# The Minimum Merrifield–Simmons Index of Unicyclic Graphs with Diameter at Most Four

Authors: Kun Zhao; Shangzhao Li; Shaojun Dai
Journal: Journal of Mathematics
 (2021)

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## Abstract

The Merrifield–Simmons indexiG of a graph G is defined as the number of subsets of the vertex set, in which any two vertices are nonadjacent, i.e., the number of independent vertex sets of G. In this paper, we determine the minimum Merrifield–Simmons index of unicyclic graphs with n vertices and diameter at most four.


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## Body

## 1. Introduction

Numerous topological and chemical indices have been used for analyzing molecular graphs ([1–4]). The Merrifield–Simmons index was introduced by Merrifield and Simmons [5] in 1989. This index is one of the topological indices whose mathematical properties turned out to be applicable to several questions of molecular chemistry. For example, the connections with physico-chemical properties such as boiling point, entropy, or heat of vaporization are well studied.LetG=VG,EG denote a graph whose set of vertices and set of edges are VG and EG, respectively. For any v∈VG, we denote the neighbors of v as NGv, and v=NGv∪v. By n, we denote the number of vertices of G. All graphs considered here are both finite and simple. We denote, respectively, by Sn, Pn, and Cn the star, path, and cycle with n vertices. For other graph-theoretical terminology and notation, we refer to [6].The Merrifield–Simmons indexiG of a graph G is defined as the number of subsets of the vertex set, in which any two vertices are nonadjacent, i.e., the number of independent vertex sets of G. Several papers deal with the Merrifield–Simmons index in several given graph classes. Usually, trees, unicyclic graphs, and certain structures involving pentagonal and hexagonal cycles are of major interest [7–15].In this paper, we investigate the Merrifield–Simmons indexiG of unicyclic graphs with diameter at most four. We will give some results in Sections 3 and 4.


## 2. Some Notations and Lemmas

In this section, we gather notations which are used throughout this paper and give some necessary lemmas which will be used to prove our main results.IfE′⊆EG and W⊆VG, then G−E′ and G−W denote the subgraphs of G obtained by deleting the edges of E′ and the vertices of W, respectively. By fn, we denote the n-th Fibonacci number, where n∈ℕ, fn+fn+1=fn+2, with initial conditions f0=0 and f1=1.Lemma 1 (see [13]).
LetG=VG,EG be a graph.(i)
IfG1,G2,…,Gm, are the components of the graph G, then iG=∏i=1miGi;(ii)
Ifx∈VG, then iG=iG−x+iG−x.Lemma 2 (see [8]).
LetG=VG,EG be a graph. Then, iSn=2n−1+1; iPn=fn+2, for any n∈ℕ; iCn=fn−1+fn+1, for any n≥3.


## 3. The Unicyclic Graphs with Diameter Three

In this section, we study the minimal Merrifield–Simmons index of unicyclic graphs withn vertices and diameter three.LetUn3 be the set of all unicyclic graphs with n vertices and diameter three. According to the length of the unique cycle and the distribution of other vertices, we may classify all the members in Un3. Let Ui be the set of unicyclic graph of the form Gi, i=1,2,3,4. It is easy to see that the graphs from U1, U2, U3, and U4 and three cycles C5, C6, and C7 are all members of the unicyclic graphs with diameter three.LetU1 be the set of unicyclic graphs of the form G1 (as depicted in Figure 1), where a+b+c=n, a,b,andc≥1, and at least two of a,b,andc are greater than 2. Let graph G1∗ be the graph of the form G1, satisfying a,b,c almost equal, that is, a−b≤1, b−c≤1, and c−a≤1.Figure 1
Unicyclic graphs with diameter threeG1.Theorem 1.
The graphG1∗ has the minimum Merrifield–Simmons index among all graphs in U1.Proof.
AssumeG1∈U1 with 1≤a≤b≤c. By Lemmas 1 and 2, we get(1)iG1=iG1−v1+iG1−v1=iP1a−1iP1b−1⋅iSc+iP1c−1+iP1b+c−2=2a−12b−1⋅2c−1+1+2c−1+2b+c−2=2n−3+2a+b−2+2a+c−2+2b+c−2≔f1a,b,c.
Ifb−a≥2, then we have(2)f1a+1,b−1,c−f1a,b,c=2b+c−32a+1−b−1<0.
Obviously,a,b,c have the same status in f1a,b,c; we infer a−b≤1; further, b−c≤1 and c−a≤1. Thus, when a,b,andc are almost equal, graph G1 has the minimal Merrifield–Simmons index.
LetU2 be the set of unicyclic graphs of the form G2 (as depicted in Figure 2), where a+b=n−2, a≥2,andb≥1. Let graph G2∗ be the graph of the form G2, satisfying 0≤a−b≤2.Figure 2
Unicyclic graphs with diameter threeG2.Theorem 2.
The graphG2∗ has the minimum Merrifield–Simmons index among all graphs in U2.Proof.
AssumeG2∈U2. By Lemmas 1 and 2, we get(3)iG2=iG2−v1+iG2−v1=iP2iP1b−1iSa+iP1a−1=32b−1⋅2a−1+1+2a−1=3⋅2n−4+3⋅2b−1+2a−1≔f2a,b.
Ifb−a≥1, then we have(4)f2a+1,b−1−f2a,b=2a−11−3⋅2b−a−1<0.
Ifa−b≥3, then we have(5)f2a−1,b+1−f2a,b=2a−23⋅2b−a+1−1<0.
Therefore, graphG2∗, where 0≤a−b≤2, is the minimum Merrifield–Simmons index among all graphs in U2.
LetU3 be the set of unicyclic graphs of the form G3 (as depicted in Figure 3), where v1 and v2 are two vertices with a−1, b−1 pendent vertices, satisfying a+b=n−2, a,b≥1; one of a and b is at least 2. Let graph G3∗ be the graph of the form G3, satisfying a,b almost equal, that is, a−b≤1.Figure 3
Unicyclic graphs with diameter threeG3.Theorem 3.
The graphG3∗ has the minimum Merrifield–Simmons index among all graphs in U3.Proof.
AssumeG3∈U3 with a≥b≥1. By Lemmas 1 and 2, we get(6)iG3=iG3−v1+iG3−v1=iP1a−1iP2iP1b−1+iP1+iP1b−1iP1=2a−13⋅2b−1+2+2b=3⋅2n−4+2b+2a≔f3a,b.
Ifa−b≥2, then we have(7)f3a−1,b+1−f3a,b=2b1−2a−b−1<0.
Obviously,a,b have the same status in f3a,b; we infer a−b≤1. Thus, when a,b are almost equal, graph G3 has the minimal Merrifield–Simmons index.
LetU4 be the set of unicyclic graphs of the form G4 (as depicted in Figure 4), where v1 and v2 are two vertices with a−1, b−1 pendent vertices, satisfying a+b=n−3, a,b≥1. Let graph G4∗ be the graph of the form G4, satisfying a,b almost equal, that is, a−b≤1.Figure 4
Unicyclic graphs with diameter threeG4.Theorem 4.
The graphG4∗ has the minimum Merrifield–Simmons index among all graphs in U4.Proof.
AssumeG4∈U4 with a≥b≥1. By Lemmas 1 and 2, we get(8)iG4=iG4−v1+iG4−v1=iP1a−1iP1b−1iP3+iP2+iP2iP1b−1=2a−15⋅2b−1+3+3⋅2b−1=5⋅2n−5+3⋅2b−1+3⋅2a−1≔f4a,b.
Ifa−b≥2, then we have(9)f4a−1,b+1−f4a,b=3⋅2b−11−2a−b−1<0.
Obviously,a,b have the same status in f4a,b; we infer a−b≤1. Thus, when a,b are almost equal, graph G4 has the minimal Merrifield–Simmons index.Theorem 5.
The graphG1∗ has the minimum Merrifield–Simmons index among all graphs in Un3 if n≥8.Proof.
We only need to compare the Merrifield–Simmons indices ofGi∗ for i=1,2,3,4.
ForG1∗, we assume that a≤b≤c. Since a,b,c are almost equal and a+b+c=n, then a≥n−2/3. We have(10)iG1∗=2n−3+2a+b−2+2a+c−2+2b+c−2≥2n−3+3⋅22a−2≥2n−3+3⋅22n−10/3≔g1n.
ForG2∗, as 0≤a−b≤2 and a+b=n−2, we have a−2≤b≤a and a≥n−2/2. Thus,(11)iG2∗=3·2n−4+3⋅2b−1+2a−1≥3⋅2n−4+3⋅2a−2−1+2a−1≥3⋅2n−4+7⋅2n−8/2≔g2n.
ForG3∗, as a−b≤1 and a+b=n−2, we have a−1≤b≤a+1 and a≥n−3/2. Thus,(12)iG3∗=3⋅2n−4+2b+2a≥3⋅2n−4+2a−1+2a≥3⋅2n−4+3⋅2n−5/2≔g3n.
ForG4∗, as a−b≤1 and a+b=n−3, we have a−1≤b≤a+1 and a≥n−4/2. Thus,(13)iG4∗=5⋅2n−5+3⋅2b−1+3⋅2a−1≥5⋅2n−5+3⋅2a−1−1+3⋅2a−1≥5⋅2n−5+9⋅2n−8/2≔g4n.
By simple calculation comparison, we obtain the following comparison:g3n−g2n=2n−8/262−7>0,forn≥8. g2n−g4n=2n−4/22n−6/2−1>0,forn≥8. g4n−g1n=2n−8/22n−2/2+9−3⋅2n−4/6>0,forn≥2. Thus, g1n≤g4n≤g2n≤g3n, for n≥8. A direct computation yields iC5=11, iC6=18, and iC7=29. Thus, the graph G1∗ has the minimum Merrifield–Simmons index among all graphs in Un3 if n≥8. From above, we obtain the result.


