On Bond Incident Degree Indices of Chemical Graphs

Abstract

By swapping out atoms for vertices and bonds for edges, a graph may be used to model any molecular structure. A graph G is considered to be a chemical graph in graph theory if no vertex of G has a degree of 5 or greater. The bond incident degree (BID) index for a chemical graph G is defined as the total of contributions $f(d_G\left(u\right),d_G\left(v\right))$ from all edges $uv$ of G, where $d_G\left(w\right)$ stands for the degree of a vertex w of G, $E\left(G\right)$ is the set of edges of G, and f is a real-valued symmetric function. This paper addresses the problem of finding graphs with extremum BID indices over the class of all chemical graphs of a fixed number of edges and vertices.

1. Introduction

Every molecular structure can be viewed as a graph, usually referred to as a chemical graph, in which vertices correspond to atoms and edges represent bonds of the considered compound [1]. In graph-theoretical notion, a graph in which no vertex has a degree of 5 or greater is referred to as a chemical graph [2]. Chemical graph theory [1,2], the part of graph theory having some chemical applications, provides useful tools in QSAR and QSPR research [3]. Such tools include topological indices [4,5]; a topological index of a chemical graph is a number that remains unchanged under graph isomorphism. Various existing well-known topological indices are defined via edge contributions of the considered chemical graph [6], and many of them have the following form:

$$BID\left(G\right)=\sum _{uv\in E\left(G\right)}f(d_G\left(u\right),d_G\left(v\right)),$$

where $d_G\left(w\right)$ represents the degree of a vertex $w\in V\left(G\right)$ of a graph G, $E\left(G\right)$ is the set of edges of G, and f is a real-valued symmetric function. These indices are named as bond incident degree indices [7,8] (for short, BID indices [8]); it is this class of topological indices with which we deal in this paper. Details regarding some of the existing properties of BID indices can be found in [9,10,11,12,13,14]. For some particular BID indices, we refer the reader to [15,16,17,18] (see also [19,20], where a novel BID index has been studied).

In the rest of this paper, by a graph we mean a connected, finite, simple, and undirected graph. We use (chemical) graph-theoretical notation and terminology from standard books, such as [1,2,21,22].

2. Statements of Main Results

Let $n_i\left(G\right)$ be the number of those vertices of a graph G that have degree i. Denote by $m_{i,j}\left(G\right)$ the number of those edges of a graph G whose end vertices have degrees i and j. In the remainder of this paper, for the sake of simplicity, we drop “$\left(G\right)$” from the notations $n_i\left(G\right)$ and $m_{i,j}\left(G\right)$.

For a graph G, its $BID$ indices can be defined as

$${\displaystyle BID\left(G\right)=\sum _{\delta \left(G\right)\le i\le j\le \Delta \left(G\right)}{m}_{i,j}\left(G\right)·{\beta }_{i,j},}$$

(1)

where ${\beta }_{i,j}$ is a non-negative real-valued (symmetric) function depending on i and j, $\delta \left(G\right)$ is the minimum degree of G, and $\Delta \left(G\right)$ is the maximum degree of G.

If G is a chemical graph of order $n\ge 5$ and size m, then the following equations hold:

$$\sum _{i=1}^4{n}_i=n,$$

(2)

$$\sum _{i=1}^{4}i·n_i=2m,$$

(3)

$$\sum _{\begin{array}{c}1\le i\le 4\\ i\ne j\end{array}}{m}_{j,i}+2m_{j,j}=j·n_j,$$

(4)

where $j=1,2,3,4$. The following values of $m_{1,4}$ and $m_{4,4}$ are obtained (see also [23]) from Equations (2)–(4):

$$m_{1,4}=\frac{4n}{3}-\frac{2m}{3}-\frac{10}{9}{m}_{1,3}-\frac{4}{3}{m}_{1,2}-\frac{2}{3}{m}_{2,2}-\frac{1}{3}{m}_{2,4}-\frac{4}{9}{m}_{2,3}-\frac{1}{9}{m}_{3,4}-\frac{2}{9}{m}_{3,3},$$

(5)

$$m_{4,4}=-\frac{4n}{3}+\frac{5m}{3}+\frac{1}{9}{m}_{1,3}+\frac{1}{3}{m}_{1,2}-\frac{1}{3}{m}_{2,2}-\frac{2}{3}{m}_{2,4}-\frac{5}{9}{m}_{2,3}-\frac{8}{9}{m}_{3,4}-\frac{7}{9}{m}_{3,3}.$$

(6)

Using Equations (5) and (6) in Equation (1), we obtain

$$\begin{array}{ccc}\hfill BID\left(G\right)& =& \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+m_{1,2}{B}_{1,2}+m_{1,3}{B}_{1,3}+m_{2,2}{B}_{2,2}\hfill \\ & & +m_{2,3}{B}_{2,3}+m_{2,4}{B}_{2,4}+m_{3,3}{B}_{3,3}+m_{3,4}{B}_{3,4},\hfill \end{array}$$

