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Article

Signed Domination Number of the Directed Cylinder

1
Department of Mathematics, Shanghai University of Electric Power, Shanghai 200090, China
2
Department of Mathematics Education, Daegu Catholic University, Gyeongsan 38430, Korea
*
Author to whom correspondence should be addressed.
Symmetry 2019, 11(12), 1443; https://doi.org/10.3390/sym11121443
Submission received: 10 September 2019 / Revised: 17 November 2019 / Accepted: 19 November 2019 / Published: 23 November 2019

Abstract

:
In a digraph D = ( V ( D ) , A ( D ) ) , a two-valued function f : V ( D ) { 1 , 1 } defined on the vertices of D is called a signed dominating function if f ( N [ v ] ) 1 for every v in D. The weight of a signed dominating function is f ( V ( D ) ) = v V ( D ) f ( v ) . The signed domination number γ s ( D ) is the minimum weight among all signed dominating functions of D. Let P m × C n be the Cartesian product of directed path P m and directed cycle C n . In this paper, the exact value of γ s ( P m × C n ) is determined for any positive integers m and n.

1. Introduction

All digraphs considered in this paper are finite without loops and multiple arcs, and we refer to [1] for notation and graph-theoretical terminology not defined here. In a digraph D with vertex set V ( D ) and arc set A ( D ) , u is said to be an in-neighbor of v if u v A ( D ) . For a vertex v V ( D ) , denote N D ( v ) = { u : u v A ( D ) } be the open in-neighborhood of v. The closed in-neighborhood of v is N D [ v ] = N D ( v ) { v } . In all cases above, we omit the subscript D when the digraph D is clear from the context. For S V ( D ) , D [ S ] denotes the subdigraph induced by S.
For two digraphs D 1 = ( V 1 , A 1 ) and D 2 = ( V 2 , A 2 ) , the Cartesian product D 1 × D 2 is the digraph with vertex set V 1 × V 2 and ( x 1 , x 2 ) ( y 1 , y 2 ) A ( D 1 × D 2 ) if and only if x 1 = y 1 and x 2 y 2 A 2 or x 2 = y 2 and x 1 y 1 A 1 , where x i , y i V i for i = 1 , 2 . We use D 1 D 2 to present that D 1 and D 2 are isomorphic. The sets of vertices of the directed path P m and the directed cycle C n are denoted by { u 1 , u 2 , , u m } and { v 1 , v 2 , , v n } , respectively. Furthermore, in the Cartesian product P m × C n (see Figure 1), let X j = i = 1 n { ( u j , v i ) } for 1 j m and let Y i = j = 1 m { ( u j , v i ) } for 1 i n . In fact, D [ X j ] and D [ Y i ] are isomorphic to the directed cycle C n and the directed path P m , respectively. Throughout this paper, for each vertex ( u j , v i ) in P m × C n , the subscript i is taken modulo n. Hence, if i 0 and k is the smallest positive integer such that 0 < k n + i n , then ( u j , v i ) = ( u j , v k n + i ) .
Domination and its variations in graphs have been widely investigated as they have many applications in the real word and other disciplines such as computer networks and location theory, see [2,3,4,5]. The signed domination is one variation of domination and is now well studied in the literature. The study of signed domination of undirected graphs was initiated by Dunbar et al. in [6]. Then, many authors paid much attention to signed domination of graphs, see [7,8,9,10,11,12,13,14,15] and elsewhere. The signed domination of graph has some applications, for example, this concept can also serve as a model of social networks in which global decisions must be made similar to that exhibited in [10]. In [16], Zelinka generalizes this concept to digraphs. For a two-valued function f : V ( D ) { 1 , 1 } , the weight of f is w ( f ) = v V ( D ) f ( v ) . Formally, a two-valued function f : V ( D ) { 1 , 1 } is said to be a signed dominating function (SDF) if f ( N [ v ] ) 1 for each vertex v V ( D ) . The signed domination number of D, denoted by γ s ( D ) , is the minimum weight among all signed dominating functions of D. Signed domination of digraphs was studied by several authors including [16,17,18]. Throughout this paper, if f is a signed dominating function of D, then we let P and M denote the sets of those vertices in D which are assigned under f the value 1 and 1 , respectively. Hence | V ( D ) | = | P | + | M | and γ s ( D ) = | P | | M | = | V ( D ) | 2 | M | .
Product graphs are considered in order to gain global information from the factor graphs [19]. Many interesting wireless networks are based on product graphs with simple factors, such as paths and cycles. In particular, any square grid (resp., torus) is the Cartesian product of two paths (resp., cycles) [20]. The domination numbers of Cartesian product of two directed paths (resp., cycles) have been recently determined by several authors [21,22,23,24,25,26,27]. Even more recently, Zhang and Shaheen [28] determined the exact value of signed domination number γ s ( P m × P n ) on Cartesian product P m and P n for any 1 m 6 and n 1 , and Wang et al. [29] solved the remaining cases.
In this paper, we continue to study the signed domination number of P m × C n and obtain its exact value for any positive integers m and n.

