On Sum--Connectivity Index of Bicyclic Graphs

TL;DR

This paper investigates the extremal sum--connectivity indices of bicyclic graphs with given vertices and matching number, identifying the graphs that maximize or minimize this index.

Contribution

It determines the minimum, maximum, and second extremal sum--connectivity indices of bicyclic graphs for specified parameters and characterizes the extremal graphs.

Findings

Identified graphs with minimum sum--connectivity index.

Identified graphs with maximum sum--connectivity index.

Characterized extremal graphs for given parameters.

Abstract

We determine the minimum sum--connectivity index of bicyclic graphs with $n$ vertices and matching number $m$, where $2\le m\le \lfloor\frac{n}{2}\rfloor$, the minimum and the second minimum, as well as the maximum and the second maximum sum--connectivity indices of bicyclic graphs with $n\ge 5$ vertices. The extremal graphs are characterized.

Equations

$$R(G)=\sum _{uv\in E(G)}\frac{1}{\sqrt{d_G(u)d_G(v)}}.$$

$$\chi (G)=\sum _{uv\in E(G)}\frac{1}{\sqrt{d_G(u)+d_G(v)}}.$$

$$\chi (G)-\chi (G-u-v)\ge \frac{d_G(w)-1}{\sqrt{d_G(w)+2}}-\frac{d_G(w)-3}{\sqrt{d_G(w)+1}}-\frac{1}{\sqrt{d_G(w)}}+\frac{1}{\sqrt{3}}$$

$$\chi (G)-\chi (G-u)\ge \frac{d_G(v)-k}{\sqrt{d_G(v)+2}}+\frac{2k-d_G(v)}{\sqrt{d_G(v)+1}}-\frac{k-1}{\sqrt{d_G(v)}}$$

$$-\frac{m+1}{\sqrt{m+4}}+\frac{m-1}{\sqrt{m+3}}+\frac{1}{\sqrt{m+2}}\ge -\frac{4}{\sqrt{7}}+\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{5}}$$

$$-\frac{m+1}{\sqrt{m+4}}+\frac{m-1}{\sqrt{m+3}}+\frac{1}{\sqrt{m+2}}$$

$$\chi (G)\ge \frac{m}{\sqrt{m+3}}+\frac{1}{\sqrt{m+2}}+\frac{m-2}{\sqrt{3}}+\frac{1}{2}$$

$$\chi (G)$$

$$\chi (G)\ge \frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1$$

$$\chi (G)$$

$$\chi (G)\ge \frac{m+1}{\sqrt{n-m+4}}+\frac{n-2m+1}{\sqrt{n-m+3}}+\frac{m-3}{\sqrt{3}}+1$$

$$\chi (G)$$

$$\chi (B_n(a+1,b-1))-\chi (B_n(a,b))$$

$$\chi (G)=\frac{2}{\sqrt{n+1}}+\frac{n-4}{\sqrt{n-1}}+\frac{1}{\sqrt{6}}+\frac{2}{\sqrt{5}}$$

$$\chi (B_{n,3})-\chi (B_n(n-1,3))$$

$$\chi (B_{n,3})-\chi (B_n(n-2,4))$$

$$\chi (H^{\prime })-\chi (H)$$

$$\chi (H^{\prime })-\chi (H)=\left(\frac{1}{2}+\frac{4}{\sqrt{6}}\right)-\left(\frac{4}{\sqrt{7}}+\frac{1}{\sqrt{6}}\right)>0.$$

$$\chi (H^{\prime })-\chi (H)=\left(1+\frac{4}{\sqrt{6}}\right)-\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{3}}\right)>0.$$

$$\chi (G_1)-\chi (G)=\frac{4}{\sqrt{5}}-\left(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{5}}\right)>0.$$

$$\chi (G_1)-\chi (G)=\left(\frac{1}{2}+\frac{4}{\sqrt{5}}\right)-\left(\frac{1}{\sqrt{7}}+\frac{3}{\sqrt{6}}+\frac{1}{\sqrt{3}}\right)>0.$$

$$\chi (G_2)-\chi (G)$$

$$\chi (H_1)=\frac{n-4}{2}+\frac{1}{\sqrt{6}}+\frac{4}{\sqrt{5}}>\chi (H_2)=\frac{n-5}{2}+\frac{6}{\sqrt{5}}.$$

$$\chi (G_1)-\chi (G)$$

$$\chi (G_2)-\chi (G)$$

$$\chi (H_1)=\frac{n-4}{2}+\frac{1}{\sqrt{6}}+\frac{4}{\sqrt{5}}>\chi (H_2)=\frac{n-5}{2}+\frac{6}{\sqrt{5}}.$$

Full text

MSC 2000: 05C90; 05C35; 05C07

Keywords: Sum–connectivity index; Randić connectivity index; Matching number; Bicyclic graphs

1 Introduction

The Randić connectivity index [8] is one of the most successful molecular descriptors in structure–property and structure–activity relationships studies, e.g., [9, 10]. Its mathematical properties as well as those of its generalizations have been studied extensively as summarized in the books [6, 5]. Recently, a closely related variant of Randić connectivity index called the sum–connectivity index was proposed in [14].

Let $G$ be a simple graph with vertex set $V(G)$ and edge set $E(G)$. For $u\in V(G)$, $d_{G}(u)$ denotes the degree of $u$ in $G$. The Randić connectivity index (or product–connectivity index [14, 7]) of the graph $G$ is defined as [8]

[TABLE]

The sum–connectivity index of $G$ is defined as [14]

[TABLE]

It has been found that the sum–connectivity index and the Randić connectivity index correlate well among themselves and with $\pi$-electronic energy of benzenoid hydrocarbons [14, 7]. Some mathematical properties of the sum–connectivity index have been established in [14, 2, 3]. Recall that an $n$-vertex connected graph is known as a tree, a unicyclic graph and a bicyclic graph if it possesses $n-1$, $n$ and $n+1$ edges, respectively. In [2], we obtained the minimum sum–connectivity indices of trees and unicyclic graphs respectively with given number of vertices and matching number, and determined the corresponding extremal graphs. The $n$-vertex trees with the first a few minimum and maximum sum–connectivity indices were determined in [14], while the $n$-vertex unicyclic graphs with the first a few minimum and maximum sum–connectivity indices were determined in [2] and [3], respectively. In this paper, we consider the sum–connectivity indices of bicyclic graphs.

