Research Article | Open Access

# Bounds on the Size of the Minimum Dominating Sets of Some Cylindrical Grid Graphs

**Academic Editor:**Liying Kang

#### Abstract

Let denote the domination number of the cylindrical grid graph formed by the Cartesian product of the graphs , the path of length *m*, and the graph , the cycle of length *n*, . In this paper we propose methods to find the domination numbers of graphs of the form with and and propose tight bounds on domination numbers of the graphs , . Moreover, we provide rough bounds on domination numbers of the graphs , and . We also point out how domination numbers and minimum dominating sets are useful for wireless sensor networks.

#### 1. Introduction

The problem of domination is one of the most widely studied topics in graph theory: the 1998 book by Haynes et al. [1] contains a bibliography with over 1200 papers on the subject. The domination problem was studied from the 1950s onwards, but the rate of research on domination significantly increased in the mid-1970s.

The signed domination number of a graph was defined in [2] and has been studied by several authors including [2, 3]. Independent dominating sets were introduced into the theory of games by Morgenstern in [4]. For an extensive survey of domination problems and comprehensive bibliography the readers are referred to [5]. The study of domination numbers of products of graphs was initiated by Vizing [6]. He conjectured that the domination number of the Cartesian product of two graphs is always greater than or equal to the product of the domination numbers of the two factors, a conjecture which is still unproven. In [7], a link is shown between the existence of tilings in Manhattan metric with , bowls and minimum total dominating sets of Cartesian products of paths and cycles. Domination numbers of Cartesian products were intensively investigated in [7â€“9].

The graphs considered here are finite, nonempty, connected, undirected without loops and without multiple edges. Besides these, any undefined terms in this paper may be found in Harary [10].

Let be a simple graph whose vertex set and edge set are and , respectively. The set of a simple graph is called a dominating set if every vertex is adjacent to some vertex . The domination number of the graph is the cardinality of a smallest dominating set of the graph ; it is usually denoted by and dominating set with smallest cardinality is called a minimum dominating set of the graph .

For any two graphs and , the Cartesian product is the graph with vertex set and with edge set such that , whenever and , or and [11].

In this paper we follow the following notations and terminologies. The numbers always denote the vertices of a path or a cycle . Let and denote the domination numbers of Cartesian product graphs and , respectively. Let , where and , where . and are called the layers of and , respectively. Moreover, layer of a dominating set means for . Next we shall define the term modified concatenation of two dominating sets of and . If and are two dominating sets of and , respectively, then the modified concatenation of and is a subset of such that , , and , ; that is, the th -layer of is coming from the th -layer of if and from the th -layer of if . The illustration of modified concatenation is shown in Figure 1.

One of the most challenging problems concerning the domination number of Cartesian products of graphs is the proof of the Vizing Conjecture, namely, [6]. Despite numerous results showing its validity in some special cases, till date the conjecture remains an open problem. Partial works have been made towards finding the domination numbers of some particular Cartesian products. This problem also seems to be difficult one and the authors of [12] proved that even for subgraphs of , this problem is NP-complete. In [13], Jacobson and Kinch established the following results:â€‰For all ,(i).(ii).(iii)

In [14], Chang and Clark established the following results:(i)(ii)

In [11], the authors established the following results regarding the Cartesian product of two cycles.(i)For , .(ii)For , .(iii)For , Furthermore, .

More works may be found in [8, 9, 15, 16].

In the paper [17], Nandi et al. dealt with the domination number of some special types of graphs, known as cylindrical grid graphs as shown in Figure 2. An alternative way of looking at the same cylindrical grid graph is also shown in Figure 2, where the leftmost column in all figures denotes the layer . In that paper the authors found the domination numbers as well as minimum dominating sets of the graphs , for and for all and provided bounds on for and for all . They pose an open problem for finding the domination numbers of the , for .

In the current paper, we deal with the above-mentioned open problem as posed in [17] and towards solving the problem, we get some partial results. We find the domination numbers as well as minimum dominating sets of the graphs , for and for all . We also give tight bounds on for and for all . Moreover, we provide rough bounds on domination numbers of the graphs , and . We also point out how domination numbers and minimum dominating sets are useful for wireless sensor networks. As a brief summary, we state the following results that we prove in the subsequent sections.â€‰For all ,(i) and for , , where (ii), , , , , and for , (iii)If and , where are natural numbers and , then

#### 2. Finding Minimum Dominating Sets of , for All

In this section we find the domination numbers as well as minimum dominating sets of particular cylindrical grid graphs of the form , for all and for . To prove the main results we state the following lemmas and theorems that are proved in [17]. Throughout the paper we use the arithmetic operations of the indices over modulo .

Lemma 1 (see [17]). *Let . Then there exists a minimum dominating set of such that for every , .*

Lemma 2 (see [17]). *There cannot be two consecutive -layers having empty intersection with a minimum dominating set of , for and .*

Lemma 3 (see [17]). *For every dominating set of , the following inequalities hold:
**
where for and . Moreover, if , then there does not exist any pair of vertices from such that they dominate a common vertex of . Finally, .*

*Remark 4. *The similar result holds for .

