The Sombor index has gained lot of attention in the recent days for its mathematical properties and chemical applicabilities. Here, we initiated the novel block number version of the classical Sombor index and its matrix representation of a graph. The Block Sombor index \(BS(G)\) is defined as the sum total of square root of the sum of squares of block numbers of adjacent vertices, where the block number of a vertex is the number of blocks to which that vertex belongs to. The main purpose of this paper is to obtain some bounds and characterizations of \(BS(G)\) and its Block Sombor energy \(E_{BS}\). Also, we estimate some properties of spectral radius of Block Sombor matrix \(A_{BS}(G)\).
The graph \(G=(V, E)\) considered are simple, finite, non-trivial and undirected with \(p\)-vertices in \(V(G)\) and \(q\)-edges in \(E(G)\). The number of vertices adjacent to \(v\) is said to be degree of vertex \(v\) and is represented as \(d_v\). The minimum and the maximum degrees of vertices are represented as \(\delta = \delta(G)\) and \(\Delta = \Delta(G)\), respectively. For the graph theoretic terminology not defined here, we refer to [1].
A vertex whose removal results in a trivial or disconnected graph is said to be the cut vertex. A graph that is connected, non-trivial, and has no cut vertices is said to be a non-separable graph. The maximal non-separable subgraph of a graph is said to be the block of that graph. Two blocks are said to be adjacent if they have a cut vertex in common. The block number \(b(G)\) represents the total number of blocks in \(G\). The concept of separable graphs play very significant role of Parsimony Haplotyping problem from computational biology, see [2]. For more details, we refer to [3,4,5,6,7].
A graph in which edges represent bonds and vertices represent atoms is said to be a molecular graph. The invariants of the form \(\sum_{}f(x,y)\) with the property \(f(x,y)=f(y,x)\) are called graphical indices. These are the real numbers derived from the structure of a graph, which are invariant under graph isomorphism. These indices reflect the chemical and physical properties of molecules. Many such invariants have been introduced so far, see [8]. Few of them are as in the Table 1.
Graphical Index | \(f(x,y)=f(d_u,d_v) \) or \(f(d_u,b_u)\) |
---|---|
Sombor Index \([SO(G)]\) (Gutman [9]) | {\( \sqrt{d^{2}_{u}+d^{2}_{v}}\)} |
First Zagreb Index [\(M_{1}(G)\)] (Gutman et al. [10]) | {\( d_{u} + d_{v}\)} |
Second Zagreb Index [\(M_{2}(G)\)] (Gutman et al. [10]) | {\( d_{u} .d_{v}\)} |
Forgotten Index[\(F(G)\)] (Furtula and Gutman [11]) | {\( d_{u}^2 + d_{v}^2\)} |
First Atom Valency Block Index [\(AVB_1 (G)\)](Chaluvaraju and Vyshnavi [12]) | {\( d_{u}+b_{u}\)} |
Second Atom Valency Block Index[\(AVB_2 (G)\)] (Chaluvaraju and Vyshnavi [12]) | {\( d_{u}.b_{u}\)} |
Theorem 1. Let \(G\) be a separable graph with \(k\)-cut vertices. Then \begin{equation}\tag{2} \label{eq2.3} \nonumber BS(G)=\sum_{i=1}^{k}\left[ \left( d_{c_{i}}-\sum_{c_{i}\sim c_{j}}1 \right) \sqrt{1+b_{c_{i}}^2} \right]+\sum_{c_{i}\sim c_{j}}\sqrt{b_{c_{i}}^2+b_{c_{j}}^2} + \left( q-\sum_{i=1}^{k}d_{c_{i}}+\sum_{c_{i}\sim c_{j}}1\right) \sqrt{2}. \end{equation}
Proof. Let \(G\) be a separable \((p,q)\)-graph and \(c_{1},c_{2},\cdots,c_{k}\) be the cut vertices of \(G\). Let their degrees be \(d_{c_{1}}, d_{c_{2}}, \cdots, d_{c_{k}}\) and block numbers be \(b_{c_{1}}, b_{c_{2}}, \cdots, b_{c_{k}}\), respectively. We have the following stages:
Stage 1. Since the cut vertices are adjacent to non-cut vertices and/or cut vertices, number of partitions of the form \((1,b_{c_{i}})\) is the difference of the degree of the cut vertex and the number of cut vertices adjacent to it.
Stage 2. Since the number of blocks adjacent to each cut vertex varies, number of partitions of the form \((b_{c_{i}},b_{c_{j}})\) depends on the adjacencies of cut vertices.
