Domination in generalized Domination in generalized Petersen graphs - - PowerPoint PPT Presentation
Domination in generalized Domination in generalized Petersen graphs Petersen graphs Advisor: Sheng-Chyang Liaw Student: Yuan-Shin Li Introduction A vertex subset S of a graph G is a dominating set g p g if each vertex in V ( G ) S is
Advisor: Sheng-Chyang Liaw Student: Yuan-Shin Li
Introduction
g p g if each vertex in V(G) − S is adjacent to at least one vertex in S.
i i d i ti t f G d t d b minimum dominating set of G, denoted by .
2
5 =
） （C γ
5）
（ γ C5
A d i i S i ll d i d d d i i
set if S is also an independent set.
cardinality of a minimum independent dominating set cardinality of a minimum independent dominating set
5
2
i C
γ = （ ） C5
5 i
γ（ ） C5
if each vertex v of G is dominated by some vertex
v ≠ u
v ≠ u
.
5
3
t C
γ = （ ） C5
5 t
γ（ ） C5
By definition γ(G) ≤ γi (G) and γ(G) ≤ γt(G). A vertex v of G is repeatedly dominated by S if p y y |N(v) ∩ S| ≥ 2.
Proposition 1.1 If G is k-regular then .
( )
( )
V G t k
G γ ⎡ ⎤ ≥ ⎢ ⎥
vertex set: ={v1, v2, ..., vn} {u1, u2, ..., un} edge set: with vn+1 = v1, un+1= u1,
1 i i i i i i k
E {v v , v u , u u }
+ +
= U
U V U
U
1 ≤ i ≤ n, and 1 ≤ k ≤
⎥ ⎦ ⎥ ⎢ ⎣ ⎢ 2 n
v6 v7 v8
V={v1, v2, ..., vn}
u8 u7 u6 u5 v5 v9
U={u1, u2, ..., un} { }
5
u4 u9 v4 v10
1 ≤ i ≤ n, and 1 ≤ k ≤
⎥ ⎦ ⎥ ⎢ ⎣ ⎢ 2 n
E= {vivi+1, viui, uiui+k}
U
u3 u10
4
v10 u2 u12 u11 v3 v11 u1 u12 u13 v2 v12
Generlized Petersen graph P(13,3)
v1 v13
Generlized Petersen graph P(13,3)
Petersen graph＝P(5 2) Petersen graph＝P(5,2)
domination number independent domination total domination number number
P(2k + 1, k) ,n≠1(mod 3)
3
3
⎡ ⎤
2n
( , ) ,
( )
,n ≡1(mod3)
P(2k k)
5 3n
5 3n
⎡ ⎤
3 2 3
1
n +
⎡ ⎤ ⎢ ⎥
P(2k, k)
2 3 n
⎡ ⎤ ⎢ ⎥
2 3 n
⎡ ⎤ ⎢ ⎥
P(n, 1) , n≡2 (mod 4)
n≡0 1 3 (mod 4)
,n ≡ 0,3 (mod 4) ,n ≡ 1,2 (mod 4)
, n≡4 (mod 6)
, n≡4 (mod 6)
⎡ ⎤ 1
2 n +
2 n
⎡ ⎤ 1
2 n +
n
⎡ ⎤
3 2n
⎡ ⎤ 1
3 2
+
n
,n≡0,1,3 (mod 4)
P(n, 2)
,n ≡ 1(mod 3)
≡ 1( d3)
2
3n
2
5 3n
⎡ ⎤
3 2n
2
1
n
⎡ ⎤ ⎢ ⎥
, n ≡ 1(mod3)
P(n, 3)
n ≡ 0,1 (mod 4) , or n=11
,n ≡ 0,1 (mod 4)
,n≡0,2,4 (mod 5)5
⎡ ⎤
2 n
5
3
1
n +
⎡ ⎤ ⎢ ⎥ ⎡ ⎤ 1
2 n +
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎪ ⎨ ⎧ )) 3 ( (
4 5 4 2 n
n ≡ 2,3 (mod 4) , n≠11 , n ≡ 2,3 (mod 4)
,n≡1,3 (mod 10) ,n≡6,8 (mod 10)2 n +
2 n
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎪ ⎩ ⎪ ⎨ + ≤ ≤ 1 )) 3 , ( (
5 4 5 4 3 2 n n t n
n P γ
Theorem 2.1 The independent domination number of p f P(2k +1, k) is for n = 2k +1.
