J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
On the Vertex - Cover Polynomials of Gear Graphs AYYAKUTTY VIJAYAN1 and BERNONDES STEPHEN JOHN2 1
Assistant Professor, Department of Mathematics, Nesamony Memorial Christian College, Marthandam Kanyakumari District, Tamil Nadu-629165, India. 2 Assistant Professor, Department of Mathematics, Annai Velankanni College, Tholayavattom Kanyakumari District, Tamil Nadu-629157, India. (Received on : March 26, 2012) ABSTRACT The vertex cover polynomial of a graph G of order n has been already introduced in3. It is defined as the polynomial, C (G, x) = | V(G) |
c (G, i) xi, where Σ i = β (G)
c (G, i) is the number of vertex covering
sets of G of size i and β (G) is the covering number of G. In this paper, we obtain the vertex cover polynomial of the Gear graph and some properties of the coefficients of the vertex cover polynomial of the Gear graph have been studied. We obtain some recurrence relations for the coefficients of the vertex cover polynomials of Gear graph. Also, it has been proved that the coefficients of the vertex cover polynomial of Gear graph are log concave. Keywords: Vertex covering sets, vertex cover number, vertex cover polynomial, Gear graph.
INTRODUCTION Let G = (V, E) be a simple graph. For any vertex v 0 V, the open neighborhood of v is the set N (v) = { u 0 V/ uv 0 E} and the closed neighborhood of v is the set N [v] = N (v) χ {v}. For a set
S φV, the open neighborhood of S is N (s) = U N (v) and the closed v∈S
neighborhood of S is N [S] = N (S) χ S . The Gear graph denoted by Gn is a graph obtained by inserting an extra vertex between each pair of adjacent vertices on the
Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)
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Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
perimeter of wheel graph. It is also known as cogwheel. Gn has 2n + 1 vertices and 3n edges. A set S φV is a vertex covering of G, if every edge uv 0 E is adjacent to atleast one vertex in S. The vertex covering number β (G) is the minimum cardinality of the minimum vertex covering sets in G. A vertex covering set with cardinality β (G) is called a β-set. Let C (G, i) be the family of vertex covering sets with cardinality i and let c (G, i) = | C (G, i)|. The | V(G) |
polynomial, C (G, x) =
Σ
c (G, i) xi is
i = β (G)
defined as the vertex cover polynomial of G. In3, many properties of the vertex cover polynomial have been studied and derived the vertex cover polynomials for some standard graphs. In this paper, we derived the vertex cover polynomial of Gear graph and some properties of the coefficients of the vertex cover polynomials of Gear graphs have been studied. Also, we obtain that the vertex cover polynomial of Gear graph is log concave. The number of edges incident to the vertex v of a graph G is called the degree of the vertex v in G. It is denoted by deg (v). δ (G) and ∆ (G) are the minimum and maximum of the degree of the vertices in G respectively. A finite sequence of real numbers (a0, a1, a2, . . . , an) is said to be Logarithmically concave (or simply, logconcave) if the inequality ai2 ∃ ai – 1 . ai + 1 is valid for every i 0 {1, 2, . . . n – 1}. A polynomial is called log-concave if the sequence of its coefficients is log concave.