Stage 3. For the non-cut vertices which belong to the same block, the number of partitions of the form \((1,1)\) is difference of total number of edges added to the number of adjacencies of cut vertices and the sum of degrees of all cut vertices.
Formulating these partitions mentioned in three stages, we get the required result.Corollary 1. Let \(G\) be a separable graph. Then,
Theorem 2. Let \(G\) be a non-separable graph with \(p\geq 2\). Then, \begin{align}\tag{3} \label{eq2.2} BS(G)=\sqrt{2}q. \end{align}
Proof. Since the block number of each vertex is exactly one in any non-separable graph \(G\). Hence the result follows.
Corollary 2.
Lemma 1. Let \(G\) be a non-trivial connected \((p,q)\)-graph. Then
Theorem 3. Let \(G\) be a non-trivial connected graph. Then $$\sqrt{2}q \leq BS(G) \leq \sqrt{2}q(p-1).$$ Left inequality holds if and only if \(G\) is non-separable.
Proof. Let \(G\) be a non-trivial connected graph. By Lemma 1(i), we have \(1 \leq b_{u} \leq p-1\). Therefore squaring up and adding the block numbers of two vertices, we have \(2 \leq b_{u}^2+b_{v}^2 \leq 2(p-1)^2\). Also, taking square root of this inequality and adding up them over the number of edges, we have the required inequality. Now, we prove the second part. If the graph \(G\) has no cut vertices, then each vertex has block number \(b_{u}=1\) as they belong to exactly one block, which leads to the partition \((1,1)\) for each edge \(uv \in E(G)\). Thus we obtain the left equality.
For existence of right equality of the above theorem, we pose the following open problem.Theorem 4. Let \(G\) be a non-trivial connected graph. Then $$ BS(G) \leq SO(G).$$Equality holds if and only if \(G\) is a non-trivial tree.
Proof.
Let \(G\) be a simple connected graph. By Lemma 1(ii), we have
\begin{align*}
BS(G)&=\sum_{uv\in E(G)} \sqrt{b_{u}^2+b_{v}^2} \leq \sum_{uv\in E(G)} \sqrt{d_{u}^2+d_{v}^2} =SO(G).
\end{align*}
Now, we prove the second part.
Since each vertex in a non-trivial tree, apart from the pendant vertices is a cut vertex, the block number of each vertex is same as the degree of that vertex. Hence the equality holds.
Conversely, suppose \(BS(G)=SO(G)\) holds for a graph which is not a tree. Then there exist at least three vertices such that every pair of vertices are adjacent forming a complete graph. This is a contradiction to our assumption as \(BS(K_p)=\dfrac{p(p-1)}{\sqrt{2}}\) and \(SO(K_{p})=\dfrac{p(p-1)^2}{\sqrt{2}}\). Hence \(BS(G)< SO(G)\) if \(G\) is not a tree.
Corollary 3. Let \(G\) be a simple connected graph with \(p\geq 2\). Then \begin{equation*} BS(G)\leq \dfrac{1}{\sqrt{2}}(M_1(G)+q(\Delta-\delta)). \end{equation*}
Theorem 5. Let \(G\) be a non-trivial connected graph. Then $$ BS(G) \leq \dfrac{\sqrt{2}M_{2}(G)}{\delta(G)}.$$
Proof. Let \(G\) be a non-trivial connected graph. Then \begin{align*} BS(G)&=\sum_{uv\in E(G)} \sqrt{b_{u}^2+b_{v}^2}=\sum_{uv\in E(G)} b_{u}b_{v}\sqrt{\dfrac{1}{b_u ^2}+\dfrac{1}{b_{v}^2}}\\ &\leq \sum_{uv\in E(G)} d_{u}d_{v}\sqrt{\dfrac{1}{\delta ^2}+\dfrac{1}{\delta^2}}\\ &=\dfrac{\sqrt{2}M_{2}(G)}{\delta(G)} . \end{align*}
