chapter 3 domination and independent domination in...
TRANSCRIPT
Chapter 3
Domination and Independent
domination in Circulant graphs
3.1 Introduction
Even though Cayley graphs are extensively dealt in various
literatures, only few authors have worked on domination in
Cayley graphs. To understand the concept of domination for
Cayley graphs one can refer to [15], [42], [34], [44]. Dejter,I.J.
and Serra,O. [15] obtained efficient dominating sets for Cayley
graphs constructed on permutation groups. The efficient dom-
ination number for vertex transitive graphs have been obtained
by Huang,J. and Jun-Ming Xu [34], where as efficient domina-
tion in Cayley graphs with two chord lengths are studied by
Obradovic,N., Peters,J. and Ruzic [44]. The existence of inde-
52
pendent perfect domination sets in Cayley graphs was studied
by Lee,J. [42]. Tamizh Chelvam and Rani [48] obtained the
domination number for certain Cayley graphs constructed on
Zn for some generating set of Zn. Further Tamizh Chelvam
and Rani [49] obtained the independent domination number
for Cayley graphs on Zn.
In order to find the existence of perfect dominating sets in
Cayley graphs, one should first attempt the question of finding
minimal dominating sets in Cayley graphs. The contents of the
second and third sections of this chapter have been published
as papers in respectively Tamkang Journal of Mathematics
and The Journal of Combinatorial Mathematics and Com-
binatorial Computing.
In this chapter, we obtain the value of the domination num-
ber of Cayley graphs constructed on Zn. Any Cayley graph
carved out of a finite cyclic group is called a circulant graph.
Since every finite cyclic group is isomorphic to Zn, in order to
find the domination parameters for circulant graph of order n,
we consider Zn. Throughout this chapter, n is a fixed posi-
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tive integer, Zn = {0, 1, 2, . . . , n − 1} and G = Cay(Zn, A),
where A is a generating set. Unless otherwise specified A
stands for the set {1, 2, . . . , k, n− k, . . . , n− 2, n− 1} where
1 ≤ k ≤ n−12 . Hereafter + stands for modulo n addition in
Zn. In the second section of this chapter, we obtain the value
of domination number for certain Cayley graphs defined on
Zn with respect to a generating set A. Actually we identify
one of the γ-set for G = Cay(Zn, A) and using the inverse
property of the underlying group we find certain other γ-sets
in G = Cay(Zn, A). This gives the way to find the domatic
partition of G = Cay(Zn, A). In the third section, the value
of independent domination number for Cay(Zn, A) has been
obtained.
3.2 Domination number for Cay(Zn, A)
Throughout this section, n is a fixed positive integer, Zn =
{0, 1, 2, . . . , n − 1} and G = Cay(Zn, A), where A is a gen-
erating set. Unless otherwise specified A stands for the set
{1, 2, . . . , k, n− k, . . . , n− 1} where 1 ≤ k ≤ n−12 . Hereafter
+ stands for modulo n addition in Zn.
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Theorem 3.2.1. Let n(≥ 3) and k be positive integers
such that 1 ≤ k ≤ n−12 . Let G = Cay(Zn, A) where A =
{1, 2, . . . , k, n−k, . . . , n−2, n−1}. Then γ(G) = ⌈ n|A|+1⌉ =
⌈ n2k+1⌉.
Proof. Note that |A| = 2k and let ℓ = ⌈ n|A|+1⌉ = ⌈ n
2k+1⌉.
Consider the set D = {0, (2k + 1), 2(2k + 1), . . . , (ℓ− 1)(2k +
1)}.
Claim: D is a dominating set of G.
It is always true that⋃ℓ−1
i=0 N [(2k + 1)i] ⊆ V (G) and so it
is enough to prove that V (G) ⊆⋃ℓ−1
i=0 N [(2k + 1)i]. Let c ∈
V (G). By division algorithm, one can write c = (2k + 1)i + h
for some i and h satisfying 0 ≤ i ≤ ℓ − 1 and 0 ≤ h ≤ 2k.
Note that, for any j, N [(2k + 1)j] = {(2k + 1)j, (2k + 1)j +
1, (2k+1)j+(n−1), (2k+1)j+2, (2k+1)j+(n−2), . . . , (2k+
1)j + k, (2k + 1)j + (n− k)}. We have the following cases:
Case 1. Suppose 0 ≤ h ≤ k, then it is easy to see that
c ∈ N [(2k + 1)i].
Case 2. Suppose k + 1 ≤ h ≤ 2k and 0 ≤ i ≤ ℓ− 2. In this
case c = (2k + 1)i + h = (2k + 1)(i + 1) + (n− (2k− h + 1)).
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By the assumption on h, 1 ≤ 2k − h + 1 ≤ k. This implies
that c ∈ N [(2k + 1)(i + 1)].
Case 3. Assume that k + 1 ≤ h ≤ 2k and i = ℓ − 1.
In this case 2k + 1 does not divide n and so we have c =
(2k + 1)(ℓ − 1) + h = (2k + 1)ℓ + (n − (2k − h + 1)) >
(2k+1) n2k+1 +(n− (2k−h+1). That is, c > n− (2k−h+1).
