On the maximum ABC index of bipartite graphs without pendent vertices

Zehui Shao; Pu Wu; Huiqin Jiang; S.M. Sheikholeslami; Shaohui Wang

doi:10.1515/chem-2020-0002

Article Open Access

On the maximum ABC index of bipartite graphs without pendent vertices

Zehui Shao , Pu Wu , Huiqin Jiang , S.M. Sheikholeslami and Shaohui Wang

Published/Copyright: March 10, 2020

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From the journal Open Chemistry Volume 18 Issue 1

Abstract

For a simple graph G, the atom–bond connectivity index (ABC) of G is defined as ABC(G) = ∑uv∈E(G)d(u)+d(v)−2d(u)d(v),where d(v) denotes the degree of vertex v of G. In this paper, we prove that for any bipartite graph G of order n ≥ 6, size 2n − 3 with δ(G) ≥ 2, ABC(G)≤2(n−6)+23(n−2)n−3+2,and we characterize all extreme bipartite graphs.

Keywords: degree (of vertex); atom–bond connectivity index; bipartite graph; extremal graph; incidence energy

1 Introduction

Let G be a simple connected graph with vertex set V = V(G) and edge set E = E(G), in which a simple graph contains no loops and duplicate edges between two vertices. The order |V| of G is denoted by n = n (G), and the size |E| of G is denoted by m = m(G). For every vertex v ∈ V, the open neighborhood N(v) is the set {u ∈ V(G) | uv ∈ E(G)}. The degree of a vertex v ∈ V is d_G(v) = d(v) = |N(v)|. The minimum degree of a graph G are denoted by δ = δ(G). A leaf or a pendant vertex of a graph G is a vertex of degree 1. For a subset S of vertices of G, we denote by G[S] the subgraph induced by S. A bipartite graph is a graph having no odd cycles.

The connectivity index, χ, is a topological index introduced in 1975 by Milan Randić [1] who has shown this index to reflect molecular branching. However, many physico-chemical properties are dependent on factors rather different than branching. In order to take this into account along with keeping the spirit of the Randić index, Ernesto Estrada et al. [2] proposed a new topological index, named the atom-bond connectivity (ABC) index. Estrada [3] proved that the ABC index provides a good model for the stability of linear and branched alkanes as well as the strain energy of cycloalkanes. For those who may not be familiar with alkanes, these are any of the series of saturated hydrocarbons having the general formula C_nH_2n₊2, including methane, ethane and propane. Additional physico-chemical applicabilities of the ABC index were presented in a few other works, including [3, 4, 5]. The atom-bond connectivity index of G is defined as

(1)ABC(G)=∑uv∈E(G)d(u)+d(v)−2d(u)d(v).

Due to its physico-chemical applicability, the ABC index has attracted significant attention from researchers in recent years, and many mathematical properties of this index have been reported [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15].

When examining a topological index, one of the fundamental questions that needs to be answered is for which graphs this index assumes minimal and maximal values. In addition, the identities of these extremal values must be uncovered. The minimal value of the ABC index has been investigated by several authors [16, 17, 18, 19, 20, 21, 22], and the maximal value of the ABC index has been studied extensively, as well [23, 24, 25, 26]. Recently, Shao et al. [15, 27] showed that for any connected graph G of order √n, size 2n − 4 ≥ 4 with δ(G)≥2,ABC(G)≤(n−2)2characterizes all extremal graphs. They also proved that for any connected graph G of order n, size 2n − 3 ≥ 5 with δ(G) ≥ 2, characterizes all extremal (n, 2n − 2) graphs without pendent vertices and with maximum ABC indices.

In this paper, we focus on bipartite graphs and show that for any bipartite graph G of order n, size 2n − 3 with δ(G)≥2,ABC(G)≤2(n−6)+23(n−2)n−3+2.We also classify all extreme bipartite graphs.

We make use of the following results and notations in this paper. To simplify notations, we define the following functions:

f(x,y)=x+y−2xy,
h(x,y)=f(x,y)−f(x+1,y).

Lemma 1

([28]) If (x, y) is strictly decreasing for x if y ≥ 3.

Lemma 2

([15]) If G is a graph with δ(G) ≥ 2 and xy ∈ E(G), thenf(x,y)≤22with equality if and only if d(x) = 2 or d(y) = 2 (or both).

The proof of the next result is straightforward and therefore omitted.

Lemma 3

([28]) For x, y ≥ 2, ifhx(x,y)=∂f(x,y)∂xandhy(x,y)=∂f(x,y)∂y,thenhx(x,y)<0.

Similarly, we can see that h_y(x, y) > 0.

Let K_3,t be the complete bipartite graph with bipartite sets X = {x₁, x₂, x₃} and Y = {y₁, . . . , y_t}. For s ≥ 6, assume Hs3,tis the bipartite graph obtained from K_3,t by adding s vertices v₁, . . . , v_s and joining them to x₁ and x₃ (see Figure 1). If s = 0, we define Hs3,tto be the graph K_3,t. The proof of next result is easy to verify by direct calculation.

$Figure 1 (a) the graph H33,3;$H_{3}^{3,3};$(b) the graph Hs3,3.$H_{s}^{3,3}.$$

Figure 1

(a) the graph H33,3;(b) the graph Hs3,3.