(7)

where

$$B_{1,2}={\beta }_{1,2}-\frac{4}{3}{\beta }_{1,4}+\frac{1}{3}{\beta }_{4,4},B_{1,3}={\beta }_{1,3}-\frac{10}{9}{\beta }_{1,4}+\frac{1}{9}{\beta }_{4,4},$$

$$B_{2,2}={\beta }_{2,2}-\frac{2}{3}{\beta }_{1,4}-\frac{1}{3}{\beta }_{4,4},B_{2,3}={\beta }_{2,3}-\frac{4}{9}{\beta }_{1,4}-\frac{5}{9}{\beta }_{4,4},$$

$$B_{2,4}={\beta }_{2,4}-\frac{1}{3}{\beta }_{1,4}-\frac{2}{3}{\beta }_{4,4},B_{3,3}={\beta }_{3,3}-\frac{2}{9}{\beta }_{1,4}-\frac{7}{9}{\beta }_{4,4},$$

$$B_{3,4}={\beta }_{3,4}-\frac{1}{9}{\beta }_{1,4}-\frac{8}{9}{\beta }_{4,4}.$$

The substitution

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,2}{B}_{1,2}+m_{1,3}{B}_{1,3}+m_{2,2}{B}_{2,2}+m_{2,3}{B}_{2,3}+m_{2,4}{B}_{2,4}+m_{3,3}{B}_{3,3}+m_{3,4}{B}_{3,4}$$

(8)

in Equation (7) gives

$$BID\left(G\right)=\frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+{\mathsf{\Gamma }}_{BID}\left(G\right).$$

(9)

Theorem 1.

Let G be a chemical graph of order n and size m such that $n-1\le m\le 2n$ and $n\ge 13$. Consider Equation (1) and let ${\beta }_{i,j}$ be a given function. Additionally, consider $B_{i,j}$ defined in Equation (7). Let all $B_{i,j}$ be positive, and let the following inequalities hold:

$$min\left\{6B_{3,4},3B_{3,3},B_{2,3}+B_{3,4},B_{3,3}+B_{3,4},2B_{1,2},B_{2,2}+B_{2,4},2B_{2,3}\right\}>2B_{2,4},$$

$$min\left\{3B_{1,3},4B_{2,4},2B_{2,2},B_{2,2}+B_{2,4},2B_{2,3},B_{2,3}+B_{2,4},B_{3,3}+B_{3,4}\right\}>3B_{3,4}.$$

(i) 

If $n+m\equiv 0(mod3)$, then

$$BID\left(G\right)\ge \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,4\}$.

(ii) 

If $n+m\equiv 1(mod3)$, then

$$BID\left(G\right)\ge \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+3B_{3,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,3,4\}$ and G has only one vertex of degree 3, which has neighbors of degree 4 only.

(iii) 

If $n+m\equiv 2(mod3)$, then

$$BID\left(G\right)\ge \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+2B_{2,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,2,4\}$ and G has only one vertex of degree 2, which has neighbors of degree 4 only.

Remark 1.

If we take ${\beta }_{i,j}=2\sqrt{ij}{(i+j)}^{-1}$, ${\beta }_{i,j}=2{(i+j)}^{-1}$, or ${\beta }_{i,j}={(i+j)}^{-1/2}$ in Equation (1), we obtain the harmonic index, the geometric-arithmetic index, or the sum-connectivity index, respectively. After elementary calculations, one deduces that for each of the aforementioned choices of ${\beta }_{i,j}$, all the conditions of Theorem 1 concerning $B_{i,j}$ are satisfied. Thus, if $BID$ is any of the aforementioned three indices, then the conclusion of Theorem 1 holds.

Theorem 2.

Let G be a chemical graph of size m and order n such that $n-1\le m\le 2n$ and $n\ge 13$. Consider Equation (1) and let ${\beta }_{i,j}$ be a given function. Additionally, consider $B_{i,j}$ defined in Equation (7). Let all $B_{i,j}$ be negative and the following inequalities hold:

$$max\left\{6B_{3,4},3B_{3,3},B_{2,3}+B_{3,4},B_{3,3}+B_{3,4},2B_{1,2},B_{2,2}+B_{2,4},2B_{2,3}\right\}<2B_{2,4}$$

and

$$max\left\{3B_{1,3},4B_{2,4},2B_{2,2},B_{2,2}+B_{2,4},2B_{2,3},B_{2,3}+B_{2,4},B_{3,3}+B_{3,4}\right\}<3B_{3,4}.$$

(i) 

If $n+m\equiv 0(mod3)$, then

$$BID\left(G\right)\le \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m,$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,4\}$.

(ii) 

If $n+m\equiv 1(mod3)$, then

$$BID\left(G\right)\le \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+3B_{3,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,3,4\}$ and G has only one vertex of degree 3, which has neighbors of degree 4 only.

(iii) 

If $n+m\equiv 2(mod3)$, then

$$BID\left(G\right)\le \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+2B_{2,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,2,4\}$ and G has only one vertex of degree 2, which has neighbors of degree 4 only.

