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Journal of Applied Mathematics
Volume 2013 (2013), Article ID 619695, 7 pages
http://dx.doi.org/10.1155/2013/619695
Research Article

On the Domination Number of Cartesian Product of Two Directed Cycles

1School of Information Science and Technology, Chengdu University, Chengdu 610106, China
2Key Laboratory of Pattern Recognition and Intelligent Information Processing, Institutions of Higher Education of Sichuan Province, Chengdu 610106, China
3School of Electronic Engineering and Computer Science, Peking University, Beijing 100871, China

Received 18 September 2013; Revised 11 October 2013; Accepted 11 October 2013

Academic Editor: Carla Roque

Copyright © 2013 Zehui Shao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Denote by the domination number of a digraph and the Cartesian product of and , the directed cycles of length . In 2010, Liu et al. determined the exact values of for . In 2013, Mollard determined the exact values of for . In this paper, we give lower and upper bounds of with for different cases. In particular, . Based on the established result, the exact values of are determined for and 10 by the combination of the dynamic algorithm, and an upper bound for is provided.

1. Introduction

All of the digraphs considered in this paper are finite and simple, that is, without multiple edges or loops. For a digraph and a vertex , and denote the set of out-neighbors and in-neighbors of . Given two vertices and in , we say dominates if or . Let . A vertex dominates all vertices in . A set is a dominating set of if dominates . The domination number of , denoted by , is the minimum cardinality of a dominating set of .

The Cartesian product of graphs and is the graph with the vertex set , and if either and or and . The Cartesian product is commutative and associative, having the one-vertex graph as a unit. The subgraph of induced by , where , is isomorphic to , called a -layer (over ) and denoted by  . For more information on the Cartesian product of graphs see [1].

We use a 0-1 matrix pattern with rows and columns to represent a dominating set of , where if and only if the value at the entry of equals 1. For example, the pattern in Figure 2 represents the dominating set (the set of black circles) shown in Figure 1. Let be a pattern; we denote by the column number of . Let and be patterns; then the pattern denotes the concatenation of patterns and , and denotes the concatenation of patterns for times.

619695.fig.001
Figure 1: A dominating set of .
619695.fig.002
Figure 2: Pattern of the dominating set shown in Figure 1.

Let denote the set . For an integer , the directed cycle of length is the graph whose vertices are and whose edges are the pairs , , where the arithmetic is done modulo . We denote by . In this paper, when we consider the Cartesian product graph of two cycles and , the edge set of is and is . Throughout the paper, when considering the vertex in the graph , we use the arithmetic operations of the index over modulo  and modulo .

The dominating set problem requires determining the domination number of a given graph. It has natural applications in numerous facility location problems. In such problems, the vertices of a graph correspond to locations, adjacency represents some notion of accessability, and the goal is to find a subset of locations accessible from all other locations at which to install fire stations, bus stops, post offices, or similar facilities [2]. Dominating sets have also been applied in coding theory [2] and social networks [3]. For more information on the history and applications of the dominating set problem, see [4].

In 1990, Faudree and Schelp [5] initially discussed the domination number of the Cartesian product of two undirected graphs. The domination number for Cartesian product of two graphs has attracted lots of attention, and there are many works on undirected graphs (see, e.g., [612]). Recently, there are also some works on the domination number of Cartesian product of two directed graphs, in particular, the Cartesian product of directed paths and cycles [1315].

In [13, 15], Liu et al. determined the exact values of for and showed the following.

Theorem 1. if    and   .

Mollard [14] determined exact values of for . This paper investigates properties on the domination of Cartesian product of and with . We give lower and upper bounds of with for different cases. In particular, . Based on the established result, the exact values of are determined for and by the combination of the dynamic algorithm, and an upper bound for is provided.

2. Main Results

2.1. Bounds on

Proposition 2 (Mollard [14]). Let be a dominating set of ; then for all considered modulo  one has .

Lemma 3. Let , . Then there exists a minimum dominating set of such that for every .

Proof. For a dominating set of , we define a function as follows: where We now assume that is a minimum dominating set of and suppose to the contrary that for some such that is maximized. Then there exists a set such that for any . Thus, . By Proposition 2, we have if . Thus is a dominating set with such that , a contradiction with the fact that is maximized.

Lemma 4. Let , . Then there exists a minimum dominating set of such that for every .

Proof. By Lemma 3, it is sufficient to consider a minimum dominating set of such that for every . If , together with Proposition 2, we have that , which completes the proof.

By Lemmas 3 and 4, there is an obvious lower bound in terms of as follows.

Lemma 5. Let and be two integers; then

Lemma 6. For two integers   and , then for any , one has(1) if ,(2) if ,(3) if ,(4) if ,(5) if ,(6) if ,(7) for .

