The evolution of the structure of ABC-minimal trees
Seyyed Aliasghar Hosseini, Bojan Mohar, Mohammad Bagher Ahmadi
TThe evolution of the structure of ABC-minimal trees
Seyyed Aliasghar Hosseini [email protected]
Department of MathematicsSimon Fraser UniversityBurnaby, BC, Canada Bojan Mohar ∗† [email protected] Department of MathematicsSimon Fraser UniversityBurnaby, BC, CanadaMohammad Bagher Ahmadi [email protected]
Department of MathematicsShiraz UniversityShiraz, IranNovember 6, 2019
Abstract
The atom-bond connectivity (ABC) index is a degree-based molecular descriptor that founddiverse chemical applications. Characterizing trees with minimum ABC-index remained anelusive open problem even after serious attempts and is considered by some as one of the mostintriguing open problems in mathematical chemistry. In this paper, we describe the exactstructure of the extremal trees with sufficiently many vertices and we show how their structureevolves when the number of vertices grows. An interesting fact is that their radius is at most 5and that all vertices except for one have degree at most 54. In fact, all but at most O (1) verticeshave degree 1, 2, 4, or 53. Let γ n = min {ABC ( T ) : T is a tree of order n } . It is shown that γ n = (cid:113) (cid:16) √
55 + 156 √ (cid:17) n + O (1) ≈ . n + O (1). Molecular descriptors [33] are mathematical quantities that describe the structure or shape ofmolecules, helping to predict the activity and properties of molecules in complex experiments. In thelast few years a number of new molecular structure descriptors has been conceived [20, 28, 29, 31].Molecular descriptors play a significant role in chemistry, pharmacology, etc. Among molecularstructure descriptors, topological indices have a prominent place. They are useful tools for modelingphysical and chemical properties of molecules, for design of pharmacologically active compounds, ∗ Supported in part by an NSERC Discovery Grant (Canada), by the Canada Research Chair program, and bythe Research Grant P1–0297 of ARRS (Slovenia). † On leave from: IMFM & FMF, Department of Mathematics, University of Ljubljana, Ljubljana, Slovenia. a r X i v : . [ m a t h . C O ] N ov or recognizing environmentally hazardous materials, etc., see [18]. One of the most importanttopological indices is the Atom Bond Connectivity index, also known as the ABC index. It wasintroduced by Estrada [21] with relation to the energy of formation of alkanes. It was quicklyrecognized that this index reflects important structural properties of graphs in general. The ABCindex was extensively studied in the last few years, from the point of view of chemical graphtheory [22, 34], and in general graphs [10]. Additionally, the physico-chemical applicability ofthe ABC index and its mathematical properties was confirmed and extended in several studies[5, 9, 11, 23, 27, 32, 35]. Some novel results about ABC index can be found in [14, 15, 16] and inthe references cited therein.Let G be a simple graph on n vertices, and let its vertex-set be V ( G ) and edge-set E ( G ). By uv we denote the edge connecting the vertices u and v . The degree of a vertex v is denoted by d v .For an edge uv in G , we consider the quantity f ( d u , d v ) = (cid:114) d u + d v − d u d v . The
Atom-Bond Connectivity index (shortly
ABC index ) of G is defined as ABC ( G ) = (cid:88) uv ∈ E ( G ) f ( d u , d v ) . When the mathematical properties of a graph-based structure descriptor are investigated, one ofthe first questions is for which graph (with a given order n ) is this descriptor minimal or maximal.It is known that adding an edge in a graph strictly increases its ABC index [12] and deleting anedge in a graph strictly decreases its ABC index [8]. According to this fact, among all connectedgraphs with n vertices, the complete graph K n has the maximum ABC index and graphs withminimum ABC index are trees. A tree is said to be ABC-minimal if no other tree on the samenumber of vertices has smaller ABC index.Although it is easy to show that the star graph S n has maximum ABC index among all treesof the same order [22], despite many attempts in the last years, it is still an open problem tocharacterize trees with minimum ABC-index (ABC-minimal trees). Eventually, a computer-aidedstudy [24] gave rise to a conjecture on the actual structure of the ABC-minimal trees. Later results[2, 3] revealed that the conjecture was false, and that the true structure of the ABC-minimal treesis more complex than the computer-aided results have indicated. In [10], the author presents lowerand upper bounds on the ABC index of general graphs and trees, and characterizes graphs forwhich these bounds are best possible.In this work we finally resolve the question on giving a precise description of ABC-minimaltrees. For small values of n , this structure is as observed in previous works (see [6] for n ≤ C -branch, emerges. When n is very large, any ABC-minimal tree just slightly deviates from being composed of one vertex oflarge degree to which the C -branches are attached. See the last section for more details.An interesting fact is that the radius of ABC-minimal trees is at most 5 (usually just 4) andthat all vertices except for one have degree at most 54. In fact, all but at most O (1) verticeshave degree 1, 2, 4, or 53. Let γ n = min {ABC ( T ) : T is a tree of order n } . It is shown that γ n = (cid:113) (cid:16) √
55 + 156 √ (cid:17) n + O (1) ≈ . n + O (1).2he proofs are simplified by introducing a natural equivalence relation, called similarity, on theset of all ABC-minimal trees and considering only those elements in each similarity class that aremaximal in certain total order on all rooted trees. They are said to be ABC-extremal . The mainstructure results that are proved along the way towards the structural description of ABC-minimaltrees are the following. For each ABC-extremal tree we let R be its vertex having maximum degree.It is proved that all edges uv have different vertex degrees, d u (cid:54) = d v , with a possible exception ofhaving one edge incident with R , whose ends both have maximum degree, and having one edgewhose ends are both of degree 2. It also has been proved that for n ≥ R has strictly decreasing degrees, with the only possible exception whenit contains an exceptional edge mentioned above (see Theorem 3.1 and Corollary 3.3). The nextmajor step is to prove that vertices at distance 2 from R have degree at most 5 (Theorem 4.1).Vertices of degree 1 are always adjacent to vertices of degree 2, and it is shown that the root cannotbe incident with vertices of degree 2 (when n ≥ B -branches,which are grouped together into C -branches (see Section 2 for definitions). Earlier works havenot expected C -branches, and it was conjectured in [24] and [13] that B -branches are the mainstructural component. It was found in [1] that C -branches will play a prominent role. A newsurprise coming from our work is that for large n only C -branches occur, see Theorem 5.3. Aninteresting final outcome is that the radius of ABC-extremal trees is at most 5 and that all verticesexcept for one have degree at most 54. In fact, all but at most O (1) vertices have degree 1, 2, 4, or53. The following known facts will be used in the paper.
Theorem 2.1 ([30]) . In every ABC-minimal tree of order at least , each vertex of degree isadjacent to a vertex of degree . Proposition 2.2 ([14]) . Let x, y ≥ be real numbers and let a ≥ and b , ≤ b < y − beconstants. Let g ( x, y ) = f ( x + a, y − b ) − f ( x, y ) . Then, g ( x, y ) is increasing in x and decreasing in y . Theorem 2.3 ([25]) . If an ABC-minimal tree has distinct vertices v , v , v such that d v ≥ d v >d v , then v cannot be adjacent to both v and v . Let T be an ABC-minimal tree of order n and let ∆ be the maximum degree of T . Let us pickone of the vertices of degree ∆ and call it a root of T . We will denote the root by R and from nowon consider any tree as a rooted tree. Thus, we can speak about descendants, predecessors, thesons of a vertex (immediate successors), etc. We also define the height function h : V ( T ) → N bytaking h ( v ) to be the distance of v from the root R in T .3or each vertex v , we denote by T v the subtree of T consisting of v and all of its descendants.If T v ∩ T v (cid:48) = ∅ , then we define another tree, T ( v, v (cid:48) ), that is obtained from T by exchanging T v and T v (cid:48) . Under certain conditions, this exchange operation reduces the ABC-index, meaning thatsuch conditions cannot occur in ABC-minimal trees. The following result was proved by Lin et al.in [7]. Lemma 2.4 ([7]) . Let uv and u (cid:48) v (cid:48) be edges of a tree T . Suppose that v is a son of u , v (cid:48) is a sonof u (cid:48) and that T v ∩ T v (cid:48) = ∅ . (a) If d u > d u (cid:48) and d v < d v (cid:48) , then ABC ( T ) > ABC ( T ( v, v (cid:48) )) . In particular, T is not ABC-minimal. (b) If d u = d u (cid:48) or d v = d v (cid:48) , then ABC ( T ) = ABC ( T ( v, v (cid:48) )) .Proof. Equality in (b) is obvious since both trees have edges with same degrees. To prove (a), weapply Proposition 2.2 with y = d v (cid:48) , a = 0 and b = d v (cid:48) − d v . Observe that: ABC ( T ) − ABC ( T ( v, v (cid:48) )) = f ( d u , d v ) + f ( d u (cid:48) , d v (cid:48) ) − f ( d u , d v (cid:48) ) − f ( d u (cid:48) , d v )= ( f ( d u , d v ) − f ( d u , d v (cid:48) )) − ( f ( d u (cid:48) , d v ) − f ( d u (cid:48) , d v (cid:48) ))= g ( d u , d v (cid:48) ) − g ( d u (cid:48) , d v (cid:48) ) , which is positive by the proposition.Part (b) of the lemma motivates the following definitions. First of all, if the assumptions ofthe lemma hold and d u = d u (cid:48) or d v = d v (cid:48) , then we say that T ( v, v (cid:48) ) is obtained from T by a similarity exchange . Further, we say that two trees T and T (cid:48) are similar (or ABC-similar ) if T (cid:48) can be obtained from T by a series of similarity exchange operations. When we treat ABC-minimaltrees as rooted trees whose root is a vertex of maximum degree, we also treat any tree obtainedby taking a different vertex of maximum degree as the root as being similar. Note that similarityis an equivalence relation that preserves the ABC-index. In order to characterize ABC-minimaltrees, it suffices to describe one tree in each similarity class. Below we will introduce some specialproperties of ABC-minimal trees that will define a subclass called ABC-extremal trees.For our next definition we will need a special linear ordering (cid:31) among the isomorphism classesof all rooted trees (with at most n vertices). For two such trees T and T (cid:48) , we first compare theirroots. If the root of T has larger degree than the root of T (cid:48) , then we set T (cid:31) T (cid:48) (and we set T (cid:48) (cid:31) T if the root of T (cid:48) has larger degree). If the degrees are the same, both equal to d ≥