Since c ≤ n − 1, n − (2k − h + 1) < c ≤ (n − 1). Since
k + 1 ≤ h < 2k, we have (2k − h + 1) ≤ k and hence
c ∈ N [0]. When h = 2k, we get a contradiction.
In all the above cases, we have proved that if c ∈ V (G),
then c ∈ N [(2k +1)i], for some i, 0 ≤ i ≤ ℓ− 1. Hence D is a
dominating set of G and so γ(G) ≤ |D| = ℓ = ⌈ n2k+1⌉. Since
∆(G) = 2k, by Theorem 2.3.5, we have ⌈ n1+2k⌉ ≤ γ(G) ≤
n− 2k. Hence we get that γ(G) = ℓ = ⌈ n2k+1⌉.
Remark 3.2.2. When |A| = n − 1, G = Cay(Zn, A) is
complete and hence γ(G) = 1.
Remark 3.2.3. γ(Cay(Zn, A)) depends upon A, the gener-
ating set of Zn. For example, consider Cay(Z10, A) where |A|
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= 4. It is obvious that {1, 2, 8, 9} and {1, 4, 6, 9} are generat-
ing sets for Z10.
When A1 = {1, 2, 8, 9}, γ(Cay(Z10, A1)) = 2 and the cor-
responding γ−set is {0, 5}(Refer Figure 3.1) and when A2 =
{1, 4, 6, 9}, γ(Cay(Z10, A2)) = 3 and the corresponding γ−set
is {0, 3, 9}(Refer Figure 3.2).
Remark 3.2.4. When n(≥ 3) is an even integer and |A| ≥ n2 ,
then γ(Cay(Zn, A)) = ⌈ n|A|+1⌉ ≤ ⌈
nn
2+1⌉ ≤ ⌈
2nn+2⌉ ≤ 2. When
n is odd and |A| ≥ ⌊n2⌋, then γ(Cay(Zn, A)) = ⌈ n
|A|+1⌉ ≤ 2.
Corollary 3.2.5. Let n, k be positive integers and G =
Cay(Zn, A) where A = {1, 2, . . . , k, n − k, . . . , n − 1} with
1 ≤ k ≤ n−12 . For n ≥ 3 and a generating set A of Zn,
1 ≤ γ(Cay(Zn, A)) ≤ ⌈n3⌉.
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Figure 3.2
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Proof. When n ≥ 3 and G = Cay(Zn, A), we have 2 ≤ |A| ≤
n− 1. When |A| = n− 1, γ(G) = 1. Further, when |A| = 2,
by Theorem 3.2.1, we have γ(G) = ⌈n3⌉.
In the following Lemma, we identify another γ-set for
Cay(Zn, A) apart from the one identified in the proof of The-
orem 3.2.1.
Lemma 3.2.6. Let n(≥ 3) and k be positive integers such
that 1 ≤ k ≤ n−12 and ℓ = ⌈ n
|A|+1⌉. Let G = Cay(Zn, A),
where A = {1, 2, . . . , k, n − k, . . . , n − 2, n − 1}. If n =
(2k + 1)(ℓ − 1) + h, for some h with 1 ≤ h ≤ 2k , then
D = {0, h, h + 2k + 1, h + 2(2k + 1), . . . , h + (ℓ− 2)(2k + 1)}
is a γ-set for G.
Proof. Note that any element v of V (G) could be written as
v − h = (2k + 1)i + r for some i and r, with 0 ≤ i ≤ ℓ − 1
and 0 ≤ r ≤ 2k. The following cases arise:
Case 1. Suppose v = h+(2k+1)i+r, 0 ≤ i ≤ ℓ−1 and 0 ≤
r ≤ k. In this case when 0 ≤ i ≤ ℓ− 2, v ∈ N [h + (2k + 1)i]
and when i = ℓ− 1, v ∈ N [0].
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Case 2. Suppose v = h + (2k + 1)i + r with 0 ≤ i ≤ ℓ − 1
and k + 1 ≤ r ≤ 2k.
Sub case 2.1. When 0 ≤ i ≤ ℓ − 3, we can write v as
v = h + (2k + 1)(i + 1) + (n − ((2k + 1) − r)) and hence
v ∈ N [h + (2k + 1)(i + 1)] with 1 ≤ (i + 1) ≤ (ℓ− 2).
Sub case 2.2 When i = ℓ − 2, v could be written as v =
h+ (2k + 1)(ℓ− 1) + (n− ((2k + 1)− r)) = n− ((2k + 1)− r)
and hence v ∈ N [0].
Sub case 2.3. When i = ℓ−1, we have v = h+(2k+1)(ℓ−
1)+r = h+(n+(r−h)), with−k+1 ≤ r−h ≤ 2k−1. Hence,
when −k + 1 ≤ r − h ≤ −1 and 0 ≤ r − h ≤ k, v ∈ N [h]
and when k + 1 ≤ r − h ≤ 2k − 1, v ∈ N [h + (2k + 1)].
In all the cases v ∈ N [D] and hence D is a γ-set.
