Domination polynomial of generalized friendship and generalized book graphs

0.28 MB·English

by  Somayeh Jahari

Domination polynomial of generalized friendship
and generalized book graphs
S. Jahari1 and S. Alikhani
5
1
Department of Mathematics, Yazd University
0
89195-741, Yazd, Iran
2

n
a
J
ABSTRACT
3
2

Let G be a simple graph of order n. The domination polynomial of G is the polynomial
]
O D(G,x) = n d(G,i)xi, where d(G,i) is the number of dominating sets of G of size i and
Pi=γ(G)
C γ(G) is the domination number of G. Let n be any positive integer and F be the Friendship
n
. graph with 2n +1 vertices and 3n edges, formed by the join of K with nK . We study the
h 1 2
t domination polynomials of generalized friendship graphs. We also consider the n-book graphs
a
m B , formed by joining n copies of the cycle graph C with a common edge and study the domi-
n 4
[ nation polynomials of some generalized book graphs. In particular we examine the domination

roots of these families, and find the limiting curve for the roots.
1
v
6 Mathematics Subject Classification: 05C60.
5
8 Keywords: Dominationpolynomial;friendshipgraph;flowergraphs;generalizedbookgraphs;family.
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0
.
1
0
5
1 Introduction
1
:
v
i
X
Let G = (V,E) be a simple graph. For any vertex v V(G), the open neighborhood of v
r ∈
a
is the set N(v) = u V(G) u,v E(G) and the closed neighborhood of v is the set
{ ∈ |{ } ∈ }
N[v] = N(v) v . For a set S V(G), the open neighborhood of S is N(S) = N(v)
∪{ } ⊆ Sv∈S
and the closed neighborhood of S is N[S] = N(S) S. A set S V(G) is a dominating
∪ ⊆
set if N[S] = V or equivalently, every vertex in V(G) S is adjacent to at least one vertex in
\
S. The domination number γ(G) is the minimum cardinality of a dominating set in G. For
a detailed treatment of these parameters, the reader is referred to [10]. The i-subset of V(G)
is a subset of V(G) of size i. Let (G,i) be the family of dominating sets of a graph G with
D
1
Correspondingauthor. E-mail: [email protected]
1
cardinality i and let d(G,i) = (G,i). The domination polynomial D(G,x) of G is defined as
|D |
D(G,x) = |V(G)| d(G,i)xi, where γ(G) is the domination number of G (see [1, 3]). A root
Pi=γ(G)
of D(G,x) is called a domination root of G. The set of distinct roots of D(G,x) is denoted by
Z(D(G,x)).
Calculating the domination polynomialof a graph G is difficult in general, as the smallest power
of a non-zero term is the domination number γ(G) of the graph, and determining whether
γ(G) k is known to be NP-complete [9]. But for certain classes of graphs, we can find a
≤
closed form expression for the domination polynomial. In [4] the domination polynomial and
thedominationrootsoffriendshipgraphshasbeenstudied. Inthispaperwewouldliketoobtain
some further results of this kind. We consider generalized friendship graph (or flower graphs),
and generalized book graphs and calculate their domination polynomials. Also we explore the
nature and location of their roots.
2 Domination polynomial of generalized friendship graphs
LetconsiderthegraphsF obtained byselectingonevertex ineach ofntriangles andidentifying
n
them (Figure 1). Some call them Dutch-Windmill graphs [15] and friendship graphs.
Figure 1: Friendship graphs F ,F ,F and F , respectively.
2 3 4 n
The generalized friendship graph F is a collection of n cycles (all of order q), meeting at a
q,n
common vertex (see Figure 2). The generalized friendship graph may also be referred to as a
flower [13].
In this section we compute the domination polynomial of the flowers F . We need some
4,n
preliminaries.
The vertex contraction G/u of a graph G by a vertex u is the operation under which all vertices
2
v
Figure 2: The flowers F , F , F and F , respectively.
4,2 4,3 4,4 4,n
in N(u) are joined to each other and then u is deleted (see[14]).
