Convexity result and trees with large Balaban index

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Abstract

Balaban index is defined as J(G)=mmn+2Σ1w(u)w(v), where the sum is taken over all edges of a connected graph G, n and m are the cardinalities of the vertex and the edge set of G, respectively, and w(u) (resp. w(v)) denotes the sum of distances from u (resp. v) to all the other vertices of G. In 2011, H. Deng found an interesting property that Balaban index is a convex function in double stars. We show that this holds surprisingly to general graphs by proving that attaching leaves at two vertices in a graph yields a new convexity property of Balaban index. We demonstrate this property by finding, for each n, seven trees with the maximum value of Balaban index, and we conclude the paper with an interesting conjecture.

Abstract

Balaban index is defined as J(G)=mmn+2Σ1w(u)w(v), where the sum is taken over all edges of a connected graph G, n and m are the cardinalities of the vertex and the edge set of G, respectively, and w(u) (resp. w(v)) denotes the sum of distances from u (resp. v) to all the other vertices of G. In 2011, H. Deng found an interesting property that Balaban index is a convex function in double stars. We show that this holds surprisingly to general graphs by proving that attaching leaves at two vertices in a graph yields a new convexity property of Balaban index. We demonstrate this property by finding, for each n, seven trees with the maximum value of Balaban index, and we conclude the paper with an interesting conjecture.

Keywords:

1 Introduction

In this paper we consider simple and connected graphs. For a graph G, by V(G) and E (G) we denote the vertex and the edge sets of G, respectively. We set n = |V(G)| and m = |E(G)|. For vertices u,v 2 V(G) we use distG(u,v) to denote the minimal distance from u to v in G, and for x ϵV(G), we have w(x)=ΣyϵV(G) distG(x,y).

The Balaban index J (G) of a graph G is defined as

J(G)=mmn+2uvE(G)1w(u)w(v),

where the sum is taken over all edges of G. It was introduced by Balaban in [2, 3] and it was used successfully in QSAR/QSPR modeling [9, 27], see also [5, 15]. Recent papers on mathematical properties of this index include [7, 12, 13, 18, 23]. For further recent topics and open problems in chemical graph theory an interested reader is referred to [1, 6, 17, 19, 25, 28].

It is known, that among graphs on n ≥ 8 vertices, Balaban index attains its maximum for the star on n vertices Sn, and for graphs on n ≤ 7 vertices, Balaban index attains its maximum for the complete graph Kn, see [11] and [21]. The problem of finding the minimum value of Balaban index among graphs of given order is much more difficult and it remains open. However, in [16] it was shown that this value is of order Θ(n-1), and the structure of candidates for extremal graphs is discussed. In [18], an upper bound for Balaban index of r-regular graphs on n vertices was given, which led to the observation that the minimum value of Balaban index in the class of graphs on n vertices tends to zero as n increases. The asymptotic behavior of J was discussed in [4] and recently in [20], where it was shown that in fact for an arbitrary positive number r there exists a sequence of graphs whose Balaban index converges to r.

From mathematical aspect, one’s aim is to determine the extremal values of indices in various classes of graphs. The class of trees is interesting especially from the point of view of chemists. It was shown by Sun [26], Dong and Guo [10], and Deng [8] that for a tree T on n vertices, n ≥ 2,

J(Pn)J(T)J(Sn)

with left (right) equality if and only if T = Pn (T = Sn), where Pn is the path on n vertices and Sn is the star on n vertices. Let Dn-2,2 be a tree obtained from Sn-1 by subdividing one edge. In [8], Deng proved that Dn-2,2 has the second maximum value of Balaban index among trees.

Theorem 1

Let T be a tree with n ≥ 4 vertices. If T is not a star, then J(T) ≤ J(Dn-2,2) with equality if and only if J(T) = J(Dn-2,2).

To prove the above mentioned results, the authors used so called edge-lifting and path-sliding transformations, which both increase Balaban index.

Theorem 2

Let G1 and G2 be two graphs with n1 and n2 vertices, respectively, n1,n2 2. If G is the graph obtained from G1 and G2 by adding an edge between a vertex u* of G1 and a vertex v* of G2, G' is the graph obtained by identifying u* of G1 to v* of G2 and adding a pendant edge to u*(v*), then G' is called the edge-lifting transformation of G and we have J(G) < J(G').

Theorem 3

Let G0 be a graph with n0 2 vertices, and P = v1v2 · · ·vr a path of length r -1 2. If G (resp. G') is the graph obtained by identifying a vertex v* of G0 to vk-1 (resp. vk) in P, 2k[r12]then G' is called the path-sliding transformation of G and we have J(G) < J(G').

Using these transformations, together with an interesting convexity property of Balaban index which we prove in the next section, we find trees with the third, fourth, . . . and seventh maximum value of Balaban index. We remark that for n big enough, in [14] the authors find first 21 trees with the largest value of Balaban index. However, their proofs are very technical, hard to follow and too many details are left to the reader. Our proofs are much more precise, less technical, we do not restrict ourselves to large n and we conclude the paper with interesting conjectures. Moreover, we expect that the convexity property which we describe in the paper, will appear useful in studying Balaban index in the future.

