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Abstract
Let G be a graph with V(G) and E(G), as vertex set and edge set, respectively. The atom-bond sum-connectivity (ABS) index is a vertex-based topological index which is defined as where
is the degree of the vertex a. In this paper, we obtain sharp upper bounds for the ABS index of molecular trees in terms of order and number of branching vertices and vertices of degree two.
1 Introduction
In this article, we are only concerned with a simple, connected, and undirected graph denoted by G that has a vertex set denoted by V(G) and an edge set denoted by E(G). Degree of a vertex u is the number of edges adjacent to the vertex u, denoted by .
Let , be the degree sequence of G. A vertex a is called a pendent vertex or a branching vertex if
or
, respectively.
Let tj
denotes the count of vertices having degree j in the graph G. Denote by the cardinality of the set consisting of the edges connecting the vertex of degree i with the vertex of degree j in the graph G.
Let be a path, denoted by vertices
internal vertex of P. A path from a branching vertex of degree a to a branching vertex of degree b in a molecular tree M, denoted by P(a, b), such that all the internal vertices (if exist) of P(a, b) have degree two.
Topological indices are numerical variables that describe the topology of a graph and are typically graph invariant. It may be analytically deduced directly from a molecule’s structural layout.
An alternative to the atom-bond connectivity index (Equation6(6)
(6) ) that incorporates the basic idea of the sum-connectivity index (Equation11
(11)
(11) ) is the atom-bond sum-connectivity (ABS) index (Equation1
(1)
(1) ). The definition of the ABS index is
The ABS index is a topological index that is based on the degree of the vertices, but it can also be thought of as a topological index that is based on the degree of the edges due to the fact that the degree of the edge is represented by the number
. Due to the fact that this index has recently been introduced, it has attracted the attention of many researchers and many articles have been published.
In (Equation13(13)
(13) ), presented the extremal ABS index of chemical trees, unicyclic, bicyclic, and tricyclic graphs. Unicyclic graphs as one of the great classes can exhibit various chemical structures as well, extremal unicyclic graphs graphs have been found in (Equation3
(3)
(3) ). Studied ABS index general graphs and trees in terms of a fixed order have been found in (Equation1
(1)
(1) ). Extremal ABS index of graphs with givwn the number of pendent vertices have been studied in (Equation5
(5)
(5) ). In (Equation9
(9)
(9) ), presented the minimum ABS index of trees in terms of number of pendent vertices. For the recent results on the mathematical properties of ABS index can be seen in (Citation4, Citation7, Citation8, Citation10, Citation12).
In this paper, we will obtain the sharp upper bounds for the ABS index of molecular trees with given branching vertices or vertices of degree two.
2 Auxiliary results
In this section, we state and prove some lemmas that will help us in proving the main results.
The proof of the following result is obvious using derivation, so we omit it from the proof here.
Lemma 1.
Suppose a and b are positive numbers, then for a > b, the following function is strictly decreasing on t and for b > a is strictly increasing on t.
It should be noted that throughout this section the molecular trees M and have the same degree sequence.
Lemma 2.
Suppose M be a molecular tree that is contains a path and an edge xy with
and
such that b1 lies on the path
. Let
be a molecular tree that obtained from M by removing the edges
and create the new edges
. Then,
.
Proof.
By the definition ABS index, we have
□
Lemma 3.
Suppose M be a molecular tree with , such that M contains the paths
and
, with
, and k < l. Let
is obtained with
. Then
.
Proof.
If with
and
. Therefore, to complete the proof, we need to check the following cases.
Case 1. .
We know that one of two ways paths P(4, 3) and have exactly one internal vertex while the other does not internal vertex. Here, we let k = 2, when l = 3, then by taking
, and when
, we setting
. It can be easily seen that in both cases, M and
have the same degree sequence with
, and
as required.
Case 2. l > 4.
When k = 2 or , with an argument similar to Case 1, the results are obtained. So, we let that
. If l = 5 then (k = 3 and) we setting
, and otherwise (that is, if
) then, we let
. It can be easily seen that in both cases, M and
have the same degree sequence, such that
, and
□
By using the definition and by direct calculation, we arrive at the proof of the following lemma, so here we state this lemma without proof.
Lemma 4.
Suppose M be a molecular tree such that with
, and y lies on one of the three paths
, and also,
. When
and
. Then,
.