⎥ ⎥ ⎤ ⎢ ⎢ ⎡ 5 3n
Proof Since γ(G) ≤ γ (G) we have γ (P(2k + 1 k)) ≥
⎥ ⎥ ⎤ ⎢ ⎢ ⎡ 3n
In the following, we choose the vertices to form an independent dominating set.
⎥ ⎥ ⎢ ⎢ 5 v7 v8 v9
S0 = {V5t+1|0 ≤ t < a}
u8 u7 u9 u10 v6 v10
|S | + |S | = a + 2a = 3a =
3n
u6 u5
10
u11 v5 v11
S1 = {U5t+3|0 ≤ t < a} ∪ {U5t+4|0 ≤ t < a}; |S0| + |S1| = a + 2a = 3a = 5
u5 u4 u12 u13 v4 v12 u3 u2 u15 u13 u14 v3 v13 u1 u15 v1 v2 v14 v15
⎣ ⎦
a
⎣ ⎦
S0 = {V5t+1|0 ≤ t < } ∪ {V5t| ≤ t ≤ a};
⎣ ⎦
2 a
⎣ ⎦
2 a
S1 = {U5t+3|0 ≤ t < a} ∪ {U5t+4|0 ≤ t < } ∪ {U5t+2| ≤ t ≤ a};
⎣ ⎦
2 a
⎣ ⎦
2 a
|S | + |S | = a + 1 + a + a + 1 = 3a + 2 = ⎡ ⎤
3n
|S0| + |S1| = a + 1 + a + a + 1 = 3a + 2 = .
⎡ ⎤
5 3n
v6 v7 v8 u5 u6 u7 u8 u9 v5 v9 u4 u9 u10 v4 v10
Therefore
u3 u11 v11
Therefore γi(P(2k+1, k)) ≤ ⎡ ⎤
5 3n
u1 u2 u12 u13 v v3
11
v12 v1 v2 v12 v13
Lemma 2.2: In the total domination number, a connected set D
vertices in G with |D| ≥ 2 . Since G is 3-regular graph
if ( d 3)
Theorem 2.3 For n = 2k + 1,
2n ⎧⎡ ⎤
1 , if (mod 3) , if (mod 3)
γt(P(2k + 1, k)) =
2 1 3 2 3 n n ⎧⎡ ⎤ + ⎪⎢ ⎥ ⎪⎢ ⎥ ⎨ ⎡ ⎤ ⎪ ⎢ ⎥ ⎪ ⎢ ⎥ ⎩
1 ≡ n 1 ≡ n
1. ⎡ ⎤
3 2n
S ≥
By Proposition 1.1
3 ⎪ ⎢ ⎥ ⎩
⎡ ⎤
3
Suppose we have the following two properties
⎡ ⎤ 1
)) , 1 2 ( (
3 2
+ ≥ +
n t
k k P γ
y p
1 ≡ n
⎡ ⎤
2
)) 1 2 ( (
n
k k P = + γ Suppose we have the following two properties. Claim 1. For any connected subset D S, |D| = d ≤ 3. Claim 2. S is formed by the union of exactly one 3-element connected
⎡ ⎤
3
)) , 1 2 ( (
t
k k P = + γ set and a − 1 2-element connected sets for n = 3a + 1. Moreover, every vertex in is dominated by exactly one vertex in S.
G S −
S Proof
⎡ ⎤ 1
)) , 1 2 ( (
3 2
+ ≥ +
n t
k k P γ
3.