The following theorem is useful for finding the recurrence relations for the domination polyno-
mials of graphs.
Theorem 1.[2, 11] Let G be a graph. For any vertex u in G we have
D(G,x) = xD(G/u,x)+D(G u,x)+xD(G N[u],x) (1+x)p (G,x),
u
− − −
where p (G,x) is the polynomial counting the dominating sets of G u which do not contain
u
−
any vertex of N(u) in G.
The following theorem gives formula for the domination polynomial of F .
n
Theorem 2. [4] For every n N,
∈
D(F ,x) = (2x+x2)n +x(1+x)2n.
n
Domination polynomial satisfies a recurrence relation for arbitrary graphs which is based on the
edge andvertex elimination operations. Therecurrenceuses composite operations, e.g. G e/u,
−
which stands for (G e)/u.
−
Theorem 3.[11] Let G be a graph. For every edge e = u,v E,
{ } ∈
x
D(G,x) = D(G e,x)+ D(G e/u,x)+D(G e/v,x)
h
− x 1 − −
−
D(G/u,x) D(G/v,x) D(G N[u],x) D(G N[v],x)
− − − − − −
+ D(G e N[u],x)+D(G e N[v],x) .
i
− − − −
3
The following theorem gives recurrence relation for the domination polynomial of F .
4,n
Theorem 4. For every n 2,
≥
D(F ,x) = ((1+x)3+x)D(F ,x) (1+3x)(x+3x2+x3)n 1
4,n 4,n 1 −
− −
+(1+x)3xn 1 (x2+x)(x3 +3x2+3x)n 1,
− −
−
where D(F ,x) = x4+4x3+6x2.
4,1
Proof. An elementary observation is that if G and G are graphs of orders n and n , respec-
1 2 1 2
tively, then
D(G G ,x) = D(G ,x)D(G ,x).
1 2 1 2
∪
Consider graph F and a vertex v in Figure 2. By Theorem 1 we have:
4,n
D(F ,x) = xD(F /v,x)+D(F v,x)+xD(F N[v],x) (1+x)p (F ,x)
4,n 4,n 4,n 4,n v 4,n
− − −
= xD(F /v,x)+D(F v,x)+xD(F ,x)
4,n 4,n 4,n 1
− −
(1+x)xD( n 1K ,x), (1)
− ∪i=−1 3
where D(K ,x) = x3+3x2+3x.
3
u u′
Figure 3: Graphs F /v and F v, respectively.
4,n 4,n
−
Now we use Theorem 3 to obtain the domination polynomial of the graph F /v = G (see
4,n
Figure 3). We have
x
D(F /v,x) = D(F v,x)+ D( n 1P ,x) D( n 1P ,x)
4,n 4,n− x 1h− ∪i=−1 3 − ∪i=−1 3
−
+ xD( n 1P ,x)+xD( n 1P ,x)
∪i=−1 3 ∪i=−1 3 i
= D(F v,x)+2xD(P ,x)n 1, (2)
4,n 3 −
−
4
where D(G e/u,x) D(G e/u,x) D(G/u,x) D(G/u,x) and (F /v) e F v
′ ′ 4,n 4,n
− ≃ − ≃ ≃ − ≃ −
in Figure 3.
Now we use Theorem 1 to obtain the domination polynomial of the graph F v = G (see
4,n
−
Figure 3). We have
D(F v,x) = xD((F v)/u,x)+D(F v u,x)+xD(F v N[u],x)
4,n 4,n 4,n 4,n
− − − − − −
(1+x)p (F v,x)
u 4,n
− −
= xD((F v)/u,x)+D(F v u,x)+x(xD( n 1P ,x))
4,n − 4,n − − ∪i=−1 3
(1+x)xD( n 1P ,x)
− ∪i=−1 3
= (1+x)D((F v)/u,x) xD(P ,x)n 1. (3)
4,n 3 −
− −
Note that we used D((F v)/u,x) D(F v u,x) (see Figure 4).