## 4. The Unicyclic Graphs with Diameter Four

In this section, we study the minimal Merrifield–Simmons index of unicyclic graphs withn vertices and diameter four.LetVn4 be the set of all unicyclic graphs with n vertices and diameter four. According to the length of the unique cycle and the distribution of other vertices, we may classify all the members in Vn4. Let Vi be the set of unicyclic graph of the form Hi, i=1,2,3,4,5,6,7,8,9. It is easy to see that the graphs from V1, V2, V3, V4, V5, V6, V7, V8, and V9 and two cycles C8 and C9 are all members of the unicyclic graphs with diameter four.LetV1 be the set of unicyclic graphs of the form H1 (as depicted in Figure 5), where v1, v4, and v5 are three vertices with c−1, b−1, and a−1 pendent vertices, a+b+c=n−2, a≥2, b≥1, and c≥1. Let graph H1∗ be the graph of the form H1, satisfying 0≤a−b≤1, 0≤a−c≤2, and 0≤b−c≤2.Figure 5
Unicyclic graphs with diameter threeH1.Theorem 6.
The graphH1∗ has the minimum Merrifield–Simmons index among all graphs in V1.Proof.
AssumeH1∈V1. By Lemmas 1 and 2, we get(14)iH1=iH1−v1+iH1−v1=iP2iP1c−1iP1b−1iSa+iP1a−1+iP1b−1iSa=3⋅2c−12b−12a−1+1+2a−1+2b−12a−1+1=3⋅2n−5+3⋅2a+c−2+3⋅2b+c−2+2b+a−2+2b−1≔f1a,b,c.
Ifb−a≥1, then we have(15)f1a+1,b−1,c−f1a,b,c=3⋅2a+c−2−3⋅2b+c−3−2b−2≤3⋅2b−1+c−2−3⋅2b+c−3−2b−2=−2b−2<0.
Ifa−b≥2, then we have(16)f1a−1,b+1,c−f1a,b,c=−3⋅2a+c−3+3⋅2b+c−2+2b−1≤−3⋅2b+2+c−3+3⋅2b+c−2+2b−1=2b−11−3⋅2c−1<0.
These imply that, ifH1 has minimum Merrifield–Simmons index among all graphs in V1, then we infer 0≤a−b≤1.
Ifc−a≥1, then we have(17)f1a+1,b,c−1−f1a,b,c=−3⋅2b+c−3+2b+a−2≤−3⋅2b+a+1−3+2b+a−2=−2a+b−1<0.
Ifa−c≥3, then we have(18)f1a−1,b,c+1−f1a,b,c=3⋅2b+c−2−2b+a−3≤3⋅2b+c−2−2b+c+3−3=−2c+b−2<0.
These imply that, ifH1 has minimum Merrifield–Simmons index among all graphs in V1, then we infer 0≤a−c≤2.
Ifc−b≥1, then we have(19)f1a,b+1,c−1−f1a,b,c=−3⋅2a+c−3+2a+b−2+2b−1≤−3⋅2a+b−2+2a+b−2+2b−1=2b−11−2a<0.
Ifb−c≥3, then we have(20)f1a,b−1,c+1−f1a,b,c=3⋅2a+c−2−2a+b−3−2b−2≤3⋅2a+b−5−2a+b−3−2b−2=−2b−2−2a+b−5<0.
These imply that, ifH1 has minimum Merrifield–Simmons index among all graphs in V1, then we infer 0≤b−c≤2.
LetV2 be the set of unicyclic graphs of the form H2 (as depicted in Figure 6), where v1, v2, v3, and v4 are four vertices with b−1, c−1, d−1, and a−1 pendent vertices, a+b+c+d=n. As the diameter is 4, we infer a≥2 and b≥1, and one value of c and d is at least 2. Let graph H2∗ be the graph of the form H2, satisfying 0≤a−b≤2, 0≤c−b≤1, 0≤c−d≤1, 0≤a−d≤2, 0≤a−c≤1, and 0≤d−b≤1.Figure 6
Unicyclic graphs with diameter threeH2.Theorem 7.
The graphH2∗ has the minimum Merrifield–Simmons index among all graphs in V2.Proof.
AssumeH2∈V2. By Lemmas 1 and 2, we get(21)iH2=iH2−v1+iH2−v1=iP1b−1iSaiP1d−1iSc+iP1c−1+iP1a+c+d−3=2b−12a−1+12d−12c−1+1+2c−1+2a+c+d−3=2n−4+2a+c+d−3+2a+b+d−3+2b+c+d−3+2a+b+c−3+2b+d−2+2b+c−2≔f2a,b,c,d.
Ifb−a≥1, then we have(22)f2a+1,b−1,c,d−f2a,b,c,d=2a+c+d−3−2b+c+d−4−2b+d−3−2b+c−3≤2b−1+c+d−3−2b+c+d−4−2b+d−3−2b+c−3=−2b+d−3−2b+c−3<0.
Ifa−b≥3, then we have(23)f2a−1,b+1,c,d−f2a,b,c,d=−2a+c+d−4+2b+c+d−3+2b+d−2+2b+c−2≤−2b+3+c+d−4+2b+c+d−3+2b+d−2+2b+c−2=2b+c+d−321−c+21−d−3<0.
These imply that, ifH2 has minimum Merrifield–Simmons index among all graphs in V2, then we infer 0≤a−b≤2.
Ifc−b≥2, then we have(24)f2a,b+1,c−1,d−f2a,b,c,d=−2a+c+d−4+2b+a+d−3+2b+d−2≤−2a+b+2+d−4+2b+a+d−3+2b+d−2=2b+d−21−2a−1<0.
Ifb−c≥1, then we have(25)f2a,b−1,c+1,d−f2a,b,c,d=2a+c+d−3−2b+a+d−4−2b+d−3≤2a+b−1+d−3−2b+a+d−4−2b+d−3=−2b+d−3<0.
These imply that, ifH2 has minimum Merrifield–Simmons index among all graphs in V2, then we infer 0≤c−b≤1.
Ifc−d≥2, then we have(26)f2a,b,c−1,d+1−f2a,b,c,d=−2a+b+c−4+2b+a+d−3+2b+d−2−2b+c−3≤−2a+b+d+2−4+2b+a+d−3+2b+d−2−2b+d+2−3=−2a+b+d−3−2b+d−2<0.
