Investigation of General Power Sum-Connectivity Index for Some Classes of Extremal Graphs

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 Y α ( G ) � 􏽐 uv ∈ E ( G ) ( d ( u ) d ( u ) + d ( v ) d ( v ) ) α , where d ( u ) and d ( v ) represent the degree of vertices u and v , respectively, and α ≥ 1. A connected graph G is called a k -generalized quasi-tree if there exists a subset V k ⊂ V ( G ) of cardinality k such that the graph G − V k is a tree but for any subset V k − 1 ⊂ V ( G ) of cardinality k − 1, the graph G − V k − 1 is not a tree. In this work, we ﬁnd a sharp lower and some sharp upper bounds for this new sum-connectivity


Introduction
In this article, all the graphs are considered as simple, connected, finite, and undirected. Let us denote a graph by G � (V(G), E(G)), where V(G) and E(G) represent the sets of vertices and edges, respectively. Degree of vertex u is the number of adjacent vertices to u and is denoted by d(u), and the set of vertices adjacent to the vertex u is denoted by N(u). e length of a shortest path between two vertices, say u and v, is termed as the distance between these vertices and is denoted by d (u, v). e maximum distance from vertex u to any other vertex is known as the eccentricity of the vertex u and is denoted by ε(u) and defined as ε(u) � max v∈V(G) d (u, v). e diameter of a graph G is max u∈V(G) � max u,v∈V(G) , and its notation is diam(G); further, see [1][2][3].
Let K n be a complete graph of order n. e complete bipartite graph K 1,n− 1 , which is also denoted as S n , represents a star of order n, while P n is the path of order n and size n − 1. A double star of order a + b is denoted by S a,b . e graph S a,b is a tree which consists of two adjacent vertices say u and v, such that u is adjacent to b − 1 pendent vertices and v is adjacent to a − 1 pendent vertices. Alternatively, the double star S a,b can be obtained by joining the centers of two stars S a+1 and S b+1 . If for a tree T the diameter of T is 2, then T is a star graph, and if the diameter of T is 3, then T is a double star. For two graphs G and H whose vertex sets are disjoint, G + H denotes its join graph having the vertex set V(G + H) � V(G) ∪ V (H) and } is its edge set. If there exists a vertex v in a graph G such that the graph G − V is a tree, then the graph G is called a quasi-tree and the vertex v 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 k-generalized quasi-tree is a graph in which there exists a subset V k ⊂ V(G) with cardinality k such that the graph G − V k is a tree but for any e vertices in the set V k are called k-quasi-vertices or quasi-vertices. We need at least k + 2 vertices for sketching k-generalized quasi-tree. Any tree is a quasi-tree which is trivial. Let the class of k-generalized quasi-trees of order n be denoted by T k (n).
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. ere 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. e 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][5][6][7][8][9][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

Main Work
is section is devoted to the main results which we proved for the new introduced sum-connectivity index for some classes of extremal graphs. e following are the main results in this regard.

Lemma 1. For any two vertices u and
(2) Theorem 1. e general power sum-connectivity index Y α (G) is minimum for a graph G if G � P n , for n ≥ 4 and α ≥ 1.
Proof. Let G be the graph of order n. To prove that Y α (G) is minimum for G � P n , where n ≥ 4, we consider two cases. □ Case 1. When n � 4, in this case, the graph is isomorphic either to a path P 4 or to a star S 4 , see Figure 1.
In this case, we have and similarly, It is obvious that Y α (P 4 ) < Y α (S 4 ). Equality holds if α � 0. is can be easily analyzed through the following plot illustrated on Figure 2.
Case 2. When n ≥ 5, in this case, suppose on contrary that Y a (P n ) is not minimum. en, there exists a graph other than P n in which at least one vertex, say y, has degree greater than 2, i.e., such that d G (y) ≥ 3. We have to discuss the following subcases .