4,n 4,n
− ≃ − −
UseTheorem1toobtainthedominationpolynomialofthegraph(F v)/u = G(seeFigure4).
4,n
−
We have
u′
Figure 4: The graph (F v)/u.
4,n
−
D(F v)/u,x) = xD((F v)/u,x)+D(F v u,x)+xD(F v N[u],x)
4,n 4,n ′ 4,n ′ 4,n ′
− − − − − −
(1+x)pu′(F4,n v,x)
− −
= xD(F ,x)+D(F ,x)+xD( n 1P ,x)
4,n−1 4,n−1 ∪i=−1 3
(1+x)(D( n 1P ,x) xn 1)
− ∪i=−1 3 − −
= (1+x)D(F ,x) D(P ,x)n 1+(1+x)(xn 1), (4)
4,n 1 3 − −
− −
5
Figure 5: Domination roots of graphs F , for 1 n 30.
4,n
≤ ≤
Consequently, by equations 1, 2, 3 and 4 have:
D(F ,x) = (1+4x+3x2+x3)D(F ,x) (1+3x)(x+3x2+x3)n 1
4,n 4,n 1 −
− −
+(1+x)3xn 1 (x2+x)(x3+3x2+3x)n 1.
− −
−
The domination roots of F exhibit a number of interesting properties (see Figure 5).
4,n
If we can find an explicit formula for the domination polynomial of a graph, there are still
interesting, difficultproblemsconcerningtheroots. For every oddnaturalnumbern, nononzero
real number is a domination root of F [4]. Also we think that for n even, F have exactly three
n n
real roots. Using Maple, we think that these are true for F . Therefore, we pose the following:
4,n
Question 1 For even n 4, does F have exactly three real roots?
n
≥
Conjecture 1 For every odd natural number n, no nonzero real number is a domination root
of F .
4,n
Question 2 What is a good upper bound on the modulus of the roots of F ?
4,n
6
It is natural to ask about the complex domination roots of F . The plot in Figure 5 suggests
4,n
that the roots tend to lie on a curve.
Conjecture 2 The limit of domination roots of F is hyperbola.
4,n
In [7] a family of graphs was produced with roots just barely in the right-half plane (showing
thatnotalldominationpolynomialsarestable), butFigure5providesanexplicitfamily(namely
the F ) whose domination roots have unbounded positive real part.
4,n
3 Domination polynomial of generalized book graphs
A book graph B , is defined as follows V(B ) = u ,u v ,w : 1 i n and E(B ) =
n n 1 2 i i n
{ }∪{ ≤ ≤ }
u u u v , u w , v w : 1 i n . We consider the generalized book graph B with
1 2 1 i 2 i i i n,m
{ }∪{ ≤ ≤ }
vertex and edge sets by V(B ) = u : 1 i m 2 v ,w : 1 i n and E(B ) =
n,m i i i n,m
{ ≤ ≤ − }∪{ ≤ ≤ }
u u : 1 i m 3 u w : 1 j n, i = m 2 u v :1 i n v w :1 i n
i i+1 i j 1 i i i
{ ≤ ≤ − }∪{ ≤ ≤ − }∪{ ≤ ≤ }∪{ ≤ ≤ }
(see Figure 6).
v
Figure 6: Graphs B and B , respectively.
n n,5
The following theorem gives formula for the domination polynomial of B .
n
Theorem 5.[4] For every n N,
∈
D(B ,x) = (x2+2x)n(2x+1)+x2(x+1)2n 2xn.
n
−
Figure 7 shows the domination roots of book graphs B for n 30.
n
≤
7
Figure 7: Domination roots of graphs B , for 1 n 30.
n
≤ ≤
In this section we compute domination polynomial of the book graphs B . We need some
n,5
preliminaries.
We begin with a graph operation. For two graphs G = (V,E) and H = (W,F), the corona
G H is the graph arising from the disjoint union of G with V copies of H, by adding edges
◦ | |
between the ith vertex of G and all vertices of ith copy of H [8]. It is easy to see that the corona
operation of two graphs does not have the commutative property. The following theorem gives
the domination polynomial of graphs of the form H K , which is needed to obtain our result.