2 3 n
S ≤ ⎡ ⎤ ⎢ ⎥
if ( d 3)
Theorem 2.3 For n = 2k + 1,
2n ⎧⎡ ⎤
1 ⎡ ⎤
3 2n
S =
, if (mod 3) , if (mod 3)
γt(P(2k + 1, k)) =
2 1 3 2 3 n n ⎧⎡ ⎤ + ⎪⎢ ⎥ ⎪⎢ ⎥ ⎨ ⎡ ⎤ ⎪ ⎢ ⎥ ⎪ ⎢ ⎥ ⎩
1 ≡ n 1 ≡ n
(2)
If n ≡ 1 (mod 3), proof
⎡ ⎤ 1
)) , 1 2 ( (
3 2
+ ≥ +
n t
k k P γ
3 ⎪ ⎢ ⎥ ⎩
Suppose we have the following two properties
⎡ ⎤
2
)) 1 2 ( (
n
k k P = + γ
Claim 1. For any connected subset
3 , ≤ = ⊂ d D S D
Suppose , we have the following two properties.
⎡ ⎤
3
)) , 1 2 ( (
t
k k P + γ
V(G) Proof.
⎡ ⎤
1 2
3 2
+ = = a S
n
S D
⎡ ⎤
1 2
3
+ a S
We have and then d ≤ 3.
n d d a 2 ) 2 2 ( ) 1 2 ( 3 ≥ + + − +
Claim 2. S is formed by the union of exactly one 3-element connected set and a − 1 2 element connected sets for n = 3a + 1 set and a − 1 2-element connected sets for n = 3a + 1. Moreover every vertex in G is dominated by exactly one vertex in S.
can dominate at most 2・(2・3+2) = 16 vertices
⎡ ⎤
1 2
3 2
+ = = a S
n
then S can dominate at most (2a+1−6)·3+16 = 6a+1 < 2n
⎡ ⎤
3
→← →← By counting each connected subset can dominated (2 · 3 + 2) + (a − 1)(2 · 2 + 2) = 2n
Let D be a connected subset of S with |D| = 3
D D N N )) ( (U
d
D N N N )) ( (U
D D N N
v
− =
∈
)) ( (
D 1
U
and
D N v N N
N v
− − =
∈ 1 2
)) ( (
1
U
vk+1 vk+2 vk+3 vk+4 vk uk+4 uk+3 uk+2 uk+1 u
uk u4 u2k+1 v4 v1 v2k+1 u4 u3 u2 u1 v3 v2
vk vk+1 vk+2 vk+3 vk 1
| | | |
vk-1 uk+3 uk+2 uk+1 uk uk-1 u3 u2k v3 v2k+1 v2k u3 u2 u1 u2k+1 v2 v1
Case3.
vk+1 vk+2 vk+3
|D ∩ U| = 3
k+1 k+3
vk+4 vk u uk+3 uk+2 uk+1 uk+4 uk
With the above cases, we know
⎡ ⎤ 1
)) , 1 2 ( (
3 2
+ ≥ +
n
k k P γ
⎡ ⎤ 1
)) , 1 2 ( (
3 +
≥ +
t
k k P γ
v2k+1 u4 u3 u2 u1 u2k+1 v4 v3 v2 v1
if ( d 3)
Theorem 2.3 For n = 2k + 1,
2n ⎧⎡ ⎤
1
F 3
v8
, if (mod 3) , if (mod 3)
γt(P(2k + 1, k)) =
2 1 3 2 3 n n ⎧⎡ ⎤ + ⎪⎢ ⎥ ⎪⎢ ⎥ ⎨ ⎡ ⎤ ⎪ ⎢ ⎥ ⎪ ⎢ ⎥ ⎩
1 ≡ n 1 ≡ n
v v7
8
v10 v9 u8 u
S0 = {V3t+1|0 ≤ t < a}; S {U |0 ≤ t < }
3 ⎪ ⎢ ⎥ ⎩
we construct a total dominating set S for required cardinality
(3) or
2 3 n
S ≤ ⎡ ⎤ ⎢ ⎥ ⎡ ⎤ 1
3 2
+
n
v6 u7 u6 u9 u10
S1 = {U3t+1|0 ≤ t < a}; S = |S0| + |S1| = 2a =
⎥ ⎥ ⎤ ⎢ ⎢ ⎡ 3 2n
we construct a total dominating set S for required cardinality.