2. Signed Domination Number of P m × C n

In this section we determine the exact values of signed domination number of Cartesian product P m × C n for any m 1 and n 2 . Notice that P 1 × C n C n . In [16], Zelinka obtained that γ s ( C n ) = n . Thus, in what follows, we may assume that m 2 in the Cartesian product P m × C n . We begin with one lemma that will be useful in the proof of our main results.
Lemma 1.
Let f be a SDF of P m × C n ( m 2 ) . The following statements are true:
(1) 
X 1 P .
(2) 
For 2 j m , | X j M | n / 2 .
(3) 
For 2 j m 1 , | ( X j X j + 1 ) M | n n / 3 .
(4) 
For 2 j m 2 , | ( X j X j + 1 X j + 2 ) M | n .
Proof. 
The statements (1) and (2) are trivial by the definition of SDF.
(3) If n 0 ( m o d 3 ) , then n = 3 k for some integer k 1 . Suppose on the contrary that there exists some 2 j m 1 such that | ( X j X j + 1 ) M | n n / 3 + 1 = 2 k + 1 . Then at least one of | X j M | k + 1 and | X j + 1 M | k + 1 holds. Without loss of generality, assume that | X j M | = k + s , where 1 s 3 k / 2 k . By the definition of SDF, | X j + 1 M | 3 k 2 ( k + s ) = k 2 s . Thus | ( X j X j + 1 ) M | ( k + s ) + ( k 2 s ) = 2 k s 2 k 1 , a contradiction. Therefore, | ( X j X j + 1 ) M | 2 k = n n / 3 . For the cases that n 1 ( m o d 3 ) or n 2 ( m o d 3 ) , one can reach the same contradictions by applying the argument similar to that above. Consequently, the statement (3) is true.
(4) If n 0 ( m o d 3 ) , then n = 3 k for some integer k 1 . Assume that | ( X j X j + 1 X j + 2 ) M | n = 3 k is false for 2 j m 2 . Then there exists some 2 j m 2 such that | ( X j X j + 1 X j + 2 ) M | 3 k + 1 . Hence it follows that at least one of | X j M | k + 1 , | X j + 1 M | k + 1 and | X j + 2 M | k + 1 is true. Suppose that | X j + 1 M | = k + s , where 1 s 3 k / 2 k . Then | X j M | 3 k 2 ( k + s ) = k 2 s and | X j + 2 M | 3 k 2 ( k + s ) = k 2 s by the definition of SDF. Thus | ( X j X j + 1 X j + 2 ) M | 3 k 3 s 3 k 3 , which is a contradiction. This implies that either | X j M | k + 1 or | X j + 2 M | k + 1 . Without loss of generality, assume that | X j M | = k + s , where 1 s 3 k / 2 k . Let | X j + 1 M | = l , where 0 l k 2 s . If l = 0 , then 2 s k . By the statement (2), | ( X j X j + 1 X j + 2 ) M | ( k + s ) + 3 k / 2 3 k , a contradiction again. So l 1 . Let X j + 1 M = a = 1 l = { ( u j + 1 , v i a ) } , X j + 2 , i a * = { ( u j + 2 , v i a 1 ) , ( u j + 2 , v i a ) } , and X j + 2 * = a = 1 l { X j + 2 , i a * } , where 1 i 1 < i 2 < < i l n . By the definition of SDF, X j + 2 * P . Since 3 k 2 l 3 k 2 ( k 2 s ) k + 4 , it follows that D [ X j + 2 X j + 2 * ] would be divided into some directed segments (that