1
◦
Theorem 6.[1] Let G be a graph. Then D(G,x) = (x2 +2x)n if and only if G = H K for
1
◦
some graph H of order n.
Given any two graphs G and H we definethe Cartesian product, denoted G(cid:3)H, to be the graph
with vertex set V(G) V(H) and edges between two vertices (u ,v ) and (u ,v ) if and only if
1 1 2 2
×
either u = u and v v E(H) or u u E(G) and v = v . This product is well known to be
1 2 1 2 1 2 1 2
∈ ∈
commutative.
Theorem 7.[12] The domination polynomial for K (cid:3)K is
n 2
D(K (cid:3)K ,x) = ((1+x)n 1)2 +2xn.
n 2
−
8
The following theorem gives formula for the domination polynomial of B .
n,5
Theorem 8. For every n N,
∈
D(B ,x) = x2(x+1)2n+1 2xn+1+(x2+2x)n(2x2+3x).
n,5
−
Proof. Consider graph B in Figure 6. By Theorems 1 and 5 we have:
n,5
D(B ,x) = xD(B /v,x)+D(B v,x)+xD(B N[v],x)
n,5 n,5 n,5 n,5
− −
(1+x)p (B ,x)
v n,5
−
= xD(B ,x)+D(B v,x)+x(D( n K ,x))
n n,5− ∪i=1 2
(1+x)[(x2 +2x)n 2xn]
− −
= x[(x2+2x)n(2x+1)+x2(x+1)2n 2xn]+D(B v,x)
n,5
− −
+x(x2+2x)n (1+x)[(x2+2x)n 2xn]
− −
= x3(x+1)2n +2xn+D(B v,x)+(x2+2x)n(2x2+x 1). (5)
n,5
− −
u
Figure 8: The graph B v.
n,5
−
Now we use Theorem 1 to obtain the domination polynomial of the graph B v = G (see
n,5
−
Figure 8). We have
D(B v,x) = xD(G/u,x)+D(G u,x)+xD(G N[u],x) (1+x)p (G,x)
n,5 u
− − − −
= xD(G/u,x)+D(G u,x)+xD(K ,x) (1+x)(xn(1+x))
1,n
− −
= xD(G/u,x)+D(G u,x)+x(xn+x(1+x)n) xn(1+x)2, (6)
− −
where D(K ,x) = xn+x(1+x)n.
1,n
9
Use Theorems 1 and 6 to obtain the domination polynomial of the graph B v u= G (see
n,5
− −
Figure 9). We have
Figure 9: Graphs (B v)/u and B v u, respectively.
n,5 n,5
− − −
D(B v u,x) = xD(G/w,x)+D(G w,x)+xD(G N[w],x)
n,5
− − − −
(1+x)p (G,x)
w
−
= xD(K K ,x)+D( n P ,x)+x(xn) (1+x)xn
n◦ 1 ∪i=1 2 −
= (2x+x2)n(x+1) xn. (7)
−
UseTheorems 1and6 and7to obtain thedomination polynomialof thegraph(B v)/u = G
n,5
−
(see Figure 9). We have
D((Bn,5 v)/u,x) = xD(G/u′,x)+D(G u′,x)+xD(G N[u′],x) (1+x)pu′(G,x)
− − − −
= xD(K (cid:3)K ,x)+D(K K ,x)+xD(K ,x) (1+x)xn
n 2 n 1 n
◦ −
= x((1+x)n 1)2+2xn)+(2x+x2)n +x((1+x)n 1) (1+x)xn
− − −
= x((1+x)n 1)(1+x)n+(2x+x2)n +xn(x 1). (8)
− −
By equations 6, 7 and 8 have:
D(B v,x) = (x2+2x)n(2x+1)+x2(x+1)2n 2xn(1+x). (9)
n,5
− −
10