2 Convexity result for Balaban index

A double star Da,b is a tree consisting of a +b vertices, two of which have degrees a and b, while the remaining ones have degree 1, see Figure 1 for Dn-2,2. By symmetry, we may assume that a > b. The following result was proved in [8].

Fig. 1

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Fig. 1

A double star Dn-2,2.

Citation: Applied Mathematics and Nonlinear Sciences 3, 2; 10.21042/AMNS.2018.2.00034

Theorem 4

J (Dn-x,x), as a function of x, is convex.

It should be noted that in the proof of Deng the function J (Dn-x,x) is defined in such a way that it is a continuous function, and the second derivative is used to show that it is convex. In what follows we will show that Theorem 4 can be generalized. For that we need the notion of a discrete convex function. This concept can be introduced in several different ways (an interested reader should consult [24]), but generally a (discrete) function f is strictly convex if for every x0 < x1 < x2 from the domain of f it holds

f(x1)<x2x1x2x0f(x0)+x1x0x2x0f(x2).

However, if the domain is the set of integers greater than or equal to b, then the above property is equivalent to

2f(x0+1)<f(x0)+f(x0+2)

for all x0 ≥ b.

Theorem 5

Let G be a graph with two distinct vertices u* and v*. Let a ≥ 2 and 0 ≤ x ≤ a. Attach x pendant edges to u*, attach a -x pendant edges to v*, and denote the resulting graph by Gx. Then J(Gx), as a (discrete) function of x, is strictly convex.

Proof. As mentioned above, it suffices to prove that J (Gx)+J(Gx+2) > 2J(Gx+1) for every x, 0 ≤ x ≤ a -2.

Therefore, it is sufficient to restrict ourselves to the case a = 2 and to prove J (G0)+J(G2) > 2J(G1).

Further, since all the graphs Gx, 0 ≤ x ≤ 2, have the same number of vertices and edges, it suffices to prove the strict convexity of uvE(Gx)1wGx(u)wGx(v).We divide the proof into two claims.

Claim 1

Let uv ϵ E (G). Then 1wGx(u)wGx(v),as a function of x, is strictly convex.

Denote c = wG1(u) and d = wG1(v). Further, denote γ = wG1(u)-wG0(u) and δ = wG1(v)-wG0(v). Then wG0(u) = c -γ and wG0(v) = d -δ. Since γ (δ, respectively) is the difference corresponding to removing a pendant edge from v* and adding it to u*, we have wG2(u) = c +γ and wG2(v) = d +δ. Hence, we need to prove

1wG0(u)wG0(v)+1wG2(u)wG2(v)>2wG1(u)wG1(v),

which is equivalent to

1(cγ)(dδ)+1(c+γ)(d+δ)>2cd.

Observe that c,d > 0, but γ and δ can be negative. Anyway, since c - γ = wG0(u) > 0 and c+γ = wG2(u) > 0, we have c - |γ| > 0. Analogously d -|δ| > 0, and so

c2d2>(c2γ2)(d2δ2)cd>(c+γ)(cγ)(d+δ)(dδ)2(c+γ)(cγ)(d+δ)(dδ)cd>(c+γ)(cγ)(d+δ)(dδ).

Now suppose that |cδ| ≥ |dγ|. (The case |dγ| ≥ |cδ| can be proved analogously since the resulting formula is symmetric, see below.) Then, regardless if is positive or negative, we have

cδ(cδ+dγ)>0.

Since d22 > 0, see above, we have

(d2δ2)γ2+cδ(cδ+dγ)>0d2γ2δ2γ2+c2δ2+cdγδ>02c2d2+2cdγδ>2c2d22c2δ22d2γ2+2γ2δ2(c+γ)(d+δ)+(cγ)(dδ)cd>2(c+γ)(d+δ)(cγ)(dδ).

Summing (2) and (3) gives

((c+γ)(d+δ)cd+(cγ)(dδ)cd)2>(2(c+γ)(d+δ)(cγ)(dδ))2,

which is equivalent to (1). This establishes the claim.

Claim 2

Function uvE(Gx)\E(G)1wGx(u)wGx(v)is strictly convex.

Let dG(u*,v*) = l. Denote c = wG(u*) and d = wG(v*). Then wG1(u*) = c+1+l +1, wG2(u*) = c+2, wG0(v*) = d +2 and wG1(v*) = d +l+2. Further, denote by u' a pendant vertex attached to u* and denote by v' a pendant vertex attached to v*. If |V(G)|=t, then wG1(u')=c+l+t+2, wG2(u')=c+t+2, wG0(v')=d+t+2 and wG1(v') = d+l+t +2. Denote g(x)=uvE(Gx)\E(G)1wGx(u)wGx(v).Then

g(0)=2(d+2)(d+t+2),

g(1)=1(c+l+2)(c+l+t+2)+1(d+l+2)(d+l+t+2),g(2)=2(c+2)(c+t+2).

Since l ≥ 1, we have

2(d+2)(d+t+2)>2(d+l+2)(d+l+t+2)2(c+2)(c+t+2)>2(c+l+2)(c+l+t+2)

and summing the two inequalities above gives g(0)+g(2) > 2g(1) as required. This establishes the claim.