Lemma 5.
Suppose M be a molecular tree such that with
and
. If
, then,
.
Proof.
Since two trees M and have the same degree sequence, then using the definition ABS index, we have
We know that , hence, there are two cases.
When
, we have
and hence
, it is easy to see that Y < 0.
When
, we have
and
(or
and
) and also,
and therefore, we get Y < 0. □
Lemma 6.
Suppose M be a molecular tree with with
, and
. We define
. Then
.
Proof.
By the definition ABS index and , we have
□
3 Main results
In this section, we obtain sharp upper bounds for the ABS index of molecular trees in terms of order and number of branching vertices and vertices of degree two.
For , let
be the set of molecular trees of order n and branching vertices tb
with
. We know that
is the path Pn
, hence, we have
and M(4, 1) is the graph M1 in and the value of ABS is shown in . Therefore, in the following we let n > 4 and
. Now, we will remind the following two families of
:
(1)
(1)
(2)
(2)
Table 1 The ABS index of the trees Mi
for .
Now here we present a lemma related to the maximum value of ABS index of molecular tree that will help us in obtaining the main results.
Lemma 7.
Suppose Mb
be the molecular tree that attain the maximum value of ABS index over the class , then
or
.
Proof.
Suppose it is not so, in other words . Therefore, there must be vertices a and b in Mb
such that
and
. Now, we let
and
with b2 and a3 lie on the path a–b. Note that
hence, we set
and
, and also it possible that a3 = b2. If
be a tree obtained from M by removing the edges aa1, aa2 and create the new edges ba1, ba2 in M, it can be easily seen that
and using Lemma 1, we have
that is contradicting with the choice of Mb
. □
The next lemma shows the relationship between the maximum value of ABS index and the number of edges .
Lemma 8.
Suppose and
be a molecular tree that possessing the maximum value of ABS index, then
.
Proof.
On the contrary, suppose that there is a path in Mb
with
and
for
. Since,
, applying Lemma 7 we get
. Moreover, we know that at least one of the neighbors of
is non-pendent vertex, since
. Setting
with
. We know that
, applying Lemma 7, we get
. Now, we define
, hence,
and
that is a contradiction. □
Note that if M be molecular tree with n > 1, we have the following results (Equation2(2)
(2) ):
(3)
(3)
(4)
(4)
(5)
(5)
The next lemma is a direct consequence of EquationEqs (3)–(5).
Lemma 9.
(Equation2(2)
(2) ) Suppose
be a molecular tree, then
If
, we have
and
.
If
, we have
and
.
if and only if
.
The next lemma shows the relationship between the maximum value of ABS index and the number of vertices of degree 2.
Lemma 10.
Suppose M be the molecular tree that attain the maximum value of ABS index over the class , then
if and only if
or
Proof.
When , then from Lemma 7, we have
, therefore,
, using Lemma 9, we have
, hence, we get
or
. On the contrary, let
or
such that
and
. Applying EquationEqs. (3)–(5), we get
, the fact
meaning
, that is a contradiction. □
Now here we need to introduce several families of chemical trees that will help us to continue the proofs.
We know that M2 and M3 of orders n = 5 (see ) are the only molecular trees in M(Citation1, Citation5), from it is clear .
For , the following families of molecular trees introduced in (Equation2
(2)
(2) ):
where Mp
and
are defined in (Equation1
(1)
(1) ) and (Equation2
(2)
(2) ). Hence, we have
(6)
(6)
In the next theorem we give an upper bound for the ABS index.
Theorem 1.
Suppose and
, then, we have
with equality if and only if
.
Proof.
Let be a molecular tree that possessing the maximum value of ABS index. Let
be the unique branching, then we have the following two situations.
Claim 1. .
On the contrary, if for
, there is a vertex
such that
and is adjacent to a pendent vertex, we called y. Here, we let
be a tree obtained from M1 by removing the edge ay and adding the new edge by. It is easy to see that
. Then, we have
where, if
and
if
Claim 2. b has exactly one non pendent neighbor.
Note that , the vertex b has at least one non pendent neighbor. Assume the opposite
and
be two paths in M1 such that
and
for
and
. We define
, we can see that
and
applying Claim 1 for , we get a contradiction. By combining these two claims, we will prove the desired result. □
In the following theorem, we present an upper bound on the ABS index based on the number of branching vertices.
