On Extended Adjacency Index with Respect to Acyclic, Unicyclic and Bicyclic Graphs

Abstract

For a (molecular) graph G, the extended adjacency index $EA\left(G\right)$ is defined as Equation (1). In this paper we introduce some graph transformations which increase or decrease the extended adjacency ($EA$) index. Also, we obtain the extremal acyclic, unicyclic and bicyclic graphs with minimum and maximum of the $EA$ index by a unified method, respectively.

1. Introduction

Molecular descriptors are playing an important role in Chemistry, Pharmacology, etc. Among them, topological indices have a prominent place. Topological indices (molecular structure descriptor) are numerical quantities of a molecular graphs (or simple graphs), that are invariant under graph isomorphism. And, are used to correlate with various physico chemical properties, chemical reactivity or biological activity. There are hundreds of topological indices that have found some applications in theoretical chemistry, especially in QSPR/QSAR research. Among all topological indices one of the most investigated are the degree based topological indices, among them, the old and widely studied topological index is Randić index [1], see the recent articles [2,3] and references cited there in. Recently researchers are studying various degree based topological indices such as Zagreb group indices [4,5,6,7,8,9], forgotten index [10,11,12,13], etc.

Let $G=(V,E)$ be a simple graph without loops and multiple edges. Let $V\left(G\right)$ and $E\left(G\right)$ be the vertex set and the edge set of G, respectively. The degree of a vertex u in G is the number of edges incident to it and is denoted by $d_G\left(u\right)$. For $v\in V\left(G\right)$ and $e\in E\left(G\right)$, let $N_G\left(v\right)$ be the set of all neighbors of v in G.

Extended adjacency index is one of the degree based topological descriptors which has been proposed by the authors Yang et al. [14] in 1994 and defined as, for any graph G extended adjacency ($EA$) index is:

$$\begin{array}{c}\hfill EA=EA\left(G\right)=\sum _{uv\in E\left(G\right)}\frac{1}{2}\left(\frac{d_G\left(u\right)}{d_G\left(v\right)}+\frac{d_G\left(v\right)}{d_G\left(u\right)}\right).\end{array}$$

(1)

In [14] Yang et al. described that $EA$ index exhibits high discriminating power and correlate well with a number of physico chemical properties and biological activities of organic compounds. There are a couple of topological indices in the literature (see [15]) which are closely related to the extended adjacency index, and they are

$$\begin{array}{c}\hfill E{A}^{*}\left(G\right)=\sum _{i<j}\left(\frac{d_i}{d_j}+\frac{d_j}{d_i}\right)=2\left|E\right|\sum _{j=1}^n\frac{1}{d_j}-n\end{array}$$

and

$$\begin{array}{c}\hfill \widehat{R}\left(G\right)=\sum _{i<j}\left(\frac{d_i}{d_j}+\frac{d_j}{d_i}\right)R_{ij}\end{array}$$

where $R_{ij}$ is the effective resistance between vertices i and j. Obviously, $EA\left(G\right)\le E{A}^{*}\left(G\right)$, and all upper bounds for the inverse degree index $\rho \left(G\right)={\sum }_{j=1}^n\frac{1}{d_j}$ can be used to furnish upper bonds for $E{A}^{*}\left(G\right)$ and $EA\left(G\right)$, even though they may not be tight for $EA\left(G\right)$.

Since 1994, neither extended adjacency matrix nor the extended adjacency index was taken into the consideration but in recent years only few articles have come out with its algebraic approach [16,17,18]. Ramane et al. determined the bounds for the $EA$ index and characterizes graphs extremal with respect to them. Also, obtained relation between $EA$ index and other well known topological indices. Moreover, determined the new results on $EA$ index from an algebraic view point [19]. As an application, one can find a unified approach for some degree based topological descriptors in [20,21,22,23,24,25]. For other undefined notations refer [26,27].

Let $S_n$, $P_n$ and $C_n$ be the star, path and cycle on n vertices, respectively. Let $G-V$ be a subgraph of graph G by deleting vertex v and $G-e$ be a subgraph of graph G by deleting edge e. Let $G_0$ be a nontrivial graph and u be its vertex. If G is obtained by $G_0$ amalgamating a tree T at u. Then we say that T is a subtree of G and u is its root. Let $u\circ v$ denote the amalgamating two vertices u and v of G.