Observation 4

If n ≥ 6, thenABC(Hn−63,3)=2(n−6)+23(n−2)n−3+2.

In the rest of paper, we employ the following notation defined in [15].

(2)ViG = {v∈V(G)|d(v)=i},

(3)V3+G = {v ∈V(G)|d(v)≥3}and pk =|V3+G|,

(4)ES,tG ={uv ∈ E(G)|d(u) = s and d(v)=t},

(5)ES,t+G ={uv ∈ E(G)|d(u) = s and d(v)≥t},

(6)ES+,t+G ={uv ∈ E(G)|d(u) ≥ s and d(v) ≥ t}

and

(7)EVG ={uv|uv ∈ E(G)}.

In the following,we will omit the superscript where no confusion can arise.

For an edge e = uv ∈ E(G), we define ABC(e|G) = d(u)+d(v)−2d(u)d(v).When no confusion can arise, we simplify ABC(e|G) to ABC(e).

2 Main results

In this section, we present an upper bound on the ABC index of a bipartite graph and characterize all extreme bipartite graphs. For any bipartite graph G, we let (X_G, Y_G) denote its bipartition.

Theorem 5

Let G be a bipartite graph of order n ≥ 6, size 2n−3 with δ(G) ≥ 2. Then ABC(G) ≤ 2(n−6)+23(n−2)n−3+2with equality if and only if G≅ Hn−63,3.

Proof: We first perform the following steps:

Step 1: Using the software geng in package nauty [29], we generate the set of all graphs of order n ∈ {6, 7, 8, 9}, size 2n − 3 with δ(G) ≥ 2.

Step 2: For each graph G in the obtained graphs, we compute ABC(G) according to formula (1). It turns out that the result holds for each n ∈ {6, 7, 8, 9}.

Now, assume that n ≥ 10. Let G_n be the family of graphs G of order n, size 2n − 3, minimum degree δ(G) ≥ 2, the maximum ABC index and different from Hn−63,3.We will show that G_n = ∅. Suppose, to the contrary, that G_n ≠∅ for some n ≥ 10.We further assume that n is as small as possible such that G_n ≠∅. Let G ∈ G_n. We proceed with establishing several claims.

Claim 1

k ≥ 6, |E_2,2| = 0 and 2k−3 = |E3+,3+|.

Proof of Claim 1

Since every edge is incident with two vertices, an edge contributes 2 to the sum of the degrees of the vertices and so

(8)∑v∈V2d(v)+∑v∈V3+d(v)=2m(G)=4n−6.

Since |V3+|=k,we have

(9)∑v∈V2d(v)=2|V2|=2(n−k).

Combining (8) and (9), we obtain

(10)∑v∈V3+d(v)=2n+2k−6.

Since

(11)∑v∈V3+d(v)=2|E3+,3+|+|E2,3+|

and m(G)=2n−3|E2,2|+|E2,3+|+|E3+,3+|,we conclude from (10) that

(12)|E3+,3+|−|E2,2|=2k−3,

and so

(13)|E3+,3+|≥2k−3.

Suppose x=|XG∩V3+|.Then k−x=|YG∩V3+|and we have

(14)|E3+,3+|≤x(k−x)≤k24.

By (13) and (14), we have k² − 8k + 12 ≥ 0, and this implies that k ≤ 2or k ≥ 6. If k = 0, then (11) and (12) imply that |E3+,3+| = |E2,3+| = 0and |E_2,2| = 3 yielding G ≅ C₃, which is a contradiction. If k = 1, then we deduce from(14) and (12) that |E3+,3+|=0and |E_2,2| = 1. Let E_2,2 = {uv} and V3+ = {w}Since d(u) = d(v) = 2 and |E_2,2| = 1,we must have uw ∈ E, vw ∈ E, which leads to a contradiction. If k = 2, then (14) and (12) imply that |E3+,3+|=1and |E_2,2| = 0. It follows that any vertex of V₂ along with two vertices in V₃⁺ forms a triangular in G, which is a contradiction. Therefore,

(15)k ≥6.

Now,we show that |E_2,2| = 0. Suppose, to the contrary, that |E_2,2| ≥ 1 and let uv ∈ E_2,2. Since k ≥ 6, it follows from (12) that there exists an edge ts ∈E3+,3+such that {s, t} ∩ {u, v} = ∅. Assume, with out loss of generality, that u, t ∈ X and v, s ∈ Y. We consider two cases.

Case 1

u ∉N(s) and v ∉N(t).

Let G^′ be the graph obtained from G by removing two edges uv, st and adding the edges us, tv. Clearly, G^′ is a bipartite graph of order n, size 2n − 3 with δ(G^′)≥ 2. By Lemma 2, we have

ABC(G′)−ABC(G)=ABC(us|G′)+ABC(tv|G′)−ABC(uv|G)−ABC(ts|G)=d(u|G′)+d(s|G′)−2d(u|G′)d(s|G′)+d(t|G′)+d(v|G′)−2d(t|G′)d(v|G′)−d(u|G)+d(v|G)−2d(u|G)d(v|G)−d(t|G)+d(s|G)−2d(t|G)d(s|G)≥22+22−22−23>0,

which is a contradiction.

