Minimum Atom-Bond Sum-Connectivity Index of Trees With a Fixed Order and/or Number of Pendent Vertices

Let $d_u$ be the degree of a vertex $u$ of a graph $G$. The atom-bond sum-connectivity (ABS) index of a graph $G$ is the sum of the numbers $(1-2(d_v+d_w)^{-1})^{1/2}$ over all edges $vw$ of $G$. This paper gives the characterization of the graph possessing the minimum ABS index in the class of all trees of a fixed number of pendent vertices; the star is the unique extremal graph in the mentioned class of graphs. The problem of determining graphs possessing the minimum ABS index in the class of all trees with $n$ vertices and $p$ pendent vertices is also addressed; such extremal trees have the maximum degree $3$ when $n\ge 3p-2\ge7$, and the balanced double star is the unique such extremal tree for the case $p=n-2$.


Introduction
A property of a graph that is preserved by isomorphism is known as a graph invariant [1].
The order and degree sequence of a graph are examples of graph invariants.The graph invariants that assume only numerical values are usually referred to as topological indices in chemical graph theory [2].
For evaluating the extent of branching of the carbon-atom skeleton of saturated hydrocarbons, Randić [3] devised a topological index and called it as the branching index, which nowadays is known as the connectivity index (also, the Randić index).
The connectivity index of a graph G is the following number: where d v and d w denote the degrees of the vertices v and w of G respectively, and E(G) denotes the set of edges of a graph G.It is believed that the connectivity index is the most-studied topological index (in both theoretical and applied aspects) [4].Detail about the study of the connectivity index can be found in the survey papers [5,6], books [7,8], and related papers cited therein.
Because of the success of the connectivity index, many modified versions of this index have been introduced in the literature.The atom-bond connectivity (ABC) index [9,10] and the sum-connectivity (SC) index [11] are among the well-studied modified versions of the connectivity index.The ABC and SC indices of a graph are defined as and The readers interested in detail about the ABC and SC indices are referred to the survey papers [12] and [13], respectively.
Using the main idea of the SC index, a modified version of the ABC index was proposed in [14] recently and it was referred to as the atom-bond sum-connectivity (ABS) index.
The ABS index of a graph G is defined as Although the ABS index is a special case of a general topological index considered in [15], no result reported in [15] covers the ABS index.The graphs possessing the maximum/minimum ABS index in the class of all (i) (molecular) trees (ii) general graphs, with a given order, were characterized in [14].Analogous results for unicyclic graphs were reported in [16], where the chemical applicability of the ABS index was also investigated.
A vertex of degree one in a tree T is called a pendent vertex and a vertex of degree at least three in T is called a branching vertex.A path P in a tree T connecting a branching vertex and a pendent vertex is called a pendent path provided that every other vertex (if exists) of P has degree two in T .A path P in a tree T is said to be an internal path if it connects two branching vertices and every other vertex (if exists) of P has degree two in T .A tree with one non-pendent vertex is called a star.A double star is a tree with exactly two non-pendent vertices.A double star tree with non-pendent vertices u and v For a general reference on graph theory see [17].
Ali et.al. [16] posed a problem asking to determine trees possessing the minimum value of the ABS index among all trees with a fixed number of pendent vertices.The main goal of the present paper is to determine trees possessing the minimum value of the ABS index in two classes of trees.For positive integers n and p, Γ p denotes the class of all trees with p pendent vertices and Γ n,p denotes the class of all trees of order n and p pendent vertices.In section 2 we give a complete solution to the problem posed by Ali et.al. [16], where we show that the star graph S p+1 uniquely attains the minimum value of the ABS index in the class Γ p .In section 3 we provide results on trees that minimize the value of the ABS index in Γ n,p .