is, these directed segments are all isomorphic to directed paths) by l pairs of vertices X j + 2 , i 1 * , X j + 2 , i 2 * ,…, X j + 2 , i l * . Without loss of generality, assume that there are t such directed segments, denoted by H b , where 1 b t l (The directed segments H b ’s of D [ X j + 2 X j + 2 * ] are shown in Figure 2). Then each segment H b contains at most ( | V ( H b ) | + 1 ) / 2 vertices which are assigned under f the value 1 by the definition of SDF, where 1 b t . Thus | X j + 2 M | ( 3 k 2 l + t ) / 2 ( 3 k l ) / 2 as b = 1 t | V ( H b ) | = 3 k 2 l . Notice that l k 2 s , that is, 2 s k l . Then | ( X j X j + 1 X j + 2 ) M | ( k + s ) + l + ( 3 k l ) / 2 3 k , which is a contradiction. Consequently, the statement (4) is true for the case when n 0 ( m o d 3 ) . By a similar argument above, one can obtain the same contradictions for the cases when n 1 ( m o d 3 ) or n 2 ( m o d 3 ) . Hence the statement (4) holds. This completes the proof of Lemma 1. □
Next we shall give our main results in this paper. The methods to prove the following theorems are similar. First we present the upper bound on γ s ( P m × C n ) by constructing signed dominating function for P m × C n with weight attaining the upper bound, then we establish the lower bound on γ s ( P m × C n ) by Lemma 1, moreover, the lower bound coincides with the upper bound. Thus the exact value of γ s ( P m × C n ) is determined.
Theorem 1.
For any integer n 2 ,
γ s ( P 2 × C n ) = n i f n 0 ( m o d 2 ) , n + 1 i f n 1 ( m o d 2 ) .
Proof. 
Let X 2 * = i = 1 n / 2 { ( u 2 , v 2 i ) } . Define f : V ( P 2 × C n ) { 1 , 1 } by assigning to each vertex of X 2 * the value 1 while to each vertex of V ( P 2 × C n ) X 2 * the value 1. It is easy to check that f is a SDF of P 2 × C n with weight w ( f ) = 2 n / 2 . If n 0 ( m o d 2 ) ( n 1 ( m o d 2 ) , respectively), then γ s ( P 2 × C n ) w ( f ) = n ( n + 1 , respectively), and γ s ( P 2 × C n ) = 2 n 2 | M | n ( n + 1 , respectively) by Lemma 1 (1) and (2). This implies that γ s ( P 2 × C n ) = n ( n + 1 , respectively) when n 0 ( m o d 2 ) ( n 1 ( m o d 2 ) , respectively). This completes the proof of Theorem 1. □
Theorem 2.
For any integers n , m 3 such that n 0 ( m o d 3 ) ,
γ s ( P m × C n ) = n ( m + 2 ) 3 .
Proof. 
Suppose that n = 3 k for some integer k 1 . We next consider the following cases to complete the proof of Theorem 2.
