On the Domination Number of Generalized Petersen Graphs P(ck,k) 1footnote 11footnote 1The research is supported by NSFC (60973014, 11001035), Specialized Research Fund for the Doctoral Program of Higher Education (200801411073) and Research Foundation of DLUT.

# On the Domination Number of Generalized Petersen Graphs P(ck,k)111The research is supported by NSFC (60973014, 11001035), Specialized Research Fund for the Doctoral Program of Higher Education (200801411073) and Research Foundation of DLUT.

Haoli Wang,  Xirong Xu,  Yuansheng Yang222corresponding author’s email : yangys@dlut.edu.cn,
Department of Computer Science
Dalian University of Technology, Dalian, 116024, P.R. China

Guoqing Wang
Center for Combinatorics, LPMC-TJKLC
Nankai University, Tianjin, 300071, P.R. China
###### Abstract

Let be a simple connected and undirected graph with vertex set and edge set . A set is a if for each either or is adjacent to some . That is, is a dominating set if and only if . The domination number is the minimum cardinalities of minimal dominating sets. In this paper, we give an improved upper bound on the domination number of generalized Petersen graphs for and . We also prove that for even , for all , and for and .

Keywords: Domination number; Generalized Petersen Graph;

## 1 Introduction

Let be a simple connected and undirected graph with vertex set and edge set . The open neighborhood and the closed neighborhood of a vertex are denoted by and , respectively. For a vertex set , and . For , let be the subgraph induced by .

A set is a if for each either or is adjacent to some . That is, is a dominating set if and only if . The domination number of , denoted by , is the minimum cardinalities of minimal dominating sets. A subset is efficient dominating set or a perfect dominating set if each vertex of is dominated by exactly one vertex in . For a more detailed treatment of domination-related parameters and for terminology not defined here, the reader is referred to [4].

In recent years, domination and its variations on the class of generalized Petersen graph have been studied extensively [1, 2, 3, 5, 6, 7, 8, 9]. The generalized Petersen graph is defined to be a graph on vertices with and , subscripts are taken modulo . In 2009, B. Javad Ebrahimi et al [2] proved a necessary and sufficient condition for the generalized Petersen graphs to have an efficient dominating set.

Lemma 1.1. [2] If has an efficient dominating set, then and (mod 4).

Theorem 1.2. [2] A generalized Petersen graph has an efficient dominating set if and only if (mod 4) and is odd.

Recently, Weiliang Zhao et al [9] have started to study the domination number of the generalized Petersen graphs , where is a constant. They obtained upper bound on for as follows:

 γ(P(ck,k))≤⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩c3⌈5k3⌉, if c≡0 (mod 3);⌈c3⌉⌈5k3⌉−⌈2k3⌉, if c≡1 (mod 3);⌈c3⌉⌈5k3⌉−⌈2k3⌉+⌈k3⌉, if c≡2 (mod 3).\par

They also determined the domination number of for and the domination number of for odd .

In this paper, we study the domination number of generalized Petersen graphs . We give an improved upper bound on the domination number of for and . We also prove that for even , for all , and for and .

Throughout the paper, the subscripts are taken modulo when it is unambiguous.

## 2 General upper bound of P(ck,k)

In this section, we shall give an improved upper bound on the domination number of for general .

Theorem 2.1. For any constant and ,

 γ(P(ck,k))≤⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ck2+α, if c≡0 (mod 4);ck2+k2−1+α, if c≡1,2 (mod 4)% and k≡0 (mod 2);ck−12+k+12+α, if c≡1 (mod 4)% and k≡1 (mod 2);ck2+k+12+α, if c≡2 (mod 4) % and k≡1 (mod 4);ck2+k−12+α, if c≡2 (mod 4) % and k≡3 (mod 4);⌊ck2⌋+⌊k4⌋+1+α, if c≡3 (mod 4) and k≠4,8;ck2+k4+α, if c≡3 (mod 4) and% k=4,8;

where

 α={0, if k≡1 (mod 2);⌊c4⌋, if k≡0 (mod 2).
###### Proof.

To show this upper bound, it suffices to give a dominating set with the cardinality equaling to the values mentioned in this theorem. Let , and mod 4. Then .

