Abstract Let , be a graph with vertex set of order and edge set . A -dominating set of is a subset such that each vertex in \ has at least neighbors in . If is a vertex of a graph , the open -neighborhood of , denoted by , is the set , . is the closed -neighborhood of . A function 1, 1 is a signed distance--dominating function of , if for every vertex , ∑ 1. The signed distance--domination number, denoted by , , is the minimum weight of a signed distance--dominating function of . In this paper, we give lower and upper bounds on , of graphs. Also, we determine the signed distance--domination number of graph , (the graph obtained from the disjoint union by adding the edges , ) when 2

B Samadi

D A Mojdeh

S M Hosseini Moghaddam

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IJST (2012) A3 (Special issue-Mathematics): 367-370 
Iranian Journal of Science & Technology 
http://www.shirazu.ac.ir/en 
 
Bounds on the signed distance- -domination number of graphs 
 
D. A. Mojdeh1, 2*, B. Samadi2 and S. M. Hosseini Moghaddam3 
 
1Department of Mathematics, University of Tafresh, Tafresh, Iran 
2School of Mathematics, Institute for Research in Fundamental 
Sciences (IPM) Tehran, Iran, P.O. Box 19395-5746  
3Shahab Danesh Institute of Higher Education, Qom, Iran 
E-mails: damojdeh@ipm.ir, b_samadi61@yahoo.com & Sm.hosseini1980@yahoo.com 
 
Abstract 
Let ,  be a graph with vertex set  of order  and edge set . A  -dominating set of  
is a subset 	  such that each vertex in 	\	  has at least  neighbors in . If  is a vertex of a graph , the 
open -neighborhood of , denoted by , is the set 	 	 			 			 , 	 . 
 is the closed -neighborhood of . A function 	 	 1, 1  is a signed distance- -dominating 
function of , if for every vertex 	 , ∑ 	 	 1. The signed distance- -domination 
number, denoted by , , is the minimum weight of a signed distance- -dominating function of . In this 
paper, we give lower and upper bounds on ,  of graphs. Also, we determine the signed distance- -domination 
number of graph , 	 	  (the graph obtained from the disjoint union  by adding the edges 	 	
, 	 ) when 	 2. 
 
Keywords: Signed distance- -dominating function; th power of a graph 
 
1. Introduction 
Let ,  be a graph with vertex set 
 of order  and edge set . For a 
subset 	 	 , we define 
⋃ 	 . If  is a vertex of a graph , the open 
-neighborhood of , denoted by , is the set 
	 	 	 			 			 , 	 	 . 
 is the closed- -neighborhood 
of . 	 	 | |; 	 	 	  and 
Δ 	 	 | |; 	 	 	 .  
A -dominating set of  is a subset 	  such 
that every vertex in 	\	  has at least  neighbors in 
. The -domination number  is the minimum 
cardinality among the -dominating sets of . A 
subset 	 	  is a total dominating set, if for every 
vertex 	  there exists a vertex 	 , such that 
 is adjacent to . Let  be a graph with no isolated 
vertex. The total domination number   is the 
minimum cardinality among the total dominating 
sets of .  
A function f 	V	 1, 1  is a signed 
distance- -dominating function of G, if for every 
vertex v	 V, f N v ∑ f u	 	N 1. The 
signed distance- -domination number, denoted 
by γ , G , is the minimum weight of a signed  
 
*Corresponding author 
Received: 5 November 2011 / Accepted: 9 January 2012 
distance- -dominating function on . A signed 
distance-1-dominating function and signed 
distance-1-domination number ,  of a graph  
are identified with the usual signed dominating 
function and signed domination number  of a 
graph  [1]. 
Let 	 2 be a positive integer. A subset 
	 	  is a  –packing if for every pair of 
vertices , 	 ,			 , 	 . The  -packing 
number 	 is the maximum cardinality of a -
packing in  [2]. The joint of simple graphs  and 
, written 	 	 , is the graph obtained from the 
disjoint union  by adding the edges 
	 	 , 	  [3]. Let  be a graph of 
order  with vertex set , , . . . , . We 
construct th power  of a graph , by 
 and  and  are adjacent in  if and only if 
0 , . 
2. Lower bounds on ,  
Observation 1. Let  be a graph of order , and  
be a positive integer. Then ,  =  	 . 
 
