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Article

Optimization in Symmetric Trees, Unicyclic Graphs, and Bicyclic Graphs with Help of Mappings Using Second Form of Generalized Power-Sum Connectivity Index

by
Muhammad Yasin Khan
1,†,
Gohar Ali
1,*,† and
Ioan-Lucian Popa
2,3,*,†
1
Department of Mathematics, Islamia College Peshawar, Peshawar 25120, Khyber Pakhtunkhwa, Pakistan
2
Department of Computing, Mathematics and Electronics, 1 Decembrie 1918 University of Alba Iulia, 510009 Alba Iulia, Romania
3
Faculty of Mathematics and Computer Science, Transilvania University of Brasov, Iuliu Maniu Street 50, 500091 Brasov, Romania
*
Authors to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2025, 17(1), 122; https://doi.org/10.3390/sym17010122
Submission received: 26 December 2024 / Revised: 11 January 2025 / Accepted: 12 January 2025 / Published: 15 January 2025
(This article belongs to the Special Issue Symmetry and Graph Theory, 2nd Edition)

Abstract

:
The topological index (TI), sometimes referred to as the connectivity index, is a molecular descriptor calculated based on the molecular graph of a chemical compound. Topological indices (TIs) are numeric parameters of a graph used to characterize its topology and are usually graph-invariant. The generalized power-sum connectivity index (GPSCI) for the graph is Ω Y α ( Ω ) = ζ ϱ E ( Ω ) ( d Ω ( ζ ) d Ω ( ζ ) + d Ω ( ϱ ) d Ω ( ϱ ) ) α , while the second form of the GPSCI is defined as Y β ( Ω ) = ζ ϱ E ( Ω ) ( d Ω ( ζ ) d Ω ( ζ ) × d Ω ( ϱ ) d Ω ( ϱ ) ) β . In this paper, we investigate Y β in the family of trees, unicyclic graphs, and bicyclic graphs. We determine optimal graphs in the desired families for Y β using certain mappings. For graphs with maximal values, two mappings are used, namely A and B, while for graphs with minimal values, mapping C and mapping D are considered.

1. Introduction

A graph is a structure formed of edges and vertices. Edges are the connections between vertices with some properties. In symbolic form, we denote a graph by Ω = ( V ( Ω ) , E ( Ω ) ) . If there is no confusion, then a graph is simply denoted by Ω . The number of vertices is termed as the order of the graph, while the total number of edges is the size of the graph. An edge between two vertices, ζ and ϱ , is denoted by ζ ϱ . The total number of edges incident to a vertex ϱ in a graph Ω is referred to as the degree of ϱ , which is denoted by d Ω ( ϱ ) , if the considered graph is Ω . The neighbors of a vertex ϱ in Ω are the vertices connected to ϱ by means of an edge and are denoted by N Ω ( ϱ ) . More generally, we denote the set of all neighbors of a vertex ϱ by N ( ϱ ) . A network is a simple connected graph. A simple graph can be defined as a graph without loops or multiple edges between the same pair of vertices. The topological index (TI) is a transformation defined by T o p : Ω R , where R represents real numbers and Ω is a simple and finite graph. The transformation satisfies the property T o p ( Ω ) = T o p ( H ) if Ω and H are isomorphic graphs. The TI is a numerical value related to the chemical constitution, used to correlate chemical structure with various properties such as chemical reactivity, biological activity, and physical properties. Various tools, such as TIs, have been developed from graph theory and provided to chemists for analytical applications. Cheminformatics is a discipline formed of a combination of mathematics, chemistry, and information science. It investigates Quantitative Structure–Activity relationships (QSARs) and Qualitative Structure–Property relationships (QSPRs), which are applied to study the biological activities and properties of chemical compounds. In QSAR/QSPR studies, the physico-chemical properties and TIs, such as the Zagreb index, Szeged index, Wiener index, Randi´c index and ABC index, are used to predict the bio-activity of various chemical compounds. “In graph theory, the structural formula of a chemical compound represents a molecular graph, where vertices represent atoms and edges represent chemical bonds”. A molecular descriptor is a numerical value that represents the properties of a chemical graph. More specially, molecular descriptors represent the analyzed chemical graph, while topological descriptors describe its physico-chemical properties and structural features. TIs have various applications in the field of nano-biotechnology and QSAR/QSPR studies. The earliest TI was introduced by Wiener [1], which he named the path number, while studying the boiling point of Paraffin. Later, it was named the Wiener index [2].
In the last few decades, the field of optimization has played a key role in various daily life problems. This field is now the most powerful tool and technique for finding solutions to complicated problems. This field has become a major area of interest for researchers due to its ability to solve complex problems effectively. This field is now one of the most modern fields and is of great interest to mathematicians.There are a variety of fields where optimization plays a crucial role, including engineering, logistics, and machine learning. A lot of work is conducted in the stated fields, and some of this work is stated in the following cited article. In [3], the authors considered k-generalized quasi trees and found a closed result for the optimal values regarding the general power-sum connectivity index. In [4], the authors considered optimal values regarding forgotten TIs in terms of graph size. In [5], the authors investigated the molecular inter-connectivity index in polymers. In [6], the author considered trees, bicyclic graphs and unicyclic graphs for optimal values of Zagreb indices. In [7], the authors investigated KBSO indices for optimal values regarding high traffic load and minimal load by improving the physical layout. In [8], the authors investigated variable Zagreb indices, the variable sum-connectivity index, the variable geometric-arithmetic index and the the variable inverse sum indeg index for optimal values and found new bounds. In [9], the authors investigated Kulli–Basava indices in generalized form for quasi trees and found some optimal values and new bounds in k-generalized quasi trees. In [10], the authors investigated the Somber index for optimal values in trees with given parameters, like matching number, diameter, branching number, etc. In [11], the authors investigated exponential Zagreb indices for new bounds in unicyclic graphs, acyclic graphs and general graphs. In [12], the authors investigated the geometric arithmetic index for graphs and found optimal bounds. In [13], the authors investigated uphill Zagreb TIs for famous families of graph and developed exact new bounds. In [14], the authors investigated the variable sum exdeg index for bicyclic graphs with a given matching number. In [15], the authors investigated Zagreb indices for minimum and maximum values in k-apex trees; for further information, we refer the reader to [16,17,18].
In this work, we considered the second form of GPSCI for obtaining optimal graphs in the families of trees, unicyclic graphs and bicyclic graphs using some mappings. The second form of the generalized power-sum connectivity index is given and defined by:
Y β ( Ω ) = ζ ϱ E ( Ω ) ( d Ω ( ζ ) d Ω ( ζ ) × d Ω ( ϱ ) d Ω ( ϱ ) ) β .
In this work, various results were compared using means plots. To better understand and implement the considered mappings, we provided graphs in the desired sections.