Case 1. m 0 ( m o d 3 ) . Then m = 3 r for some integer r 1 . We first show that γ s ( P m × C n ) n ( m + 2 ) / 3 by constructing a SDF of P m × C n with weight n ( m + 2 ) / 3 . Let X 3 j 1 * = t = 0 k 1 { ( u 3 j 1 , v 3 t + 1 ) } , X 3 j * = t = 0 k 1 { ( u 3 j , v 3 t + 2 ) } for j = 1 , 2 , , r , and X 3 j + 1 * = t = 0 k 1 { ( u 3 j + 1 , v 3 t + 3 ) } for j = 1 , 2 , , r 1 . Define f : V ( P 3 r × C 3 k ) { 1 , 1 } by assigning to each vertex of ( j = 1 r 1 { X 3 j 1 * , X 3 j * , X 3 j + 1 * } ) { X 3 r 1 * , X 3 r * } the value 1 while to each vertex of V ( P 3 r × C 3 k ) ( j = 1 r 1 { X 3 j 1 * , X 3 j * , X 3 j + 1 * } ) { X 3 r 1 * , X 3 r * } the value 1. It is easy to verify that f is a SDF of P 3 r × C 3 k with weight w ( f ) = k ( 3 r + 2 ) = n ( m + 2 ) / 3 (The SDF f of P 9 × C 9 is shown in Figure 3). Thus γ s ( P 3 r × C 3 k ) w ( f ) = n ( m + 2 ) / 3 . By Lemma 1 (3) and (4), we have | ( X 2 X 3 ) M | 2 k and | ( X 3 j 2 X 3 j 1 X 3 j ) M | 3 k for every 2 j r , and so | M | 2 k + 3 k ( r 1 ) = k ( 3 r 1 ) from Lemma 1 (1). Then γ s ( P 3 r × C 3 k ) = 9 r k 2 | M | 3 r k + 2 k = n ( m + 2 ) / 3 , which implies that γ s ( P 3 r × C 3 k ) = n ( m + 2 ) / 3 .
Case 2. m 1 ( m o d 3 ) . Then m = 3 r + 1 for some integer r 1 . We first prove that γ s ( P m × C n ) n ( m + 2 ) / 3 by establishing a SDF of P m × C n with weight n ( m + 2 ) / 3 . Set X 3 j 1 * = t = 0 k 1 { ( u 3 j 1 , v 3 t + 1 ) } , X 3 j * = t = 0 k 1 { ( u 3 j , v 3 t + 2 ) } and X 3 j + 1 * = t = 0 k 1 { ( u 3 j + 1 , v 3 t + 3 ) } for j = 1 , 2 , , r . Define g : V ( P 3 r + 1 × C 3 k ) { 1 , 1 } as follows: each vertex of j = 1 r { X 3 j 1 * , X 3 j * , X 3 j + 1 * } is assigned the value 1 while each vertex of V ( P 3 r + 1 × C 3 k ) j = 1 r { X 3 j 1 * , X 3 j * , X 3 j + 1 * } is assigned the value 1. It is not hard to see that g is a SDF of P 3 r + 1 × C 3 k with weight w ( g ) = 3 k ( r + 1 ) = n ( m + 2 ) / 3 (The SDF g of P 7 × C 9 can be obtained by restricting f defined in Case 1 on P 7 × C 9 , which is depicted in Figure 3). Thus γ s ( P 3 r + 1 × C 3 k ) w ( g ) = n ( m + 2 ) / 3 . For each 1 j r , it follows from Lemma 1 (4) that | ( X 3 j 1 X 3 j X 3 j + 1 ) M | 3 k . According to Lemma 1 (1), we have | M | 3 r k . Hence γ s ( P 3 r + 1 × C 3 k ) = 3 k ( 3 r + 1 ) 2 | M | 3 r k + 3 k = n ( m + 2 ) / 3 . This implies that γ s ( P 3 r + 1 × C 3 k ) = n ( m + 2 ) / 3 .
Case 3. m 2 ( m o d 3 ) . Then m = 3 r + 2 for some integer r 1 . We first demonstrate that γ s ( P m × C n ) n ( m + 2 ) / 3 by giving a SDF of P m × C n with weight n ( m + 2 ) / 3 . We write X 3 j 1 * = t = 0 k 1 { ( u 3 j 1 , v 3 t + 1 ) } for j = 1 , 2 , , r + 1 , X 3 j * = t = 0 k 1 { ( u 3 j , v 3 t + 2 ) } and X 3 j + 1 * = t = 0 k 1 { ( u 3 j + 1 , v 3 t + 3 ) } for j = 1 , 2 , , r . Assigning to all vertices of ( j = 1 r { X 3 j 1 * , X 3 j * , X 3 j + 1 * } ) { X 3 r + 2 * } the value 1 and to all other vertices