Theorem 6. Let \(G\) be a non-trivial connected graph. Then $$ BS(G) \leq \sqrt{qF(G)}.$$
Proof. Let \(G\) be a non-trivial connected graph. Then \begin{align*} BS(G)&=\sum_{uv\in E(G)} \sqrt{b_{u}^2+b_{v}^2}=\sum_{uv\in E(G)} 1.\sqrt{b_{u}^2+b_{v}^2}\\ &\leq \sqrt{ \sum_{uv\in E(G)} 1^2. \sum_{uv\in E(G)} (d_{u}^2+d_{v}^2)} =\sqrt{qF(G)}. \end{align*}
In [14], it was proven for a non-trivial connected graph that, \begin{equation*} F(G)\leq (\Delta+\delta)M_1(G)-2q\Delta\delta. \end{equation*} From the above and Theorem 6, we obtain the following result:Corollary 4. Let \(G\) be a simple connected graph with \(p \geq 2\). Then \begin{equation*} BS(G)\leq \sqrt{q[(\Delta+\delta)M_1(G)-2q\Delta\delta]}. \end{equation*}
In [15], it was proven for a non-trivial connected graph that, \begin{equation*} F(G)\leq \Delta M_1(G)-\dfrac{(2q\Delta-M_1(G))^2}{n\Delta-2q}. \end{equation*} From the above and Theorem 6, we obtain the following result:Corollary 5. Let \(G\) be a simple connected graph. Then \begin{equation*} BS(G)\leq \sqrt{q\left[\Delta M_1(G)-\dfrac{(2q\Delta-M_1(G))^2}{n\Delta-2q}\right]}. \end{equation*}
Theorem 7. Let \(G\) be a connected regular graph with \(p\geq2\). Then $$BS(G) \leq \dfrac{\sqrt{2} \, AVB_{2}(G)}{\delta(G)}.$$
Proof.Let \(G\) be a \((p,q)\)-regular graph with \(p\geq2\). Then \begin{align*} BS(G)&=\sum_{uv\in E(G)} \sqrt{b_{u}^2+b_{v}^2}=\sum_{uv\in E(G)} b_{u}b_{v}\sqrt{\dfrac{1}{b_u ^2}+\dfrac{1}{b_{v}^2}}\\ &\leq \sum_{uv\in E(G)} b_{u}d_{u}\sqrt{\dfrac{1}{\delta ^2}+\dfrac{1}{\delta^2}}\\ &=\dfrac{\sqrt{2}AVB_{2}(G)}{\delta(G)}. \end{align*}
In [12], it was proven for a non-trivial connected graph that, \begin{equation*} AVB_2(G)\leq \dfrac{1}{4} AVB_1(G)^2. \end{equation*} From the above and Theorem 7, we obtain the following result:Corollary 6. Let \(G\) be a simple connected graph with \(p\geq2\). Then \begin{equation*} BS(G)\leq \dfrac{AVB_{1}(G)^2}{2\sqrt{2}\delta(G)}. \end{equation*}
Lemma 2. Let \(\lambda_{1} \geq \lambda_{2}\geq \cdots \geq \lambda_{p}\) be the eigen values of \(A_{BS}\). Then
Proof.Let \(\lambda_{1} \geq \lambda_{2} \geq \lambda_{3} \cdots \geq \lambda_{p}\) be the eigen values of \(A_{BS}\).
Theorem 8. Let \(G\) be any non-trivial connected \((p,q)\)-graph. Then, \begin{equation*} \lambda_{1} \leq \sqrt{\dfrac{2(p-1)}{(p-2)}F(G)}. \end{equation*}
Proof.Let \(G\) be non-trivial connected \((p,q)\)-graph. Then, taking \(a_{i}=\lambda_{i}\) and \(b_{i}=1\) for \(i=1,2,\cdots,p\) in Cauchy-Schwarz inequality, we get, \begin{align*} \left(\sum_{i=2}^{p}\lambda_{i} \right)^2 &\leq (p-1)\sum_{i=2}^{p}\lambda_{i}^2.\\ \implies \left(\sum_{i=2}^{p}\lambda_{i} \right)^2 -\lambda_{1}^2 &\leq (p-1) \left[ \sum_{i=2}^{p}\lambda_{i}^2-\lambda_{1}^2 \right]. \end{align*} Solving this, we get the required inequality.
