The minimal-ABC trees with B1-branches

The atom-bond connectivity index (or, for short, ABC index) is a molecular structure descriptor bridging chemistry to graph theory. It is probably the most studied topological index among all numerical parameters of a graph that characterize its topology. For a given graph G = (V, E), the ABC index of G is defined as ABC(G)=∑ij∈E(di+dj-2)/(didj), where di denotes the degree of the vertex i, and ij is the edge incident to the vertices i and j. A combination of physicochemical and the ABC index properties are commonly used to foresee the bioactivity of different chemical composites. Additionally, the applicability of the ABC index in chemical thermodynamics and other areas of chemistry, such as in dendrimer nanostars, benzenoid systems, fluoranthene congeners, and phenylenes is well studied in the literature. While finding of the graphs with the greatest ABC-value is a straightforward assignment, the characterization of the tree(s) with minimal ABC index is a problem largely open and has recently given rise to numerous studies and conjectures. A B1-branch of a graph is a pendent path of order 2. In this paper, we provide an important step forward to the full characterization of these minimal trees. Namely, we show that a minimal-ABC tree contains neither 4 nor 3 B1-branches. The case when the number of B1-branches is 2 is also considered.


Introduction
The atom-bond connectivity index, widely known as ABC index, of a graph is a thoroughly studied vertex-degree-based graph invariant both in chemistry and mathematical communities. For a given simple graph G = (V, E), let us denote by d u the degree of vertex u, and uv the edge incident to the vertices u and v. The atom-bond connectivity index (or, simply, ABC index) is a vertex-degree-based graph topological index, which is a variation of the Randić graph-theoretic invariant [1], and is defined as From the previous considerations, different types of branches will play a crucial role in our quest. Namely, the B k -branches, with k ! 1, and the B Ã k -branches, with k ! 1, are illustrated in Fig 1. In this regard, the most relevant results on minimal-ABC trees are listed next.  3 -branches and one B 2 , including a pendent path of length 3 that may belong to a B Ã 3 -branch or B Ã 2 -branch. As a consequence of Proposition 6, we get the following proposition immediately. Proposition 7. A minimal-ABC tree cannot contain a B 2 -branch and a B Ã 1 -branch simultaneously.

Lemmas
First we recall some technical lemmas.
Due to the symmetry of the function g(x, y), we can also get an equivalent version of Lemma 10.

The root of B 1 -branches
A Kragujevac tree is a tree comprising of a single central vertex, B k -branches, with k ! 1, and at most one B Ã k -branch. Lemma 14 ([28,Theorem 11]). If T is a Kragujevac tree with minimal ABC index, and the degree of the central vertex of T is at least 19, then T contains no B 1 -branch.
Taking into account Lemma 14, we can establish the main result in this section. Proposition 15. If T is a minimal-ABC tree on more than 122 vertices containing B 1branches, then the B 1 -branches cannot be attached to the root vertex of T.
proof. Observe that the B 1 -branches of T are attached to the same vertex, say u, otherwise, there are at least two T k -branches, which is a contradiction to Proposition 8. Suppose to the contrary that u is the root vertex of T.
First, by Proposition 3, u contains no B k -branch with k > 4. Next by Proposition 5, u contains no B 4 -branch, and by Propositions 4 and 7, u contains no B Ã 1 -branch, no matter u has B 3branches or B 2 -branches. Now we may deduce that the branches attached to u must be B 3 -, B 2or B 1 -branches, i.e., T is of the structure as depicted in Fig 2. Notice that T is actually a Kragujevac tree. Denote by d u the degree of u in T.
If d u ! 19, then from Lemma 14, T contains no B 1 -branch, which is a contradiction to the assumption for the existence of B 1 -branches in T.
If d u 18, then recall that every branch attached to u in T is a B k -branch with k = 1, 2, 3, and thus the order of T is at most which is a contradiction to the assumption for the order of T. Now the result follows. Since all the minimal-ABC trees of order up to 300 are completely determined in [29], we may assume that the trees considered in our main results have more than 300 vertices.

Switching transformation
Before we proceed with the main results of this paper, we present the so-called switching transformation explicitly stated by Lin, Gao, Chen, and Lin [30].
Lemma 16 (Switching transformation). Let G = (V, E) be a connected graph with uv, xy 2 E (G) and uy, xv The switching transformation was used in the proofs of some characterizations of the minimal-ABC trees, and the following observation that will be applied in the further analysis. Observation 1. Let G be a minimal-ABC tree with the root vertex v 0 and let v 0 , v 1 , . . ., v n be the sequence of vertices obtain by the breadth-first search of G.