Theorem 2.
For with
, we have
with equality if and only if
.
Proof.
Let be a molecular tree that possessing the maximum value of ABS index, when
. Note that
, using Lemma 10, we get
and with combination Lemma 7, we get
, and applying Lemma 9(b), we have
and
. Since,
, by Lemma 8, we get guaranties that
(7)
(7)
taking in (Equation6
(6)
(6) ) for i = 1, we have
(8)
(8)
Since or
, we have
, therefore, we get
(9)
(9)
Now here we claim that . Contrarily, assume that
. From, (Equation6
(6)
(6) ) with i = 2, we obtain that
(10)
(10)
Using (Equation3(3)
(3) ) and (Equation4
(4)
(4) ), we get
(11)
(11)
For (Equation6(6)
(6) ) if i = 4, we have
(12)
(12)
Applying (Equation8(8)
(8) )–(Equation11
(11)
(11) ) in (Equation12
(12)
(12) ), we get
that is a contradiction. Therefore, the claim is true. Now, we claim that
(13)
(13)
Assume the opposite . Setting
with
. Note that
, by setting
with
and
. Let
be a tree that obtained by applying the transformation mentioned in the statement of Lemma 6, hence, by Lemma 6 we get,
, that is a contradiction to the definition of Mb
. Hence,
. Applying (Equation6
(6)
(6) )-(Equation8
(8)
(8) ) and (Equation13
(13)
(13) ), we have
and
. Then,
□
The next theorem is related to the relationship between ABS index branching vertices.
Theorem 3.
For with
, we have
the equality occurs if and only if
.
Proof.
Let be a molecular tree that possessing the maximum value of ABS index, when
. Applying Lemma 10, we get
with
and using Lemma 7 we have
or
. Applying part (b) of Lemma 9, we have
and
and by Lemma 8, Equationequations (7)
(7)
(7) and Equation(8)
(8)
(8) are hold. We know that
is obtained from
With an argument similar to the proof of Theorem 2, we have that
and
. Therefore, we have
□
Theorem 4.
For with
, we have
Proof.
Applying Lemma 9 part (c), we get , this meaning
. Hence,
and
. □
The next result is related to ABS index and branching vertices such that .
Theorem 5.
For with
, we have
and the equality occurs if and only if
.
Proof.
Let be a molecular tree that possessing the maximum value of ABS index, when
. Applying Lemma 10, we get
(as
), that is
. From Lemma 9(a), we have
and
. Note that
is equal to
, and this means that
. Applying Lemmas 2–4 and using the fact
, we get
(14)
(14)
and
(15)
(15)
Applying Lemmas 2–4 and (Citation6, 14), and (Equation15(15)
(15) ), we get
. Therefore, we obtain that
□
The next theorem is related to ABS index and branching vertices such that .
Theorem 6.
For with
, we have
and the equality occur if and only if
.
Proof.
Let be a molecular tree that possessing the maximum value of ABS index, when
. Applying Lemma 10, we can obtain that
as
, this is what it means
and applying Lemma 9(a), we have
and
. We know that
, hence,
, this means
.
Using Lemmas 2–4 it is clear that we have to place the vertices as described in the proof of Theorem 5. We know that and using (Citation6, 14), and (Equation15
(15)
(15) ), we get
and
. Therefore, we have
. □
The next theorem gives an upper bound for ABS index with the condition .
Theorem 7.
For with
, we have
and the equality occurs if and only if
.
Proof.
Let be a molecular tree that possessing the maximum value of ABS index, when
. Applying Lemma 10, we have
as
, that is
and from Lemma 9 (a), we have
and
. Since,
, we can write that
, that is
. Applying Lemmas 2–4 it is clear that we have to place the vertices of degree 4 between the pendent vertices and the vertices of degree 3. We know that
, hence, we have
(16)
(16)
and
(17)
(17)
Applying (Citation6, 16), and (Equation17(17)
(17) ), we get
and
. Therefore, we have
□
Suppose that be the trees with n vertices, from which r vertices have degree 2.
It is obvious that is the path graph and there is no graph with value
in the class
. Therefore, we suppose that
.
Lemma 11.
(Equation2(2)
(2) ) Let
, then
if and only if
and
if and only if
and
,
if and only if
and
The next theorem examines the relationship between the maximum value of ABS index and the number of vertices of degree 2.