Case 2

u ∈ N(s) (the case v ∈ N(t) is similar).

We distinguish two subcases.

Subcase 2.1

V₂∩ N(v) ≠∅.

Let w ∈ V₂∩ N(v). Since k ≥ 6, |E3+,3+|≥ 2k − 3 and |E_s ∩E3+,3+|≤ k − 1, there exists an edge pq ∈E3+,3+such that s ∉{p, q}. Assume without loss of generality that p ∈ X_G and q ∈ Y_G.We deduce from d(u) = d(v) = 2, N(u) = {v, s} and N(v) = {u, w} that u ∉ N(q) and p ∉N(v). Now, as Case 1, we obtain a contradiction.

Subcase 2.2

V₂∩ N(v) = ∅.

Assume that w ∈V3+∩ N(v). Since k ≥ 6, |E3+,3+|≥ 2k − 3 and |E_s ∩E3+,3+| + |E_w ∩E3+,3+| ≤ k, we conclude that there exists an edge pq ∈E3+,3+such that {p, q} ∩ {s, w} = ∅. Assume without loss of generality that p ∈ X_G and q ∈ Y_G. Since d(u) = d(v) = 2, N(u) = {v, s} and N(v) = {u, w}, we have u ∉N(q) and v ∉N(p). As in Case 1, we obtain a contradiction again.

Thus |E_2,2| = 0 and by (12) we have |E3+ ,3+| = 2k−3.This completes the proof of Claim 1.

Claim 2

|E_2,3| = 0.

Proof of Claim 2

Suppose, to the contrary, that E_2,3 ≠ ∅ and let uv ∈ E_2,3. Assume without loss of generality that d(v) = 2 and d(u) = 3. Since E_2,2 = ∅, the other neighbor of v, say t, is in V3+.We distinguish the following cases.

(2.a). N(u) ∩V3+≠∅.

Let w ∈ N(u) ∩V3+and let G^′ = G − v. Clearly, G^′ is a bipartite graph of order n(G^′) = n − 1, size m(G^′) = (2n − 3) − 2 = 2(n − 1) − 3 and δ(G^′) ≥ 2. By the choice of n, we have ABC(G′)≤2(n−7)+23(n−3)n−4+2.We also have ABC(uw|G′)=22,ABC(uv|G)=ABC(vt|G)=22and ABC(uw|G)=d(w)+13d(w).Using inequality

23(n−3)n−4−23(n−2)n−3<22−23,

we obtain

ABC(G)≤ABC(G′)−ABC(uw|G′) + ABC(uw|G)+ABC(uv|G)+ABC(vt|G) ≤ 2(n−7)+23(n−3)n−4+2−22+23+22 + 22 = 2(n−6)+2+23(n−3)n−4−22+23 < 2(n−6)+2+23(n−2)n−3=ABC(Hn−63,3),

which is a contradiction.

(2.b). N(u) ∩V3+≠∅.

Since |E3+,3+| = 2k −3 ≥9 ,there is an edge sh ∈E3+,3+.Assume without loss of generality that u, s ∈ Y and v, h ∈ X. First let s ∉N(v). Suppose G^′ is the graph obtained from G by removing the edges vu, sh and adding the edges vs and uh. Clearly, G^′ is bipartite of order n, size 2n − 3 and δ(G^′)≥ 2. Since ABC(uh|G^′)≥ ABC(sh|G) and ABC(vs|G^′) = ABC(vu|G), we have ABC(G^′) ≥ ABC(G) yielding ABC(G^′) = ABC(G). Hence, G^′ ∈ G_n. By Claim 1, we have k ≥ 6 , |E2,2G′| =0 and 2k −3 = |E3+,3+G′|. If w ∈ N_G(u) − {v}, then by assumption we have uw ∈ E2,3G′ ∩ EuG′.Considering the edge uw ∈ G^′, we are in situation (a) and we obtain a contradiction.

Now let s ∈ N(v). Using an argument similar to that described in Case 2 of Claim 1, we deduce that there exists an edge pq ∈E3+,3+such that s ∉{p, q}. Using the proof of Case (b), we obtain a contradiction again.

This completes the proof of Claim 2.

Claim 3

There exists a bipartite graph G^′ = (X_G′ , Y_G′) of order n, size 2n − 3 and δ(G^′)≥ 2, which satisfies the following conditions:

ABC(G^′)≥ ABC(G),
|∪u∈V2G′∩XG′N(u)|=2 or V2G′∩XG′=∅,
|∪u∈V2G′∩YG′N(u)|=2 or V2G′∩YG′=∅.