the value 1, we produce a SDF h of P 3 r + 2 × C 3 k with weight w ( f ) = k ( 3 r + 4 ) = n ( m + 2 ) / 3 (The SDF h of P 8 × C 9 can be obtained by restricting f defined in Case 1 (see Figure 3) to P 8 × C 9 ). So γ s ( P 3 r + 2 × C 3 k ) n ( m + 2 ) / 3 . According to Lemma 1 (3) and (4), we obtain | ( X 2 X 3 ) M | 2 k , | ( X 3 r + 1 X 3 r + 2 ) M | 2 k and | ( X 3 j 2 X 3 j 1 X 3 j ) M | 3 k for every 2 j r . This means that | M | 4 k + 3 k ( r 1 ) = k ( 3 r + 1 ) by Lemma 1 (1). Therefore, γ s ( P 3 r + 2 × C 3 k ) = 3 k ( 3 r + 2 ) 2 | M | 3 r k + 4 k = n ( m + 2 ) / 3 , implying that γ s ( P 3 r + 2 × C 3 k ) = n ( m + 2 ) / 3 . This completes the proof of Theorem 2. □
Theorem 3.
For any positive integers n 4 , m 3 such that n 1 ( m o d 3 ) ,
γ s ( P m × C n ) = m ( n + 2 ) + 2 n 8 3 i f m 0 ( m o d 2 ) , m ( n + 2 ) + 2 n 2 3 i f m 1 ( m o d 2 ) .
Proof. 
Suppose that n = 3 k + 1 for some integer k 1 . We first prove that Theorem 3 is true when k = 1 . Let X 2 j * = i = 1 2 { ( u 2 j , v 2 i ) } for 1 j m / 2 . Define f : V ( P m × C 4 ) { 1 , 1 } by assigning to each vertex of j = 1 m / 2 X 2 j * the value 1 while to each vertex of V ( P m × C 4 ) j = 1 m / 2 X 2 j * the value 1. It is easy to check that f is a SDF of P m × C 4 with weight w ( f ) = 4 m / 2 . Then γ s ( P m × C 4 ) w ( f ) = 2 m or 2 m + 2 when m 0 ( m o d 2 ) or m 1 ( m o d 2 ) . By Lemma 1 (1) and (3), if m 0 ( m o d 2 ) ( m 1 ( m o d 2 ) , respectively), then we have | M | m ( m 1 , respectively), and so γ s ( P m × C 4 ) = 4 m 2 | M | 2 m ( 2 m + 2 , respectively). This implies that γ s ( P m × C 4 ) = 2 m or 2 m + 2 when m 0 ( m o d 2 ) or m 1 ( m o d 2 ) . We next assume that k 2 and proceed the proof of Theorem 3 by considering two cases.
Case 1. m 0 ( m o d 2 ) . Then m = 2 r for some integer r 2 . We first show that γ s ( P m × C n ) ( m ( n + 2 ) + 2 n 8 ) / 3 by constructing a SDF of P m × C n with weight ( m ( n + 2 ) + 2 n 8 ) / 3 . Set X 2 j 1 * = t = 0 k 2 { ( u 2 j 1 , v n + 2 j 3 t ) } for j = 2 , 3 , , r and X 2 j * = t = 0 k 1 { ( u 2 j , v n + 3 j 3 t ) } { ( u 2 j , v 5 j ) } for j = 1 , 2 , , r . Define g : V ( P 2 r × C 3 k + 1 ) { 1 , 1 } as follow: each vertex of j = 2 r { X 2 j 1 * , X 2 j * } ( t = 0 k 1 { ( u 2 , v n + 2 3 t ) } { ( u 2 , v 4 ) } ) is assigned the value 1 while each vertex of V ( P 2 r × C 3 k + 1 ) ( j = 2 r { X 2 j 1 * , X 2 j * } ( t = 0 k 1 { ( u 2 , v n + 2 3 t ) } { ( u 2 , v 4 ) } ) ) is assigned the value 1. It is not hard to verify that g is a SDF of P 2 r × C 3 k + 1 with weight w ( g ) = 2 r ( k + 1 ) + 2 k 2 = ( m ( n + 2 ) + 2 n 8 ) / 3 (The SDF g of P 8 × C 10 is shown in Figure 4). Thus