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International Journal of Combinatorics

Volume 2011 (2011), Article ID 389369, 14 pages

http://dx.doi.org/10.1155/2011/389369

## Minimum 2-Tuple Dominating Set of an Interval Graph

^{1}Department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Midnapore 721102, India^{2}Department of Mathematics, Raja N. L. Khan Women's College, Vidyasagar University, Midnapore 721102, India

Received 8 September 2011; Revised 4 November 2011; Accepted 17 November 2011

Academic Editor: Johannes Hattingh

Copyright © 2011 Tarasankar Pramanik 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

The -tuple domination problem, for a fixed positive integer , is to find a minimum size vertex subset such that every vertex in the graph is dominated by at least vertices in this set. The case when is called 2-tuple domination problem or double domination problem. In this paper, the 2-tuple domination problem is studied on interval graphs from an algorithmic point of view, which takes time, is the total number of vertices of the interval graph.

#### 1. Introduction

An undirected graph is an *interval graph* if the vertex set can be put into one-to-one correspondence with a set of intervals on the real line such that two vertices are adjacent in if and only if their corresponding intervals have nonempty intersection. The set is called an *interval representation* of and is referred to as the *intersection graph* of [1]. Let , where for , be the interval representation of the graph , is the left endpoint, and is the right end point of the interval . Without any loss of generality, assume the following:(a)an interval contains both its endpoints and that no two intervals share a common endpoint [1],(b)intervals and vertices of an interval graph are one and the same thing,(c)the graph is connected, and the list of sorted endpoints is given(d)the intervals in are indexed by increasing right endpoints, that is, .

In a graph , a vertex is said to *dominate* itself and all of its neighbors. A *dominating set* of is a subset of such that every vertex in is dominated by at least one vertex in . The *domination number * is the minimum size of a dominating set of . For a fixed positive integer , a -*tuple dominating set* of is a subset of such that every vertex in is dominated by at least vertices of . As introduced by Harary and Haynes [2], a -*tuple dominating set * is a set for which for every , where is the *closed neighborhood* of the vertex . Note that we must have the minimum degree greater than or equal to for a -tuple dominating set to exist. The -*tuple domination number * is the minimum cardinality of -tuple dominating set of . When , this is called *double domination* [3].

A 2-tuple dominating set is said to be *minimal *if there does not exist any such that is a 2-tuple dominating set of . A 2-tuple dominating set , denoted by , is said to be *minimum*, if it is minimal as well as it gives 2-tuple domination number.

In graph theory, a connected component of an undirected graph is a subgraph in which any two vertices are connected to each other by paths. For a graph , if the subgraph itself is a connected component, then the graph is called *connected*, else the graph is called *disconnected* and each connected component subgraph is called its *component*. Removal of a vertex from a graph means the removal of vertex and edges incident to . A *cut vertex* of a connected graph is a vertex of whose removal renders disconnected. Pal et al. [4] described an algorithm for computing cut vertices and blocks on interval graphs.

A graph is *vertex domination-critical* if for any vertex of , the domination number of is less than the domination number of . If such a graph has domination number , it is called -*critical*. Brigham et al. [5] studied -critical graphs and posed the following questions.(1)If is a -critical graph, is ?(2)If a -critical graph has vertices, is regular?(3)Does for all -critical graphs?(4)Let be the diameter of the -critical graph . Does always hold?

Later in this paper, it has been proved that for some vertex (or cut vertex) of , and have the same domination number .

##### 1.1. Survey of Related Works

Various works have been found on interval graphs. Interval graphs are useful in modeling resource allocation problems in operations research. A. Pal and M. Pal [6] have studied about interval graphs. So many algorithms and results of various parameters on interval graphs have been found in [4, 7–12]. The *domination* is one of the parameters in graphs which has a great importance in modern circuit designing systems. Chang et al. [13] have extensively studied about domination in graphs. Also domination and its variations can be found in [14–17]. Another type of dominating set has been widely studied in [18] which is a *total dominating set*. Henning has worked on graphs with *large total domination number* in [19]. For a domination number, Sumner and Blitch [20] studied graphs where the addition of any edge changed the domination number. They called graphs with this property *domination edge critical*. Brigham et al. [5] and Fulman et al. [21] have worked on *vertex domination-critical graphs*. Wojcicka [22] have found some results on Hamiltonian properties of *domination-critical graphs*. The total domination edge critical graphs, that is, graphs where the addition of any edge decreased the total domination number were studied by Haynes et al. in [23–26]. Among the variations of domination, the -tuple domination was introduced in [3]. The case when was called double domination in [3], where exact values of the double domination numbers for some special graphs are obtained. In the same paper, various bounds of double and -tuple domination numbers are available in terms of the other parameters.