Proof of Claim 3

For any bipartite graph G₁ of order n, size 2n − 3 and δ(G₁) ≥ 2, we define and L(G₁) = |A(G₁)|. Let H be the family of bipartite graphs H≠Hn−63,3such that |V(H)| = n, m(H) = 2n−3, δ(H) ≥ 2andABC(H) ≥ ABC(G). Since G ∈ H, we establish that H ≠ ∅. Choose a graph G^′ ∈ H such that L(G^′) is as small as possible. Clearly, G^′ satisfies in the condition (i). Then G^′ ∈ G_n and so G^′ satisfies Claims 1 and 2. If G^′ satisfies in the conditions (ii) and (iii) of Claim 3, then we are done. Assume that G^′ does not satisfy in the condition (ii) or (iii). Assume without loss of generality that G^′ does not satisfy condition (ii). Then |V2G′∩XG′|∉{0,1}.It follows that |∪u∈V2G′∩XG′N(u)|≥3.Then there are two vertices u1,u2∈V2G′∩XG′such that N_G (u₁) ≠ N_G^′ (u₂). Let N_G′ (u₁) = {v₁, v₂}. Since E2,2G′=∅,we have d(v₁|G^′) ≥ 3 and d(v₂|G^′) ≥ 3. We consider the following cases.

Case A

|N_G′ (u₁) ∩ N_G′ (u₂)| = 1.

Suppose without loss of generality that N_G′ (u₂) = {v₂, v₃} and that d(v₁|G^′)≤ d(v₃|G^′). Let

U1={v∈V2G′∩XG′|NG′(v)={v1,v2}} ={a1(=u1),…,ak1},

U2={v∈V3+G′|v3∈NG′(v)and v1∉NG′(v)} ={b1,b2,…,bk2},

and |U₁| = k₁, |U₂| = k₂. Clearly, k₁ ≥ 1, k₂ ≥ 0. We distinguish the following sub-cases.

Subcase A.1

k₂ ≥ k₁.

Let G_s be the graph obtained from G^′ by removing the edges v₁a_i , v₃b_i(1 ≤ i ≤ k₁) and adding the edges v₃a_i , v₁b_i(1 ≤ i ≤ k₁). Clearly, G_s is bipartite, and ABC(a_iv₁|G^′) = ABC(a_iv₃|G_s) for 1 ≤ i ≤ k₁. Since d(v₁|G^′)≤ d(v₃|G^′), we have d(v₁|G_s) ≤ d(v₃|G_s), and this implies that ABC(b_iv₃|G^′)≤ ABC(b_iv₁|G_s) for 1 ≤ i ≤ k₁. Hence ABC(G_s) ≥ ABC(G^′). By definition of Hn−63,3,we have Gs≠Hn−63,3and so G_s ∈ H. This contradicts the choice of G^′ since L(G_s) = L(G^′)− 1.

Subcase A.2

k₂ = 0 and d(v₁|G^′)≤ k₁ + 1.

In this case we have 1 ≤ d(v₁|G^′)− 2 ≤ k₁ − 1. Let G_s be the graph obtained from G^′ by removing the edges v₁a_i(1 ≤ i ≤ d(v₁|G^′)− 2) and adding the edges v₃a_i(1 ≤ i ≤ d(v₁|G^′)− 2). Clearly, G_s is a bipartite graph of order n,size 2n−3withδGs≥2andGs≠Hn−63,3because deg_Gs (a_k1) = deg_Gs (u₂) = 2 and NG′ak1∩NG′u2=1.Since d(v₁|G^′)≤ k₁ + 1, we have |NG′(v3)∩V3+G′|≤1.

h1=f(2,d(c))−f(d(v1|G′),d(c)) =∑i=2d(v1|G′)−1((fi,d(c))−f(i+1,d(c)))

and

h2=f(d(v3|G′),d(c))−f(d(v3|G′)+d(v1|G′)−2,d(c)) =∑i=d(v3|G′)d(v3|G′)+d(v1|G′)−3(f(i,d(c))−f(i+1,d(c))).

Thus,

By Lemma 1, we have h₁ − h₂ > 0 and so ABC(G_s) − ABC(G^′)> 0. Hence ABC(G_s) > ABC(G^′), which is a contradiction.

Subcase A.3

k₂ = 0 and d(v₁|G^′)≥ k₁ + 2.

Let G_s be the graph obtained from G^′ by removing the edges v₁a_i(1 ≤ i ≤ k₁) and adding the edges v₃a_i(1 ≤ i ≤ k₁). Clearly, G_s is a bipartite graph of order n, size 2n − 3 with δ(G_s) ≥ 2.

First let NG′(v3)∩V3+G′=∅.Then d(v₁|G_s) < d(v₁|G^′), ABC(a_iv₁|G^′) = ABC(a_iv₃|G_s) for each 1 ≤ i ≤ k₁, and ABC(e|Gs)=ABC(e|G′)=22for any e∈Ev3G′.This implies that ABC(G_s) ≥ ABC(G^′). Because L(G_s) = L(G^′)− |{{v₁, v₂}}| = L(G^′)− 1, we obtain a contradiction with the choice of G^′.

Now let NG′(v3)∩V3+G′≠∅and let NG′(v3)∩V3+G′={c1,c2,…,ct}. Since d(c_i|G^′) = d(c_i|G_s) for i ∈ {1, 2, . . . , t}, we write d(c_i) = d(c_i|G^′). For i ∈ {1, 2, · · · , t}, let

and

Assume that l = ABC(G_s) − ABC(G^′). We have

We conclude fromLemma1 and the fact d(v₃|G^′)−d(v₁|G^′)+ k₁ ≥ 1 that l = ABC(G_s) − ABC(G^′) > 0, which leads to a contradiction as above.