Remark 2. 

If we take${\beta }_{i,j}=i{\left(j\right)}^{-1}+j{\left(i\right)}^{-1}$, ${\beta }_{i,j}=\sqrt{i^2+j^2}$, ${\beta }_{i,j}=\sqrt{{(i-1)}^2+{(j-1)}^2}$, ${\beta }_{i,j}=(i+j){\left(4ij\right)}^{-1/2}$, ${\beta }_{i,j}=(i+j){\left(4ij\right)}^{-1/2}-2\sqrt{ij}{(i+j)}^{-1}$, ${\beta }_{i,j}=ln(i+j)$, or ${\beta }_{i,j}={1-2{(i+j)}^{-1}}^{1/2}$in Equation (1), we obtain the symmetric division deg index, the Sombor index, the reduced Sombor index, the arithmetic-geometric index, the difference between arithmetic-geometric index and geometric-arithmetic index, the natural logarithm of the multiplicative sum Zagreb index, or the atom-bond sum-connectivity index, respectively. After elementary calculations, one deduces that for each of the aforementioned choices of${\beta }_{i,j}$, all the conditions of Theorem 2 concerning$B_{i,j}$are satisfied, and hence if$BID$is any of these graph invariants, then the conclusion of Theorem 2 holds.

Example 1.

Let us consider the class $\mathcal{T}$ of graphs representing all octane isomers. Certainly, the class $\mathcal{T}$ consists of 18 chemical graphs, each having 8 vertices and 7 edges. Note that there is only one graph in $\mathcal{T}$ having the degree set $\{1,4\}$, and that graph corresponds to the chemical compound $2,2,3,3$-tetramethylbutane. As indicated in Remark 2, the conclusion of Theorem 2 holds also for the atom-bond sum-connectivity (ABS) index. Therefore, by Theorem 2(i), we conclude that the chemical compound $2,2,3,3$-tetramethylbutane has the maximum value of the ABS index among all octane isomers.

Theorem 3.

Let G be a chemical graph of size m and order n such that $4\le n-1\le m\le 2n$. Consider Equation (1) and let ${\beta }_{i,j}$ be a given function. Additionally, consider $B_{i,j}$ defined in Equation (7). Let all $B_{i,j}$ be negative and the following inequalities hold:

$$max\left\{B_{2,4}+B_{3,4},2B_{3,4},2B_{2,4},3B_{3,3},B_{1,3}+B_{3,3},B_{1,2}+B_{2,2},2B_{2,2}\right\}<min\{B_{1,2}+B_{2,4},2B_{1,3}+B_{3,4}\},$$

(10)

$$max\left\{2B_{2,3},B_{1,2}+B_{2,3}\right\}<B_{1,2}+B_{2,4},$$

(11)

$$max\left\{3B_{2,3},B_{1,3}+B_{2,3}\right\}<2B_{1,3}+B_{3,4},$$

(12)

$$max\{B_{2,2},B_{2,3},B_{2,4}\}<B_{1,2},$$

(13)

and

$$max\{B_{2,3},B_{3,3},B_{3,4}\}<B_{1,3}.$$

(14)

(i) 

If $n+m\equiv 0(mod3)$, then

$$BID\left(G\right)\le \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m,$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,4\}$.

(ii) 

If $n+m\equiv 1(mod3)$, then

$$BID\left(G\right)\le \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+2B_{1,3}+B_{3,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,3,4\}$ and G possesses exactly one vertex of degree 3, which has one neighbor of degree 4 and two neighbors of degree 1.

(iii) 

If $n+m\equiv 2(mod3)$, then

$$BID\left(G\right)\le \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+B_{1,2}+B_{2,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,2,4\}$ and G possesses only one vertex of degree 2, which has one neighbor of degree 4 and one neighbor of degree 1.

Remark 3.

The choice ${\beta }_{i,j}={(i+j)}^2$ in Equation (1) gives the hyper Zagreb index. It can be easily checked that all the conditions of Theorem 3 concerning $B_{i,j}$ are satisfied for the choice ${\beta }_{i,j}={(i+j)}^2$. Thus, if $BID$ is the hyper Zagreb index, then the conclusion of Theorem 3 holds.

Theorem 4.

Let G be a chemical graph of order n and size m such that $4\le n-1\le m\le 2n$. Consider Equation (1) and let ${\beta }_{i,j}$ be a given function. Additionally, consider $B_{i,j}$ defined in Equation (7). Let all $B_{i,j}$ be positive and the following inequalities hold:

$$min\left\{B_{2,4}+B_{3,4},2B_{3,4},2B_{2,4},3B_{3,3},B_{1,3}+B_{3,3},B_{1,2}+B_{2,2},2B_{2,2}\right\}>max\{B_{1,2}+B_{2,4},2B_{1,3}+B_{3,4}\},$$

$$min\left\{2B_{2,3},B_{1,2}+B_{2,3}\right\}>B_{1,2}+B_{2,4},$$

$$min\left\{3B_{2,3},B_{1,3}+B_{2,3}\right\}>2B_{1,3}+B_{3,4},$$

$$min\{B_{2,2},B_{2,3},B_{2,4}\}>B_{1,2},$$

and

$$min\{B_{2,3},B_{3,3},B_{3,4}\}>B_{1,3}.$$

(i) 

If $n+m\equiv 0(mod3)$, then

$$BID\left(G\right)\ge \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m,$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,4\}$.