Claims 1 and 2 assure that uvE(Gx)1wGx(u)wGx(v)is a strictly convex function, since the sum of strictly convex functions is a strictly convex function. Thus the claim of the theorem follows.

We remark that the fact that Sn and Dn-2,2 are trees of order n with the largest and second largest, respectively, Balaban index is a direct consequence of Theorem 5.

Let ν be an invariant on the class of trees. We say that ν is star-convex, if for every tree G with two specific vertices u* and v* and for every a ≥ 2, the invariant ν is strictly convex on Gx, which is obtained from G by attaching x pendant edges to u* and a -x pendant edges to v*. Further, if n(T') > ν(T) when T' is obtained from T by path-sliding or edge-lifting, and moreover ν is star-convex, then we say that ν has SLC-property (which is a short notation for sliding-lifting-convexity property). By Theorems 2, 3 and 5, Balaban index has the SLC-property. We remark that also the sum-Balaban index has the SLC-property, see [22].

3 Trees with large Balaban index

In what follows, by Tivwe denote a tree which has the i-th greatest value of invariant n. We omit the superscript ν in the case when ν is the Balaban index. As described in the introduction, T1 =Sn and T2 =Dn-2,2. Here we determine T3, T4, T5, T6 and T7. In every case, using the fact that Balaban index has SLC-property we first reduce the class of possible candidates, and then we determine the optimum one.

As we will see later, a double star Dn-t,t is one of the candidates in every case. There are four orbits of vertices in Dn-t,t with values of w equal to 2n+t -4, n+t -2, 2n-t -2 and 3n-t -4, thus the following holds.

Lemma 6

For t ≥ 2 we have

J(Dnt,t)=(n1)(nt1(2n+t4)(n+t2)+1(n+t2)(2nt2)+1(2nt2)(3nt4)).

A caterpillar Ha1,a2,…,ad-1 is a tree consisting of a diametric path of length d (i.e., with d +1 vertices) and a couple of pendant edges, such that the degrees of vertices of the diametric path are 1,a1,a2, · · · ,ad-1, 1. Due to symmetry, we may assume that a1 ≥ ad-1 in Ha1,a2,⋯,ad-1 . See Figure 2 for an example.

Fig. 2

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Fig. 2

A caterpillar H3,7,2.

Citation: Applied Mathematics and Nonlinear Sciences 3, 2; 10.21042/AMNS.2018.2.00034

First we find T3. Notice that there is only one tree on 3 vertices and only two trees on 4 vertices. Thus only the cases when n ≥ 5 are considered when searching for T3.

Lemma 7

Let ν be an invariant with SLC-property. If {T1v,T2v}={Sn,Dn2,2}, then T3v{Dn3,3,H2,n3,2}.

Proof. Since edge-lifting reduces the diameter at most by 1, it suffices to consider trees with diameters 3 and 4. Since ν is star-convex, the only candidate among trees of diameter 3 is Dn-3,3.

As regards trees of diameter 4, due to edge-lifting it suffices to consider those whose vertices of degrees at least 2 form a path, i.e., the caterpillars. Hence, we have to consider caterpillars Ha,b,c. Observe that all a, b and c are at least 2. However, if at least two from a, b and c are at least 3, then ν(Ha,b,c) < n(Dn-3,3) due to edge-lifting. Moreover, ν(Hn-3,2,2) < n(H2,n-3,2) due to path-sliding if n ≥ 6, while if n = 5 then H2,n-3,2 = Hn-3,2,2. Therefore, T3v{Dn3,3,H2,n3,2}.

Now we can state the result for T3.

Proposition 8

The following holds:

T3={P5,ifn=5Dn3,3,ifn6.

Proof. Since {Ti; 1 ≤ i ≤ 2} = {Sn,Dn-2,2} and the Balaban index has SLC-property, it suffices to compare J (Dn-3,3) and J (H2,n-3,2), by Lemma 7. Since for n = 5 the graph Dn-3,3 with n -3 3 does not exist (observe that Dn-3,3 should be D3,2, which is T2 in this case), we have T3 = H2,2,2 = P5. In what follows we assume n ≥ 6.

By Lemma 6 we have

J(Dn3,3)=(n1)(2(3n7)(2n5)+1(2n5)(n+1)+n4(n+1)(2n1)).

In H2,n-3,2 there are also four orbits of vertices with value of w equal to 3n-5, 2n-3, n +1 and 2n-1.

Therefore

J(H2,n3,2)=(n1)(2(3n5)(2n3)+2(2n3)(n+1)+n5(n+1)(2n1)).

Since f(x)=1xis a convex function, we have 12n5+12n1>22n3,and so

1(2n5)(n+1)+1(2n1)(n+1)>2(2n3)(n+1).

Since 2(3n7)(2n5)>2(3n5)(2n3),we get J(Dn-3,3) > J(H2,n-3,2). Thus for n ≥ 6, Dn-3,3 is a tree with the third maximum value of Balaban index.

In the sequel, we will refer to the following two facts from calculus.

Lemma 9

Let a, b, c, d, and e, be constants such that b,d > 0. Then,

limnn+a(bn+c)(dn+e)=1bd.

Lemma 10

Let a, b, c, d, e, and f be constants. Then,

limn(n+a)(n+b(n+c)(2n+d)n+b(n+e)(2n+f))=2(ec)+(fd)42.