2. Mapping for Largest Values of Y β

Mapping A: Let ζ ϱ E ( Ω ) such that d Ω ( ϱ ) 2 and N ( ζ ) = { ϱ , w 1 , w 2 , w 3 , , w t } . Here d ( w 1 ) = d ( w 2 ) = d ( w 3 ) = = d ( w t ) = 1 and Ω = G + { ϱ w 1 , ϱ w 2 , ϱ w 3 , , ϱ w t } { ζ w 1 , ζ w 2 , ζ w 3 , , ζ w t } . Refer to Figure 1 for an illustration.
Lemma 1. 
Let Ω be obtained from Ω by means of mapping A and d Ω ( ϱ ) d Ω ( ζ ) ; then,
Y β ( Ω ) > Y β ( Ω ) .
Proof. 
Let Y = G { ζ , w 1 , w 2 , w 3 , , w t } ; the value of Y β for Ω is expressed as
Y β ( Ω ) = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( w 1 ) d Ω ( w 1 ) ) β + + ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( w t ) d Ω ( w t ) ) β . = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β + + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β , = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( t + 1 ) ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β .
For Ω, the value of Y β is given by
Y β ( Ω ) = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β + ( d Ω ( ζ ) d Ω ( ζ ) × d Ω ( w 1 ) d Ω ( w 1 ) ) β + ( d Ω ( ζ ) d Ω ( ζ ) × d Ω ( w 2 ) d Ω ( w 2 ) ) β + + ( d Ω ( ζ ) d Ω ( ζ ) × d Ω ( w t ) d Ω ( w t ) ) β , = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β + t ( d Ω ( ζ ) d Ω ( ζ ) × 1 1 ) β .
From (2) and (3), it follows that
Y β ( Ω ) Y β ( Ω ) = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( t + 1 ) ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β t ( d Ω ( ζ ) d Ω ( ζ ) × 1 1 ) β , = δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β + 2 ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β + ( t 1 ) ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β t ( d Ω ( ζ ) d Ω ( ζ ) × 1 1 ) β , > 0 .
For the inequality in (4), we consider
δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β > δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( δ ) d Ω ( δ ) ) β , 2 ( d Ω ( ϱ ) d Ω ( ϱ ) ) β × 1 1 ) + ( t 1 ) ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 1 ) β > ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β + t ( d Ω ( ζ ) d Ω ( ζ ) × 1 1 ) β .
It follows that
d Ω ( ϱ ) d Ω ( ϱ ) > d Ω ( ϱ ) d Ω ( ϱ ) , d Ω ( ϱ ) d Ω ( ϱ ) > d Ω ( ζ ) d Ω ( ζ ) , d Ω ( ϱ ) d Ω ( ϱ ) × 1 > d Ω ( ϱ ) d Ω ( ϱ ) , d Ω ( ϱ ) d Ω ( ϱ ) × 1 > d Ω ( ζ ) d Ω ( ζ ) , 2 ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 ) β > ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( ζ ) d Ω ( ζ ) ) β .
Using (5) and (6), we determine that Y β ( Ω ) > Y β ( Ω ) . □
Remark 1. 
As a remark on mapping A, every tree can be changed to a star, and any unicyclic and bicyclic graph can be changed to a unicyclic and bicyclic graph in which all noncyclic edges have a degree of one.
Mapping B: Let Ω be a graph and ζ , ϱ E ( Ω ) . Adjacent vertices with a degree of one to vertex ζ and ϱ are u 1 , u 2 , , u r and φ 1 , φ 2 , , φ t , respectively. We obtain Ω and Ω , where Ω = G + { ϱ u 1 , ϱ u 2 , , ϱ u r } { ζ u 1 , ζ u 2 , , ζ u r } , and Ω = G { ϱ φ 1 , ϱ φ 2 , , ϱ φ t } + { ζ φ 1 , ζ φ 2 , , ζ φ t } . Figure 2 demonstrates this.
Lemma 2. 
Let Ω and Ω be obtained from Ω using mapping B; then, either Y β ( Ω ) < Y β ( Ω ) or Y β ( Ω ) < Y β ( Ω ) .
Proof. 
Let Y = G { φ 1 , φ 2 , , φ t , u 1 , u 2 , , u r } ; further, we consider the following cases:
Let d Y ( ζ ) = θ and d Y ( ϱ ) = ϕ .
Case 1. If ζ and ϱ are not neighbors in Ω , then it follows that
Y β ( Ω ) = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( d Ω ( ζ ) d Ω ( ζ ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Y ( δ ) d Y ( δ ) ) β + ( d Ω ( ζ ) d Ω ( ζ ) × 1 ) β + ( d Ω ( ζ ) d Ω ( ζ ) × 1 ) β + + ( d Ω ( ζ ) d Ω ( ζ ) × 1 ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 ) β + + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 ) β , = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β + r ( ( θ + r ) θ + r × 1 ) β + t ( ( ϕ + t ) ϕ + t × 1 ) β .
For Ω , it follows that
Y β ( Ω ) = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( d Y ( ζ ) d Y ( ζ ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Y ( δ ) d Y ( δ ) ) β + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 ) β + + ( d Ω ( ϱ ) d Ω ( ϱ ) × 1 ) β , = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ϕ + r + t × 1 ) β
For Ω , it follows that
Y β ( Ω ) = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( d Ω ( ζ ) d Ω ( ζ ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( d Y ( ϱ ) d Y ( ϱ ) × d Y ( δ ) d Y ( δ ) ) β + ( d Ω ( ζ ) d Ω ( ζ ) × 1 ) β + + ( d Ω ( ζ ) d Ω ( ϱ ) × 1 ) β , = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r + t ) θ + r + t × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) θ + r + t × 1 ) β .
Let Δ 1 = Y β ( Ω ) Y β ( Ω ) and Δ 2 = Y β ( Ω ) Y β ( Ω ) ; we consider the following:
Δ 1 = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ϕ + r + t × 1 ) β δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β , = δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ϕ + r + t × 1 ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β
Here Δ 1 > 0 if
δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ϕ + r + t × 1 ) β > [ δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β + r ( ( θ + r ) θ + r × 1 ) β + t ( ( ϕ + t ) ϕ + t × 1 ) β ] , δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ϕ + r + t × 1 ) β > [ δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β + r ( ( θ + r ) θ + r × 1 ) β + t ( ( ϕ + t ) ϕ + t × 1 ) β ] .
If Δ 1 > 0 then, Δ 2 < 0 , we proceed as follows:
Δ 2 = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r + t ) θ + r + t × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) θ + r + t × 1 ) β δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β + r ( ( θ + r ) θ + r × 1 ) β + t ( ( ϕ + t ) ϕ + t × 1 ) β , = δ N Y ( ζ ) ( ( θ + r + t ) θ + r + t × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) θ + r + t × 1 ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β , < δ N Y ( ζ ) ( ( θ + r + t ) θ + r + t × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) θ + r + t × 1 ) β δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β ( r + t ) ( ( ϕ + r + t ) ϕ + r + t × 1 ) β = δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β < 0 .
Case 2. If ζ , ϱ V ( Ω ) are adjacent, then
Y β ( Ω ) = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β + ( ( θ + r ) θ + r × 1 ) β + + ( ( θ + r ) θ + r × 1 ) β + ( ( ϕ + t ) ϕ + t × 1 ) β + + ( ( ϕ + t ) ϕ + t × 1 ) β ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β , = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β + r ( ( θ + r ) θ + r × 1 ) β + t ( ( ϕ + t ) ϕ + t × 1 ) β ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β .
For Ω , it follows that
Y β ( Ω ) = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( ( ϕ + r + t ) ( ϕ + r + t ) × 1 ) β + + ( ( ϕ + r + t ) ϕ + r + t × 1 ) β ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β , = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ ) θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × 1 ) β ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β .
For Ω , we consider the following:
Y β ( Ω ) = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( ( θ + r + t ) ( θ + r + t ) × 1 ) β + + ( ( θ + r + t ) θ + r + t × 1 ) β ( ( θ + r + t ) ( θ + r + t ) × ϕ ϕ ) β , = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) ( θ + r + t ) × 1 ) β ( ( θ + r + t ) ( θ + r + t ) × ϕ ϕ ) β .
Consider Δ 1 = Y β ( Ω ) Y β ( Ω ) and Δ 2 = Y β ( Ω ) Y β ( Ω ) . For Δ 1 , it follows that
Δ 1 = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ ) θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × 1 ) β ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β
Δ 1 = δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × 1 ) β ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β , = δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β .
If Δ 1 > 0 , it follows that
δ N Y ( ζ ) ( θ θ × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β > ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β .
δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × ( θ + r + t ) ( θ + r + t ) ) β + ( ( θ + r ) θ + r × ( ϕ + r + t ) ϕ + r + t ) β > ( θ θ × ( ϕ + r + t ) ( ϕ + r + t ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β .
If condition (17) holds, then
Δ 2 = δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β + δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) ( θ + r + t ) × 1 ) β + ( ( θ + r + t ) ( θ + r + t ) × ϕ ϕ ) β δ y E ( Ω { ζ , ϱ } ) ( d Y ( δ ) d Y ( δ ) × d Y ( y ) d Y ( y ) ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β .
Δ 2 = δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) ( θ + r + t ) × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β ( ( θ + r + t ) ( θ + r + t ) × ϕ ϕ ) β δ N Y ( ζ ) ( ( θ + r ) θ + r × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + t ) ϕ + t × d Y ( δ ) d Y ( δ ) ) β r ( ( θ + r ) θ + r × 1 ) β t ( ( ϕ + t ) ϕ + t × 1 ) β .
Δ 2 < δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β + δ N Y ( ϱ ) ( ϕ ϕ × d Y ( δ ) d Y ( δ ) ) β + ( r + t ) ( ( θ + r + t ) ( θ + r + t ) × 1 ) β + ( ( θ + r ) θ + r × ( ϕ + t ) ϕ + t ) β δ N Y ( ζ ) ( ( θ + r + t ) ( θ + r + t ) × d Y ( δ ) d Y ( δ ) ) β δ N Y ( ϱ ) ( ( ϕ + r + t ) ϕ + r + t × d Y ( δ ) d Y ( δ ) ) β ( r + t ) ( ( ϕ + r + t ) ( ϕ + r + t ) × ( θ + r + t ) ( θ + r + t ) ) β ( ( θ + r ) θ + r × ( ϕ + r + t ) ϕ + r + t ) β , < 0 .
Hence, it follows that if Δ 1 > 0 , then Δ 2 < 0 . □
Remark 2. 
Using mapping B repeatedly, any unicyclic graph can be turned to a unicyclic graph in which all the edges with a degree of 1 are connected directly to a unique vertex. Any graph with two cycles can be changed to a bicyclic graph in which all the pendent edges are attached to a unique vertex.