Trees with a Fixed Number of Pendent Vertices
We will need the next already known result.Proof: Suppose to the contrary that T * has at least one vertex of degree 2. Take v ∈ V (T * ) such that N (v) = {u, w} and d u ≥ d w ≥ 1.Let T ′ be the tree formed by removing the vertex v (and its incident edges) and adding the edge uw (see Figure 1).In what follows, by d x we denote the degree of a vertex x in T * .Using the definition of the ABS index, we have a contradiction to the assumption that T * attains the minimum value of the ABS index among all trees with p pendent vertices. where Certainly, the function g is strictly increasing on y because y ≥ 3. Hence, g(y) < g(x + y − 2) as x ≥ 3. Consequently, Equation (1) gives Since the function ψ defined by and define the function ψ(x, y; s) as where x, y ≥ 1 and s > 0. Then ψ(x, y; s) is strictly decreasing on x and on y.
it suffices to show the case of x.Note that the first partial derivatives of f are calculated which are both strictly decreasing in x.This implies that Thus ψ(x, y; s) is strictly decreasing on x and on y. □ Theorem 1 Let p ≥ 2 be an integer.Then for every T ∈ Γ p , with equality holds if and only if T ∼ = S p+1 .
Proof: Let T be a graph attaining the minimum ABS value among all trees with p pendent vertices.By Lemma 2, T has no vertex of degree 2. We claim that T has only one vertex of degree greater than 2. Contrarily, we assume that T contains at least two vertices of degree greater than 2.Among the vertices of degrees at least 3, we pick H 1 In what follows, by d w we denote the degree of a vertex w in By Lemma 4, the following inequalities hold for all i = 1, ..., d u − 1 and j = 1, ..., d v − 1: Thus, Equation (3) yields Since d u ≥ d v , by Lemma 3, Inequality (4) gives By Lemma 3 the function g(s) defined by with s ≥ 3, is strictly decreasing, and Therefore, the right-hand side of ( 5) is positive for d u ≥ 3.This completes the proof.

□
In this section, we characterize trees attaining the minimum value of the ABS index in Γ n,p , where p ≥ 3 and n ≥ 3p − 2.
Lemma 5 If y is a fixed real number greater than or equal to 3 then the function f , defined in Lemma 4, is strictly increasing in x.
For a tree T , denote by W 1 (T ) the set of pendent vertices of and A vertex in T of degree at least three is called a branching vertex.A path is called an internal path, if its end vertices are branching vertices and every other vertex has degree two.A path is called a pendent path if one of the end vertices is pendent and every other vertex has degree two.
Again this contradicts the minimality of ABS(T * ).Thus T * does not have an internal path of length greater than one.□ Proof: Let y be a pendent vertex adjacent to u and let x be a non-pendent vertex adjacent to v and not on the u In what follows, by d x we denote the degree of a vertex x in T * .Clearly, T ′ ∈ Γ n,p , and so ABS(T ′ ) ≥ ABS(T * ).Thus we have, If n = 3p − 2 + t for some t ≥ 0, then the following assertions hold. (i) Then by Lemma 7, Then Consider the function We have where A 1 = (s 2 +3s−2)(s+3) 2 (s + 1)(s + 2) and B 1 = (s 2 +5s+2)(s+2) 2 s(s + 3).
The function is strictly decreasing for s ≥ 1.Thus ABS(T ′ ) − ABS(T * ) < h 4 (6) < 0, a contradiction.Calculations show that the right hand side of this inequality is negative when d w ∈ {4, 5} and k ∈ {1, ..., d w }.This also yields a contradiction.Therefore the maximum degree of T * is 3 as desired.□ Now we are ready to characterize the trees that minimize ABS index in Γ n,p where p ≥ 3 and n ≥ 3p − 2. For p ≥ 3 and n = 3p − 2 + t with t ≥ 0, let Γ * n,p ⊂ Γ n,p such that an arbitrary tree T * belonging to the class Γ * n,p is defined as follows.
Figure 3 gives a tree of the class Γ * n,p .

Figure 2 .
Figure 2. The graph transformation used in the proof of Theorem 1.For i ∈ {1, . . ., d u − 1} and j ∈ {1, . . ., d v − 1}, the subtree J i may or may not consist of only one vertex u i and the subtree H j may or may not consist of only one vertex v j , respectively.