0, welexicographically compare their subtrees T , . . . , T d and T (cid:48) , . . . , T (cid:48) d rooted by the sons of their roots,and we set T (cid:31) T (cid:48) if the subtrees of T are lexicographically larger. Note that these subtrees arelexicographically the same if and only if T and T (cid:48) are isomorphic. (This can be easily proved byinduction.) We write T (cid:23) T (cid:48) if either T (cid:31) T (cid:48) or T and T (cid:48) are isomorphic as rooted trees.An ABC-minimal tree T is said to be ABC-extremal if it is (cid:31) -largest in its similarity classwhich is the same as greedy tree for a given degree sequence. Note that this also included the bestchoice of the root among the vertices of maximum degree. The ABC-extremal trees have someadditional properties that will be useful for us. Let us summarize some of them.Let T be an ABC-extremal tree. Then T has the following properties: It will be shown later that there are at most two such vertices in any ABC-minimal tree. u, v ∈ V ( T ). Suppose that h ( u ) < h ( v ) and if u is not the root then v / ∈ T u . Then T u (cid:23) T v and, in particular, d u ≥ d v .(P2) Let h = min { h ( v ) | d v = 2 } . If T contains an edge uu (cid:48) such that d u = d u (cid:48) = 2, then either h ( u ) = h or h ( u (cid:48) ) = h .(P3) Suppose that u, v are non-root vertices with the same degree, d u = d v , and let r = d u − u , . . . , u r be the sons of u and let v , . . . , v r be the sons of v . Suppose that T u (cid:23) T u (cid:23)· · · (cid:23) T u r and T v (cid:23) T v (cid:23) · · · (cid:23) T v r . If h ( u ) < h ( v ), then T u r (cid:23) T v . If h ( u ) = h ( v ), theneither T u r (cid:23) T v or T v r (cid:23) T u . Assuming that T u r (cid:23) T v , then we have, in particular, that d u ≥ d u ≥ · · · ≥ d u r ≥ d v ≥ d v ≥ · · · ≥ d v r . Proof. (P1) The proof is by induction on h ( u ). If u is the root, then the property is clear by thedefinition of similarity which includes exchanging the root with another vertex of maximum degreeif that rooted tree is (cid:31) -larger. Therefore we may assume that u is not the root. Let ˆ u and ˆ v be thepredecessors of u and v , respectively. By the induction hypothesis, we have that d ˆ u ≥ d ˆ v . Suppose,for a contradiction, that T u ≺ T v . In particular, d u ≤ d v . Since v / ∈ T u , we have T u ∩ T v = ∅ . ByLemma 2.4(a), we conclude that either d ˆ u = d ˆ v or d u = d v . Therefore, T ( u, v ) is obtained from T by a similarity exchange. Since T v (cid:31) T u , we conclude that T ( u, v ) (cid:31) T , which contradicts theassumption that T is ABC-extremal.(P2) As proved in [26] (see Lemma 2.5), there is at most one such 2-2 edge in T . Suppose that h ( u ) < h ( u (cid:48) ), and let v be a degree-2 vertex with h ( v ) = h . If h ( u ) > h , then (P1) implies that T v (cid:23) T u . By Theorem 2.3 we see that the son of v has degree 1 and since the son u (cid:48) of u has degree2, we have that T u (cid:31) T v , a contradiction.(P3) If T u r ≺ T v and either h ( u ) < h ( v ) or T v r ≺ T u , then one of the similarity exchanges T ( u r , v ) or T ( v r , u ) would yield a (cid:31) -larger tree. This contradiction shows that (P3) holds.ABC-extremal trees and their properties (P1)–(P3) have been used frequently in previous worksand were sometimes called “greedy trees”.At the end of the next section, we will show that (P1) holds also when v is a successor of u . (cid:124) (cid:123)(cid:122) (cid:125) k . . . B k B B ∗ B B − Figure 1: Definition of B k -branches.Suppose that all sons of a vertex v ∈ V ( T ) are of degree 2 and all second descendants are ofdegree 1. Then the subtree T v is said to be a B k -branch with root v , where k is the number of sons In some earlier papers, B − -branches are called B -branches. v (which is d v − v (cid:54) = R ). See Figure 1. In our later figures, we will represent each B k -branchby a triangle with the number k next to it. If k = 3, the number may be omitted.Simulations and existing results [14, 15, 16] show that ABC-minimal trees have lots of B -branches and only a small number of B k -branches for k (cid:54) = 3; see also our Corollary 4.6.If T v can be obtained from a B k -branch by adding a new vertex and joining it to one of verticesof degree 1, then we say that this is a B ∗ k -branch. This kind of subtrees may appear in ABC-minimaltrees, but cannot occur more than once (see [26]). The reason is the following result about (edges whose both ends have degree 2, also known as pendent path of length three). Lemma 2.5 ([26, 17]) . Any ABC-minimal tree has at most one 2-2 edge and, if there is one, it ispart of a B -branch. Furthermore, ABC-minimal trees of order ≥ contain no 2-2 edges. For a tree T , let B ( T ) be the set of those vertices different from the root that are of degreeat least 3 that have a son of degree 2, whose son is of degree 1. Note that for every k ≥ B ( T )contains all roots of B k -branches and B ∗ k -branches ( k ≥ u ∈ B ( T ) is not a root ofsome B k -branch, then it is referred to as an exceptional vertex in B ( T ) and its subtree T u is called B -exceptional branch . Note that the root of any B ∗ k -branch with k ≥ B ( T ). Lemma 2.6.
Any ABC-extremal tree T has at most one exceptional vertex in B ( T ) . Moreover, if v is exceptional, then d v = max { d u | u ∈ B ( T ) } .Proof. Theorem 2.1 implies that a vertex in B ( T ) cannot have a son of degree 1. Thus, an excep-tional vertex either has a son of degree at least 3, or it has a son of degree 2 that is not incident toa vertex of degree 1.Suppose that there are two exceptional vertices, v and v , where v i has a son v (cid:48) i of degree2 (whose son is a degree-1 vertex) and also has a son u i that is not the father of a vertex ofdegree 1 ( i = 1 , u i to be of degree more than 2 if possible. Suppose that d v ≥ d v .Lemma 2.4 shows that ABC ( T ( v (cid:48) , u )) ≤ ABC ( T ), where the inequality is strict unless d v = d v or d u = d v (cid:48) = 2. Since T is ABC-minimal, we have one of the two equalities. In either case, replacing T with the tree T ( v (cid:48) , u ) is a similarity exchange. If d v > d v , then this exchange gives a (cid:31) -largertree, contradicting extremality of T . The same may give a contradiction if d v = d v ; but if it doesnot, then we consider T ( v (cid:48) , u ), and it is easy to see that this yields a (cid:31) -larger tree. This showsthat there is at most one exceptional vertex.Suppose now that T has precisely one exceptional vertex v ∈ B ( T ). If d v is not the largest in { d u | u ∈ B ( T ) } , doing a similar exchange with the vertex u in B ( T ) of maximum degree gives usa contradiction to the extremality of T .We will show in Lemma 4.5 that in addition to B ∗ k ( k ≥
2) only one type of B -exceptionalbranches may exist in any ABC-minimal tree.Our next goal is to show that B k -branches may occur only for k ≤ Lemma 2.7 ([14, 19]) . If an ABC-minimal tree contains a B k -branch, then k ≤ . If it containsa B ∗ k -branch, then k ≤ . B ∗ k -branchescan be found in [19]. We include a sketch of our own proof, some of whose easier details are left tothe reader. Proof.
Let u be the root of a B k -branch ( B ∗ k -branch) considered. We may assume that u is not theroot. Let ˆ u be the father of u . For B k ( k ≥ T u with B k − and B ∗ , both attached toˆ u . Note that the degree of ˆ u increases by one. Let T (cid:48) be the resulting tree. Now it is easy to seethat ABC ( T ) − ABC ( T (cid:48) ) >
0, which is a contradiction.Similarly, for B ∗ k ( k ≥ T u with B k − and B attached to ˆ u . And for B ∗ , wereplace T u with the tree B ∗∗ shown in Figure 12. Details are omitted.Dimitrov [14] also proved that B -branches can be excluded under the assumption that thereis a B or B -branch as a sibling . Below we give a slightly stronger result. Lemma 2.8.
Let T be an ABC-minimal tree. If a B k -branch and a B l -branch are siblings, then | k − l | ≤ .Proof. Suppose that k ≥ l and let t = k − l . Let us assume that a B k -branch and a B k − t -branchexist as siblings in T . Let the parent of B k and B k − t be a vertex of degree d . Theorem 2.3 impliesthat for every path starting at the root, the vertex-degrees along the path never increase, thus wehave d ≥ k + 1. By detaching one vertex of degree 2 from B k and attaching it to B k − t we obtain atree T (cid:48) in which B k is replaced by B k − and B l with B l +1 . Since f (2 , x ) = √ / x , we have ABC ( T ) − ABC ( T (cid:48) ) = f ( d, k + 1) + f ( d, k − t + 1) − f ( d, k ) − f ( d, k − t + 2) . For fixed d and k , this difference is decreasing in terms of k − t + 1 (by Proposition 2.2 used on thesecond and the last term with a = 0, b = 1 and y = k − t + 2). This means that the difference is(strictly) increasing in terms of t . Since ABC ( T ) − ABC ( T (cid:48) ) = 0 when t = 1, we conclude that for t ≥ T .As in the above proof, we will frequently compare the ABC-index of a tree T with that of amodified tree T (cid:48) . To make the notation shorter we will write∆( T, T (cid:48) ) =
ABC ( T ) − ABC ( T (cid:48) ) . We define a C k -branch as a subtree T v , in which v has precisely k sons v , . . . , v k , and theirsubtrees T v , . . . , T v k are all B -branches. In our figures, we will represent a C k -branch as a squarewith k written inside the square. Lemma 2.9 ([17]) . Let T be an ABC-minimal tree. If there are C k -branch and C l -branch assiblings, then | k − l | ≤ . Lemma 2.10.