(ii) 

If $n+m\equiv 1(mod3)$, then

$$BID\left(G\right)\ge \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+2B_{1,3}+B_{3,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,3,4\}$ and G possesses only one vertex of degree 3, which has one neighbor of degree 4 and two neighbors of degree 1.

(iii) 

If $n+m\equiv 2(mod3)$, then

$$BID\left(G\right)\ge \frac{4({\beta }_{1,4}-{\beta }_{4,4})}{3}n+\frac{5{\beta }_{4,4}-2{\beta }_{1,4}}{3}m+B_{1,2}+B_{2,4},$$

where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,2,4\}$ and G possesses only one vertex of degree 2, which has one neighbor of degree 4 and one neighbor of degree 1.

3. Proofs

We start this section by proving the following following lemma, which is useful in proving Theorem 1.

Lemma 1.

Let G be a chemical graph such that $n_2+n_3\ge 2$.

(a). 

If ${\mathsf{\Gamma }}_{BID}$ is a graph invariant defined in Equation (8) such that every $B_{i,j}$ used there is positive and that the inequalities

$$min\left\{6B_{3,4},3B_{3,3},B_{2,3}+B_{3,4},B_{3,3}+B_{3,4},2B_{1,2},B_{2,2}+B_{2,4},2B_{2,3}\right\}>2B_{2,4}$$

(15)

and

$$min\left\{3B_{1,3},4B_{2,4},2B_{2,2},B_{2,2}+B_{2,4},2B_{2,3},B_{2,3}+B_{2,4},B_{3,3}+B_{3,4}\right\}>3B_{3,4}$$

(16)

hold, then

$${\mathsf{\Gamma }}_{BID}\left(G\right)>2B_{2,4}\mathrm{and}{\mathsf{\Gamma }}_{BID}\left(G\right)>3B_{3,4}.$$

(17)

(b). 

If ${\mathsf{\Gamma }}_{BID}$ is a graph invariant defined in Equation (8) such that every $B_{i,j}$ used there is negative and that the inequalities

$$max\left\{6B_{3,4},3B_{3,3},B_{2,3}+B_{3,4},B_{3,3}+B_{3,4},2B_{1,2},B_{2,2}+B_{2,4},2B_{2,3}\right\}<2B_{2,4}$$

and

$$max\left\{3B_{1,3},4B_{2,4},2B_{2,2},B_{2,2}+B_{2,4},2B_{2,3},B_{2,3}+B_{2,4},B_{3,3}+B_{3,4}\right\}<3B_{3,4}$$

hold, then the inequalities’ signs in (17) are reversed.

Proof. 

Since the proofs of two parts are similar to each other, we provide the proof of part (a) only. We prove the desired result by considering the four possible cases: (i) $m_{2,2}\ge 1$, (ii) $m_{2,3}\ge 1$, (iii) $m_{3,3}\ge 1$, (iv) $m_{2,2}=m_{2,3}=m_{3,3}=0$.

Case (i). 

$m_{2,2}\ge 1$.

From Equations (8) and (15), it follows that

$$\begin{array}{c}\hfill {\mathsf{\Gamma }}_{BID}\left(G\right)\ge m_{2,2}{B}_{2,2}+(m_{1,2}+m_{2,3}+m_{2,4})B_{2,4}.\end{array}$$

(18)

Equation (4) with $j=2$ this becomes $m_{1,2}+2m_{2,2}+m_{2,3}+m_{2,4}=2n_2$, which gives the following two possibilities (because $n_2\ge 2$ in this case):

• $m_{2,2}=1$ and $m_{1,2}+m_{2,3}+m_{2,4}>1$;
• $m_{2,2}\ge 2$.

If $m_{2,2}=1$ and $m_{1,2}+m_{2,3}+m_{2,4}>1$, then by using (15) and (16) in (18), we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)>B_{2,2}+B_{2,4}>\left\{\begin{array}{cc}3B_{3,4},\hfill & \\ 2B_{2,4}.\hfill \end{array}\right.$$

If $m_{2,2}\ge 2$, then again by using (15) and (16) in (18), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge 2B_{2,2}>\left\{\begin{array}{cc}3B_{3,4},\hfill & \\ 2B_{2,4}.\hfill \end{array}\right.$$

Case (ii). 

$m_{2,3}\ge 1$.