Proof. Starting from the left side, we derive

limn(n+a)(n+b(n+c)(2n+d)n+b(n+e)(2n+f))=limn(n+a)(n+b)(n+c)(2n+d)(n+e)(2n+f)limn((n+e)(2n+f)(n+c)(2n+d))=12limn(n+e)(2n+f)(n+c)(2n+d)(n+e)(2n+f)+(n+c)(2n+d)=2(ec)+(fd)42.

When searching for T4,T5 and T6, we assume that n is at least 6, since otherwise there are less than 4 non-isomorphic trees and T4 does not exist.

Lemma 11

Let ν be an invariant with SLC-property. If {T1v,T2v,T3v}={Sn,Dn2,2,Dn3,3},thenT4v{Dn4,4,H2,n3,2}.

Proof. Analogously as in Lemma 7, when looking for T4vone has to consider Dn-4,4 and caterpillars Ha,b,c. However, if at least two from a,b,c are at least 4, then ν(Ha,b,c) < ν(Dn-4,4) due to edge-lifting. Similarly, if a ≥ 4 and one of b,c is at least 3, then ν(Ha,b,c) < n(Dn-4,4) due to edge-lifting. Therefore it suffices to consider Dn-4,4, H2,n-3,2, H3,n-4,2, H3,n-5,3 and Hn-3,2,2. However, ν(Hn-3,2,2) < ν(H2,n-3,2) due to path-sliding. Further, ν(D4,n-4)+ν(H3,n-4,2) > 2ν(H3,n-4,2) due to star-convexity, and so n(Dn-4,4) > ν(H3,n-4,2). Finally, 2n(H4,n-5,2) > 2n(H3,n-5,3) due to star-convexity and ν(Dn-4,4) > ν(H4,n-5,3) due to edge-lifting, which means that ν(Dn-4,4) > ν(H3,n-5,3). Therefore, T4v2 {Dn-4,4,H2,n-3,2}.

Proposition 12

The following holds:

T4={D4,4,ifn=8H2,n3,2,ifn6,n8.

Proof. Since for n ≥ 6 we have {Ti; 1 ≤ i ≤ 3} = {Sn,Dn-2,2,Dn-3,3} and the Balaban index has SLC-property, it suffices to compare J(H2,n-3,2) and J(Dn-4,4), by Lemma 11. As described in the proof of Proposition 8,

J(H2,n3,2)=(n1)(2(3n5)(2n3)+2(2n3)(n+1)+n5(n+1)(2n1)),

and by Lemma 6,

J(Dn4,4)=(n1)(3(3n8)(2n6)+1(2n6)(n+2)+n5(n+2)2n).

Since there is no Dn-4,4 for n ≤ 7, we have T4 = H2,n-3,2 if 6 ≤ n ≤ 7. For n ≥ 9, the following holds:

J(H2,n3,2)>J(Dn4,4),

which implies that T4 = H2,n-3,2, for n ≥ 9. Note that showing (6) analytically would be extremely long and tedious and we would like to omit such a proof here. In order to convince the reader we show that the difference J (H2,n-3,2)-J(Dn-4,4) tends to a positive constant. By (4) and (5) we have

limn(J(H2,n3,2)J(Dn4,4))=26+223612+342=73446=·0.8291885.

Let us remark that the difference J (H2,n-3,2)-J(Dn-4,4) is increasing with n, and we obtain the first four decimals 0:8291 of the above limit when n is of order 8 · 104.

Finally, (6) does not hold for n = 8, so for this order T4 = D4,4.

Let n ≥ 7. Then by Rn we denote the graph obtained from a star on n -3 vertices by subdividing three distinct edges, see Figure 3 for R11. Now we determine T5.

Fig. 3

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Fig. 3

The graph R11.

Citation: Applied Mathematics and Nonlinear Sciences 3, 2; 10.21042/AMNS.2018.2.00034

Lemma 13

Let ν be an invariant with SLC-property. If {Tiv;1i4}={Sn,Dn2,2,Dn3,3,H2,n3,2},then T5v{Dn4,4,Hn3,2,2}.

Proof. Notice that T5vmust be of diameter 3, 4 or 5, since H2,n-3,2 has diameter 4. So, first consider trees of diameter 5. If T5vhas diameter 5, it must have the property that every edge-lifting of T5vresults in H2,n-3,2.Consequently, T5vhas a unique vertex of degree at least 3 and even in this case one can choose edge-lifting resulting in a tree of diameter 4 other than H2,n-3,2. Hence, it suffices to consider Dn-4,4 and trees of diameter 4.

If T5vhas diameter 4 and is a caterpillar, then analogously as in Lemma 11 the candidates are H3,n-4,2, H3,n-5,3 and Hn-3,2,2. However, in Lemma 11 we have shown that ν(Dn-4,4) > ν(H3,n-4,2) and ν(Dn-4,4) > ν(H3,n-5,3). If T5vhas diameter 4 and is not a caterpillar, then the central vertex of a diametric path of T5vis adjacent with three distinct vertices of degree at least 2, and every edge-lifting of T5vmust result in H2,n-3,2. So T5vis Rn. We have ν(H3,n-4,2)+ν(H3,n-4,2) > 2ν(Rn) by star-convexity. Since ν(Dn-4,4) > ν(H3,n-4,2), we have ν(Dn-4,4) > ν(Rn). Therefore, T5vϵ {Dn-4,4,Hn-3,2,2}.