v5 v11 u5 u11 v4 v12 u4 u3 u12 v3 v13 u1
3
u2 u u13 u14 v v2 v14 v
1
u15 v1 v15
S0 ={V3 |1 ≤ t ≤ }∪ {V3 +1| < t ≤ a};
⎣ ⎦
2 a
⎣ ⎦
2 a
S0 {V3t|1 ≤ t ≤ }∪ {V3t+1| < t ≤ a};
⎣ ⎦
2
⎣ ⎦
2
S1 ={U3t|1 ≤ t ≤ }∪ {U3t+1| < t ≤ a}∪{U1,Uk};
⎣ ⎦
2 a
⎣ ⎦
2 a
v6
S |S | + |S | + + 2
⎤ ⎡2n
u5 u6 u7 v5 v7
S = |S0| + |S1| = a + a + 2 = .
⎥ ⎥ ⎤ ⎢ ⎢ ⎡ 3 2n
u4 u7 u8 v4 v8 u3 u9 v9 u2 u9 u10 v3 v9 u1 u11 v2 v10 v11 v1
Theorem 2.4 For n = 2k the domination number of P(2k, k) is . ⎡ ⎤
3 2n
For choose i = 1 n we define a S to be induced subgraph of G For choose i = 1, ..., n we define a Si to be induced subgraph of G
P(10,5)
Si
vi+k-1 vi+k vi+k+1 ui+k+1 ui+k ui+k-1 ui-1 ui ui+1 vi+1 vi vi-1
Si
Th f k 4 ≥ D And each vertex of G belongs to exactly 6 S We have ∑ =
≥
n i i
n S V D
1
4 ) ( I
Therefore we know . 4 ≥
i
D And each vertex of G belongs to exactly 6 Si So,
∑ =
=
n i i
D D
1
6
Hence 6|D| ≥ 4n and ⎡ ⎤
3 2n
D ≥
Hence 6|D| ≥ 4n and ⎡ ⎤
3
D ≥
Show that γ(P(2k, k)) ≤ ⎡ ⎤
3 2n
⎡ ⎤
3
(1) n = 3a S0 = {v3t+1|0 ≤ t < a} { 3t+1| }
|S0| + |S1| = a + a = 2a = 3 2n
v8 v7 v6 v5 u u6
S1 = {u3t+2|0 ≤ t < } ∪ {u3t|0 ≤ t < }
2 a 2 a
v9 u8 u7 u6 u5 v9 v4 u9 u4 u3 v3 v10 u11 u10 u3 u2 v2 v11 u12 u1 v1 v12
(2) n = 3a + 1 ( )
S0 = {v3t+1|0 ≤ t ≤ } ∪ {v3t| < t ≤ a}
⎣ ⎦
2 a
⎣ ⎦
2 a
S1 = {u3t+2|0 ≤ t < a}
v5 v6
⎡ ⎤
v4 v7 u6 u5 u4
|S0| + |S1| = 2a + 1 =⎡ ⎤
3 2n
u7 u4 v3 v8 u8 u3 u9 u2 v2 v9 u9 u10 u1 v1 v10
Lemma 2.5 For n = 2k the independent domination number of P(2k k) is ⎡ ⎤
2n
P(2k, k) is .
⎡ ⎤
3
Theorem 2 6 For n = 2k the total domination number of P(2k k) Theorem 2.6 For n = 2k the total domination number of P(2k, k) is n P f Th b i b d f (P(2k k)) i
Let D be the minimum total dominating set of G = P(2k, k).
Define a Si={vi,vi+k,ui,ui+k} for i = 1, ..., n
vi+k ui+k ui vi
Si
Let Di = |D ∩ V(Si)| now we show that Di ≥ 2 since D is a total dominating set
vi+k
now we show that Di ≥ 2.
ui+k
Therefore we know Di ≥ 2.
ui
∑ ∑
= =
≥ =
n i n i i i
n S V D D
1 1
2 | ) ( | I
vi
Si
∑ ∑
= = i i 1 1
and each vertex of G belongs to exactly 2 Si So
∑
=
n D
D | | 2
So,
∑ =
=
i i
D D
1
| | 2
Hence 2|D| ≥ 2n and |D| ≥ n
3 domination numbers of P(n,1) and P(n,2) Note: For n ≥ 3,
h ≡ 0 1 3 ( d 4)
⎡ ⎤
⎧
, when n ≡ 0, 1, 3 (mod 4);
, when n ≡ 2 (mod 4),
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 ) P(n,1) (
2 2 n n
γ
⎡ ⎤
⎩ 2
Theorem 3.1 For n ≥ 3,
⎡ ⎤
⎧
, if n ≡ 0, 3 (mod 4); , if n ≡ 1, 2 (mod 4).