A matching $M$ of the graph $G$ is a subset of $E(G)$ such that no two edges in $M$ share a common vertex. A matching $M$ of $G$ is said to be maximum, if for any other matching $M^{\prime}$ of $G$, $|M^{\prime}|\leq|M|$. The matching number of $G$ is the number of edges of a maximum matching in $G$.

If $M$ is a matching of a graph $G$ and vertex $v\in V(G)$ is incident with an edge of $M$, then $v$ is said to be $M$-saturated, and if every vertex of $G$ is $M$-saturated, then $M$ is a perfect matching.

In this paper, we obtain the minimum sum–connectivity index in the set of bicyclic graphs with $n$ vertices and matching number $m$, where $2\leq m\leq\lfloor n/2\rfloor$. We also determine the minimum and the second minimum, as well as the maximum and the second maximum sum–connectivity indices in the set of bicyclic graphs with $n\geq 5$ vertices. The extremal graphs are characterized.

Study on the Randić connectivity indices of bicyclic graphs may be found in [6, 15, 1, 12], and in particular, the minimum and the maximum Randić connectivity indices in the set of bicyclic graphs with $n\geq 5$ vertices were determined in [12] and [1], respectively.

We note that some other graph invariants based on end–vertex degrees of edges in a graph have been studied recently, see, e.g., [4, 11, 13].

2 Preliminaries

For $2\leq m\leq\lfloor n/2\rfloor$, let $\mathcal{B}(n,m)$ be the set of bicyclic graphs with $n$ vertices and matching number $m$.

For $3\leq m\leq\lfloor n/2\rfloor$, let $B_{n,m}$ be the graph obtained by identifying a vertex of two triangles, and attaching $n-2m+1$ pendent vertices (vertices of degree one) and $m-3$ paths on two vertices to the common vertex of the two triangles, see Fig.

1. Obviously, $B_{n,m}\in\mathcal{B}(n,m)$.

\left\}\begin{array}[]{c}\\ \\ \\ \end{array}\right.$$n-2m+1$$\left\{\begin{array}[]{c}\\ \\ \\ \end{array}\right.$$m-3Fig. 1. The graph $B_{n,m}$.

Let $C_{n}$ ba a cycle on $n\geq 3$ vertices. Let $\widetilde{\mathbb{B}}(n)$ be the set of $n$-vertex bicyclic graphs without pendent vertices, where $n\geq 4$. Let ${\bf B}_{1}^{(1)}(n)$ be the set of bicyclic graphs obtained by joining two vertex–disjoint cycles $C_{a}$ and $C_{b}$ with $a+b=n$ by an edge, where $n\geq 6$. Let ${\bf B}_{1}^{(2)}(n)$ be the set of bicyclic graphs obtained by joining two vertex–disjoint cycles $C_{a}$ and $C_{b}$ with $a+b<n$ by a path of length $n-a-b+1$, where $n\geq 7$. Let ${\bf B}_{2}(n)$ be the set of bicyclic graphs obtained by identifying a vertex of $C_{a}$ and a vertex of $C_{b}$ with $a+b=n+1$, where $n\geq 5$. Let ${\bf B}_{3}^{(1)}(n)$ be the set of bicyclic graphs obtained from $C_{n}$ by adding an edge, where $n\geq 4$. Let ${\bf B}_{3}^{(2)}(n)$ be the set of bicyclic graphs obtained by joining two non-adjacent vertices of $C_{a}$ with $4\leq a\leq n-1$ by a path of length $n-a+1$, where $n\geq 5$. Obviously, $\widetilde{\mathbb{B}}(n)={\bf B}_{1}^{(1)}(n)\cup{\bf B}_{1}^{(2)}(n)\cup{\bf B}_{2}(n)\cup{\bf B}_{3}^{(1)}(n)\cup{\bf B}_{3}^{(2)}(n)$.

Let $\mathbb{B}(n)$ be the set of bicyclic graphs on $n\geq 4$ vertices.

3 Minimum sum–connectivity index of bicyclic graphs with given matching number

First we give some lemmas that will be used.

For a graph $G$ with $u\in V(G)$, $G-u$ denotes the graph resulting from $G$ by deleting the vertex $u$ (and its incident edges).

Lemma 3.1

[2]** Let $G$ be an $n$-vertex connected graph with a pendent vertex $u$, where $n\geq 4$. Let $v$ be the unique neighbor of $u$, and let $w$ be a neighbor of $v$ different from $u$.

1. (i)

If $d_{G}(v)=2$ and there is at most one pendent neighbor of $w$ in $G$, then

[TABLE]

with equality if and only if one neighbor of $w$ has degree one, and the other neighbors of $w$ are of degree two. 2. (ii)

If there are at most $k$ pendent neighbors of $v$ in $G$, then

[TABLE]

with equality if and only if $k$ neighbors of $v$ have degree one, and the other neighbors of $v$ are of degree two.

Lemma 3.2

[2]** (i) The function $\frac{x-1}{\sqrt{x+2}}-\frac{x-3}{\sqrt{x+1}}-\frac{1}{\sqrt{x}}$ is decreasing for $x\geq 2$.

(ii) For integer $a\geq 1$, the function $\frac{x-a}{\sqrt{x+2}}+\frac{2a-x}{\sqrt{x+1}}-\frac{a-1}{\sqrt{x}}$ is decreasing for $x\geq a+1$.

Lemma 3.3

[2]** Let $G$ be a connected graph with $uv\in E(G)$, where $d_{G}(u)$, $d_{G}(v)\geq 2$, and $u$ and $v$ have no common neighbor in $G$. Let $G_{1}$ be the graph obtained from $G$ by deleting the edge $uv$, identifying $u$ and $v$, which is denoted by $w$, and attaching a pendent vertex to $w$. Then $\chi(G)>\chi(G_{1})$.