2.VERTEX COVER POLYNOMIAL OF GEAR GRAPH Definition 2.1 Let C (G, i) be the family of all vertex covering sets of G with cardinality i and let c (G, i) = | C (G, i)|. The vertex cover polynomial of G is defined as | V(G) |
C (G, x) =
Σ c (G, i) i = β (G)
xi, where β(G) is
the vertex covering number. In3 the vertex cover polynomial of paths, cycles and wheels are obtained as n
Σ
C (Pn , x) =
n −1 i = 2
n
Σ
C (Cn, x) =
n i = 2
n i+1 i n − i
n i i n − i
i x ;
i x
C (Wn, x) = xn – 1 n
Σ
+
n+1 i = 2
n − 1 i − 1 i x . i − 1 n − i
Theorem: 2.2 Let Gn be any Gear graph with 2n + 1 vertices and 3n edges then its vertex cover polynomial is C (Gn, x) = xn + n
Σ
i=1
n + 1 + i n n + i 2n + 1 +x x i i −1
Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)
Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
Proof: The Gear graph with 2n + 1 vertices and 3n edges is given below in figure 1. The only minimum vertex covering set with n elements is S1 = { v1, v2, v3, . . ., vn } Therefore, c (Gn, n) = 1 The vertex covering sets with n + 1 elements are
199
Similarly, c (Gn , n + 3) = (n + 1) C3 + nC2 . . . . , Continuing this way the vertex covering sets having n + k elements are S1 with any k elements from S2 or S2 with any k – 1 elements from S1 [Since S1 and S2 are independent, |S1| = n and | S2 | = n + 1 ] Therefore, c (Gn, n + k) = (n + 1) Ck + n Ck – 1 , for k < n ; and c (Gn , 2n + 1) = 1
S1 ∪ { ui }, i = 1, 2 , . . . n + 1 and S2 = {u1, u2, u3, . . . , un + 1} Therefore, c (Gn , n + 1) = n + 2 The vertex covering sets with n + 2 elements are S1 ∪ {ui , uj }, ui , uj ∈ S2 S2 ∪ {vi} , vi ∈ S1 ;
or
Therefore, c (Gn , n + 2) = (n + 1) C2 + nC1
Figure 1
Therefore, c (Gn , x) = xn + [n + 1 + 1] xn + 1 + [ (n + 1) C2 + nC1] xn + 2 + . . . + [(n + 1) C3 + n C2] xn + 3 + . . . + [ (n + 1) Ck + n Ck – 1 ] xn + k + . . . + [ (n + 1) Cn + nCn – 1 ] x2n + x2n + 1
(A)
n + 1 + 2 n+2 n + 1 + 1 . 1 xn + 1 + + n x 1 2 n + 1 + 3 n (n −1) n + 3 n! (n + 1)! n+ k + . . .+ + x x 3 2 k! (n + 1 − k)! (k −1)! (n +1 − k)! n + 1 + n 2n 2n + 1 +...+ n x +x n
= xn +
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n + 1 + 1 n n+1 n + 1 + 2 n + x 1 2 0 1 n +1+ 3 n n+3 n + 1+ k n + . . .+ + x k 3 k − 2 n + 1 + n n 2n 2n + 1 +...+ x + x n − 1 n n n + 1 + i n n+i Therefore C (Gn , x) = xn + Σ + x 2n + 1 x i − 1 i i=1 = xn +
n+2 x n+k x 1
Remark 2.3 C (Gn , x) is also written as C (Gn , x) = xn [1 + x]n + 1 + xn + 1 [1 + x]n – x2n + 1. (A) ⇒ C (Gn , x) = xn + (n + 1) xn + 1 + (n + 1) C2 xn + 2 + . . . + (n + 1) Cn x2n + x2n + 1 + xn + 1 + n C1 xn + 2 + n C2 xn + 3 + . . . nCn – 1 x2n. = xn [1 + (n + 1) C1 x + (n + 1) C2 x2 + . . . + (n + 1) Cn xn + (n + 1) Cn + 1 xn + 1 ] + xn + 1 [1 + nC1 x + nC2 x2 + . . . + nCn – 1 xn – 1 + n Cn xn ] – x2n + 1 Therefore, C (Gn , x) = xn [1 + x]n + 1 + xn + 1 [1 + x]n – x2n + 1. Theorem 2.4
Proof
The coefficients of the vertex cover polynomials of the Gear graph satisfy the following:
By the definition of C (Gn , x) , (i) and (ii) are trivial.