Lemma 3. [27] Consider a class of real polynomials \(\mathscr{P}(a_{1},a_{2})\), of the form $$P_{n}(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+b_{3}x^{n-3}+\cdots+b_{n},$$ where \(a_{1}\) and \(a_{2}\) are given real numbers. Let \(x_{1} \geq x_{2} \geq \cdots \geq x_{n}\) be roots of \(P_{n}(x) \in \mathscr{P}(a_{1},a_{2})\). Then, \begin{align} \label{eq4.1} \overline{x}+\dfrac{1}{n}\sqrt{\dfrac{\theta}{n-1}} & \leq x_{1} \leq \overline{x}+\dfrac{1}{n}\sqrt{\theta (n-1)},\tag{3}\\ \label{eq4.2} \overline{x}+\dfrac{1}{n}\sqrt{\dfrac{\theta (i-1)}{n-i+1}} & \leq x_{i} \leq \overline{x}+\dfrac{1}{n}\sqrt{\dfrac{\theta (n-i)}{i}};\quad i=2,3,\cdots,n-1,\tag{4}\\ \label{eq4.3} \overline{x}-\dfrac{1}{n}\sqrt{\theta (n-1)}& \leq x_{n} \leq \overline{x}-\dfrac{1}{n}\sqrt{\dfrac{\theta}{n-1}},\tag{5} \end{align} where \(\overline{x}=\dfrac{1}{n}\sum_{i=1}^{n}x_{i}\) and \( \theta=n\sum_{i=1}^{n}x_{i}^2 – \left( \sum_{i=1}^{n}x_{i}\right) ^{2}\).
Lemma 4. The following inequalities hold for \(\sigma_{1} \geq \sigma_{2} \geq \cdots \geq \sigma_{p}\) \begin{align*} & \dfrac{E_{BS}}{p}+\dfrac{1}{p}\sqrt{\dfrac{p\, tr(A_{BS}^2)-E_{BS}^2}{p-1}} \leq \sigma_{1} \leq \dfrac{E_{BS}}{p}+\dfrac{1}{p}\sqrt{(p-1)(p\, tr(A_{BS}^2)-E_{BS}^2)},\\ & \dfrac{E_{BS}}{p}+\dfrac{1}{p}\sqrt{ \dfrac{(i-1)\left[p\, tr(A_{BS}^2)-E_{BS}^2 \right]}{p-i+1}} \leq \sigma_{i} \leq \dfrac{E_{BS}}{p}+ \dfrac{1}{p}\sqrt{ \dfrac{(p-i)\left[p\, tr(A_{BS}^2)-E_{BS}^2 \right]}{i}};\quad i=2,3,\cdots,p-1,\\ & \dfrac{E_{BS}}{p}-\dfrac{1}{p}\sqrt{(p-1)(p\, tr(A_{BS}^2)-E_{BS}^2)} \leq \sigma_{p} \leq \dfrac{E_{BS}}{p}- \dfrac{1}{p}\sqrt{\dfrac{p\, tr(A_{BS}^2)-E_{BS}^2}{p-1}}. \end{align*}
Proof.
Consider the polynomial,
\begin{equation*}
P_{p}(x)=\prod_{i=1}^{p}(x-\sigma_{i})=x^p+a_{1}x^{p-1}+a_{2}x^{p-2}+b_{3}x^{p-3}+\cdots+b_{p}.
\end{equation*}
Since \(a_{1}=-\sum_{i=1}^{p}\sigma_{i}=-E_{BS} \) and \(a_{2}=\frac{1}{2}\left( \left( \sum_{i=1}^{p}\sigma_{i}\right) ^{2} – \sum_{i=1}^{p}\sigma_{i}^2\right)=\frac{1}{2}\left[ E_{BS}^2 -tr(A_{BS})^2 \right], \)
polynomial \(P_{p}(x)\) belongs to the class of real polynomials of the form \(\mathscr{P}(-E_{BS}, \frac{1}{2}E_{BS}^2-\frac{1}{2}tr(A_{BS})^2)\).
By Lemma 3, we have,
\begin{align*}
\overline{x}&=\dfrac{1}{p}\sum_{i=1}^{p}\sigma_{i}=\dfrac{E_{BS}}{p}.\\
\theta &=p\sum_{i=1}^{p}\sigma_{i}^2 – \left( \sum_{i=1}^{p}\sigma_{i}\right) ^{2}=p\, tr(A_{BS}^2)-E_{BS}^2.
\end{align*}
Substituting these in inequalities (\ref{eq4.1}), (\ref{eq4.2}) and (\ref{eq4.3}), we get the required results.
Theorem 9. Let \(G\) be a connected \((p,q)\)-graph with \(p \geq 2\). Then \begin{equation} E_{BS} \leq k+ \sqrt{(p-1)(tr(A_{BS}^{2})-k^2)}, \end{equation} for any real number \(k\) with the property \(\sigma_{1} \geq k \geq \sigma_{p}\).