Subcase A.4

0 < k₂ < k₁.

Let G_s be the graph obtained from G^′ by removing the edges v₁a_i , v₃b_i(1 ≤ i ≤ k₂) and adding the edges v₁b_i , v₃a_i(1 ≤ i ≤ k₂). Clearly, G_s is a bipartite graph of order n, size 2n−3 with δ(G_s) ≥ 2. Note that d(v|G_s) = d(v|G^′) for any v ∈ G_s and ABC(b_iv₁|G_s) ≥ ABC(b_iv₃|G^′) for any i = 1,2, . . . , k₂. It is not hard to verify that ABC(G_s) ≥ ABC(G^′) and L(G_s) ≤ L(G^′). Considering G_s instead of G^′ and applying the argument similar to that described in Subcases A.2 and A.3, we obtain a contradiction.

Case B

|N_G^′ (u₁) ∩ N_G^′ (u₂)| = 0.

Let NG′ (u₂) = {v₃, v₄} and assume without loss of generality that d(v₂|G^′)≤ d(v₄|G^′). Define

W1=v∈V2G′NG′v=v1,v2=a1=u1,…,aq1,W2=v∈V3+G′v4∈NG′vandv2∉NG′v=b1,b2,…,bq2,

q₁ = |W₁| and q₂ = |W₂|.

Obviously, q₁ ≥ 1 and q₂ ≥ 0. Now, we consider the following subcases.

Subcase B.1

q₂ ≥ q₁.

Let G_s be the graph obtained from G^′ by removing the edges v₂a_i , v₄b_i(1 ≤ i ≤ q₁) and adding the edges v₄a_i , v₂b_i(1 ≤ i ≤ q₁). Clearly, G_s is a bipartite graph of order n, size 2n − 3 with δ(G_s) ≥ 2 and ABC(a_iv₂|G^_′) = ABC(a_iv₄|G_s) for each i ∈ {1, 2, . . . , q₁}. Since d(v₂|G^′)≤ d(v₄|G^′),we have d(v₂|G_s) ≤ d(v₄|G_s) and this implies that ABC(b_iv₄|G^_′)≤ ABC(b_iv₂|G_s) for any i ∈ {1, 2, . . . , q₁}.

Thus ABC(G_s) ≥ ABC(G^′). Clearly, L(G_s) ≤ L(G^′) and we deduce from the choice of G^′ that L(G_s) = L(G^′). By definition, we have Gs≠Hn−63,3.Thus G_s ∈ H and we obtain a contradiction as in Case A.

Subcase B.2

q₂ = 0 and d(v₂|G^′)≤ q₁ + 1.

Now let |NG′(v4)∩V3+G′|=1and let NG′(v4)∩V3+G′={c}.Clearly, we have ABC(a_iv₂|G^′) = ABC(a_iv₄|G_s) for each 1 ≤ i ≤ d(v₂|G^′)− 2 and d(v₂|G_s) = 2. Since d(c|G^′) = d(c|G_s), we write d(c) = d(c|G^′). Assume that

and l = ABC(G_s) − ABC(G^′). Then we have

We deduce from Lemma 1 that ABC(G_s)−ABC(G^′)> 0. This implies that ABC(G_s) > ABC(G^′), which is a contradiction.

Subcase B.3

q₂ = 0 and d(v₂|G^′)≥ q₁ + 2.

Let G_s be the graph obtained from G^′ by removing the edges v₂a_i(1 ≤ i ≤ q₁) and adding the edges v₄a_i(1 ≤ i ≤ q₁). Clearly, G_s is a bipartite graph of order n, size 2n − 3 with δ(Gs)≥2 and Gs≠Hn−63,3.

First let NG′(v4)∩V3+G′=∅.Then we have ABC(a_iv₂|G^′) = ABC(a_iv₄|G_s) for any 1 ≤ i ≤ q₁, d(v₂|G_s) < d(v₂|G^′), and ABC(e|Gs)=ABC(e|G′)=22for any edge e∈Ev4G′.Hence ABC(G_s) ≥ ABC(G^′) and L(G_s) ≤ L(G^′). We deduce from the choice of G^′ that L(G_s) = L(G^′). Now, we consider G_s instead of G^′ and proceed as Case A to obtain a contradiction.

Now let NG′(v4)∩V3+G′≠∅and let NG′(v4)∩V3+G′={c1,c2,…,ct}. Since d(c_i|G^′) = d(c_i|G_s) for each i ∈ {1, 2, . . . , t}, we write d(c_i) = d(c_i|G^′). Let

and l = ABC(G_s) − ABC(G^′). Then we have

It follows from Lemma 1 and the fact d(v₄|G^′) − d(v₂|G^′) + k₁ ≥ 1 that ABC(G_s) − ABC(G^′) > 0. Thus ABC(G_s) > ABC(G^′), which is a contradiction.

Subcase B.4

0 < q₂ < q₁.

