)(n−<!-- - -->1)(2m+γ<!-- ? -->c(G)−<!-- - -->x2)≤<!-- = -->0x≥<!-- = -->2m+γ<!-- ? -->c(G)n. $$\begin{array}{} \begin{split} \displaystyle f'(x) & \le & 0 \\ \Rightarrow 1-\frac{x(n-1)}{\sqrt{(n-1)(2m+\gamma_c(G)-x^2)}} & \le & 0 \\ x & \ge & \sqrt{\frac{2m+\gamma_c(G)}{n}}. \end{split} \end{array}$$

Since (2m + k) ≥ n, we have 2m+γ<!-- ? -->c(G)n≤<!-- = -->2m+γ<!-- ? -->c(G)n≤<!-- = -->λ<!-- ? -->1. $\begin{array}{} \sqrt{\frac{2m+\gamma_{c}(G)}{n}} \le \frac{2m+\gamma_c(G)}{n} \le \lambda_1. \end{array}$ Also f(λ<!-- ? -->1)≤<!-- = -->f(2m+γ<!-- ? -->c(G)n). $\begin{array}{} f(\lambda_1) \le f\bigg(\frac{2m+\gamma_c(G)}{n} \bigg). \end{array}$

i.eEDc(G)≤<!-- = -->f(λ<!-- ? -->1)≤<!-- = -->f(2m+γ<!-- ? -->c(G)n).i.eEDc(G)≤<!-- = -->f(2m+γ<!-- ? -->c(G)n) $$\begin{array}{c} \displaystyle \text{i.e}~~ E_{D_c}(G) \le f(\lambda_1) \le f\bigg( \frac{2m+\gamma_c(G)}{n} \bigg). \\\displaystyle \text{i.e}~~ E_{D_c}(G) \le f\bigg( \frac{2m+\gamma_c(G)}{n} \bigg) \end{array}$$

Hence by (12), the result follows.□

Let G = T be a tree with at least three vertices, then E_D(G) = E_Dc(G) if and only if every internal vertex of T is a support vertex.

Let G = T be a tree of order at least 3. Let F = {u₁, u₂, ⋯, u_k} be the set of internal vertices of T. Then clearly F is the minimal dominating set of G. Therefore in A_D(G) the values of u_i = 1 in the diagonal entries. Further, observe that 〈F〉 is connected. Hence F is the minimal connected dominating set. Therefore, A_D(G) = A_Dc(G). In general, A_D(G) = A_Dc(G) is true if every minimum dominating set is connected. In other words, A_D(G) = A_Dc(G) if γ(G) = γ_c(G). Therefore, the result follows from Theorem 2.1 in [2].□

Let G be a unicyclic graph with cycle C = u₁u₂ ⋯, u_nu₁ n ≥ 5 and let X = {v ∈ C : d_G(v) ≥ 2}. Then E_D(G) = E_Dc(G) if the following conditions hold:

(a). Every v ∈ V – N[X] with d_G(V) ≥ 2 is a support vertex.

Let G be unicyclic graph with |V(G)| ≥ 4 containing a cycle C = C₃, and let X = {v ∈ C : d_G(v) = 2}. Then E_D(G) = E_Dc(G) if the following conditions hold:

(a). Every v ∈ V – N[X] with d_G(V) ≥ 2 is a support vertex.

Let G be unicyclic graph with |V(G)| ≥ 5 containing a cycle C = C₄, and let X = {v ∈ C : d_G(v) = 2}. Then E_D(G) = E_Dc(G) if the following conditions hold:

(a). Every v ∈ V – N[X] with d_G(V) ≥ 2 is a support vertex.

Let G be a connected cubic graph of order n, Then E_D(G) = E_Dc(G) if G ≅ K₄, C₆,K_3,3, G₁ or G₂ where G₁ and G₂ are given in Fig. 4.

Let G be a block graph of with l ≥ 2. Then E_D(G) = E_Dc(G) if every cutvertex of G is an end block cutvertex.

Since E_D(G) = E_Dc(G) if γ(G) = γ_c(G). Therefore, the result follows from Theorem 2 in [7].□