Lemma 3.4

For $m\geq 3$, $m+\frac{4}{\sqrt{6}}-\frac{3}{2}>\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1$, and for $m\geq 5$, $(\frac{1}{2}+\frac{1}{\sqrt{6}})m-\frac{1}{2}-\frac{2}{\sqrt{6}}+\sqrt{2}>\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1$.

Proof. Let $f(m)=\left(m+\frac{4}{\sqrt{6}}-\frac{3}{2}\right)-\left(\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1\right)$ for $m\geq 3$, and let $g(m)=\left[\left(\frac{1}{2}+\frac{1}{\sqrt{6}}\right)m-\frac{1}{2}-\frac{2}{\sqrt{6}}+\sqrt{2}\right]-\left(\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1\right)$ for $m\geq 5$. Note that $f^{\prime\prime}(m)=g^{\prime\prime}(m)=-\frac{3}{4}(m+3)^{-5/2}+(\frac{1}{4}m+\frac{13}{4})(m+4)^{-5/2}>0$. Then $f^{\prime}(m)\geq f^{\prime}(3)>0$, implying that $f(m)\geq f(3)>0$, and $g^{\prime}(m)\geq g^{\prime}(5)>0$, implying that $g(m)\geq g(5)>0$. $\Box$

Lemma 3.5

For $m\geq 3$,

[TABLE]

with equality if and only if $m=3$.

Proof. Let $f(m)=(m+2)^{-1/2}+m\cdot(m+3)^{-1/2}$ for $m\geq 3$. Then $f^{\prime\prime}(m)=\frac{3}{4}(m+2)^{-5/2}-(\frac{1}{4}m+3)(m+3)^{-5/2}<0$, implying that $f(m)-f(m+1)$ is increasing on $m$. It is easily seen that

[TABLE]

with equality if and only if $m=3$. $\Box$

Let $H_{6}$ be the graph obtained by attaching a pendent vertex to every vertex of a triangle. For $2\leq m\leq\lfloor n/2\rfloor$, let $U_{n,m}$ be the unicyclic graph obtained by attaching $n-2m+1$ pendent vertices and $m-2$ paths on two vertices to one vertex of a triangle.

Lemma 3.6

[2]** Let $G$ be a unicyclic graph with $2m$ vertices and perfect matching, where $m\geq 3$. Suppose that $G\neq H_{6}$. Then

[TABLE]

with equality if and only if $G=U_{2m,m}$.

For an edge $uv$ of the graph $G$ (the complement of $G$, respectively), $G-uv$ ($G+uv$, respectively) denotes the graph resulting from $G$ by deleting (adding, respectively) the edge $uv$.

Lemma 3.7

Let $G\in\mathcal{B}(2m,m)$ and no pendent vertex has neighbor of degree two, where $m\geq 3$. Then $\chi(G)\geq\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1$ with equality if and only if $m=3$ and $G=B_{6,3}$.

Proof. Let $f(m)=\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1$.

Since $G\in\mathcal{B}(2m,m)$ and no pendent vertex has neighbor of degree two, $G$ is obtainable by attaching some pendent vertices to a graph in $\widetilde{\mathbb{B}}(k)$, where $m\leq k\leq 2m$, and any two pendent vertices have no common neighbor (if $k=2m$, then no pendent vertex is attached).

Case 1. There is no vertex of degree two in $G$. Then either $k=m$, $G$ is obtainable by attaching a pendent vertex to every vertex of a graph in $\widetilde{\mathbb{B}}(m)$, or $k=m+1$, $G$ is obtainable by attaching a pendent vertex to every vertex with degree two of a graph in ${\bf B}_{1}^{(1)}(m+1)\cup{\bf B}_{3}^{(1)}(m+1)$. By direct calculation, we find that $\chi(G)=\frac{5}{\sqrt{6}}+1>f(3)$ for $m=3$, $\chi(G)\geq\frac{1}{\sqrt{8}}+\frac{4}{\sqrt{7}}+\frac{2}{\sqrt{5}}+1>f(4)$ for $m=4$, and $\chi(G)\geq\left(\frac{1}{2}+\frac{1}{\sqrt{6}}\right)m-\frac{1}{2}-\frac{2}{\sqrt{6}}+\sqrt{2}$ for $m\geq 5$. Thus by Lemma 3.4, we have $\chi(G)>f(m)$.

Case 2. There is a vertex, say $u$, of degree two in $G$. Denote by $v$ and $w$ the two neighbors of $u$ in $G$. Then one of the two edges incident with $u$, say $uv\in M$, where $M$ is a perfect matching of $G$. Suppose that there is no vertex of degree two in any cycle of $G$. Since no pendent vertex has neighbor of degree two in $G$, $u$ lies on the path joining the two disjoint cycles of $G$. For $G_{1}=G-uw+vw\in\mathcal{B}(2m,m)$, the difference of the numbers of vertices of degree two outside any cycle of $G$ and $G_{1}$ is equal to one, and thus by Lemma 3.3, $\chi(G_{1})<\chi(G)$. Repeating the operation from $G$ to $G_{1}$, we finally get a graph $G^{\prime}\in\mathcal{B}(2m,m)$, which has no vertex of degree two, such that $\chi(G)>\chi(G^{\prime})$, and thus the result follows from Case 1. Now suppose that $u$ lies on some cycle of $G$. Consider $G^{\prime}=G-uw$, which is a unicyclic graph with perfect matching. If $G^{\prime}=H_{6}$, then $G$ is obtained from $H_{6}$ by adding an edge either between two pendent vertices, and thus $\chi(G)=\frac{3}{\sqrt{6}}+\frac{2}{\sqrt{5}}+1$, or between two neighbors of a vertex of degree three, one of which being a pendent vertex, and thus $\chi(G)=\frac{2}{\sqrt{7}}+\frac{2}{\sqrt{6}}+\frac{2}{\sqrt{5}}+\frac{1}{2}$. In either case, $\chi(G)>f(3)$. Suppose that $G^{\prime}\neq H_{6}$. Then by Lemma 3.6, $\chi(G^{\prime})\geq\frac{m}{\sqrt{m+3}}+\frac{1}{\sqrt{m+2}}+\frac{m-2}{\sqrt{3}}+\frac{1}{2}$. Note that $2\leq d_{G}(v),d_{G}(w)\leq 5$ and $w$ has at most one pendent neighbor. By Lemmas 3.2 (i) and 3.5, we have

[TABLE]

with equalities if and only if $d_{G}(v)=2$, $d_{G}(w)=5$, $G^{\prime}=U_{2m,m}$ and $m=3$, i.e., $G=B_{6,3}$.