Let G_s be the graph obtained from G^′ by removing the edges v₂a_i , v₄b_i(1 ≤ i ≤ q₂) and adding the edges v₂b_i , v₄a_i(1 ≤ i ≤ q₂). Clearly, G_s is a bipartite graph of order n, size 2n − 3 with δ(Gs)≥2 and Gs≠Hn−63,3.It is easy to verify that ABC(G_s) ≥ ABC(G^′) and L(G_s) ≤ L(G^′). Considering G_s instead of G^′ and applying the argument similar to that described in Subcases A.2 and A.3, we obtain a contradiction. This completes the proof of Claim 3.

Let G′≠Hn−63,3be the graph satisfying the conditions of Claim 3. Then G^′ ∈ G_n. By exchanging G and G^′, we may assume that the graph G satisfies the conditions of Claim 3. Let

A_X = {{u, v}| there is a vertex w ∈ Y such that N(w) = {u, v}},

A_Y = {{u, v}| there is a vertex w ∈ X such that N(w) = {u, v}}.

By Claim 3, |A_X| ≤ 1 and |A_Y | ≤ 1. If A_X ≠ ∅ (resp. A_Y ≠ ∅), then let A_X = {{u_x , v_x}} (resp. A_Y = {{u_y , v_y}}).

Claim 4

(a) If AX≠∅,|Eux∩E3+,3+|+|Evx∩E3+,3+|≤5.

(b) If Ay≠∅,|Euy∩E3+,3+|+|Evy∩E3+,3+|≤5.

Proof of Claim 4

We only prove (a). Let A_X ≠ ∅. Suppose, to the contrary, that |Eux∩E3+,3+|+|Evx∩E3+,3+|≥6.Since G is a bipartite graph of size 2n−3,we conclude that d(v) ≤ n − 3 for any v ∈ V(G). Assume, without loss of generality, that |Eux∩E3+,3+|≥1.Then there are two vertices t, s ∈ Y such that tux∈E3+,3+and s ∈ N(u_x) ∩ N(v_x) ∩ V₂. Let E1=Eux∩E3+,3+and E2=Evx∩E3+,3+.Suppose G^′ = G − s. Clearly, G^′ is a bipartite graph of order n − 1, size 2(n − 1) − 3 with δ(G^′)≥ 2. By the choice of n, we have ABC(G′)≤2(n−7)+23(n−3)n−4+2.Also, we have

ABC(tux|G′)−ABC(tux|G)=f(d(ux)−1,d(t))−f(d(ux),d(t))=h(d(ux)−1,d(t))≥h(n−4,d(t))≥h(n−4,3).

Similarly, for any edge e ∈ E₁∪ E₂, we have ABC(e|G^′)− ABC(e|G) ≥ h(n − 4, 3). Hence

ABC(G′)+2−ABC(G)=∑e∈E1∪E2(ABC(e|G′)−ABC(e|G)) ≥(|E1|+|E2|)h(n−4,3) ≥6h(n−4,3) =ABC(Hn−73,3)+2−ABC(Hn−63,3).

Since ABC(G)≥ABC(Hn−63,3),we have

ABC(Hn−73,3)+2−ABC(Hn−63,3)≥ABC(G′)+2−ABC(G) ≥ABC(Hn−73,3)+2 −ABC(Hn−63,3)

and so the inequalities occurring above become equality, that is, ABC(G)=ABC(Hn−63,3)and ABC(G′)=ABC(Hn−73,3).It follows that G′≅Hn−73,3and ABC(e|G^′)−ABC(e|G) = h(n− 4, 3) for any e ∈ E₁∪ E₂. Hence d(u_x) = d(v_x) = n − 3 and so G≅Hn−63,3,a contradiction.

Claim 5

k ≤ 9.

Proof of Claim 5

Suppose, to the contrary, that k ≥ 10. Let q = |A_X| + |A_Y |. By Claim 4, we have q ∈ {0, 1, 2}. Let B = {u_x , v_x , u_y , v_y} (possibly B = ∅). Then |B| = 2q. We consider the following cases.

Case C.I

There exists a vertex v ∉B such that d(v) ≥ 6.

Then

ABC(G)=∑e∈E2,3+ABC(e)+∑e∈EvABC(e)+∑e∈{E3+,3+∖Ev}ABC(e)≤(n−k)2+d(v)d(v)+13d(v)+(2k−3−d(v))23=(n−k)2+(2k−3)23−d(v)(23−d(v)+13d(v))≤(n−k)2+(2k−3)23−6(23−718)≤(n−10)2+(20−3)23−6(23−718)≤(n−6)2+2+23<(n−6)2+2+23(n−2)n−3=ABC(Hn−63,3),

a contradiction.

Case C.II

There exist two vertices v₁, v₂∉B such that d(v₁) = 5 and d(v₂) ≥ 4.

Let t₁ = |E_v1∩ E_v2|. Then t₁∈ {0, 1} and

h1=t1(3+d(v2)5d(v2)−615−d(v2)+13d(v2)+23) =t1(f(3,3)−f(d(v2),3)−(f(3,5)−f(d(v2),5))) =t1(∑j=1d(v2−1)h(j,3)−∑j=1d(v2−1)h(j,5)) ≤0.

