Abstract

For a graph , its general sum-connectivity is usually denoted by and is defined as the sum of the numbers over all edges of , where represent degrees of the vertices , respectively, and is a real number. This paper addresses the problem of finding graphs possessing the minimum value over the class of all trees with a fixed order and fixed number of pendent vertices for . This problem is solved here for the case when and , by deriving a lower bound on for trees in terms of their orders and number of pendent vertices.

1. Introduction

We only discuss simple connected graphs in this study. For a graph , its edge set and vertex set are represented by and , respectively. Let indicate the degree of a vertex . When there is no uncertainty regarding the graph under consideration, we write instead of . Those (chemical-) graph-theoretical concepts and notation utilized in this article but not described here may be found in related books, such as [14].

For a graph , its general sum-connectivity index , devised in [5], is defined as,where is a real number. The special cases and of yield the first Zagreb index (for example, see [6]) and sum-connectivity index of [7].

A pendent vertex in a graph is a vertex of degree 1. In this paper, we are concerned with the following problem concerning the general sum-connectivity index.

Problem 1. Find the graphs having the maximum and minim general sum-connectivity indexamong all trees of a fixed order and number of pendent vertices.
Cui and Zhong [8] studied the maximal part of problem 1 for . The minimal part of problem 1 for was solved in [9]. Tache and Tomescu [10] reported the solution to the maximal part of problem 1 when ; the minimal part was solved in [11] for . Additional information about the known mathematical aspects of the general sum-connectivity index can be found, for example, in [1216], in [17] (where several general results give special cases for the general sum-connectivity index), and in the related references given therein.
In this paper, a lower bound on for trees of order and number of pendent vertices is derived for and . As a consequence of the obtained bound, the minimal part of problem 1 is solved when and .

2. Main Results

Before proving the main results, we give some definitions and notation that are used in the rest of this section. The star and path graphs with vertices are represented by and , respectively. A branching vertex of a graph is a vertex of degree greater than 2. For a graph and a vertex , denote by the set of all those vertices of that are adjacent to . The elements of the set are known as the neighbors of . A pendent path of a graph is a non-trivial path in such that one of the vertices is pendent and the other is branching, and each (if ) has degree 2.

Transformation 1. Let be a tree containing at least two branching vertices. Let be a pendent path of such that , where and . Choose a vertex lying on the unique path between and another branching vertex of . Suppose that is formed from by inserting the two edges , , and dropping the two edges , . The trees and are depicted in Figure 1.

Lemma 1. Ifandare the trees specified in Transformation 1, thenfor.

Proof. Since and , by using the definition of , one hasfor .
An internal path of a graph is a non-trivial path in such that both the vertices are branching, and each (if ) has degree 2. Denote by the class consisting of all trees with order , pendent vertices , and maximum degree 3 such that every pendent (internal) path of has length one (at least two, respectively). If then after simple calculations, one gets.

Theorem 1. Letbe a tree withvertices, among whichare pendent vertices. Thenwhere the equality sign in the inequality holds if and only if .

Proof. For , the result is straightforwardly verified. Next, suppose that . Note that the function defined by , is strictly increasing for and because . Thus, for we have and hencefor and . In what follows, we assume that with and we prove the result by using mathematical induction on . If , then andfor . Next, assume that and that the theorem holds for , where and .

Case 1. The tree does not contain any pendent path of length greater than 1.
In this case, we consider its several subcases.

Subcase 1. The tree contains at least one pendent vertex whose neighbor has degree greater than 3.
Let be a vertex of degree having pendent neighbors and non-pendent neighbors , where and . Since , one has . By keeping in mind the inequalities , , , and inductive hypothesis, we have.

Subcase 2. Every pendent vertex of is adjacent to a vertex of degree 3 and there is at least one vertex of degree 3 having only one pendent neighbor.
Let be a vertex of degree 3 and be a pendent neighbor of . Assume that and are the non-pendent neighbors of . By utilizing the induction hypothesis, we get,where if and only if and ; e.g., if and only if .

Subcase 3. Every pendent vertex of is adjacent to a vertex of degree 3 and there is at least one vertex of degree 3 having one branching neighbor and two pendent neighbors.
Let be a vertex of degree 3 and be its two pendent neighbors. Assume that is the non-pendent neighbor of . By utilizing the induction hypothesis, we get

Subcase 4. Every pendent vertex of is adjacent to a vertex of degree 3, which has one neighbor of degree 2 and two pendent neighbors.
If has the maximum degree 3 and if contains no pair of adjacent vertices of degree 3, then and we get .
Now, assume that has the maximum degree 3 and such that . Let be a vertex of degree 3 having two pendent vertices and a vertex of degree 2. Take . Let be the tree formed from by deleting the edges , , , and adding the edges , , . Then, is a tree with vertices and pendent vertices. Also, note that both the trees and have the same degree sequence. Since the maximum degree of is 3, it holds that or 3. On the other hand, we haveNote that if then and the tree contains a vertex (namely ) of degree 3 having two pendent neighbors and a neighbor of degree 3, and hence from Subcase 3 it follows that .
If then . If then we are done. If contains at least one pair of adjacent branching vertices then we repeat the above process until we obtain the desired result.
It remains to prove the desired result in the considered case (that is, Subcase 4) when the maximum degree of is greater than 3. Let be a vertex of maximum degree , where . Take and let be a vertex of degree 3 having two pendent neighbors and a neighbor of degree 2. Let be a pendent neighbor of . Without loss of generality, we assume that lies on the unique - path. Let be the tree deduced from by deleting the edge and inserting the edge . Observe that the tree has vertices among which are pendent. Thus, by using the the inequalities , , (because of the considered case), and inductive hypothesis, we havefor .

Case 2. contains at least one pendent path of length greater than 1.
First, we suppose that contains only one branching vertex. Assume also that contains at least two pendent paths of lengths at least 2. Let and be two such pendent paths, where are branching vertices and are pendent vertices. Let be the tree formed by deleting the edge and adding the edge . Since and , we haveThus, from the above discussion, we conclude that if contains only one branching vertex thenwhere is a tree formed by attaching pendent vertices to one of the pendent vertices of the path graph . For , by direct calculations it is verified that the inequalityholds for Also, for and , one hasthe right hand side of this inequality is positive becauseTherefore,Next, we suppose that contains at least two branching vertices. By Transformation 1 and Lemma 1 there exists a tree of order with pendent vertices such that contains no pendent path of length greater than 1 and . However, by Case 1, it holds that . Therefore, . This completes the proof.
Observe that the class is non-empty whenever and . Thus, the next result immediately follows from Theorem 1.

Corollary 1. Forand, among all trees with orderand number of pendent vertices, the elementof the classuniquely minimize the general sum-connectivity index.

3. Concluding remarks

For a graph , its general Platt index [18] is defined aswhere is a real number. The graph invariant is known as the reformulated first Zagreb index of [19]. Because of the similarity between the definitions of and , one obtains Theorem 2 (corresponding to Theorem 1) and Corollary 1 (corresponding to Corollary 2).

Theorem 2. Letbe a tree withvertices, among whichare pendent vertices. Thenwhere the equality sign in the inequality holds if and only if .

Corollary 2. Forand, among all trees with orderand number of pendent vertices, the elementof the classuniquely minimize the general Platt index.

Data Availability

Data about this study may be requested from the authors.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This research has not been funded by any source.