From Observation 1, we may assume that the trees considered are all greedy trees.

The existence of four B 1 -branches
In this section we will prove our first main result: Any minimal-ABC tree cannot contain four B 1 -branches.
The following result is recent and establishes a forbidden configuration for minimal-ABC trees. proof. Suppose to the contrary that T is a minimal-ABC tree containing exactly four B 1branches. Observe that the four B 1 -branches are attached to the same vertex, say u, otherwise, there are at least two T k -branches, which is a contradiction to Proposition 8. Moreover, by Proposition 15, u is not the root vertex of T. Let us denote by v the parent of u.
First, by Proposition 3, u contains no B k -branch with k > 4. Next by Proposition 5, u contains no B 4 -branch, and by Propositions 4 and 7, u contains no B Ã 1 -branch, no matter u has B 3branches or B 2 -branches. Now we may deduce that the branches attached to u must be B 3 -, B 2or B 1 -branches, i.e., T is of the structure depicted in Fig 3. Denote by s the number of B 3 -branches attached to u, and t the number of B 2 -branches attached to u. Clearly, s + t ! 1, and s + t 6 from Proposition 17.
Let d x be the degree of vertex x in T.
In this case, we apply the transformation T 2 depicted in Fig 5. After applying T 2 , the degree of vertex v increases by s + t, while the degree of vertex u decreases to 4, and a child of u in T belonging to a B 2 -branch increases its degree from 3 to 4.
If r = 2, then by (4), we have Moreover, by Lemma 12, we know that 3f So for d v ! 11, we get that i.e., ABC(T 1 ) < ABC(T). For the remaining cases that 7 d v 10, by virtue of Mathematica, the right-hand side of (4) is negative, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily. Subcase 2.3. r = 3. If r = 3, then by (4), we have Moreover, by Lemma 12, we know that 4f So for d v ! 20, we get that i.e., ABC(T 1 ) < ABC(T). For the remaining cases that 8 d v 19, by virtue of Mathematica, the right-hand side of (4) is negative, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily. Subcase 2.4. r = 4. If r = 4, then by (4), we have Moreover, by Lemma 12, we know that 5f(d So for d v ! 31, we get that i.e., ABC(T 1 ) < ABC(T). For the remaining cases that 9 d v 30, by virtue of Mathematica, the right-hand side of (4) is negative, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily. Subcase 2.5. r = 5. If r = 5, then by (4), we have Moreover, by Lemma 12, we know that 6f So for d v ! 42, we get that The minimal-ABC trees with B 1 -branches i.e., ABC(T 1 ) < ABC(T). For the remaining cases that 10 d v 41, by virtue of Mathematica, the right-hand side of (4) is negative, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily. Subcase 2.6. r = 6. If r = 6, then by (4), we have Moreover, by Lemma 12, we know that 7f So for d v ! 56, we get that i.e., ABC(T 1 ) < ABC(T). For the cases that 18 d v 55, by virtue of Mathematica, the righthand side of (4) is negative, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily.
As to the remaining cases that 11 d v 17, let us be a bit more precisely in (2) Notice that the degree of every neighbor of v in " T is at least 3 from Proposition 2. Furthermore, by Lemma Now together with (4), it follows that By virtue of Mathematica, the right-hand side of (5) is negative, equivalently ABC(T 1 ) < ABC(T), for 11 d v 17, follows from direct calculation easily.
Combining the above cases, the result follows easily.
proof. Let d x be the degree of vertex x in T. First we apply the transformation T 1 illustrated in Fig 7. After applying T 1 , the degree of vertex u decreases by 3, while the degrees of three children of u in T belonging to a B 2 -branch increase from 3 to 4, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T 1 is r : The minimal-ABC trees with B 1 -branches Now it follows that By virtue of Mathematica, the right-hand side of (6) is negative, equivalently ABC(T 1 ) < ABC (T), follows from direct calculation easily, except the case t = 3 and s = 0. In such case, we apply the transformation T 2 illustrated in   The minimal-ABC trees with B 1 -branches After applying T 2 , the degree of vertex v increases by 2, the degrees of three children of u in T belonging to a B 2 -branch increase from 3 to 4, a pendent vertex in T belonging to a B 2branch increases its degree from 1 to 2, the degree of u decreases from 7 to 1, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T 2 is Clearly Note that d v ! d u = 7 from Proposition 2, and from Lemma 12, we know that 3f(d v + 2, 4) − f(d v , 7) decreases in d v ! 7. Therefore, for d v ! 20, we get that ABCðT 1 Þ À ABCðTÞ < 3f ð20 þ 2; 4Þ À f ð20; 7Þ þ f ð2; 1Þ À 3f ð7; 3Þ < 0 : For the remaining cases that 7 d v 19, let us be a bit more precisely in (7) Note that every neighbor of v in " T has degree at least three from Proposition 2. By Lemma 10, Now together with (7), it follows that By virtue of Mathematica, the right-hand side of (8) is negative, equivalently ABC(T 1 ) < ABC (T), for 7 d v 19, follows from direct calculation easily. Then the result follows.