Lemma 12.
Let be a tree that possessing the maximum value of ABS index and
, then,
.
Proof.
Since there is a unique branching vertex with
if
, and
if
. Suppose that for
, there are paths
and
in M with
for all
and
, and
. Let
be a tree with
, hence,
taking
or
in the above inequality, we get
, that is a contradiction. □
The proof of the next lemma is similar to Lemma 8, so here we omit the proof of this lemma.
Lemma 13.
Let be a tree that possessing the maximum value of ABS index and
, then
.
Since and
contain unique trees M1 and M2, respectively, with the ABS index values in . Moreover, from Lemma 12, we know that M5 and M6 are the molecular trees with maximum ABS index values among the graphs in
and
, respectively. Among all the molecular trees
with
, if we defined
and
(given in ), the next result is obtained.
Theorem 8.
Let be a molecular tree with
, then:
If
, then
.
If
, then
.
Proof.
Applying Lemma 12, we get that M3 attains the maximum value of ABS index among the class and M4 attain the maximum value of ABS index among the class
(see ), which completes the proof. □
Next, we need to remember the families of subsets of as:
and
with
and
and
with
and
this means that
moreover if
for
then
with
so
and
and whenever
along with
hence,
moreover
where
and
.
The next theorem gives two upper bounds for ABS index based on the number of vertices with degree two.
Theorem 9.
For and
with
, we have
Equalities hold if and only if .
Proof.
Let be a molecular tree that attains the maximum value of ABS index with
. Applying Lemma 11, we have
(mod 3),
and
. Therefore, vertices of degree two were placed according to the conditions of proof of Lemmas 6 and 13, this means that all the vertices of degree 1 are connected to vertices of degree four, i.e., by Lemma 13, we have
(18)
(18)
this means that
(19)
(19)
Moreover, the vertices of degree 2 are to be placed between the vertices of degree 4 in such a way that if there is an edge connecting the vertices of degree 4, then no two vertices of degree 2 are adjacent. Therefore, . Here, we need to check the following two cases as:
Case i. or
. Using Lemma 6, we have
(20)
(20)
Applying (Citation6, 18–20), we have and
. Therefore, we have
.
Case ii. or
. Again by using Lemma 6, we get
(21)
(21)
Applying (Citation6, 18–21), we obtain that and
. Hence, we have
. □
In the next results, we present upper bounds for ABS index which has the number of vertices of degree 3 equal to one.
Theorem 10.
For with
, we have
Equalities hold if and only if .
Proof.
Let be a molecular tree that possessing the maximum value of ABS index with
, then applying Lemma 11 we get
(mod 3),
. Using Lemmas 2, 13, and 5 we know that
. This means
(22)
(22)
and the vertices of degree 4 are to be placed in the three neighbors of the vertex of degree 3 in such a way that if a pendent vertex is present in
which is adjacent to the vertex of degree 3, then
. To continue the proof, we need to check the following cases:
Case 1.
Subcase 1.1. r = 0.
Here, we let for all
and
. Therefore, the desired graph is M6, which is described in , and its ABS index value is also shown in .
Subcase 1.2. r > 0.
Here, for , from Lemmas 5 and 6, and Equationequations (6)
(6)
(6) and Equation(22)
(22)
(22) , we get
and
. Therefore, we get
.
Case 2. .
Subcase 2.1. r = 0.
For n = 10 and using Lemma 2, we have
(23)
(23)
Applying (Citation6, 22), and (Equation23(23)
(23) ), we get
and
, hence, by the definition ABS index, we have
.
Subcase 2.2. r = 1
Now, for n = 11, from (Citation6, 22), we get and
, hence, we have
.
Subcase 2.3. .
Here, when , and from Lemmas 5, 6 and Equationequations (6)
(6)
(6) and Equation(22)
(22)
(22) , we obtain that
and
. Therefore,
.
Case 3.
Using Lemma 2, we have
(24)
(24)
Subcase 3.1.
Applying Lemmas 5, 6 and Equationequations (6)(6)
(6) , Equation(22)
(22)
(22) , and Equation(24)
(24)
(24) , we get
and
. Therefore, we have
.
Subcase 3.2. r > 2.
Applying Lemmas 5, 6 and Equationequations (6)(6)
(6) , Equation(22)
(22)
(22) , and Equation(24)
(24)
(24) , we get
and
. Then, we have,
.