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 2 n n i
n P γ
if ≡ 2 ( d 4)
n 2
( ( ,1)) 1
i n
P n γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ ≥ ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩ , if n ≡ 2 (mod 4).
2
1
i n +
⎡ ⎤ ⎪⎢ ⎥ ⎩
Since
) ( ) ( G G
i
γ γ ≤
, if n ≡ 1, 2 (mod 4).
n 2 2
( ( ,1)) 1
i n
P n γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ ≤ ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩ , if n 1, 2 (mod 4).
2
1 + ⎡ ⎤ ⎪⎢ ⎥ ⎩
By construct a total dominating set S for the required cardinality.
i 2
( ( ,1)) 1
n
P n γ ≥ + ⎡ ⎤ ⎢ ⎥
Theorem 3.1 For n ≥ 3,
⎡ ⎤
⎧
, if n ≡ 0, 3 (mod 4); , if n ≡ 1, 2 (mod 4).
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 2 n n i
n P γ
S = {v |0 ≤ t < a};
v5 v4
S0 = {v4t+1|0 ≤ t < a}; S1 = {v4t+3|0 ≤ t < a};
v6 v u4 u5
|S0| + |S1| = 2a = 2
n
v6 v3 u6 u3 u7 u2 v7 v2 u8 u1 v8 v1
Theorem 3.1 For n ≥ 3,
⎡ ⎤
⎧
, if n ≡ 0, 3 (mod 4); , if n ≡ 1, 2 (mod 4).
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 2 n n i
n P γ
v6 v5
S0 = {v4t+1|0 ≤ t ≤ a};
v7 v4 u6 u5
S1 = {u4t+3|0 ≤ t < a} ∪ un; |S0| + |S1| = 2a + 2 =⎡ ⎤ 1
n +
u7 u4
|S0| |S1| 2a 2 ⎡ ⎤ 1
2 +
v8 u9 u8 u3 u v3 v2 u1 v9 u10 u9 u2 v1 v10
Theorem 3.1 For n ≥ 3,
⎡ ⎤
⎧
, if n ≡ 0, 3 (mod 4); , if n ≡ 1, 2 (mod 4).
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 2 n n i
n P γ
i 2
( ( ,1)) 1
n
P n γ ≥ + ⎡ ⎤ ⎢ ⎥
Suppose that , for n ≡ 1 (mod 4).
i 2
( ( ,1))
n
P n γ ≥ ⎡ ⎤ ⎢ ⎥
⎡ ⎤
1 2 S + = = a
n
⎡ ⎤
1 2 S
2
+ = = a
n a a a 2 2 8 4 8 ) 1 2 ( 4 = + > + = +
u
there are some vertices which are repeatedly dominated
n a a 2 1 8 3 ) 1 2 ( 4 < + = − + →←
Theorem 3.1 For n ≥ 3,
⎡ ⎤
⎧
, if n ≡ 0, 3 (mod 4); , if n ≡ 1, 2 (mod 4).
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 2 n n i
n P γ
v5 v4 v3
If N(u1)∩S = {u2, un}
S = {v4t+1|0 ≤ t < a} ∪
5
u5 u4 u3
If N(u1) ∩ S = {v1, un} { 4t+1| ≤ } ∪ {u4t+3|0 ≤ t < a} ∪ un
v6 v2 u6 u5 u2 u1 v1 u1 un un-1 un-2 vn-1 vn vn-2
Theorem 3.2 For n ≥ 3, Theorem 3.2 For n ≥ 3, , if (mod 6) , if
(mod 6) .