3. Graphs Having Largest Value of Y β ( Ω )

Let U n k be a unicyclic graph obtained from C k , where n k pendent edges are attached to the same vertex on C k by means of Lemmas 1 and 2. It follows that Theorem 1 is true.
Theorem 1. 
Let Ω be a unicyclic graph of order n and girth k. If G U n k ; then, Y β ( Ω ) < Y β ( U n k ) .
Consider a bicyclic graph of size n + 1 and order n and denoted by ( n , n + 1 ) . Let G ( n , n + 1 ) be a set of simple connected graphs of order n and size n + 1 . For any G G ( n , n + 1 ) , consider two cycles, C θ and C ϕ . All ( n , n + 1 ) g r a p h s are divided into three classes as follows.
(1)
A ( θ , ϕ ) is a set containing G G ( n , n + 1 ) in which C θ and C ϕ have a common vertex.
(2)
B ( θ , ϕ ) is a set containing G G ( n , n + 1 ) in which C θ and C ϕ have no vertices in common.
(3)
C ( θ , ϕ , l ) is a set containing G G ( n , n + 1 ) in which C θ and C ϕ have a path of length l in common.
Induced sub-graphs of G A ( θ , ϕ ) , B ( θ , ϕ ) , and C ( θ , ϕ , l ) ) are shown in Figure 3i–iii.

4. Largest Value of Y β in A ( θ , ϕ )

This section is devoted to the investigation of bicyclic graphs for the greatest value of Y β in A ( θ , ϕ ) . Let S n ( θ , ϕ ) belong to A ( θ , ϕ ) , where n + 1 θ ϕ vertices with a degree of one are connected to a vertex which is common between C θ and C ϕ . Figure 4 demonstrates this.
Theorem 2. 
In A ( θ , ϕ ) , the graph S n ( θ , ϕ ) has greatest value of Y β .
Proof. 
If we use mapping A and mapping B on Ω , we obtain Ω , where all edges not on the cycles are attached to vertex ϱ . Using Lemmas 1 and 2, we obtain Y β ( Ω ) > Y β ( Ω ) with equality if and only if all the edges not on the cycles are attached to the same vertex in Ω . If Ω S n ( θ , ϕ ) , then vertices ζ and ϱ are different and ζ is the common vertex of C θ and C ϕ . Let ϱ be on C θ ; we consider the following cases:
Case 1. If vertices ζ and ϱ are not neighbors, then it follows that
Y β ( S n ( θ , ϕ ) ) = ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + + ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β , = ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β .
For Ω , it follows that
Y β ( Ω ) = ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 ) β + + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 ) β + 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 2 2 ) β + ( θ 4 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + 4 ( 4 4 × 2 2 ) β , = ( n + 1 θ ϕ ) ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 ) β + 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 2 2 ) β + ( θ 4 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 4 ( 1024 ) β .
Consider the following difference:
Y β ( S n ( θ , ϕ ) ) Y β ( Ω ) = ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β ( n + 1 θ ϕ ) ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 ) β 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 2 2 ) β ( θ 4 ) ( 16 ) β 4 ( 1024 ) β , = ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + 2 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β + 2 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β ( n + 1 θ ϕ ) ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 ) β 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 2 2 ) β ( θ 4 ) ( 16 ) β 4 ( 1024 ) β > 0 .
The difference is positive, while the equality holds if and only if Ω S n ( θ , ϕ ) .
Case 2. If the vertices ζ and ϱ are adjacent in Ω , then
Y β ( Ω ) = ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 1 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 1 ) β + + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 1 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 2 2 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 4 ) β + 3 ( 4 4 × 2 2 ) + ( θ 3 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β , = ( n + 1 θ ϕ ) ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 1 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 256 ) β + 3 ( 1024 ) β + ( θ 3 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β .
Now, consider the following difference:
Y β ( S n ( θ , ϕ ) ) Y β ( Ω ) = ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β ( n + 1 θ ϕ ) ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 1 ) β ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 ) β ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 256 ) β 3 ( 1024 ) β ( θ 3 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β , = ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + 2 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β + ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β ( n + 1 θ ϕ ) ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 1 1 ) β ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 ) β ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 256 ) β ( θ 3 ) ( 16 ) β 3 ( 1024 ) β > 0 .
The difference is positive with equality if and only if Ω S n ( θ , ϕ ) . □
For θ 3 and ϕ 3 , according to the previous Theorem, it follows that S n ( θ , ϕ ) is the unique graph with the greatest value of Y β in A ( θ , ϕ ) .
Lemma 3. 
(i) If 3 < θ , then Y β ( S n ( θ 1 , ϕ ) ) > Y β ( S n ( θ , ϕ ) ) ;
(ii) If 3 < ϕ , then Y β ( S n ( θ , ϕ 1 ) ) > Y β ( S n ( θ , ϕ ) ) .
Proof. 
(i) We have
Y β ( S n ( θ , ϕ ) ) = ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β + 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β .
If we replace θ with θ 1 , we obtain
Y β ( S n ( θ 1 , ϕ ) ) = ( n + 1 θ + 1 ϕ ) ( ( n + 5 θ + 1 ϕ ) ( n + 5 θ + 1 ϕ ) × 1 ) β + 4 ( ( n + 5 θ + 1 ϕ ) ( n + 5 θ + 1 ϕ ) × 2 2 ) β + ( θ 1 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β , = ( n + 2 θ ϕ ) ( ( n + 6 θ ϕ ) ( n + 6 θ ϕ ) × 1 ) β + 4 ( ( n + 6 θ ϕ ) ( n + 6 θ ϕ ) × 2 2 ) β + ( θ 3 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β
Consider the difference
Y β ( S n ( θ 1 , ϕ ) ) Y β ( S n ( θ , ϕ ) ) = ( n + 2 θ ϕ ) ( ( n + 6 θ ϕ ) ( n + 6 θ ϕ ) × 1 ) β + 4 ( ( n + 6 θ ϕ ) ( n + 6 θ ϕ ) × 2 2 ) β + ( θ 3 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β ( θ 2 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β > 0 .
Hence, Y β ( S n ( θ 1 , ϕ ) ) > Y β ( S n ( θ , ϕ ) ) .
As in (i), we obtain the following for (ii):
Y β ( S n ( θ , q 1 ) ) Y β ( S n ( θ , ϕ ) ) = ( n + 2 θ ϕ ) ( ( n + 6 θ ϕ ) ( n + 6 θ ϕ ) × 1 ) β + 4 ( ( n + 6 θ ϕ ) ( n + 6 θ ϕ ) × 2 2 ) β + ( θ 2 ) ( 16 ) β + ( q 3 ) ( 16 ) β ( n + 1 θ ϕ ) ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 1 ) β 4 ( ( n + 5 θ ϕ ) ( n + 5 θ ϕ ) × 2 2 ) β ( θ 2 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β > 0 .
Hence, Y β ( S n ( θ , q 1 ) ) > Y β ( S n ( θ , ϕ ) ) . □
From Theorem 2 and Lemma 3, it follows that Theorem 3 is true.
Theorem 3. 
For any θ 3 and ϕ 3 , Y β is S n ( 3 , 3 ) in A ( θ , ϕ ) and is the unique graph with greatest value of Y β .