By combining Cases 1 and 2, the result follows. $\Box$

Lemma 3.8

Let $G\in\mathcal{B}(6,3)$. Then $\chi(G)\geq\frac{4}{\sqrt{7}}+\frac{1}{\sqrt{6}}+1$ with equality if and only if $G=B_{6,3}$.

Proof. If $G$ has a pendent vertex whose neighbor is of degree two, then $G$ is the graph obtained from the unique $4$-vertex bicyclic graph by attaching a path on two vertices to either a vertex of degree three, or a vertex of degree two, and thus it is easily seen that $\chi(G)>\frac{4}{\sqrt{7}}+\frac{1}{\sqrt{6}}+1$. Otherwise, by Lemma 3.7, $B_{6,3}$ is the unique graph with the minimum sum–connectivity index. $\Box$

Now we consider the bicyclic graphs with perfect matching. There is a unique bicyclic graph with four vertices, and its matching number is two.

Theorem 3.1

Let $G\in\mathcal{B}(2m,m)$, where $m\geq 3$. Then

[TABLE]

with equality if and only if $G=B_{2m,m}$.

Proof. Let $f(m)=\frac{m+1}{\sqrt{m+4}}+\frac{1}{\sqrt{m+3}}+\frac{m-3}{\sqrt{3}}+1$. We prove the result by induction on $m$. If $m=3$, then the result follows from Lemma 3.8.

Suppose that $m\geq 4$ and the result holds for graphs in $\mathcal{B}(2m-2,m-1)$. Let $G\in\mathcal{B}(2m,m)$ with a perfect matching $M$.

If there is no pendent vertex with neighbor of degree two in $G$, then by Lemma 3.7, $\chi(G)>f(m)$. Suppose that $G$ has a pendent vertex $u$ whose neighbor $v$ is of degree two. Then $uv\in M$ and $G-u-v\in\mathcal{B}(2m-2,m-1)$. Let $w$ be the neighbor of $v$ different from $u$. Since $|M|=m$, we have $d_{G}(w)\leq m+2$. Note that there is at most one pendent neighbor of $w$ in $G$. Then by Lemma 3.1 (i), Lemma 3.2 (i) and the induction hypothesis,

[TABLE]

with equalities if and only if $G-u-v=B_{2m-2,m-1}$ and $d_{G}(w)=m+2$, i.e., $G=B_{2m,m}$. $\Box$

In the following we consider the sum–connectivity indices of graphs in the set of bicyclic graphs with $n$ vertices and matching number $m$. We first consider the case $m\geq 3$.

Lemma 3.9

[15]** Let $G\in\mathcal{B}(n,m)$ with $n>2m\geq 6$, and $G$ has at least one pendent vertex. Then there is a maximum matching $M$ and a pendent vertex $u$ such that $u$ is not $M$-saturated.

Theorem 3.2

Let $G\in\mathcal{B}(n,m)$, where $3\leq m\leq\lfloor n/2\rfloor$. Then

[TABLE]

with equality if and only if $G=B_{n,m}$.

Proof. Let $f(n,m)=\frac{m+1}{\sqrt{n-m+4}}+\frac{n-2m+1}{\sqrt{n-m+3}}+\frac{m-3}{\sqrt{3}}+1$. We prove the result by induction on $n$. If $n=2m$, then the result follows from Theorem 3.1. Suppose that $n>2m$ and the result holds for graphs in $\mathcal{B}(n-1,m)$. Let $G\in\mathcal{B}(n,m)$.

Suppose that there is no pendent vertex in $G$. Then $G\in\widetilde{\mathbb{B}}(n)$ and $n=2m+1$. It is easily seen that there are exactly three values for $\chi(G)$, and thus we have $\chi(G)\geq\chi(H)=m-1+\frac{4}{\sqrt{6}}$ with $H\in{\bf B}_{2}(2m+1)$. Let $g(m)=\left(m-1+\frac{4}{\sqrt{6}}\right)-f(2m+1,m)=\left(m-1+\frac{4}{\sqrt{6}}\right)-\left(\frac{m+1}{\sqrt{m+5}}+\frac{2}{\sqrt{m+4}}+\frac{m-3}{\sqrt{3}}+1\right)$ for $m\geq 3$. Then $g^{\prime\prime}(m)=(\frac{1}{4}m+\frac{17}{4})(m+5)^{-5/2}-\frac{3}{2}(m+4)^{-5/2}>0$, and thus $g^{\prime}(m)\geq g^{\prime}(3)>0$, implying that $g(m)\geq g(3)>0$, i.e., $m-1+\frac{4}{\sqrt{6}}>f(2m+1,m)$. Then $\chi(G)>f(2m+1,m)$.

Suppose that there is at least one pendent vertex in $G$. By Lemma 3.9, there is a maximum matching $M$ and a pendent vertex $u$ of $G$ such that $u$ is not $M$-saturated. Then $G-u\in\mathcal{B}(n-1,m)$. Let $v$ be the unique neighbor of $u$. Since $M$ is a maximum matching, $M$ contains one edge incident with $v$. Note that there are $n+1-m$ edges of $G$ outside $M$. Then $d_{G}(v)-1\leq n+1-m$, i.e., $d_{G}(v)\leq n-m+2$. Let $s$ be the number of pendent neighbors of $v$ in $G$. Since at least $s-1$ pendent neighbors of $v$ are not $M$-saturated, we have $s-1\leq n-2m$, i.e., $s\leq n-2m+1$. By Lemma 3.1 (ii) with $k=n-2m+1$, Lemma 3.2 (ii) and the induction hypothesis,

[TABLE]

with equalities if and only if $G-u=B_{n-1,m}$, $s=n-2m+1$ and $d_{G}(v)=n-m+2$, i.e., $G=B_{n,m}$. $\Box$

Now we consider the sum–connectivity indices of graphs bicyclic graphs matching number two. Let $B_{n}(a,b)$ be the graph obtained by attaching $a-3$ and $b-3$ pendent vertices to the two vertices of degree three of the unique $4$-vertex bicyclic graph, respectively, where $a\geq b\geq 3$, $a+b=n+2$ and $n\geq 4$.