Proof. For any , let It can be seen that if , ; otherwise, . Let ; then and there are at most vertices, of of which each is not dominated by . Denoted by , we consider the following cases.
(1)  .  In this case, it is easy to see that . By adding at most vertices to , we get a dominating set of as required.
(2)  .  Since , we will obtain a desired dominating set of by adding at most vertices to .
(3)  . Since , we have to add at most vertices so that a dominating set of is constructed based on .
(4)  .  Since is an even and , we have to add at most vertices to so that a dominating set of will be formed.
(5)  . Since , we will obtain a dominating set of by adding at most vertices to .
(6)  .  Let , when ; otherwise, let , when ; let when ; otherwise, let , when ; let ,, for . In this way, if we let , then is a dominating set of . In addition, , , and . So, .
(7)  .  Let . Then is a dominating set of with cardinality .

The result of Lemmas 5 and 6 show the following.

Corollary 7. Let and be two integers; then

3. On the Domination of

Lemma 8. Let and be two integers. Suppose is a minimum dominating set of . Let and for . (a)If there is an integer such that (considered modulo ), then one of is and the other is .(b)If there is an integer such that , , , and , then for any distinct vertices , in and , in , , ; for any distinct vertex , in and , in , or and or , where , , , , and they are taken modulo .

Proof. It follows immediately that (a) holds by Lemmas 3 and 4. In addition, since ,  ,  ,  and , each of vertices of and is dominated by only one vertex; also, only one vertex of is dominated by two vertices. Now we will first show that it is impossible that there are three consecutive vertices in . Otherwise, suppose that , , and (modulo ) are three consecutive vertices. We can deduce that vertices and are in and , respectively, so is dominated by both and , a contradiction.
Suppose that there are , (or , ) in (or ) such that (or ). Then it is easy to show that there exist three consecutive vertices or three consecutive vertices , , and (modulo ), thus , , and (modulo ) can only be dominated by , and since it is impossible that there are four consecutive vertices in by . So, it is an easy task to prove that either () is not dominated by or () is dominated by two vertices, a contradiction. By using the same approach as above, we also can prove the cases of vertices in and .

Lemma 9. Let and . If is a minimum dominating set of such that , where and for some , , then if and only if for any and .

Proof. By Lemma 8 and the symmetrical structure of , we assume without loss of generality . Since each vertex in is not dominated by , we have . By observing that each vertex in needs to be dominated, we have the desired set . Repeating this process until is determined, we have the desired result.

Theorem 10. Let . Then

Proof. Consider the following.
Case 1.  .
The lower bound follows from Lemma 5, and the upper bound follows from Lemma 6 with in Case 1.
Case 2.  .
By using Lemmas 3 and 4, we obtain that if and only if . Therefore, the desired lower bound is established. Upper bound can be obtained by applying Lemma 6 with in the following cases:(i)  , in case 4;(ii) , in case 1;(iii) , in case 2;(iv) , in case 5.
We give the upper bound for the case , by constructions.
Let . The pattern depicted in Figure 2 is a dominating set of with 46 vertices. Moreover, the leftmost 14 columns induce a dominating set of with 35 vertices. By repeating the leftmost 14 columns for times, we get a dominating set of with vertices. Therefore, , and thus, .
Let . The pattern depicted in Figure 3 is a dominating set of with 61 vertices. Moreover, the leftmost 14 columns induce a dominating set of with 35 vertices. By repeating the leftmost 14 columns for times, we get a dominating set of with vertices. Therefore, , and thus, .
Case 3.  .
The upper bound follows from Lemma 6. We then show the lower bound. Let . If there exists a minimum dominating set of with cardinality , then by Lemmas 3 and 4, we have such a dominating set with , where and . We without loss of generality assume . Then and so . We now consider , , and with and known. Then there are four vertices in not dominated by , a contradiction. Therefore, if . By using the same approach, we can obtain the desired lower bound if and the proof is omitted.

619695.fig.003
Figure 3: A dominating set of .

4. On the Domination of 

We use the dynamic algorithm to provide the result on for and . We first describe the concept needed to describe our computer checking. The idea is introduced in [16] in a more general framework and extended in [17], but for our purpose the following description will be sufficient.

For a graph and , we define a function such that if and if . We call such a function the 2-coloring of . If for there exists a vertex such that , then we say is dominated. We can see that this notation is equivalent to the dominating set and we will use it to describe the dynamic algorithm for this problem.

Let be a directed cycle in a weighted digraph . We say that the sum of the weights of all the edges in to be the weight of .

Let be a 2-coloring of ; is said to be an -vertex if the number of ones in the first copy ofequals and the second equals (see, e.g., the left side of the vertex in Figure 4 is a -vertex). An -vertex is said to be redundant if it is dominated and any removal of one vertex in the first column destroys the domination.

619695.fig.004
Figure 4: Two vertices of .