No ABC-minimal tree contains a C k -branch with k ≥ . For example, the case where the root has only C k branches and one B as its children is not considered in [14]. roof. Assume that T is an ABC-minimal tree with a C k -branch, where k ≥ k is odd. For the even case only some small modifications are needed. We can replace the C k -branch with two C k (cid:48) -branches, where k (cid:48) = k − , see Figure 2. More precisely, the B -brancheswithin C k are divided evenly between the two C k (cid:48) -branches, and the remaining B is replaced bythree paths attached to three B branches (which turns them into B ) as indicated in Figure 2.Let T (cid:48) be the resulting tree. We have: R z d R − . . .. . . z z d R − . . . z . . . . . . (cid:124) (cid:123)(cid:122) (cid:125) k − (cid:124) (cid:123)(cid:122) (cid:125) k − − (cid:124) (cid:123)(cid:122) (cid:125) k R (cid:48) Figure 2: Suggested change when there exists a C k branch with odd k ≥ T, T (cid:48) ) = f ( d R , k + 1) + kf ( k + 1 ,
4) + d R − (cid:88) i =1 f ( d R , d z i ) − f ( d R + 1 , k +12 ) − ( k − f ( k +12 , − f ( k +12 , − d R − (cid:88) i =1 f ( d R + 1 , d z i ) . Using Proposition 2.2 we can see that f ( d R , d z i ) − f ( d R + 1 , d z i ) is increasing in d z i . Thus, tohave the worst case we may consider the lowest possible values for the degrees d z i . Note that thereare B -branches in C k and since k < d R , Lemma 2.4 shows that d z i ≥
4. So:∆(
T, T (cid:48) ) ≥ f ( d R , k + 1) + kf ( k + 1 ,
4) + ( d R − f ( d R , − f ( d R + 1 , k +12 ) − ( k − f ( k +12 , − f ( k +12 , − ( d R − f ( d R + 1 , . Now, let us rewrite this inequality as follows:∆(
T, T (cid:48) ) ≥ f ( d R , k + 1) − f ( d R + 1 , k +12 ) +( d R − f ( d R , − f ( d R + 1 , k ( f ( k + 1 , − f ( k +12 , f ( k +12 , − f ( k +12 , f ( k +12 , − f ( k +12 , d R + 1) . Again, using Proposition 2.2 we can see that the value in each line except the first one is8ncreasing in k . Regarding the first line, observe the following: f ( d R , k + 1) − f ( d R + 1 , k +12 ) = f ( d R , k + 1) − f ( d R + 1 , k + 1)+ f ( d R + 1 , k + 1) − f ( d R + 1 , k )+ f ( d R + 1 , k ) − f ( d R + 1 , k − . . . + f ( d R + 1 , k +32 ) − f ( d R + 1 , k +12 ) . Here, each line is increasing in k , therefore f ( d R , k + 1) − f ( d R + 1 , k +12 ) is also increasing in k . If wesubstitute k by 143, then the above lower bound only depends on one variable, d R , and it is easy tocheck that ∆( T, T (cid:48) ) > d R ≥ k . Therefore, ∆( T, T (cid:48) ) > d R ≥ k ≥ d z i = 4 and also considered only one copy of a C k branch. . . . . . . Ru u r v v s k k k r u v Figure 3: The basic structure (after deletion of a small number of vertices). When n is small,we have only B -branches ( r = 0). As n grows, a combination of both occurs and when n issufficiently large, only C k i -branches remain ( s = 0), eventually with all k i being equal to 52 and r = n/ − O (1).In this paper it will be proved that ABC-minimal trees have the structure close to that shownin Figure 3 in the sense that there is a small number of vertices whose deletion gives us this form.Moreover, the following transition occurs. Let us denote by r the number of C k -branches (whoseroots u , . . . , u r are adjacent to the root R ) and by s the number of B -branches, whose roots v , . . . , v s are adjacent to R . When n is relatively small, we have no C k i -branches ( r = 0). In theintermediate range between around a 1000 and several thousands, we have a combination of bothextremes, depending on the remainder of n divided by 365. When n is sufficiently large, it turnsout that B -branches disappear ( s = 0) and all values k i stabilize at 52, with a few exceptions (forwhich k i = 51 or 53; see Lemma 2.9). Let T be an ABC-minimal tree of order n and let ∆ be the maximum degree of T . Theorem 2.3implies that for every path starting at the root, the vertex-degrees along the path never increase.9he goal of this section is to prove that the degrees are strictly decreasing, with two sporadicexceptions. Theorem 3.1.
Let T be an ABC-minimal tree of order greater than 9 and maximum degree ∆ .For every k ≥ , T contains at most one edge, whose end vertices both have degree k . Moreover, ifsuch an edge exists, then k is either or ∆ .Proof. ABC-minimal trees of order ≤ ≥
415 there are no 2-2 edges in ABC-minimal trees (see Lemma2.5). Since Theorem 3.1 holds for trees of smaller order we may assume that k > uv whose end vertices have the same degree, d u = d v = k . Suppose that u is closerto the root than v . If there is more than one such edge, consider the one with the highest value of k and if there is more than one such k - k edge, consider one which is farthest from the root. Thenall descendants of v have degree smaller than k . Detach the child x of v with the largest degree(together with its subtree T x ) and connect it to u as shown in Figure 4. (The subtree T x is notshown in the figure.) Let T (cid:48) be the resulting tree. Note that by selection of the edge uv , we knowthat d x < k . We have two cases. If u is the root, then if uv is the only edge whose end verticeshave degree k , then Theorem 3.1 holds for k = ∆. If there is another vertex w with d w = k , thenit should be adjacent to u and we can consider w to play the role of R in Figure 4, so we have d R ≥ d u . If u is not the root, then let R be the parent of u and therefore d R ≥ d u . Rzyx Rzyx (a) (b) uv u v Figure 4: Changing the tree when 2 < d u = d v ≤ ∆.Let y , . . . , y k − be the children of v different from x and let z , . . . , z k − be the children of u different from v . We have selected x so that d x ≥ d y j . Note that if there exist i such that d x ≥ d z i then we can exchange the branch rooted at z i with the branch rooted at x without changingthe ABC index of the tree. So without loss of generality we can assume that d z i ≥ d x ≥ d y j (1 ≤ i, j ≤ k − α ( T ) and α ( T (cid:48) ) be10he contribution of all these adges to ABC ( T ) and ABC ( T (cid:48) ), respectively. Clearly, α ( T ) = f ( d R , d u ) + f ( d u , d v ) + k − (cid:88) i =1 f ( d u , d z i ) + k − (cid:88) j =1 f ( d v , d y j ) + f ( d v , d x )= f ( d R , k ) + f ( k, k ) + k − (cid:88) i =1 f ( k, d z i ) + k − (cid:88) j =1 f ( k, d y j ) + f ( k, d x ) . Similarly, α ( T (cid:48) ) = f ( d R , k + 1) + f ( k + 1 , k −
1) + k − (cid:88) i =1 f ( k + 1 , d z i ) + k − (cid:88) j =1 f ( k − , d y j ) + f ( k + 1 , d x ) . Note that there exists z ∈ { z , . . . , z k − } such that f ( k, d z i ) − f ( k +1 , d z i ) ≥ f ( k, d z ) − f ( k +1 , d z )for all i = 1 , . . . , k −
2. Considering a similar inequality for y ∈ { y , . . . , y k − } , we have: α ( T ) − α ( T (cid:48) ) ≥ f ( d R , k ) − f ( d R , k + 1) + f ( k, k ) − f ( k + 1 , k −
1) + f ( k, d x ) − f ( k + 1 , d x ) + ( k − f ( k, d z ) − ( k − f ( k + 1 , d z ) +( k − f ( k, d y ) − ( k − f ( k − , d y ) . We have discussed that d R ≥ k ≥ d z ≥ d x ≥ d y and we would like to show that α ( T ) − α ( T (cid:48) ) > f ( d R , k ) − f ( d R , k + 1) is increasingin d R . Since d R ≥ k , this implies that f ( d R , k ) − f ( d R , k + 1) ≥ f ( k, k ) − f ( k, k + 1) . Similarly, we have: f ( k, d x ) − f ( k + 1 , d x ) is increasing in d x , f ( k, d z ) − f ( k + 1 , d z ) is increasingin d z , and f ( k, d y ) − f ( k − , d y ) is decreasing in d y . Therefore we may replace d z and d y by d x =: m < k , so we have: α ( T ) − α ( T (cid:48) ) ≥ f ( k, k ) − f ( k, k + 1) + f ( k, k ) − f ( k + 1 , k −
1) + f ( k, m ) − f ( k + 1 , m ) + ( k − f ( k, m ) − ( k − f ( k + 1 , m ) +( k − f ( k, m ) − ( k − f ( k − , m ) . (1)If m = 1, then it follows by Theorem 2.1 that k = 2 and we have settled this case before. Sowe may assume that m ≥
2. Therefore 1 < m < k . Using computer, we have calculated the valuesof the right-hand side of (1) for all pairs ( m, k ), where 1 < m < k ≤ , and k ≥
5. The same waschecked for k = 4 when m = 2. In all cases the computation confirms that ∆( T, T (cid:48) ) > k > . Let m = ck where k < c <
1. Using Taylor series we can expandthe right-hand side of (1) in terms of k (factor out 1 / √ k and then use Taylor series of order 5 to11 RRRR R (cid:48) R (cid:48) R (cid:48) R (cid:48) R (cid:48) v ux Figure 5: Changing the tree when k = 4 and m (cid:54) = 2.expand). We also made the substitution A = (cid:113) cc +1 and have obtained the following: α ( T ) − α ( T (cid:48) ) ≥ √ k (cid:32) − √ A + √ A (cid:33) +32 √ k (cid:18) Ac − A − A − Ac + 1 + 32 A c + 1 (cid:19) + O ( k − / ) . (2) The expansion was produced with the help of the software platform Maple, version 18. A is bounded, 0 < A < k − is positive. When k < c < k +1 < cc +1 < . This implies that 1 − √ A + √ A > − √ / >
0. By rewriting (2), we obtain: k / ( α ( T ) − α ( T (cid:48) )) > √ −
28 + Am + O ( k ) . (3)The order of the middle term in (3) is O ( k − / ) and is positive. Note that c +1 and cc +1 are boundedand the only case that can cause problem is when c is in the denominator and c ∼ k , which means m = O (1). For k ≥ , the terms are negligible in comparison with the constant value of the firstterm. This shows that α ( T ) − α ( T (cid:48) ) > k > and any value of m < k .There are some remaining cases for small values of k ( k = 4 and k = 3). When k = 4 and m = 2 it is easy to check that (1) is positive. So we may assume that k = 4 and m (cid:54) = 2, or k = 3.For these cases note that since equation (1) is not positive, we cannot only consider the worst caseand need to discuss all possible values of d z i , d x and d y i . Figure 5 deals with the case when k = 4.One can check that using the suggested change of the tree, the ABC-index becomes smaller. Notethat since u has a neighbor of degree 3 ( m (cid:54) = 2), all neighbors of R have degree ≥