Lemma 3.10

Among the graphs in $\mathcal{B}(n,2)$ with $n\geq 6$, $B_{n}(n-1,3)$ and $B_{n}(n-2,4)$ are respectively the unique graphs with the minimum and the second minimum sum–connectivity indices, which are equal to $\frac{1}{\sqrt{n+2}}+\frac{n-4}{\sqrt{n}}+\frac{2}{\sqrt{n+1}}+\frac{2}{\sqrt{5}}$ and $\frac{1}{\sqrt{n+2}}+\frac{2}{\sqrt{n}}+\frac{n-5}{\sqrt{n-1}}+\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{5}}$, respectively.

Proof. Let $G\in\mathcal{B}(n,2)$. Then $G$ may be of three types:

(a) $G=B_{n}(a,b)$ with $a\geq b\geq 3$. Suppose that $a\geq b\geq 4$. Let $f(x)=(x-4)x^{-1/2}+2(x+1)^{-1/2}$ for $x\geq 3$. Then $f^{\prime\prime}(x)=-(\frac{1}{4}x+3)x^{-5/2}+\frac{3}{2}(x+1)^{-5/2}<0$, implying that $f(x+1)-f(x)$ is decreasing for $x\geq 3$. It is easily seen that

[TABLE]

and thus, $\chi(B_{n}(a,b))>\chi(B_{n}(a+1,b-1))$ for $a\geq b\geq 4$. It follows that $B_{n}(n-1,3)$ and $B_{n}(n-2,4)$ are respectively the unique graphs with the minimum and the second minimum sum–connectivity indices, which are equal to $\frac{1}{\sqrt{n+2}}+\frac{2}{\sqrt{n+1}}+\frac{n-4}{\sqrt{n}}+\frac{2}{\sqrt{5}}$ and $\frac{1}{\sqrt{n+2}}+\frac{2}{\sqrt{n}}+\frac{n-5}{\sqrt{n-1}}+\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{5}}$, respectively.

(b) $G$ is the graph obtained by attaching $n-4$ pendent vertices to a vertex of degree two of the unique $4$-vertex bicyclic graph. Then

[TABLE]

since $\chi(G)-\chi(B_{n}(n-2,4))=[g(n-1)-g(n)]+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}>0$, where $g(x)=\frac{1}{\sqrt{x+2}}+\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}$ is decreasing for $x\geq 5$.

(c) $G$ is the graph obtained by attaching some pendent vertices to one or two vertices of degree three of the unique $5$-vertex bicyclic graph in ${\bf B}^{(2)}_{3}(5)$, and by Lemma 3.3 and the arguments in case (a), $\chi(G)>\chi(B_{n}(n-2,4))$.

Now the result follows easily. $\Box$

4 Minimum sum–connectivity index of bicyclic graphs

In this section, we determine the minimum and the second minimum sum–connectivity indices of bicyclic graphs with $n\geq 5$ vertices.

Theorem 4.1

Among the graphs in $\mathbb{B}(n)$ with $n\geq 5$, $B_{n}(n-1,3)$ is the unique graph with the minimum sum–connectivity index, which is equal to $\frac{1}{\sqrt{n+2}}+\frac{n-4}{\sqrt{n}}+\frac{2}{\sqrt{n+1}}+\frac{2}{\sqrt{5}}$, the graph obtained by attaching a pendent vertex to a vertex of degree two of the unique $4$-vertex bicyclic graph for $n=5$ is the unique graph with the second minimum sum–connectivity index, which is equal to $\frac{3}{\sqrt{6}}+\frac{2}{\sqrt{5}}+\frac{1}{2}$, $B_{n}(n-2,4)$ for $n=6,7$ is the unique graph with the second minimum sum–connectivity index, which is equal to $\frac{1}{\sqrt{n+2}}+\frac{2}{\sqrt{n}}+\frac{n-5}{\sqrt{n-1}}+\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{5}}$, and $B_{n,3}$ for $n\geq 8$ is the unique graph with the second minimum sum–connectivity index, which is equal to $\frac{4}{\sqrt{n+1}}+\frac{n-5}{\sqrt{n}}+1$.

Proof. There are five graphs in $\mathbb{B}(5)$. Thus, the case $n=5$ may be checked directly. Suppose in the following that $n\geq 6$.

Let $G\in\mathbb{B}(n)$ and $m$ the matching number of $G$, where $2\leq m\leq\lfloor n/2\rfloor$. If $m=2$, then by Lemma 3.10, $\chi(G)\geq\chi(B_{n}(n-1,3))$ with equality if and only if $G=B_{n}(n-1,3)$. If $m=3$, then by Theorem 3.2, $\chi(G)\geq\chi(B_{n,3})$ with equality if and only if $G=B_{n,3}$. If $m\geq 4$, then by Theorem 3.2 and Lemma 3.3, $\chi(G)\geq\chi(B_{n,m})>\chi(B_{n,m-1})>\cdots>\chi(B_{n,3})$. Let $f(x)=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}$ for $x\geq 6$. Then $f^{\prime\prime}(x)=\frac{3}{4}x^{-5/2}-\frac{3}{4}(x+1)^{-5/2}>0$, implying that $f(x+1)-f(x)$ is increasing for $x\geq 6$. Note that

[TABLE]

Thus $B_{n}(n-1,3)$ is the unique graph with the minimum sum–connectivity index.