Let . We define a weighted digraph (when no confusion can arise, the set is omitted and the digraph is denoted by ) as follows.

The vertices of consist of all the nonredundant 2-colorings of . Let be a vertex of . Then and represent the 2-colorings of restricted to the first and second copies of . Let and be two vertices of . Then denotes the 2-coloring of obtained by applying and to the consecutive copies of . For each vertex of , we make an arc from to in if and only if the following conditions are fulfilled:(i) equals ,(ii)each of the vertices of is dominated (by ),(iii)for each -vertex of , .

Finally, the weight of is the number of ones in of .

Let . Figure 4 shows two vertices of . We can see that the 2-coloring (0010010) of the second copy of of the first vertex equals the first copy of of the second vertex. Moreover, the weight of the second copy of of the first vertex is 2. It follows that has an arc from the first (left) vertex to the second (right) with weight 2.

Lemma 11. Let . Then possesses a directed cycle of length with weight if and only if contains a minimal dominating set with .

Proof. If is a minimal dominating set of , then restricted to is a 2-coloring of . Since is minimal, we have that is nonredundant.
If possesses a directed cycle of length , we concatenate the second cycle of each vertex of the cycle and then obtain a dominating set of . Since each vertex in the cycle is nonredundant, we have that is minimal.

We define a weighted adjacency matrix of as follows. Let . Then where is the weight of the edge .

Let ( rows, columns) and ( rows, columns) be two weighted adjacency matrices. We define multiplied by as follows where for and .

We denote by the th power of ; we have the following result.

Theorem 12. Let and be the weighted adjacency matrix of with vertices. Then one has

Lemma 13. Let , , and a dominating set with cardinality . (a)If is odd, then for any .(b)If is even, then for any but at most one integer such that .

Proof. (a) Suppose that there exists such that . We can deduce that Thus, by the pigeonhole principle it can be seen that there exist some such that , which contradicts with Lemma 4.
(b) The proof is similar to (a).

Theorem 14. Let . Then

Proof. When   , the result follows from Lemmas 6, 8, and 9. Otherwise, let . We implemented the algorithm as described above, and the digraph was created and it has 4905 vertices. By Lemma 13, it can be seen that there may exist at most one -vertex such that in the auxiliary graph . We consider each possible auxiliary graph , whose vertex set is the union of that of and one -vertex with . Note that has only 4906 vertices; the matrix multiplication can be done easily. By using matrix multiplication, we found a formula as stated in Theorem 14. (Such parameters and exist; see [17].) If (for some ) is greater than , then .

Remark 15. Note that when Lemmas 4 and 13 are applied, we can operate matrix multiplication on the graph whose row number is 4906. It can be clearly seen that for any with the condition of Lemma 13, and it is natural to consider the graph , where . However, has 24845 vertices, whose vertex number is much more than that of . Therefore, Lemmas 4 and 13 play the key role in computing the exact values of , and this helps to make many hard instances become easy to solve.

5. Bounds on the Domination Number of 

We will conclude this work with upper bounds for the domination number of .

Theorem 16. Let . Then

Proof. We show the desired weight for all cases in Table 1, where and stands for case with in Lemma 6. For example, the result of the case follows from case 1 with in Lemma 6.
Also, patterns with 13 rows are presented (see Figures 5 and 6). By repeating the pattern of for times and concatenating , we will obtain a pattern of , where . From the above, it can be seen that all the desired upper bounds are established.

tab1
Table 1: Constructions for upper bounds of .
619695.fig.005
Figure 5: : dominating set of ; : dominating set of ; : dominating set of .
619695.fig.006
Figure 6: : dominating set of ; : dominating set of ; : dominating set of .

Disclosure

The authors confirm that they have given due consideration to the protection of intellectual property associated with this work and that there are no impediments to publication, including the timing of publication, with respect to intellectual property. In so doing, the authors confirm that they have followed the regulations of their institutions concerning intellectual property. The authors understand that the corresponding author is the sole contact for the editorial process (including editorial manager and direct communications with the office). He/she is responsible for communicating with the other authors about progress, submissions of revisions, and final approval of proofs. The authors confirm that they have provided current, correct email addresses (fnlang56@163.com and zshao@cdu.edu.cn) which is accessible by the corresponding author and which has been configured to accept email from.

Conflict of Interests

The authors wish to confirm that there is no known conflict of interests associated with this paper and there has been no significant financial support for this work that could have influenced its outcome. The authors confirm that the paper has been read and approved by all named authors and that there are no other persons who satisfied the criteria for authorship but are not listed. The authors further confirm that the order of authors listed in the paper has been approved by all of them.

Acknowledgments

The authors would like to acknowledge the referees for their valuable comments and suggestions. This work was supported by the National Natural Science Foundation of China under Grants nos. 61309015 and 61173121.

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