5. Largest Value of Y β in B ( θ , ϕ )

In this section, we investigate Y β in B ( θ , ϕ ) . Suppose that T n r ( θ , ϕ ) is a graph with order n and size n + 1 , which is obtained by connecting C θ and C ϕ with a path of length r. The remaining n + 1 θ ϕ r edges are connected to the common vertex of the path and cycle C θ (see Figure 5a). Similarly the graph T n r ( ϕ , θ ) is shown in Figure 5b, while T n ( θ , ϕ ) is a graph of order n and size n + 1 and is obtained by connecting C θ and C ϕ with a path ζ ϱ w of length 2; all the others are attached to ϱ . Figure 5c demonstrates this.
Theorem 4. 
Let G B ( θ , ϕ ) , and let cycles C θ and C ϕ be connected by a path with a length of r; then,
Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) with equality if and only , if G T n r ( θ , ϕ ) Y β ( T n r ( ϕ , θ ) ) Y β ( Ω ) with equality if and only , if G T n r ( q , θ ) .
Proof. 
Let P = φ 1 φ 2 φ t φ r + 1 be the shortest path between C θ and C ϕ in Ω . Let φ 1 be a vertex which is common in C θ and P. Let φ r + 1 be a common vertex of C ϕ and P. By mapping A and B on Ω , we obtain Ω , as given in Figure 5, in which the edges not on the cycles with a degree of 1 are attached to ϱ . Using Lemmas 1 and 2, we obtain Y β ( Ω ) > Y β ( Ω ) , with equality if and only if all the edges not on the cycle are pendant and attached to the same vertex in Ω . Now, we have two cases:
Case 1. If ϱ lies on C θ , vertices ϱ and φ 1 are not adjacent (see Figure 5d); then,
Y β ( Ω ) = ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + 2 ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 2 2 ) β + ( 3 3 × d ( φ 2 ) d ( φ 2 ) ) β + ( θ 4 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + ( r 2 ) ( 2 2 × 2 2 ) β + 3 ( 2 2 × 3 3 ) β + 2 ( 2 2 × 3 3 ) β , = ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + 2 ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 4 ) β + ( 27 × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) 8 ( θ 4 ) + ( 16 ) β ( ϕ 2 ) + 8 β ( r 2 ) + 5 ( 108 ) β .
Similarly for Y β ( T n r ( θ , ϕ ) ) , it follows that
Y β ( T n r ( θ , ϕ ) ) = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 2 2 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β + ( r 2 ) ( 2 2 × 2 2 ) β + ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + 3 ( 2 2 × 3 3 ) β , = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) β ( r 2 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β .
Consider the following difference:
Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) β ( r 2 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β 2 ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 4 ) β ( 27 × d ( φ 2 ) d ( φ 2 ) ) β ( 16 ) β ( θ 4 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β ( r 2 ) 5 ( 108 ) β , = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β 2 ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 4 ) β ( 27 × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) β ( θ 2 ) + 3 ( 108 ) β ( 16 ) β ( θ 4 ) 5 ( 108 ) β 0 .
Hence, Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) > 0 , with equality iff Ω T n r ( θ , ϕ ) . The inequality can be observed in Figure 6.
If ϱ and φ 1 are adjacent in Ω , then
Y β ( Ω ) = ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 2 2 ) β + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 3 3 ) β + ( 3 3 × d ( φ 2 ) d ( φ 2 ) ) β + 4 ( 2 2 × 3 3 ) β + ( θ 3 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + ( r 2 ) ( 2 2 × 2 2 ) β , = ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 4 ) β + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 27 ) β + ( 3 3 × d ( φ 2 ) d ( φ 2 ) ) β + 4 ( 108 ) β + ( θ 3 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + ( r 2 ) ( 16 ) β .
Consider Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) in the following:
Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) β ( r 2 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 4 ) β ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 27 ) β ( 3 3 × d ( φ 2 ) d ( φ 2 ) ) β ( θ 3 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β ( r 2 ) ( 16 ) β 4 ( 108 ) β , = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) β ( θ 2 ) + 3 ( 108 ) β ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 4 ) β ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 27 ) β ( 3 3 × d ( φ 2 ) d ( φ 2 ) ) β ( θ 3 ) ( 16 ) β 4 ( 108 ) β
Hence, Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) > 0 , with equality iff Ω ( T n r ( θ , ϕ ) ) . The difference can be observed in Figure 7.
Case 2. If vertex ϱ is on C ϕ , then the proof is similar to Case 1. Figure 5e demonstrates this.
Case 3. If ϱ lies on P (Figure 5f), and if Ω T n ( θ , ϕ ) , then 3 r , and the following is true:
If r > t > 2 , then 3 < r , and we have
Y β ( T n ( θ , ϕ ) ) = ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + + ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 3 3 ) β + 4 ( 2 2 × 3 3 ) β + ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β , = ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 3 3 ) β + 4 ( 108 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β .
For Ω , we obtain
Y β ( Ω ) = ( ( n + 1 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + + ( ( n + 1 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + 2 ( ( n + 1 θ ϕ r ) ( n + 1 θ ϕ r ) × 2 2 ) β + ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + ( r 4 ) ( 2 2 × 2 2 ) β + 6 ( 2 2 × 3 3 ) , = ( n + 1 θ ϕ r ) ( ( n + 1 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + 2 ( ( n + 1 θ ϕ r ) ( n + 1 θ ϕ r ) × 4 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + ( r 4 ) ( 16 ) β + 6 ( 108 ) β .
Consider the following difference
Y β ( T n ( θ , ϕ ) ) Y β ( Ω ) = ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 3 3 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 4 ( 108 ) β ( n + 1 θ ϕ r ) ( ( n + 1 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β 2 ( ( n + 1 θ ϕ r ) ( n + 1 θ ϕ r ) × 4 ) β ( θ 2 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β ( r 4 ) ( 16 ) β 6 ( 108 ) β , = ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 3 3 ) β + 4 ( 108 ) β ( n + 1 θ ϕ r ) ( ( n + 1 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β 2 ( ( n + 1 θ ϕ r ) ( n + 1 θ ϕ r ) × 4 ) β ( r 4 ) ( 16 ) β 6 ( 108 ) β .
Figure 8 illustrates the difference in (29).
If t = r or t = 2 , then for Ω , we obtain
Y β ( Ω ) = 5 ( 2 2 × 3 3 ) β + ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + ( 3 3 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β + ( 2 2 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + + ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β ( r 3 ) ( 2 2 × 2 2 ) β = ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + ( 3 3 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β + ( 2 2 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β + ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β + ( r 3 ) ( 2 2 × 2 2 ) β + 5 ( 108 ) β , > 0 .
We consider the following difference in this case:
Y β ( T n ( θ , ϕ ) ) Y β ( Ω ) = ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 27 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 4 ( 108 ) β ( 27 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β ( 4 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β ( θ 2 ) ( 16 ) β ( ϕ 2 ) ( 16 ) β ( r 3 ) ( 16 ) β 5 ( 108 ) β = ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 27 ) β ( 27 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β ( 4 × ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) ) β ( n + 1 θ ϕ r ) ( ( n + 3 θ ϕ r ) ( n + 3 θ ϕ r ) × 1 ) β ( r 3 ) ( 16 ) β ( 108 ) β , > 0 .
The above last difference can be observed in Figure 9. □
Lemma 4. 
Y β ( T n ( θ , ϕ ) ) Y β ( T n ( 3 , 3 ) ) with equality if and only if θ = 3 = ϕ .
Proof. 
As we have
Y β ( T n ( θ , ϕ ) ) = ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 3 3 ) β + 4 ( 108 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β .
for Y β ( T n ( 3 , 3 ) ) , we obtain;
Y β ( T n ( 3 , 3 ) ) = 4 ( 3 3 × 2 2 ) β + 2 ( 2 2 × 2 2 ) β + 2 ( ( n 5 ) ( n 5 ) × 3 3 ) β + ( n 7 ) ( ( n 5 ) ( n 5 ) × 1 ) β , = 2 ( 16 ) β + 4 ( 108 ) β + 2 ( ( n 5 ) ( n 5 ) × 1 ) β + ( n 7 ) ( ( n 5 ) ( n 5 ) × 1 ) β .
We thus obtain the following difference:
Y β ( T n ( 3 , 3 ) ) Y β ( T n ( θ , ϕ ) ) = 4 ( 108 ) β + 2 ( 16 ) β + 2 ( ( n 5 ) ( n 5 ) × 27 ) β + ( n 7 ) ( ( n 5 ) ( n 5 ) × 1 ) β ( n 1 θ ϕ ) ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 1 ) β + 2 ( ( n + 1 θ ϕ ) ( n + 1 θ ϕ ) × 3 3 ) β , + 4 ( 108 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β 0 .
Figure 10 illustrates the inequality. The equality holds if and only if θ = 3 = q . Figure 11 demonstrates this. □
Lemma 5. 
For 2 r , Y β ( T n r 1 ( θ , ϕ ) ) > Y β ( T n r ( θ , ϕ ) ) .
Proof. 
Consider the following:
If r > 2 , then Y β ( T n r ( θ , ϕ ) ) is applied in the following:
Y β ( T n r ( θ , ϕ ) ) = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × d ( φ 2 ) d ( φ 2 ) ) β + ( 16 ) β ( r 2 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β .
Here, d ( φ 2 ) = 2 ; then, it follows that
Y β ( T n r ( θ , ϕ ) ) = ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β + 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β + ( 16 ) β ( r 2 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β .
If we plug r by r 1 into (35), we obtain
Y β ( T n r 1 ( θ , ϕ ) ) = ( n + 2 θ ϕ r ) ( ( n + 5 θ ϕ r ) ( n + 5 θ ϕ r ) × 1 ) β + 2 ( ( n + 5 θ ϕ r ) ( n + 5 θ ϕ r ) × 4 ) β + ( ( n + 5 θ ϕ r ) ( n + 5 θ ϕ r ) × 4 ) β + ( 16 ) β ( r 3 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β .
Consider the following difference:
Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r ( θ , ϕ ) ) = ( n + 2 θ ϕ r ) ( ( n + 5 θ ϕ r ) ( n + 5 θ ϕ r ) × 1 ) β + 2 ( ( n + 5 θ ϕ r ) ( n + 5 θ ϕ r ) × 4 ) β + ( ( n + 5 θ ϕ r ) ( n + 5 θ ϕ r ) × 4 ) β + ( 16 ) β ( r 3 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β ( n + 1 θ ϕ r ) ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 1 ) β 2 ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β ( ( n + 4 θ ϕ r ) ( n + 4 θ ϕ r ) × 4 ) β ( 16 ) β ( r 2 ) ( 16 ) β ( θ 2 ) ( 16 ) β ( ϕ 2 ) 3 ( 108 ) β .
Figure 12 demonstrates this.
If r = 2 , then it follows that
Y β ( T n r ( θ , ϕ ) ) = ( n 1 θ ϕ ) ( ( n + 2 θ ϕ ) ( n + 2 θ ϕ ) × 1 ) β + 2 ( ( n + 2 θ ϕ ) ( n + 2 θ ϕ ) × 4 ) β + ( ( n + 2 θ ϕ ) ( n + 2 θ ϕ ) × 4 ) β + ( 16 ) β ( 2 2 ) + ( 16 ) β ( θ 2 ) + ( 16 ) β ( ϕ 2 ) + 3 ( 108 ) β .
Similarly,
Y β ( T n r 1 ( θ , ϕ ) ) = ( n θ ϕ ) ( 1 × ( n θ ϕ + 3 ) ( n θ ϕ + 3 ) ) β + 2 ( 2 2 × ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) ) β + ( 3 3 × ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) ) β + ( θ 2 ) ( 2 2 × 2 2 ) β + ( ϕ 2 ) ( 2 2 × 2 2 ) β + 2 ( 2 2 × 3 3 ) β , = ( n θ ϕ ) ( 1 × ( n θ ϕ + 3 ) ( n θ ϕ + 3 ) ) β + 2 ( 4 × ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) ) β + ( 27 × ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 2 ( 108 ) β .
Consider the difference Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r ( θ , ϕ ) ) , as in the following:
Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r ( θ , ϕ ) ) = ( n θ ϕ ) ( 1 × ( n θ ϕ + 3 ) ( n θ ϕ + 3 ) ) β + 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 27 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 2 ( 108 ) β ( n 1 θ ϕ ) ( ( n + 2 θ ϕ ) ( n + 2 θ ϕ ) × 1 ) β 2 ( ( n + 2 θ ϕ ) ( n + 2 θ ϕ ) × 4 ) β ( ( n + 2 θ ϕ ) ( n + 2 θ ϕ ) × 4 ) β ( 16 ) β ( 2 2 ) ( 16 ) β ( θ 2 ) ( 16 ) β ( ϕ 2 ) 3 ( 108 ) β , > 0 .
The above difference can be observed in Figure 13. □
Lemma 6. 
Y β ( T n 1 ( 3 , 3 ) ) Y β ( T n 1 ( θ , ϕ ) ) with equality if and only if θ = 3 = q .
Proof. 
We have
Y β ( T n 1 ( θ , ϕ ) ) = ( n θ ϕ ) ( 1 × ( n θ ϕ + 3 ) ( n θ ϕ + 3 ) ) β + 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 ) β + ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 27 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 2 ( 108 ) β .
If we plug θ = 3 = q into (36), we obtain
Y β ( T n 1 ( 3 , 3 ) ) = ( n 6 ) ( ( n 3 ) ( n 3 ) × 1 ) β + 2 ( ( n 3 ) ( n 3 ) × 4 ) β + ( ( n 3 ) ( n 3 ) × 27 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 2 ( 108 ) β .
We consider the following difference:
Y β ( T n 1 ( 3 , 3 ) ) Y β ( T n 1 ( θ , ϕ ) ) = ( n 6 ) ( ( n 3 ) ( n 3 ) × 1 ) β + 2 ( ( n 3 ) ( n 3 ) × 4 ) β + ( ( n 3 ) ( n 3 ) × 27 ) β ( n θ ϕ ) ( 1 × ( n θ ϕ + 3 ) ( n θ ϕ + 3 ) ) β 2 ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 4 ) β ( ( n + 3 θ ϕ ) ( n + 3 θ ϕ ) × 27 ) β 0 .
Figure 14 and Figure 15 illustrate this, with equality if and only if θ = 3 = q . □
It is not difficult to verify that Y β ( T n 1 ( 3 , 3 ) ) > Y β ( T n ( 3 , 3 ) ) . Further, it follows that Theorem 5 is true:
Theorem 5. 
In B ( θ , ϕ ) , T n 1 ( 3 , 3 ) is the unique tree having greatest value of Y β for all ϕ , θ 3 .