Proposition 14

The following holds:

T5={H2,5,2,ifn=8Hn4,4,if9n17Hn3,2,2,ifn{6,7}orn18.

Proof. Since for n ≥ 6 and n ≠ 8 we have {Ti; 1 ≤ i ≤ 4} = {Sn,Dn-2,2,Dn-3,3,H2,n-3,2} and the Balaban index has SLC-property, it suffices to compare J (Hn-3,2,2) and J (Dn-4,4) if n ≠ 8, by Lemma 13. In Hn-3,2,2 there are five orbits of vertices with value of w equal to 4n-10, 3n-8, 2n-4, n +2 and 2n. Therefore

J(Hn3,2,2)=(n1)(1(4n10)(3n8)+1(3n8)(2n4)+1(2n4)(n+2)+n4(n+2)2n).

We already know that

J(Dn4,4)=(n1)(3(3n8)(2n6)+1(2n6)(n+2)+n5(n+2)2n).

Since there is no Dn-4,4 for n ≤ 7, we have T5 = Hn-3,2,2 if 6 ≤ n ≤ 7. The case n = 8 was checked by a computer and it was found that T5 = H2,5,2. (In fact, it suffices to check the graphs Ha,b,c, and the star convexity rules out all of them except H2,5,2 and H5,2,2, for which J (H2,5,2) > J(H5,2,2) by path-sliding.) Hence, it suffices to consider n ≥ 9.

For n ≥ 18, the following holds

J(Hn3,2,2)>J(Dn4,4),

which implies that T5 = Hn-3,2,2 for n ≥ 18. Again including a rigorous proof of (7) would be too complicated.

We only show that the difference J (Hn-3,2,2)-J(Dn-4,4) tends to a positive constant. By (4) and (5) we have

limn(J(Hn3,2,2)J(Dn4,4))=112+16+12+123612=1+62212=·0.179285>0.

The difference J (Hn-3,2,2)-J(Dn-4,4) is increasing with n, and we obtain the first four decimals 0:1792 of the above limit when n is of order 37 · 103.

For 9 ≤ n ≤ 17 it holds J(Dn-4,4) > J(Hn-3,2,2), and so T5 = Dn-4,4 in these cases.

Now we concentrate on T6.

Lemma 15

Let ν be an invariant with SLC-property. If {Tiv;1i5}={Sn,Dn2,2,Dn3,3,H2,n3,2,Hn3,2,2},then T6v=Dn4,4.

Proof. Analogously as in Lemma 13, T6vmay be of diameter 3, 4 or 5. By star-convexity Dn-4,4 is the only candidate among trees of diameter 3. Further, H3,n-4,2 and H3,n-5,3 are the only candidates among caterpillars of diameter 4, but in Lemma 11 we have already found that ν(Dn-4,4) > ν(H3,n-4,2) and ν(Dn-4,4) > ν(H3,n-5,3).

If T6vhas diameter 4 and is not a caterpillar, then the central vertex of T6vis adjacent with three distinct vertices of degree at least 2, and every edge-lifting of T6vmust result in H2,n-3,2 or Hn-3,2,2. Consequently, every edge-lifting of T6vmust result in H2,n-3,2 and T6vis Rn. Since in Lemma 13 we derived ν(Dn-4,4) > ν(Rn), also Rn is out of consideration.

Finally, if T6vis a tree of diameter 5, it must be a caterpillar since every edge-lifting of T6vresults in a caterpillar of diameter 4. If T6vhas a unique vertex of degree at least 3 then it is H2,n-4,2,2. However, an edge-lifting of H2,n-4,2,2 results in H3,n-4,2. Since ν(Dn-4,4) > ν(H3,n-4,2), we have ν(Dn-4,4) > ν(H2,n-4,2,2). On the other hand, if T6vhas at least two vertices of degree at least 3 then there is an edge-lifting of T6vwhich results in a caterpillar of diameter 4 with at least two vertices of degree at least 3. Since such a caterpillar is not in {Tiv;1i5},also this tree is out of consideration. Therefore, T6v=Dn4,4.

Proposition 16

The following holds:

T6={P6ifn=6H3,3,2,ifn=7D5,5,ifn=10Hn3,2,2,ifn{8,9}or11n17Dn4,4,ifn18.

Proof. Since for n ≥ 18 we have {Ti; 1 ≤ i ≤ 5} = {Sn, Dn-2,2,Dn-3,3,H2,n-3,2,Hn-3,2,2} and the Balaban index has SLC-property, T6 = Dn-4,4 if n ≥ 18, by Lemma 15. The other cases were solved by a computer. (Analogously as for the case n = 8 in the proof of Proposition 14, for each n it suffices to check the Balaban index of a few graphs.)

In what follows we assume that n ≥ 7, as otherwise T7 does not exist.