3, otherwise wecan exchange them and get a tree with smaller ABC-index ( d R ≥ d u ).As a case in point we will discuss the first case shown in Figure 5 and leave the rest to thereader. As before, we let α ( T ) be the contribution to ABC ( T ) of all edges that are shown in thefigure. We have: α ( T ) = f ( d R ,
4) + 5 f (4 ,
3) + f (4 ,
4) + 10 f (3 ,
2) + 10 f (2 ,
1) and α ( T (cid:48) ) = 3 f ( d R + 2 ,
5) + 12 f (5 ,
2) + 12 f (2 , . The change in ABC-index when passing from T to T (cid:48) is equal to α ( T ) − α ( T (cid:48) ) plus all differences f ( d R , d x ) − f ( d R + 2 , d x ) for each neighbor x of R different from u . By Proposition 2.2, and since d x ≥
3, we have f ( d R , d x ) − f ( d R + 2 , d x ) ≥ f ( d R , − f ( d R + 2 , . Consequently, ∆(
T, T (cid:48) ) ≥ α ( T ) − α ( T (cid:48) ) + ( d R −
1) ( f ( d R , − f ( d R + 2 , . (4)Using (4), it is easy to check that ∆( T, T (cid:48) ) > d R ≥ k = 3. In this case d y = d x = 2and d z = 2 or 3. To solve this case we will first show that a vertex of degree 3 cannot have twodescendants of degree 3. For a contradiction assume u has two degree-3 descendants. Since theedge uv is taken farthest from the root, the descendants of these degree-3 vertices have degree 2,and all further descendants have degree 2 or 1. Having additional descendants of degree 2 does notaffect the computation in the sequel, thus we may assume that the situation is as shown in Figure6. Note that if u has only one descendent of degree three (and one descendent of degree two) andthere is another 3-3 edge in the graph, then we have two cases. If d R = 3, then another descendentof R is either of degree two or three and in both of these cases by an exchange we will get thestructure shown in Figure 6. If there is another disjoint 3-3 edge in the graph, then one of those13egree three vertices will be the root of a B that can be exchanged with z (without changing theABC-index), and we can then make the change shown in Figure 6. Also note that d R ≥
3; otherwisewe would get a small tree and small trees are known to satisfy our theorem [6]. Now it is easy tocheck that the suggested structure will have smaller ABC-index, thus yielding a contradiction.
R R (cid:48)
Figure 6: Changing the tree when two sons of u have degree 3. R R
Figure 7: Changing the tree when d R ≥ Rw w (cid:48)
Figure 8: Changing the tree when d R = 4.The only remaining case is when the only degree-3 neighbor of u is v and there is no other 3-3edge in the graph. Then the second descendant of u has degree two (by Theorem 2.1) and we havethe situation that is depicted in Figure 7. The suggested change improves the ABC-index when14 R ≥
5. We showed above that we cannot have two 3-3 edges, therefore d R (cid:54) = 3 and we may assumethat d R = 4. If R is the root, then again we have a small tree and, as mentioned before, smalltrees are known to satisfy our theorem (see [6]). We may assume that there is a vertex w withdegree > d w = 4, then we should have selected the edge wR for our process). Since d R > d u ,descendants of R should have degree at least 3, otherwise we could have changed them with the B branch and get a smaller ABC-index. Figure 8 presents this case and the suggested structureimproves the ABC-index for d w ≥
5. This completes the proof.Combining Theorems 2.3 and 3.1, we get the following corollaries.
Corollary 3.2.
In any ABC-minimal tree with maximum vertex degree ∆ , there are at most twovertices whose degrees are equal to ∆ , and if there are two, they are adjacent. Corollary 3.3.
In any ABC-minimal tree with maximum vertex degree ∆ , the degree sequenceon any path starting from a vertex of maximum degree is strictly decreasing with the followingexceptions: • When two consecutive vertices on the path have degree 2 and the tree is less than 415 vertices. • When the path starts with two vertices whose degrees are equal to ∆ . Another consequence of these results is that the property (P1) holds also when v is a successorof u .(P1’) Let T be an ABC-extremal tree. Suppose that u, v ∈ V ( T ) and h ( u ) < h ( v ). Then T u (cid:23) T v and, in particular, d u ≥ d v . Proof. If v is not a successor of u or when u is the root, the property is just (P1). Therefore wemay assume that u is not the root and T v ⊂ T u . By Corollary 3.3, we have that d u ≥ d v . Suppose,for a contradiction, that T u ≺ T v . Then we have d u ≤ d v , which implies that d u = d v . ApplyingCorollary 3.3 again, we conclude that v must be a son of u and that their degree is either 2 or ∆.Clearly, d u = d v = 2 gives that T u (cid:31) T v ; in the other case, u must be the root, a contradiction. In this section we will show that vertices at distance at least 2 from the root have degree at most5. Combining this with the results in the previous section, we will be able to conclude that thediameter of ABC-minimal trees is bounded. Along the way we will prove several other propertiesof ABC-extremal trees.
Theorem 4.1.
In any ABC-extremal tree, every vertex of degree at least 6 is either the root or isadjacent to the root.
The proof of Theorem 4.1 consists of two parts. First, we prove a weaker statement (Lemma4.2 below) which has a possible exception when the result may not hold. In the rest of the sectionwe shall then prove that such an anomaly does not occur (Lemma 4.9).15 .1 Exceptional branches
To evolve terminology, let us say that a son u of the root is a U -exceptional vertex if it has a sonof degree at least 6; its subtree T u is said to be a U -exceptional branch . Lemma 4.2.
Let T be an ABC-extremal tree. Then T has at most one U -exceptional vertex, andif u is such a vertex, then u has largest degree among the sons of the root and u also has a neighborof degree at most .Proof. Let R be the root of T and u be the neighbor of R with the (cid:31) -largest subtree T u . Notethat this implies that u is a son of R with maximum degree. We may assume that d u >
2. We willcontract the edge Ru and add a neighbor to some vertex w of degree one instead. By Theorem2.1, the neighbor of w has degree 2 and is thus different from u . The change is shown in Figure 9,where y i ( i = 1 , . . . , d u −
1) are sons of u and z j ( j = 1 , . . . , d R −
1) are the neighbors of R differentfrom u . y i y i Ruw R (cid:48) z j z j w Figure 9: Suggested change for the proof of Lemma 4.2.Note that d R ≥ d u , d z j ≥ d y i ( j = 1 , . . . , d R − i = 1 , . . . , d u − α ( T ) ( α ( T (cid:48) )) be the sum of f -values of those edges in T ( T (cid:48) ) whose contribution to the ABC-indexhas changed: α ( T ) = f (2 ,
1) + f ( d R , d u ) + d R − (cid:88) j =1 f ( d R , d z j ) + d u − (cid:88) i =1 f ( d u , d y i )and α ( T (cid:48) ) = f (2 ,
1) + f (2 ,
2) + d R − (cid:88) j =1 f ( d R + d u − , d z j ) + d u − (cid:88) i =1 f ( d R + d u − , d y i ) . We consider a vertex z ∈ { z j | j = 1 , . . . , d R − } such that f ( d R , d z j ) − f ( d R + d u − , d z j ) ≥ f ( d R , d z ) − f ( d R + d u − , d z ) for all j = 1 , . . . , d R −
1. Considering a similar inequality for the sonsof u (and denoting by y the corresponding vertex where the minimum is attained), we have: α ( T ) − α ( T (cid:48) ) ≥ f ( d R , d u ) − f (2 ,
2) + ( d R − f ( d R , d z ) − f ( d R + d u − , d z ))+( d u − f ( d u , d y ) − f ( d R + d u − , d y )) . (5)16roposition 2.2 shows that the two differences within the parentheses in (5) are increasing interms of parameters d z and d y . Because it suffices to consider the worst case, we may consider theirsmallest value. If d z , d y ≥
6, we take the value 6. Then (5) changes to: α ( T ) − α ( T (cid:48) ) ≥ f ( d R , d u ) − f (2 ,
2) + ( d R − f ( d R , − f ( d R + d u − , d u − f ( d u , − f ( d R + d u − , . (6)For every d R ≥ d u = d R . By considering the values when d u = d R , we see that the values decrease when d R grows and that the values are always positive. This gives a contradiction when d R ≥ d u ≤ d R ≤
100 by computerand it turns out that we always have ∆(
T, T (cid:48) ) >
0. We conclude that
ABC ( T ) > ABC ( T (cid:48) ) for allvalues of d R . This contradiction completes the proof when d z , d y ≥ y i or z j are less than 6 and degreesof some of them are ≥
6. By property (P1) of ABC-extremal trees, we see that d z j ≥ d y i for all j ∈ { , . . . , d R − } and i ∈ { , . . . , d u − } .Suppose first that d z <
6. Then d y i < i ∈ { , . . . , d u − } . Next, consider any otherneighbor z j of R . Recall that T u (cid:23) T z j . If d u = d z j , property (P3) implies that all sons of z j havedegree at most 5. If d u > d z j , the same conclusion follows by Lemma 2.4(a). This shows that allvertices at distance 2 or more from R have degree at most 5.We may now assume that d z j ≥ j ∈ { , . . . , d R − } and that y has degree less than6. In the same way as above, we see that all sons of z , . . . , z d R − have degree ≤
5. Hence, at mostone neighbor of the root, namely u , may have sons of degree ≥ ≤ d u > d y i (by Theorem 3.1) and u has a descendant of degree ≤
5, the degrees of alldescendants of y i are less than 6 by (P1). This completes the proof.In the remainder of this section we will show that U -exceptional branches do not exist in ABC-extremal trees, see Lemma 4.9. This will make the proof of Theorem 4.1 complete. Lemma 4.3.