We are now prepared to establish the main result of this section. Theorem 22. A minimal-ABC tree cannot contain three B 1 -branches. proof. Similarly to Theorem 18, let us suppose to the contrary that T is a minimal-ABC tree containing exactly three B 1 -branches. Observe that the three B 1 -branches are attached to the same vertex, say u, otherwise, there are at least two T k -branches, which is a contradiction to Proposition 8. Moreover, by Proposition 15, u is not the root vertex of T. Denote by v the parent of u.
First, by Proposition 3, u contains no B k -branch with k > 4. Next by Proposition 5, u contains no B 4 -branch, and by Propositions 4 and 7, u contains no B Ã 1 -branch, no matter u has B 3branches or B 2 -branches. Now we may deduce that the branches attached to u must be B 3 -, B 2or B 1 -branches, i.e., T is of the structure depicted in Fig 6. Let us denote by s the number of B 3 -branches attached to u, and by t the number of B 2branches attached to u. Clearly, s + t ! 1, and s + t 8, from Proposition 19.
We apply the transformation T depicted in Fig 9. And let d x be the degree of vertex x in T.
After applying T , the degree of vertex v increases by s + t, while the degree of vertex u decreases by s + t, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T is Clearly T Þ, and thus On one hand, from Lemma 10, So it follows that The minimal-ABC trees with B 1 -branches On the other hand, note that d v ! d u = s + t + 4 from Proposition 2, and both f(d v + s + t, 4) and f(d v + s + t, 3) decrease in d v ! s + t + 4, i.e., the right-hand side of (10) also decreases in d v ! s + t + 4.
Besides the upper bound about ABC(T 1 ) − ABC(T) as (10), by considering a bit precisely in (9) for the term X we may get a somewhat stricter upper bound about ABC(T 1 ) − ABC(T). Note that, from and from Proposition 2, every neighbor of v in " T has degree at least three, thus X Now together with (9), it follows that Case 1. t = 0. In this case, note that 1 s 8, and d v ! s + 4. By direct calculation, we may deduce that the right-hand side of (10) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: we would turn to use (11), and negative upper bounds, equivalently ABC(T 1 ) < ABC(T), follow from direct calculation. Case 2. t = 1.
In this case, note that 0 s 7, and d v ! s + 5.
By direct calculation, we may deduce that the right-hand side of (10) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: • s = 7 and 12 d v 32, we would turn to use (11), and negative upper bounds, equivalently ABC(T 1 ) < ABC(T), follow from direct calculation easily. Case 3. t = 2. In this case, note that 0 s 6, and d v ! s + 6. By direct calculation, we may deduce that the right-hand side of (10) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: On the other hand, by direct calculation, we may deduce that the right-hand side of (10) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: Besides that, by direct calculation, we may deduce that the right-hand side of (10) is negative, equivalently ABC(T 1 ) < ABC(T), for s = 1 and d v ! 145. For the remaining cases s = 1 and 9 d v 144, we would turn to use (11), and a negative upper bound, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily. Case 6. t = 5. In this case, note that s = 0, 1, 2, 3. The contradiction for the cases s = 0 and s = 1, 2, 3 may be, respectively, deduced from Propositions 20 and 21.
In this case, note that s = 0, 1. The contradiction for the cases s = 0 and s = 1 may be, respectively, deduced from Propositions 20 and 21.
Case 9. t = 8. In this case, note that s = 0. The contradiction may be deduced from Proposition 20 directly.
Combining the above cases, the result follows.

The existence of two B 1 -branches
This last section is devoted to the analysis of the existence of two B 1 -branches in a minimal-ABC tree. The first two propositions are known results establishing forbidden configurations in such cases. • t = 4 and s = 0, 1, 2, 3, 4, 5, 6;
proof. First we apply the transformation T 1 illustrated in Fig 11. Let d x be the degree of vertex x in T.