In the present work, we obtain extremal properties of the $EA$ index. In Section 2, we present some graph transformations which increase or decrease $EA$ index. In Section 3, we obtain extremal acyclic, unicyclic and bicyclic graphs with minimum and maximum $EA$ index by a unified method, respectively.

2. Some Graph Transformations

In this section, we present some graph transformations which increase or decrease the $EA$ index and these graph transformations play an important role to determine the extremal graphs of the $EA$ index among acyclic, unicyclic and bicyclic graphs, respectively.

Transformation I. Let $G_0$ be a non-trivial connected graph and v is a given vertex in $G_0$. Let $G_1$ be a graph obtained from $G_0$ by attaching at v two paths $p:v{u}_1{u}_2\dots u_k$ of length k and $Q:v{w}_1{w}_2\dots w_l$ of length l. Let $G_2$ be a graph which is obtained from the graph $G_1$, by Transformation I, $G_2=G_1-v{w}_1+u_k{w}_1$.

Lemma 1.

Let $G_2$ be a graph obtained from $G_1$ by Transformation I as shown in Figure 1, then

$$\begin{array}{c}\hfill EA\left(G_1\right)>EA\left(G_2\right).\end{array}$$

Proof. 

In Transformation I degree of the vertex v is decreased and the degrees of its neighbor vertices $N_{G_0}\left(v\right)$ remains same value. Let us assume that $d_{G_1}\left(v\right)>0$. Then by the definition of $EA$ index, we have

$$\begin{array}{cc}\hfill EA\left(G_1\right)-EA\left(G_2\right)& >\left[\begin{array}{c}\frac{1}{2}\left(\frac{d_{G_1}\left(v\right)}{d_{G_1}\left(w_1\right)}+\frac{d_{G_1}\left(w_1\right)}{d_{G_1}\left(v\right)}\right)+\frac{1}{2}\left(\frac{d_{G_1}\left(u_1\right)}{d_{G_1}\left(v\right)}+\frac{d_{G_1}\left(v\right)}{d_{G_1}\left(u_1\right)}\right)\hfill \\ +\frac{1}{2}\left(\frac{d_{G_1}\left(u_{k-1}\right)}{d_{G_1}\left(u_k\right)}+\frac{d_{G_1}\left(u_k\right)}{d_{G_1}\left(u_{k-1}\right)}\right)\hfill \end{array}\right]\hfill \\ & -\left[\begin{array}{c}\frac{1}{2}\left(\frac{d_{G_2}\left(u_k\right)}{d_{G_2}\left(w_1\right)}+\frac{d_{G_2}\left(w_1\right)}{d_{G_2}\left(u_k\right)}\right)+\frac{1}{2}\left(\frac{d_{G_2}\left(u_1\right)}{d_{G_2}\left(v\right)}+\frac{d_{G_2}\left(v\right)}{d_{G_2}\left(u_1\right)}\right)\hfill \\ +\frac{1}{2}\left(\frac{d_{G_2}\left(u_{k-1}\right)}{d_{G_2}\left(u_k\right)}+\frac{d_{G_2}\left(u_k\right)}{d_{G_2}\left(u_{k-1}\right)}\right)\hfill \end{array}\right]\hfill \\ & =\frac{{(2+d_{G_0}\left(v\right))}^2+4}{2(2+d_{G_0}\left(v\right))}+\frac{5}{4}-\left[\frac{{(1+d_{G_0}\left(v\right))}^2+4}{4(1+d_{G_0}\left(v\right))}+2\right]\hfill \\ & =\frac{d_{G_0}\left(v\right)}{4(1+d_{G_0}\left(v\right))}\left[\frac{d_{G_0}{\left(v\right)}^2}{(2+d_{G_0}\left(v\right))}+3\right]>0.\hfill \end{array}$$

□

Remark 1.

By continuing the process of Transformation I, any tree T of size t connected to a graph $G_1$ can be changed into a path P with size $(t+1)$ (i.e., $P_{t+1}$). From this process, we infer that $EA$ index is strictly decreases.

Transformation II. Let $G_1$ be a connected graph with an edge $uv$ and $d_{G_1}\left(v\right)\ge 2$. Suppose that $N_{G_1}\left(u\right)=\{v,w_1,w_2,\dots ,w_t\}$ and $w_1,w_2,\dots ,w_t$ are pendent vertices. Let $G_2=G_1-\{u{w}_1,u{w}_2,\dots ,u{w}_t\}+\{v{w}_1,v{w}_2,\dots ,v{w}_t\}$.