Every ABC-minimal tree has at most eleven vertices of degree 3 and at most fourvertices of degree 5. Moreover, there is at most one B -branch.Proof. Theorem 3.1 shows that the only way to have a vertex v of degree 3 is when T v is a B ora B ∗ -branch. Note that the 2-2 edge of a possible B ∗ -branch can be moved to other branches by asimilarity exchange, so we may assume that there is no B ∗ in our ABC-extremal tree T (unless allvertices of degree 1 are within B -branches and one B ∗ -branch). It is shown in [12, 15, 16] that inany ABC-minimal tree there are at most 11 B - or B ∗ -branches, so there are at most 11 verticesof degree 3.Let us assume for a contradiction that we have at least 5 vertices of degree 5 and let v be oneof them with u i ( i = 1 , , ,
4) as its descendants. Corollary 3.3 indicates that 2 ≤ d u i <
5. If all ofthem have degree 2 it means that we have a B (since B ∗ does not exist by Lemma 2.7) and it isknown that at most four B -branches can exist [14]. If every u i has degree 4, the change indicatedin Figure 10 gives us an ABC-smaller tree, and if all u i have degree 3 the change shown in Figure The same holds when d R < The lower bound becomes √ / − √ / > d R → ∞ .
171 gives us an ABC-smaller tree. Note that the sons of different degree-5 vertices can be reshuffledby similarity exchanges (without changing the value of the ABC-index). Thus we could group theirsons of degree 4 or 3 together. This implies that the vertices of degree 5 cannot have more thanthree sons of degree 3 all together, and at most three sons of degree 4 all together. Therefore wehave at least 13 sons of degree 2, which gives us 3 copies of B . If we have more than one vertexof degree 3 or 4 as children of vertices of degree 5, then by similarity exchanges we can get at least6 copies of B − (sons of degree 2 that are not part of a B k branches) which is not possible in anyABC-minimal tree (see [15]). And if we have at most one vertex of degree > T contains two copies of B . We replace one of them with a B ∗ and the otherone with two copies of B (so the degree of the father of one of B -branches increases by 1). It iseasy to see that this decreases the ABC index. This completes the proof. u u w w Figure 10: The change of a tree when a 5-vertex has four sons of degree 4. u u Figure 11: The change of a tree when a 5-vertex has four sons of degree 3.
Lemma 4.4. If T is an ABC-extremal tree with at least 40 vertices, then no neighbor of the rootcan have degree .Proof. As mentioned before, all ABC-minimal trees with at most 1100 vertices are known [6]. Itturns out that the largest one among them having a degree-2 neighbor of the root has 39 vertices.Thus, we may assume that n > R has a neighbor u of degree 2. By (P1), all vertices at distance 2 from R havedegree 2 or 1. If T has a 2-2 edge, then we may assume that it is incident with u by (P2). Thus,all neighbors of the root form B − -branches and B k -branches, where 1 ≤ k ≤ B k (1 ≤ k ≤
5) has at most 11vertices. Therefore d R ≥ (cid:100) (cid:101) = 100.If there are four B − -branches, we replace them by one B ∗ -branch. Let x , . . . , x d R − be theother neighbors of R . Then we have∆( T, T (cid:48) ) ≥ f (2 , − f (4 , d R −
3) + d R − (cid:88) i =1 ( f ( d x i , d R ) − f ( d x i , d R − ≥ f (2 , − f (4 , d R −
3) + ( d R − f (6 , d R ) − f (6 , d R − d R ≥
12, a contradiction.So we have at most three B − -branches. We may have up to four B and one B . Thus, theremust exist a B k for k ∈ { , } . Now we replace B k and B − with a B k +1 . Again, it is easy to seethat for d R ≥
17, this change decreases the ABC-index, a contradiction.Recall that a vertex u is in B ( T ) if there exist vertices v and w such u is the parent of v , v is the parent of w , d v = 2 and d w = 1; and T u is a B -exceptional branch if u also has a son ofdegree more than 2 or a son of degree 2 that is incident with a 2-2 edge. In the following lemmawe will show that in addition to B ∗ k -branches only one type of B -exceptional branches can exist inABC-minimal trees. Lemma 4.5.
Each ABC-extremal tree either contains no B -exceptional branches, or contains asingle B -exceptional branch which is isomorphic to B ∗ , B ∗ , or to the tree B ∗∗ depicted in Figure12. B ∗∗ Figure 12: The only possibility for a B -exceptional branch different from B ∗ k -branches. Proof.
Suppose first that there is a B -exceptional branch that is different from B ∗ k ( k ≥ u be its root. Let x be the child of u with the smallest degree, subject to the condition that d x ≥
3. By the definition of B ( T ) we know that u also has a child of degree 2. Let y , . . . , y d u − be the remaining sons of u . By definition of B ( T ), u is not the root, and thus it has a predecessor r . Consider the change shown in Figure 13.Since u has a child of degree 2, (P1) implies that all children of x and all children of y i are ofdegree ≤
2. And since vertices of degree one are adjacent only to vertices of degree 2, T x and each19 uy i x x y i u r Figure 13: Changing the tree having a B -exceptional branch.of T y i is a B − or B k -branch for some 2 ≤ k ≤ B and B − cannot happenat the same time; replace them with one B and one B ∗ ). Note that for this conclusion we also use(P2). We will assume that there are no 2-2 edges. If there is one, we first contract it and make thesame changes as in the continuation of this proof. After the changes, we uncontract the 2-2 edge.The fact is that the difference ∆( T, T (cid:48) ) will be exactly the same as when there are no 2-2 edges.We will return to this at the end of the proof.Since the edges incident to vertices of degree 2 have their f -value constant, we have:∆( T, T (cid:48) ) = f ( d r , d u ) + f ( d u , d x ) + d u − (cid:88) i =1 f ( d u , d y i ) − f ( d r , d u − − f ( d u − , d x + 1) − d u − (cid:88) i =1 f ( d u − , d y i ) . Using Proposition 2.2 we see that the above difference is decreasing in d r , d x and d y i . Thereforeit suffices to show that the difference is positive when replacing d r , d x and d y i by largest valuesthat are allowed for these degrees. We can replace f ( d r , d u ) − f ( d r , d u −
1) with the limit when d r tends to infinity, which is equal to (cid:112) /d u − (cid:112) / ( d u − d x = 5, then u has at most threedescendants y i whose degree is more than 2, since they would all be of degree 5 by our choice of x .In this case we would have:∆( T, T (cid:48) ) > √ d u + 4 f ( d u , − √ d u − − f ( d u − , − f ( d u − , d u ≥ d x ≤
4, then we similarly have:∆(
T, T (cid:48) ) > √ d u + 4 f ( d u ,
5) + ( d u − f ( d u , − √ d u − − f ( d u − , − ( d u − f ( d u − , d u ≥
15. Thus, we may assume from now on that d u ≤ u are copies of B − and B i -branches for i = 2 , ,
4. Let k i be the number of children whose subtrees are isomorphic to B i (for i = 2 , ,
4) or B − (for i = 1).20e know that 4 ≤ d u ≤
14, 1 ≤ k ≤ d u −
2, 0 ≤ k ≤ min { d u − , } , 0 ≤ k ≤ d u − ≤ k ≤ min { d u − , } . Also, either k or k is zero; moreover, k , k and k cannotall be zero. Let K be the set of all 4-tuples ( k , k , k , k ) satisfying all these conditions with d u = k + k + k + k , and let k = 1 + 2 k + 5 k + 7 k + 9 k be the number of vertices in the B -exceptional branch. Note that we will have at least one B − andat least one B i branch and therefore k ≥
8. Also k = 8 cannot happen in the ABC-minimal treebecause we will have a 3-3 edge. The only way to get k = 10 (without contradicting the previouslymentioned properties) is the B ∗∗ branch which we are claiming to be the only B-exceptional branchin the absence of a 2-2 edge. Observe that k = 9 or 11 is not possible, thus we have k ≥
12. Forevery possible ( k , k , k , k ) ∈ K , we replace the k vertices in the B -exceptional branch with copiesof B -branches attached to R as follows: . . .. . . . . . . . .u R (cid:124) (cid:123)(cid:122) (cid:125) k (cid:124) (cid:123)(cid:122) (cid:125) k (cid:124) (cid:123)(cid:122) (cid:125) k (cid:124) (cid:123)(cid:122) (cid:125) k z i Figure 14: A tree having a B -exceptional branch when d u ≤ • If k ≡ k copies of B . • If k ≡ k − − B and one copy of B ∗ . • If k ≡ k − − B and one copy of B . • If k ≡ k − − B and one copy of B ∗∗ . • If k ≡ k − − B and two copies of B . • If k ≡ k − copies of B and one copy of B . • If k ≡ k − copies of B and one copy of B ∗ . For the remaining case, when d R ≤
94 we can again replacethem with two copies of B and one copy of B ∗ , but for larger d R we replace them with onecopy of B and one copy of B . 21igure 15: The B -exceptional branch that needs special treatment for d R ≥ k ≥
12 and ( k , k , k , k ) ∈K , the degree of R increases and therefore the change in the ABC-index is increasing in term of d z i . Thus, to consider the worst case, we let d z i = 3 (for i = 1 , . . . , d R − d R ) and it is easy to check by computer that the change is possible andit improves the ABC-index. Therefore the only B -exceptional branch different from B ∗ k for k ≥ B ∗∗ .Let us now return to the case when we had a 2-2 edge. As mentioned above, the same proofworks and we conclude that all ending branches (those based at vertices in B ( T )) are B , B , B , B , and B ∗∗ . Now we put the 2-2 edge back. It can be added to any degree-2 vertex. By Lemma2.7, there are no B ∗ k for k = 4 ,
5. If there is no B ∗∗ , then we obtain a single B ∗ or B ∗ , as claimed.On the other hand, if B ∗∗ (with its root u ) is present, then we uncontract the 2-2 edge within thisbranch and then replace the whole branch with a B . It is easy to see that this change decreasesthe ABC-index (which is a contradiction) if the degree of the father r of u is at least 9. Thus, wemay assume that 5 ≤ d r ≤
8. In this case we can replace B ∗∗ together with the expanded 2-2 edgeby B and B ∗ attached to r . Then we have∆( T, T (cid:48) ) ≥ f (3 ,
4) + f (4 , d R ) − f (3 , d R + 1) + d R − (cid:88) i =1 ( f ( d x i , d R ) − f ( d x i , d R + 1)) ≥ f (3 ,
4) + f (4 , d R ) − f (3 , d R + 1) + ( d R − f (3 , d R ) − f (3 , d R + 1))which is positive for d r ∈ { , , , } .Let us observe that B ∗∗ cannot be excluded in all cases. In fact, the tree in Figure 16 with k = 43 B -branches and one B ∗∗ is an ABC-minimal tree that contains B ∗∗ . It has n = 312vertices. This is the smallest ABC-minimal tree containing a B ∗∗ , see [2]. However, we believe that B ∗∗ cannot occur when n is sufficiently large.By Lemmas 2.6 and 4.5 the following corollary is immediate. Corollary 4.6.