Lemma 3.2.7. Let n(≥ 3) be a positive integer, A1 =
{1, n − 1} and A2 = {1, n − 1, ...}. Let G1 = Cay(Zn, A1)
and G2 = Cay(Zn, A2). Suppose D is a γ-set for G1, then
D is a dominating set for G2.
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Proof. Since A1 ⊂ A2, G1 = Cay(Zn, A1) is a spanning sub-
graph of G2 = Cay(Zn, A2). Hence D is a dominating set of
G2.
Theorem 3.2.8. Let n(≥ 3) and k be positive integers
such that 1 ≤ k ≤ n−12 . Let G = Cay(Zn, A) where A =
{1, 2, . . . , k, n − k, . . . , n − 2, n − 1}. Then Cayley graph
G = Cay(Zn, A) is excellent.
Proof. By the Theorem 3.2.1, we have γ(G) = ⌈ n|A|+1⌉ =
⌈ n2k+1⌉ and D = {0, 2k + 1, 2(2k + 1), . . . , (ℓ − 1)(2k + 1)}
is a γ-set. Suppose c ∈ V (G), then c = (2k + 1)i + h for
some i and h with 0 ≤ i ≤ ℓ − 1 and 0 ≤ h ≤ 2k. Consider
D+h = {h, (2k+1)+h, 2(2k+1)+h, . . . , (ℓ−1)(2k+1)+h}.
Clearly D + h is a γ-set containing c and hence G is excel-
lent.
Theorem 3.2.9. Let n(≥ 3) and k be positive integers such
that 1 ≤ k ≤ n−12 and n = (2k + 1)t + 1 for some positive
integer t. Let A = {1, 2, . . . , k, n − k, . . . , n − 2, n − 1}.
Then G = Cay(Zn, A) is 2-excellent.
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Proof. Let p = 2k + 1. Note that |A| = 2k. By the Theorem
3.2.1, γ(G) = ⌈ n|A|+1⌉ = ⌈(2k+1)t+1
2k+1 ⌉ = t + 1. Let x, y ∈ V (G).
Without loss of generality one may assume that x < y. Let us
find a γ-set containing x and y.
Case 1. Suppose y − x ≤ p. Let D = {x, y, y + p, . . . , y +
(t− 1)p}. Suppose we prove that D is a dominating set, then
|D| = t + 1 ensures that D is a γ-set containing x and y.
One can easily see that the elements of V (G) are of the form
V (G) = {y, y + 1, . . . , y + 2k, y + p, y + p + 1, . . . y + p +
2k, . . . , y+(t−1)p, y+(t−1)p+1, . . . , y+(t−1)p+2k, y+tp}.
Note that N [y+ip] = {y+ip, y+ip+1, y+ip+(n−1), . . . , y+
ip + k, y + ip + (n− k)}. Let v ∈ V (G).
Sub case 1.1. Suppose v = y + ip + c for some i and c
satisfying 0 ≤ i ≤ t− 1 and 0 ≤ c ≤ k, then v ∈ N [y + ip] ⊆
N [D].
Sub case 1.2. Suppose v = y + ip + c for some i and c with
0 ≤ i ≤ t−2 and k+1 ≤ c ≤ 2k, then v = y+(i+1)p− (p−
c) = y+(i+1)p+(n−p+c) = y+(i+1)p+n+c′ where c′ = c−p
and −k ≤ c′ ≤ −1. Thus v ∈ N [y + (i + 1)p] ⊆ N [D].
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Sub case 1.3. When v = y + (t − 1)p + (k + 1), then
v = y + n − 1 − k ≤ x + (2k + 1) + n − (k + 1) = x + k.
Therefore v ≤ x + k. Since x < y and y < v, we have v > x.
Hence v ∈ N [x] ⊆ N [D].
Sub case 1.4. Suppose v = y + (t − 1)p + c for some
c, with k + 2 ≤ c ≤ 2k, then 2 ≤ p − c + 1 ≤ k. Now
v = y +(n− 1)−p+ c = y +n− (p− c+1) ∈ N [y] ⊆ N [D].
Sub case 1.5. Suppose v = y + tp then v = y + (n− 1) ∈
N [y] ⊆ N [D]. Hence in this case V = N [D].
Case 2. Assume that y − x > p and y − x is divisible by
p. When this is the case y = x + hp for some h. Consider
D = {x, x+p, . . . , x+hp, x+(h+1)p, x+(h+2)p, . . . , x+tp}.
As in Case 1, let us prove that D is a dominating set. Elements
of V (G) can be listed out as {x, x + 1, . . . , x + 2k, x + p, x +
p + 1, . . . , x + p + 2k, . . . , x + hp, x + hp + 1, . . . , x + hp +
2k, x+(h+1)p, x+(h+1)p+1, . . . , x+(h+1)p+2k, . . . , x+
(t− 1)p, x + (t− 1)p + 1, . . . , x + (t− 1)p + 2k, x + tp}. Let
v ∈ V (G).