First, we prove the inequality ${\mathsf{\Gamma }}_{BID}\left(G\right)>2B_{2,4}$. From Equations (8) and (16), it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge m_{2,3}{B}_{2,3}+(m_{1,3}+m_{3,3}+m_{3,4})B_{3,4}.$$

(19)

Equation (4) with $j=3$, this becomes $m_{1,3}+m_{2,3}+2m_{3,3}+m_{3,4}=3n_3$, which gives the following two possibilities (as $n_3\ge 1$ in this case):

• $m_{2,3}=1$ or 2, and $m_{1,3}+m_{3,3}+m_{3,4}\ge 1$;
• $m_{2,3}\ge 3$.

If $m_{2,3}=1$ or 2, and $m_{1,3}+m_{3,3}+m_{3,4}\ge 1$, then by using (15) in (19), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge B_{2,3}+B_{3,4}>2B_{2,4}.$$

If $m_{2,3}\ge 3$, then again by using (15) in (19), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge 3B_{2,3}>2B_{2,4}.$$

Next, in the remaining part of this case, we prove the inequality ${\mathsf{\Gamma }}_{BID}\left(G\right)>3B_{3,4}$. From Equations (8) and (15), it follows that

$$\begin{array}{ccc}\hfill {\mathsf{\Gamma }}_{BID}\left(G\right)& \ge & m_{2,3}{B}_{2,3}+(m_{1,2}+m_{2,2}+m_{2,4})B_{2,4}.\hfill \end{array}$$

(20)

Equation (4) with $j=2$, this becomes $m_{1,2}+2m_{2,2}+m_{2,3}+m_{2,4}=2n_2$, which gives the following two possibilities (because $n_2\ge 1$ in this case):

• $m_{2,3}=1$ and $m_{1,2}+m_{2,2}+m_{2,4}\ge 1$;
• $m_{2,3}\ge 2$.

If $m_{2,3}=1$ and $m_{1,2}+m_{2,2}+m_{2,4}\ge 1$, then by using (16) in (20), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge B_{2,3}+B_{2,4}>3B_{3,4}.$$

If $m_{2,3}\ge 2$, then again by using (16) in (20), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge 2B_{2,3}>3B_{3,4}.$$

Case (iii). 

$m_{3,3}\ge 1$.

From (8) and (16), it follows that

$$\begin{array}{c}\hfill {\mathsf{\Gamma }}_{BID}\left(G\right)\ge m_{3,3}{B}_{3,3}+(m_{1,3}+m_{2,3}+m_{3,4})B_{3,4}.\end{array}$$

(21)

Equation (4) with $j=3$, this becomes $m_{1,3}+m_{2,3}+2m_{3,3}+m_{3,4}=3n_3$. Note that $n_3\ge 2$ in this case and hence there are two subcases that need to be considered:

• $m_{3,3}=1$ or 2, and $m_{1,3}+m_{2,3}+m_{3,4}>1$;
• $m_{3,3}\ge 3$.

If $m_{3,3}=1$ or 2, and $m_{1,3}+m_{2,3}+m_{3,4}>1$, then by using (15) and (16) in (21), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)>B_{3,3}+B_{3,4}>\left\{\begin{array}{cc}3B_{3,4},\hfill & \\ 2B_{2,4}.\hfill \end{array}\right.$$

If $m_{3,3}\ge 3$, then again by using (15) and (16) in (21), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge 3B_{3,3}>\left\{\begin{array}{cc}3B_{3,4},\hfill & \\ 2B_{2,4}.\hfill \end{array}\right.$$

Case (iv). 

$m_{2,2}=m_{2,3}=m_{3,3}=0$.

By using (15) and (16) in Equation (8), we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge (m_{1,2}+m_{2,4})B_{2,4}+(m_{1,3}+m_{3,4})B_{3,4}.$$

Equation (4) with $j=2,3$, gives $m_{1,2}+m_{2,4}=2n_2$ and $m_{1,3}+m_{3,4}=3n_3$. Thus,

$${\mathsf{\Gamma }}_{BID}\left(G\right)\ge 2n_2{B}_{2,4}+3n_3{B}_{3,4}.$$

(22)

The assumption $n_2+n_3\ge 2$ gives three possibilities: (i) $n_2\ge 1$ and $n_3\ge 1$, (ii) $n_2=0$ and $n_3\ge 2$, (iii) $n_2\ge 2$ and $n_3=0$; for each of these possibilities, by using (15) and (16) in (22), we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)>\left\{\begin{array}{cc}2B_{2,4},\hfill & \\ 3B_{3,4}.\hfill \end{array}\right.$$

The proof is completed.   □

Lemma 1 has now enabled us to prove Theorem 1.

Proof of Theorem 1.

The following congruence follows from Equations (2) and (3) (see also the paper [23]):

$$n+m\equiv n_3-n_2(\mathrm{mod}3).$$

(23)

(i) 

In this part, we have $n+m\equiv 0(mod3)$. From Equation (8), it follows that ${\mathsf{\Gamma }}_{BID}\left(G\right)\ge 0$, where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,4\}$ and hence from Equation (9) the desired conclusion is deduced.