Lemma 17

Let ν be an invariant with SLC-property. If {Tiv;1i6}={Sn,Dn2,2,Dn3,3,Dn4,4,H2,n3,2,Hn3,2,2,},then

T7v{Dn5,5,H3,n4,2,Hn4,3,2,Hn4,2,3}.

Proof. Analogously as in Lemmas 13 and 15, T7vmay be of diameter 3, 4 or 5. Moreover, in the very same way as in Lemma 15 one can derive that there is no candidate among trees of diameter 5, and the only candidate among trees of diameter 4 which is not a caterpillar is Rn. Further, Dn-5,5 is the only candidate among trees of diameter 3, so it remains to consider caterpillars Ha,b,c of diameter 4. However, if two from a,b,c are at least 5, then ν(Ha,b,c) < ν(Dn-5,5) due to edge-lifting. Analogously, if a ≥ 5 and b +c ≥ 6, then ν(Ha,b,c) < ν(Dn-5,5) due to edge-lifting. Therefore, we have 9 candidates, namely Dn-5,5, H3,n-4,2, H4,n-5,2, H3,n-5,3, H4,n-6,3, H4,n-7,4, Hn-4,3,2, Hn-4,2,3 and Rn.

In Lemma 13 we have already shown that ν(H3,n-4,2) > ν(Rn), so Rn is out of consideration. The function ν(Hx,n-5,6-x) is strictly convex due to star-convexity, and so ν(Dn-5,5) > ν(H4,n-5,2) and ν(Dn-5,5) > n(H3,n-5,3). Analogously, ν(Dn-5,5) > ν(Dn-6,6) > ν(H4,n-6,3) and ν(Dn-5,5) > ν(Dn-7,7) > ν(H4,n-7,4).

Hence, T7v{Dn5,5,

H3,n-4,2,Hn-4,3,2,Hn-4,2,3g.

Proposition 18

The following holds:

T7={R7,ifn=7H7,2,2,ifn=10D6,5,ifn=11H3,n4,2,ifn{8,9}orn12.

Proof. Since for n ≥ 8 and n ≠ 10 we have {Ti; 1 ≤ i ≤ 6} = {Sn,Dn-2,2,Dn-3,3,Dn-4,4, H2,n-3,2,Hn-3,2,2} and the Balaban index has SLC-property, T7 2 {Dn-5,5,H3,n-4,2,Hn-4,3,2, Hn-4,2,3} if n ≥ 8 and n 6= 10, by Lemma 17. In these cases it suffices to compare J(H3,n-4,2), J (Hn-4,2,3), J(Hn-4,3,2) and J(Dn-5,5).

By Lemma 6,

J(Dn5,5)=(n1)(4(3n9)(2n7)+1(2n7)(n+3)+n6(n+3)(2n+1)).

In H3,n-4,2 there are six orbits of vertices with values of w equal to 3n-4, 2n-2, 3n-6, 2n-4, n +2 and 2n.

Therefore

J(H3,n4,2)=(n1)(1(3n4)(2n2)+1(2n2)(n+2)+2(3n6)(2n4)+1(2n4)(n+2)+n6(n+2)2n).

In Hn-4,3,2 there are also six orbits of vertices with values of w equal to 4n-11, 3n-9, 2n-5, 3n-7, n +3 and 2n+1. Therefore

JHn4,3,2=n114n113n9+13n92n5+12n53n7+12n5n+3+n5n+32n+1.

Finally, in Hn-4,2,3 there are five orbits of vertices with values of w equal to 4n-12, 3n-10, 2n-4, n +4 and 2n+2. Therefore

J(Hn4,2,3)=(n1)(2(4n12)(3n10)+1(3n10)(2n4)+1(2n4)(n+4)+n5(n+4)(2n+2)).

For n ≥ 8, the following holds

J(H3,n4,2)>J(Hn4,3,2)>J(Hn4,2,3).

Again giving a rigorous proof of (8) would be tedious. We only show that each of the differences J (H3,n-4,2) − J (Hn-4,3,2) and J (Hn-4,3,2)-J(Hn-4,2,3) tends to the same positive constant. By (4) and (5) we have

limn(J(H3,n4,2)J(Hn4,3,2))=16+12+26+1211216161212+342=422+3346=·0.649903>0

and

limnJHn4,2,3JHn4,2,3=112+16+16+122121616+342=422+3346=0.649903>0.

Let us remark that both differences J(H3,n-4,2)-J(Hn-4,3,2) and J(Hn-4,3,2)-J(Hn-4,2,3) are increasing with n, and we obtain the first four decimals 0:6499 of the above limit when n is of order 8 · 104.

Analogously, for n ≥ 12 we have

J(H3,n4,2)>J(Dn5,5).

Again we only find the limit

limn(J(H3,n4,2)J(Dn5,5))=26+12+16+124612+342=73446=·0.829188>0.

The difference J (H3,n-4,2)-J(Dn-5,5) is increasing with n, and we obtain the first four decimals 0:8291 of the above limit when n is of order 8 · 104.

Consequently, T7 = H3,n-4,2 if n ≥ 12. We have T7 = H3,n-4,2 also if n ϵ {8,9} since in these cases Dn-5,5 does not exists. For n = 11 the inequality (9) does not hold so in this case T7 = Dn-5,5, and the cases n ϵ {7,10} were found by a computer.