After applying T 1 , the degree of vertex u decreases by 2, while the degrees of two children of u in T belonging to a B 2 -branch increase from 3 to 4. The rest of the vertices do not change their degrees. The change of the ABC index after applying T 1 is Now it follows that The minimal-ABC trees with B 1 -branches The right-hand side of (12) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: • t = 4 and s = 3, 4, 5, 6; • t = 5 and s = 2, 3, 4, 5; • t = 6 and s = 1, 2, 3, 4; • t = 7 and s = 1, 2, 3; • t = 8 and s = 1, 2; • t = 9 and s = 1.
Next for the following cases: • t = 2 and s = 0; • t = 3 and s = 1, 2; • t = 4 and s = 1, 2, we apply the transformation T 2 illustrated in Fig 12. After applying T 2 , the degree of vertex v increases by s + t − 1, the degrees of two children of u in T belonging to a B 2 -branch increase from 3 to 4, a pendent vertex in T belonging to a B 3branch increases its degree from 1 to 2, the degree of u decreases from s + t + 3 to 1, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T 2 is The minimal-ABC trees with B 1 -branches Now together with (13), it follows that By direct calculation, we may deduce that the right-hand side of (14) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: • t = 3, s = 1, and d v ! 64; • t = 3, s = 2, and d v ! 44; • t = 4, s = 1, and d v ! 4015; • t = 4, s = 2, and d v ! 116.
At this point, there are still two remaining cases: t = 4, s = 0, and t = 5, s = 1.
For the case t = 4 and s = 0, we apply the transformation T 3 illustrated in Fig 13. After applying T 3 , the degree of vertex v increases by 2, the degrees of two children of u in T belonging to a B 2 -branch increase from 3 to 5, one child of u in T belonging to another B 2branch increases its degree from 3 to 4, the remaining child of u in T belonging to a B 2 -branch decreases its degree from 3 to 2, the degree of u decreases from 7 to 1, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T 3 is Clearly, Note that d v ! d u = 7 from Proposition 2, and by Lemma 11, f(d v + 2, 5) − f(d v + 2, 4) decreases in d v ! 7. On the other hand, by Lemma 12, 3f(d v + 2, 4) − f(d v , 7) decreases in d v ! 7. So the right-hand side of (17) also decreases in d v ! 7.
Besides the upper bound about ABC(T 1 ) − ABC(T) as (17), by considering in (16) the term we may get a somewhat stricter upper bound about ABC(T 1 ) − ABC(T). Note that, from Lemma 10, and from Proposition 2, every neighbor of v in " T has degree at least three, thus Now together with (16), it follows that þ2ðf ð1; 2Þ À f ð7; 3ÞÞ : For d v ! 20, by (17), we have i.e., ABC(T 1 ) < ABC(T). For the remaining cases 7 d v 19, we would turn to use (18), and a negative upper bound, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation straightforwardly.
As to the last case t = 5 and s = 1, we apply the transformation T 4 illustrated in Fig 14. After applying T 4 , the degree of vertex v increases by 4, three children of u in T belonging to a B 2 -branch increase its degrees from 3 to 4, the degree of one child of u in T belonging to another B 2 -branch increases from 3 to 5, the remaining child of u in T belonging to a B 2branch decreases its degree from 3 to 2, the degree of u decreases from 9 to 1, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T 4 is ð19Þ Note that d v ! d u = 9 from Proposition 2, and by Lemma 11,f(d v + 4,5) − f(d v + 4, 4) decreases in d v ! 9. On the other hand, by Lemma 12, 5f(d v + 4, 4) − f (d v , 9) decreases in d v ! 9. So the right-hand side of (20) also decreases in d v ! 9. The minimal-ABC trees with B 1 -branches Besides the upper bound about ABC(T 1 ) − ABC(T) as (20), by considering in (19) the term we may get a somewhat stricter upper bound about ABC(T 1 ) − ABC(T). Note that, from and from Proposition 2, every neighbor of v in " T has degree at least three, thus Now together with (19), it follows that For d v ! 15, by (20), we have i.e., ABC(T 1 ) < ABC(T). For the remaining cases 9 d v 14, we would turn to use (21), and a negative upper bound, equivalently ABC(T 1 ) < ABC(T), follows from direct calculation easily. Combining the above arguments, the result follows. Our main result is stated next. As we will see, the configuration depicted in Fig 10 is very important since, minimal-ABC trees may contain two B 1 -branches only in two very particular configurations.