(ii) 

In this part, we have $n+m\equiv 1(mod3)$.

If $n_2+n_3\ge 2$, then Lemma 1 guarantees that ${\mathsf{\Gamma }}_{BID}\left(G\right)>3B_{3,4}$.

Next, assume that $n_2+n_3\le 1$. Then, Equation (23) yields $n_2=0$ and $n_3=1$. Thereby, Equation (4) (with $j=3$) gives $m_{1,3}+m_{3,4}=3n_3=3$ and hence by using (16) in Equation (8) we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,3}{B}_{1,3}+m_{3,4}{B}_{3,4}\ge m_{1,3}{B}_{3,4}+m_{3,4}{B}_{3,4}=3B_{3,4}.$$

The equality ${\mathsf{\Gamma }}_{BID}\left(G\right)=3B_{3,4}$ holds if and only if $m_{1,3}=0$; that is, the unique vertex of degree 3 is adjacent to three vertices of degree 4.

Now, by combining the conclusions of both cases considered in this part and by using Equation (9), we arrive at the desired result.

(iii) 

In this part, we have $n+m\equiv 2(mod3)$.

If $n_2+n_3\ge 2$, then Lemma 1 guarantees that ${\mathsf{\Gamma }}_{BID}\left(G\right)>2B_{2,4}$.

Now, suppose that $n_2+n_3\le 1$. Then, Equation (23) yields $n_2=1$ and $n_3=0$. Thereby, Equation (4) (with $j=2$) gives $m_{1,2}+m_{2,4}=2n_2=2$, and hence by using (15) in Equation (8) we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,2}{B}_{1,2}+m_{2,4}{B}_{2,4}\ge m_{1,2}{B}_{2,4}+m_{2,4}{B}_{2,4}=2B_{2,4}.$$

The equality ${\mathsf{\Gamma }}_{BID}\left(G\right)=2B_{2,4}$ holds if and only if $m_{1,2}=0$; that is, the unique vertex of degree 2 has two neighbors of degree 4.

By combining the conclusions of both cases considered in this part and by using Equation (9), we arrive at the desired result.   □

The proof of Theorem 2 is fully analogous to that of Theorem 1, and is hence omitted. In order to prove Theorem 3, we prove the following lemma first.

Lemma 2.

Let G be a chemical graph such that $n_2+n_3\ge 2$.

(a). 

If ${\mathsf{\Gamma }}_{BID}$ is a graph invariant defined in Equation (8) such that every $B_{i,j}$ used there is negative and that the inequalities

$$max\left\{B_{2,4}+B_{3,4},2B_{3,4},2B_{2,4},3B_{3,3},B_{1,3}+B_{3,3},3B_{2,3},B_{1,3}+B_{2,3},2B_{2,2},B_{1,2}+B_{2,2}\right\}<2B_{1,3}+B_{3,4},$$

(24)

$$max\left\{B_{2,4}+B_{3,4},2B_{3,4},2B_{2,4},2B_{2,2},B_{1,2}+B_{2,2},2B_{2,3},B_{1,2}+B_{2,3},3B_{3,3},B_{1,3}+B_{3,3}\right\}<B_{1,2}+B_{2,4},$$

(25)

$$max\{B_{2,2},B_{2,3},B_{2,4}\}<B_{1,2},$$

(26)

and

$$max\{B_{2,3},B_{3,3},B_{3,4}\}<B_{1,3}$$

(27)

hold, then

$${\mathsf{\Gamma }}_{BID}\left(G\right)<2B_{1,3}+B_{3,4}and{\mathsf{\Gamma }}_{BID}\left(G\right)<B_{1,2}+B_{2,4}.$$

(28)

(b). 

If ${\mathsf{\Gamma }}_{BID}$ is a graph invariant defined in Equation (8) such that every $B_{i,j}$ used there is positive and that the inequalities

$$min\left\{4B_{1,2},6B_{1,3},2B_{1,2}+3B_{1,3},3B_{3,3},B_{1,3}+B_{3,3},3B_{2,3},B_{1,3}+B_{2,3},B_{2,2}\right\}>2B_{1,3}+B_{3,4},$$

$$min\left\{4B_{1,2},6B_{1,3},2B_{1,2}+3B_{1,3},2B_{2,2},B_{1,2}+B_{2,2},2B_{2,3},B_{1,2}+B_{2,3},B_{3,3}\right\}>B_{1,2}+B_{2,4},$$

$$min\{B_{2,2},B_{2,3},B_{2,4}\}>B_{1,2},$$

and

$$min\{B_{2,3},B_{3,3},B_{3,4}\}>B_{1,3}$$

hold, then the inequalities’ signs in (28) are reversed.

Proof. 

Since the proof of two parts are similar to each other, we prove only part (a). We prove the desired result by considering the four possible cases: (i) $m_{2,2}\ge 1$, (ii) $m_{2,3}\ge 1$, (iii) $m_{3,3}\ge 1$, (iv) $m_{2,2}=m_{2,3}=m_{3,3}=0$.