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Sub case 2.1.Suppose v = x + ip + c for some i and c
satisfying 0 ≤ i ≤ t − 1 and 0 ≤ c ≤ k. In this case v ∈
N [x + ip] ⊆ N [D].
Sub case 2.2. Suppose v = x + ip + c where 0 ≤ i ≤ t− 1
and k + 1 ≤ c ≤ 2k then v = x + (i + 1)p + (n + c − p) =
x + (i + 1)p + (n + c′) where c′ = c − p and −k ≤ c′ ≤ −1.
Thus v ∈ N [x + (i + 1)p] ⊆ N [D] for 0 ≤ i ≤ t− 1.
Sub case 2.3. When v = x+tp then v ∈ N [x+tp] ⊆ N [D].
Case 3. Suppose y − x > p and y − x is not divisible
by p, then y = x + hp + r, 1 ≤ r ≤ p − 1. Consider D =
{x, x+p, x+2p, . . . , x+hp, y, y+p, y+2p, . . . , y+(t−h−1)p}.
As in the Cases 1 and 2 , let us prove that D is a dominating
set. Elements of V (G) can be listed out as {x, x + 1, . . . , x +
2k, x + p, x + p + 1, . . . , x + p + 2k, . . . , x + hp, . . . , x + hp +
2k, x+(h+1)p, x+(h+1)p+1, . . . , x+(h+1)p+2k, . . . , x+
(t− 1)p, x + (t− 1)p + 1, . . . x + (t− 1)p + 2k, x + tp}. Let
v ∈ V (G).
Sub case 3.1. Suppose v = x + ip + c for some h and
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c satisfying 0 ≤ i ≤ h and 0 ≤ c ≤ k. In this case v ∈
N [x + ip] ⊆ N [D].
Sub case 3.2. Suppose v = x+ip+c where 0 ≤ i ≤ h−1 and
k+1 ≤ c ≤ 2k then v = x+ip+c = x+(i+1)p+(n+c−p) ∈
N [x + (i + 1)p] ⊆ N [D] for 0 ≤ i ≤ h− 1.
Sub case 3.3. When v = x + hp + c for some c satisfying
k + 1 ≤ c ≤ 2k, then −(k − 1) ≤ c − r ≤ 2k − 1 and
v = y + (c− r). When −(k − 1) ≤ c− r ≤ k, v ∈ N [y] and
when k+1 ≤ c−r ≤ 2k−1, then v = y+p+(n+(c−r−p))
and −k ≤ c− r − p ≤ −2 implies v ∈ N [y + p] ⊆ N [D].
Sub case 3.4. Suppose v = x + ip + c, h + 1 ≤ i ≤ t− 1,
0 ≤ c ≤ k, then −2k ≤ c − r ≤ k − 1. When −2k ≤
c − r ≤ −(k + 1), v ∈ N [y + (i − h − 1)p] ⊂ N [D]. When
−k ≤ c− r ≤ k − 1, v ∈ N [y + (i− h)p] ⊆ N [D].
Sub case 3.5. Suppose v = x + ip + c, h + 1 ≤ i ≤
t− 1, k +1 ≤ c ≤ 2k, then −(k− 1) ≤ c− r ≤ 2k− 1. When
(k − 1) ≤ c − r ≤ k, v ∈ N [y + (i − h)p] ⊂ N [D]. When
k +1 ≤ c− r ≤ 2k−1, then v ∈ N [y +(i−h+1)p] ⊆ N [D].
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Sub case 3.6. Suppose v = x + tp = x + (n− 1) ∈ N [x] ⊆
N [D].
In all the cases D is a dominating set.
Theorem 3.2.10. Let n(≥ 3) and k be positive integers
such that 1 ≤ k ≤ n−12 , A = {1, 2, . . . , k, n − k, . . . , n −
2, n−1} and n = (ℓ−1)(2k +1)+ j with 1 ≤ j ≤ 2k. Then
G = Cay(Zn, A) is 2-excellent if and only if j = 1.
Proof. Suppose j = 1. Then by Theorem 3.2.9, G is
2−excellent. Suppose G is 2−excellent and n = (ℓ− 1)(2k +
1)+j, 2 ≤ j ≤ 2k. Consider the vertices 0 and 1, and a γ−set
D such that {0, 1} ⊆ D. Then {0, 1} dominates exactly 2k+2
vertices of G and |N [u]| = 2k + 1,∀u ∈ V (G). Hence D can
dominate at most (ℓ − 2)(2k + 1) + 2k + 2 vertices, whereas
(ℓ− 2)(2k + 1) + 2k + 2 = (ℓ− 1)(2k + 1) + 1 < n which is a
contradiction to D is a dominating set. Hence j = 1.
Lemma 3.2.11. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and G = Cay(Zn, A) where A = {1, 2, . . . , k,
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n− k, . . . , n− 2, n− 1}. Suppose 2k + 1 does not divide n.
Then D + h is a γ−set for any positive integer h, 1 ≤ h ≤
n− 1 whenever D is a γ−set of G.
Lemma 3.2.12. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and G = Cay(Zn, A) where A = {1, 2, . . . , k,
n − k, . . . , n − 2, n − 1}. Suppose 2k + 1 divides n. Then
D + h is a γ−set for any positive integer h, 1 ≤ h ≤ 2k
whenever D is a γ−set of G.