Ifd−c≥1, then we have(27)f2a,b,c+1,d−1−f2a,b,c,d=2a+b+c−3−2b+a+d−4−2b+d−3+2b+c−2≤2a+b+d−1−3−2b+a+d−4−2b+d−3+2b+d−1−3=−2b+d−4<0.
These imply that, ifH2 has minimum Merrifield–Simmons index among all graphs in V2, then we infer 0≤c−d≤1.
Ifd−a≥1, then we have(28)f2a+1,b,c,d−1−f2a,b,c,d=−2b+c+d−4+2a+b+c−3−2b+d−3≤−2b+c+a+1−4+2a+b+c−3−2b+a+1−3=−2b+a−2<0.
Ifa−d≥3, then we have(29)f2a−1,b,c,d+1−f2a,b,c,d=2b+c+d−3−2a+b+c−4+2b+d−2≤2b+c+a−3−3−2a+b+c−4+2b+a−3−2=2b+a−51−3⋅2c−1<0.
These imply that, ifH2 has minimum Merrifield–Simmons index among all graphs in V2, then we infer 0≤a−d≤2.
Ifc−a≥1, then we have(30)f2a+1,b,c−1,d−f2a,b,c,d=2a+b+d−3−2b+c+d−4−2b+c−3≤2c−1+b+d−3−2b+c+d−4−2b+c−3=−2b+c−3<0.
Ifa−c≥2, then we have(31)f2a−1,b,c+1,d−f2a,b,c,d=−2a+b+d−4+2b+c+d−3+2b+c−2≤−2c+2+b+d−4+2b+c+d−3+2b+c−2=2b+c−21−2d−1<0.
These imply that, ifH2 has minimum Merrifield–Simmons index among all graphs in V2, then we infer 0≤a−c≤1.
Ifb−d≥1, then we have(32)f2a,b−1,c,d+1−f2a,b,c,d=2a+c+d−3−2a+b+c−4−2b+c−3≤2a+c+b−1−3−2a+b+c−4−2b+c−3=−2b+c−3<0.
Ifd−b≥2, then we have(33)f2a,b+1,c,d−1−f2a,b,c,d=−2a+c+d−4+2a+b+c−3+2b+c−2≤−2a+c+b+2−4+2a+b+c−3+2b+c−2=2b+c−21−2a−1<0.
These imply that, ifH2 has minimum Merrifield–Simmons index among all graphs in V2, then we infer 0≤d−b≤1.
LetV3 be the set of unicyclic graphs of the form H3 (as depicted in Figure 7), where v1, v2, v3, and v4 are four vertices with a−1, b−1, c−1, and d−1 pendent vertices, a+b+c+d=n, a,b,c,andd≥1. As the diameter is 4, a and c, or b and d, are at least 2. Let graph H3∗ be the graph of the form H3, satisfying a,b,c,andd almost equal, that is, ∀x,y∈a,b,c,d, x≠y, and x−y≤1.Figure 7
Unicyclic graphs with diameter threeH3.Theorem 8.
The graphH3∗ has the minimum Merrifield–Simmons index among all graphs in V3.Proof.
AssumeH3∈V3. By Lemmas 1 and 2, we get(34)iH3=iH3−v1+iH3−v1=iP1a−1iSbiSdiP1c−1+iP1b+d−2+iP1b+d−2iSc=2a−12c−12b−1+12d−1+1+2b+d−2+2b+d−22c−1+1=2n−4+2a+c+d−3+2a+b+d−3+2b+c+d−3+2a+b+c−3+2b+d−2+2a+c−2≔f3a,b,c,d.
Ifa−b≥2, then we have(35)f3a−1,b+1,c,d−f3a,b,c,d=−2a+c+d−4+2b+c+d−3−2a+c−3+2b+d−2≤−2b+2+c+d−4+2b+c+d−3−2b+2+c−3+2b+d−2=−2b+d−22c−1−1−2b+c−1<0.
Ifb−a≥2, then we have(36)f3a+1,b−1,c,d−f3a,b,c,d=2a+c+d−3−2b+c+d−4+2a+c−2−2b+d−3≤2a+c+d−3−2a+2+c+d−4+2a+c−2−2a+2+d−3=−2a+c−22d−1−1−2a+d−1<0.
These imply that, ifH3 has minimum Merrifield–Simmons index among all graphs in V3, then we infer a−b≤1. Obviously, as f3a,b,c,d=f3c,b,a,c, we can get c−b≤1. Similarly, we have c−d≤1 and a−d≤1 by f3a,b,c,d=f3a,d,c,b.
Ifa−c≥2, then we have(37)f3a−1,b,c+1,d−f3a,b,c,d=2a+b+d−4+2b+c+d−2−2a+b+d−3−2b+c+d−3=2b+c+d−3−2a+b+d−4=2b+c+d−31−2a−c−1<0.
Ifc−a≥2, then we have(38)f3a+1,b,c−1,d−f3a,b,c,d=2a+b+d−2+2b+c+d−4−2a+b+d−3−2b+c+d−3=2a+b+d−3−2b+c+d−4=2a+b+d−31−2c−a−1<0.
These imply that, ifH3 has minimum Merrifield–Simmons index among all graphs in V3, then we infer a−c≤1.
Ifb−d≥2, then we have(39)f3a,b−1,c,d+1−f3a,b,c,d=2a+b+c−4+2a+c+d−2−2a+b+c−3−2a+c+d−3=2a+c+d−3−2a+b+c−4=2a+c+d−31−2b−d−1<0.
Ifd−b≥2, then we have(40)f3a,b+1,c,d−1−f3a,b,c,d=2a+b+c−2+2a+c+d−4−2a+b+c−3−2a+c+d−3=2a+b+c−3−2a+c+d−4=2a+b+c−31−2d−b−1<0.
These imply that, ifH3 has minimum Merrifield–Simmons index among all graphs in V3, then we infer b−d≤1.
LetV4 be the set of unicyclic graphs of the form H4 (as depicted in Figure 8), where v1, v2, v4, and v5 are four vertices with b−1, c−1, d−1, and a−1 pendent vertices, a+b+c+d=n−1, a≥2, b,c,andd≥1. Let graph H4∗ be the graph of the form H4, satisfying 0≤a−b≤2, 0≤c−b≤1, c−d≤1, 0≤a−d≤1, 0≤a−c≤1, and d−b≤1.Figure 8
Unicyclic graphs with diameter threeH4.Theorem 9.