6. Largest Value of Y β in C ( θ , ϕ , l )

C ( θ , ϕ , l ) represents the bicyclic graph of cycles C θ and C ϕ . C θ and C ϕ are connected by a path of length l. Suppose L n l ( θ , ϕ ) is obtained from Figure 3iii by connecting n + l + 1 θ ϕ vertices to the vertex of degree 3, as shown in Figure 16i.
Theorem 6. 
Let G C ( θ , ϕ , l ) ; then, Y β ( Ω ) Y β ( Y ) with equality if and only if G Y .
Proof. 
If we apply mappings A and B to Ω , we obtain Ω , in which all the pendant edges not on the cycles are connected to v 0 , i.e, Ω is one (Figure 16). Using Lemmas 1 and 2, Y β ( Ω ) Y β ( Ω ) with equality if and only if all the edges not on the cycle are pendant edges attached to the same vertex in Ω .
Let Q 1 = ϱ x t 1 x t 2 x 2 x 1 ζ be the common path of C θ and C ϕ in Ω 1 (Figure 16). Q 2 = ϱ y r y r 1 y 2 y 1 ζ and Q 3 = ϱ z t z t 1 z 2 z 1 ζ are other paths from ζ to ϱ on the cycles C θ and C ϕ , respectively; r = θ l 1 , t = ϕ l 1 , 0 r , 0 l e q t , 1 l , and 3 l + r + t .
If u v E ( Ω 1 ) , where d Ω 1 ( u ) = d Ω 1 ( v ) = 2 , we obtain Ω 1 by deleting edge u v and add a new pendant edge u u = e to ζ , then, Y β ( Ω 1 ) > Y β ( Ω 1 ) , as ( d Ω 1 ( u ) d Ω 1 ( u ) × d Ω 1 ( v ) d Ω 1 ( v ) ) β = ( 2 2 × 2 2 ) β = 8 β and ( d Ω 1 ( ζ ) d Ω 1 ( ζ ) × d Ω 1 ( ζ ) d Ω 1 ( ζ ) ) β > 8 β . Hence, Y β ( Y ) Y β ( Ω 1 ) with equality if and only if Ω 1 Y .
Similarly, if there are two edges in the graph Ω 2 , where the degree of their end vertices is 2, we obtain Ω 2 by removing these edges and add new edges to vertex x i ; then, Y β ( Ω 2 ) Y β ( Ω 2 ) ; therefore, Y β ( Ω 1 ) Y β ( Ω 1 ) and Y β ( Ω 2 ) Y β ( Ω 4 ) . Based on this, it is not difficult to show that Y β ( Y ) > Y β ( Ω 3 ) and Y β ( Y ) > Y β ( Ω 4 ) . □
Theorem 7. 
In all bicyclic graphs, Y is the unique graph with the greatest value of Y β .
Proof. 
Using Theorems 3, 5 and 6, we compare the value of Y β for S n ( 3 , 3 ) , T n 1 ( 3 , 3 ) , and Y . Hence, it follows that
Y β ( T n 1 ( 3 , 3 ) ) < Y β ( S n ( 3 , 3 ) ) < Y β ( Y ) .
Figure 17 and Figure 18 illustrate this inequality, where
Y β ( T n 1 ( 3 , 3 ) ) = ( n 6 ) ( ( n 3 ) ( n 3 ) × 1 ) β + 2 ( ( n 3 ) ( n 3 ) × 4 ) β + ( ( n 3 ) ( n 3 ) × 27 ) β + ( θ 2 ) ( 16 ) β + ( ϕ 2 ) ( 16 ) β + 2 ( 108 ) β .
Y β ( S n ( 3 , 3 ) ) = 2 ( 16 ) β + 4 ( ( n θ ϕ + 5 ) ( n θ ϕ + 5 ) × 4 ) β + ( n θ ϕ + 1 ) ( 1 × ( n θ ϕ + 5 ) n θ ϕ + 5 ) β .
Y β ( Y ) = 2 ( 108 ) β + 2 ( ( n + 2 + 3 θ ϕ ) ( n + 2 + 3 θ ϕ ) × 4 ) β + ( ( n θ ϕ + 2 + 3 ) n θ ϕ + 2 + 3 × 27 ) β + ( n θ ϕ + 2 ) ( ( n θ ϕ + 2 + 3 ) n θ ϕ + 2 + 3 × 1 ) β .