#### 2. Interval Graph and Some of Its Properties

Let , , , be a connected interval graph in which the vertices are given in the sorted order of the right endpoints of the interval representation of the graph. Intervals are labeled according to increasing order of their endpoints. This labeling is referred to as IG ordering. Let or denote the existence of an adjacency relation between two vertices , . It is assumed that is always true, that is, . If and are two end points of the vertices and , respectively, then are adjacent if at least one of the following conditions hold:(i),(ii),(iii),(iv).

The following lemma is true for a given interval graph, .

Lemma 2.1 (see [27]). *If the vertices are such that in the IG ordering and is adjacent to , then is also adjacent to .*

For each vertex , let represent the highest numbered adjacent vertex of . If no adjacent vertex of exists with higher IG number than , then is assumed to be . In other words, .

Throughout this paper, we use the notation for 2-tuple dominating set. For the purpose to find of the interval graph , we consider a function which is defined by if , otherwise, . We define the function so that for , . The weight of the function is . Also, is defined as , for all .

#### 3. Algorithm for 2-Tuple Domination

In a connected interval graph, the vertices are ordered by IG ordering. First of all, we treat none of a vertex of as a member of dominating set . Then, insert vertices one by one by testing their consistency. If a vertex is dominated by at least two vertices then leave it, otherwise, take the highest numbered adjacent vertex (vertices) from as member(s) of dominating set .

Let us associate a new term for a vertex , for all () to each adjacent vertices of in order to set IG ordering of intervals in the following way:

Basically, , the highest numbered adjacent vertex of [28]. In connection with the name of , we call this as the th* numbered adjacent vertex of * through Definition 3.1.

*Definition 3.1 (th numbered adjacent vertex). *Let *. *If for some , such that , then is called the th numbered adjacent vertex of *. *

From the definition, it is easily seen that, for a vertex , exists for maximum possible , that is, degree of the vertex . Therefore, in a graph, the maximum possible occurs in the degree of the graph, that is, . An illustration of the computations of all for the graph of Figure 1 are shown in Table 1.

Now, we describe an algorithm to find two sets of vertices and depending only on and .

Actually, the Algorithm 2DIG (Algorithm 1) gives the set which is the minimum 2-tuple dominating set and , the 2-tuple domination number of the interval graph . Before going to prove this result, we first verify Algorithm 2DIG in Figure 2. Here, we denote the set as the set of leading vertices corresponding to the 2-tuple dominating set .

In Algorithm 2DIG, at th iteration, if , then is a member of and is said to be the *leading vertex of order *2 corresponding to the vertices and of , and if , then is said to be the *leading vertex of order *1 corresponding to the vertex or of , otherwise, does not belong to .

Therefore, we conclude that if , then is adjacent to exactly two vertices of .

##### 3.1. Verification of the Algorithm

Suppose we are to find 2-tuple dominating set and 2-tuple domination number of the interval graph , where shown in Figure 1. First, set , for all . In Step 2, set , and , that is, initially and are empty. Step 2 repeats for times. Here, , number of vertices in the graph. We illustrate the iterations in the following way.

*Iteration 1. *For the first iteration , . Calculate . The first condition of *if-end if* is satisfied. Since , we find and . Then, set and . Also, set , , and .

*Iteration 2. *. . The vertex 2 is dominated by two vertices 4 and 5 of . So, in this iteration, could not be calculated. Hence, and remain the same and is being increased to 3.

*Iteration 3. *. . In this iteration, also and remain unchanged. The iteration number is being increased to 4.

*Iteration 4. *Here, and . So, and are the same as the previous iteration. Set .

