# Two Classes of Topological Indices of Phenylene Molecule Graphs

Authors: Tingzeng Wu
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

A phenylene is a conjugated hydrocarbons molecule composed of six- and four-membered rings. The matching energy of a graphG is equal to the sum of the absolute values of the zeros of the matching polynomial of G, while the Hosoya index is defined as the total number of the independent edge sets of G. In this paper, we determine the extremal graph with respect to the matching energy and Hosoya index for all phenylene chains.


---

## Body

## 1. Introduction

Phenylenes are a class of conjugated hydrocarbons composed of six- and four-membered rings, where the six-membered rings (hexagons) are adjacent only to four-membered rings, and every four-membered ring is adjacent to a pair of hexagons. They are nanostructures that can be precisely designed and manufactured for a wide variety of applications; see [1–3] and the references therein.A topological index is a numerical quantity derived in an unambiguous manner from the structure graph of a molecule, as a graph structural invariant; that is, it does not depend on the labeling or the pictorial representation of a graph. Various topological indices usually reflect molecular size and shape. One topological index is Hosoya index, which was first introduced by Hosoya [4]. It plays an important role in the so-called inverse structure-property relationship problems. For details of Hosoya index and its applications, the readers are suggested to refer to [5, 6]. A new topological index in chemistry, matching energy, is first introduced by Gutman and Wagner [7] in 2012 to study topological resonance energy of conjugated molecules, which has received a lot of attention from researchers in recent years. For more background and applications about matching energy, see [8–16].In this paper, our aim is to determine the phenylenes with minimum and maximum matching energy (Hosoya index) among all the phenylenes withn hexagons.In the following we present some definitions and notations.LetG
=
(
V
,
E
) be a graph with the vertex set V
(
G
) and the edge set E
(
G
). Let e and v be an edge and a vertex in G, respectively. We denote by G
-
e the graph obtained from G by removing edge e and by G
-
v the graph obtained from G by deleting vertex v.Bym
(
G
,
k
) we denote the number of k-matchings of a graph G. The matching polynomial of a graph G with n vertices is fined as(1)
α
G
,
x
=
∑
k
≥
0
-
1
k
m
G
,
k
x
n
-
2
k
,where m
(
G
,
0
)
=
1 and m
(
G
,
k
)
≥
0 for all k
=
1,2
,
…
,
[
n
/
2
]. This expression α
(
G
,
x
) induces a quasi-order relation (i.e., reflexive and transitive relation) on the set of all graphs with n vertices. If G and H are two graphs with matching polynomial in the form (1), then the quasi-order ⪰ is defined by(2)
G
⪰
H
⟺
m
G
,
k
≥
m
H
,
k
∀
k
=
0,1
,
…
,
n
/
2
. Particularly, if G
⪰
H and there exists some k such that m
(
G
,
k
)
>
m
(
H
,
k
), then we write G
≻
H.Gutman and Wagner in [7] first proposed the concept of thematching energy of a graph, denoted by M
E
(
G
), as(3)
M
E
=
M
E
G
=
2
π
∫
0
∞
x
-
2
ln
⁡
∑
k
≥
0
m
G
,
k
x
2
k
d
x
.Meanwhile, they gave also another form of definition of matching energy of a graph. That is,(4)
M
E
G
=
∑
i
=
1
n
μ
i
,where μ
i denotes the root of matching polynomial of G. By (2) and (3), we easily obtain the fact as follows:(5)
G
⪰
H
⟹
M
E
G
≥
M
E
H
,
G
≻
H
⟹
M
E
G
>
M
E
H
.TheZ-counting polynomial was defined by Hosoya [4] as(6)
Z
G
=
Z
G
,
x
=
∑
k
m
G
,
k
x
k
.Particularly, set x
=
1; then Z
(
G
,
1
)
=
∑
k
‍
m
(
G
,
k
) is calledHosoya index of G. Furthermore, The Z-counting polynomial of graphs has the property as follows.Lemma 1 (see [4]).
(a) LetG be a graph consisting of two components G
1 and G
2. Then Z
(
G
)
=
Z
(
G
1
)
Z
(
G
2
)
.
(b) Letu
v
∈
E
(
G
) be an edge of G. Then Z
(
G
)
=
Z
(
G
-
u
v
)
+
x
Z
(
G
-
u
-
v
).Aphenylene chain containing n
(
n
≥
2
) hexagons, denoted by P
H
B
n, is a phenylene with the properties that (a) no vertex is incident with two hexagons or squares and (b) no hexagon is adjacent to more than two squares. We denote by B
n the set of all phenylene chains with n hexagons. Let P
H
B
n
∈
B
n. If the subgraph P
H
B
n induced by the vertices with degree 3 is the union of n
-
1 disjoint copies of a square, then P
H
B
n is called alinear phenylene chain and denoted by P
H
L
n (see Figure 1). If the subgraph P
H
B
nof induced by the vertices with degree 3 is isomorphic to the graph S
n
-
1 having n
-
1 squares (see Figure 1), then P
H
B
n is called azigzag phenylene chain and is denoted by P
H
Z
n (see Figure 1). It is easy to see that B
2
=
{
P
H
L
2
}
=
{
P
H
Z
2
} and B
3
=
{
P
H
L
3
,
P
H
Z
3
}. Finally, by the definition of a phenylene chain, any element P
H
B
n in B
n can be obtained from an appropriately chosen graph P
H
B
n
-
1
∈
B
n
-
1 by attaching to it a new graph θ, where θ is obtained from an edge of a square attaching an edge of a hexagon; see Figure 2.Figure 1
A linear phenylene chainP
H
L
n and a zigzag phenylene chain P
H
Z
n.Figure 2
A phenylene chainP
H
B
n.