7. Mappings for Decreasing Y β

In this section, we provide a mapping which is used for decreasing the value of Y β ; the mapping is given in the following.
Mapping C: Let Ω be a simple and connected graph other than P 1 , and select vertex ζ from V ( Ω ) . Suppose Ω 1 is the graph in which we mark ζ with φ k of the path φ 1 φ 2 φ 3 φ n , where 1 < k < n . Let Ω 2 be obtained from Ω 1 by deleting φ k 1 φ k and adding φ n φ k 1 . Figure 19 demonstrates this.
Lemma 7. 
Let Ω 1 and Ω 2 be as in Figure 19; then, Y β ( Ω 2 ) < Y β ( Ω 1 ) .
Proof. 
Using the definition of Y β , we obtain
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω 1 ( φ k 1 ) ) d Ω 1 ( φ k 1 ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω 1 ( φ k + 1 ) ) d Ω 1 ( φ k + 1 ) ) β + ( ( d Ω 1 ( φ n 1 ) ) d Ω 1 ( φ n 1 ) × ( d Ω 1 ( φ n ) ) d Ω 1 ( φ n ) ) β .
For Ω 2 , it follows that
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × ( d Ω 2 ( φ k + 1 ) ) d Ω 2 ( φ k + 1 ) ) β + ( ( d Ω 2 ( φ n 1 ) ) d Ω 2 ( φ n 1 ) × ( d Ω 2 ( φ n ) ) d Ω 2 ( φ n ) ) β + ( ( d Ω 2 ( φ n ) ) d Ω 2 ( φ n ) × ( d Ω 2 ( φ k 1 ) ) d Ω 2 ( φ k 1 ) ) β .
Let us investigate the cases in the following.
Case 1: If k = 2 and n = 3 , then
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ 1 ) d Ω 1 ( φ 1 ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ 3 ) d Ω 1 ( φ 3 ) ) β , = δ N Ω ( ζ ) ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 1 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 1 ) β .
For Ω 2 , we obtain
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω 2 ( φ 3 ) d Ω 2 ( φ 3 ) ) β + ( ( d Ω 2 ( φ 3 ) ) d Ω 2 ( φ 3 ) × d Ω 2 ( φ 1 ) d Ω 2 ( φ 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β × ( 4 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) as given below:
Y β ( Ω 1 ) Y β ( Ω 2 = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 1 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 1 ) β δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β ( 4 ) β > 0 .
So Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Case 2: If k = 2 and n > 3 , then we obtain
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ 1 ) d Ω 1 ( φ 1 ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ 3 ) d Ω 1 ( φ 3 ) ) β + ( ( d Ω 1 ( φ n 1 ) ) d Ω 1 ( φ n 1 ) × ( d Ω 1 ( φ n ) ) d Ω 1 ( φ n ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 1 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 1 ) β + ( 4 ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ 3 ) d Ω 1 ( φ 3 ) ) β + ( ( d Ω 2 ( u n 1 ) ) d Ω 2 ( φ n 1 ) × d Ω 2 ( φ n ) d Ω 2 ( φ n ) ) β + ( ( d Ω 2 ( φ n ) ) d Ω 2 ( φ n ) × ( d Ω 2 ( φ 1 ) ) d Ω 2 ( φ 1 ) ) β , = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β + ( 4 × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β + ( 16 ) β + ( 4 ) β .
The difference in Y β ( Ω 1 ) and Y β ( Ω 2 ) is considered in the following:
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( ( 2 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 1 ) β + ( 4 ) β δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × ( d Ω ( δ ) ) d Ω ( δ ) ) β ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β ( 16 ) β ( 4 ) β , > 0 .
Using term-by-term comparison, it follows that Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Case 3: If k > 2 and n = k + 1 , then
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ k + 1 ) d Ω 1 ( φ k + 1 ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ k 1 ) d Ω 1 ( φ k 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β
For Ω 2 , it follows that
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω 2 ( φ k + 1 ) d Ω 2 ( φ k + 1 ) ) β + ( ( d Ω 2 ( φ k + 1 ) ) d Ω 2 ( φ k + 1 ) × d Ω 2 ( φ k 1 ) d Ω 2 ( φ k 1 ) ) β , = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β + ( 4 × 4 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β + ( 16 ) β .
We consider the difference in Y β ( Ω 1 ) and Y β ( Ω 2 ) in the following:
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × 4 ) β + ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × 4 ) β δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 4 ) β ( 16 ) β > 0 .
Using term-by-term comparison, we obtain Y β ( Ω 1 ) Y β ( Ω 2 ) > 0 ; hence, Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Case 4: If n > k + 1 and k > 2 , for Ω 1 and Ω 2 we have
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ k 1 ) d Ω 1 ( φ k 1 ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω 1 ( φ k + 1 ) d Ω 1 ( φ k + 1 ) ) β + ( ( d Ω 1 ( φ n 1 ) ) d Ω 1 ( φ n 1 ) × d Ω 1 ( φ n ) d Ω 1 ( φ n ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 2 ) d Ω ( ζ ) + 2 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( 4 ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω 2 ( φ k + 1 ) d Ω 2 ( φ k + 1 ) ) β + ( ( d Ω 2 ( φ n 1 ) ) d Ω 2 ( φ n 1 ) × d Ω 2 ( φ n ) d Ω 2 ( φ n ) ) β + ( ( d Ω 2 ( φ n ) ) d Ω 2 ( φ n ) × d Ω 2 ( φ k 1 ) d Ω 2 ( φ k 1 ) ) β , = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β + ( 4 × 4 ) β + ( 4 × 4 ) β , = δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β + ( 16 ) β + ( 16 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given below:
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( 4 ) β δ N Ω ( ζ ) ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β ( ( 1 + d Ω ( ζ ) ) 1 + d Ω ( ζ ) × 4 ) β ( 16 ) β ( 16 ) β , > 0 .
Hence, Y β ( Ω 1 ) Y β ( Ω 2 ) > 0 . For the last two terms, it follows that ( ( 2 + d Ω ( ζ ) ) 2 + d Ω ( ζ ) × 4 ) β + ( 4 ) β ( 16 ) β ( 16 ) β > 0 . □
Remark 3. 
By using mapping  C repeatedly, any tree attached to a graph can be transformed into a path as in Figure 20. The value of Y β decreases.
Mapping D: Consider graph Ω and ζ , ϱ E ( Ω ) , and let Ω 1 be obtained by marking ζ with u 0 , ϱ with v 0 for the two paths u 0 u 1 u r 1 u r and v 0 φ 1 φ t 1 φ t , respectively. Suppose Ω 2 is obtained from Ω 1 by deleting u 0 u 1 and add φ t u 1 (Figure 21).
Lemma 8.
Consider Ω 1 and Ω 2 from Figure 21 with 1 < d Ω ( ϱ ) d Ω ( ζ ) , r 1 , and t 0 ; then, it follows that
(1) 
If t > 0 , then Y β ( Ω 1 ) > Y β ( Ω 2 ) ;
(2) 
If t = 0 and y N Ω ( ϱ ) { ζ } d Ω ( y ) < δ N Ω ( ζ ) { ϱ } d Ω ( δ ) , then Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Proof. 
(1) Here d Ω ( ζ ) > 1 and t > 0 , we consider;
Case 1. If t = 1 and r = 1 , then;
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 1 ( φ 1 ) d Ω 1 ( φ 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) × 1 + 1 ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 1 ) β .
For Y β ( Ω 2 ) , it follows that
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 2 ( φ 1 ) d Ω 2 ( φ 1 ) ) β + ( ( d Ω 2 ( φ 1 ) ) d Ω 2 ( φ 1 ) × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 4 ) β + ( 4 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 1 ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 1 ) β δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 4 ) β ( 4 ) β > 0 , d Ω ( ζ ) d Ω ( ϱ ) .
Hence, Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Case 2. If t > 1 and r = 1 , then
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + ( ( d Ω 1 ( φ t 1 ) ) d Ω 1 ( φ t 1 ) × d Ω 1 ( φ t ) d Ω 1 ( φ t ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 1 ) β + ( 4 × 1 ) β = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 1 ) β + ( 4 ) β
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω 2 ( φ t 1 ) ) d Ω 2 ( φ t 1 ) × d Ω 2 ( φ t ) d Ω 2 ( φ t ) ) β + ( ( d Ω 2 ( φ t ) ) d Ω 2 ( φ t ) × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( 4 × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( 16 ) β + ( 4 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 1 ) β + ( 4 ) β δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β ( 16 ) β ( 4 ) β , > 0 .
So Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Case 3. If t = 1 and 1 < r , then
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 1 ( φ 1 ) d Ω 1 ( φ 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 4 ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 1 ) β .
For Y β ( Ω 2 ) , it follows that
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 2 ( φ 1 ) d Ω 2 ( φ 1 ) ) β + ( ( d Ω 2 ( φ 1 ) ) d Ω 2 ( φ 1 ) × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 4 ) β + ( 4 × 4 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 4 ) β + ( 16 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 1 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 4 ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 1 ) β δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × 4 ) β ( 16 ) β , > 0 .
So Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Case 4. If t > 1 and 1 < r , then
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + ( ( d Ω 1 ( φ t 1 ) ) d Ω 1 ( φ t 1 ) × d Ω 1 ( φ t ) d Ω 1 ( φ t ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 4 ) β + ( 4 ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω 2 ( φ t 1 ) ) d Ω 2 ( φ t 1 ) × d Ω 2 ( φ t ) d Ω 2 ( φ t ) ) β + ( ( d Ω 2 ( φ t ) ) d Ω 2 ( φ t ) × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( 4 × 4 ) β + ( 4 × 1 ) β , = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( 16 ) β + ( 4 ) β .
Consider the following:
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × 4 ) β + ( 4 ) β δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( 16 ) β + ( 4 ) β , > 0 .
Hence, Y β ( Ω 1 ) Y β ( Ω 2 ) > 0 so Y β ( Ω 1 ) > Y β ( Ω 2 ) .
(2). If t = 0 and y N Ω ( ϱ ) { ζ } d Ω ( y ) < δ N Ω ( ζ ) { ϱ } d Ω ( δ ) , if ζ and ϱ are adjacent vertices, then
Y β ( Ω 1 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + y N Ω ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( y ) d Ω ( y ) ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β + y N Ω ( ϱ ) ( ( d Ω ( ϱ ) + 1 ) ( d Ω ( ϱ ) + 1 ) × d Ω ( y ) d Ω ( y ) ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + y N Ω ( ϱ ) ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( y ) d Ω ( y ) ) β δ N Ω ( ζ ) ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β . ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β y N Ω ( ϱ ) ( ( d Ω ( ϱ ) + 1 ) ( d Ω ( ϱ ) 1 ) × d Ω ( y ) d Ω ( y ) ) β , > 0 . d Ω 1 ( u 1 ) = d Ω 2 ( u 1 )
Hence, Y β ( Ω 1 ) > Y β ( Ω 2 ) .
If ζ and ϱ are adjacent, then we have
Y β ( Ω 1 ) = δ N Ω ( ζ ) { ϱ } ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + y N Ω ( ϱ ) { ζ } ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( y ) d Ω ( y ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( ϱ ) d Ω ( ϱ ) ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = δ N Ω ( ζ ) { ϱ } ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β + y N Ω ( ϱ ) { ζ } ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω ( y ) d Ω ( y ) ) β + ( d Ω ( ζ ) d Ω ( ζ ) × ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = δ N Ω ( ζ ) { ϱ } ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( δ ) d Ω ( δ ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω 1 ( u 1 ) d Ω 1 ( u 1 ) ) β + y N Ω ( ϱ ) { ζ } ( d Ω ( ϱ ) d Ω ( ϱ ) × d Ω ( y ) d Ω ( y ) ) β + ( ( d Ω ( ζ ) + 1 ) d Ω ( ζ ) + 1 × d Ω ( ϱ ) d Ω ( ϱ ) ) β δ N Ω ( ζ ) { ϱ } ( ( d Ω ( ζ ) ) d Ω ( ζ ) × d Ω ( δ ) d Ω ( δ ) ) β ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω 2 ( u 1 ) d Ω 2 ( u 1 ) ) β y N Ω ( ϱ ) { ζ } ( ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 × d Ω ( y ) d Ω ( y ) ) β ( d Ω ( ζ ) d Ω ( ζ ) × ( d Ω ( ϱ ) + 1 ) d Ω ( ϱ ) + 1 ) β , > 0 . d Ω 1 ( u 1 ) = d Ω 2 ( u 1 )
Hence, Y β ( Ω 1 ) Y β ( Ω 2 ) > 0 , so Y β ( Ω 1 ) > Y β ( Ω 2 ) . □
Remark 4. 
Using mapping  C and mapping D continuously, every tree can be transformed into a path, any unicyclic graph can be transformed into a unicyclic graph such that the path is connected to a cycle, and every bicyclic graph can be changed into a bicyclic graph cycle in which the path is connected to one of the graphs provided in Figure 22 (Lemma 8(i)). Generally, any bicyclic graph can be changed into a bicyclic graph where the path is connected to a vertex of degree 2 (Lemma 8(ii)), and the value of Y β decreases.
Lemma 9. 
Consider the graph Ω 1 given in Figure 23 if the path x 1 , x 2 , , x k with 1 < k is connected to vertex x 1 . Suppose Ω 2 is obtained from Ω 1 by removing ϱ x 1 and add ϱ x k ; then, Y β ( Ω 1 ) > Y β ( Ω 2 ) .
Proof. 
If k = 2 , then x k and x 1 are different; it follows that
Y β ( Ω 1 ) = ( d ( ζ ) d ( ζ ) × d Ω 1 ( x 1 ) d Ω 1 ( x 1 ) ) β + ( d Ω 1 ( x 1 ) d Ω 1 ( x 1 ) × d Ω 1 ( x 2 ) d Ω 1 ( x 2 ) ) β + ( d Ω 1 ( x 1 ) d Ω 1 ( x 1 ) × d Ω 1 ( ϱ ) d Ω 1 ( ϱ ) ) β , = ( d ( ζ ) d ( ζ ) × 27 ) β + ( 27 ) β + ( 27 × d Ω 1 ( ϱ ) d Ω 1 ( ϱ ) ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = ( d ( ζ ) d ( ζ ) × d Ω 2 ( x 1 ) d Ω 2 ( x 1 ) ) β + ( d Ω 2 ( x 1 ) d Ω 2 ( x 1 ) × d Ω 2 ( x 2 ) d Ω 2 ( x 2 ) ) β + ( d Ω 2 ( x 2 ) d Ω 2 ( x 2 ) × d Ω 2 ( ϱ ) d Ω 2 ( ϱ ) ) β , = ( d ( ζ ) d ( ζ ) × 4 ) β + ( 16 ) β + ( 4 × d Ω 2 ( ϱ ) d Ω 2 ( ϱ ) ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = ( d ( ζ ) d ( ζ ) × 27 ) β + ( 27 ) β + ( 27 × d Ω 1 ( ϱ ) d Ω 1 ( ϱ ) ) β ( d ( ζ ) d ( ζ ) × 4 ) β ( 16 ) β ( 4 × d Ω 2 ( ϱ ) d Ω 2 ( ϱ ) ) β , > 0 .
For k > 2 , it follows that;
Y β ( Ω 1 ) = ( d ( ζ ) d ( ζ ) × d Ω 1 ( x 1 ) Ω 1 ( x 1 ) ) β + ( d Ω 1 ( x 1 ) d Ω 1 ( x 1 ) × d Ω 1 ( x 2 ) Ω 1 d ( x 2 ) ) β + ( d Ω 1 ( x 1 ) d Ω 1 ( x 1 ) × d Ω 1 ( ϱ ) d Ω 1 ( ϱ ) ) β + ( d Ω 1 ( x k 1 ) d Ω 1 ( x k 1 ) × d Ω 1 ( x k ) d Ω 1 ( x k ) ) β , = ( d ( ζ ) d ( ζ ) × 27 ) β + ( 108 ) β + ( 27 × d Ω 1 ( ϱ ) d Ω 1 ( ϱ ) ) β + ( 4 ) β .
For Y β ( Ω 2 ) , we obtain
Y β ( Ω 2 ) = ( d ( ζ ) d ( ζ ) × d Ω 2 ( x 1 ) Ω 2 ( x 1 ) ) β + ( d Ω 2 ( x 1 ) d Ω 2 ( x 1 ) × d Ω 2 ( x 2 ) Ω 2 d ( x 2 ) ) β + ( d Ω 2 ( x k ) d Ω 2 ( x k ) × d Ω 2 ( ϱ ) d Ω 2 ( ϱ ) ) β + ( d Ω 2 ( x k 1 ) d Ω 2 ( x k 1 ) × d G 2 ( x k ) d Ω 2 ( x k ) ) β , = ( d ( ζ ) d ( ζ ) × 4 ) β + ( 16 ) β + ( 4 × d Ω 2 ( ϱ ) d Ω 2 ( ϱ ) ) β + ( 16 ) β .
Consider Y β ( Ω 1 ) Y β ( Ω 2 ) , as given by
Y β ( Ω 1 ) Y β ( Ω 2 ) = ( d ( ζ ) d ( ζ ) × 27 ) β + ( 27 × d Ω 1 ( ϱ ) d Ω 1 ( ϱ ) ) β ( d ( ζ ) d ( ζ ) × 4 ) β ( 4 × d Ω 2 ( ϱ ) d Ω 2 ( ϱ ) ) β + ( 108 ) β + ( 4 ) β 2 ( 16 ) β , > 0 .
Hence, Y β ( Ω 1 ) Y β ( Ω 2 ) > 0 in both cases if k = 2 and k > 2 . This confirms that Y β ( Ω 1 ) > Y β ( Ω 2 ) . □