The graphH4∗ has the minimum Merrifield–Simmons index among all graphs in V4.Proof.
AssumeH4∈V4. By Lemmas 1 and 2, we get(41)iH4=iH4−v1+iH4−v1=iP1b−1iSaiSciSd+iP1c+d−2+iP1a+c+d−2=2b−12a−1+12c−1+12d−1+1+2c+d−2+2a+c+d−2=2n−4+2a+b+d−3+2a+b+c−3+2b+c+d−2+2a+c+d−2+2a+b−2+2b+c−2+2b+d−2+2b−1≔f4a,b,c,d.
Ifa−b≥3, then we have(42)f4a−1,b+1,c,d−f4a,b,c,d=−2a+c+d−3+2b+c+d−2+2b+d−2+2b+c−2+2b−1≤−3⋅2b+c+d−2+2b+d−2+2b+c−2+2b−1=2b−11−2c+d−1+2d−11−2c+2c−11−2d<0.
Ifb−a≥1, then we have(43)f4a+1,b−1,c,d−f4a,b,c,d=2a+c+d−2−2b+c+d−3−2b+d−3−2b+c−3−2b−2≤2b−1+c+d−2−2b+c+d−3−2b+d−3−2b+c−3−2b−2=−2b+d−3−2b+c−3−2b−2<0.
These imply that, ifH4 has minimum Merrifield–Simmons index among all graphs in V4, then we infer 0≤a−b≤2.
Ifc−b≥2, then we have(44)f4a,b+1,c−1,d−f4a,b,c,d=2a+b+d−3−2a+c+d−3+2b+a−2+2b+d−2+2b−1≤2a+b+d−3−2a+b+2+d−3+2b+a−2+2b+d−2+2b−1=2b−11−2a+d−1+2a−11−2d−1+2d−11−2a−1<0.
Ifb−c≥1, then we have(45)f4a,b−1,c+1,d−f4a,b,c,d=−2a+b+d−4+2a+c+d−3−2b+d−3−2b+a−3−2b−2≤−2a+b+d−4+2a+b−1+d−3−2b+d−3−2b+a−3−2b−2=−2b+d−3−2b+a−3−2b−2<0.
These imply that, ifH4 has minimum Merrifield–Simmons index among all graphs in V4, then we infer 0≤c−b≤1.
Ifc−d≥2, then we have(46)f4a,b,c−1,d+1−f4a,b,c,d=2a+b+d−3−2a+b+c−4+2b+d−2−2b+c−3≤2a+b+c−2−3−2a+b+c−4+2b+c−2−2−2b+c−3=−2a+b+c−5−2b+c−4<0.
Ifd−c≥2, then we have(47)f4a,b,c+1,d−1−f4a,b,c,d=−2a+b+d−4+2a+b+c−3−2b+d−3+2b+c−2≤−2a+b+c+2−4+2a+b+c−3−2b+c+2−3+2b+c−2=−2a+b+c−3−2b+c−2<0.
These imply that, ifH4 has minimum Merrifield–Simmons index among all graphs in V4, then we infer c−d≤1.
Ifa−d≥2, then we have(48)f4a−1,b,c,d+1−f4a,b,c,d=−2a+b+c−4+2b+c+d−2+2b+d−2−2b+a−3≤−2d+2+b+c−4+2b+c+d−2+2b+d−2−2b+d+2−3=−2b+d−2<0.
Ifd−a≥1, then we have(49)f4a+1,b,c,d−1−f4a,b,c,d=2a+b+c−3−2b+c+d−3−2b+d−3+2b+a−2≤2a+b+c−3−2b+c+a+1−3−2b+a+1−3+2b+a−2=−2a+b+c−3<0.
These imply that, ifH4 has minimum Merrifield–Simmons index among all graphs in V4, then we infer 0≤a−d≤1. Obviously, as f4a,b,c,d=f4a,b,d,c, we can get 0≤a−c≤1.
Ifd−b≥2, then we have(50)f4a,b+1,c,d−1−f4a,b,c,d=2a+c+b−3−2a+c+d−3+2b+a−2+2b+c−2+2b−1≤2a+c+b−3−2a+c+b+2−3+2b+a−2+2b+c−2+2b−1=2b−11−2c+a−1+2a−11−2c−1+2c−11−2a−1<0.
Ifb−d≥2, then we have(51)f4a,b−1,c,d+1−f4a,b,c,d=−2a+c+b−4+2a+c+d−2−2b+a−3−2b+c−3−2b−2≤−2a+c+b−4+2a+c+b−2−2−2b+a−3−2b+c−3−2b−2=−2b+a−3−2b+c−3−2b−2<0.
These imply that, ifH4 has minimum Merrifield–Simmons index among all graphs in V4, then we infer d−b≤1.
LetV5 be the set of unicyclic graphs of the form H5 (as depicted in Figure 9), where v1, v2, v5, and v6 are four vertices with b−1, c−1, d−1, and a−1 pendent vertices, a+b+c+d=n−2, a≥2, b,c,andd≥1. Let graph H5∗ be the graph of the form H5, satisfying 0≤a−b≤2, −1≤c−b≤2, −1≤d−b≤2, c−d≤1, 0≤a−c≤1, and 0≤a−d≤1.Figure 9
Unicyclic graphs with diameter threeH5.Theorem 10.
The graphH5∗ has the minimum Merrifield–Simmons index among all graphs in V5.Proof.
AssumeH5∈V5. By Lemmas 1 and 2, we get(52)iH5=iH5−v1+iH5−v1=iP1b−1iSaiSciSd+1+iP1c−1iSd+iP2iP1a+c+d−3=2b−12a−1+12c−1+12d+1+2c−12d−1+1+3⋅2a+c+d−3=3⋅2n−6+2a+b+d−2+2a+b+c−2+3⋅2b+c+d−3+3⋅2a+c+d−3+2a+b−2+2b+c−1+2b+d−1+2b−1≔f5a,b,c,d.
Ifa−b≥3, then we have(53)f5a−1,b+1,c,d−f5a,b,c,d=3⋅2b+c+d−3−3·2a+c+d−4+2b+c−1+2b+d−1+2b−1≤3⋅2b+c+d−3−3·2b+3+c+d−4+2b+c−1+2b+d−1+2b−1=2b−11−2c+d−2+2d1−2c+2c1−2d<0.
Ifb−a≥1, then we have(54)f5a+1,b−1,c,d−f5a,b,c,d=−3⋅2b+c+d−4+3⋅2a+c+d−3−2b+c−2−2b+d−2−2b−2≤−3⋅2b+c+d−4+3⋅2b−1+c+d−3−2b+c−2−2b+d−2−2b−2=−2b+c−2−2b+d−2−2b−2<0.
These imply that, ifH5 has minimum Merrifield–Simmons index among all graphs in V5, then we infer 0≤a−b≤2.
Ifc−b≥3, then we have(55)f5a,b+1,c−1,d−f5a,b,c,d=2a+b+d−2−3⋅2a+c+d−4+2b+a−2+2b+d−1+2b−1≤2a+b+d−2−3⋅2a+b+3+d−4+2b+a−2+2b+d−1+2b−1=2b−11−2a+d+2a−11−2d+2+2d1−2a+1<0.