## 2. Main Results

Theorem 2.
LetB
n be the set of all phenylene chains with n hexagons. For any PH
B
n
∈
B
n, then (7)
ME
PH
L
n
≤
ME
PH
B
n
≤
ME
PH
Z
n
,where the equalities on the left side hold only if PH
B
n
≅
PH
L
n, and the equalities on the right side hold only if PH
B
n
≅
PH
Z
n.By (2) and (5), we know that Theorem 2 holding only needs to prove the following result.Theorem 3.
For anyPH
B
n
∈
B
n and for each k
≥
0,(8)
m
PH
L
n
,
k
≤
m
PH
B
n
,
k
≤
m
PH
Z
n
,
k
,where the equalities on the left side hold only if PH
B
n
≅
PH
L
n and the equalities on the right side hold only if PH
B
n
≅
PH
Z
n.Letf
(
x
)
=
∑
k
=
0
n
‍
a
k
x
k and g
(
x
)
=
∑
k
=
0
n
‍
b
k
x
k be two polynomials of x. We say f
(
x
)
⪯
g
(
x
) if a
k
≤
b
k for all k. If f
(
x
)
⪯
g
(
x
) and there exists some k such that a
k
<
b
k, then we say f
(
x
)
≺
g
(
x
). By (6), it is easy to obtain the following result which is equivalent to Theorem 3.Theorem 4.
For anyPH
B
n
∈
B
n
(
n
≥
2
),(I)
ifPH
L
n
≠
PH
B
n, then Z
(
PH
L
n
)
≺
Z
(
PH
B
n
),
(II)
ifPH
Z
n
≠
PH
B
n, then Z
(
PH
B
n
)
≺
Z
(
PH
Z
n
).In the following we will use the notationG for Z
(
G
), when it would lead to no confusion.Proof.
Checking Figure2, by Lemma 1, we obtained that(9)
P
H
B
n
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
B
n
-
1
+
x
+
4
x
2
+
3
x
3
P
H
B
n
-
1
-
s
2
n
-
3
+
P
H
B
n
-
1
-
t
2
n
-
3
+
x
2
+
3
x
3
+
x
4
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
,
(10)
P
H
B
n
-
y
=
1
+
4
x
+
3
x
2
P
H
B
n
-
1
+
x
+
2
x
2
P
H
B
n
-
1
-
t
2
n
-
3
+
x
+
3
x
2
+
x
3
P
H
B
n
-
1
-
s
2
n
-
3
+
x
2
+
2
x
3
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
1
n
,
1
+
4
x
+
3
x
2
P
H
B
n
-
1
+
x
+
2
x
2
+
x
3
P
H
B
n
-
1
-
t
2
n
-
3
+
x
+
2
x
2
P
H
B
n
-
1
-
s
2
n
-
3
+
x
2
+
x
3
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
2
n
,
1
+
4
x
+
3
x
2
P
H
B
n
-
1
+
x
+
2
x
2
+
x
3
P
H
B
n
-
1
-
s
2
n
-
3
+
x
+
2
x
2
P
H
B
n
-
1
-
t
2
n
-
3
+
x
2
+
x
3
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
3
n
,
1
+
4
x
+
3
x
2
P
H
B
n
-
1
+
x
+
2
x
2
P
H
B
n
-
1
-
s
2
n
-
3
+
x
+
3
x
2
+
x
3
P
H
B
n
-
1
-
t
2
n
-
3
+
x
2
+
2
x
3
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
4
n
,
(11)
P
H
B
n
-
y
-
z
=
1
+
3
x
+
x
2
P
H
B
n
-
1
+
x
+
x
2
P
H
B
n
-
1
-
t
2
n
-
3
+
x
+
2
x
2
P
H
B
n
-
1
-
s
2
n
-
3
+
x
2
+
x
3