Proof. By Lemma 4, we have \begin{align*} &k \leq \sigma_{1} \leq \dfrac{E_{BS}}{p}+\dfrac{1}{p}\sqrt{(p-1)(p\, tr(A_{BS}^{2})-E_{BS}^2)}\\ &\implies (pk-E_{BS})^2 \leq (p-1)p\,tr(A_{BS}^{2})-pE_{BS}^2+E_{BS}^2\\ &\implies p^2k^2+E_{BS}^2-2pkE_{BS}+pE_{BS}^2-E_{BS}^2 \leq (p-1)p\,tr(A_{BS}^{2})\\ &\implies (E_{BS}-k)^2\leq (p-1)tr(A_{BS}^{2})-(p-1)k^2\\ &\implies E_{BS} \leq k+ \sqrt{(p-1 )(tr(A_{BS}^{2})-k^2)}. \end{align*}
By Lemma 4 and Theorem 9, we have the following result:Corollary 7. Let \(G\) be a \((p,q)\)-graph with \(p \geq 2\). Then, \begin{equation*} E_{BS} \leq min \left\lbrace \sigma_{1}+\sqrt{(p-1 ) tr(A_{BS}^2)-\sigma_{1}^2)},\sigma_{p}+\sqrt{(p-1)tr(A_{BS}^2)-\sigma_{p}^2)} \right\rbrace. \end{equation*}
Corollary 8. Let \(G\) be a \((p,q)\)-graph with \(p \geq 2\). Then,$$ E_{BS} \leq \sqrt{2pF(G)}.$$
Proof. For \(k=\sqrt{\dfrac{tr(A_{BS}^{2})}{p}}\), \(\sigma_{1} \geq \sqrt{\dfrac{tr(A_{BS}^{2})}{p}} \geq \sigma_{p}\), by Theorem 9, we have, \begin{align*} E_{BS}&\leq \sqrt{\dfrac{tr(A_{BS}^{2})}{p}}+\sqrt{(p-1)\left[ tr(A_{BS}^{2})-\dfrac{tr(A_{BS}^{2})}{p} \right]}\\ &=\sqrt{\dfrac{tr(A_{BS}^{2})}{p}}+\sqrt{\dfrac{(p-1)^2}{p}tr(A_{BS}^{2})}\\ &=p\sqrt{\dfrac{tr(A_{BS}^{2})}{p}}\leq \sqrt{2pF(G)}. \end{align*}
Lemma 5.[28] For a sequence of non-negative real numbers \(b_{1} \geq b_{2} \geq \cdots \geq b_{n} \geq 0,\) \begin{equation} \tag{7}\label{eq4.5} \sum_{i=1}^{n}b_{i}+n(n-1)\left( \prod_{i=1}^{n}b_{i} \right) ^\frac{1}{n} \leq \left( \sum_{i=1}^{n} \sqrt{b_{i}} \right) ^2 \leq (n-1) \sum_{i=1}^{n}b_{i}+n\left( \prod_{i=1}^{n}b_{i} \right) ^\frac{1}{n} . \end{equation}
Theorem 10. Let \(G\) be a (p,q)-graph with \(p\geq 2\). Then \begin{equation*} \sqrt{tr(A_{BS}^{2})+p(p-1)(|det A|)^\frac{2}{p}} \leq E_{BS} \leq \sqrt{(p-1)tr(A_{BS}^{2})+p(|det A|)^\frac{2}{p}}. \end{equation*}
Proof. Substituting \(b_{i}=\sigma_{i}^2 (i=1,2,\cdots,p),\) in equation \ref{eq4.5} of Lemma 5, we obtain \begin{align*} \sum_{i=1}^{p} \sigma_{i}^2 +p(p-1)\left( \prod_{i=1}^{p}\sigma{i}^2 \right) ^\frac{1}{p} \leq \left( \sum_{i=1}^{p} \sqrt{\sigma{i}^2} \right) ^2 \leq (p-1) \sum_{i=1}^{p}\sigma{i}^2+\quad p\left( \prod_{i=1}^{p}\sigma{i}^2 \right) ^\frac{1}{p}. \end{align*} $$\implies tr(A_{BS}^{2})+p(p-1)(|det A|)^\frac{2}{p} \leq E_{BS} ^2 \leq (p-1)tr(A_{BS}^{2})+p(|det A|)^\frac{2}{p}.$$ Thus, we obtain the above inequality.
Theorem 11. Let \(G\) be a (p,q)-graph with \(p\geq 2\). Then \begin{equation*} \sqrt{2tr(A_{BS}^{2})}\leq E_{BS}\leq \sqrt{p\, tr(A_{BS}^{2})}. \end{equation*} Left equality holds if and only if \(\lambda_{1}=-\lambda_{p},\, \lambda_{2}=\lambda_{3}=\cdots=\lambda_{p-1}=0\). Right equality holds if and only if \(\sigma_{1}=\sigma_{2}= \cdots =\sigma_{p}\).