i) ii) iii) iv) v)
(iii)
vi) vii)
c (Gn , n) = 1 c (Gn , 2n +1) = 1 c (Gn , 2n) = 2n + 1 c (Gn , n + 1) = n + 2 c (Gn + 1, n + 3) = c (Gn , n + 2) +n+2; n>2 c (Gn + 1 , 2 n + 1) = c (Gn , 2n) + c (Gn , 2n + 1) + 1 c (Gn , 2n – 1) = n2
By theorem (2.1) ,
c (Gn , 2n) =
n + 1 + n n n n −1
= 2n + 1 (iv) c (Gn, n +1) =
n + 1 + 1 n 1 1 −1 n 0
= (n + 2) = n+2
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Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
(v)
R.H.S. = c (Gn , n + 2) + n + 2 , for every n > 2
=
= =
n+1+2 n + n + 2 [Q i = 2] 2 2 −1 n + 3 n + n+2 2 1 n (n + 3) +n+2
2 n + 5n + 4
=
n n − 1 n−2 2n n = n − 1 2 2n n (n − 1) = . n − 1 2 =
2n
= n2 Theorem 2.4
2
=
201
2 (n + 4) (n + 1)
2 n + 4 n + 1 = 2 1 n + 1 + 1 + 2 n + 1 = 2 2 −1 = c (Gn + 1 , n + 1 + 2) = c (Gn + 1 , n + 3)
The coefficients of the vertex cover polynomials of the Gear graph satisfy the relation, c (Gn + 1 , i + 2) = c (Gn , i) + c (Gn , i + 1) , for every i = n to 2n – 1. Proof : Proof is by induction on 'i'. when i = n , c (Gn , n) + c (Gn , n + 1) = 1 + n + 2 [Q by (i) & (iv) of theorem 2] = n+3
(vi) R.H.S = c (Gn, 2n) + c (Gn , 2n + 1) + 1 = 2n + 1 + 1 + 1 [Q by (ii) & (iii)]
n + 3 n + 1 n + 1 + 1 + 1 n + 1 = 1 0 1 1 −1 = c (G n + 1 , n + 1 + 1)
=
= c (Gn + 1 , n + 2).
2 (n + 1) + 1 n + 1 n+1 1 n+1 + 1+n+1 n+1 = n+1 n + 1−1 = c (G n + 1 , 2 n + 1) = c (Gn + 1 , 2n + 2) vii) c (Gn,2n–1) =
n + 1 +n − 1 n n − 1 n−1−1
=
Therefore, the result is true for i = n Assume the result is true for all i less than or equal to 2n – 2 and prove it for i = 2n – 1. Take i = 2n – 1, c (Gn , 2n – 1) + c (Gn , 2n) =
n + 1+n − 1 n + 2n + 1 n − 1 n −1 −1
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Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
Therefore, the result is true for i = 2n – 1, Hence the result is true for all ‘i’.
n + 2n + 1 n − 1 n −2 2n n = + 2n + 1 n − 1 2 =
= =
2n
Theorem 2.5 The coefficients of the vertex cover polynomial of Gear graph satisfy the following relation :
n2 + 2n + 1 (n + 1) (n + 1)
n + 1 . n 2 2n + 2 n + 1 = n n + 1− 2 n + 1 + 1 + n n + 1 = n n −1 =
2 (n + 1)
∞
Σ
i =3
∞
c (Gn + 1 , i) = 2
Σ c (Gn
, i) + 1
i =3
Proof : By the definition of c (Gn , i), c (Gn , i) = 0, if i < n or i > 2n + 1 ; Also, c (Gn + 1 , i) = 0 if i < n + 1 or i > 2n + 3
= c (G n + 1 , 2n + 1) = c (Gn + 1 , 2n – 1 + 2) Hence, the above equality reduces to 2n + 3
Σ
i = n +1
2n +1
c (Gn + 1 , i) = 2
Σ c (Gn , i)
+1
i=n
2n +1
R.H.S = 2
Σ c (Gn , i)
+ 1
i=n
= 2 [ c (Gn , n) + c (Gn , n + 1) + c (Gn , n + 2) + . . . + c (Gn , 2n) + c (Gn , 2n + 1)] + 1 = c (Gn , n) + [c (Gn , n) + c (Gn , n + 1)] + [c (Gn , n + 1) + c (Gn , n + 2)] + . . . . + [c (Gn , 2n) + c (Gn , 2n + 1) + 1] + c (Gn , 2n + 1)]
= 1 + [1 + n + 2] + n + 2 +