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
1
n
,
z
=
b
2
n
;
1
+
3
x
+
x
2
P
H
B
n
-
1
+
x
+
x
2
P
H
B
n
-
1
-
t
2
n
-
3
+
x
+
x
2
P
H
B
n
-
1
-
s
2
n
-
3
+
x
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
2
n
,
z
=
b
3
n
;
1
+
3
x
+
x
2
P
H
B
n
-
1
+
x
+
x
2
P
H
B
n
-
1
-
s
2
n
-
3
+
x
+
2
x
2
P
H
B
n
-
1
-
t
2
n
-
3
+
x
2
+
x
3
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
i
f
y
=
b
3
n
,
z
=
b
4
n
.
By (10) and (11), we have(i)
P
H
B
n
-
b
3
n
≺
P
H
B
n
-
b
1
n and P
H
B
n
-
b
2
n
≺
P
H
B
n
-
b
4
n;
(ii)
P
H
B
n
-
b
2
n
-
b
3
n
≺
P
H
B
n
-
b
1
n
-
b
2
n and P
H
B
n
-
b
2
n
-
b
3
n
≺
P
H
B
n
-
b
3
n
-
b
4
n.
Particularly, ifP
H
B
n
=
P
H
L
n, then(i′)
P
H
L
n
-
a
2
n
=
P
H
L
n
-
a
3
n
≺
P
H
L
n
-
a
1
n
=
P
H
L
n
-
a
4
n,
(ii′)
P
H
L
n
-
a
2
n
-
a
3
n
≺
P
H
L
n
-
a
1
n
-
a
2
n
=
P
H
L
n
-
a
3
n
-
a
4
n,
(iii′)
(
P
H
L
n
-
a
2
n
)
+
(
P
H
L
n
-
a
3
n
)
≺
(
P
H
L
n
-
a
2
n
)+(
P
H
L
n
-
a
1
n
)
=
(
P
H
L
n
-
a
3
n
)
+
(
P
H
L
n
-
a
4
n
).
We prove Theorem4(I) by mathematical induction.First we considern
=
3. In this case, B
n
=
{
P
H
L
3
,
P
H
Z
3
}.By (9), we have (12)
P
H
L
3
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
L
2
+
x
+
4
x
2
+
3
x
3
P
H
L
2
-
a
2
2
+
P
H
L
2
-
a
3
2
+
x
2
+
3
x
3
+
x
4
P
H
L
2
-
a
2
2
-
a
3
2
,
P
H
Z
3
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
Z
2
+
x
+
4
x
2
+
3
x
3
P
H
Z
2
-
d
1
2
+
P
H
Z
2
-
d
2
2
+
x
2
+
3
x
3
+
x
4
P
H
Z
2
-
d
1
2
-
d
2
2
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
L
2
+
x
+
4
x
2
+
3
x
3
P
H
L
2
-
a
4
2
+
P
H
L
2
-
a
3
2
+
x
2
+
3
x
3
+
x
4
P
H
L
2
-
a
3
2
-
a
4
2
. By (i′)–(iii′), we have P
H
L
3
≺
P
H
Z
3.Suppose that Theorem4(I) is right for all phenylene chains with few n hexagons. Let P
H
B
n be a phenylene chain with n
≥
4 hexagons, which is obtained from P
H
B
n
-
1
∈
B
n
-
1 by attaching to it a new θ (see Figure 2). We show that if P
H
L
n
≠
P
H
B
n, then P
H
L
n
≺
P
H
B
n. By (9) we obtain that (13)
P
H
L
n
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
L
n
-
1
+
x
+
4
x
2
+
3
x
3
P
H
L
n
-
1
-
u
2
n
-
3
+
P
H
L
n
-
1
-
v
2
n
-
3
+
x
2
+
3
x
3
+
x
4
P
H
L
n
-
1
-
u
2
n
-
3
-
v
2
n
-
3
,
P
H
B
n
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
B
n
-
1
+
x
+
4
x
2
+
3
x
3
P
H
B
n
-
1
-
s
2
n
-
3
+
P
H
B
n
-
1
-
t
2
n
-
3
+
x
2
+
3
x
3
+
x
4