We now show that Transformation II strictly increases the $EA$ index of a graph.

Lemma 2.

Let $G_2$ be a graph obtained from $G_1$ by Transformation II as shown in Figure 2. Then

$$\begin{array}{c}\hfill EA\left(G_2\right)>EA\left(G_1\right).\end{array}$$

Proof. 

Let $d_{G_0}\left(v\right)>0$. In Transformation II $d_{G_2}\left(v\right)>d_{G_1}\left(v\right)$. So similar to the proof of Lemma 1, we have

$$\begin{array}{ccc}\hfill EA\left(G_2\right)-E\left(G_1\right)& >& \left[\sum _{i=1}^t\frac{1}{2}\left(\frac{d_{G_2}\left(v\right)}{d_{G_2}\left(w_i\right)}+\frac{d_{G_2}\left(w_i\right)}{d_{G_2}\left(v\right)}\right)+\frac{1}{2}\left(\frac{d_{G_2}\left(u\right)}{d_{G_2}\left(v\right)}+\frac{d_{G_2}\left(v\right)}{d_{G_2}\left(u\right)}\right)\right]\hfill \\ & -& \left[\sum _{i=1}^t\frac{1}{2}\left(\frac{d_{G_1}\left(u\right)}{d_{G_1}\left(w_i\right)}+\frac{d_{G_1}\left(w_i\right)}{d_{G_1}\left(u\right)}\right)+\frac{1}{2}\left(\frac{d_{G_1}\left(u\right)}{d_{G_1}\left(v\right)}+\frac{d_{G_1}\left(v\right)}{d_{G_1}\left(u\right)}\right)\right]\hfill \\ & =& \frac{1}{2}\sum _{i=1}^t\left[\left(\frac{d_{G_2}\left(v\right)}{d_{G_2}\left(w_i\right)}+\frac{d_{G_2}\left(w_i\right)}{d_{G_2}\left(v\right)}\right)-\left(\frac{d_{G_1}\left(u\right)}{d_{G_1}\left(w_i\right)}+\frac{d_{G_1}\left(w_i\right)}{d_{G_1}\left(u\right)}\right)\right]\hfill \\ & >& 0.\hfill \end{array}$$

□

Remark 2.

By continuing the process of Transformation II, any tree T of size t connected to a graph $G_1$ can be changed into a star $S_{t+1}$. And from this process $EA$ index increases.

Transformation III. Let $G_1$ be a non-trivial connected graph, u and v be two vertices of $G_1$. Let $P_l=v_1(=u)v_2\dots v_l(=v)$ is a non-trivial path of length t connected to the vertices u and v in $G_1$. If $G_2=G_1-\{v_1{v}_2,v_2{v}_3,\dots ,v_{l-1}{v}_l\}+\{w(=u\circ v)v_1,w{v}_2,\dots ,w{v}_l\}$, see the Figure 3.

Lemma 3.

Let $G_2$ be a connected graph obtained from $G_1$ by Transformation III as shown in Figure 3. Then

$$\begin{array}{c}\hfill EA\left(G_2\right)>EA\left(G_1\right).\end{array}$$

Proof. 

Let $d_{H_1}\left(u\right)=x$ and $d_{H_2}\left(v\right)=y$, while w be the new vertex by merging u and v with $d_{G_2}\left(w\right)=x+y+l-1$, with $l\ge 2$. We can easily get that $EA\left(G_2\right)-EA\left(G_1\right)>0$, for $l=2$. We now show that $EA\left(G_2\right)-EA\left(G_1\right)>0$, for $l>2$. From (1), we have

$$\begin{array}{ccc}\hfill EA\left(G_2\right)-EA\left(G_1\right)& >& \frac{1}{2}\sum _{i=1}^{l-1}\left(\frac{d_{G_2}\left(w\right)}{d_{G_2}\left(v_i\right)}+\frac{d_{G_2}\left(v_i\right)}{d_{G_2}\left(w\right)}\right)-\left[\frac{1}{2}\left(\frac{x}{2}+\frac{2}{x}\right)+\frac{1}{2}\left(\frac{y}{2}+\frac{2}{y}\right)+(l-3)\right]\hfill \\ & =& (l-1)\frac{1}{2}\left(\frac{(x+y+l-1)}{1}+\frac{1}{(x+y+l-1)}\right)\hfill \\ & -& \left(\frac{x^2+4}{4x}\right)-\left(\frac{y^2+4}{4y}\right)-(l-3)\hfill \\ & >& \left[\frac{{(x+y+l-1)}^2+1}{2(x+y+l-1)}-\frac{x^2+4}{4x}\right]+\left[\frac{{(x+y+l-1)}^2+1}{2(x+y+l-1)}-\frac{y^2+4}{4y}\right]>0.\hfill \end{array}$$