It follows that

ABC(G)=∑e∈E2,3+ABC(e)+∑e∈Ev1ABC(e)+∑e∈Ev2∖Ev1ABC(e)+∑e∈{E3+,3+∖(Ev1∪Ev2)}ABC(e)≤(n−k)2+t13+d(v2)5d(v2)+(5−t1)615+(d(v2)−t1)d(v2)+13d(v2)+(2k−8−d(v2)+t1)23=h1+(n−k)2+5615+d(v2)d(v2)+13d(v2)+(2k−8−d(v2))23≤(n−k)2+5615−d(v2)(23−d(v2)+13d(v2))+(2k−8)23≤(n−10)2+5615−4(23−4+112)+(2k−8)23≤(n−6)2+2+23<(n−6)2+2+23(n−2)n−3=ABC(Hn−63,3),

which is a contradiction.

Case C.III

There exist two vertices u, v ∉B such that d(v) = 5 and d(x) = 3 for any x ∈V3+\ {v}.

By Claim 4, we have

∑v∈V3+d(v|G[V3+])=∑v∈V3+\Bd(v|G[V3+])+∑v∈Bd(v|G[V3+]) ≤5+3(k−2q−1)+5q =2+3k−q ≤2+3k.

By Claim 1, we have ∑v∈V3+d(v|G[V3+])=∑v∈V3+d(v)=2(2k−3)≤2+3kyielding k ≤ 8, which is a contradiction.

Case C.IV

There are three vertices v₁, v₂, v₃∉B such that d(v₁) = d(v₂) = d(v₃) = 4.

LetW be the set of edges with two ends in {v₁, v₂, v₃}, t₁ = |W| and W₁ = (E_v1∪ E_v2∪ E_v3 )\ W. Clearly, t₁∈ {0, 1, 2} and we have

ABC(G)=∑e∈E2,3+ABC(e)+∑e∈WABC(e)+∑e∈W1ABC(e)+∑e∈{E3+,3+∖{W∪W1}}ABC(e)≤(n−k)2+t1616+(12−2t1)512+(2k−3−12+t1)23=t1(616−2512+23)+(n−k)2+12512+(2k−15)23≤(n−k)2+12512+(2k−15)23≤(n−10)2+12512+(20−15)23≤(n−6)2+2+23<(n−6)2+2+23(n−2)n−3=ABC(Hn−63,3),

a contradiction.

Case C.V

For any vertex v ∉B, d(v) ≤ 4 and |{v|d(v) = 4} ∩ (V3+\ B)| ≤ 2.

By claim 4, we have

∑v∈V3+d(v|G[V3+])=∑v∈V3+\Bd(v|G[V3+])+∑v∈Bd(v|G[V3+]) ≤8+3(k−2q−2)+5q =2+3k−q ≤2+3k.

Claim 1 implies that ∑v∈V3+d(v|G[V3+])=∑v∈V3+d(v)=2(2k−3)≤2+3kyielding k ≤ 8, a contradiction. This completes the proof of Claim 5.

Now, for any graph G^′′,we define d′(u)={d(u)d(u)≥34d(u)≤2,ABC′(uv)=d′(u)+d′(v)−2d′(u)d′(v) and ABC′(G″)=∑uv∈E(G″)ABC′(uv).

By Claims 1 and 2, |E2,2G|=|E2,3G|=0.We have d(v) ≥ 4 with d(v|G[V3+])≤ 2 for any v ∈V3+.Thus ABC(G) ≤ (n − k) 2+ABC′(G[V3+]).

Claim 6

k = 6.

Proof of Claim 6

Suppose, to the contrary, that k ≥ 7. First let k = 9. Then ABC(G)≤(n−9)2+ABC′(G[V3+]).Since ABC(G)≥ABC(Hn−63,3)>(n−6)2+23+2,we obtain ABC′(G[V3+])>32+23+2.Hence, |V(G[V3+])|=9,|E(G[V3+])|=15 and ABC′(G[V3+])>32+23+2.Acomputer search shows that there are exactly two bipartite graphs G^′′ satisfying the conditions |V(G^′′)| = 9, |E(G^′′)| = 15 and ABC′(G″)>32+23+2(these two graphs G91 and G92illustrated in Figure 2). It is easy to verify that d(u) + d(v) ≥ 6 for every two vertices u,v∈V(G91)or u,v∈V(G92),a contradiction with Claim 4.

Figure 2

Graphs used in the proof of Claim 6.

Now let k = 8. Then ABC(G)≤(n−8)2+ABC′(G[V3+]).Since ABC(G)≥ABC(Hn−63,3)>(n−6)2+23+2,we have ABC′(G[V3+])>22+23+2.Thus G[V3+]satisfies the following |V(G[V3+])|=8,|E(G[V3+])|=13and ABC′(G[V3+])>22+23+2. A computer search shows that there are exactly two bipartite graphs G^′′ satisfying the conditions |V(G^′′)| = 8,|E(G^′′)| = 13 and ABC′(G′′)>22+23+2(these two graphs G91andG92illustrated in Figure 2). If G[V3+]=G81,then it is easy to verify d(u) + d(v) ≥ 6 for every two vertices u,v∈V(G81),a contradiction with Claim 4. If G[V3+]=G82,then let u be the vertex of degree 2, and assume without loss of generality that u ∈ X. Suppose v ∈ X be a vertex of degree 3. By Claim 4, we conclude that G is a graph obtained from G₈² by adding vertices a₁, a₂, . . . , a_n₋₈ and the edges a_iu, a_iv for each i ∈ {1, 2, . . . , n − 8} (See graph F₈ in Figure 3). It is not hard to verify that ABC(G)<ABC(Hn−63,3),a contradiction.