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
.By the inductive hypotheses we haveP
H
L
n
-
1
⪯
P
H
B
n
-
1, (
P
H
L
n
-
1
-
u
2
n
-
3
)
+
(
P
H
L
n
-
1
-
v
2
n
-
3
)
⪯
(
P
H
B
n
-
1
-
s
2
n
-
3
)
+
(
P
H
B
n
-
1
-
t
2
n
-
3
), and (
P
H
L
n
-
1
-
u
2
n
-
3
-
v
2
n
-
3
)
⪯
(
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
). Since P
H
L
n
≠
P
H
B
n, either P
H
L
n
-
1
≠
P
H
B
n
-
1 or {
s
n
-
1
,
t
n
-
1
}
≠
{
u
2
n
-
3
,
v
2
n
-
3
}, and hence at least one of the three inequalities is strict. Therefore, we get that P
H
L
n
≺
P
H
B
n.In the following we prove Theorem4(II) by induction.By the proof of Theorem4(I), we know that P
H
B
3
≺
P
H
Z
3.Similarly, suppose that Theorem4(II) is right for all phenylene chains with few n hexagons. Let P
H
B
n be a phenylene chain with n
≥
4 hexagons, which is obtained from P
H
B
n
-
1
∈
B
n
-
1 by attaching to it a new θ (see Figure 2). We show that if P
H
B
n
≠
P
H
Z
n, then P
H
B
n
≺
P
H
Z
n. By (9) we have (14)
P
H
B
n
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
B
n
-
1
+
x
+
4
x
2
+
3
x
3
P
H
B
n
-
1
-
s
2
n
-
3
+
P
H
B
n
-
1
-
t
2
n
-
3
+
x
2
+
3
x
3
+
x
4
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
P
H
Z
n
=
1
+
6
x
+
9
x
2
+
2
x
3
P
H
Z
n
-
1
+
x
+
4
x
2
+
3
x
3
P
H
Z
n
-
1
-
u
2
n
-
3
+
P
H
Z
n
-
1
-
v
2
n
-
3
+
x
2
+
3
x
3
+
x
4
P
H
Z
n
-
1
-
u
2
n
-
3
-
v
2
n
-
3
.By the inductive hypotheses we have P
H
B
n
-
1
⪯
P
H
Z
n
-
1, (
P
H
B
n
-
1
-
s
2
n
-
3
)
+
(
P
H
B
n
-
1
-
t
2
n
-
3
)
⪯
(
P
H
Z
n
-
1
-
u
2
n
-
3
)
+
(
P
H
Z
n
-
1
-
v
2
n
-
3
), and (
P
H
B
n
-
1
-
s
2
n
-
3
-
t
2
n
-
3
)
⪯
(
P
H
Z
n
-
1
-
u
2
n
-
3
-
v
2
n
-
3
). Since P
H
B
n
≠
P
H
Z
n, either P
H
B
n
-
1
≠
P
H
Z
n
-
1 or {
s
n
-
1
,
t
n
-
1
}
≠
{
u
2
n
-
3
,
v
2
n
-
3
}, and hence at least one of the three inequalities is strict. Therefore, we get that P
H
B
n
≺
P
H
Z
n.The proof is complete.By the definition of Hosoya index and Theorem4, we can obtain the following result.Theorem 5.
LetB
n be the set of all phenylene chains with n hexagons. For any PH
B
n
∈
B
n, then (15)
Z
PH
L
n
,
1
≤
Z
PH
B
n
,
1
≤
Z
PH
Z
n
,
1
,where the equalities on the left side hold only if PH
B
n
≅
PH
L
n and the equalities on the right side hold only if PH
B
n
≅
PH
Z
n.



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*Source: 8421396-2016-04-21.xml*