Proof: Let  be a signed distance- -dominating 
function of . It is easy to see that for every 
	 , 	 . Hence 
. Therefore  is a signed distance- -
dominating function of  if and only if  is a signed 
 
 
IJST (2012) A3 (Special issue-Mathematics): 1-12   368 
 
distance dominating set of . Thus 
, . 
Let  be a graph of order , and  be a positive 
integer. 	 δ  and Δ Δ . 
 
Theorem 2. [4] For any graph  with 	 	 2, 
. 
As an immediate result from Observation 1 and 
Theorem 2 we have. 
 
Corollary 3. For any graph  with δ 2, 
,  . 
 
Proposition 4. Let  be a graph of order . Then 
2	 	 	 	 . 
 
Proof Let  be a minimum signed dominating 
function of . Let 	 	 1	  and 
	 1	 . If , then the 
proof is clear. 
If 	  since 1, then  has at 
least two adjacent in . Therefore  is a 2-
dominating set for  and | | . Since 
| | | | and | | 	 | |, then 
2| |  and finally we have 
2 . 
 
Proposition 5. Let  be a graph of order  and with 
no isolated vertex. Then 2 . 
 
Proof: The proof is similar to the Proposition 4. 
3. Upper bounds on ,  
Theorem 6. Let  be a positive integer. If  is a 
simple graph of order  and minimum degree 
	δ 2 and  is a maximum value of 1-
packing sets. Then , 2 , and this 
bound is sharp. 
 
Proof: Let  be a 1-packing set with | |
. We define : 	 1, 1  by, 
 
1		 	
1			 	 . 
 
It is easy to show that 2	 . 
Therefore, it is sufficient to show that  is a signed 
distance- -dominating function on . Let  be a 
vertex in . Since 	  2, then | | 3 since  
is a 1–packing set. Hence 	 	  	 	 , 
and 	 1. Now let  be a vertex in . 
There are two cases. 
 
Case 1. . Since  is a 1 -packing 
set in graph , then | 	 	 | 1 and let 
	  . Otherwise let  be a vertex in 
	 	  different from . This shows that 
, 	 1. This is a contradiction. Since 
	 2 therefore 	 1. 
 
Case 2. 	 . If 	 , then 
1. Let 	 	  and let 
, , . . . , 	 . Since , 	 	
2, there exists a vertex  on the  path which 
is distinct from the vertex  on the  path. 
Thus there exist at least  distinct vertices in 
. Suppose  be a vertex in  such 
that  is adjacent to  for each 1	 	 . 
Therefore, ∑ f 	∑ f
	 	 	1. And  is a signed- -dominating function on 
 with weight | | | | 2	 . Hence 
, 2 . 
Now, we show that the bound is sharp. The desired 
graph  will be the union  copies of . Then 
, 2  and . Therefore ,
2 	 4 2 2 . This completes the 
proof. 
 
Corollary 7. Let  be a positive integer. If  is a 
simple graph of order  and minimum degree 	  
2 and  is a maximum value of 1 -
packing sets, then 
1 . 
 
Proof: By Theorem 6 and Corollary 3 the proof is 
clear. 
 
Theorem 8. Let  be a connected graph of order . 
Let  and  be the sets of vertices degree 1 (leaves) 
and  (support vertices) respectively. If  is a 
maximum 2-packing set in  , then   
2| |, and this bound is sharp. 
 
Proof: We define : 	 1, 1  by, 
		
1		 	
1		 	 . 
It is easy to show that 2	| |. 
Therefore, it is sufficient to show that  is a signed 
distance-2-dominating function on . For each 
vertex , if  is a vertex in  then 
2	 1. Let 	 , if 
	  then obviously 1. If 
, since  is a 2-packing set in 
 then | | 1, and let 	
. Otherwise if  be a vertex in 	 	  
different from  then , 	 2. This is a 
contradiction. Since 	 2 then 	
2. Finally, if 	 , since 	 2 and 
	  then 	 1. Therefore  is 
 
 
 
369                             IJST (2012) A3 (Special issue-Mathematics): 1-12 
 
a signed dominating function of  with weight 
2| |. Hence 2| |. 
Now, we show that the bound is sharp. Let 
,  	 2 . Then  and . 
Thus 2| |. 
In Theorem 6 the graph  can be a simple 
disconnected graph of order  and 	 1. Since 
, ,...,  are components of , then 
. . . . By a similar reason 
we can prove  	 	2| |, where  and  
are the sets of vertices of degree 1 and  
respectively. 
But there exists a natural question here. What 
would happen if 1	? We are going to answer 
this question by concept of th  of the graph . 
Firstly, we have the following lemma. 
 
