ON THE INDEPENDENCE POLYNOMIALS OF CERTAIN MOLECULAR GRAPHS
MOHAMMAD H. REYHANI, SAEID ALIKHANIa, ROSLAN HASNIb*
Department of Mathematics, Faculty of Science, Islamic Azad University, Yazd Branch, Yazd, Iran
aDepartment of Mathematics, Yazd University 89195-741, Yazd, Iran
bSchool of Mathematical Sciences, Universiti Sains Malaysia 11800 USM, Penang, Malaysia
The independence polynomial of a molecular graph G is the polynomial k
x
ki x G
I ( , ) = , where ik denote the number of independent sets of cardinality k in G. In this paper, we consider specific graphs denoted by G(m) and G1(m)G2 and obtain formulas for their independence polynomials which are in terms of Jacobsthal polynomial. Also we compute the independece polynomal of another kind of graphs.
(Recieved November 17, 2011; Accepted February 10, 2012)
Keywords: Independence polynomial; Jacobsthal polynomial; Graph
1. Introduction
A simple graph
G = ( V , E )
is a finite nonempty set V(G) of objects called vertices together with a (possibly empty) set E(G) of unordered pairs of distinct vertices of G called edges. In chemical graphs, the vertices of the graph correspond to the atoms of the molecule, and the edges represent the chemical bonds.An independent set of a graph G is a set of vertices where no two vertices are adjacent. The independence number is the size of a maximum independent set in the graph. For a graph G with independence
, let ikdenote the number of independent sets of cardinality k inG (
k = 0,1,
,
). The independence polynomial of G,( , ) = ,
0
= k k k
i x x
G
I is the generating
polynomial for the independent sequence
( i
0, i
1, i
2,
, i
)
([3]). The path P4 on 4 vertices, for example, has one independent set of cardinality 0 (the empty set), four independent sets of cardinality 1, and three independent sets of cardinality 2; its independence polynomial is then4, )=1 4 3 2
(P x x x
I .
Hoede and Li [5] obtained the following recursive formula for the independence polynomial of a graph.
Theorem 1 . For any vertex v of a graph G, I(G,x)=I(Gv,x)xI(G[v],x) where
[v ]
is the closed neighberhood of v , contains of v , together with all vertices incident withv .
*Corresponding author: [email protected]
Let us observe that if G and H are isomorphic, then I(G,x)=I(H,x). The converse is not generally true. For Two graphs G and H are independent equivalent, written G ~ H, if
) , (
= ) ,
(G x I H x
I . A graph G is independent unique, if I(H,x)=I(G,x) implies that H G. Let
[G ]
denote the independent equivalence class determined by the graph G under the equivalence relation~. Clearly, G is independent unique if and only if[ G ] =
{G}. A zero of) , (G x
I is called a independence zero of G.
The corona of two graphs G1 and G2, as defined by Frucht and Harary in [4], is the graph G1G2formed from one copy of G1 and |V(G1)|copies of G2, where the ith vertex of G1 is adjacent to every vertex in the ith copy of G2. The corona GK1, in particular, is the graph constructed from a copy of G, where for each vertex
v V
(G), a new vertex v and a pendant edge vv' are added.In Section 2, we study Jacobsthal polynomial and introduce two graphs with specific structures denoted by G(m) and
G
1( m ) G
2. Using the results related to Jacobsthal polynomial, we compute the independence polynomials of G(m) andG
1( m ) G
2 in Section 3.2. Jacobsthal polynomial
Jacobsthal polynomials,
J
n(x )
, named after the German mathematician E. Jacobsthal are related to Fibonacci polynomials. They are defined by) ( )
(
= )
( x J
1x xJ
2x J
n n
nwhere
J
1( x ) = J
2( x ) = 1
(see [6], p.469).In this section, we shall find the zeros of
J
n(x )
. First, we need the following two lemmas to obtain a solution of Jacobsthal polynomials.
Lemma 1 . For any real number u,
( ) = (1 ) ( )
1.
1 0
=
2 n i n i
i
n
u u u u
J
Proof. It is clear that the result holds when n=2. Now let n3. By induction, we have
)
( ) ( ) (
= )
( u
2u J
1u
2u u
2u J
2u
2u
J
n
n
n
i n n i
i i
n n i
i
u u u
u u
u
3 30
= 2
2 2 0
=
) ( ) (1 ) ( )
( ) (1
=
i n n i
i i n n i
i
u u
u
u
3 1 20
= 2 2
0
=
) ( ) (1 )
( ) (1
=
i n n i
i
i n n i
i n
u u
u u u
2 3 1
0
=
3 2 0
= 2
) ( ) (1
) ( ) (1 )
(1
=
i n n i
i
n
u u
u
3 1
0
=
2
(1 ) ( ) )
(1
=
. ) ( ) (1
=
11 0
=
i n n i
i
u
u
▄Corollary 1 . For any real number u ,
(2 u 1) J
n( u
2 u ) = (1 u )
n ( u )
n.
Proof. The result follows from Lemma 1 by using the identity
,
) (
=
11 0
=
n
i n ii n
n
b a b a b
a
for
a = 1 u
, b=u. ▄Lemma 2 . ( [2], p.64) For real numbers a, b and positive integer n ,
. 2 ;
2 )
)(
(
, 2 ;
2 )
(
=
2 2 2
2
1
= 2 2 2
1
1
=
even is n n
ab s b a b a b a
odd is n n
ab s b a b a b
a n
s n
n s n
cos cos
Theorem 2 . For any positive integer n ,
2 .