(7)
Obviously, from (6) and (7), we have Y α (G) > Y α (P n ) which can be observed through the following plots, see Figure 6, for various values of n, which is a contradiction to the minimality of Y α (G). 2 Complexity en, there is a vertex f which is not adjacent to w. en, by means of Lemma 1, we have which is a contradiction to the maximality of Y α (G). Hence, we are forced to accept that d G (w) � n − 1.
us, if i < k, then z i ≥ z k which implies that y Figure 3: A graph G with a vertex y adjacent to two leaves. Complexity is is due the decreasing function. We can also obtain that We obtained that F(z * ) − F(z 1 , z 2 , . . . , z j ) > 0. We can easily deduce the following result.
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 A1 transformation.
Proof. Let T ∈ T n , such that Y α (T) is maximum. We need to prove that T � S n . For this proof, suppose on contrary that Y α (T) is maximum and T is neither a star S n nor a double star. en, there is a vertex in T, say t, which is adjacent to two vertices x and y, such that d T (x) � a, d T (y) � b, and d T (t) � c, where a and b ≥ 2. Without loss of generality, let a ≥ b, see Figure 7.
Furthermore, for the neighbors of x and y, we have N(x) − t � x 1 , x 2 , . . . , x a− 1 and N(y) − t � y 1 , y 2 , . . . , y b− 1 }. We can construct another tree T * from T in such a way that we delete the edges yy 1 , yy 2 , . . . , yy b− 1 and we insert new edges xy 1 , xy 2 , . . . , xy b− 1 as in [11]. Now, we have the following: Similarly, for T, we have y Figure 5: A graph G with a vertex y adjacent to one leaf.