Ifb−c≥2, then we have(56)f5a,b−1,c+1,d−f5a,b,c,d=−2a+b+d−3+3⋅2a+c+d−3−2b+d−2−2b+a−3−2b−2≤−2a+b+d−3+3⋅2a+b−2+d−3−2b+d−2−2b+a−3−2b−2=−2a+b+d−5−2b+d−2−2b+a−3−2b−2<0.
These imply that, ifH5 has minimum Merrifield–Simmons index among all graphs in V5, then we infer −1≤c−b≤2. Obviously, as f5a,b,c,d=f5a,b,d,c, we can get −1≤d−b≤2.
Ifc−d≥2, then we have(57)f5a,b,c−1,d+1−f5a,b,c,d=2a+b+d−2−2a+b+c−3+2b+d−1−2b+c−2≤2a+b+d−2−2a+b+d+2−3+2b+d−1−2b+d+2−2=−2a+b+d−2−2b+d−1<0.
Ifd−c≥2, then we have(58)f5a,b,c+1,d−1−f5a,b,c,d=−2a+b+d−3+2a+b+c−2−2b+d−2+2b+c−1≤−2a+b+c+2−3+2a+b+c−2−2b+c+2−2+2b+c−1=−2a+b+c−2−2b+c−1<0.
These imply that, ifH5 has minimum Merrifield–Simmons index among all graphs in V5, then we infer c−d≤1.
Ifc−a≥1, then we have(59)f5a+1,b,c−1,d−f5a,b,c,d=2a+b+d−2−3⋅2b+c+d−4+2a+b−1−2b+c−2≤2a+b+d−2−3⋅2b+a+1+d−4+2a+b−1−2b+a+1−2=−2a+b+d−3<0.
Ifa−c≥2, then we have(60)f5a−1,b,c+1,d−f5a,b,c,d=−2a+b+d−3+3⋅2b+c+d−3−2a+b−3+2b+c−1≤−2c+2+b+d−3+3⋅2b+c+d−3−2c+2+b−3+2b+c−1=−2b+c+d−3<0.
These imply that, ifH5 has minimum Merrifield–Simmons index among all graphs in V5, then we infer 0≤a−c≤1. Obviously, as f5a,b,c,d=f5a,b,d,c, we can get 0≤a−d≤1.
LetV6 be the set of unicyclic graphs of the form H6 (as depicted in Figure 10), where v1, v2, v3, v4, and v5 are five vertices with a−1, b−1, c−1, d−1, and e−1 pendent vertices, a+b+c+d+e=n, a,b,c,d,ande≥1. As the diameter is 4, from symmetry, we may assume a and c, or a and d, are at least 2. Let graph H6∗ be the graph of the form H6, satisfying a,b,c,d,ande almost equal, that is, ∀x,y∈a,b,c,d,e, x≠y, and x−y≤1.Figure 10
Unicyclic graphs with diameter threeH6.Theorem 11.
The graphH6∗ has the minimum Merrifield–Simmons index among all graphs in V6.Proof.
AssumeH6∈V6. By Lemmas 1 and 2, we get(61)iH6=iH6−v1+iH6−v1=iP1a−1iP1c−1iSbiP1d−1iSe+iP1e−1+iP1b+d−2iSe+iP1b+e−2iP1c−1iSd+iP1d−1=2a−12c−12b−1+12d−12e−1+1+2e−1+2b+d−22e−1+1+2b+e−22c−12d−1+1+2d−1=2n−5+2a+b+c+d−4+2a+b+c+e−4+2a+c+d+e−4+2a+b+d+e−4+2b+c+d+e−4+2a+c+d−3+2a+c+e−3+2a+b+d−3+2b+c+e−3+2b+d+e−3≔f6a,b,c,d,e.
Ifa−b≥2, then we have(62)f6a−1,b+1,c,d,e−f6a,b,c,d,e=−2a+c+d+e−5+2b+c+d+e−4−2a+c+d−4−2a+c+e−4+2b+c+e−3+2b+d+e−3≤−2b+2+c+d+e−5+2b+c+d+e−4−2b+2+c+d−4−2b+2+c+e−4+2b+c+e−3+2b+d+e−3=−3⋅2b+c+e−3+2b+d+e−31−2c−1<0.
Ifb−a≥2, then we have(63)f6a+1,b−1,c,d,e−f6a,b,c,d,e=2a+c+d+e−4−2b+c+d+e−5+2a+c+d−3+2a+c+e−3−2b+c+e−4−2b+d+e−4≤2a+c+d+e−4−2a+2+c+d+e−5+2a+c+d−3+2a+c+e−3−2a+2+c+e−4−2a+2+d+e−4=−2a+c+d+e−4+2a+c+d−3−2a+c+e−3−2a+d+e−2=2a+c+d−31−2e−1−2a+c+e−3−2a+d+e−2<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer a−b≤1.
Ifb−c≥2, then we have(64)f6a,b−1,c+1,d,e−f6a,b,c,d,e=2a+c+d+e−4−2a+b+d+e−5−2a+b+d−4−2b+d+e−4+2a+c+d−3+2a+c+e−3≤2a+b−2+d+e−4−2a+b+d+e−5−2a+b+d−4−2b+d+e−4+2a+b−2+d−3+2a+b−2+e−3=−2b+d+e−4−2a+d+b−5+2a+b+e−51−2d−1<0.
Ifc−b≥2, then we have(65)f6a,b+1,c−1,d,e−f6a,b,c,d,e=−2a+c+d+e−5+2a+b+d+e−4+2a+b+d−3+2b+d+e−3−2a+c+d−4−2a+c+e−4≤−2a+b+2+d+e−5+2a+b+d+e−4+2a+b+d−3+2b+d+e−3−2a+b+2+d−4−2a+b+2+e−4=−2a+b+e−2−2a+b+d−3+2b+e+d−31−2a−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer c−b≤1.
Ifc−d≥2, then we have(66)f6a,b,c−1,d+1,e−f6a,b,c,d,e=−2a+b+c+e−5+2a+b+d+e−4−2a+c+e−4−2b+c+e−4+2a+b+d−3+2b+d+e−3≤−2a+b+d+2+e−5+2a+b+d+e−4−2a+d+2+e−4−2b+d+2+e−4+2a+b+d−3+2b+d+e−3=−2b+d+e−3−2a+d+e−2+2a+b+d−31−2e−1<0.
Ifd−c≥2, then we have(67)f6a,b,c+1,d−1,e−f6a,b,c,d,e=2a+b+c+e−4−2a+b+d+e−5+2a+c+e−3+2b+c+e−3−2a+b+d−4−2b+d+e−4≤2a+b+d−2+e−4−2a+b+d+e−5+2a+d−2+e−3+2b+d−2+e−3−2a+b+d−4−2b+d+e−4=−2b+d+e−5−2a+b+d−4+2a+e+d−51−2b−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer c−d≤1.
Ife−d≥2, then we have(68)f6a,b,c,d+1,e−1−f6a,b,c,d,e=2a+b+c+d−4−2a+b+c+e−5+2a+c+d−3+2a+b+d−3−2a+c+e−4−2b+c+e−4≤2a+b+c+e−2−4−2a+b+c+e−5+2a+c+e−2−3+2a+b+e−2−3−2a+c+e−4−2b+c+e−4=−2b+c+e−4−2a+c+e−5+2a+b+e−51−2c−1<0.