Suppose that $G\neq B_{n}(n-1,3)$. If $m=2$, then by Lemma 3.10, $\chi(G)\geq\chi(B_{n}(n-2,4))$ with equality if and only if $G=B_{n}(n-2,4)$. By the arguments as above, the second minimum sum–connectivity index of graphs in $\mathbb{B}(n)$ is precisely achieved by the minimum one of $\chi(B_{n,3})$ and $\chi(B_{n}(n-2,4))$. If $n=6,7$, then $\chi(B_{n,3})>\chi(B_{n}(n-2,4))$. Suppose that $n\geq 8$. Let $g(x)=\frac{1}{\sqrt{x+1}}-\frac{3}{\sqrt{x}}-\frac{x-5}{\sqrt{x-1}}$ for $x\geq 8$. Then $g^{\prime\prime}(x)=\frac{3}{4}(x+1)^{-5/2}+\left[(\frac{1}{4}x+\frac{11}{4})(x-1)^{-5/2}-\frac{9}{4}x^{-5/2}\right]>0$, implying that $g(x)-g(x+1)$ is decreasing for $x\geq 8$. Note that

[TABLE]

and then $\chi(B_{n,3})<\chi(B_{n}(n-2,4))$. Thus $B_{n}(n-2,4)$ for $n=6,7$ and $B_{n,3}$ for $n\geq 8$ are the unique graphs with the second minimum sum–connectivity index among graphs in $\mathbb{B}(n)$. $\Box$

5 Maximum sum–connectivity index of bicyclic graphs

In this section, we determine the maximum and the second maximum sum–connectivity indices of bicyclic graphs with $n\geq 5$ vertices. Let $P_{n}$ be the path on $n$ vertices.

Lemma 5.1

[14]** For a connected graph $Q$ with at least two vertices and a vertex $u\in V(Q)$, let $G_{1}$ be the graph obtained from $Q$ by attaching two paths $P_{a}$ and $P_{b}$ to $u$, $G_{2}$ the graph obtained from $Q$ by attaching a path $P_{a+b}$ to $u$, where $a\geq b\geq 1$. Then $\chi(G_{1})<\chi(G_{2})$.

Lemma 5.2

Suppose that $M$ is a connected graph with $u\in V(M)$ and $2\leq d_{M}(u)\leq 4$. Let $H$ be the graph obtained from $M$ by attaching a path $P_{a}$ to $u$. Denote by $u_{1}$ and $u_{2}$ the two neighbors of $u$ in $M$, and $u^{\prime}$ the pendent vertex of the path attached to $u$ in $H$. Let $H^{\prime}=H-uu_{2}+u^{\prime}u_{2}$.

1. (i)

If $d_{M}(u)=2$ and the maximum degree of $M$ is at most five, then $\chi(H^{\prime})>\chi(H)$. 2. (ii)

If $d_{M}(u)=3$, and there are at least two neighbors of $u$ in $M$ with degree two and $d_{M}(u_{2})=2$, then $\chi(H^{\prime})>\chi(H)$. 3. (iii)

If $d_{M}(u)=4$ and all the neighbors of $u$ in $M$ are of degree two, then $\chi(H^{\prime})>\chi(H)$.

Proof. (i) If $a=1$, then

[TABLE]

If $a\geq 2$, then

[TABLE]

(ii) There are two neighbors of $u$ with degree two, let $d_{1}$ be the degree of the third neighbor of $u$ in $M$. If $a=1$, then since $\frac{1}{2}+\frac{1}{\sqrt{5}}-\frac{2}{\sqrt{6}}>0$, we have

[TABLE]

If $a\geq 2$, then since $1+\frac{2}{\sqrt{5}}-\frac{3}{\sqrt{6}}-\frac{1}{\sqrt{3}}>0$, we have

[TABLE]

(iii) If $a=1$, then

[TABLE]

If $a\geq 2$, then

[TABLE]

The proof is now completed. $\Box$

Let $\mathbb{B}_{1}(n)$ be the set of connected graphs on $n\geq 6$ vertices with exactly two vertex–disjoint cycles. Let $\mathbb{B}_{2}(n)$ be the set of connected graphs on $n\geq 5$ vertices with exactly two cycles of a common vertex. Let $\mathbb{B}_{3}(n)$ be the set of connected graphs on $n\geq 4$ vertices with exactly two cycles with at least one edge in common. Obviously, $\mathbb{B}(n)=\mathbb{B}_{1}(n)\cup\mathbb{B}_{2}(n)\cup\mathbb{B}_{3}(n)$. For $u,v\in V(G)$, let $d_{G}(u,v)$ be the distance between $u$ and $v$ in $G$.

Lemma 5.3

Among the graphs in $\mathbb{B}_{1}(n)$ with $n\geq 7$, the graphs in ${\bf B}_{1}^{(1)}(n)$ and the graphs in ${\bf B}_{1}^{(2)}(n)$ are respectively the unique graphs with the maximum and the second maximum sum–connectivity indices, which are equal to $\frac{n-4}{2}+\frac{1}{\sqrt{6}}+\frac{4}{\sqrt{5}}$ and $\frac{n-5}{2}+\frac{6}{\sqrt{5}}$, respectively.

Proof. Suppose that $G$ is a graph in $\mathbb{B}_{1}(n)\setminus\left\{{\bf B}_{1}^{(1)}(n)\right\}$ with the maximum sum–connectivity index, and $C^{(1)}$ and $C^{(2)}$ are its two cycles. Let $x_{1}\in V\left(C^{(1)}\right)$ and $y_{1}\in V\left(C^{(2)}\right)$ be the two vertices such that $d_{G}(x_{1},y_{1})=\min\{d_{G}(x,y):x\in V\left(C^{(1)}\right),y\in V\left(C^{(2)}\right)\}$. Let $Q$ be the path joining $x_{1}$ and $y_{1}$. By Lemma 5.1, the vertices outside $C^{(1)}$, $C^{(2)}$ and $Q$ are of degree one or two, the vertices on $C^{(1)}$, $C^{(2)}$ and $Q$ different from $x_{1}$ and $y_{1}$ are of degree two or three, and $d_{G}(x_{1}),d_{G}(y_{1})=3$ or $4$.