Let T be an ABC-extremal tree and u be a non-root vertex with d u ≥ and witha son of degree 2. Then T u is isomorphic to one of the following: B , B ∗ , B , B ∗ , B ∗∗ , B , or B . Any B -exceptional branch can occur at most once, B can occur at most eleven times, B canoccur up to four times, and B can occur at most once. There are additional restrictions that follow from (P1), but they are not needed for the proof. . . (cid:124) (cid:123)(cid:122) (cid:125) k Figure 16: An ABC-minimal tree containing a B ∗∗ -branch.Note that if there is B ∗ or B ∗ , then it is just one of them and there cannot be any of B , or B , because in that case the 2-2 edge could be used to make a B ∗ or B ∗ , contrary to Lemma 2.7.Also, we cannot have a copy of B ∗∗ together with a B or a B since this would contradict (P1)( B ∗∗ contains a 4-3 edge and B and B have a 5-2 and 6-2 edge, respectively). C k -branches Lemma 4.7. If T is an ABC-minimal tree with a vertex u adjacent to (at least) roots of C k -branches, then k ≤ .Proof. Suppose that there are 365 = 7 ×
52 + 1 copies of C k adjacent to u . Let us consider theremaining neighbors of u , and if there are any, let R be one of them that has the highest degree. (If u is not the root then R is the parent of u ). We can apply the change shown in Figure 17 to obtaina tree T (cid:48) . Note that the two trees have the same number of vertices and that d u (cid:48) = d u + 7 k − . . . . . . (cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125) k +1 Ru
52 52 52 k k k u (cid:48) R Figure 17: Suggested change when there are 365 copies of C k .Let d = d u −
366 and if d (cid:54) = −
1, let a , . . . , a d be the degrees of the sons of u in the shadedpart. If k >
52, we have d u < d u (cid:48) and by Proposition 2.2, the differences f ( R, u ) − f ( R, u (cid:48) ) and f ( u, a i ) − f ( u (cid:48) , a i ) are increasing in terms of d R and each a i . Suppose first that R exists. Then itsuffices to consider the case when u is the root and d R = 4 and each a i = 4 (since d u > k + 1).Then we have:∆( T, T (cid:48) ) ≥ (cid:16) f ( d u , k + 1) + k (cid:16) f ( k + 1 ,
4) + 6 √ (cid:17)(cid:17) + ( d + 1) f ( d u , − (7 k + 1) (cid:16) f ( d u (cid:48) ,
53) + 52 (cid:16) f (53 ,
4) + 6 √ (cid:17)(cid:17) − ( d + 1) f ( d u (cid:48) , . (7)23ote that by Lemma 2.10, k < k the equation only depends on d u and it is easy check by computer that the suggested change decreases the ABC-index for all53 ≥ k ≥
142 and d u ≥ R does not exist, then d = − T, T (cid:48) ) has the same lower bound (7). From this weobtain the same conclusion. This completes the proof.
Lemma 4.8.
Let u be a vertex in an ABC-minimal tree and let k be a positive integer. If k ≤ ,then there are at most k + 7 copies of C k -branches whose roots are the sons of u . If d u is greaterthan
474 (874 and , respectively ) , then there are at most k + 7 copies of C k for k = 49 ( k = 50 and k = 51 , respectively ) .Proof. Let u be a vertex that has 7 k + 8 copies of C k as his children. Let x , . . . , x m be the sons of u that are not in the considered C k -branches. First we will discuss the degree of the vertices x i . Let I = { i : d x i > k + 2 } . As discussed at the end of Lemma 4.2 when u is not the root, the degree ofall grandchildren of u is at most 5. Since d x i > k + 2 for i ∈ I and there are B -branches attachedto vertices of degree k + 1, then x i cannot have children of degree less than 4 (by Lemma 2.4(a).Therefore each x i ( i ∈ I ) has children of degree 4 or 5 only. Recall that (by Lemma 4.3) we haveat most 4 vertices of degree 5, and therefore all but at most one of the x i are roots of C k (cid:48) -branches(after possible similarity exchanges). Lemma 2.9 shows that k (cid:48) ≤ k + 1 and therefore |I| ≤ u is the root, there is at most one U -exceptional branch (we can assumethat it is R ) and using the same argument as above, the degree of all but at most one of x i ’s is atmost k + 2. k k k Ru x i | {z } k +8 Ru x i k + 1 k + 1 | {z } k +1 . . . . . . Figure 18: Compactifying C k -branches when k ≤ T, T (cid:48) ) = f ( d R , d u ) − f ( d R , d u −
7) + (7 k + 8) f ( d u , k + 1) − (7 k + 1) f ( d u − , k + 2)+ d u − k − (cid:88) i =1 f ( d u , d x i ) − d u − k − (cid:88) i =1 f ( d u − , d x i )+ k (7 k + 8) f ( k + 1 , − (7 k + 1)( k + 1) f ( k + 2 , − f (2 , . By Proposition 2.2 this difference is decreasing in d R and d x i .24herefore it suffices to consider the limit when d R → ∞ and to let d x = d u − d x i = k + 2for i = 2 , . . . , d u − k − u is the root the worst possible case for d R is d u − d R , considering d R → ∞ is enough and covers other cases as well). Wehave:∆( T, T (cid:48) ) ≥ √ d u − √ d u − k + 8) f ( d u , k + 1) − (7 k + 1) f ( d u − , k + 2)+( d u − k −
10) ( f ( d u , k + 2) − f ( d u − , k + 2)) + f ( d u , d u − − f ( d u − , d u − k (7 k + 8) f ( k + 1 , − (7 k + 1)( k + 1) f ( k + 2 , − f (2 , . Since for a fixed value of k ( ≤
48) the right hand side of this inequality only depends on d u ( ≥ k + 8), one can check that ∆( T, T (cid:48) ) > d u and for all k ≤ k = 49 ,
50 and 51 if d u is at least 474, 874 and 3273,respectively. U -exceptional branches Lemma 4.9.
ABC-extremal trees have no U -exceptional branches.Proof. Let u be the root of a U -exceptional branch and let w be the child of u with the highestdegree ( d w ≥ x l ( l = 1 , . . . , m ) be the other children of u with d x l ≥ y j be childrenof u with d y j ≤ j = 1 , . . . , d u − m − R R u z i z i y j y j ww u x l x l Figure 19: Suggested change when there is a U -exceptional branch.First note that since d y j ≤ d u ≥ d z i ≥ d x l , all children of w , of each x l and of each z i havedegree at most 5 and since we have at most 4 vertices of degree 5 and 11 vertices of degree 3 and atmost one B -exceptional branch, all but at most two subtrees of x l and at most two subtrees of z i are C k -branches. To see this, observe that vertices of degree 3 are roots of B or B ∗ branches andto have a vertex of degree 4 (which is not B , B ∗ or B ∗∗ ) we should use B branches as childrenof a vertex of degree 4. Since we have at most 11 B branches, all but at most 4 vertices of degree4 are roots of B branches (this can be improved to 1, see Lemma 4.11). Therefore at most two z i branch will contain vertices of degree 5 (that are not among y j ’s) and vertices of degree 4 that arenot B branches. Also at most two x l branches will contain any remaining B branches.25he difference between the ABC-indices of trees in Figure 19 is:∆( T, T (cid:48) ) = f ( d R , d u ) + d R − (cid:88) i =1 f ( d R , d z i ) + f ( d u , d w ) + d u − m − (cid:88) j =1 f ( d u , d y j ) + m (cid:88) l =1 f ( d u , d x l ) − f ( d R + 1 , d u − − d R − (cid:88) i =1 f ( d R + 1 , d z i ) − f ( d R + 1 , d w ) (8) − d u − m − (cid:88) j =1 f ( d u − , d y j ) − m (cid:88) l =1 f ( d u − , d x l ) . Using Proposition 2.2 we can see that this difference is increasing in d z i and d w and decreasing in d y j and d x l . In order to verify that the difference is positive, it suffices to prove it when d y j = 5and d z i = d x l = d w =: d (cid:48) for i = 1 , . . . , d R − j = 1 , . . . , d u − m − l = 1 , . . . , m . We have:∆( T, T (cid:48) ) ≥ f ( d R , d u ) − f ( d R + 1 , d u −
1) + ( d R − (cid:0) f ( d R , d (cid:48) ) − f ( d R + 1 , d (cid:48) ) (cid:1) + f ( d u , d (cid:48) ) − f ( d R + 1 , d (cid:48) ) + ( d u −
2) ( f ( d u , − f ( d u − , m (cid:0) f ( d u , d (cid:48) ) − f ( d u − , d (cid:48) ) − f ( d u ,
5) + f ( d u − , (cid:1) . Note that the coefficient of m is a negative number by Proposition 2.2 and therefore the right-hand-side of this inequality is decreasing in m . Therefore to consider the worst case, we will let m = d u − m by the definition of U -exceptional branches). We have:∆( T, T (cid:48) ) ≥ f ( d R , d u ) − f ( d R + 1 , d u −
1) + ( d R − (cid:0) f ( d R , d (cid:48) ) − f ( d R + 1 , d (cid:48) ) (cid:1) + f ( d u , d (cid:48) ) − f ( d R + 1 , d (cid:48) ) + ( f ( d u , − f ( d u − , d u − (cid:0) f ( d u , d (cid:48) ) − f ( d u − , d (cid:48) ) (cid:1) . Let g be the value of the right-hand side of this inequality. Lemmas 2.8, 4.8 and 4.7 show thatwhen d R ≥ C k -branches adjacent to R are C , with at most 364 exceptions of C or C (if we have d u ≥ x l ). Notethat if we have some vertices of degree 54, then by Lemma 2.4 they are among z i and since theright-side value in (8) is increasing in each d z i , the worst case occurs when we have no vertices ofdegree 54. Also if we have vertices of degree 52, then they are among x l and since g is decreasingin terms of d x l , the worst case happens when we have no vertices of degree 52. Observe that evenif m or d u is small and we do not have enough C k branches as children of u , since the worst casefor the degree of each z i is 53 and because d x l ≤ d z i and the equation is decreasing in terms of d x l ,the worst case for the degree of each x l is 53.A simple computer search shows that when d (cid:48) ≤ d u < d R < g is positive. So the onlyremaining case to be considered is when d R ≥ d (cid:48) = 53 (i.e. when each x l andeach z i is the root of a C -branch). The Taylor expansion (in terms of d R ) of ∂g∂d u shows that forlarge values of d R , g is decreasing in d u and therefore it suffices to show that g > d u = d R (the highest possible value). By substituting d u with d R , the function will only depend on d R andit is easy to check that g >
0. By a simple computer verification we have also checked that g > d u ≤ d R ≤ d (cid:48) = 53.26o summarize, in all cases we have g >