Ifd−e≥2, then we have(69)f6a,b,c,d−1,e+1−f6a,b,c,d,e=−2a+b+c+d−5+2a+b+c+e−4−2a+c+d−4−2a+b+d−4+2a+c+e−3+2b+c+e−3≤−2a+b+c+e+2−5+2a+b+c+e−4−2a+c+e+2−4−2a+b+e+2−4+2a+c+e−3+2b+c+e−3=−2a+c+e−3−2a+b+e−2+2b+c+e−31−2a−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer e−d≤1.
Ife−a≥2, then we have(70)f6a+1,b,c,d,e−1−f6a,b,c,d,e=2a+b+c+d−4−2b+c+d+e−5+2a+c+d−3+2a+b+d−3−2b+c+e−4−2b+d+e−4≤2e−2+b+c+d−4−2b+c+d+e−5+2e−2+c+d−3+2e−2+b+d−3−2b+c+e−4−2b+d+e−4=−2b+c+e−4−2b+d+e−5+2e+c+d−51−2b−1<0.
Ifa−e≥2, then we have(71)f6a−1,b,c,d,e+1−f6a,b,c,d,e=−2a+b+c+d−5+2b+c+d+e−4−2a+c+d−4−2a+b+d−4+2b+c+e−3+2b+d+e−3≤−2e+2+b+c+d−5+2b+c+d+e−4−2e+2+c+d−4−2e+2+b+d−4+2b+c+e−3+2b+d+e−3=−2c+d+e−2−2b+d+e−3+2e+c+b−31−2d−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer e−a≤1.
Ifa−c≥2, then we have(72)f6a−1,b,c+1,d,e−f6a,b,c,d,e=−2a+b+d+e−5+2b+c+d+e−4−2a+b+d−4+2b+c+e−3≤−2c+2+b+d+e−5+2b+c+d+e−4−2c+2+b+d−4+2b+c+e−3=−2b+c+d−2+2b+c+e−31−2d−1<0.
Ifc−a≥2, then we have(73)f6a+1,b,c−1,d,e−f6a,b,c,d,e=2a+b+d+e−4−2b+c+d+e−5+2a+b+d−3−2b+c+e−4≤2a+b+d+e−4−2b+a+2+d+e−5+2a+b+d−3−2b+a+2+e−4=−2a+b+e−2+2a+b+d−31−2e−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer a−c≤1.
Ifb−d≥2, then we have(74)f6a,b−1,c,d+1,e−f6a,b,c,d,e=−2a+b+c+e−5+2a+d+c+e−4+2a+c+d−3−2c+b+e−4≤−2a+d+2+c+e−5+2a+d+c+e−4+2a+c+d−3−2c+d+2+e−4=−2d+c+e−2+2a+c+d−31−2e−1<0.
Ifd−b≥2, then we have(75)f6a,b+1,c,d−1,e−f6a,b,c,d,e=2a+b+c+e−4−2a+d+c+e−5−2a+c+d−4+2c+b+e−3≤2a+b+c+e−4−2a+b+2+c+e−5−2a+c+b+2−4+2c+b+e−3=−2a+b+c−2+2b+c+e−31−2a−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer b−d≤1.
Ifa−d≥2, then we have(76)f6a−1,b,c,d+1,e−f6a,b,c,d,e=−2a+b+c+e−5+2b+d+c+e−4+2b+e+d−3−2c+a+e−4≤−2d+2+b+c+e−5+2b+d+c+e−4+2b+e+d−3−2c+d+2+e−4=−2d+c+e−2+2b+e+d−31−2c−1<0.
Ifd−a≥2, then we have(77)f6a+1,b,c,d−1,e−f6a,b,c,d,e=2a+b+c+e−4−2b+d+c+e−5−2b+e+d−4+2c+a+e−3≤2a+b+c+e−4−2b+a+2+c+e−5−2b+e+a+2−4+2c+a+e−3=−2a+b+e−2+2a+e+c−31−2b−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer a−d≤1.
Ife−b≥2, then we have(78)f6a,b+1,c,d,e−1−f6a,b,c,d,e=2a+b+c+d−4−2a+c+d+e−5+2a+b+d−3−2a+c+e−4≤2a+b+c+d−4−2a+c+d+b+2−5+2a+b+d−3−2a+c+b+2−4=−2b+c+a−2+2a+b+d−31−2c−1<0.
Ifb−e≥2, then we have(79)f6a,b−1,c,d,e+1−f6a,b,c,d,e=−2a+b+c+d−5+2a+c+d+e−4−2a+b+d−4+2a+c+e−3≤−2a+e+2+c+d−5+2a+c+d+e−4−2a+e+2+d−4+2a+c+e−3=−2d+e+a−2+2a+c+e−31−2d−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer e−b≤1.
Ife−c≥2, then we have(80)f6a,b,c+1,d,e−1−f6a,b,c,d,e=2a+b+c+d−4−2a+b+d+e−5+2a+c+d−3−2b+d+e−4≤2a+b+c+d−4−2a+b+d+c+2−5+2a+c+d−3−2b+d+c+2−4=−2b+c+d−2+2a+c+d−31−2b−1<0.
Ifc−e≥2, then we have(81)f6a,b,c−1,d,e+1−f6a,b,c,d,e=−2a+b+c+d−5+2b+a+d+e−4−2a+c+d−4+2e+b+d−3≤−2a+b+e+2+d−5+2b+a+d+e−4−2a+e+2+d−4+2e+b+d−3=−2a+d+e−2+2e+d+b−31−2a−1<0.
These imply that, ifH6 has minimum Merrifield–Simmons index among all graphs in V6, then we infer e−c≤1.
LetV7 be the set of unicyclic graphs of the form H7 (as depicted in Figure 11), where v1, v2, and v3 are three vertices with a−1, b−1, and c−1 pendent vertices, a+b+c=n−3, a,b,andc≥1, and one of a,b,andc is at least 2. Let graph H7∗ be the graph of the form H7, satisfying a,b,andc almost equal, that is, a−b≤1, b−c≤1, and c−a≤1.Figure 11
Unicyclic graphs with diameter threeH7.Theorem 12.
The graphH7∗ has the minimum Merrifield–Simmons index among all graphs in V7.Proof.
AssumeH7∈V7. By Lemmas 1 and 2, we get(82)iH7=iH7−v1+iH7−v1=iP1a−1iP3iP1c−1iSb+iP1b−1iP2+iP1b−1iP2iP1c−1+iP1=2a−15⋅2c−12b−1+1+3⋅2b−1+2b−13⋅2c−1+2=5⋅2n−6+5⋅2a+c−2+3⋅2a+b−2+3⋅2b+c−2+2b≔f7a,b,c.
Ifa−b≥2, then we have(83)f7a−1,b+1,c−f7a,b,c=−5⋅2a+c−3+3⋅2b+c−2+2b≤−5⋅2b+2+c−3+3⋅2b+c−2+2b=2b1−7·2c−2<0.
Ifb−a≥2, then we have(84)f7a+1,b−1,c−f7a,b,c=5⋅2a+c−2−3⋅2b+c−3−2b−1≤5⋅2b−2+c−2−3⋅2b+c−3−2b−1=−2b+c−4−2b−1<0.