Proof. We have, \begin{equation*} \left( \sum_{i=1}^{p} \lambda{i} \right)^2 =0= \sum_{i=1}^{p} \lambda{i}^2 +2\sum_{i< j} \lambda_{i}\lambda_{j}. \end{equation*} Thus, \begin{equation*} \sum_{i=1}^{p} \lambda_{i}^2 =-2 \sum_{i< j} \lambda_{i}\lambda_{j} =2 \left| \sum_{i< j} \lambda_{i}\lambda_{j} \right|. \end{equation*} Therefore, \begin{align*} E_{BS}^2&=\left( \sum_{i=1}^{p} \left| \lambda{i} \right| \right)^2 = \sum_{i=1}^{p} \left| \lambda{i} \right| ^2 +2\sum_{i< j} \left| \lambda_{i} \right| \left| \lambda_{j} \right|\\ & \geq \sum_{i=1}^{p} \left| \lambda{i} \right| ^2 +2 \left| \sum_{i< j} \lambda_{i}\lambda_{j} \right|=2\sum_{i=1}^{p} \left| \lambda{i} \right| ^2 =2tr(A_{BS}^{2}). \end{align*} Thus the left inequality is proved. The equality holds if and only if \(\sum_{i< j} \left| \lambda_{i} \right| \left| \lambda_{j} \right|=\left| \sum_{i< j} \lambda_{i}\lambda_{j} \right| \), that is when \(\lambda_{1}=-\lambda_{p},\, \lambda_{2}=\lambda_{3}=\cdots=\lambda_{p-1}=0\). The Lagrange's identity says, for \((a)=(a_{1}, a_{2},\cdots, a_{n})\) and \((b)=(b_{1}, b_{2},\cdots, b_{n})\), the two sets of real numbers, \begin{equation*} \sum_{i=1}^{n} a_{i}^2 \sum_{i=1}^{n} b_{i}^2-\left( \sum_{i=1}^{n} a_{i}b_{i} \right) ^2=\sum_{1 \leq i< j\leq n} (a_{i}b_{j}-a_{j}b_{i})^2. \end{equation*} Substituting \(a_{i}=\sigma{i}, b_{i}=1,(i=1,2,\cdots,p)\) in the above identity, we get, \begin{align*} p\sum_{i=1}^{p} \sigma_{i}^2-\left( \sum_{i=1}^{p} \sigma{i} \right) ^2 = \sum_{1 \leq i< j \leq p}(\sigma_{i}-\sigma_{j})^2. \end{align*} \(\sum_{1 \leq i< j \leq p}(\sigma_{i}-\sigma_{j})^2 \geq 0\) with equality if and only if \(\sigma_{1}=\sigma_{2}= \cdots =\sigma_{p}\). Thus we have, \begin{align*} p\sum_{i=1}^{p} \sigma_{i}^2-\left( \sum_{i=1}^{p} \sigma{i} \right) ^2 \geq 0 \implies p\, tr(A_{BS}^{2})\geq E_{BS}^{2}. \end{align*} Thus the right inequality is obtained.
Corollary 9. Let \(G\) be a (p,q)-graph with \(p\geq 2\). If \(\sqrt{b_{u}^2+b_{v}^2}\geq c> 0,\) then \begin{equation*} 2c\sqrt{q} \leq 2\sqrt{c{BS}(G)} \leq E_{BS}. \end{equation*}
Theorem 12. For any connected \((p,q)\)-graph with \(p\geq 2\), \begin{equation*} \dfrac{\sqrt{2}}{(p-1)} BS(G)\leq E_A(G) \leq \left(\sqrt{2}p BS(G)\right)^{\frac{1}{2}}. \end{equation*}
Proof.
Let \(G\) be a connected \((p,q)\)-graph with \(p\geq 2\).
By Theorem 3, we have
\begin{equation}\tag{8}\label{5.1}
q \leq \dfrac{BS(G)}{\sqrt{2}}\qquad and \qquad q \geq \dfrac{BS(G)}{\sqrt{2}(p-1)}.
\end{equation}
McClelland inequality is \(E_A(G) \leq \sqrt{2pq}\). Thus from first inequality of (\ref{5.1}) and McClelland inequality, we have the required left inequality. Also, the required right inequality follows from the second inequality of (\ref{5.1}).