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 3 2 t n n
n P γ
4 ≡ n 4 ≡ n
1 Proof
2
( ( 1))
n
P n γ ≥ ⎡ ⎤ ⎢ ⎥
, if
(mod 6) .
⎡ ⎤
⎩ +1
3
t 3
( ( ,1)) P n γ ≥ ⎡ ⎤ ⎢ ⎥
By Proposition 1.1
⎡ ⎤
4 ≡ n
⎡ ⎤ 1
)) 1 , ( (
3 2
+ =
n t
n P γ
if (mod 6)
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + ≤ 1 )) 1 , ( (
2 3 2 t n n
n P γ
4 ≡ n 4 ≡ n , if (mod 6) .
⎡ ⎤
⎩ +1
3
4 ≡ n
Theorem 3.2 For n ≥ 3, Theorem 3.2 For n ≥ 3, , if (mod 6) , if
(mod 6) .
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
2 3 2 t n n
n P γ
4 ≡ n 4 ≡ n (2) If n ≡ 4 (mod 6) which also n ≡ 1 (mod 3)
, if
(mod 6) .
⎡ ⎤
⎩ +1
3
( ) ( ) ( ) let n = 3a + 1, a be odd, and
⎡ ⎤
3 2n
S =
we have the same two properties as in the proof of we have the same two properties as in the proof of Theorem 2.3 Claim 1 For any connected subset
3 ≤ = ⊂ d D S D
Claim 2. S is formed by the union of exactly one 3-element connected t d 1 2 l t t d t f 3 + 1 Claim 1. For any connected subset
3 , ≤ = ⊂ d D S D
set and a − 1 2-element connected sets for n = 3a + 1. Moreover every vertex in G is dominated by exactly one vertex in S. vertex in S.
Let D be the only 3-element connected subset of S.
u3 un-2 v3 u1 u2 un-1 vn-2 v2 vn-1 u1 un v1 vn
v5 v4 v6
v6t+1, v6t+2 S S
v3 u5
u6t+4, u6t+5 S
u6 u4 u3 v2 u u2 u
4
vn-5 un-5 v1 u1 un u
1
un-2 un-3 un-4 vn-4 vn un-1 vn-3 vn-1 vn-2
Theorem 3.2 For n ≥ 3,
⎡ ⎤
⎧
, if (mod 6) , if
(mod 6) .
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
3 2 3 2 t n n
n P γ
4 ≡ n 4 ≡ n
v6 v5 v4
⎩
For n 3a :
S0 = {v3t+1|0 ≤ t < a}; S { |0 ≤ }
u6 u5 u4
S1 = {u3t+1|0 ≤ t < a}; |S0| + |S1| = 2a =
3 2n
v7 v3 u7
6
u3 u8 u2 v8 v2 u1 u9 v1 v9
Theorem 3.2 For n ≥ 3,
⎡ ⎤
⎧
, if (mod 6) , if
(mod 6) .
⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = 1 )) 1 , ( (
3 2 3 2 t n n
n P γ
4 ≡ n 4 ≡ n F 3 2
v7 v6 v5
⎩
S0 = {v3t+1|0 ≤ t ≤ a};
v8 v4 u7 u6 u5
S1 = {u3t+1|0 ≤ t ≤ a};
|S | + |S | = 2a + 2 =⎡ ⎤
2n
v8 v4 u8 u4
|S0| + |S1| = 2a + 2 =⎡ ⎤
3
v3 v9 u9 u3 u1 v u10 u2 u11 v2 v1 v10 v11
N t F ≥ 3
⎡ ⎤
3
)) 2 ( (
n
P
Note: For n ≥ 3, .
⎡ ⎤
5 3
)) 2 , ( (
n
n P = γ
exact value ofγ(P(n, 2))
Theorem 3.3 For n ≥ 3, .
⎡ ⎤
5 3 i
)) 2 , ( (
n
n P = γ
Proof.
⎡ ⎤
5 3
)) 2 , ( (
n i
n P ≥ γ
Since
) ( ) ( G G
i
γ γ ≤ ) ( ) (
i
γ γ
⎡ ⎤
5 3
)) 2 , ( (
n i
n P ≤ γ
We construct the independent dominating set S = S0 ∪ S1 for the following cases.