These imply that, ifH7 has minimum Merrifield–Simmons index among all graphs in V7, then we infer a−b≤1. Obviously, as f7a,b,c=f7c,b,a, we can get c−b≤1.
Ifa−c≥2, then we have(85)f7a−1,b,c+1−f7a,b,c=−3⋅2b+a−3+3⋅2b+c−2≤−3⋅2b+c+2−3+3⋅2b+c−2=−3⋅2b+c−2<0.
Ifc−a≥2, then we have(86)f7a+1,b,c−1−f7a,b,c=3⋅2b+a−2−3⋅2b+c−3≤3⋅2b+c−2−2−3⋅2b+c−3=−3⋅2b+c−4<0.
These imply that, ifH7 has minimum Merrifield–Simmons index among all graphs in V7, then we infer a−c≤1.
LetV8 be the set of unicyclic graphs of the form H8 (as depicted in Figure 12), where v1, v2, and v3 are three vertices with a−1, b−1, and c−1 pendent vertices, a+b+c=n−4, a,b,andc≥1, and one of a,b,andc is at least 2. Let graph H8∗ be the graph of the form H8, satisfying a,b,andc almost equal, that is, a−b≤1, b−c≤1, and c−a≤1.Figure 12
Unicyclic graphs with diameter threeH8.Theorem 13.
The graphH8∗ has the minimum Merrifield–Simmons index among all graphs in V8.Proof.
AssumeH8∈V8. By Lemmas 1 and 2, we get(87)iH8=iH8−v1+iH8−v1=iP1a−1iP4iP1c−1iSb+iP1b−1iP3+iP1b−1iP3iP1c−1+iP2=2a−18⋅2c−12b−1+1+5⋅2b−1+2b−15⋅2c−1+3=2n−4+2a+c+1+5⋅2a+b−2+5⋅2b+c−2+3⋅2b−1≔f8a,b,c.
Ifa−b≥2, then we have(88)f8a−1,b+1,c−f8a,b,c=−2a+c+5⋅2b+c−2+3⋅2b−1≤−2b+2+c+5⋅2b+c−2+3⋅2b−1=2b−13−11⋅2c−1<0.
Ifb−a≥2, then we have(89)f8a+1,b−1,c−f8a,b,c=2a+c+1−5⋅2b+c−3−3⋅2b−2≤2b−2+c+1−5⋅2b+c−3−3⋅2b−2=−2b+c−3−3⋅2b−2<0.
These imply that, ifH8 has minimum Merrifield–Simmons index among all graphs in V8, then we infer a−b≤1. Obviously, as f8a,b,c=f8c,b,a, we can get c−b≤1.
Ifa−c≥2, then we have(90)f8a−1,b,c+1−f8a,b,c=−5⋅2b+a−3+5⋅2b+c−2≤−5⋅2b+a−3+5⋅2b+a−2−2=−5⋅2b+a−4<0.
Ifc−a≥2, then we have(91)f8a+1,b,c−1−f8a,b,c=5⋅2b+a−2−5⋅2b+c−3≤5⋅2b+a−2−5⋅2b+a+2−3=−5⋅2b+a−2<0.
These imply that, ifH8 has minimum Merrifield–Simmons index among all graphs in V8, then we infer a−c≤1.
LetV9 be the set of unicyclic graphs of the form H9 (as depicted in Figure 13), where v1, v3, and v5 are three vertices with a−1, b−1, and c−1 pendent vertices, a+b+c=n−3, a,b,andc≥1, and one of a,b,andc is at least 2. Let graph H9∗ be the graph of the form H9, satisfying a,b,andc almost equal, that is, a−b≤1, b−c≤1, and c−a≤1.Figure 13
Unicyclic graphs with diameter threeH9.Theorem 14.
The graphH9∗ has the minimum Merrifield–Simmons index among all graphs in V9.Proof.
AssumeH9∈V9. By Lemmas 1 and 2, we get(92)iH9=iH9−v1+iH9−v1=iP1a−1iSc+1iSb+1+iP1b+c+iSciSb+iP1b+c−2=2a−12c+12b+1+2b+c+2c−1+12b−1+1+2b+c−2=2n−3+2a+b−1+2b+c−1+2a+c−1+2a−1+2b−1+2c−1+1≔f9a,b,c.
Ifa−b≥2, then we have(93)f9a−1,b+1,c−f9a,b,c=2b+c−1−2a+c−2−2a−2+2b−1≤2a−2+c−1−2a+c−2−2a−2+2a−2−1=−2a+c−3−2a−3<0.
Obviously,a,b,andc have the same status in f9a,b,c. Thus, if H9 has minimum Merrifield–Simmons index among all graphs in V9, then we infer a−b≤1; further, b−c≤1 and c−a≤1. This is the above desired result.Theorem 15.
The graphH6∗ has the minimum Merrifield–Simmons index among all graphs in Vn4 if n≥29.Proof.
We only need to compare the Merrifield–Simmons indices ofHi∗ for i=1,2,3,4,5,6,7,8,9.
ForH1∗, as 0≤a−b≤1 and 0≤a−c≤2, we have b≤a≤b+1 and c≤a≤c+2. Since a+b+c=n−2, we get a≥n−2/3. Therefore,(94)iH1∗=3⋅2n−5+3⋅2a+c−2+3⋅2b+c−2+2b+a−2+2b−1≥3⋅2n−5+3⋅22a−4+3⋅22a−5+22a−3+2a−2=3⋅2n−5+13⋅22a−5+2a−2≥3⋅2n−5+13⋅22n−19/3+2n−8/3≔g1n.
ForH2∗, as 0≤a−b≤2, 0≤a−d≤2, and 0≤a−c≤1, we have b≤a≤b+2, c≤a≤c+1, and d≤a≤d+2. Since a+b+c+d=n, we get a≥n/4. Therefore,(95)iH2∗=2n−4+2a+c+d−3+2a+b+d−3+2b+c+d−3+2a+b+c−3+2b+d−2+2b+c−2≥2n−4+11⋅23a−8+3⋅22a−6≥2n−4+11⋅23n−32/4+3⋅2n−12/2≔g2n.
ForH3∗, as a,b,c,andd are almost equal, we have b−1≤a≤b+1, c−1≤a≤c+1, and d−1≤a≤d+1. Since a+b+c+d=n, we get a≥n−3/4. Therefore,(96)iH3∗=2n−4+2a+c+d−3+2a+b+d−3+2b+c+d−3+2a+b+c−3+2b+d−2+2a+c−2≥2n−4+7⋅23a−6+3⋅22a−4≥2n−4+7⋅23n−33/4+3⋅2n−11/2≔g3n.