n+1+2 n + ... 2 2−1
+ [2n + 1 + 1 + 1] + 1
= 1 + [n + 3] + n + 2 +
n+3 n + . . . + [2 (n + 1) + 1] + 1 2 1
Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)
Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
203
2 (n + 2) + n (n + 3) + . . . + [ 2 (n + 1) + 1] + 1 2 n + 3 n + 1 (n + 1) (n + 4) = 1+ + + . . .+ [2 (n + 1) + 1 ] + 1 2 1 0 n + 3 n + 1 n + 4 n + 1 = 1 + + + . . . + [2 (n + 1) + 1] + 1 1 0 2 1 n + 1+ 1 + 1 n + 1 n + 1+ 1 + 2 n + 1 = 1 + + + ... 1 2 1 − 1 2 − 1 = 1 + [n + 3] +
+ [ 2 (n + 1) + 1] + 1 = c (Gn + 1 , n + 1) + c (G n + 1 , n + 1 + 1) + (G n + 1 , n + 1 + 2) + . . . + c (G n + 1 , 2n + 1) +c (Wn + 1 , 2n + 1 + 1) 2n +1+1
=
Σ
(G n + 1 , i)
Σ
c (Gn + 1 , i)
i = n+1 2n +3
=
i = n+1
Theorem 2.6 The polynomial x–-n [C (Gn , x)] is log -concave. Proof : By the definition of C (Gn , x) = xn +
n
Σ i =1
n + 1 + i n n+i + x2n + 1 x i i−1
Case (i) : i = 1. 2
we have
Therefore,
n +1 +1 n 1 1− 1 n (n + 3) 2 [n + 2] ≥
n+1+2 n ≥ 1. 2 2− 1
2
2 (n2 + 4n + 4) ≥ n (n + 3) Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)
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Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
2(n 2 + 4n + 4)
> 1
n 2 + 3n
for all ‘n’
Therefore, when i = 1, the result is true Case (ii) : i = n 2
n + 1 + n −1 n ≥ . 1 n − 1 n − 1 −1 n ≥ . n −1 2
n+1+n n we prove n n − 1 2n 2 That is to prove [2n + 1]
2n
That is to prove [2n + 1]2 ≥
n −1
.
n(n − 1) 2
2
That is to prove
[2n + 1] n2
> 1
for all ‘n’
But, this is always true for all n. Therefore, the result is true for i = n. Case (iii) : 1 < i < n We have to prove c (Gn , i + 1)]2 ≥ c (Gn , i) . c (Gn , i + 2) , It is enough to prove : n+1+i+1 n i+1 i + 1− 1
2
n + 1 + i n ≥ i i − 1
1 < i < n.
n + 1 + i + 2 n i+2 i + 2 − 1
That is to prove
n+2+i n i+1 i
2
n + 1 + i n ≥ i i − 1
n + 3 + i n i + 2 i + 1
ie. to prove 2
n! n! n! n+2+i n +1+ i n+3+i . ≥ i!(n − i)! i (i − 1)! (n − i + 1)! i + 2 (i + 1)!(n − i − 1) i+1
ie. to prove Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)
!
Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (2), 197-205 (2012)
n+2+i (i + 1) (n − i)!
That is to prove,
ie, to prove
2
205
n+3+i n+1+i ≥ (n − i + 1)! (i + 2) (n − i − 1) !
[n + 2 + i]2 (n − i) (i + 1)
≥
(n + 1 + i) (n − i + 1)
[n + 2 + i]2 . (n − i + 1) (i + 2) (n + 1 + i) (n + 3 + i) (n − i) (i + 1)
.
(n + 3 + i) (i + 2) ≥ 1
(A)
[Q n + 2 + i]2 > (n + 1 + i) (n + 3 + i) for every i > 1] Clearly,
[n + 2 + i]2 (n + 1 + i) (n + 3 + i)
Therefore
> 1 and
(n − i + 1) (i + 2) (n − i) (i + 1)
(n + 2 + i) 2 (n − i + 1) (i + 2) (n + 1 + i) (n + 3 + i) (n − i) (i + 1)
Therefore x– n c (Gn, x)
≥ 1
> 1 for every i > 1
for all i > 1
is log concave.
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Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)