Theorem 3.3 For n ≥ 3, .
⎡ ⎤
5 3 i
)) 2 , ( (
n
n P = γ
5
v
S0 = {v5t+2|0 ≤ t < a} ∪ {v5t+4|0 ≤ t < a};
u6 u5 v5 v6
S1 = {u5t+3|0 ≤ t < a};
u7
6
u4 v4 v7
|S0| + |S1| = 3a = 5
3n
u3 v3 v8 u8 u2 u1 u9 u10 v2 v9 v1 v10
Theorem 3.3 For n ≥ 3, .
⎡ ⎤
5 3 i
)) 2 , ( (
n
n P = γ
5
v v
S0 = {v5t+2|0 ≤ t ≤ a} ∪ {v5t+4|0 ≤ t < a};
v5 v6 v7 v8
S1 = {U5t+3|0 ≤ t < a} ∪ un−1; |S0| + |S1| = 3a + 2 =⎡ ⎤
5 3n
u8 u7 u6 u5
5
u4 u9 v4 v9 u3 u u10 v3 v10 u1 u2 u11 v u12 v1 v2 v11 v12
N F 3 Note: For n ≥ 3, ⎡ ⎤ ⎡ ⎤
⎨ ⎧ = )) 2 , ( (
3 2 t n
n P γ
, if n ≡ 0, 2 (mod 3); if 1 ( d 3) ⎡ ⎤
⎩ ⎨ +1 )) 2 , ( (
3 2 t n
n P γ
, if n ≡ 1 (mod 3).
g , g , y g [ ] proved this
4 Varied domination numbers of P(n,3) N F 7 Note: For n ≥ 7,
, when n ≡ 0, 1 (mod 4) or n=11
2 n
⎧ ⎡ ⎤ ⎪ ⎢ ⎥ ⎨
( ) , when n ≡ 2,3 (mod 4), n≠11
2 2
(P(n,3)) 1
n
γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ = ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩
, [ ] g
Theorem 4.1 For n ≥ 6, , when n ≡ 0, 1 (mod 4); , when n ≡ 2, 3 (mod 4).
2 i 2
(P(n,3)) 1
n n
γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ = ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩
, , ( )
2
1 + ⎡ ⎤ ⎪⎢ ⎥ ⎩
Proof
1 h 0 1 ( d 4) 11
n
⎧ ⎡ ⎤ ⎪ ⎢ ⎥ 1. , when n ≡ 0, 1 (mod 4) or n=11 , when n ≡ 2, 3 (mod 4), n≠11
2 2
(P(n,3)) 1
n i n
γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ ≥ ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩ Si
) ( ) ( G G ≤
Since
) ( ) ( G G
i
γ γ ≤
2. , when n ≡ 0, 1 (mod 4);
2
(P( 3))
n
⎧ ⎡ ⎤ ⎪ ⎢ ⎥ ≤ ⎨ 2. , when n ≡ 2, 3 (mod 4).
2 2
(P(n,3)) 1
i n
γ ⎪ ⎢ ⎥ ≤ ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩ We construct the independent dominating set S = S0 ∪ S1 3.
⎡ ⎤ 1
)) 3 , 11 ( (
2 +
=
n i P
γ
Theorem 4.1 For n ≥ 6, , when n ≡ 0, 1 (mod 4); , when n ≡ 2, 3 (mod 4).
2 i(P(n,3))
1
n n
γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ = ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩
, , ( )
2
1 + ⎡ ⎤ ⎪⎢ ⎥ ⎩
v5 v4
S0 = {v4t+1|0 ≤ t < a} S = {u |0 ≤ t < a}
u5 u4
S1 = {u4t+3|0 ≤ t < a} |S0| + |S1|= 2a = 2
n
v6 v3 u6 u3 u2 v7 v2 u1 u7 v8 v1 u8
Theorem 4.1 For n ≥ 6, , when n ≡ 0, 1 (mod 4); , when n ≡ 2, 3 (mod 4).