For , let , where

 A={v4i:0≤i≤m−1} \ \ and \ \ B={u4i+2:0≤i≤m−1},

and let

 S=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩S0,                                                  if c≡0 (mod 4);S0∪{un−2−4i,un−4−4i:0≤i≤⌊k4⌋−1}∪{un−1},                                                       if c≡1 (mod 4) and k≡1 (mod 4);S0∪{un−2−4i,un−4−4i:0≤i≤⌈k4⌉−1}∪{vn−3},                                                       if c≡1 (mod 4) and k≡3 (mod 4);S0∪{un−2−4i,un−5−4i:0≤i≤⌊k4⌋−1}∪{un−1,un−3},                                                       if c≡2 (mod 4) and k≡1 (mod 4);S0∪{un−2−4i,un−3−4i:0≤i≤⌈k4⌉−1},                                                       if c≡2 (mod 4) and k≡3 (mod 4);S0∪{un−2−4i:0≤i≤⌊k4⌋}∪{vn−3},  if c≡3 (mod 4) and k≡1 (mod 4);S0∪{un−2−4i:0≤i≤⌊k4⌋},               if c≡3 (mod 4) and k≡3 (% mod 4).

It is not hard to check that

 |S|=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ck2,if c≡0 (mod 4);2×⌊ck4⌋+2×⌊k4⌋+1=ck−12+k+12,  if c≡1 (mod 4) and k≡1 (mod 4);2×⌊ck4⌋+2×⌈k4⌉+1=ck−12+k+12,  if c≡1 (mod 4) and k≡3 (mod 4);2×⌊ck4⌋+2×⌊k4⌋+2=ck2+k+12,     if c≡2 (mod 4) and k≡1 (mod 4);2×⌊ck4⌋+2×⌈k4⌉=ck2+k−12,          if c≡2 (mod 4) and% k≡3 (mod 4);2×⌊ck4⌋+⌊k4⌋+2=⌊ck2⌋+⌊k4⌋+1,   if c≡3 (mod 4) and k≡1 (mod 4);2×⌊ck4⌋+⌊k4⌋+1=⌊ck2⌋+⌊k4⌋+1,   if c≡3 (mod 4) and k≡3 (mod 4).

For , let and mod 4. Denote

, where

 A40={v4kj+2+4i,u4kj+4i:0≤i≤k4−1,  0≤j≤m2−1},B40={v4kj+k+1+4i,u4kj+k+3+4i:0≤i≤k4−1,  0≤j≤m2−1},C40={v4kj+2k+4i,u4kj+2k+2+4i:0≤i≤k4−1,  0≤j≤m2−1},D40={v4kj+3k+3+4i,u4kj+3k+1+4i:0≤i≤k4−1,  0≤j≤m2−1},E40={v4kj+3k:0≤j≤m2−1},

, where

 A42={v4kj+4i,u4kj+2+4i:0≤i≤k−24−1,  0≤j≤m2−1},   B42={v4kj+k+1+4i,u4kj+k−1+4i:0≤i≤k−24−1,  0≤j≤m2−1},   C42={v4kj+2k+2+4i,u4kj+2k+4i:0≤i≤k−24−1,  0≤j≤m2−1},   D42={v4kj+3k−1+4i,u4kj+3k+1+4i:0≤i≤k−24−1,  0≤j≤m2−1},   E42={v4kj+k−2,v4kj+2k−2,v4kj+4k−3,u4kj+2k−3,u4kj+4k−2,:0≤j≤m2−1},

and

 S4={S40, if k≡0 (mod 4);S42, if k≡2 (mod 4).

Then

 |S4|=⎧⎪⎨⎪⎩2×k4×c−r4×4+c−r4=(c−r)k2+c−r4, if k≡0 (mod 4);2×k−24×c−r4×4+5×c−r4=(c−r)k2+c−r4, if k≡2 (mod 4).

If , let . Then .

If , let

 S={S4∪{ui:n−k+1≤i≤n−1},if k≡0 (mod 4);S4∪{ui:n−k+1≤i≤n−4}∪{vn−k,vn−3,un−1},% if k≡2 (mod 4).

Then

 |S|=⎧⎪⎨⎪⎩(c−1)×k2+c−14+k−1=ck2+k2+⌊c4⌋−1,% if k≡0 (mod 4);(c−1)×k2+c−14+k−4+3=ck2+k2+⌊c4⌋−1, if k≡2 (mod 4).

If , let

 S=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩S4∪{vn−2k+2+4i,un−2k+4i:0≤i≤k4−1}    ∪{ui:n−k≤i≤n−1}∖{un−2k}, if k≡0 (mod 4);S4∪{vn−2k+4i,un−2k+2+4i:0≤i≤k−24−1}    ∪{ui:n−k−3≤i≤n−5}∪{vn−3}, if k≡2 (mod 4).