□

Transformation IV. Let $G_1$ be a non-trivial connected graph and $x>3$, $y>3$ are two neighbors of vertex $v_1$. Assume that a pendent path $P=v_1{v}_2,v_2{v}_3,\dots ,v_{t-1}{v}_t$ is attached at $v_1$ in graph $G_1$, then $G_2=G_1-x{v}_1+x{v}_t$, see Figure 4.

Lemma 4.

Let $G_2$ be a connected graph obtained from $G_1$ by Transformation IV. Then

$$\begin{array}{c}\hfill EA\left(G_2\right)>EA\left(G_1\right).\end{array}$$

(2)

Proof. 

By the definition of $EA$ index, we have

$$\begin{array}{ccc}\hfill EA\left(G_2\right)-E\left(G_1\right)& >& \left[\begin{array}{c}\frac{1}{2}\left(\frac{d_{G_2}\left(x\right)}{d_{G_2}\left(v_1\right)}+\frac{d_{G_2}\left(v_1\right)}{d_{G_2}\left(x\right)}\right)+\frac{1}{2}\left(\frac{d_{G_2}\left(v_1\right)}{d_{G_2}\left(v_2\right)}+\frac{d_{G_2}\left(v_2\right)}{d_{G_2}\left(v_1\right)}\right)\hfill \\ +\frac{1}{2}\left(\frac{d_{G_2}\left(v_{t-1}\right)}{d_{G_2}\left(v_t\right)}+\frac{d_{G_2}\left(v_t\right)}{d_{G_2}\left(v_{t-1}\right)}\right)+\frac{1}{2}\left(\frac{d_{G_2}\left(v_t\right)}{d_{G_2}\left(y\right)}+\frac{d_{G_2}\left(y\right)}{d_{G_2}\left(v_t\right)}\right)\hfill \end{array}\right]\hfill \\ & -& \left[\begin{array}{c}\frac{1}{2}\left(\frac{d_{G_1}\left(x\right)}{d_{G_1}\left(v_1\right)}+\frac{d_{G_1}\left(v_1\right)}{d_{G_1}\left(x\right)}\right)+\frac{1}{2}\left(\frac{d_{G_1}\left(v_1\right)}{d_{G_1}\left(v_2\right)}+\frac{d_{G_1}\left(v_2\right)}{d_{G_1}\left(v_1\right)}\right)\hfill \\ +\frac{1}{2}\left(\frac{d_{G_1}\left(v_{t-1}\right)}{d_{G_1}\left(v_t\right)}+\frac{d_{G_1}\left(v_t\right)}{d_{G_1}\left(v_{t-1}\right)}\right)+\frac{1}{2}\left(\frac{d_{G_1}\left(v_1\right)}{d_{G_1}\left(y\right)}+\frac{d_{G_1}\left(y\right)}{d_{G_1}\left(v_1\right)}\right)\hfill \end{array}\right]\hfill \\ & =& \left[\frac{1}{2}\left(\frac{x}{2}+\frac{2}{x}\right)+2+\frac{1}{2}\left(\frac{2}{y}+\frac{y}{2}\right)\right]\hfill \\ & -& \left[\frac{1}{2}\left(\frac{x}{3}+\frac{3}{x}\right)+\frac{13}{12}+\frac{5}{4}+\frac{1}{2}\left(\frac{3}{y}+\frac{y}{3}\right)\right]\hfill \\ & =& \frac{x^2+4}{4x}-\left(\frac{x^2+9}{6x}\right)+\frac{y^2+4}{4x}-\left(\frac{y^2+9}{6y}\right)-\frac{1}{3}>0.\hfill \end{array}$$

□

Transformation V: Let $G_0$ be a non-trivial connected graph. Let u and v be a pair of equivalent vertices in $G_0$ with $d_{G_0}\left(u\right)=d_{G_0}\left(v\right)=x$ and $G_1$ be a graph obtained by attaching $S_{k+1}$ and $S_{l+1}$ at the vertices u and v of $G_0$ with $k\ge l$, respectively. If $G_2$ is the graph obtained by deleting the l pendent vertices at v in $G_1$ and connecting them to the vertex u of G, respectively, see Figure 5.