Suppose that $d_{G}(x_{1},y_{1})\geq 2$. If there is some vertex, say $x$, on $C^{(1)}$, $C^{(2)}$ or $Q$ different from $x_{1}$ and $y_{1}$ with degree three, then making use of Lemma 5.2 (i) to $H=G$ by setting $u=x$, we may get a graph in $\mathbb{B}_{1}(n)\setminus\left\{{\bf B}_{1}^{(1)}(n)\right\}$ with larger sum–connectivity index, a contradiction. Thus the vertices on $C^{(1)}$, $C^{(2)}$ and $Q$ different from $x_{1}$ and $y_{1}$ are of degree two. If $d_{G}(x_{1})=4$, then making use of Lemma 5.2 (ii) to $H=G$ by setting $u=x_{1}$, we may get a graph in $\mathbb{B}_{1}(n)\setminus\left\{{\bf B}_{1}^{(1)}(n)\right\}$ with larger sum–connectivity index, a contradiction. Thus $d_{G}(x_{1})=3$. Similarly, we have $d_{G}(y_{1})=3$. It follows that $G\in{\bf B}_{1}^{(2)}(n)$.

Suppose that $d_{G}(x_{1},y_{1})=1$. Suppose that one of $x_{1}$ and $y_{1}$, say $x_{1}$, is of degree four. Then by Lemma 5.2 (i), the vertices on $C^{(1)}$ and $C^{(2)}$ different from $x_{1}$ and $y_{1}$ are of degree two. If $d_{G}(y_{1})=4$, then making use of Lemma 5.2 (ii) to $H=G$ by setting $u=y_{1}$, we may get a graph in $\mathbb{B}_{1}(n)\setminus\left\{{\bf B}_{1}^{(1)}(n)\right\}$ with larger sum–connectivity index, a contradiction. Thus $d_{G}(y_{1})=3$. Denote by $x_{2}$ the pendent vertex of the path attached to $x_{1}$. Consider $G_{1}=G-x_{1}y_{1}+x_{2}y_{1}\in{\bf B}_{1}^{(2)}(n)$. If $d_{G}(x_{1},x_{2})=1$, then

[TABLE]

If $d_{G}(x_{1},x_{2})\geq 2$, then

[TABLE]

In either case, $\chi(G_{1})>\chi(G)$ with $G_{1}\in{\bf B}_{1}^{(2)}(n)$, a contradiction. Thus $d_{G}(x_{1})=d_{G}(y_{1})=3$. Note that $G\not\in{\bf B}_{1}^{(1)}(n)$ and by Lemma 5.2 (i), there is exactly one vertex, say $x_{3}\in V\left(C^{(1)}\right)$, on $C^{(1)}$ and $C^{(2)}$ different from $x_{1}$ and $y_{1}$ with degree three. Denote by $x_{4}$ the pendent vertex of the path attached to $x_{3}$. Consider $G_{2}=G-x_{1}y_{1}+x_{4}y_{1}\in{\bf B}_{1}^{(2)}(n)$. Let $d_{1}$ be the degree of the neighbor of $x_{4}$, one neighbor of $x_{1}$ on $C^{(1)}$ is of degree two, and we denote by $d_{2}$ the other degree of the neighbor of $x_{1}$ on $C^{(1)}$, where $d_{1},d_{2}=2$ or $3$. We have

[TABLE]

and thus, $\chi(G_{2})>\chi(G)$ with $G_{2}\in{\bf B}_{1}^{(2)}(n)$, which is also a contradiction.

Now we have shown that the graphs in ${\bf B}_{1}^{(2)}(n)$ are the unique graphs in $\mathbb{B}_{1}(n)\setminus\left\{{\bf B}_{1}^{(1)}(n)\right\}$ with the maximum sum–connectivity index. Note that for $H_{1}\in{\bf B}_{1}^{(1)}(n)$ and $H_{2}\in{\bf B}_{1}^{(2)}(n)$,

[TABLE]

The result follows. $\Box$

Lemma 5.4

Among the graphs in $\mathbb{B}_{3}(n)$ with $n\geq 5$, the graphs in ${\bf B}_{3}^{(1)}(n)$ and the graphs in ${\bf B}_{3}^{(2)}(n)$ are respectively the unique graphs with the maximum and the second maximum sum–connectivity indices, which are equal to $\frac{n-4}{2}+\frac{1}{\sqrt{6}}+\frac{4}{\sqrt{5}}$ and $\frac{n-5}{2}+\frac{6}{\sqrt{5}}$, respectively.

Proof. Suppose that $G$ is a graph in $\mathbb{B}_{3}(n)\setminus\left\{{\bf B}_{3}^{(1)}(n)\right\}$ with the maximum sum–connectivity index. Then $G$ has exactly three cycles, let $C^{(1)}$ and $C^{(2)}$ be its two cycles such that the remaining one is of the maximum length. Let $A$ be the set of the common vertices of $C^{(1)}$ and $C^{(2)}$. Let $v_{1}$ and $v_{2}$ be the two vertices in $A$ such that $d_{G}(v_{1},v_{2})=\max\left\{d_{G}(x,y):x,y\in A\right\}$. By Lemma 5.1, the vertices outside $C^{(1)}$ and $C^{(2)}$ are of degree one or two, the vertices on $C^{(1)}$ and $C^{(2)}$ different from $v_{1}$ and $v_{2}$ are of degree two or three, and $d_{G}(v_{1}),d_{G}(v_{2})=3$ or $4$. Denote by $v_{1}^{\prime}$ ($v_{2}^{\prime}$, respectively) the neighbor of $v_{1}$ on $C^{(1)}$ ($v_{2}$ on $C^{(2)}$, respectively) different from the vertices in $A$.

If $d_{G}(v_{1},v_{2})\geq 2$, then by Lemma 5.2 (i) and (ii), we have $G\in{\bf B}_{3}^{(2)}(n)$.

Suppose that $d_{G}(v_{1},v_{2})=1$. Suppose that the lengths of $C^{(1)}$ and $C^{(2)}$ are at least four. Consider $G_{1}=G-\{v_{1}v_{1}^{\prime},v_{2}v_{2}^{\prime}\}+\{v_{1}^{\prime}v_{2},v_{1}v_{2}^{\prime}\}\in\mathbb{B}_{1}(n)\setminus\left\{{\bf B}_{1}^{(1)}(n)\right\}$. Note that

[TABLE]

If $d_{G}(v_{1})=d_{G}(v_{2})$, then $\chi(G_{1})=\chi(G)$. If $d_{G}(v_{1})\neq d_{G}(v_{2})$, then by Lemma 5.2 (i), we have $d_{G}(v_{1}^{\prime})=d_{G}(v_{2}^{\prime})=2$, and thus $\chi(G_{1})=\chi(G)$. In either case, we have $\chi(G_{1})=\chi(G)$. By Lemma 5.3, we have $\chi(G)=\chi(G_{1})\leq\chi(H)=\frac{n-5}{2}+\frac{6}{\sqrt{5}}$ for $H\in{\bf B}_{1}^{(2)}(n)$ with equality if and only if $G_{1}\in{\bf B}_{1}^{(2)}(n)$, i.e., $G\in{\bf B}_{3}^{(2)}(n)$.