γ s ( P 2 r × C 3 k + 1 ) w ( g ) = ( m ( n + 2 ) + 2 n 8 ) / 3 . By Lemma 1 (3) and (4), we have | ( X 2 X 3 X 4 ) M | 3 k + 1 and | ( X 2 j 1 X 2 j ) M | 2 k for every 3 j r , and so | M | ( 3 k + 1 ) + 2 k ( r 2 ) = k ( 2 r 1 ) + 1 from Lemma 1 (1). Then γ s ( P 2 r × C 3 k + 1 ) = 2 r ( 3 k + 1 ) 2 | M | 2 r ( k + 1 ) + 2 k 2 = ( m ( n + 2 ) + 2 n 8 ) / 3 , which implies that γ s ( P 2 r × C 3 k + 1 ) = ( m ( n + 2 ) + 2 n 8 ) / 3 .
Case 2. m 1 ( m o d 2 ) . Then m = 2 r + 1 for some integer r 1 . We first demonstrate that γ s ( P m × C n ) ( m ( n + 2 ) + 2 n 2 ) / 3 by establishing a SDF of P m × C n with weight ( m ( n + 2 ) + 2 n 2 ) / 3 . We write X 2 j 1 * = t = 0 k 2 { ( u 2 j 1 , v n + 2 j 3 t ) } for j = 2 , 3 , , r + 1 and X 2 j * = t = 0 k 1 { ( u 2 j , v n + 3 j 3 t ) } { ( u 2 j , v 5 j ) } for j = 1 , 2 , , r . Assigning to all vertices of j = 2 r { X 2 j 1 * , X 2 j * } ( t = 0 k 1 { ( u 2 , v n + 2 3 t ) } { ( u 2 , v 4 ) } ) ( t = 0 k 2 { ( u 2 r + 1 , v n + 1 r 3 t ) } ) the value 1 , and to all other vertices the value 1, we produce a SDF h of P 2 r + 1 × C 3 k + 1 with weight w ( h ) = 2 r ( k + 1 ) + 3 k + 1 = ( m ( n + 2 ) + 2 n 2 ) / 3 (The SDF h of P 7 × C 10 can be obtained by restricting g defined in Case 1 (see Figure 4) to P 7 × C 10 ). So γ s ( P 2 r + 1 × C 3 k + 1 ) w ( h ) = ( m ( n + 2 ) + 2 n 2 ) / 3 . For every 1 j r , it follows from Lemma 1 (3) that | ( X 2 j X 2 j + 1 ) M | 2 k . This derives that | M | 2 k r by Lemma 1 (1). Therefore, γ s ( P 2 r + 1 × C 3 k + 1 ) = ( 2 r + 1 ) ( 3 k + 1 ) 2 | M | 2 r ( k + 1 ) + 3 k + 1 = ( m ( n + 2 ) + 2 n 2 ) / 3 , implying that γ s ( P 2 r + 1 × C 3 k + 1 ) = ( m ( n + 2 ) + 2 n 2 ) / 3 . This completes the proof of Theorem 3. □
Theorem 4.
For any positive integers n 2 , m 3 such that n 2 ( m o d 3 ) ,
γ s ( P m × C n ) = m ( n + 1 ) + 2 n 4 3 i f m 0 ( m o d 2 ) , m ( n + 1 ) + 2 n 1 3 i f m 1 ( m o d 2 ) .
Proof. 
Suppose that n = 3 k + 2 for some integer k 0 . We first show that Theorem 4 is true when k = 0 . Define f : V ( P m × C 2 ) { 1 , 1 } by assigning to each vertex of j = 1 m / 2 { ( u 2 j , v 2 ) } the value 1 while to each vertex of V ( P m × C 2 ) j = 1 m / 2 { ( u 2 j , v 2 ) } the value 1. It is easy to check that f is a SDF of P m × C 2 with weight w ( f ) = 2 m / 2 . Then γ s ( P m × C 2 ) w ( f ) = m or m + 1 when m 0 ( m o d 2 ) or m 1 ( m o d 2 ) . By Lemma 1 (1) and (3), if m 0 ( m o d 2 ) ( m 1 ( m o d 2 ) , respectively), then we have | M | m / 2 ( ( m 1 ) / 2 , respectively), and so γ s ( P m × C 4 ) = 2 m 2 | M | m ( m + 1 , respectively). This implies that γ s ( P m × C 2 ) = m or m + 1 when m 0 ( m o d 2 ) or m 1 ( m o d 2 ) . We next assume that k 1 and consider two cases to complete the proof of Theorem 4.