ForH4∗, as 0≤a−b≤2, 0≤a−d≤1, and 0≤a−c≤1, we have b≤a≤b+2, c≤a≤c+1, and d≤a≤d+1. Since a+b+c+d=n−1, we get a≥n−1/4. Therefore,(97)iH4∗=2n−4+2a+b+d−3+2a+b+c−3+2b+c+d−2+2a+c+d−2+2a+b−2+2b+c−2+2b+d−2+2b−1≥2n−4+7⋅23a−6+22a−3+2a−3≥2n−4+7⋅23n−27/4+2n−7/2+2n−13/4≔g4n.
ForH5∗, as 0≤a−b≤2, 0≤a−c≤1, and 0≤a−d≤1, we have b≤a≤b+2, c≤a≤c+1, and d≤a≤d+1. Since a+b+c+d=n−2, we get a≥n−2/4. Therefore,(98)iH5∗=3⋅2n−6+2a+b+d−2+2a+b+c−2+3⋅2b+c+d−3+3⋅2a+c+d−3+2a+b−2+2b+c−1+2b+d−1+2b−1≥3⋅2n−6+23⋅23a−7+3·22a−4+2a−3≥3⋅2n−6+23⋅23n−34/4+3⋅2n−10/2+2n−14/4≔g5n.
ForH6∗, as a,b,c,d,ande are almost equal, we have b−1≤a≤b+1, c−1≤a≤c+1, d−1≤a≤d+1, and e−1≤a≤e+1. Since a+b+c+d+e=n, we get a≥n−4/5. Therefore,(99)iH6∗=2n−5+2a+b+c+d−4+2a+b+c+e−4+2a+c+d+e−4+2a+b+d+e−4+2b+c+d+e−4+2a+c+d−3+2a+c+e−3+2a+b+d−3+2b+c+e−3+2b+d+e−3≥2n−5+9⋅24a−8+23a−3≥2n−5+9⋅24n−56/5+23n−27/5≔g6n.
ForH7∗, as a,b,andc are almost equal, we have b−1≤a≤b+1 and c−1≤a≤c+1. Since a+b+c=n−3, we get a≥n−5/3. Therefore,(100)iH7∗=5⋅2n−6+5⋅2a+c−2+3⋅2a+b−2+3⋅2b+c−2+2b≥5⋅2n−6+19⋅22a−4+2a−1≥5⋅2n−6+19⋅22n−22/3+2n−8/3≔g7n.
ForH8∗, as a,b,andc are almost equal, we have b−1≤a≤b+1 and c−1≤a≤c+1. Since a+b+c=n−4, we get a≥n−6/3. Therefore,(101)iH8∗=2n−4+2a+c+1+5⋅2a+b−2+5⋅2b+c−2+3⋅2b−1≥2n−4+31⋅22a−4+3⋅2a−2≥2n−4+31⋅22n−24/3+3⋅2n−12/3≔g8n.
ForH9∗, as a,b,andc are almost equal, we have b−1≤a≤b+1 and c−1≤a≤c+1. Since a+b+c=n−3, we get a≥n−5/3. Therefore,(102)iH9∗=2n−3+2a+b−1+2b+c−1+2a+c−1+2a−1+2b−1+2c−1+1≥2n−3+5·22a−3+2a+1≥2n−3+5⋅22n−19/3+2n−5/3+1≔g9n.
By simple calculation comparison, we obtain the following comparison:(103)g2n−g3n=23n−33/411⋅21/4−7+3⋅2n−12/21−21/2>3⋅23n−29/4−3⋅2n−14/2=3⋅2n−14/22n−1/4−1>0.
Thus, we getg2n>g3n, for n≥12.(104)g4n−g3n=7⋅23n−33/423/2−1+2n−11/222−3+2n−13/4>7⋅23n−33/4+2n−11/2+2n−13/4>0.
Thus, we getg4n>g3n for n≥13,(105)g3n−g8n=22n−24/37⋅2n−3/12−31+3⋅2n−12/32n−9/6−1.
Ifn≥29, then 7⋅2n−3/12−31>0 and 2n−9/6−1>0. Thus, we get g3n>g8n for n≥29.(106)g9n−g8n=2n−4+22n−24/35⋅25/3−31+2n−12/327/3−3+1>2n−4−22n−12/3+2n−9/3+1=22n−12/32n/3−1+2n−9/3+1>0.
Thus, we getg9n>g8n for n≥12.(107)g7n−g8n=2n−6+22n−24/319⋅22/3−31+2n−12/324/3−3=2n−6−22n−24/3−2n−15/3=2n−15/322n−3/3−2n−9/3−1>0.
Ifn≥12, let 2n−9/3=x≥2, then 22n−3/3−2n−9/3−1=32x2−x−1>0. Thus, we get g7n>g8n for n≥12.(108)g1n−g5n=3⋅2n−6+13⋅22n−19/3+2n−8/3−23⋅23n−34/4−3⋅2n−10/2−2n−14/4=23n−34/43⋅2n+10/4−23+2n−10/213⋅2n−8/6−3+2n−14/42n+10/12−3.
Ifn≥14, then 3⋅2n+10/4−23>0, 13⋅2n−8/6−3>0, and 2n+10/12−3>0. Thus, we get g1n>g5n for n≥14.(109)g5n−g6n=2n−6+23⋅23n−34/4+3⋅2n−10/2+2n−14/4−9⋅24n−56/5−23n−27/5>24n−56/52n+26/5−9+23n−27/523⋅23n−62/20−1.
Ifn≥21, then 2n+26/5−9>0 and 23⋅23n−62/20−1>0. Thus, we get g5n>g6n for n≥21.(110)g8n−g6n=2n−5+31⋅22n−24/3+3⋅2n−12/3−9⋅24n−56/5−23n−27/5>24n−56/52n+31/5−9+23n−27/531⋅2n−39/15−1.
Ifn≥14, then 2n+31/5−9>0 and 31⋅2n−39/15−1>0. Thus, we get g8n>g6n for n≥14.
By above calculation comparison, we getg6n<gin, where i=1,2,3,4,5,7,8,9, for n≥29. A direct computation yields iC8=18, iC9=29. Thus, the graph H6∗ has the minimum Merrifield–Simmons index among all graphs in Vn4 if n≥29.


## 5. Summary and Conclusions

In this paper, we investigate extremal properties of the Merrifield–Simmons index for unicyclic graphs with diameter at most four. There is no doubt that topological indices have been proven useful for analyzing molecules by means of their graph structure. Special analytical results for the Merrifield–Simmons index contribute to a better understanding of molecular topology when using this measure.



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