In view of Lemma3.2.6, we can prove the following lemma
Lemma 3.2.13. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and G = Cay(Zn, A) where A = {1, 2, . . . , k,
n−k, . . . , n−2, n−1}. Suppose n = (2k+1)t+h for some
positive integers t and h with 1 ≤ h ≤ 2k and ℓ = ⌈ n|A|+1⌉.
Then D = {0, c+h, c+h+(2k +1), . . . , c+h+(ℓ− 2)(2k +
1)}, 1 ≤ c ≤ 2k − h + 1 is a γ−set for G.
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Lemma 3.2.14. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and G = Cay(Zn, A) where A = {1, 2, . . . , k,
n − k, . . . , n − 2, n − 1}. Suppose 2k + 1 divides n. Then
G is just excellent.
Proof. By Lemma 3.2.12, we have D + h is a γ−set for any
positive integer h, 1 ≤ h ≤ 2k whenever D is a γ−set of G.
Let D = {0, 2k + 1, . . . , (ℓ− 1)(2k + 1)} be the γ−set of G.
It is clear that h ∈ D + h. To prove that the γ−set is unique.
Suppose h ∈ D + m where 1 ≤ m ≤ 2k and h 6= m. Then
h−m = (2k+1)i, 0 ≤ i ≤ (ℓ−1) which is possible only when
i = 0 leading to a contradiction. Hence G is just excellent.
Lemma 3.2.15. Let n be a positive even integer and k an
integer such that 1 ≤ k ≤ n−12 . Suppose D is a γ−set for
G1 = Cay(Zn, A1) where A1 = {1, . . . , k, n− k, . . . , n− 1}.
Then D is a dominating set for G2 = Cay(Zn, A2) where
A2 = {1, . . . , k, n− k, . . . , n− 1, n2}.
Proof. Since A1 ⊂ A2, we have G1 is a spanning subgraph of
G2. Hence D is a dominating set of G2.
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3.3 Domination number for Cay(Zn, B)
In this section, we attempt to find the value of domination
number for G = Cay(Zn, B) where B = {1, 3, . . . , 2k−1, n−
(2k− 1), . . . , n− 3, n− 1} with 1 ≤ k ≤ n−12 and (2k− 1) 6=
n−(2k−1) (to ensure that elements of B are distinct). Further
we have proved that Cay(Zn, B) is excellent.
Theorem 3.3.1. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and (2k − 1) 6= n − (2k − 1). Let B =
{1, 3, . . . , 2k − 1, n − (2k − 1), . . . , n − 3, n − 1} and G =
Cay(Zn, B). If 2k + 1 divides n, then γ(G) = n2k+1.
Proof. Let ℓ = n2k+1 and D = {0, 2k +1, . . . , (ℓ− 1)(2k +1)}.
To prove that D is a dominating set, it is enough to show
that V ⊆ N [D]. Let v ∈ V . Then v could be written as
v = (2k + 1)i + j, 0 ≤ i ≤ ℓ − 1 and 0 ≤ j ≤ 2k. We know
that N [(2k + 1)i] = {(2k + 1)i, (2k + 1)i + 1, (2k + 1)i + (n−
1), (2k + 1)i + 3, (2k + 1)i + (n − 3), . . . , (2k + 1)i + (2k −
1), (2k + 1)i + (n − (2k − 1))}, 0 ≤ i ≤ ℓ − 1. We have the
following cases:
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Case 1. Suppose 0 ≤ i ≤ ℓ − 1 and j is odd. Then v ∈
N [(2k + 1)i].
Case 2. Suppose 0 ≤ i ≤ ℓ − 1 and j = 0. Then v ∈
N [(2k + 1)i].
Case 3. Suppose 0 ≤ i ≤ ℓ − 2 and j is even. That is
j = 2t, 1 ≤ t ≤ k. Then v = (2k + 1)i + j = (2k + 1)(i +
1) + (n + j − (2k + 1)) = (2k + 1)(i + 1) + (n + 2(t− k)− 1).
Hence v ∈ N [(2k + 1)(i + 1)] with 1 ≤ i + 1 ≤ ℓ− 1.
Case 4. Suppose i = ℓ− 1 and j is even. That is j = 2t, 1 ≤
t ≤ k. Then v = (2k + 1)(i + 1) + (n + j − (2k + 1)) =
(2k + 1)ℓ + (n + 2(t − k) − 1) = n + 2(t − k) − 1. Hence
v ∈ N [0].
In all the above cases, we have V ⊆ N [D]. Hence D is a
dominating set and |D| = n2k+1. Therefore D is a γ−set for
G.
Similary one can prove the following.
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Theorem 3.3.2. Let n and k be positive integers. Suppose
G = Cay(Zn, B) where B = {1, 3, . . . , 2k − 1, n − (2k −
1), . . . , n−3, n−1}, 1 ≤ k ≤ n−12 with (2k−1) 6= n−(2k−1)
and elements of B distinct. Then when 2k+1 divides n, G
is excellent.