Case (i). 

$m_{2,2}\ge 1$.

From (8) and (26), it follows that

$$\begin{array}{c}\hfill {\mathsf{\Gamma }}_{BID}\left(G\right)\le m_{2,2}{B}_{2,2}+(m_{1,2}+m_{2,3}+m_{2,4})B_{1,2}.\end{array}$$

(29)

Equation (4) with $j=2$ becomes $m_{1,2}+2m_{2,2}+m_{2,3}+m_{2,4}=2n_2$, which gives the following two possibilities (because $n_2\ge 2$ in this case):

• $m_{2,2}=1$ and $m_{1,2}+m_{2,3}+m_{2,4}>1$;
• $m_{2,2}\ge 2$.

If $m_{2,2}=1$ and $m_{1,2}+m_{2,3}+m_{2,4}>1$, then from (24), (25), and (29), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)<B_{2,2}+B_{1,2}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

If $m_{2,2}\ge 2$, then again from (24), (25), and (29), we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le 2B_{2,2}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

Case (ii). 

$m_{2,3}\ge 1$.

First, we prove the inequality ${\mathsf{\Gamma }}_{BID}\left(G\right)<2B_{1,3}+B_{3,4}$. From (8) and (27), it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le m_{2,3}{B}_{2,3}+(m_{1,3}+m_{3,3}+m_{3,4})B_{1,3}.$$

(30)

Equation (4) with $j=3$ becomes $m_{1,3}+m_{2,3}+2m_{3,3}+m_{3,4}=3n_3$, which gives the following two possibilities (as $n_3\ge 1$ in this case):

• $m_{2,3}=1$ or 2, and $m_{1,3}+m_{3,3}+m_{3,4}\ge 1$;
• $m_{2,3}\ge 3$.

If $m_{2,3}=1$ or 2, and $m_{1,3}+m_{3,3}+m_{3,4}\ge 1$, then from (24) and (30) it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le B_{2,3}+B_{1,3}<2B_{1,3}+B_{3,4}.$$

If $m_{2,3}\ge 3$, then again from (24) and (30), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le 3B_{2,3}<2B_{1,3}+B_{3,4}.$$

Next, in the remaining part of this case, we prove the inequality ${\mathsf{\Gamma }}_{BID}\left(G\right)<B_{1,2}+B_{2,4}$. From (8) and (26), it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le m_{2,3}{B}_{2,3}+(m_{1,2}+m_{2,2}+m_{2,4})B_{1,2}.$$

(31)

Equation (4) with $j=2$ becomes $m_{1,2}+2m_{2,2}+m_{2,3}+m_{2,4}=2n_2$, which gives the following two possibilities (because $n_2\ge 1$ in this case):

• $m_{2,3}=1$ and $m_{1,2}+m_{2,2}+m_{2,4}\ge 1$;
• $m_{2,3}\ge 2$.

If $m_{2,3}=1$ and $m_{1,2}+m_{2,2}+m_{2,4}\ge 1$, then from (25) and (31), it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le B_{2,3}+B_{1,2}<B_{1,2}+B_{2,4}.$$

If $m_{2,3}\ge 2$, then again from (25) and (31), we obtain

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le 2B_{2,3}<B_{1,2}+B_{2,4}.$$

Case (iii). 

$m_{3,3}\ge 1$.

From (8) and (27), we have

$$\begin{array}{c}\hfill {\mathsf{\Gamma }}_{BID}\left(G\right)\le m_{3,3}{B}_{3,3}+(m_{1,3}+m_{2,3}+m_{3,4})B_{1,3}.\end{array}$$

(32)

Equation (4) with $j=3$ becomes $m_{1,3}+m_{2,3}+2m_{3,3}+m_{3,4}=3n_3$. Note that $n_3\ge 2$ in this case and hence there are two subcases that need to be considered:

• $m_{3,3}=1$ or 2, and $m_{1,3}+m_{2,3}+m_{3,4}>1$;
• $m_{3,3}\ge 3$.

If $m_{3,3}=1$ or 2, and $m_{1,3}+m_{2,3}+m_{3,4}>1$, then from (24), (25), and (32), it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)<B_{3,3}+B_{1,3}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

If $m_{3,3}\ge 3$, then again from (24), (25), and (32), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le 3B_{3,3}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

Case (iv). 

$m_{2,2}=m_{2,3}=m_{3,3}=0$.