We remark that it is possible to continue in this manner, but the situation is more and more complicated. It would be helpful to obtain a further structural statement, not to increase the number of graphs one has to compare.

Our results are summarized in Table 1. We remark that for all n ≤ 17 we found the seven trees with the maximum Balaban index also using the computer program Sage. We first generate all trees on n vertices using the computer program Nauty, and then determine the corresponding extremes in Sage. This is doable only for small values of n (not much bigger than 20) on our computer resources.

Table 1

First seven trees with maximal values of Balaban index.

nT1T2T3T4T5T6T7
2-3

4

5
P5



6
P6

7
H2,n-3,2
Hn-3,2,2
H3,3,2
R
7
8
S nD4,4
H2,5,2
9
Dn-2,2Hn-3,2,2
H3,n-4,2
10
D n-3,3D5,5
Hn-3,2,2
11
H2,n-3,2D n-4,4D6,5
12-17

H n-3,2,2
18-¥Hn-3,2,2Dn-4,4H3,n-4,2

Observe that

J(Sn)=(n1)(n1(n1)(2n3)).

By (4) and (5) we have

limn(J(Sn)J(H3,n4,2))=5216122612+942=2131246=·2.487565.

Hence, although J(Sn)n2,the difference J (T1)-J(T7) is tending to a constant. The same can be noted for the difference J (T1)-J(Dn-k,k). Namely, using (4), (5) and Lemma 6 we derive

limn(J(Sn)J(Dnk,k))=(k1)73446.

Since the star-convexity implies J (Tk) ≥ J (Dn-k,k), we obtain that J (T1)-J(Tk) is at most J (Sn)-J(Dn-k,k), which tends to a constant. Thus we have the following.

Corollary 19

For every k, J(T1)-J(Tk) is bounded by a constant depending on k but not on n.

In fact, we believe that the difference J (T1)-J(Tk) is not just bounded by a constant, but it tends to a real number. Hence, we have the following conjecture.

Conjecture 20

For every k, limn(J(T1)J(Tk))is a constant.

Communicated by Juan L.G. Guirao

Acknowledgements

The first author acknowledges partial support by Slovak research grants VEGA 1/0026/16, VEGA 1/0042/17, APVV-15-0220 and APVV-17-0428. The research was partially supported by Slovenian research agency ARRS, program no. P1–0383 and project no. L1–4292.

References

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    A. T. Balaban, Topological indices based on topological distances in molecular graphs, Pure Appl. Chem. 55 (1983) 199–206. 10.1351/pac198855020199

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    A. T. Balaban, N. Ionescu-Pallas, T. S. Balaban, Asymptotic values of topological indices J and J’ (average distance sum connectivities) for infinite cyclic and acyclic graphs, MATCH Commun. Math. Comput. Chem. 17 (1985) 121–146.

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    A. T. Balaban, P. V. Khadikar, C. T. Supuran, A. Thakur, M. Thakur, Study on supramolecular complexing ability vis‘a–vis estimation of pK a of substituted sulfonamides: Dominating role of Balaban index J Bioorg. Med. Chem. Letters 15 (2005) 3966–3973.

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    B. Basavanagoudb, W. Gao, S. Patil, V. R. Desai, K. G. Mirajkar, B. Pooja, Computing First Zagreb index and F-index of New C-products of Graphs, Applied Mathematics and Nonlinear Sciences 2 (2017) 285–298. 10.21042/AMNS.2017.1.00024

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    H. Deng, On the Balaban index of trees, MATCH Commun. Math. Comput. Chem. 66 (2011) 253–260.

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    X. Dong, L. You, The maximum Balaban Index (sum–Balaban index) of Unicyclic Graphs, J. Math. Res. Appl. 34 (2014) 392–402.

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    W. Fang, Y. Gao, Y. Shao, W. Gao, G. Jing, Z. Li, Maximum Balaban index and Sum-Balaban index of bicyclic graphs, MATCH Commun. Math. Comput. Chem. 75 (2016) 129–156.

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    D. Hawei, G. Xiaofeng, Ordering of trees by Balaban index, J. of Mathematical Study 45(3) (2012) 213–232.

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    P. V. Khadikar, C. T. Supuran, A. Thakur, M. Thakur, QSAR study on benzene-sulphonamide carbonic anhydrase inhibitors: topological approach using Balaban index, Bioorg. Med. Chem. 12 (2004) 789–793. 10.1016/j.bmc.2003.10.058

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    M. Knor, J. Kranjc, R. Škrekovski, A. Tepeh, A search for the minimum value of Balaban index, Appl. Math. Comput. 286 (2016) 301–310. 10.1016/j.amc.2016.04.023

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[1]

V. Andova, F. Kardoš, R. Škrekovski, Mathematical aspects of fullerenes, Ars Math. Contemp. 11 (2016) 353–379. 10.26493/1855-3974.834.b02

[2]

A. T. Balaban, Highly discriminating distance based numerical descriptor, Chem. Phys. Lett. 89 (1982) 399–404. 10.1016/0009-2614(82)80009-2

[3]

A. T. Balaban, Topological indices based on topological distances in molecular graphs, Pure Appl. Chem. 55 (1983) 199–206. 10.1351/pac198855020199

[4]

A. T. Balaban, N. Ionescu-Pallas, T. S. Balaban, Asymptotic values of topological indices J and J’ (average distance sum connectivities) for infinite cyclic and acyclic graphs, MATCH Commun. Math. Comput. Chem. 17 (1985) 121–146.