Case 1. m 0 ( m o d 2 ) . Then m = 2 r for some integer r 2 . We first prove that γ s ( P m × C n ) ( m ( n + 1 ) + 2 n 4 ) / 3 by constructing a SDF of P m × C n with weight ( m ( n + 1 ) + 2 n 4 ) / 3 . Let X 2 j 1 * = t = 0 k 1 { ( u 2 j 1 , v n + 2 j 3 t ) } for j = 2 , 3 , , r and X 2 j * = t = 0 k { ( u 2 j , v n + 3 j 3 t ) } for j = 1 , 2 , , r . Define g : V ( P 2 r × C 3 k + 2 ) { 1 , 1 } as follow: each vertex of j = 2 r { X 2 j 1 * , X 2 j * } ( t = 0 k { ( u 2 , v n + 2 3 t ) } ) is assigned the value 1 while each vertex of V ( P 2 r × C 3 k + 2 ) j = 2 r { X 2 j 1 * , X 2 j * } ( t = 0 k { ( u 2 , v n + 2 3 t ) } ) is assigned the value 1. It is not hard to verify that g is a SDF of P 2 r × C 3 k + 2 with weight w ( g ) = 2 r ( k + 1 ) + 2 k = ( m ( n + 1 ) + 2 n 4 ) / 3 (The SDF g of P 8 × C 11 is shown in Figure 5). Thus γ s ( P 2 r × C 3 k + 2 ) w ( g ) = ( m ( n + 1 ) + 2 n 4 ) / 3 . By Lemma 1 (3) and (4), we have | ( X 2 X 3 X 4 ) M | 3 k + 2 and | ( X 2 j 1 X 2 j ) M | 2 k + 1 for every 3 j r . So | M | ( 3 k + 2 ) + ( 2 k + 1 ) ( r 2 ) = k ( 2 r 1 ) + r according to Lemma 1 (1). Then γ s ( P 2 r × C 3 k + 2 ) = 2 r ( 3 k + 2 ) 2 | M | 2 r ( k + 1 ) + 2 k = ( m ( n + 1 ) + 2 n 4 ) / 3 , which implies that γ s ( P 2 r × C 3 k + 2 ) = ( m ( n + 1 ) + 2 n 4 ) / 3 .
Case 2. m 1 ( m o d 2 ) . Then m = 2 r + 1 for some integer r 1 . We first demonstrate that γ s ( P m × C n ) ( m ( n + 1 ) + 2 n 1 ) / 3 by establishing a SDF of P m × C n with weight ( m ( n + 1 ) + 2 n 1 ) / 3 . We write X 2 j 1 * = t = 0 k 1 { ( u 2 j 1 , v n + 2 j 3 t ) } for j = 2 , 3 , , r + 1 and X 2 j * = t = 0 k { ( u 2 j , v n + 3 j 3 t ) } for j = 1 , 2 , , r . Assigning to all vertices of j = 2 r { X 2 j 1 * , X 2 j * } ( t = 0 k 1 { ( u 2 , v n + 2 3 t ) } ) ( t = 0 k 1 { ( u 2 r + 1 , v n + 1 r 3 t ) } ) the value 1 , and to all other vertices the value 1, we produce a SDF h of P 2 r + 1 × C 3 k + 2 with weight w ( h ) = 2 r ( k + 1 ) + 3 k + 2 = ( m ( n + 1 ) + 2 n 1 ) / 3 (The SDF h of P 7 × C 11 can be obtained by restricting g defined in Case 1 (see Figure 5) to P 7 × C 11 ). So γ s ( P 2 r + 1 × C 3 k + 2 ) w ( h ) = ( m ( n + 1 ) + 2 n 1 ) / 3 . For every 1 j r , it follows from Lemma 1 (3) that | ( X 2 j X 2 j + 1 ) M | 2 k + 1 . This means that | M | r ( 2 k + 1 ) by Lemma 1 (1). Therefore, γ s ( P 2 r + 1 × C 3 k + 2 ) = ( 2 r + 1 ) ( 3 k + 2 ) 2 | M | 2 r ( k + 1 ) + 3 k + 2 = ( m ( n + 1 ) + 2 n 1 ) / 3 , implying that γ s ( P 2 r + 1 × C 3 k + 2 ) = ( m ( n + 1 ) + 2 n 1 ) / 3 . This completes the proof of Theorem 4. □