Theorem 26. A minimal-ABC tree cannot contain two B 1 -branches, unless the two B 1branches belong to the configuration depicted in Fig 10 with s = 0, t = 1, or s = 0, t = 3.
proof. Suppose to the contrary that T is a minimal-ABC tree containing exactly two B 1branches. Observe that the two B 1 -branches are attached to the same vertex, say u, otherwise, there are at least two T k -branches, which is a contradiction to Proposition 8. Moreover, by Proposition 15, u is not the root vertex of T. Denote by v the parent of u.
First, by Proposition 3, u contains no B k -branch with k > 4. Next by Proposition 5, u contains no B 4 -branch, and by Propositions 4 and 7, u contains no B Ã 1 -branch, no matter u has B 3branches or B 2 -branches. Now we may deduce that the branches attached to u must be B 3 -, B 2or B 1 -branches, i.e., T is of the structure depicted in Fig 10. Set s and t for the numbers of B 3 -and B 2 -branches attached to u, respectively. Clearly, s + t ! 1, and s + t 10 from Proposition 23.
We apply the transformation T depicted in Fig 15. And let d x be the degree of vertex x in T. After applying T , the degree of vertex v increases by s + t, while the degree of vertex u decreases by s + t, and the rest of the vertices do not change their degrees. The change of the ABC index after applying T is ð22Þ T Þ, and thus On one hand, from Lemma 10, So it follows that Furthermore, note that d v ! d u = s + t + 3 from Proposition 2, and both f(d v + s + t, 4) and f (d v + s + t, 3) decrease in d v ! s + t + 3, i.e., the right-hand side of (23) also decreases in d v ! s + t + 3. The minimal-ABC trees with B 1 -branches Besides the upper bound about ABC(T 1 ) − ABC(T) as (23), by considering a bit precisely in (22) we may get a somewhat stricter upper bound about ABC(T 1 ) − ABC(T). Note that, from decreases in d x , and from Proposition 2, every neighbor of v in " T has degree at least three, thus X Now together with (22), it follows that Case 1. t = 0. In this case, note that 1 s 10, and d v ! s + 3 from Proposition 2.
By direct calculation, we may deduce that the right-hand side of (23) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: In this case, note that 0 s 7, and d v ! s + 6 from Proposition 2. On one hand, the contradiction for s = 1, 2 follows from Proposition 25.
On the other hand, by direct calculation, we may deduce that the right-hand side of (23) is negative, equivalently ABC(T 1 ) < ABC(T), holds for the following cases: In this case, note that 0 s 6. The contradiction may be deduced from Proposition 25. Case 6. t = 5. In this case, note that 0 s 5. The contradiction for the cases that s = 0 and s = 1, 2, 3, 4, 5 may be deduced from Propositions 24 and 25, respectively. Case 7. t = 6. In this case, note that 0 s 4. The contradiction for the cases that s = 0 and s = 1, 2, 3, 4 may be deduced from Propositions 24 and 25, respectively. Case 8. t = 7.
In this case, note that 0 s 3. The contradiction for the cases that s = 0 and s = 1, 2, 3 may be deduced from Propositions 24 and 25, respectively. Case 9. t = 8. In this case, note that s = 0, 1, 2. The contradiction for the cases that s = 0 and s = 1, 2 may be deduced from Propositions 24 and 25, respectively. Case 10. t = 9.
In this case, note that s = 0, 1. The contradiction for the cases that s = 0 and s = 1 may be deduced from Propositions 24 and 25, respectively. Case 11. t = 10.
In this case, note that s = 0. The contradiction may be deduced from Proposition 24 directly.
Combining the above arguments, the result finally follows.

Discussion
The characterization of minimal-ABC trees is a rather active topic in chemical graph theory these years, which has led to a lot of structural properties and potential conjectures. It is known that every pendent vertex of minimal-ABC trees belongs to some B k -branch. As a strengthening, this paper proves that a minimal-ABC tree contains at most two B 1 -branches. Moreover, we claim that a minimal-ABC tree can not contain two B 1 -branches simultaneously, except for two cases that s = 0, and t = 1 or 3.
During the investigation of this paper, we also considered the two unsolved cases. However, to the best of our knowledge, until now we only get a solution under some particular degree conditions. In future research, the key point is to construct a more perfect graph transformation involve in general cases, which lead to a desired solution finally.
Actually, our ultimate goal is to show that the minimal-ABC trees contain no B 1 -branch, when the order of that tree is large sufficiently.