8. Smallest Value of Y β in Unicyclic Graphs, Trees, and Bicyclic Graphs

Here, we found unicyclic graphs, trees, and bicyclic graphs with the smallest value of Y β ; using Lemma 7, we obtain the following:
Theorem 8. 
Consider a T on n vertices; if T is not P n , then Y β ( T n ) > Y β ( P n ) .
Let U n k be a unicyclic graph which is obtained by connecting a path to C k of length n k . According to Lemmas 7 and 8, it follows that
Theorem 9. 
Let Ω be a unicyclic graph of girth k and order n; if Ω is not U n k , then Y β ( F n ) k < Y β ( Ω ) .
Using Lemma 9, it follows that
Theorem 10. 
C n is the unique graph on n vertices with smallest value for Y β .
Let F 1 , F 2 , and F 3 be the graphs in Figure 22; from Lemmas 7 and 9, the bicyclic graph with a minimum value of Y β is one from F 1 , F 2 , and F 3 , where
Y β ( F 1 ) = 4 ( 4 × 256 ) β + ( n 3 ) ( 4 × 4 ) β , = ( n 3 ) ( 16 ) β + 4 ( 1024 ) β .
.
Y β ( F 2 ) = Y β ( F 3 ) = 4 ( 4 × 27 ) β + ( 27 × 27 ) β + ( n 4 ) ( 4 × 4 ) β , = ( n 4 ) ( 16 ) β + 4 ( 108 ) β + ( 729 ) β , if ζ and ϱ are adjacent .
Y β ( F 2 ) = Y β ( F 3 ) = 6 ( 4 × 27 ) β + ( n 5 ) ( 4 × 4 ) β , = ( n 5 ) ( 16 ) β + 6 ( 108 ) β , if ζ and ϱ are not adjacent .
The difference is considered below:
Y β ( F 1 ) Y β ( F 2 ) = ( 16 ) β + 4 ( 1024 ) β 4 ( 108 ) β ( 729 ) β > 0 .
The inequality in (38) can be observed in Figure 24.
Y β ( F 1 ) Y β ( F 3 ) = 2 ( 16 ) β + 4 ( 1024 ) β 6 ( 108 ) β > 0 .
Y β ( F 2 ) Y β ( F 3 ) = ( 16 ) β + 4 ( 108 ) β + ( 729 ) β 6 ( 108 ) β > 0 .
The inequalities in (39) and (40) can be observed in Figure 25 and Figure 26.
From the above investigation, the following is true:
Theorem 11. 
Consider a family Ω containing bicyclic graphs on n vertices; then, the greatest value of Y β is for F 1 .
Theorem 12. 
Consider a family Ω containing bicyclic graphs on n vertices; then, F 3 is the graph with the smallest value for Y β . Here vertices with a degree of 3 are not adjacent, i.e., Y β ( F 3 ) < Y β ( F 2 ) < Y β ( F 1 ) .