Case 4.
Subcase 4.1. .
It is easy to see that this means
or
. Applying Lemmas 5, 6 and Equationequations (6)
(6)
(6) , Equation(22)
(22)
(22) and Equation(23)
(23)
(23) , we get
and
. Therefore, we have
.
Subcase 4.2. .
We have, , this means
(25)
(25)
Applying Lemmas 5, 6 and taking Equationequations (22)(22)
(22) , Equation(24)
(24)
(24) , and Equation(25)
(25)
(25) in (Equation6
(6)
(6) ), we get
and
. Therefore,
.
Subcase 4.3. .
This means , and applying Lemmas 5 and 6, we get
(26)
(26)
taking (22, 24–26) in (Equation6(6)
(6) ), we obtain that
and
. Therefore,
□
In the next results, we present upper bounds for ABS index which has the number of vertices of degree 3 equal to two.
Theorem 11.
Let such that
. Then
Equalities hold if and only if .
Proof.
Let be a molecular tree that possessing the maximum value of ABS index with
, hence, applying Lemma 11, we get
(mod 3),
. From Lemma 13, we have
(27)
(27)
Using Lemmas 2–6, we know that . Now, to continue the proof, we need to check the following cases:
Case 1. or
.
Subcase 1.1. r = 0
Applying (Equation6(6)
(6) ), we get
and
, that is
Subcase 1.2. r > 0.
Applying (27) in (Equation6(6)
(6) ), we obtain that
and
, that is
.
Case 2. or
.
From Lemmas 2–4 and (27), we get
(28)
(28)
(29)
(29)
and
(30)
(30)
Subcase 2.1. r = 0.
Applying (27)–(30) in (Equation6(6)
(6) ), we get that
and
, that is
Subcase 2.2. .
Note that here, we have , therefore, using Lemmas 5 and 6 and taking Equationequations (27)
(27)
(27) , Equation(30)
(30)
(30) in (Equation6
(6)
(6) ), we will arrive
and
, this means
.
Subcase 2.3. .
Now, when , and using Subcase 2.2, we have,
and
. Hence,
Case 3. or
.
Applying Lemmas 2–4 and (27) show that (30) holds and
(31)
(31)
also
(32)
(32)
Now we consider the following subcases:
Subcase 3.1. .
From Lemmas 5 and 6, and (Equation6(6)
(6) ), (27) and (30)–(32), we get that
and
. Therefore,
Subcase 3.2. r > 4.
With a similar argument Subcase 3.1, we reach and
. Therefore,
Case 4. or
.
From Lemmas 2–4 we know that (31) and (32) hold, hence, we have
(33)
(33)
Subcase 4.1. and
Since, (30) holds in this case, applying Lemmas 5, 6 and Equationequations (6)(6)
(6) , Equation(27)
(27)
(27) , Equation(30)
(30)
(30) and Equation(31)–(33), we get
and
. Then,
Subcase 4.2. and r > 6.
With a similar argument Subcase 3.1, we reach and
. Then,
Subcase 4.3. and
.
This case means , using Lemma 5, we have
. Applying Lemma 6 and (Equation6
(6)
(6) ), (27) and (31)–(33), we get
and
. Therefore,
.
Subcase 4.4. and
.
Since (30) holds in this case and applying (Equation6(6)
(6) ), (27), and (31)–(33), we will arrive
and
. Therefore,
.
Subcase 4.5. and
.
Applying Lemmas 5, 6, and (Equation6(6)
(6) ), (27), (30), and (31)–(33), we get
and
. Therefore,
□
4 Conclusions
In the paper, we determined the maximum value of the atom-bond sum-connectivity (ABS) index among molecular trees of a given order and a fixed number of (i) vertices of degree greater than 2. (ii) vertices of degree 2. But the problem of minimum values of the ABS index among the considered classes of molecular trees has not been investigated yet, so we present the following problem.
Problem 1. Characterize graphs that attain the minimum atom-bond sum-connectivity index among molecular trees of a given order and a fixed number of (i) vertices of degree greater than 2. (ii) vertices of degree 2.
Acknowledgments
The authors are grateful to the anonymous referee for his insightful comments.
Disclosure statement
The authors declare that they have no competing interests.
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Funding
References
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