Abstract
In this work, we introduce a new topological index called a general power sum-connectivity index and we discuss this graph invariant for some classes of extremal graphs. This index is defined by , where and represent the degree of vertices and , respectively, and . A connected graph is called a -generalized quasi-tree if there exists a subset of cardinality such that the graph is a tree but for any subset of cardinality , the graph is not a tree. In this work, we find a sharp lower and some sharp upper bounds for this new sum-connectivity index.
1. Introduction
In this article, all the graphs are considered as simple, connected, finite, and undirected. Let us denote a graph by , where and represent the sets of vertices and edges, respectively. Degree of vertex is the number of adjacent vertices to and is denoted by , and the set of vertices adjacent to the vertex is denoted by . The length of a shortest path between two vertices, say and , is termed as the distance between these vertices and is denoted by . The maximum distance from vertex to any other vertex is known as the eccentricity of the vertex and is denoted by and defined as . The diameter of a graph is , and its notation is ; further, see [1–3].
Let be a complete graph of order . The complete bipartite graph , which is also denoted as , represents a star of order , while is the path of order and size . A double star of order is denoted by . The graph is a tree which consists of two adjacent vertices say and , such that is adjacent to pendent vertices and is adjacent to pendent vertices. Alternatively, the double star can be obtained by joining the centers of two stars and . If for a tree the diameter of is 2, then is a star graph, and if the diameter of is 3, then is a double star. For two graphs and whose vertex sets are disjoint, denotes its join graph having the vertex set and is its edge set.
If there exists a vertex in a graph such that the graph is a tree, then the graph is called a quasi-tree and the vertex is called a quasi-vertex. Obviously, every tree is a quasi-tree, as by deletion of any vertex in a tree, the resulting graph is again a tree. Furthermore, a -generalized quasi-tree is a graph in which there exists a subset with cardinality such that the graph is a tree but for any subset with cardinality , the graph is not a tree. The vertices in the set are called -quasi-vertices or quasi-vertices. We need at least vertices for sketching -generalized quasi-tree. Any tree is a quasi-tree which is trivial. Let the class of -generalized quasi-trees of order be denoted by .
In this modern world, the network structure plays the basic role in the field of chemistry, technology, and communication. Every network is distinguished by means of numerical quantity under some parameter. Such roles are called topological indices. A numerical quantity that is invariant under graph automorphisms is termed as the topological index. There are many topological indices such as degree-based, distance-based, and counting-related topological indices. In all the said indices, degree-based topological indices are one of the basic indices and play the key role in chemistry and chemical graph theory. The mostly studied topological indices are the atom-bond (ABC) connectivity index, the harmonic (H) index, the Zagreb index, and various others; further we refer to [4–10].
On the basis of the previously defined indices, we introduce a new sum-connectivity index called a general power sum-connectivity index defined such that
2. Main Work
This section is devoted to the main results which we proved for the new introduced sum-connectivity index for some classes of extremal graphs. The following are the main results in this regard.
Lemma 1. For any two vertices , where , then
Theorem 1. The general power sum-connectivity index is minimum for a graph if , for and .
Proof. Let be the graph of order . To prove that is minimum for , where , we consider two cases.
Case 1. When , in this case, the graph is isomorphic either to a path or to a star , see Figure 1.
In this case, we haveand similarly,It is obvious that . Equality holds if . This can be easily analyzed through the following plot illustrated on Figure 2.
Case 2. When , in this case, suppose on contrary that is not minimum. Then, there exists a graph other than in which at least one vertex, say , has degree greater than 2, i.e., such that . We have to discuss the following subcases .
Subcase B1: when is adjacent to two leaves. A graph for this case is shown in Figure 3. Here, we have and for , we have From (5) and (6), this is obvious that which can be determined through the following plots illustrated on Figure 4 for various values of . This is a contradiction to the minimality of . Hence, in this case, is minimal. Subcase B2: when is adjacent to only one leaf. In this case, the graph is illustrated in Figure 5. Here, is given below: Obviously, from (6) and (7), we have which can be observed through the following plots, see Figure 6, for various values of , which is a contradiction to the minimality of .
Lemma 2. Let , where is the class of -generalized quasi-trees. If is maximum and is a quasi-vertex of , then .
Proof. Consider that is a -generalized quasi-tree from , where is maximum for and let be a quasi-vertex of . We show that . On contrary, suppose that is maximum for and . Then, there is a vertex which is not adjacent to . Then, by means of Lemma 1, we havewhich is a contradiction to the maximality of . Hence, we are forced to accept that .
Roughly speaking that, for , we define the following function:which is defined for , where is a set containing vectors having coordinates from positive integers and satisfying the following:Further, for and , we replace the components of by , rearranging the replaced components in decreasing order asLet ; then, obviously . Let this transformation be denoted by . Next, we haveWe define another function:This function is a strictly decreasing function for and . Thus, if , then which implies that , and thus, . This is due the decreasing function. We can also obtain thatWe obtained that . We can easily deduce the following result.
Lemma 3. If there is an having , then is strictly decreasing on having .
In other words, we observe that the function is strictly increased if we push one unity to the left in the degree sequence and arrange this new sequence in decreasing order using the transformation.
Theorem 2. Let be a tree in , where is the class of trees of order . Then, is maximum if is the star graph .
Proof. Let , such that is maximum. We need to prove that . For this proof, suppose on contrary that is maximum and is neither a star nor a double star. Then, there is a vertex in , say , which is adjacent to two vertices and , such that , , and , where . Without loss of generality, let , see Figure 7.