Lemma 5.

Let $G_2$ be a connected graph obtained from $G_1$ by Transformation V. Then

$$\begin{array}{c}\hfill EA\left(G_2\right)>EA\left(G_1\right).\end{array}$$

Proof. 

Let $k\ge l\ge 1$. By (1), we have

$$\begin{array}{ccc}\hfill EA\left(G_2\right)-EA\left(G_1\right)& >& \frac{1}{2}\left(\sum _{i=1}^k\frac{d_{G_2}\left(u\right)}{d_{G_2}\left(u_i\right)}+\frac{d_{G_2}\left(u_i\right)}{d_{G_2}\left(u\right)}\right)-\frac{1}{2}\left(\sum _{i=1}^k\frac{d_{G_1}\left(u\right)}{d_{G_1}\left(u_i\right)}+\frac{d_{G_1}\left(u_i\right)}{d_{G_1}\left(u\right)}\right)\hfill \\ & +& \frac{1}{2}\left(\sum _{i=1}^l\frac{d_{G_2}\left(u\right)}{d_{G_2}\left(v_i\right)}+\frac{d_{G_2}\left(v_i\right)}{d_{G_2}\left(u\right)}\right)-\frac{1}{2}\left(\sum _{i=1}^l\frac{d_{G_1}\left(v\right)}{d_{G_1}\left(v_i\right)}+\frac{d_{G_1}\left(v_i\right)}{d_{G_1}\left(v\right)}\right)\hfill \\ & =& k\left(\frac{1}{2}\left(\frac{d_{G_2}\left(u\right)}{1}+\frac{1}{d_{G_2}\left(u\right)}\right)\right)-\frac{1}{2}\left(\frac{d_{G_1}\left(u\right)}{1}+\frac{1}{d_{G_1}\left(u\right)}\right)\hfill \\ & +& l\left(\frac{1}{2}\left(\frac{d_{G_2}\left(u\right)}{1}+\frac{1}{d_{G_2}\left(u\right)}\right)\right)-\frac{1}{2}\left(\frac{d_{G_1}\left(v\right)}{1}+\frac{1}{d_{G_1}\left(v\right)}\right)\hfill \\ & =& k+l>0.\hfill \end{array}$$

□

Remark 3.

From Lemmas 3–5, we can say that Transformation III, Transformation IV and Transformation V increases the $EA$ index of a graph respectively.

3. Main Results

In this section, we determine the extremal $EA$ index of graphs from ${\mathcal{A}}_n$, ${\mathcal{U}}_n$ and ${\mathcal{B}}_n$, respectively by a unified method.

Let ${\mathcal{A}}_n$, ${\mathcal{U}}_n$ and ${\mathcal{B}}_n$ are the set of connected acyclic, unicyclic and bicyclic graphs of order n respectively. Let $C_n(p,q)$ be the graph contains two cycles $C_p$ and $C_q$ having a common vertex with $p+q-1=n$, $P_n^{k,l,m}$ be the graph obtained by connecting two cycles $C_k$ and $C_m$ with a path $P_l$ with $k+l+m-2=n$ and $C_n(r,l,t)$ be the graph obtained by joining two triples of pendent vertices of three paths $P_l$, $P_r$ and $P_t$ to two vertices with $l+r+t-4=n$. (without loss of generality, we set $2\le l\le r\le t$). If a bicyclic graph contains one of the three graphs which are depicted in Figure 6 as its subgraph then we have three subsets of ${\mathcal{B}}_n$ as ${\mathcal{B}}_n^1=\{C_n(p,q):p+q-1=n\}$, ${\mathcal{B}}_n^2=\{C_n(r,l,t):l+r+t-4=n\}$ and ${\mathcal{B}}_n^3=\{P_n^{k,l,m}:k+l+m-2=n\}$. So the set ${\mathcal{B}}_n$ can be partitioned into three subsets ${\mathcal{B}}_n^1$, ${\mathcal{B}}_n^2$ and ${\mathcal{B}}_n^3$.