Proof. By Theorem 3.3.1, we have γ(G) = n2k+1 and D =
{0, 2k + 1, . . . , (ℓ − 1)(2k + 1)} where ℓ = n2k+1 is a γ−set.
Suppose c ∈ V (G). Then c = (2k + 1)i + h for some i
and h with 0 ≤ i ≤ ℓ − 1 and 0 ≤ h ≤ 2k. Consider
D + h = {h, (2k + 1) + h, . . . , (ℓ− 1)(2k + 1) + h}. Clearly
D + h is a γ−set containing c and hence G is excellent.
3.4 Independent domination in Cay(Zn, A)
In this section, we attempt to find the independent dom-
ination number, perfect domination number and independent
dominating sets of Cay(Zn, A), for a specified generating set A
of Zn. The contents of this section has been published as a pa-
per entitled ‘Independent domination number for Cay(Zn, A)’
in The Journal of Combinatorial Mathematics and Com-
binatorial Computing.
70
Theorem 3.4.1. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and 2k + 1 divides n. Let G = Cay(Zn, A),
where A = {1, 2, . . . , k, n − k, . . . , n − 1}. Then i(G) =
γp(G) = n2k+1.
Proof. By Theorem 3.2.1, D = {0, (2k+1), 2(2k+1), 3(2k+
1), . . . , (ℓ − 1)(2k + 1)} where ℓ = ⌈ n2k+1⌉ is a γ-set of G.
Since (2k + 1) divides n we have ℓ = n2k+1. Note that, for any
j, N [(2k + 1)j] = {(2k + 1)j, (2k + 1)j + 1, (2k + 1)j + (n−
1), (2k+1)j +2, (2k+1)j +(n−2), . . . , (2k+1)j +k, (2k+
1)j + (n− k)}.
Now, let v1, v2 ∈ D and v1 6= v2. We claim that N [v1] ∩
N [v2] = ∅. Let v1 = (2k + 1)i and v2 = (2k + 1)j with
0 ≤ i < j ≤ ℓ − 1. Suppose N [v1] ∩ N [v2] 6= φ and let x ∈
N [v1]∩N [v2]. Then x = (2k+1)i+t1 or (2k+1)i+(n−t1) and
x = (2k+1)j + t2 or (2k+1)j +(n− t2) where 0 ≤ t1, t2 ≤ k.
Clearly 1 ≤ j−i ≤ ℓ−1, −k ≤ t1−t2 ≤ k, 0 ≤ t1+t2 ≤ 2k
and 2k + 1 ≤ (2k + 1)(j − i) ≤ (2k + 1)(ℓ− 1).
If (2k+1)i+t1 = (2k+1)j+t2, then (2k+1)(j−i) = t1−t2.
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If (2k+1)i+(n−t1) = (2k+1)j+t2, then (2k+1)(j−i) =
n− (t1 + t2).
If (2k+1)i+t1 = (2k+1)j+(n−t2), then (2k+1)(j−i) =
t1 + t2.
If (2k + 1)i + (n − t1) = (2k + 1)j + (n − t2), then (2k +
1)(j − i) = n− (t1 − t2).
In each of these cases, the left hand side is a multiple of
2k + 1, whereas the right hand side is not so, which is a con-
tradiction. Hence N [v1] ∩N [v2] = φ. Thus D is independent
and |D| = ℓ = n2k+1 and hence i(G) = n
2k+1. Also every ele-
ment of V − D is adjacent to exactly one element in D and
hence D is perfect. Therefore γp(G) = n2k+1.
Remark 3.4.2. In the above theorem 3.4.1, the set D is an
E−set.
Corollary 3.4.3. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and 2k + 1 divides n. Let G = Cay(Zn, A),
where A = {1, 2, . . . , k, n−k, . . . , n−1} and D = {0, (2k+
1), 2(2k + 1), 3(2k + 1), . . . , (⌈ n2k+1⌉ − 1)(2k + 1)}. Then
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for each h, 1 ≤ h ≤ 2k, D + h is both an independent and
perfect dominating set of G with minimum cardinality.
Corollary 3.4.4. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and 2k + 1 divides n. Let G = Cay(Zn, A),
where A = {1, 2, . . . , k, n − k, . . . , n − 1}. Then di(G) =
dp(G) = 2k + 1.
Proof. Any element of V is of the form (2k +1)t+h, 0 ≤ t ≤
⌈ n2k+1⌉ − 1, 0 ≤ h ≤ 2k. Also if D = {0, (2k + 1), 2(2k +
1), 3(2k + 1), . . . , (⌈ n2k+1⌉ − 1)(2k + 1)}, then (D + h1) ∩
(D + h2) = φ for 0 ≤ h1, h2 ≤ 2k and h1 6= h2. If not, let
x ∈ (D + h1) ∩ (D + h2). Then h1 − h2 = (2k + 1)(j − i)
for some i and j with i < j, and 0 ≤ i < j ≤ ℓ − 1. Since
0 < h1−h2 < 2k, it cannot be a multiple of 2k + 1 and hence
a contradiction. Hence V =⋃2k
h=0(D + h) and each D + h is
both an independent and perfect dominating set of minimum
cardinality. Therefore di(G) = dp(G) = 2k + 1.