Figure 3

Graphs constructed in the proof of Claim 6.

Finally let k = 7. A computer search shows that there are exactly two bipartite graphs G^′′ satisfying the conditions |V(G^′′)| = 7, |E(G^′′)| = 11. We deduce from Claim 4 that G is a graph obtained from G^′′ by adding new vertices v₁, v₂, . . . , v_n₋₇ and joining then to two vertices u, v ∈ X_G^′′ (see the graph F₇ illustrated in Figure 3). It is easy to verify that ABC(F7)<ABC(Hn−63,3),a contradiction. This completes the proof of Claim 6.

Claim 7

For any graph G_s with |V(G_s)| = n, |E(G_s)| = 2n − 3, |δ(G_s)| ≥ 2, and ABC(G_s) ≥ ABC(G), we have Gs[V3+]≅K3,3and|⋃u∈V2Gs∩YGsNGs(u)|∈{0,2}and |⋃u∈V2Gs∩XGsN(u)|∈{0,2}.

Proof of Claim 7

Note that E2,2Gs=E2,3Gs=∅.Then by Claim 6,we have Gs[V3+]≅K3,3.By above Claims,we only need to show |⋃u∈V2Gs∩YGsNGs(u)|∈{0,2}.Since E2,2Gs=E2,3Gs=∅,we have NGs(u)∩V3+Gs≠∅for any u ∈ V(G). Since Gs[V3+]≅K3,3,we have NGs(u)∩V3+Gs=NGs(v)∩V3+Gs≠∅for any u,v∈X∩V3+Gs.Then, by a close look at the proof of Case A in Claim 3, we have |⋃u∈V2Gs∩YGsNGs(u)|∈{0,2}.

Let t=|⋃u∈V2∩YN(u)|+|⋃u∈V2∩XN(u)|.We conclude from Claim 7 and the fact n ≥ 10 that t ∈ {2, 4}. If t = 2, then G≅ABC(Hn−63,3),a contradiction. Let t = 4. Assume that {ux,vx}=⋃u∈V2∩YN(u),{uy,vy}=⋃u∈V2∩XN(u),U₁ = {v|N(v) = {u_x , v_x}}, U₂ = {v|N(v) = {u_y , v_y}}, {u₁} = X − ({u_x , v_x} ∪ V₂), {u₂} = Y − ({u_y , v_y} ∪ V₂) and d(u_y) ≤ d(u_x). Let t ∈ U₂, and let G^′ be the bipartite graph obtained from G−{tu_y , tv_y} by adding the edges tu_x and tv_x. Let h₁ = 4ABC(u_xu_y|G^′)− 4ABC(u_xu_y|G) and

h2=2ABC(u1uy|G′)−2ABC(u1uy|G)+2ABC(u2ux|G′)−2ABC(u2ux|G).

Then h1=4d(ux)+d(uy)−2(d(ux)+1)(d(uy)−1)−d(ux)+d(uy)−2d(ux)d(uy)and

h2=2ABC(u1uy|G′)−2ABC(u1uy|G)+2ABC(u2ux|G′)−2ABC(u2ux|G)=2(f(d(uy)−1,3)−f(d(uy),3)+f(d(ux),3)−f(d(ux)+1,3))=2(h(d(uy)−1,3)−h(d(ux),3)).

Since d(u_y) ≤ d(u_x), we have h₁ > 0 and h₂ > 0. Thus

a contradiction.

This completes the proof.

By applying the approach used above, we also obtain the following result:

Theorem 6

Let −3 ≤ l ≤ −1 and let G be a bipartite graph of order n ≥ l + 9, size 2n + l and δ(G) ≥ 2. Then ABC(G) ≤ 2(n−9−l)+(l+6)(l+7)3+2(l+6)n−23(n−3)with equality if and only if G≅Hn−l−93,l+6.

3 Discussions

We studied the atom–bond connectivity (ABC) index on bipartite graphs, in which the ABC index has an important application in rationalizing the stability of linear and branched alkanes as well as the strain energy of cycloalkanes. All extremal graphs attained at these extremal values are characterized. Our method is to use monotonic functions and combine them with graph operations. Some special graphs and their values are obtained by computational searches. Our results extend the previous outcomes and deduce all bounds.

Acknowledgement

This work was supported by the National Key R & D Program of China (Grant No. 2019YFA0706402). The authors would like to express their sincere gratitude to the anonymous referees and the editor for many friendly and helpful suggestions, which led to a great deal of improvement of the original manuscript.

Conflict of Interest
Conflict of Interests: Authors declare no conflict of interest.

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Received: 2018-11-04

Accepted: 2019-11-12

Published Online: 2020-03-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/chem-2020-0002

Keywords for this article

degree (of vertex); atom–bond connectivity index; bipartite graph; extremal graph; incidence energy

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