Then

 |S|=⎧⎪⎨⎪⎩(c−2)×k2+c−24+2×k4+k−1=ck2+k2+⌊c4⌋−1,         if k≡0 (mod 4);(c−2)×k2+c−24+2×k−24+k−1+1=ck2+k2+⌊c4⌋−1, if k≡2 (mod 4).

If , let

 S=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩S4∪{vn−ik,vn−ik+3:1≤i≤3}    ∪{un−2k+2,un−k+1}∖{vn−k},if k=4;S4∪{vn−ik,vn−ik+3,vn−ik+6:1≤i≤3}    ∪{un−3k+4,un−2k+2,un−2k+7,un−k+1,un−k+5},if% k=8;S4∪{vn−3k+6+4i,un−3k+4+4i:0≤i≤k4−2}    ∪{vn−2k+9+4i,un−2k+11+4i:0≤i≤k4−3}    ∪{vn−k+8+4i,un−k+9+4i,un−k+10+4i:0≤i≤k4−3}    ∪{vn−ik,vn−ik+3:1≤i≤3}    ∪{vn−2k+6,vn−k+6,vn−1}    ∪{un−2k+2,un−2k+7,un−k+1,un−k+5},if k≡0 (mod 4)and k≠4,8;S4∪{vn−3k+4i,un−3k+2+4i:0≤i≤k−24−1}    ∪{vn−2k+1+4i,un−2k−1+4i:0≤i≤k−24}    ∪{vn−k+3+4i,un−k+4i,un−k+1+4i:0≤i≤k−24−1}    ∪{vn−2k−2,un−2},if k≡2 (mod 4).

Then

 |S|=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(c−3)×k2+c−34+8−1=ck2+k4+⌊c4⌋,      if k=4;(c−3)×k2+c−34+14=ck2+k4+⌊c4⌋,          if k=8;(c−3)×k2+c−34+2×(k4−1)+5×(k4−2)+13=ck2+k4+⌈c4⌉,                                                            if k≡0 (mod 4) and k≠4,8;(c−3)×k2+c−34+2×(k−24+1)+5×k−24+2=ck2+k−24+⌈c4⌉,                                                            if k≡2 (mod 4).

It is not hard to verify that is a dominating set of with cardinality equaling to the values mentioned in this theorem. ∎

In Figure 2.1 and Figure 2.2, we show the dominating sets of for and , where the vertices of dominating sets are in dark.

As an immediate consequence of Lemma 1.1, Theorem 1.2 and Theorem 2.1, we have the following

Theorem 2.2. For ,

 γ(P(4k,k))={2k, % if k≡1 (mod 2);2k+1, if k≡0 (mod 2).

## 3 The domination number of P(5k,k)

In this section, we shall determine the exact domination number of for .

From Theorem 2.1, we have the following upper bound for .

Lemma 3.1. For , .

To prove the lower bound, we need some further notations. In the rest of the paper, let be an arbitrary dominating set of . For convenience, let

 Ai={vi+jk:0≤j≤c−1},   Bi={ui+jk:0≤j≤c−1},   Di(j)={vi+jk,ui+jk},  0≤j≤c−1,

for , where the vertices of are on the outer cycle and those of are on the inner cycle(s). For , let be the th subgraph induced by and .

Lemma 3.2. Let . If there exists two vertices such that , then .

###### Proof.

Suppose to the contrary that . Without loss of generality, we may assume and , i.e., (see Figure 3.1). Then at least one vertex of would not be dominated by , a contradiction. ∎

Lemma 3.3. For any , . Moreover, if there exists an integer such that , then , is an independent set, and the following statements hold.

(i) If , then . Moreover, the equality holds only if ;
(ii) If , then . Moreover, the equality holds only if ;
where the subscripts are taken modulo .

###### Proof.

Since is isomorphic to and every vertex of must be dominated by , we have that for any .

Suppose that there exists an integer such that .

Assume to the contrary that , or and is not an independent set. Then at least one vertex of would not be dominated by , a contradiction. Hence, and is an independent set.

(i) Suppose . Then , and is an independent set. Without loss of generality, we may assume . Since , to dominate , we have . It follows from Lemma 3.2 that .

If , to dominate , then , which implies that and . Since , to dominate , we have (see Figure 3.2 (1)). It follows from Lemma 3.2 that .

(ii) Suppose . If , then and . To dominate all the vertices in , we have that for every . It follows that , a contradiction with . Hence, .

Now suppose . It is easy to see that there exist at least two different index such that and .

If , that is, , say and , since , to dominate , we have that . It follows from Lemma 3.2 that , a contradiction with . Hence, we conclude that and for .

Without loss of generality, we may assume and . To dominate , , we have that and , . Since , to dominate , we have . Since