3. Conclusions and Future Work

This paper is a contribution to the theory of signed domination of graphs. In particular, we determine the exact value of signed domination number of the Cartesian product P m × C n for any positive integers m and n. It is also interesting to study the signed domination number of the Cartesian product of directed cycles in future.

Author Contributions

All authors contributed equally to the manuscript typed, read and approved the final manuscript.

Funding

This work is supported by the Natural Science Foundation of Shanghai (No. 14ZR1417900) and Basic Science Research Program, the National Research Foundation of Korea, the Ministry of Education, (NRF-2018R1D1A1B07049584).

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. The Cartesian product P m × C n .
Figure 1. The Cartesian product P m × C n .
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Figure 2. The directed segments H b ’s in D [ X j + 2 X j + 2 * ] .
Figure 2. The directed segments H b ’s in D [ X j + 2 X j + 2 * ] .
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Figure 3. The SDF f of P 9 × C 9 . The red solid vertices are assigned the value 1 under f.
Figure 3. The SDF f of P 9 × C 9 . The red solid vertices are assigned the value 1 under f.
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Figure 4. The SDF g of P 8 × C 10 . The red solid vertices are assigned the value 1 under g.
Figure 4. The SDF g of P 8 × C 10 . The red solid vertices are assigned the value 1 under g.
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Figure 5. The SDF g of P 8 × C 11 . The red solid vertices are assigned the value 1 under g.
Figure 5. The SDF g of P 8 × C 11 . The red solid vertices are assigned the value 1 under g.
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Wang, H.; Kim, H.K. Signed Domination Number of the Directed Cylinder. Symmetry 2019, 11, 1443. https://doi.org/10.3390/sym11121443

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Wang H, Kim HK. Signed Domination Number of the Directed Cylinder. Symmetry. 2019; 11(12):1443. https://doi.org/10.3390/sym11121443

Chicago/Turabian Style

Wang, Haichao, and Hye Kyung Kim. 2019. "Signed Domination Number of the Directed Cylinder" Symmetry 11, no. 12: 1443. https://doi.org/10.3390/sym11121443

APA Style

Wang, H., & Kim, H. K. (2019). Signed Domination Number of the Directed Cylinder. Symmetry, 11(12), 1443. https://doi.org/10.3390/sym11121443

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