[5]

A. T. Balaban, P. V. Khadikar, C. T. Supuran, A. Thakur, M. Thakur, Study on supramolecular complexing ability vis‘a–vis estimation of pK a of substituted sulfonamides: Dominating role of Balaban index J Bioorg. Med. Chem. Letters 15 (2005) 3966–3973.

[6]

B. Basavanagoudb, W. Gao, S. Patil, V. R. Desai, K. G. Mirajkar, B. Pooja, Computing First Zagreb index and F-index of New C-products of Graphs, Applied Mathematics and Nonlinear Sciences 2 (2017) 285–298. 10.21042/AMNS.2017.1.00024

[7]

Z. Chen, M. Dehmer, Y. Shi, H. Yang, Sharp upper bounds for the Balaban index of bicyclic graphs, MATCH Commun. Math. Comput. Chem. 75 (2016), 105–128.

[8]

H. Deng, On the Balaban index of trees, MATCH Commun. Math. Comput. Chem. 66 (2011) 253–260.

[9]

J. Devillers, A. T. Balaban (Eds.), Topological Indices and Related Descriptors in QSAR and QSPR Gordon & Breach, Amsterdam, 1999.

[10]

H. Dong, X. Guo, Character of graphs with extremal Balaban index, MATCH Commun. Math. Comput. Chem. 63 (2010) 799–812.

[11]

H. Dong, X. Guo, Character of trees with extreme Balaban index, MATCH Commun. Math. Comput. Chem. 66 (2011) 261–272.

[12]

X. Dong, L. You, The maximum Balaban Index (sum–Balaban index) of Unicyclic Graphs, J. Math. Res. Appl. 34 (2014) 392–402.

[13]

W. Fang, Y. Gao, Y. Shao, W. Gao, G. Jing, Z. Li, Maximum Balaban index and Sum-Balaban index of bicyclic graphs, MATCH Commun. Math. Comput. Chem. 75 (2016) 129–156.

[14]

D. Hawei, G. Xiaofeng, Ordering of trees by Balaban index, J. of Mathematical Study 45(3) (2012) 213–232.

[15]

P. V. Khadikar, C. T. Supuran, A. Thakur, M. Thakur, QSAR study on benzene-sulphonamide carbonic anhydrase inhibitors: topological approach using Balaban index, Bioorg. Med. Chem. 12 (2004) 789–793. 10.1016/j.bmc.2003.10.058

[16]

M. Knor, J. Kranjc, R. Škrekovski, A. Tepeh, A search for the minimum value of Balaban index, Appl. Math. Comput. 286 (2016) 301–310. 10.1016/j.amc.2016.04.023

[17]

M. Knor, R. Škrekovski, Wiener index of line graphs, in M. Dehmer and F. Emmert-Streib (Eds.), Quantitative Graph Theory: Mathematical Foundations and Applications CRC Press (2014) 279–301. 10.1201/b17645-10

[18]

M. Knor, R. Škrekovski, A. Tepeh, Balaban index of cubic graphs, MATCH Commun. Math. Comput. Chem. 73 (2015) 519–528.

[19]

M. Knor, R. Škrekovski, A. Tepeh, Mathematical aspects of Wiener index, Ars Math. Contemp. 11 (2016) 327–352. 10.26493/1855-3974.795.ebf

[20]

M. Knor, R. Škrekovski, A. Tepeh, A note on accumulation points of Balaban index, MATCH Commun. Math. Comput. Chem. 78 (2017) 163–168.

[21]

M. Knor, J. Komorník, R. Škrekovski, A. Tepeh, Some remarks on Balaban and Sum-Balaban index, manuscript, 2017.

[22]

M. Knor, R. Škrekovski, A. Tepeh, Trees with large sum-Balaban index, manuscript, 2017.

[23]

M. Knor, R. Škrekovski, A. Tepeh, Mathematical aspects of Balaban index, MATCH Commun. Math. Comput. Chem. 79 (2018) 685–716.

[24]

K. Murota, Discrete Convex Analysis, SIAM Monographs on Discrete Mathematics and Applications Vol. 10, SIAM, Philadelphia, 2003. doi 10.1007/BF02680565

[25]

H. S. Ramane and R. B. Jummannaver, Note on forgotten topological index of chemical structure in drugs, Applied Mathematics and Nonlinear Sciences 1 (2016) 369–374. 10.21042/AMNS.2016.2.00032

[26]

L. Sun, Bounds on the Balaban index of trees, MATCH Commun. Math. Comput. Chem. 63 (2010) 813–818.

[27]

R. Todeschini, V. Consonni, Handbook of Molecular Descriptors Wiley CH, Weinheim, 2000. 10.1002/9783527613106

[28]

B. Zhao, J. Gan, H. Wu, Redefined Zagreb indices of Some Nano Structures Applied Mathematics and Nonlinear Sciences 2 (2016) 291–300. 10.21042/AMNS.2016.1.00024

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