Abstract
Let be a graph the atom-bond connectivity (ABC) index is defined as the sum of weights over all edges of , where denotes the degree of a vertex of . In this paper, we determined a few structural features of the trees with minimal ABC index also we characterized the trees with and minimal ABC index, where is induced by the vertices of degree greater than 2 in and is the diameter of .
1. Introduction
Let be a finite, simple, and undirected graph. The degree of a vertex is denoted by . The atom-bond connectivity (ABC) index is defined as the sum of weights over all edges of ; that is, The index of a graph was defined by Estrada et al. [1] and it has many chemical applications [1, 2].
When examining a topological index, one of the fundamental questions that needs to be answered is for which graphs this index assumes minimal and maximal values and what are these extremal values. In the case of the ABC index, finding the tree for which this index is maximal was relatively easy [3]; it is the star. Eventually, also the trees with second-maximal, third-maximal, and so forth ABC index were determined [4].
We [5] have shown that by deleting an edge from any graph, the ABC index decreases. This result implies that among all -vertex graphs, the complete graph has maximal ABC value. Further, among all connected -vertex graphs, minimal ABC is achieved by some tree. Thus the -vertex trees with minimal ABC index are also the vertex connected graphs with minimal ABC index. But the problem of characterizing the -vertex trees with minimal ABC index turned out to be much more difficult, and a complete solution of this problem is not known. For more results on ABC index see [6–13].
In a recent work [6] a combination of computer search and mathematical analysis was undertaken, aimed at elucidating the structure of the minimal trees. And some structural features of the trees with minimal index are given in [7].
Lemma 1 (see [6]). If , then the -vertex tree with minimal ABC index contains at most one pendent path of length .
Lemma 2 (see [7]). If , then each pendent vertex of the -vertex tree with minimal ABC index belongs to a pendent path of length , .
By inspecting the structural features of these trees, in [8] the branches and were given. Let be a branch of tree formed by attaching pendant path of length to the vertex such that the degree of in is . Let be a branch of tree formed by attaching pendant path of length and a pendant path of length to the vertex such that the degree of in is (see Figure 1). Denote by the union of the branches and by the number of branches in . From Lemmas 1 and 2, we know all branches in a tree with minimal index must be of the type or , and , . According to Lemma 1, in the following we assume that and , for all .
In [9] the -vertex minimal trees were determined up to and then a conjecture about the trees with minimal index was presented.
Conjecture 3 (see [9]). Let be a tree with minimal index among all trees of size . Let , , , , , , and be the structures depicted in Figure 2.(i)If , and , then has the structure .(ii)If , and , then has the structure .(iii)If , and , then has the structure .(iv)If , and , then has the structure .(v)If , and , then has the structure .(vi)If , and , then has the structure .(vii)If , and , then has the structure .
In this paper, we determined a few structural features of the trees with minimal ABC index, also we characterized the trees with and minimal ABC index, where is the diameter of , which was induced by the vertices of degree greater than 2 in .
2. The Structural Features of the Trees with Minimal ABC Index
Now, we are going to determine a few structural features of the trees with minimal index.
Theorem 4. The -vertex tree with minimal index does not contain branches and ().
Proof. Suppose that is a tree with minimal index, possessing a branch , . Let be a vertex of , adjacent to the vertex , and the degree of is . Consider the tree (see Figure 3).
By direct calculation, we have
If , it can be easily checked by computer that
For the case , if the inequality holds, it implies that
By elementary calculation, this inequality can be transformed to
That is,
By squaring the above relation and rearranging, we get
Since the function
and it holds that
thus, we have
In the same way, we can prove that the -vertex tree with minimal index does not contain branch ().
The proof is complete.
Note that Theorem 4 holds for all -vertex trees with minimal index.
Theorem 5. Let be a tree with minimal index, then every vertex of must not be connected with both and .
Proof. Suppose that is a tree with minimal index; let be a vertex of , which is connected with both and , and the degree of is . Construct the tree by deleting the edge and connecting with (see Figure 4).
The transformation causes the following change of the index:
If the inequality holds, it implies that
By elementary calculation, this inequality can be transformed to
Thus we have , for .
The proof is complete.
Theorem 6. Let be a tree with minimal index; then every vertex of must not be connected with both and .
Proof. Suppose that is a tree with minimal index; let be a vertex of , which is connected with both and , and the degree of is (obviously ). Construct the tree by deleting the edges , and adding the edges , (see Figure 5).
The transformation causes the following change of the ABC index:
If the inequality holds, it implies that
That is
By squaring the above relation and rearranging, we get
Thus we have , for , and the proof is complete.
Theorem 7. Let be a tree with minimal index; then every vertex of must not be connected with .
Proof. Suppose that is a tree with minimal ABC index, possessing a vertex in connected with (see Figure 6). Let be the set of adjacent vertices to . Let be the degree of , respectively. We consider the tree shown in Figure 6.
Here we are going to show that , for any .
Consider
for any , .
Now we are going to show that ; that is
By elementary calculation, this inequality can be transformed to
The largest root of the above polynomial is ; therefore, the value of the above polynomial is positive for .
Thus we have , and the proof is complete.
Theorem 8. Let be a tree with minimal index; then every vertex of must not be connected with .
Proof. Suppose that is a tree with minimal index, possessing a vertex connected with . Let be the vertex of , adjacent to the vertices and , and the degree of is (). Consider the tree (see Figure 7).
The transformation causes the following change of the index:
It can be easily checked by computer that , for .
The proof is complete.
3. The Minimal ABC Indices of Trees with Order and
Denote by the subgraph of induced by its vertices of degree greater than . For a connected graph , the diameter of , denoted by , is the length of a longest path of .