Lemma 5 (see [17]). *For , there exists a minimum dominating set of such that for every , .*

Lemma 6 (see [17]). *For , there exists a minimum dominating set of such that for every , either (a) or (b) with and for and for all , for some .*

Lemma 7 (see [17]). *For , there cannot be a dominating set with five consecutive -layers having exactly one vertex in common with .*

Lemma 8 (see [17]). *Let be a minimum dominating set with the property as stated in Lemma 6. Again let and be two layers having two vertices in common with and , . Then either , , , , or and for all .*

Theorem 9 (see [17]). *For , .*

*Note 1. *Let us call the collection of -layers as a block, where and are as in Lemma 8.

Then , where denotes the number of -layers having and vertices, respectively, in common with , denotes the number of blocks in which every -layer has exactly one vertex in common with , denotes the number of blocks in which every -layer has either or vertices in common with , and denotes the number of blocks, where , that is, when the block contains no -layer.

The above note will be useful to prove Theorem 13.

Theorem 10 (see [17]). *Consider the following:**, , and .*

Using the above-mentioned lemmas and theorems, we are going to prove the following main results.

Theorem 11. *For , , where
*

*Proof. *Consider the dominating sets of and as in Theorem 10 and the dominating sets of for as shown in Figure 3.

Now using modified concatenation of the dominating set for repeatedly with suitably choosing one of these ten dominating sets, we get dominating sets beyond . For example, to find a dominating set for , we use modified concatenation of the dominating set with . Again to find a dominating set for , we use modified concatenation of the dominating set with .

Corollary 12. *For ,
*

*Proof. *Using Theorems 9, 10, and 11 we get the desired result.

Theorem 13. *For ,*(1)*, if ,*(2)*, if ,*(3)*, if .*

*Proof. *To prove the theorem we first recall Note 1. Here for any minimum dominating set with the property of , as stated in Lemma 8, we get , where the symbols are as in Note 1.

(a) Let ; that is, , .

We claim that for any minimum dominating set of with the property as stated in Lemma 8, . Otherwise, there will exist a minimum dominating set of with the same property such that . Then and (since ).

Hence has the following properties:(1)each -layers has exactly or vertices in common with ,(2)each block has exactly -layers; that is, without loss of generality we can write
Since and and , the number of vertices dominated by each of the vertices of and will be (excluding themselves) and no vertex will be dominated by both of them simultaneously. Therefore or are the only two possibilities (as shown in Figure 4) of .

Now, without loss of generality let . Then .

Therefore , and ; for all . Consider and for all .

Hence cannot be dominated by the vertices of , contradicting that is a dominating set.

Therefore .

Hence for any minimum dominating set , ; that is, . Therefore . The rest of the proof of (a) follows from Theorem 11.

We note the following observation.*Observation **1.* Ifâ€‰â€‰ = = , = = = = , then (i) and imply and (ii) and imply .

(b) Let ; that is, .

We claim that for any minimum dominating set of with the property as stated in Lemma 8.

If possible let there exist a minimum dominating set with the property as stated in Lemma 8 and . Then and which will be impossible since .

If possible let there exist a minimum dominating set with the property as stated in Lemma 8 and . Then and . Therefore , , and or and .

Now if and then let denote the number of -layers having vertices in common with and .

By Lemma 8, is the number of -layers having vertices in common with and which is a contradiction.

Again if and then , which is also a contradiction.

If possible let there exist a minimum dominating set with the property as stated in Lemma 8 and . Then (otherwise, implies and , contradicting ) and .

Now there are the following three cases.*Case **1. *Consider and .

Since each block contains only those -layers which have exactly one vertex in common with .

Again since , there are two sub cases. *Subcase **1.1.* When only one block contains two -layers and other block contains four -layers then without loss of generality let for and , otherwise. Now or . If possible let . Then and for . But this is not possible since and and for . Note that .

Similar contradiction will be arrived for . *Subcase **1.2.* When only two blocks contain three -layers and other blocks contain four -layers then if(i)the blocks containing three -layers, each occurs consecutively then without loss of generality let for and , otherwise. Now if then which is a contradiction.Similar contradiction will be arrived for ,(ii)the blocks containing three -layers do not occur consecutively then among them one of the blocks consist of -layer for some and either and or and . In each case, we arrive at a contradiction. Hence we conclude that Case 1 cannot occur. *Case **2. *Consider, and .

In this case only two blocks contain three -layers and other blocks contain four -layers. Among these two blocks one contains only those -layers which have exactly one vertex in common with and other block contains three layers whose number of vertices are consecutively.

Now we note the following observations. *Observation **2.* If , and then implies and implies .*Observation **3.* If then implies or and implies or .

Without loss of generality we now assume that . Therefore by Observation 2 and Observation 1 in Case (a) we have either and or and which is again a contradiction by Observation 3.*Case **3. *Consider, and .

In this case only two blocks contain three -layers and other blocks contain four -layers and these two blocks contain three layers whose number of vertices is consecutively. Therefore by the Observation 2, as in Case 2, we arrive at a contradiction.

Then .