2 i 2
(P(n,3)) 1
n n
γ ⎧ ⎡ ⎤ ⎪ ⎢ ⎥ = ⎨ + ⎡ ⎤ ⎪⎢ ⎥ ⎩
v7 v6 v5
, , ( )
2
1 + ⎡ ⎤ ⎪⎢ ⎥ ⎩
S0 = {v4t+1|0 ≤ t ≤ a} S { |0 ≤ < } ∪
v u7 u6 u5
S1 = {u4t+3|0 ≤ t < a} ∪ {u2, un−3} |S | + |S | = 2a + 3 =
n
1 + ⎡ ⎤ ⎢ ⎥
v8 v4 u8 u4
|S0| + |S1| 2a + 3
2
1 + ⎡ ⎤ ⎢ ⎥
v3 v9 u9 u3 v v10 u11 u10 u2 u1 v2 v1
10
v11 u11
Theorem 4.2 For n ≥ 10,
⎡ ⎤
⎧ , if n ≡ 0, 2, 3, 4, 7, 9 (mod 10) , if n ≡ 1, 5, 6, 8 (mod 10)
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = ≤ 1 )) 3 , n ( (
5 4 5 4 3 2 n n t n
P γ
following cases. following cases. S0 = {v10t+2|0 ≤ t < a} ∪
v6 v5 u5
S0 {v10t+2|0 ≤ t a} ∪ {v10t+3|0 ≤ t < a} ∪ {v10t+5|0 ≤ t < a} ∪ {v |0 ≤ t < a} ∪
v7 v4 u7 u6 u5 u4
{v10t+6|0 ≤ t < a} ∪ {v10t+8|0 ≤ t < a} ∪ {v10t+9|0 ≤ t < a}
v8 v3 u8 u3
S1 = {u10t+4|0 ≤ t < a} ∪ {u10t+7|0 ≤ t < a}
u u9 u2
|S0| + |S1| = 8a = 5
4n
v2 v1 u1 v10 v9 u10
|S0| + |S1| 8a
5
Theorem 4.2 For n ≥ 10,
⎡ ⎤
⎧ , if n ≡ 0, 2, 3, 4, 7, 9 (mod 10) , if n ≡ 1, 5, 6, 8 (mod 10)
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎩ ⎨ ⎧ + = ≤ 1 )) 3 , ( (
5 4 5 4 3 2 n n t n
n P γ
S { |0 ≤ t < } ∪
v7 v8 v9
S0 = {v10t+2|0 ≤ t < a} ∪ {v10t+3|0 ≤ t < a} ∪ {v10t+5|0 ≤ t < a} ∪
u8 u7 u6 u9 v6 v10
{v10t+6|0 ≤ t < a} ∪ {v10t+8|0 ≤ t < a} ∪ {v10t+9|0 ≤ t < a} ∪
u5 u10 u11 v5 v11
S1 = {u10t+4|0 ≤ t < a} ∪ {u |0 ≤ t < a} { 10t 9| } vn∪vn−1 ∪vn−3∪vn−3 ∪ vn−4
u4 u12 v4 v12
{u10t+7|0 ≤ t < a} |S0| + |S1| = 8a + 5 = ⎡ ⎤ 1
5 4
+
n
u3 u2 u13 u14 v3 v13 u1 u15 v1 v2 v14 v15
In [3],[4],[5],they determined the exact values
domination number P(2k + 1, k)
3n
P(n, 1) , when n ≡ 2 (mod 4)
5 3n
⎡ ⎤ 1
2 n +
, when n ≡ 0,1,3 (mod 4)
P(n, 2)
⎡ ⎤
2
2 n
P(n, 2) P(n 3)
hen n ≡ 0 1 (mod 4) or n 11
5 3n
⎡ ⎤
n
P(n, 3) , when n ≡ 0,1 (mod 4) or n=11
, when n ≡ 2,3 (mod 4),n≠11
2 n +
⎡ ⎤
2
Note: For n = 2k + 1, γ(P(2k+1, k) )= .
⎡ ⎤
5 3n
⎡ ⎤ thatγ(P(2k+1, k)) ≤ in 2008. ⎡ ⎤
5 3n
γ(P(2k+1, k)) ≥ in 2009. ⎡ ⎤
5 3n