Lemma 9 (see [7]). Let be an -vertex tree with minimal index; then is a tree.
Note that the structures in Conjecture 3 have (). Let be the set of -vertex trees with .
Now we will characterize the trees with and minimal ABC index, which partially solve Conjecture 3.
Lemma 10. Let be a tree with minimal index; then does not contain .
Proof. Let be a tree with minimal ABC index. By Theorems 7 and 8, contains at most , ; hence, for , must contain . Suppose that possesses ; then we can construct the tree by deleting the edge and adding the edge (see Figure 8).
The transformation causes the following change of the ABC index:
It can be easily checked that , for .
The proof is complete.
Lemma 11. Let be a tree with minimal index; if the maximum degree , then must not contain ().
Proof. Suppose that is a tree with minimal ABC index, possessing (see Figure 9). Let be the vertex with maximum degree of , adjacent to the vertices , , and , and the degree of is . Construct the tree by deleting the edges and and adding the edges and . Let be a set of adjacent vertices to . Let be the degree of , respectively.
The transformation causes the following change of the ABC index:
By Theorems 4 and 5, Lemma 10, and noticing that , we know that or .
And consider
Putting this in the above expression, we get
If the inequality holds, it implies that
By elementary calculation, this inequality can be transformed to
By squaring the above relation and rearranging for two times, we get
The largest root of the above polynomial is ; therefore, the value of the above polynomial is positive for . Thus we have for , and the proof is complete.
Lemma 12. Let be a tree with minimal index; if the maximum degree , then must not contain .
Proof. Suppose that is a tree with maximum degree and minimal ABC index, possessing (see Figure 10). By Theorem 6 and Lemma 11, we have and . Thus in must be contained.
Let be the vertex with maximum degree in and let be a set of adjacent vertices to . Let be the degree of , respectively. Since and by Theorem 4 and Lemma 10, we have , or . Construct the tree by deleting the edge and adding the edge to .
The transformation causes the following change of the index:
If the inequality holds, it implies that
Note that , , , and
We have, .
Now we are going to show that
That is,
By squaring the above relation and rearranging for two times, we get
Since (for ) and the largest root of the following polynomial is ,
Therefore the value of the above polynomial is positive for . Thus we have ABC ABC for and the proof is complete.
Theorem 13. Let be a tree with minimal ABC index and the maximum degree . Let , , , , , and be the structures depicted in Figure 2.(i)If , and , then has the structure .(ii)If , and , then has the structure .(iii)If , and , then has the structure .(iv)If , and , then has the structure .(v)If , and , then has the structure depicted in Figure 11(b).(vi)If , and , then has the structure .(vii)If , and , then has the structure .
| (a) The minimal ABC tree in |
| (b) |
Proof. Let be a tree with minimal ABC index. From Lemmas 1 and 2, we know all branches of must be of the type or , . And from Theorem 4 and Lemmas 10 and 12, we know all branches of must be of the type or , .
Then the minimal tree has
vertices, where counts the pendent paths of length 3.
From above we see that the tree with minimal index and must possess the structure as shown in Figure 11(a). It is easy to see that
Putting (36) in the above equation, we have
where and .
From Lemma 11 and Theorems 5 and 7, we get
Note that the parameters , , , , and in (36) are nonnegative integers.
Consider(i) and .
Case 1.1. If , then .
Thus, , we get , or , .
That is, or .
Case 1.2. If , then , ; we get .
That is, .
Case 1.3. If , then and ; there is no solution.
Comparing the values , , and and noting that , we get that
is the smallest one (); the result (i) follows.
Consider(ii) and .
Case 2.1. If , then .
We get or .
That is, or .
Case 2.2. If , then and .
Thus ; there is no solution.
Case 2.3. If , then and .
Thus ; there is no solution.
Comparing the values , , we get that
is the smallest one (); the result (ii) follows.
Consider(iii) and .
Case 3.1. If , then .
We get , or , .
That is, or .
Case 3.2. If , then and .
Thus ; there is no solution.
Case 3.3. If , then and .
Thus ; there is no solution.
Comparing the values , , we get that
is the smallest one (); the result (iii) follows.
Consider(iv) and .
Case 4.1. If , then .
We get , or , .
That is, or .
Case 4.2. If , then and .
Thus ; there is no solution.
Case 4.3. If , then and .
Thus ; there is no solution.
Comparing the values , , we get that
is the smallest one (); the result (iv) follows.
Consider(v) and .
Case 5.1. If , then .
We get , or , .
That is, or .
Case 5.2. If , then , .
Thus ; there is no solution.
Case 5.3. If , then and .
Thus , we get .
That is, .
Comparing the values , , and , we get that
is the smallest one () and that
is the smallest one (). The result (v) follows.
Consider(vi) and .
Case 6.1. If , then .
We get , or , .
That is, or .
Case 6.2. If , then and .
Thus ; there is no solution.
Case 6.3. If , then and .
Thus , we get .
That is, .
Comparing the values , , and , we get that
is the smallest one (); the result (vi) follows.
Consider(vii) and .
Case 7.1. If , then .
We get , or , .
That is, or .
Case 7.2. If , then and .
Thus , we get .
That is, .
Case 7.3. If , then and .
Thus ; there is no solution.
Comparing the values , , and , we get that
is the smallest one (); the result (vii) follows.
In [13], the authors also gave a similar result as Theorem 13 (in [13], a tree without pendent path of length 3 is called a proper Krag tree), but we do it independently and the methods are also different.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The project was supported by the National Nature Science Foundation of China (nos. 11101087 and 11301440), by the Nature Science Foundation of Fujian Province (no 2013J05006), and by the Foundation to the Educational Committee of Fujian (nos. JA13025 and JA13034).