Lemma 6
Let T * ∈ Γ n,p attaining minimum ABS value.Then every internal path of T * has length one.Proof: For a contradiction, assume T * contains an internal path v = v 0 −v 1 −v 2 −..−v r = u of length r ≥ 2. Let w ∈ W 2 (T ) and let y be a pendent vertex adjacent to w.Let T ′ = T * − {vv 1 , uv r−1 } + {vu, yv 1 }.Clearly T ′ ∈ Γ n,p .In what follows, by d x we denote the degree of a vertex x in T * .If r > 2, then

Lemma 7
Let T * ∈ Γ n,p attaining minimum ABS value.Then d u ≤ d v for every u ∈ W 2 (T * ) and v ∈ W 3 (T * ).

(Theorem 2
ii) We conclude from part (i) that |W 2 (T )| = p.So p + 2p + u∈W 3 (T ) d u = 6p − 6 + 2t, which yields u∈W 3 (T ) d u = 3p − 6 + 2t.□ 7 that every vertex of T * of degree two is on a pendent path.Assume there are two vertices u, v ∈ W 3 (T * ) such that d u = d v = 2 and u and v do not belong to the same pendent path.Since v ∈ W 3 (T * ) and d v = 2, there are two vertices x and y such that d x ≤ d y = 2 and v − y − x.Let z be the pendent vertex at the end of the pendent path containing u and let T ′ = T * − xy + zy.Clearly T ′ ∈ Γ n,p and ABS(T ′ ) = ABS(T * ).We conclude that it is possible to obtain a tree T * 1 ∈ Γ n,p , in which all vertices of degree two in W 3 (T * 1 ) belong to the same pendent path, such that ABS(T * 1 ) = ABS(T * ).Moreover, T * and T * 1 have the same maximum degree and the subtrees of T * and T * 1 induced on their sets of branching vertices are isomorphic.Let p ≥ 3 and n ≥ 3p − 2. If T * ∈ Γ n,p attaining the minimum ABS value, then the maximum degree of T * is 3. Proof: For a contradiction, assume T * has a vertex of degree at least 4. Let β = min{d v | v ∈ W 3 (T )}.Then, by Lemma 8 (ii), 4 + β(|W 3 (T * )| − 1) ≤ 3p − 6 + 2t.Consequently, β ≤ 3p−10+2t p−3+t < 3.So there is a vertex u ∈ W 3 (T * ) such that d u = 2.We select u so that one of its neighbors has degree at least three.Now let v be the farthest vertex from u that has degree at least 4. Clearly, for each vertex x ∈ N (v) that does not lie on the u − v path, we have d x ≤ 3.In what follows, d a denotes the degree of a vertex a in T * .Case 1. u ∈ N (v).In this case choose y ∈ N (v) \ {u} and let T ′ = T * − {yv} + {yu}.

Case 2 .
u ̸ ∈ N (v).Let w ∈ N (v) and z ∈ N (u) such that w and z lie on the u − v path.By Case 1, me may suppose that d z ≥ 3. Now we divide this case to four subcases.Subcase 2.1.d v ≥ 5. Let y ∈ N (v) \ {w} and take T ′ = T * − {yv} + {yu}.Then T ′ ∈ Γ n,p .Moreover, from Lemma 4, it follows that

Subcase 2 . 3 :
d v = 4 and d w ≥ 6.Note that d x ≤ 4 for each x ∈ N w \ {w 1 , v}, because otherwise we reach a contradiction as in Subcase 2.1, where w 1 is the unique neighbor of w in the u − v path.Let T ′ = T * − {vw} + {vu}.Then ABS

Figure 3 .
Figure 3.A tree attaining the minimum value of the ABS index in the class Γ n,p where p ≥ 3 and n ≥ 3p − 2.
Lemma 1 [14,Corollary 8]Let u and v be non-adjacent vertices in a connected graph G. Then ABS(G + uv) > ABS(G), where G + uv is the graph obtained from G by adding the edge uv.Let p ≥ 2 be an integer.If T * is a tree attaining the minimum value of the ABS index in the class Γ p , then T * has no vertex of degree 2.
Theorem 3 Let p ≥ 3 and n = 3p − 2 + t where t ≥ 0. If T ∈ Γ n,p then with equality if and only if T ∈ Γ * n,p .