Abstract
The modified first Zagreb connection index is a graph invariant that appeared about fifty years ago within a study of molecular modeling, and after a long time, it has been revisited in two papers ((Ali and Trinajstić, 2018) and (Naji et al., 2017)) independently. For a graph , this graph invariant is defined as , where is the degree of the vertex and is the connection number of (that is, the number of vertices having distance 2 from ). In this paper, the graphs with maximum/minimum value are characterized from the class of all -vertex trees with fixed number of pendent vertices (that are the vertices of degree 1).
1. Introduction
Throughout this paper, we consider only simple and connected graphs. The vertex set and edge set of a graph are denoted by and , respectively. The degree of a vertex is the number of edges incident to and is denoted by or simply by if the graph under consideration is clear.
Let be the collection of all graphs. A mapping is called a graph invariant or a topological index, if for every graph isomorphic to , it holds that , where is the set of all real numbers. In chemical graph theory, there are many topological indices having different applications in isomer discrimination, QSAR/QSPR investigation, pharmaceutical drug design, etc. There are various topological indices that are extensively studied by a number of researchers. The first Zagreb index and the second Zagreb index are among these much studied topological indices. These Zagreb indices for a graph are defined as
To the best of our knowledge, the first Zagreb index firstly appeared in a formula derived in [1] and the second Zagreb index was firstly introduced in [2]. These two Zagreb indices have several chemical applications, for example, see the recent papers [3, 4]. Detail about the mathematical properties of the indices and can be found in the recent survey papers [5–8], recent papers [9–22], and related references listed therein.
The following topological index is known as the modified first Zagreb connection index [23]:where is the connection number of the vertex (that is, the number of vertices having distance 2 from , see [24]). Actually, this index initially appeared within a certain formula, derived by Gutman and Trinajstić [1]. The index was referred as the third leap Zagreb index in [25]. After the publications of the papers [23, 25], the modified first Zagreb connection index has attracted a considerable attention from researchers, for example, see [25–39].
The main idea of the present paper comes from [40]. In the present paper, the sharp lower and upper bounds on the modified first Zagreb connection index of trees in terms of order and number of pendent vertices are derived and the corresponding extremal trees are characterized.
2. Some Definitions and Notations
For , let be a path in a graph with unless . If and , then is called a pendent path of and is called the length of this pendent path. If , then is called an internal path of . A tree containing exactly one vertex of degree greater than 2 is called a starlike tree. is used to denote the starlike tree of order which is obtained by attaching paths of lengths to the pendent vertices of the star where and for all .
is used to denote the set of all trees of order and with pendent vertices. Since the path graph is the only member of and the star graph is the unique element of , we assume in the remaining part of the paper. For any , we assume is a pendent vertex of , , and . Taking and , we assume that . Then, , , and (see Figure 1).
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Let has vertices of degree 3, where , further . Let be a set of trees of order obtained from by replacing each edge of by a path with length at least 2.
3. On the Minimum Modified First Zagreb Connection Index of Trees with Fixed Number of Pendent Vertices
Lemma 1. (see [41]). Let and , then(i),(ii) which implies that is a starlike tree.
Lemma 2. Let and be considered as a suspended path of such that and . Considering and for and for and let for , then(a)If , then (i) and (ii) .(b)If , then (i) and (ii) .
Proof. (a)See [41].(b)(i) As contains subtrees containing , respectively, where each has at least pendent vertices of . Therefore, or .(ii)Since for the result is obvious, so let , and we observe that and also as .Hence, or .
Lemma 3. If is a tree such that is as small as possible, then contains at most one pendent path of length greater than 1.
Proof. We contrarily assume that and are two pendent paths of such that and . If , then and we havewhich is a contradiction to the choice of .
Let us denote and .
Lemma 4. For any tree ,Equality in the above expression holds if and only if .
Proof. Let be the tree with minimal among all the members of . Since , therefore it contains at least one pendent path of length greater than 1. By using Lemma 4, we conclude that contains exactly one pendent path of length greater than 1. Therefore, . Since is a starlike tree, and for any ,
and equality in the above expression holds if and only if .