0. This gives a contradiction and completes theproof.
Corollary 4.10.
In every ABC-extremal tree the vertices of degree one are at distance at most 5from the root.Proof.
We will use the main results proved above. First of all, every path from the root hasdecreasing degrees with possible exception of the first edge on the path, or when the path containsa 2-2 edge. Recall that there is at most one 2-2 edge in T (Lemma 2.5). By Theorem 4.1, wemay assume that vertices at distance 2 from the root have degree 5 or less. Suppose that a pathstarting at the root has length 6 or more. Consider degrees of vertices on the path starting at thevertex at distance 2 from R . The only possible cases are the following degree sequences: 4-3-2-2-1,5-4-3-2-2-1, 5-4-3-2-1, 5-4-2-2-1, 5-3-2-2-1.We will first prove that h = min { h ( v ) | d v = 2 } ≤
3, i.e. there is a degree-2 vertex at distanceat most 3 from R . If there is no vertex of degree 5 at distance 2 from the root (a bad 5-vertex),this is easy to see (similar arguments as in the more complicated case below), and we omit details.So let us suppose that there is a bad 5-vertex v . By (P1), all sons of the root have degree at least5 in this case. Since there are at most 4 vertices of degree 5 (Lemma 4.3), it follows that thereare more than 11 vertices at distance 2 from the root that are not of degree 5. Thus one of themleads to a leaf without including a vertex of degree 3 (Lemma 4.3). The predecessor of this leaf isa degree-2 vertex at distance at most 3 from R . Therefore h ≤ h , which is a contradiction to the existence of 4-3-2-2-1, 5-4-3-2-2-1, 5-4-2-2-1 and5-3-2-2-1.Now observe that since h ≤
3, then 5-4-3-2-1 cannot also happen because it has a vertex ofdegree 3 at distant 4 from the root, which is a contradiction.In fact, distance 5 from the root in the last corollary can be reduced to 4 (if there are no B ∗ , B ∗ and B ∗∗ ). As this is not important for the main structure results, we do not intend to deal withthis improvement here. Lemma 4.11.
Suppose that T is an ABC-minimal tree. (a) The only non-root vertices of degrees 3, 4 or 5 are roots of B , B ∗ , B , B ∗ , B ∗∗ or B branches. (b) There are no non-root vertices of degree k for ≤ k ≤ .Proof. First note that ABC-minimal trees of order ≤ n ≥ d R ≥
5. We will use Corollary4.6 throughout.Let v be a non-root vertex of degree k . If k = 3 then T v is either B or B ∗ . If k = 4 and T v is not B , B ∗ or B ∗∗ , then all children of v are B or B ∗ branches ( T v will have 16 or 17vertices). It is easy to check that by exchanging T v with one B and one B (16 vertices) or one27 and two B branches (17 vertices) will improve the ABC-index. If k = 5 and T v is not a B ,by Lemmas 4.5 and 2.7 all children of v have degrees 3 or 4. So in this case all subtrees of T v are B , B ∗ , B , B ∗ or B ∗∗ . Let k , k ∗ , k , k ∗ and k ∗∗ represent the number of each of these branches,respectively. Therefore, k + k ∗ + k + k ∗ + k ∗∗ = 4. From previous arguments we know that someof these branches cannot occur at the same time and also we know that k ∗ + k ∗ + k ∗∗ ≤
1. Let n v = 1 + 5 k + 6 k ∗ + 7 k + 8 k ∗ + 10 k ∗∗ be the number of vertices in T v (21 ≤ n v ≤ T v by the subtrees used in the proof of Lemma 4.5 (treating seven different cases of thevalue of k modulo 7). As an example, when n v = 21 ( n v ≡ T v by 3 copiesof B . Note that there are several cases but in all of them the degree of the parent of v , say R ,increases and therefore the difference between ABC-indices of these two trees (in the worst case)only depends on d R . It is easy to check by computer that the difference is positive for all values of d R ≥
5. This completes the proof of part (a).When k = 6, similar arguments as in part (a) can be used to show that every vertex of degree6 is a root of B . We will prove later that B can be excluded as well.For k ≥
7, we give a proof iteratively starting with k = 7 , , etc. Then we may assume thatchildren of v have degrees 3, 4, 5 or 6 and all subtrees of T v are B , B ∗ , B , B ∗ , B ∗∗ , B or B branches. Let k , k ∗ , k , k ∗ , k ∗∗ , k and k represent the number of each of these branches,respectively. Therefore, k + k ∗ + k + k ∗ + k ∗∗ + k + k = k −
1. We also know that some ofthese branches cannot occur at the same time and also we know that k ≤ , k ∗ + k ∗ + k ∗∗ ≤ k ≤
4. Let n v = 1 + 5 k + 6 k ∗ + 7 k + 8 k ∗ + 10 k ∗∗ + 9 k + 11 k be the number of vertices in T v . Again, we will replace T v by the subtrees used in the proof of Lemma 4.5 (with seven differentcases for the value of k modulo 7). The difference between ABC-indecies of these two trees (in theworst case) only depends on d R and it is easy to check that the difference is positive for all valuesof d R ≥ k which completes the proof of part (b) when k (cid:54) = 6.Let v be the root of a B branch and let w be its father. By the above, d w ≥
16. If there is atleast one B or B attached to w , Lemma 2.8 applies and we are done. Since we can have up to 4vertices of degree 5 (and v has more than 4 siblings), there is a sibling u of v of degree at least 16(by the previous paragraph). It is easy to see that there is a B (or B ) as a child of u . Now applythe following change: detach a B − from B (thus changing it to B ) and attach it to this B (or B ). The change of ABC-index is increasing in d u and decreasing in d w . So the worst case is whenthe degree of w goes to infinity and d u = 16 for which it is easy to check that the change of theindex is positive. This contradiction shows that there is no B and thus completes the proof. Corollary 4.12.
In any ABC-extermal tree, all B branches (if any) are attached to the root.Proof. Let u be a non-root vertex which has 1,2,3 or 4 B branches as its children. By Lemma4.11, the degree of u is at least 16 and by Theorem 4.1, it is adjacent to the root. We will firstprove an upper bound on the degree of u when the degree of the root is large enough. Considera change similar to Figure 2 in Lemma 2.10 (we have 1, 2, 3 or 4 copies of B as well). One cancheck that in all of these four cases, degree of u is bounded above by 107 when degree of the rootis at least 952.Now that we have the desired upper bound, consider the following change. Detach all B branches from u and attach them to the root. Degree of the root increases and therefore the This was checked by computer. The program and its output are available from the authors. R (except u ). Therefore forthe worst case we can only consider their smallest possible degree (some of them can have degree5 and the rest will have degree 4). A simple computer search shows that in different cases of smallvalues of d R ≤
951 (and all d u ≤ d R ) the suggested change improves the ABC index. Hence we canassume that d R ≥
952 and therefore d u ≤ d u the worst case of theequation only depends of d R and it is easy to check that this change decreases the ABC-index. Corollary 4.13.
Any ABC-minimal tree is similar to a tree where at most one non-root vertex ofdegree ≥ exists that is not a root of a C k -branch.Proof. We may assume T is an ABC-extremal tree. Let u be a non-root vertex which is not theroot of a C k branch and has d u ≥
6. Since there are no such vertices of degree 6 , . . . ,
15 we have d u ≥
16 and therefore u is a son of the root. Since d u ≥ u has some B branches as subtrees(see Corollary 4.6) and by (P1), (P2) and (P3), either T u contains all B branches (and possibly a B ∗ ) or T u contains B ∗ or B ∗∗ branches. If there is a B (or B ∗ ) and also one B ∗ or B ∗∗ in the tree,then by a similarity exchange we can move them and make all of them sons of u . Therefore, allother vertices at distance two from the root will be B branches and all children of the root (withdegree at least 6) except u are C k branches. Corollary 4.14.
In any ABC-minimal tree of order at least , the degree of the root is at least .Proof. If the root has a child u that is none of B , B ∗ , B , B ∗ , B ∗∗ or B , then by Lemma 4.11 d u ≥
16 and therefore the degree of the root is also at least 16 and we are done. So we may assumethat all children of the root are B , B ∗ , B , B ∗ , B ∗∗ or B . In [6], all ABC-minimal trees of orderup to 1100 have been determined and the largest one where the degree of the root is less than 16is a tree with 110 vertices (contains 2 B and 13 B branches). Lemma 4.15.