Furthermore, for the neighbors of and , we have and . We can construct another tree from in such a way that we delete the edges and we insert new edges as in [11]. Now, we have the following:Similarly, for , we haveNext, we get the following:As a consequence of Lemma 3, for and , , , and applying the transformation several times, we get a contradiction to the maximality of . Now, we show that, for and , we have the following:For and , we obtainTheir difference holdsThe reason of the above last inequality can be easily determined as and which implies that and . This further implies that and where . Hence, . This completes the proof.
In a graph , if the weight of every vertex is fixed, let such weight be ; then, we define
Theorem 3. For any and , the unique tree of order having maximum value of is .
Proof. On contrary base, let be a tree having maximum value of , where is neither a star nor a double star say . Then, is greater or equal to 4. Then, there is a vertex in , say , that is adjacent to two vertices say and , such that , , and without loss of generality let . Such a tree is shown in Figure 7 while considering that every vertex has the fixed weight. Furthermore, for and , we can have that and . We can easily construct another tree from by eliminating the edges and adding new edges ; for this, we refer [11]. For our proof, we proceed as follows:On similar way for , we haveAs in Theorem 2, we can easily show that there is a contradiction to the maximality of , where . Also, the following holds:The last inequality can easily be determined as and which implies that , . This further implies that and , where . Hence, .
Theorem 4. For , let . If is minimum and be a quasi-vertex of , then .
Proof. Let such that is minimum. We have to show that where is quasi-vertex. On contrary base, suppose that . In this case, cannot be quasi-vertex because will be pendent vertex and will make no difference in the formation of quasi-tree. On the contrary, let , and this shows that is adjacent to more than two vertices in . Therefore, for elimination of any edge , we have where . This is contradiction to the minimality of . This contradiction is due to our wrong supposition that . Hence, in both cases, there is contradiction due to which we are forced to accept that .
Theorem 5. Let , where and . holdswhere the equality is satisfied if and only if .
Proof. Let and be such that is maximum as possible. Let the set of quasi-vertex be represented by . By means of Lemmas 1 and 2, we have , where is a tree of order . We have to prove that . For this, we have, from [4],We calculate each sum separately:Similarly, the middle sum in (26) can be obtained asUsing Theorem 2, the sum in (28) attains its maximum if . For this, the above last sum becomesSimilarly, the last sum in (26) is calculated as follows:By means of Theorem 2, the above last sum will obtain its maximum if . For this, we have the following:Combining (27), (29), and (31), we have , and consequently, for , we get the following:
Theorem 6. Let and .(i)If and , then . The equality holds if and only if , where is a cycle of order .(ii)If and , then . If and , then . Furthermore, the equality holds if and only if consists of two cycles of length three having a common edge for or two cycles with a common path having length at least two for or two cycles joined by a path of length at least two for .
Proof. Suppose that , where is as small as possible. Let the set of quasi-vertices be ; then, by the definition of a -generalized quasi-tree, is a tree having order . By Theorem 5, we have for all . Furthermore, for every vertex in , is adjacent to two vertices of . From this discussion, we deduce that is a connected graph with edges. It means that has internal -cycles. First of all, we have to show that has no pendent vertex. For this, suppose on contrary base that has a pendent vertex . Let there be another vertex where there is a path between and , and let also belong to a cycle, say , in . Let this path be . We denote the vertices adjacent to different from by , and let their degrees be , . Then, we have . Obviously, and at least one degree . We can define a new -cyclic graph having order , by deleting the edge between and and then arranging between two consecutive vertices, say and , on the cycle . Let and ; clearly, . Then, in the new graph, say , the degree of is , , , and for every . We can write the difference of asFor , we have the following:Also, ; by using Lemma 3, we have , which shows that . This is a contradiction to the minimality of , that is, is another -cyclic graph having smaller than that of .
Now, for , it implies that is a pendent vertex of ; by doing the same process for instead of , we will observe that there is a pendent vertex that is adjacent to on cycle ; then, in this case. We will again get a contradiction, and this contradiction is due to our wrong supposition; hence, we are forced to accept that .
Next, we have to suppose that, for all , .(i)Then, for , this implies that is a connected graph having one cycle and edges. Obviously, ; thus, each vertex of has degree 2, and then, Otherwise, will be greater than that of .(ii)For , it shows that is a 2-cyclic graph and has edges. Now, the sum of degrees of is given in the following: Let us denote these degree sequences by or , that is, or If the degree sequence of is , then(a)The graph contains a common path of length , or(b)The graph has two cycles joined by a path of length In these cases, if , then such graph is shown in Figure 8. For this, we have If , then we have a graph illustrated in Figure 9. For this graph, we obtain If degree sequence of is , then in this case has two cycles having a vertex in common, as shown in Figure 10. For this graph, we get From , , and , we have which is equivalent to and is obvious which can be easily determined through Figure 11. Now, which implies that is positive. This can be seen in Figure 12.
Data Availability
The data used to support the findings of the study are included within this paper.
Conflicts of Interest
The authors declare that there are no conflicts of interest.
Authors’ Contributions
Rui Cheng wrote the final version of this paper and approved the results, Gohar Ali supervised this work, Gul Rahmat proved the results, Muhammad Yasin Khan wrote the paper, Andrea Semanicova-Fenovcikova plotted the results, and Jia-Bao Liu analyzed the results and arranged the funding for this paper.
Acknowledgments
This work was supported by the Slovak Research and Development Agency under the contract no. APVV-19-0153 and by VEGA 1/0233/18.