Remark 3.4.5. For any v ∈ G = Cay(Zn, A), |N [v]| =
2k + 1. In view of Theorem 3.4.1 and Corollary 3.4.3, the
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efficient dominating set of G exists only when 2k + 1 divides
n.
Corollary 3.4.6. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and 2k + 1 divides n. Let G = Cay(Zn, A),
where A = {1, 2, . . . , k, n − k, . . . , n − 1}. Then G is do-
matically full.
Proof. Since G is a 2k regular graph, by using Corollary 3.4.4,
we have G is domatically full.
Theorem 3.4.7. Let n and k be positive integers such that
1 ≤ k ≤ n−12 and 2k + 1 does not divide n. Let G =
Cay(Zn, A), where A = {1, 2, . . . , k, n−k, . . . , n−1}. Then
i(G) = ⌈ n2k+1⌉.
Proof. Let ℓ = ⌈ n2k+1⌉ and t be the least positive integer sat-
isfying n ≡ t(mod (2k + 1)). Take D = {0, 2k + 1, 2(2k +
1), . . . , (ℓ − 2)(2k + 1), n − (k + ⌈ t2⌉)}. Let v1, v2 ∈ D and
such that v1 6= v2. Without loss of generality, one can assume
that v1 = (2k + 1)i and v2 = (2k + 1)j or n − (k + ⌈ t2⌉)
for some i, j with 0 ≤ i, j ≤ (ℓ − 2) and i 6= j. Suppose
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v1 ∈ N(v2). Then v1 = v2 + s or v1 = v2 + (n− s) for some s
with 1 ≤ s ≤ k.
Now the following cases arise:
Case 1. Let v1 = (2k + 1)i and v2 = (2k + 1)j. The non-
adjacency of v1 and v2 can be proved as in Theorem 3.2.1.
Case 2. Let v1 = (2k + 1)i and v2 = n− (k + ⌈ t2⌉). We have
the following sub cases:
Sub case 2.1. Suppose v1 = v2 + s. Then (2k + 1)i =
n− (k + ⌈ t2⌉) + s and so n = (2k + 1)i + (k + ⌈ t
2⌉)− s. Since
(k + ⌈ t2⌉) − s < (2k + 1), we get that n = (2k + 1)i + (k +
⌈ t2⌉)− s < n, which is a contradiction.
Sub case 2.2. Suppose v1 = v2 + n − s. Then (2k + 1)i =
n−(k+⌈ t2⌉)+(n−s) and so n−s = (2k+1)i+(k+⌈ t
2⌉) < n−s,
which is a contradiction.
Hence in both the cases we have v1 /∈ N [v2]. Therefore D
is independent. It follows from Theorem 3.2.1 that D is a
dominating set and hence i(G) ≤ ⌈ n2k+1⌉. Now the reverse
inequality follows from i(G) ≥ γ(G) ≥ ⌈ n2k+1⌉.
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Corollary 3.4.8. Let n and k be positive integers such
that 1 ≤ k ≤ n−12 . Let G = Cay(Zn, A), where A =
{1, 2, . . . , k, n− k, . . . , n− 1}. For n ≥ 3, 1 ≤ i(G) ≤ ⌈n3⌉.
Proof. When n ≥ 3, we have 2 ≤ |A| ≤ (n − 1). When
|A| = n − 1, i(G) = 1. Further, when |A| = 2, by Theorem
3.4.1 and Theorem 3.4.7, we have i(G) = ⌈n3⌉.
Corollary 3.4.9. Let n, k be positive integers such that
1 ≤ k ≤ n−12 and (2k + 1) does not divide n. Let G =
Cay(Zn, A), where A = {1, 2, . . . , k, n−k, . . . , n−1}. If D
is an independent dominating set of G, then D + u is an
independent dominating set for any u with 1 ≤ u ≤ n− 1.
In view of Corollary 3.4.3 and Corollary 3.4.9, we have the
following corollary.
Corollary 3.4.10. Let n, k be positive integers such that
1 ≤ k ≤ n−12 and G = Cay(Zn, A), where A = {1, 2, . . . , k,
n− k, . . . , n− 1}. Then G is γi−excellent.
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Theorem 3.4.11. Let n and k be positive integers such
that 1 ≤ k ≤ n−12 . Then ii(G) = γi(G) = γγ(G) = 2⌈ n
2k+1⌉,
where G = Cay(Zn, A) and A = {1, 2, . . . , k, n−k, . . . , n−
1}.
Proof. Let ℓ = ⌈ n2k+1⌉. When 2k + 1 divides n, the result
follows from Theorem 3.4.1. When 2k + 1 does not divide n,
by Theorem 3.4.7, D = {0, 2k + 1, 2(2k + 1), . . . , (ℓ− 2)(2k +
1), n − (k + ⌈ t2⌉)} is an independent dominating set. When
the points of Zn are represented as n equi-distant points on the
circumference of a circle, then D is simply a successive set of
points starting from 0 and distance 2k+1, except the last point,
whose circular distance from 0 is less than 2k+1. Actually this
circular distance is simply the remainder t when n is divided by
2k+1. If t > 1, then D and D+1 are disjoint dominating sets
and if t = 1, then D and D + 2 are disjoint dominating sets.