Any ABC-minimal tree is similar to a tree T in which one can remove a set of atmost vertices so that the resulting tree has no vertices of degree 3 or 5 and all ending branchesare B . The removed vertices are all from at most four B -branches adjacent to the root and fromat most one subtree corresponding to a son of the root.Proof. We may assume T is ABC-minimal. By removing at most one vertex, we eliminate possible2-2 edge. By removing 3 vertices, B ∗∗ can be changed into a B . Similarly, by removing 2 vertices,a B can be changed into a B . These operations do not change degrees of vertices except for theroots of newly formed B -branches. Since a 2-2 edge or B ∗∗ cannot coexist with a B , this stepremoves at most 8 vertices all together.By Lemma 4.11 (a), any vertex of degree 4 is a B -branch (or B ∗ or B ∗∗ ) and any non-rootvertex of degree 5 is a B adjacent to the root. Thus, all remaining vertices of degree 3 have fathersof degree at least 16 (by Lemma 4.11 and Corollary 4.14). Each such father w has only B and B as subtrees. By (P1)–(P3), all B -branches are descendants of a single such vertex w . By removingall ( ≤ B -branches, all ending branches are B . By Corollary 4.13 we may assume that possible2-2 edge or B ∗∗ is also a subtree of T w . this complete the proof.29 The main structure
Lemma 5.1.
Let T be an ABC-extremal tree whose root R has degree d R ≥ . Then there areat least d R − C k - and C k +1 -branches attached to the root for some k ∈ { , , } .Proof. Let x , . . . , x d R be the sons of R and T i = T x i their subtrees ( i = 1 , . . . , d R ). By removingone of these subtrees and at most four B -branches (Lemma 4.15), all the remaining subtrees arecopies of B and copies of C k i . Lemma 2.10 shows that each such k i is at most 142.Suppose first that there are at most 322 B -branches among them. Then we have at least d R − > C k i -branches. By Lemma 2.9, the values k i take only two consecutive values. ByLemmas 4.7 and 4.8, all but at most 364 of these C -branches are C or C .Suppose now that there are at least 323 B -branches adjacent to R . Now consider s = 323copies of these B branches and apply the change outlined in Figure 20, where we replace these B -branches with seven C ( s − / -branches (there may be more B branches attached to the root inthe shaded part). . . . | {z } s R R s − s − s − . . . | {z } Figure 20: Improving ABC-index if there are s B -branches, where s ≡ z i ( i = 1 , . . . , d R − s ) be neighbors of R of in the shaded part of Figure 20. Then thechange of ABC-contribution of the corresponding edge is f ( d R , d z i ) − f ( d R − s + 7 , d z i ), which isdecreasing in d z i by Proposition 2.2. We know that d z i ≤
143 but it is not possible to have thedegree of all of them equal to 143. So we have two extreme cases here. If we have at most 2 × C k branches, then we can have 364 of z i ’s to be the root of C and the rest to be the root of C branches. Note that in this case the other children of the root will be (at most 4) B and B branches. Therefore the change in the ABC-index will be bounded below as follows:∆( T, T (cid:48) ) ≥ f ( d R , f ( d R , f ( d R ,
5) + ( d R − × − f ( d R ,
4) + 6 √ / − f ( d R − s + 7 , − f ( d R − s + 7 , − f ( d R − s + 7 , ( s + 6) / − f ( d R − s + 7 , − ( d R − s − × − f ( d R − s + 7 , − ( s − f (( s + 6) / , . It can be shown that this is positive for s = 323 when d R ≥ × C k branches, then all but 364 of them are C . So in the worst case the root has 364 children of degree 54 and d R − − s − and 4 children of degree 5. Therefore we have:∆( T, T (cid:48) ) ≥ f ( d R ,
54) + ( d R − − f ( d R ,
53) + 4 f ( d R ,
5) + s × f ( d R ,
4) + 6 √ / − f ( d R − s + 7 , − ( d R − − f ( d R − s + 7 , − f ( d R − s + 7 , − f ( d R − s + 7 , ( s + 6) / − ( s − f (( s + 6) / , . It can be shown that this is positive for s = 323 when d R ≥ Lemma 5.2.
Let T be an ABC-extremal tree of order n ≥ . Then there are no B -branchesattached to the root and there are at least n − C -branches.Proof. We start as in the previous proof. By removing at most 63 vertices, we end up with asubtree T whose main subtrees are all B and C k i ( i = 1 , . . . , r ), where k ≥ k ≥ · · · ≥ k r . Firstwe claim that T contains a C and C -branch. If d R ≥ C or C among the main subtrees, so one of them is also contained in T .Suppose now that d R < B -branches adjacent to R . Removingthem, leaves at least n − − · n − T are contained in at most four B -branches and all degree-3 vertices are contained in a singlesubtree T x of some son x of R . By (P1)–(P3), x has smallest degree among all sons of R (when itexists).Thus T x , . . . , T x r − are also subtrees of T . In particular, each such k i is at most 143. Thus, d R − ≥ ( n − − (1 + 7 k r )) / (1 + 7 · ≥ k i (2 ≤ i < r )is either 51, 52, or 53, and at least one of them is 51 or 52.We conclude that there is a C k -branch in T with k ∈ { , } . Suppose that a B -branch isattached to the root. Now consider the change suggested in Figure 21. R R k k + 1 Figure 21: Merging C k and B attached to the root.If a neighbor of R in the shaded part of Figure 21 has degree a , Lemmas 2.9 and 4.7 implythat 4 ≤ a ≤
54. Then the change of the ABC-contribution of the corresponding edge is f ( d R , a ) − f ( d R − , a ), which is decreasing in a by Proposition 2.2. Therefore the worst case happens when k = 52 and a = 54. Also note that by Lemma 4.7 there can be at most 364 copies of C in T . Sowe have: ∆( T, T (cid:48) ) ≥ f ( d R ,
54) + ( d R − f ( d R ,
53) + f ( d R ,
4) + 52 f (53 , − f ( d R − , − ( d R − f ( d R − , − f (54 , . The same can be concluded for the subtrees containing vertices of degree 3 because of (P1)–(P3). d R it is easy to check that for d R ≥ ABC ( T ) − ABC ( T (cid:48) ) > d R ≥ k ≤
53. By repeating the calculation from above: we first remove up to 63 vertices, next weremove T x r which has at most 1 + 7 ·
53 = 365 vertices. For possible C -branches we remove oneof their B -subtrees. Each of the resulting branches then contains at most 365 vertices. Thus d R ≥ n − − − · / > n − ≥ . The last inequality combined with Lemma 4.8 also shows that most of the C -branches are C and that at most 364 of these are C or C .We proved that there are no B -branches attached to the root. This means that there are atleast d R − − − ≥ n − C -branches. Theorem 5.3.
For every ABC-extremal tree of order n , one can delete O (1) vertices to obtain asubtree T r shown in Figure 22, where r = (cid:98) n/ (cid:99) − O (1) . More precisely after deleting at most
63 + o (1) vertices we are left with a tree whose root has degree d satisfying n − (13 + o (1)) ≤ d ≤ n + (1 + o (1)) , and all its sons are C -branches together with at most C k -branches where k is either or and possibly one additional C l -branch where l ≤ .Proof. The o (1) notation takes care of small values of n , so we may assume that n ≥ o (1) will be 0. Then Lemma 5.2 applies (in which case we remove at most 36 verticesin B branches and at most 365 vertices in a subtree of the root of smallest degree in which someending branches would be different from B ). . . . R u u r
52 52 52 u T r Figure 22: The structure of large ABC-minimal trees after deleting O (1) vertices.The estimates on the number of vertices of the tree used in the above theorems are very liberal.The transition to the desired form with mostly C -branches occurs much earlier. Let n be theorder of an ABC-minimal tree. As introduced in Figure 3, let r be the number of C k -branchesadjacent to the root, and let s be the number of B -branches adjacent to the root. For small valuesof n the main structure is when r = 0 which gives us the structure that was conjectured by Gutman[24]. Simulation show that for larger n we have positive values of r . To be more precise, we havepositive r when the following holds: n ≡ n ≥ n ≡ n ≥ n ≡ n ≥ n ≡ n ≥ n ≡ n ≥ n ≡ n ≥ n ≡ n ≥ n , Lemma 5.2 shows that s = 0, which is the structure conjectured in [1].32 orollary 5.4. Let γ n = min {ABC ( T ) : | V ( T ) | = n } and let c = (cid:113) (cid:0) √
55 + 156 √ (cid:1) .Then c n − c + (cid:113) − O ( n − ) ≤ γ n ≤ c n + 365 c + (cid:113) + O ( n − ) and hence lim n →∞ n γ n = c ≈ . . Proof.
Let us first establish the upper bound. Let r = (cid:98) ( n − / (cid:99) , and t = n − r − T r +1 with r + 1 C -branches, and let T be a tree of order n obtained from T r +1 by removing 365 − t vertices from the last C -branch. The removal of the vertices can be done insuch a way that ABC ( T r ) ≤ ABC ( T ) ≤ ABC ( T r +1 ) . (9)To see this, note first that deleting any subset of vertices of degree 1 from the C -branch decreasesthe ABC-index. The same holds when removing any subset of B -branches and some degree-1vertices. Finally, by removing the last vertex – the root of the C -branch – a short calculationshows that the ABC-index drops as long as r is large enough (which we may assume).Thus, it suffices to estimate ABC ( T r +1 ) (and ABC ( T r )), which we do next. Clearly, r +1 ABC ( T r +1 ) = f (53 , r + 1) + 52 f (53 ,
4) + 156 f (4 ,
2) + 156 f (2 , (cid:113) (cid:113) r +1 + 26 (cid:113) + 156 √ (cid:113) (cid:16) r +1) + O ( r − ) (cid:17) + 26 (cid:113) + 156 √ . Since r ≤ n/ ABC ( T ) ≤ ABC ( T r +1 )= (cid:16)(cid:113) + 26 (cid:113) + 156 √ (cid:17) r + (cid:16)(cid:113)
153 512 + (cid:113) + 26 (cid:113) + 156 √ O ( r − ) (cid:17) ≤ c n + 365 c + (cid:113) + O ( n − ) . The lower bound follows from Theorem 5.3 and using the same calculation as above (for r instead r + 1), and using the fact that r ≥ n/ − Acknowledgement.
The authors are grateful to an anonymous referee for carefully checking the results and for providingadditional information and insight that simplified some of our proofs.
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