The following theorem gives the minimum and maximum value of the $EA$ index.

Theorem 1.

Let G be a acyclic connected graph with order n. Then

$$\begin{array}{c}\hfill EA\left(P_n\right)\le EA\left(G\right)\le EA\left(S_n\right).\end{array}$$

The lower bound and upper bound is attained iff $G\cong P_n$ and $G\cong S_n$ respectively.

Proof. 

By using Lemmas 1 and 2 above inequalities holds good. □

The graphs which are depicted in Figure 7 will be used in the following proof.

Theorem 2.

Let G be a unicyclic graph with order n. Then

$$EA\left(C_n\right)\le EA\left(G\right)\le EA\left(S_n^1\right),$$

where the lower bound and upper bound is attained iff $G\cong C_n$ and $G\cong S_n^1$ respectively.

Proof. 

Let G contains a uniquely cycle $C_l$ and by Lemma 3 we obtain the graph $G_2$ in which the size of the cycle is three and its $EA$ index is strictly increased. Moreover, from Lemma 5, we can get the uniquely maximum graph $S_n^1$ with respect to $EA$ index (see Figure 7 ). On the other hand, by Lemma 1 we conclude that the minimum graph is $C_n$. □

Theorem 3.

Let G be a bicyclic graph with n vertices. Then

$$\begin{array}{c}\hfill n+\frac{3}{2}\le EA\left(G\right)\le \frac{[{(n-1)}^2+1][3n-8]+17}{6(n-1)}+\frac{13}{6},\end{array}$$

(3)

where the lower bound and upper bound is attained iff $G\in \{P_n^{k,l,m}:l\ge 3\}\cup \{C_n(r,l,t):l\ge 2\}$ and $G\cong S_n^2$ respectively.

Proof. 

Firstly, we have to prove the upper bound for the bicyclic graph with respect to $EA$ index. Suppose G is isomorphic to $S_n^2$$(orG\cong S_n^2)$, then from (1), we get

$$EA\left(G\right)=\frac{[{(n-1)}^2+1][3n-8]+17}{6(n-1)}+\frac{13}{6}.$$

Next, we show that $EA\left(G\right)<EA\left(S_n^2\right)$ for G is not isomorphic to $S_n^2$.

Case 1:$K_4-e$ is the subgraph of G.

If $K_4-e$ is the subgraph of a graph G, then from Lemmas 2 and 5 we obtain $G^{\prime }$ as a new (bicyclic) graph whose $EA$ index is more than that of G (see Figure 7). One can easily check that $EA\left(G\right)=\frac{[{(n-1)}^2+1][3n-8]+17}{6(n-1)}+\frac{13}{6}$, equality attains iff $G\cong S_n^2$.

Case 2:$K_4-e$ is not the subgraph of G.

From Lemma 3 we can say that may be there are a bicyclic graph whose $EA$ index is more than that of graph G has the subgraph $K_4-e$. Hence following two subcases exist.

Subcase 2.1:G contains $C_s(3,2,m)$ as a subgraph.

By Lemma 3 Subcase 2.1 deduce to Case 1.

Subcase 2.2:$C_s(3,2,m)$ is not a subgraph of G.

If $C_s(3,2,m)$ is not a subgraph of G, then from Lemmas 2, 3 and 5, we will have a new graph $G^{″}$ whose $EA$ index is more than that of G, see Figure 7. It is easy to verify that $EA\left(G\right)<\frac{[{(n-1)}^2+1][3n-8]+17}{6(n-1)}+\frac{13}{6}$.

Furthermore, We have to show the lower bound. By Lemmas 1, 2 and 4, we infer that the extremal graph of the minimum $EA$ index in bicyclic graphs must be the element which belongs to the set $\{{\mathcal{B}}_n^1,{\mathcal{B}}_n^2,{\mathcal{B}}_n^3\}$.

We easily get that $EA\left(C_n(p,q)\right)=n+2$; $EA\left(P_n^{k,l,m}\right)=n+\frac{17}{12}$ if $l=2$ and $EA\left(P_n^{k,l,m}\right)=n+\frac{3}{2}$, otherwise; $EA(C_n(r,l,t)=n+\frac{3}{2}$ if $l\ge 2$. Hence the lower bound and the equality attains iff $G\in \{P_n^{k,l,m}:l\ge 3\}\cup \{C_n(r,l,t):l\ge 2\}$. □