Hence ii(G) = i(G) + i(G) = 2ℓ, γi(G) = γ(G) + i(G) = 2ℓ
and γγ(G) = γ(G) + γ(G) = 2ℓ.
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Lemma 3.4.12. Let n and k be positive integers such
that 1 ≤ k ≤ n−12 . Let G = Cay(Zn, A) where A =
{1, 2, . . . , k, n− k, . . . , n− 1}. Then G is γγ−minimum.
Proof. By Theorem 3.4.11, we have γγ(G) = 2γ(G) and hence
the lemma is true.
Now we attempt to find the value of independent domination
number for G = Cay(Zn, B) where B = {1, 3, . . . , 2k−1, n−
(2k− 1), . . . , n− 3, n− 1} with 1 ≤ k ≤ n−12 and (2k− 1) 6=
n− (2k − 1) (to ensure that elements of B are distinct).
Theorem 3.4.13. Let n and k be positive integers such
that 1 ≤ k ≤ n−12 and (2k − 1) 6= n − (2k − 1). Let G =
Cay(Zn, B) where B = {1, 3, . . . , 2k−1, n−(2k−1), . . . , n−
3, n− 1}. When 2k + 1 divides n, i(G) = γp(G) = n2k+1.
Proof. By Theorem 3.3.1, we know that D = {0, 2k + 1, . . . ,
(ℓ − 1)(2k + 1)} where ℓ = n2k+1 is a γ−set of G. Note that,
for any j, N [(2k +1)j] = {(2k +1)j, (2k +1)j +1, (2k +1)j +
78
(n − 1), (2k + 1)j + 3, (2k + 1)j + (n − 3), . . . , (2k + 1)j +
(2k − 1), (2k + 1)j + (n− (2k − 1))}.
Claim. D is independent and perfect.
Equivalently, we have to prove that for v1, v2 ∈ D with v1 6=
v2, N [v1]∩N [v2] = φ. Let v1, v2 ∈ D. Then v1 = (2k+1)i and
v2 = (2k+1)j with 0 ≤ i, j ≤ ℓ−1. Without loss of generality,
let us assume that i < j. Suppose N [v1] ∩ N [v2] 6= φ and
let x ∈ N [v1] ∩ N [v2]. Then x = (2k + 1)i + (2t1 − 1) or
(2k + 1)i + (n − (2t1 − 1)) and x = (2k + 1)i + (2t2 − 1) or
(2k + 1)i + (n− (2t2− 1)) where 1 ≤ t1, t2 ≤ k. We find that
1 ≤ j − i ≤ ℓ − 1,−k ≤ t1 − t2 ≤ k, 0 ≤ t1 + t2 ≤ 2k and
2k + 1 ≤ (2k + 1)(j − i) ≤ (2k + 1)(ℓ − 1). The following
cases arise:
Case 1. Suppose (2k+1)i+(2t1−1) = (2k+1)j +(2t2−1).
Then (2k + 1)(j − i) = 2(t1 − t2). Here left hand side is a
multiple of 2k + 1 but the right hand side is not a multiple of
2k + 1, which is a contradiction.
Case 2. Suppose (2k + 1)i + (n− (2t1 − 1)) = (2k + 1)j +
(2t2 − 1). Then (2k + 1)(j − i) = n− 2((t1 + t2)− 1).
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Case 3. Suppose (2k + 1)i + (2t1 − 1) = (2k + 1)j + (n −
(2t2 − 1)). Then (2k + 1)(j − i) = 2((t1 + t2)− 1).
Case 4. Suppose (2k + 1)i + (n− (2t1 − 1)) = (2k + 1)j +
(n− (2t2 − 1)). Then (2k + 1)(j − i) = n− 2(t1 − t2).
In each of these cases, the left hand side is a multiple of 2k+
1, whereas the right hand side is not so. Hence a contradiction
as in Case 1.
Hence N [v1] ∩N [v2] = φ. Therefore D is independent and
|D| = ℓ = n2k+1 implies that i(G) = n
2k+1. Also every element
of V − D is adjacent to exactly one element in D and hence
D is perfect. Therefore γp(G) = n2k+1.
Remark 3.4.14. In the above theorem 3.4.13, the set D is
an E−set.
In view of the above Theorem 3.4.13, one can prove the
following corollary.
Corollary 3.4.15. Let n, k be positive integers such that
1 ≤ k ≤ n−12 , (2k−1) 6= n− (2k−1) and (2k +1) divides n.
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Let G = Cay(Zn, B) where B = {1, 3, . . . , 2k− 1, n− (2k−
1), . . . , n−3, n−1}. If D is an independent dominating set
of G, then D+u is an independent and perfect dominating
set of minimum cardinality for any u with 1 ≤ u ≤ 2k.
81