Lemma 9. Let  be a simple graph of order n and 
 be the th power of the graph . Then 
 is a maximum set of -packing vertices if 
and only if 	 	  is a maximum set of -
packing vertices. 
 
Proof: Since every edge in  is equal to a path 
with length 	  we have  and , two vertices in 
 such that there is no path between them with 
length 	  if and only if  and  are two vertices 
in  such that there is no path between them 
with length . This shows that  is a 
set of -packing vertices if and only if 	 	  
is a set of -packing vertices. Also, it is easy to see 
that  is maximum if and only if 	
	  is maximum. This completes the proof. 
 
Theorem 10. Let 	 2 be a positive integer. If  
is a simple graph and each component is order 
3, with minimum degree 	  1 and  is a 
maximum 2 -packing set, then ,
2	 , where | |, and this bound is sharp. 
 
Proof: Let 	be the th power of the graph . By 
Observation 1 we have , . Since 
	 3 then 	 2. Therefore, by Theorem 6 
we have , 	 . Finally 
by Lemma 9 we have ,  2  	. 
Now we show that the bound is sharp. The 
desired graph  will be the union  copies of star 
, . Then , , 3 , and  . 
Therefore , 3 2 2 . This 
completes the proof. 
 
Observation 11. Let  and  be two simple 
graphs. If 	 2 then  
, 	 	
		
1				 			| | | |			 			
	2				 			| | | |		 				 .
 
 
Now we show that for any integer  we can find a 
simple graph  such that  . 
 
Theorem 12. For any integer	 , there exists a 
connected graph  with  . 
 
Proof: We consider four cases. 
 
Case 1. Let 	0. We consider the star , | |  
with vertices , , ....,	 	 | |  and central vertex 
. We add vertices  1	 	 2| | 2 		be 
adjacent to  and  in modulo 2| | 2. Then 
we add edges 1	 	 2| | 2  in modulo 
2| | 2. Finally, we add vertices  1	 	
| | 1  adjacent to  and  (when 3,  
is illustrated in Figure 1). 
We define : 	 1, 1  by, 
 
1				 			 , , . . . . , 	 | |
	 1		 , , . . . . , 	 | | 	 , , . . . . , 	| | 	 .
 
 
In the following, we prove that  is a signed 
dominating function of . By symmetry it is 
sufficient to show that 	 1 for 	 , ,
, . 	∑ | | 2| |
3	 1 . 
| |
| | 1 1. 
1	
1.	 1	 1. 
Therefore  is a signed dominating function of  
with weight 
1 2| | 2 | | 1 2| | 2
| | . Hence . 
On the other hand, let  be a minimum signed 
dominating function on  such that 
, we have, ∑ 	
	∑ | | 		 ∑| |
∑ 1| | 	 		∑ 1| | 1= | | 	 . Therefore 
	 . 
 
 
 
Fig. 1. example of Theorem 12 for 	 	 3	
 
 
IJST (2012) A3 (Special issue-Mathematics): 1-12   370 
 
Case 2. If 0. We consider the Hajos graph  
(Fig. 2). 
We define : 	 1, 1  by, 
 
	 		
1				 			 , , 	
			 1				 			 , , 	 	 .
 
 
It is easy to see that f is a signed dominating 
function of , with weight 0. Therefore 
0. On the other hand, let  be a minimum signed 
dominating function of  such that (
. We have,  ( )
	∑ 	 V 	 1 1
0. Therefore,  ( ) 0. Hence  0. 
 
 
 
Fig. 2. Hajous graph	
 
Case 3. If 1. Obviously for the complete graph 
 we have  ( )=1. 
 
Case 4. If 	 2. We consider the star , . It is 
easy to see that  ( , . This completes the 
proof. 
References 
[1] Xing, H., Sun, L. & Chen, X. (2006). On signed 
distance- k -domination in graphs. Czechoslovak 
Mathematical Journal, 56(1), 229-238 
[2] Haynes, T. W., Hedetniemi, S. T. & Slater, P. J. 
(1998). Fundamentals of Domination in Graphs. New 
York, Marcel Dekker. 
[3] West, D. B. (2001). Introduction to Graph Theory 
(Second Edition). USA: Prentice Hall. 
[4] Haas, R. & Wexler, T. B. (2002). Bounds on the 
signed domination number of a graph. Electronic Notes 
in Discrete Mathematics, 11, 742-750. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 


Bounds on the signed distance--domination number of graphs

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Bounds on the signed distance--domination number of graphs

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