Lemma 5. If is a tree such that is as small as possible, then does not contain a pendent path of length greater than 1.
Proof. We contrarily assume that be a pendent path of such that and . As , so there must be a vertex , with . Also, there must be a path between and . Let be a vertex in this path, adjacent to , and also, let . If , then and we havewhich is a contradiction to the minimality of .
Theorem 1. If for , then
In the above inequality (7), equality holds if and only if . In (8), equality holds if and only if and .
Proof. Let we denote . If we take , then by Lemma 4, and the equality holds if and only if . So, the above theorem holds. Now, we assume that and . We observe that if , then and equality in equation (8) can be obtained by a simple elementary calculation. Now, by applying induction on , we show that if , then (8) holds and the equality in (8) holds only if . Let us choose such that is as small as possible.
If , then by Lemma 5 when , or if . Hence, and . Therefore, equality in (8) holds for only if and . We assume that and the result is true for all smaller values of .
Let and denote the degree of vertex by . Considering and as the pendent and nonpendent neighbors of , respectively, then (because ). Lemma 5 ensures that , and we consider the following cases: Case I: . Let . So, and we have Case II: . If , then we take and let for . If , then and and equality holds only if and . Further, by induction hypothesis, . As , so there must be an internal path of length at least 4, connecting and in and . Hence, and belongs to . If we take , then and let . Assuming that be an internal path of with and , having , we consider the following cases: Subcase I. If , we consider , then and Subcase II. If , we can obtain a tree such as and To get equality, all the relations considered above should be reduced to equalities. So, we get , , and . Further, by induction hypothesis, and . Therefore, and which completes the proof.
4. On the Maximum Modified First Zagreb Connection Index of Trees with Fixed Number of Pendent Vertices
Lemma 6. Let be a tree that maximizes , then(a)For , contains at least one pendent path of length greater than 1,(b)For , contains at least one pendent path of length 1.
Proof. (a)Let , and we assume that every pendent path of has length at most 1, so we have for all . Now, we show that for all . Otherwise, there would be a path , such that for some , , , and , where . Let , . Then, and for . If , then and we have which is a contradiction to the choice of . Hence, we have the result that for all . Therefore, which gives , a contradiction.(b)Now for , if we assume that each pendent vertex of is adjacent to a vertex of degree 2, then . Since , we therefore have . Hence, , a contradiction.
Theorem 2. Let be a tree such that , and if , then
Equalities in (14) and (15) hold if and only if and , respectively.
Proof. We observe that if and , then, respectively, equalities (14) and (15) hold by using simple elementary calculation.
Let us denote and . Now, by applying induction on , we show that if for , then (14) and (15) hold and the equalities in (14) and (15) hold only if and , respectively. Let , then is a starlike tree and . It can be easily verified that and (see Figure 2).
Note that , , , and , . Therefore, Theorem 2 holds for , so we assume that or , and we find the following results:
So, now we have to consider , as the results hold for the smaller values of . Let if , then . Therefore, and
We observe that equality in (17) holds if . Also, if , then . Now, we consider the case that and for .
Let be a pendent path of such that and . Considering and . Then, , and . Now, we consider the following two cases: Case I. . Here, we choose such that is as large as possible. Therefore, by Lemma 6, contains at least one pendent path (say) of length greater than 1. Let us consider , so . Now, for , Lemma 2 implies that and where the equality holds if all the inequalities mentioned in the above argument turn into equalities. Thus, we have . By the induction hypothesis, . Here, contains a unique vertex of degree greater than 2, and hence . Now if , then Case II: By using Lemma 2, we may choose with . Let , then . So. Now if , then and we get where the equality holds only if , , and . As and contain a unique vertex with degree greater than 2, so we have . If we have , then The above inequality follows from Lemma 2. Equality shows that all the above relations are also equalities. Particularly . Therefore, by induction hypothesis, . We observe that contains a unique vertex of degree greater than 2 and , hence . By this result the proof of Theorem 2 is complete.
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Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the National University of Computer and Emerging Sciences, Lahore, Pakistan.