9. Conclusions

Topological indices are numerical values associated with a graph and remain unchanged for isomorphic graphs. The field of indices is of considerable interest to researchers. In this article, we considered the second form of GPSCI and investigated its use on trees, unicyclic graphs, and bicyclic graphs. In the desired families, graphs with optimal values were found by means of mapping. Two mappings were considered for maximum values, and two mappings were considered for minimum values. The considered bi-cyclic graphs are of three types: cycles with no vertices in common, cycles with one vertex in common, and cycles with a path in common.

Author Contributions

Investigation, G.A.; Resources, I.-L.P.; Writing—original draft, M.Y.K.; Writing—Review and Editing, M.Y.K.; Supervision, G.A.; Funding acquisition, I.-L.P. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare no conflicts of interest.

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Figure 1. Diagram representing mapping A.
Figure 1. Diagram representing mapping A.
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Figure 2. Diagram presenting mapping B.
Figure 2. Diagram presenting mapping B.
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Figure 3. Diagram presenting bicyclic graphs. (i) Bicyclic graph having a common vertex, (ii) Bi cyclic graphs having no vertex in common, (iii) Bicyclic graphs having a path in common.
Figure 3. Diagram presenting bicyclic graphs. (i) Bicyclic graph having a common vertex, (ii) Bi cyclic graphs having no vertex in common, (iii) Bicyclic graphs having a path in common.
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Figure 4. Diagram presenting bicyclic graph S n ( θ , ϕ ) .
Figure 4. Diagram presenting bicyclic graph S n ( θ , ϕ ) .
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Figure 5. (a) The graph T n r ( θ , ϕ ) ; (b) the graph T n r ( ϕ , θ ) ; (c) the graph T n ( θ , ϕ ) . (d) All the pendant edges are connected to vertex ϱ on the cycle C θ . (e) All the pendant edges are connected to vertex ϱ on the cycle C ϕ ; (f) All the pendant edges are connected to vertex ϱ on the cycle on the path.
Figure 5. (a) The graph T n r ( θ , ϕ ) ; (b) the graph T n r ( ϕ , θ ) ; (c) the graph T n ( θ , ϕ ) . (d) All the pendant edges are connected to vertex ϱ on the cycle C θ . (e) All the pendant edges are connected to vertex ϱ on the cycle C ϕ ; (f) All the pendant edges are connected to vertex ϱ on the cycle on the path.
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Figure 6. Plot for Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) , where φ 1 and ϱ are not adjacent.
Figure 6. Plot for Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) , where φ 1 and ϱ are not adjacent.
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Figure 7. Plot for Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) , where φ 1 and ϱ are adjacent.
Figure 7. Plot for Y β ( T n r ( θ , ϕ ) ) Y β ( Ω ) , where φ 1 and ϱ are adjacent.
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Figure 8. Plot for Y β ( T n ( θ , ϕ ) ) Y β ( Ω ) , where 3 < r .
Figure 8. Plot for Y β ( T n ( θ , ϕ ) ) Y β ( Ω ) , where 3 < r .
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Figure 9. Plot for Y β ( T n ( θ , ϕ ) ) Y β ( Ω ) , where t = r or t = 2 .
Figure 9. Plot for Y β ( T n ( θ , ϕ ) ) Y β ( Ω ) , where t = r or t = 2 .
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Figure 10. Plot for Y β ( T n ( 3 , 3 ) ) Y β ( T n ( θ , ϕ ) ) .
Figure 10. Plot for Y β ( T n ( 3 , 3 ) ) Y β ( T n ( θ , ϕ ) ) .
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Figure 11. Plot for Y β ( T n ( 3 , 3 ) ) Y β ( T n ( θ , ϕ ) ) , where θ = 3 = q .
Figure 11. Plot for Y β ( T n ( 3 , 3 ) ) Y β ( T n ( θ , ϕ ) ) , where θ = 3 = q .
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Figure 12. Plot for Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r 1 ( θ , ϕ ) ) , where r > 2 .
Figure 12. Plot for Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r 1 ( θ , ϕ ) ) , where r > 2 .
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Figure 13. Plot for Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r 1 ( θ , ϕ ) ) , where r = 2 .
Figure 13. Plot for Y β ( T n r 1 ( θ , ϕ ) ) Y β ( T n r 1 ( θ , ϕ ) ) , where r = 2 .
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Figure 14. Plot for Y β ( T n 1 ( 3 , 3 ) ) Y β ( T n 1 ( θ , ϕ ) ) , where θ and ϕ , and both are not 3.
Figure 14. Plot for Y β ( T n 1 ( 3 , 3 ) ) Y β ( T n 1 ( θ , ϕ ) ) , where θ and ϕ , and both are not 3.
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Figure 15. Plot for Y β ( T n 1 ( 3 , 3 ) ) Y β ( T n 1 ( θ , ϕ ) ) , where θ = 3 = ϕ .
Figure 15. Plot for Y β ( T n 1 ( 3 , 3 ) ) Y β ( T n 1 ( θ , ϕ ) ) , where θ = 3 = ϕ .
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Figure 16. The graphs (i) Ω 1 , (ii) Ω 2 , (iii) Ω 3 , (iv) Ω 4 , (v) Y .
Figure 16. The graphs (i) Ω 1 , (ii) Ω 2 , (iii) Ω 3 , (iv) Ω 4 , (v) Y .
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Figure 17. Sketch for Y β ( S n ( 3 , 3 ) ) Y β ( T n 1 ( 3 , 3 ) ) .
Figure 17. Sketch for Y β ( S n ( 3 , 3 ) ) Y β ( T n 1 ( 3 , 3 ) ) .
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Figure 18. Plot for Y β ( Y ) Y β ( S n ( 3 , 3 ) ) .
Figure 18. Plot for Y β ( Y ) Y β ( S n ( 3 , 3 ) ) .
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Figure 19. Diagram for mapping C.
Figure 19. Diagram for mapping C.
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Figure 20. Diagram for remark using mapping C.
Figure 20. Diagram for remark using mapping C.
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Figure 21. Diagram for mapping D.
Figure 21. Diagram for mapping D.
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Figure 22. Diagram for F 1 , F 2 , and F 3 .
Figure 22. Diagram for F 1 , F 2 , and F 3 .
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Figure 23. Diagram for Lemma 9.
Figure 23. Diagram for Lemma 9.
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Figure 24. Plot for Y β ( F 1 ) Y β ( F 2 ) .
Figure 24. Plot for Y β ( F 1 ) Y β ( F 2 ) .
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Figure 25. Plot for Y β ( F 2 ) Y β ( F 3 ) .
Figure 25. Plot for Y β ( F 2 ) Y β ( F 3 ) .
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Figure 26. Plot for Y β ( F 2 ) Y β ( F 3 ) .
Figure 26. Plot for Y β ( F 2 ) Y β ( F 3 ) .
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Khan, M.Y.; Ali, G.; Popa, I.-L. Optimization in Symmetric Trees, Unicyclic Graphs, and Bicyclic Graphs with Help of Mappings Using Second Form of Generalized Power-Sum Connectivity Index. Symmetry 2025, 17, 122. https://doi.org/10.3390/sym17010122

AMA Style

Khan MY, Ali G, Popa I-L. Optimization in Symmetric Trees, Unicyclic Graphs, and Bicyclic Graphs with Help of Mappings Using Second Form of Generalized Power-Sum Connectivity Index. Symmetry. 2025; 17(1):122. https://doi.org/10.3390/sym17010122

Chicago/Turabian Style

Khan, Muhammad Yasin, Gohar Ali, and Ioan-Lucian Popa. 2025. "Optimization in Symmetric Trees, Unicyclic Graphs, and Bicyclic Graphs with Help of Mappings Using Second Form of Generalized Power-Sum Connectivity Index" Symmetry 17, no. 1: 122. https://doi.org/10.3390/sym17010122

APA Style

Khan, M. Y., Ali, G., & Popa, I.-L. (2025). Optimization in Symmetric Trees, Unicyclic Graphs, and Bicyclic Graphs with Help of Mappings Using Second Form of Generalized Power-Sum Connectivity Index. Symmetry, 17(1), 122. https://doi.org/10.3390/sym17010122

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