Complexity
Next, we get the following: As a consequence of Lemma 3, for j � 2 and z 1 � a a + c c , and applying the transformation A1 several times, we get a contradiction to the maximality of Y α (T). Now, we show that, for a and b ≥ 2 and a + b � n, we have the following: For S n and S a,b , we obtain eir difference holds e reason of the above last inequality can be easily determined as n − 1 > a and n − 1 > b which implies that (S a,b ). is completes the proof.
In a graph G, if the weight of every vertex is fixed, let such weight be r ≥ 1; then, we define (21) □ Theorem 3. For any α ≥ 1 and r ≥ 1, the unique tree of order n having maximum value of Y α,r (G) is S n .
Proof. On contrary base, let T be a tree having maximum value of Y α,r (G), where T is neither a star nor a double star say S a,b . en, diam(T) is greater or equal to 4. en, there is a vertex in T, say t, that is adjacent to two vertices say x and y, such that d T (x) � a, d T (y) � b, and d T (t) � c without loss of generality let a and b ≥ 2. Such a tree is shown in Figure 7 while considering that every vertex has the fixed weight. Furthermore, for x and y, we can have that N(x) − t � x 1 , x 2 , . . . , x a− 1 and N(y) − t � y 1 , y 2 , . . . , y b− 1 . We can easily construct another tree T * from T by eliminating the edges yy 1 , yy 2 , . . . , yy b− 1 and adding new edges xy 1 , xy 2 , . . . , xy b− 1 ; for this, we refer [11]. For our proof, we proceed as follows: On similar way for T, we have As in eorem 2, we can easily show that there is a contradiction to the maximality of Y α,r (G), where G � T. Also, the following holds: (24) e last inequality can easily be determined as n − 1 > a and n − 1 > b which implies that (n − 1) n− 1 > (a − 1) a− 1 , is further implies that and w be a quasi-vertex of G, then d(w) � 2. Proof. Let G ∈ T k (n) such that Y α (G) is minimum. We have to show that d(w) � 2 where w is quasi-vertex. On contrary base, suppose that d(w) � 1. In this case, w cannot be quasivertex because w will be pendent vertex and w will make no difference in the formation of quasi-tree. On the contrary, let d(w) ≥ 2, and this shows that w is adjacent to more than two erefore, for elimination of any edge is is contradiction to the minimality of Y α (G).
is contradiction is due to our wrong supposition that d(w) ≥ 2. Hence, in both cases, there is contradiction due to which we are forced to accept that d(w) � 2.
where the equality is satisfied if and only if G � K k + S n− k .
Proof. Let G ∈ T k (n) and α ≥ 1 be such that Y α (G) is maximum as possible. Let the set of quasi-vertex be represented by V k ⊂ V(G). By means of Lemmas 1 and 2, we have G � K k + T n− k , where T n− k is a tree of order n − k.
We have to prove that T n− k � S n− k . For this, we have, from [4], We calculate each sum separately: Similarly, the middle sum in (26) can be obtained as Using eorem 2, the sum in (28) attains its maximum if T n− k � S n− k . For this, the above last sum becomes Similarly, the last sum in (26) is calculated as follows: By means of eorem 2, the above last sum will obtain its maximum if T n− k � S n− k . For this, we have the following: Combining (27), (29), and (31), we have T n− k � S n− k , and consequently, for G � K k + S n− k , we get the following: (32) Theorem 6. Let G ∈ T k (n) and α ≥ 1.
(i) If k � 1 and n ≥ 3, then Y α (G) ≥ n · 8 α . e equality Y α (G) � n · 8 α holds if and only if G � C n , where C n is a cycle of order n. (ii) If k � 2 and n � 4, then Y α (G) ≥ 54 α + 4 · 31 α . If k � 2 and n ≥ 5, then Y α (G) ≥ 6 · 31 α + (n − 5) · 8 α . Furthermore, the equality holds if and only if G consists of two cycles of length three having a common edge for n � 4 or two cycles with a common path having length at least two for n ≥ 5 or two cycles joined by a path of length at least two for n ≥ 7.
Proof. Suppose that G ∈ T k (n), where Y α (G) is as small as possible. Let the set of quasi-vertices be V k ; then, by the definition of a k-generalized quasi-tree, G − V k is a tree having order n − k. By eorem 5, we have d(s) � 2 for all s ∈ V k . Furthermore, for every vertex s in V k , s is adjacent to two vertices of V − V k . From this discussion, we deduce that G is a connected graph with n + k − 1 edges. It means that G has internal k-cycles. First of all, we have to show that G has no pendent vertex. For this, suppose on contrary base that G has a pendent vertex x. Let there be another vertex z where there is a path between x and z, and let also z belong to a cycle, say C, in G. Let this path be x, v, . . . , z. We denote the vertices adjacent to v different from x by x 1 , x 2 , . . . , x t , and let their degrees be d( en, we have d G (v) � t + 1. Obviously, t ≥ 1 and at least one degree d i ≥ 2. We can define a new k-cyclic graph having order n, by deleting the edge between x and v and then arranging x between two consecutive vertices, say g and h, on the cycle C.
Let d G (g) � a and d G (h) � b; clearly, a and b ≥ 2. en, in the new graph, say G * , the degree of v is t, For t ≥ 2, we have the following: Also, ((t + 1) (t+1) + 1) α ≥ 28 α ; by using Lemma 3, we have (((t + 1) (t+1) + 1) α + (a a + b b )) α − (a a + 2 2 ) α − (b b + 2 2 ) α > 0, which shows that Y α (G) − Y α (G * ) > 0. is is a contradiction to the minimality of G, that is, G * is another k-cyclic graph having Y α (G * ) smaller than that of Y α (G). Now, for t � 1, it implies that v is a pendent vertex of G * ; by doing the same process for v instead of x, we will observe that there is a pendent vertex v that is adjacent to z on cycle C; then, d(z) ≥ 3 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 d(s) ≠ 1.
Next, we have to suppose that, for all s ∈ V(G), d(s) ≥ 2 .   Complexity (i) en, for k � 1, this implies that G is a connected graph having one cycle and n edges. Obviously, G � C n ; thus, each vertex of G has degree 2, and then, Otherwise, Y α (G) will be greater than that of n.8 α . (ii) For k � 2, it shows that G is a 2-cyclic graph and has n + 1 edges. Now, the sum of degrees of G is given in the following: Let us denote these degree sequences by d * or d * * , that is, If the degree sequence of G is d * , then (a) e graph G contains a common path of length L ≥ 1, or (b) e graph G has two cycles joined by a path of length L ≥ 1 In these cases, if L � 1, then such graph is shown in Figure 8. For this, we have Y α (G) � 54 α + 4 · 31 α +(n − 4) · 8 α � A.
If degree sequence of G is d * * , then in this case G has two cycles having a vertex in common, as shown in Figure 10.
For this graph, we get From A, B, and C, we have A > B which is equivalent to 8 α + 54 α > 2 · 31 α and is obvious which can be easily determined through Figure 11. Now, C > A which implies that C − A is positive. is can be seen in Figure 12.

Data Availability
e data used to support the findings of the study are included within this paper.

Conflicts of Interest
e 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.