Therefore . This completes the proof of .

(c) Let ; that is, .

We claim that for any minimum dominating set of with the property as stated in Lemma 8, .

If possible let . Then as in (b), . As a result, we have the same contradiction.

If possible let . Then and therefore the following two cases arise. *Case **1. *Consider and .*Case **2. *Consider, , and .

In these two cases contradiction can be shown similarly as in (b), where .

Hence .

Therefore . This completes the proof of (c).

*Remark 14. *For and ,
and . Moreover, for , .

#### 3. Bounds on Domination Numbers of for All

In this section we give upper and lower bounds of the domination numbers of . Towards this direction, we first prove the following lemmas.

Lemma 15. *Let . Then there exists a minimum dominating set of with .*

*Proof. *Let be a minimum dominating set of with the property . Such a exists by Lemma 1. Suppose further that for many layers.

Assume that for some , then ; otherwise is a dominating set with , which is a contradiction.

Similar contradiction shows that . Now construct . Then and hence is a minimum dominating set with many layers having vertices in common with . Repeating this construction we get the desired minimum dominating set and hence the lemma follows.

Lemma 16. *For any minimum dominating set , implies and .*

*Proof. *The proof of the lemma follows from Remark 4.

Lemma 17. *For , there exists a minimum dominating set of with the property (i) if , , for some , then ;**(ii) if , , for some , then and (iii) .*

*Proof. *Let be a minimum dominating set of with the property . Such a exists by Lemma 15. Let .

When , then , and imply that since by Remark 4, . Note that in this case, and hence .

Let . If , and then . Also, ; otherwise will be a dominating set with , a contradiction. Clearly, . Also , because if , then or , each of which will be a contradiction, since and cannot be dominated by one vertex from and one vertex from . Similar contradiction will be arrived for . Let . Then . Hence we have . Therefore we can construct . In a similar way we can construct when . Repeating this construction for all for which , and , , , , we get a minimum dominating set with the desired property.

Lemma 18. *Consider a minimum dominating set of with the property as stated in Lemma 17. Let .**Then .*

*Proof. **Case **1*. Consider ; this implies (since ).*Subcase*â€‰â€‰*1**.**1*. Considerâ€‰; this implies .*Subcase*â€‰â€‰*1.2*.Consider; this implies .*Subcase*â€‰â€‰*1.3*.Consider.

For every subcase we get the desired result.*Case **2.* Consider; this implies .*Subcaseâ€‰â€‰**2**.1*. Consider; this implies .*Subcase*â€‰â€‰*2.2*. Consider; this implies which implies (since if then by Lemma 17â€‰â€‰).*Subcase*â€‰â€‰*2.3*. Consider.

For every subcase we get the desired result.*Case **3*. Consider .*Subcase*â€‰â€‰*3**.1*. Consider; this implies .*Subcase*â€‰â€‰*3.2*. Consider; this implies which implies (by Lemma 17).*Subcase*â€‰â€‰*3.3*. Consider.

For every subcase we get the desired result.*Case **4*. Consider .*Subcase 4.1*. Consider; this implies as in Subcase 1.2.*Subcase 4.2*. Consider.

For every subcase we get the desired result.*Case **5*. Consider .

In this case also .

Based on Lemma 17, we prove the following theorem which provides a lower bound on , for .

Theorem 19. *For , .*

*Proof. **Case **1*. When , for any dominating set (since the total number of vertices is and one vertex of can dominate at most vertices including itself). Therefore . Hence the theorem is true for .*Case **2*. For consider a minimum dominating set of with the property as stated in Lemma 17. Then . Therefore that implies that implies ; that is, , hence the theorem.

*Remark 20. *For simplification, henceforth, in all the figures, we are only considering the grid structure avoiding the circular arks to represent the circular grid as shown in Figure 5.

The following theorem provides domination numbers and dominating sets for some particular cyclic grid graphs which will help in providing the upper bounds for , for as addressed in Theorem 23.

Theorem 21. *Consider the following:**â€‰â€‰ , â€‰â€‰ , â€‰â€‰ , â€‰â€‰ , and â€‰â€‰.*

*Proof. *(a) A dominating set for is , as shown in Figure 6. As , . We want to show .

If possible let there exist a dominating set , of with .

Let .

Therefore,(i),(ii),(iii),(iv),(v),(vi),(vii).

We claim that ; otherwise, if possible let ; then .

Therefore, .

This implies and hence (from (vii)).

Therefore, and then implies , contradicting (v), hence the claim.

Similarly, we can show that .

Therefore, (using (iv) and (v)).

Hence, which implies , contradicting (iii).

Therefore, .

(b) A dominating set for is , as shown in Figure 6. As , . We want to show .

If possible let there exist a dominating set of with .

Let .

Therefore,(i),(ii),(iii),(iv),(v),(vi),(vii).

We claim that ; otherwise, if possible let . Then , which implies , contradicting (vii), hence the claim.

Therefore, . Similarly, we can show that . Next we claim that . If possible let . Therefore, , contradicting (vii), hence the claim. Therefore . Similarly, , contradicting (i).

Therefore,