In this case, Equation (8) becomes

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,2}{B}_{1,2}+m_{1,3}{B}_{1,3}+m_{2,4}{B}_{2,4}+m_{3,4}{B}_{3,4}.$$

(33)

In this case, note also that $m_{1,2}\le m_{2,4}$ and $m_{1,3}\le 2m_{3,4}$. Moreover, Equation (4) with $j=2,3$ gives $m_{1,2}+m_{2,4}=2n_2$ and $m_{1,3}+m_{3,4}=3n_3$. Thus,

$$m_{2,4}\ge n_2\mathrm{and}{m}_{3,4}\ge n_3.$$

(34)

If $n_2\ge 1$ and $n_3\ge 1$, then from (34) one obtains $min\{m_{2,4},m_{3,4}\}\ge 1$, and hence from (24), (25), and (33), one obtains

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le B_{2,4}+B_{3,4}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

If $n_2=0$ and $n_3\ge 2$, then from (34) the inequality $m_{3,4}\ge 2$ follows, and hence from (24), (25), and (33), it follows that

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le 2B_{3,4}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

Finally, if $n_2\ge 2$ and $n_3=0$, then from (34) we have $m_{2,4}\ge 2$, and hence by using (24), (25), and (33), we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)\le 2B_{2,4}<\left\{\begin{array}{cc}2B_{1,3}+B_{3,4},\hfill & \\ B_{1,2}+B_{2,4}.\hfill \end{array}\right.$$

The proof is completed.   □

Now, we are in position to prove Theorem 3.

Proof of Theorem 3.

(i) In this part, we have $n+m\equiv 0(mod3)$. From Equation (8), it follows that ${\mathsf{\Gamma }}_{BID}\left(G\right)\le 0$, where the necessary and sufficient condition for the equality is that the degree set of G is $\{1,4\}$, and hence from Equation (9) the desired conclusion is deduced.

(ii) 

In this part, we have $n+m\equiv 1(mod3)$.

If $n_2+n_3\ge 2$, then Lemma 2 guarantees that ${\mathsf{\Gamma }}_{BID}\left(G\right)<2B_{1,3}+B_{3,4}$ (because the system of inequalities (10)–(14) is equivalent to the system of inequalities (24)–(27)).

Next, assume that $n_2+n_3\le 1$. Then, Equation (23) yields $n_2=0$ and $n_3=1$. Thereby, Equation (4) (with $j=3$) gives $m_{1,3}+m_{3,4}=3n_3=3$. Additionally, Equation (8) becomes

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,3}{B}_{1,3}+m_{3,4}{B}_{3,4}.$$

Note that $m_{3,4}\ge 1$ (because $n\ge 5$), and hence by using (14) we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,3}{B}_{1,3}+m_{3,4}{B}_{3,4}\le 2B_{1,3}+B_{3,4}.$$

The equality ${\mathsf{\Gamma }}_{BID}\left(G\right)=2B_{1,3}+B_{3,4}$ holds if and only if $m_{1,3}=2$ and $m_{3,4}=1$; that is, the unique vertex of degree 3 has one neighbor of degree 4 and two neighbors of degree 1.

Now, by combining the conclusions of both cases considered in this part and by using Equation (9), we arrive at the desired result.

(iii) 

In this part, we have $n+m\equiv 2(mod3)$.

If $n_2+n_3\ge 2$, then Lemma 2 guarantees that ${\mathsf{\Gamma }}_{BID}\left(G\right)<B_{1,2}+B_{2,4}$ (because the system of inequalities (10)–(14) is equivalent to the system of inequalities (24)–(27)).

Now, suppose that $n_2+n_3\le 1$. Then, Equation (23) yields $n_2=1$ and $n_3=0$. Thereby, Equation (4) (with $j=2$) gives $m_{1,2}+m_{2,4}=2n_2=2$. Additionally, Equation (8) gives

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,2}{B}_{1,2}+m_{2,4}{B}_{2,4}.$$

Note that $m_{2,4}\ge 1$ (because $n\ge 4$), and hence by using (13) we have

$${\mathsf{\Gamma }}_{BID}\left(G\right)=m_{1,2}{B}_{1,2}+m_{2,4}{B}_{2,4}\le B_{1,2}+B_{2,4}.$$

The equality ${\mathsf{\Gamma }}_{BID}\left(G\right)=B_{1,2}+B_{2,4}$ holds if and only if $m_{1,2}=m_{2,4}=1$ (i.e., the unique vertex of degree 2 has one neighbor of degree 4 and one neighbor of degree 1).

By combining the conclusions of both cases considered in this part and by using Equation (9), we arrive at the desired result.   □

The proof of Theorem 4 is fully analogous to that of Theorem 3, and is hence omitted.

4. Concluding Remarks

If we take ${\beta }_{i,j}={\left(ij\right)}^{-1/2}$ or ${\beta }_{i,j}=ij$ in Equation (1), we obtain the Randić index or the second Zagreb index, respectively. Although Theorems 1–4 cover several well-known topological indices, there are some renowned topological indices for which the conclusions of these theorems hold but the conditions of the theorems are not fully satisfied—for example, the Randić index obeys the conclusion of Theorem 2 (see [23]) but not its conditions completely, and the second Zagreb index obeys the conclusion of Theorem 3 (see [24]) but not its conditions completely. Thereby, as a future work, it would be interesting to relax the conditions of Theorems 1–4 provided that their conclusions remain unchanged (that is, to give stronger versions of these theorems) so that some additional indices may also be covered.