Suppose that at least one of $C^{(1)}$ and $C^{(2)}$, say $C^{(1)}$, is of length three. Since $G\not\in{\bf B}_{3}^{(1)}(n)$, there are some vertices outside $C^{(1)}$ and $C^{(2)}$. By Lemma 5.2 (i) and (ii), the subgraph induced by the vertices outside $C^{(1)}$ and $C^{(2)}$ is a path, say $P_{k}$, which is attached to $x\in V\left(C^{(1)}\right)\cup V\left(C^{(2)}\right)$. Suppose that $x\neq v_{1}^{\prime}$. Denote by $v_{3}$ the neighbor of $x$ outside $C^{(1)}$ and $C^{(2)}$. Consider $G_{2}=G-xv_{3}+v_{1}^{\prime}v_{3}\in\mathbb{B}_{3}(n)\setminus\left\{{\bf B}_{3}^{(1)}(n)\right\}$. If $x=v_{1}$ or $v_{2}$, then

[TABLE]

and thus $\chi(G_{2})>\chi(G)$, a contradiction. Hence $x\in V\left(C^{(2)}\right)\setminus\{v_{1},v_{2}\}$, and the length of $C^{(2)}$ is at least four. Note that one neighbor of $x$ on $C^{(2)}$ is of degree two. Denote by $d_{1}$ the degree of the other neighbor of $x$ on $C^{(2)}$, where $d_{1}=2$ or $3$. Then

[TABLE]

and thus $\chi(G_{2})>\chi(G)$, which is also a contradiction. Thus, $x=v_{1}^{\prime}$. If $k=1$, then $\chi(G)=\frac{n-5}{2}+\frac{3}{\sqrt{6}}+\frac{2}{\sqrt{5}}+\frac{1}{2}$, and if $k\geq 2$, then $\chi(G)=\frac{n-6}{2}+\frac{3}{\sqrt{6}}+\frac{3}{\sqrt{5}}+\frac{1}{\sqrt{3}}$. In either case, we have $\chi(G)<\frac{n-5}{2}+\frac{6}{\sqrt{5}}$.

Now we have shown that the graphs in ${\bf B}_{3}^{(2)}(n)$ are the unique graphs in $\mathbb{B}_{3}(n)\setminus\left\{{\bf B}_{3}^{(1)}(n)\right\}$ with the maximum sum–connectivity index. Note that for $H_{1}\in{\bf B}_{3}^{(1)}(n)$ and $H_{2}\in{\bf B}_{3}^{(2)}(n)$,

[TABLE]

The result follows. $\Box$

Theorem 5.1

Among the graphs in $\mathbb{B}(n)$ with $n\geq 5$, the graph in ${\bf B}_{3}^{(1)}(5)$ and the graph in ${\bf B}_{3}^{(2)}(5)$ for $n=5$ are respectively the unique graphs with the maximum and the second maximum sum–connectivity indices, the graphs in ${\bf B}_{1}^{(1)}(6)\cup{\bf B}_{3}^{(1)}(6)$ and the graph in ${\bf B}_{3}^{(2)}(6)$ for $n=6$ are respectively the unique graphs with the maximum and the second maximum sum–connectivity indices, the graphs in ${\bf B}_{1}^{(1)}(n)\cup{\bf B}_{3}^{(1)}(n)$ and the graphs in ${\bf B}_{1}^{(2)}(n)\cup{\bf B}_{3}^{(2)}(n)$ for $n\geq 7$ are respectively the unique graphs with the maximum and the second maximum sum–connectivity indices, where $\chi(G)=\frac{n-4}{2}+\frac{1}{\sqrt{6}}+\frac{4}{\sqrt{5}}$ for $G\in{\bf B}_{1}^{(1)}(n)\cup{\bf B}_{3}^{(1)}(n)$ and $\chi(H)=\frac{n-5}{2}+\frac{6}{\sqrt{5}}$ for $H\in{\bf B}_{1}^{(2)}(n)\cup{\bf B}_{3}^{(2)}(n)$.

Proof. Suppose that $G$ is a graph in $\mathbb{B}_{2}(n)$ with the maximum sum–connectivity index, and $C^{(1)}$ and $C^{(2)}$ are its two cycles. Let $u$ be the unique common vertex of $C^{(1)}$ and $C^{(2)}$. By Lemma 5.1, the vertices outside $C^{(1)}$ and $C^{(2)}$ are of degree one or two, the vertices on $C^{(1)}$ and $C^{(2)}$ different from $u$ are of degree two or three, and $d_{G}(u)=4$ or $5$. Moreover, by Lemma 5.2 (i), the vertices on $C^{(1)}$ and $C^{(2)}$ different from $u$ are of degree two. If $d_{G}(u)=5$, then making use of Lemma 5.2 (iii) to $H=G$, we may get a graph in $\mathbb{B}_{2}(n)$ with larger sum–connectivity index, a contradiction. Thus $d_{G}(u)=4$, i.e., $G\in{\bf B}_{2}(n)$.

Note that for $H_{1}\in{\bf B}_{1}^{(1)}(n)$, $H_{1}^{\prime}\in{\bf B}_{1}^{(2)}(n)$, $H_{2}\in{\bf B}_{2}(n)$, $H_{3}\in{\bf B}_{3}^{(1)}(n)$ and $H_{3}^{\prime}\in{\bf B}_{3}^{(2)}(n)$,

[TABLE]

Then the result follows from Lemmas 5.3 and 5.4. $\Box$

Acknowledgement. This work was supported by the National Natural Science Foundation of China (Grant No. 10671076).