2 1 2
= ) (
2 1
1
=
n x s x
x J
n
s n
cos
Proof. If put
a = 1 u
, b=u, we havea b = a
2 b
2= 1 2 u
, therefore by using Lemma 2 and Corollary 1, for any real number2
1
u
,2 . ) 2(
1 2 2
= )
(
2 22 1
1
=
2
n u s
u u
u u
u J
n
s n
cos
Observe that for any real number x with
4
> 1
x
, there is a real number2
1
u
such thatx u
u
2 =
. Thus for each real number with4
> 1
x
,2 .
2 1 2
= ) (
2 1
1
=
n x s x
x J
n
s n
cos
Since Jn(x) is a polynomial with degree less than n, the above equality also holds for any real number
4
1
x
. Thus the result is obtained. ▄3. Independence polynomial of certain graphs
In this section we consider some specific graphs and compute their independence polynomial (see [1]). Let Pm1 be a path with vertices labeled by y0,y1,,ym, for
m 0
and let G be any graph. Denote by( )
0
m
G
v (or simply G(m), if there is no likelihood of confusion) a graph obtained from G by identifying the vertexv
0 of G with an end vertexy
0 of Pm1 (see Figure 1). For example, if G is a path P2, then G(m)=P2(m) is the path Pm2. Also, we denote the graph obtained from graphs G1 and G2 by adding a path Pm from a vertex in G1 to a vertexof G2, by G1(m)G2. (Figure 1).
Fig.1. Graphs G(m) and G1(m)G2, respectively.
Theorem 3 . Let n2 be integer. Then, the independence polynomial of G(n) is
).
, ( ) ( )
(1), ( ) (
= ) ), (
( G n x J x I G x xJ
1x I G x
I
n
n
Proof. Proof by induction on n. Since
J
1( x ) = J
2( x ) = 1
, the result is true forn = 2
by Theorem 1. Now suppose that the result is true for all natural numbers less than n and prove it for n. By using Theorem 1 forv = y
n, and induction hypothesis, we have= ) 2), ( ( ) 1), ( (
= ) ), (
(G n x I G n x xI G n x
I
) , ( ) ( )
(1), ( ) (
= J
n1x I G x xJ
n2x I G x ) , ( ) ( )
(1), ( ) (
( J
2x I G x xJ
3x I G x
x
n
n
) , ( )) ( )
( ( ) (1), ( )) ( )
( (
= J
n1x xJ
n2x I G x x J
n2x xJ
n3x I G x ).
, ( ) ( )
(1), ( ) (
= J
nx I G x xJ
n1x I G x
▄The following theorem gives the formula for computing the independence polynomial of graphs 2
1
( m ) G
G
as shown in Figure 1 :Theorem 4 . Let n5 be integer. The independence polynomial of
G
1( n ) G
2 is)
, ( ) (1), ( ( ) ( ) (1), ( ) (1), (
= ) , ) ( (
2 1
2 2
1 2 1
x G I x G I x x J x G I x G I
x G n G I
n
).
( ) , ( ) , ( ) ( )) (1), ( ) ,
(G1 x I G2 x J 3 x x2I G1 x I G2 x J 4 x
I n n
Proof. Proof by induction on n. If n=5, then by Theorems 1 and 3, and induction hypothesis, we have
= ) (2), ( ) , ( ( ) (3), ( ) (1), (
= ) , (5)
( G
1G
2x I G
1x I G
2x x I G
1x I G
2x
I
) , ( ) (1), ( ( ) (1), ( ) (1), ( ) (1
= x I G
1x I G
2x x I G
1x I G
2x
)., ( ) , ( )) (1), ( ) ,
(G1 x I G2 x x2I G1 x I G2 x
I
So the theorem is true for n=5. Now suppose that the result is true for less than n and we prove it for n. By Theorems 1 and 3, and induction hypothesis, we have
= ) 3), ( ( ) , ( ( ) 2), ( ( ) (1), (
= ) , ) ( (
2 1
2 1
2 1
x n G I x G I x x n G I x G I
x G n G I
) , ( ) (1), ( ( ) ( ) (1), ( ) (1),
( G
1x I G
2x J
2x x I G
1x I G
2x
I
n
).
( ) , ( ) , ( ) ( )) (1), ( ) ,
(G1 x I G2 x J 3 x x2I G1 x I G2 x J 4 x
I n n
▄
Theorem 3 implies that all forms of G1(m)G2 have the same independence polynomials.
As application of Theorem 3, we obtain the following formula:
Corollary 2 .
1. The independence polynomial of path
P
n is2 . 2 2
1 2
= ) (
= ) , (
2 1
1
=
2
n
x s x
x J x P I
n
s n
n
cos
2. The independence polynomial of cycle Cn (n2) is
).
( )
(
= ) ,
( C x J
1x xJ
1x I
n n
n Proof.1. By using Theorem 1, for
G =K
1, we have) , ( ) ( )
(1), ( ) (
= ) ), ( (
= ) ,
( P
1x I K
1n x J x I K
1x xJ
1x I K
1x
I
n n
n) ( )
( )
( ) 2 (1
= x Jn x xJn1 x x2Jn1 x
)
( )
(
= J
n2x xJ
n1x ).
(
= J
n3x
So we have the result.2. It follows from Theorems 1 and Part 1. ▄
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