*Iteration 5. *In this iteration, and , and hence no change occurs. is being increased to 6.

*Iteration 6. * and . So, domination criteria are not satisfied here. The *else-if* condition of *if-end if* is satisfied. Now, we check either or not. We see that , and hence set . Update by and by . is being increased to 7.

*Iteration 7. * and . Here, also domination criteria are not satisfied. As , set and with . is being increased to 8.

*Iteration 8. *In this iteration, that is, for , . . Hence, and remain unchanged and is being increased to 9.

*Iteration 9. *At ninth iteration, . Here, . Then, and with and . Set .

*Iteration 10. *For , . Hence, and remain unchanged. As there are 10 vertices in Figure 3, so, this is the last iteration.

So, by the Algorithm 2DIG, we get , that is, and . Therefore, cardinality of . In Figure 3, thick lines represent the members of .

#### 4. Proof of Correctness and Time Complexity

Here, we will prove that is a minimum 2-tuple dominating set.

Theorem 4.1. *The set is a minimal 2-tuple dominating set.*

*Proof. *Let be the 2-tuple dominating set obtained by Algorithm 2DIG. We are to prove that this is minimal 2-tuple dominating set, that is, there does not exist any such that is a 2-tuple dominating set.

Suppose, there exists a such that is a 2-tuple dominating set. Since , there must exist at least one member of , say , such that . Let the leading vertex corresponding to be , then . Again, since is a 2-tuple dominating set and , and with respect to the 2-tuple dominating set . Therefore, is dominated by only one vertex of , which is a contradiction of our assumption that is a 2-tuple dominating set. Thus, is minimal 2-tuple dominating set.

Theorem 4.2. *The 2-tuple domination number of the given interval graph is the cardinality of the 2-tuple dominating set , that is, . *

*Proof. *Let be the set of leading vertices corresponding to the minimal 2-tuple dominating set of . Suppose there exists another minimal 2-tuple dominating set such that .

Without loss of generality, we assume that is the leading vertex of order 2 corresponding to the two vertices and of . Then, is adjacent to exactly two vertices , () of . Also , the highest numbered adjacent vertex to . So, there does not exist any vertex in such that is adjacent to . If , then there exist at least two vertices, say, such that , where each and are adjacent to . If , then we have . So, . For , consider the following two cases.*Case 1. *Let be the leading vertex of order 1 corresponding to a vertex . Since is of order 1, either or (by Algorithm 2DIG) and is adjacent to but not adjacent to . If is adjacent to , then is adjacent to three vertices , and of (not exactly two), a contradiction. Hence, is not adjacent to the vertices and . As, or implies . Therefore, is not dominated by at least two vertices of and hence there exist at least two vertices , where each and are adjacent to . Hence, . *Case 2. *Let be the leading vertex of order 2 corresponding to the vertices . Then, is not adjacent to any vertex higher than . Also, is not adjacent to and as is not adjacent to . Therefore, if is a 2-tuple dominating set, must be dominated by at least two vertices of , say and . Hence, . Thus, there does not exist any such that , that is, is minimum and hence .

Henceforth, means the minimum 2-tuple dominating set and is the set of leading vertices corresponding to .

Theorem 4.3. *The 2-tuple dominating set of an interval graph can be computed sequentially in time. *

*Proof. *Let the processor take unit time to perform a single instruction. Step 1 of Algorithm 2DIG takes time. The algorithm consists of a loop from Step 2.1 to Step 2.3. This loop carry over times. Within this loop, we see that a loop occurs, which is terminated after times. It is clear that , is the upper bound of , for fixed . In the worst case, we assume the loop runs over times. So the total time complexity of Step 2 is . Hence, the overall time complexity of the Algorithm 2DIG is of .

#### 5. Some Important Results Related to Minimum 2-Tuple Domination

In this section, we present some important results related to minimum 2-tuple domination on interval graphs. For a given interval graph , let a tree be defined such that let be the root of . This tree is called the interval tree. The various properties of interval tree are available in [6, 10, 28].

The following lemma is true for every connected interval graph.

Lemma 5.1 (see [28]). *For a connected interval graph, there exists a unique interval tree . *

For each vertex of interval tree, is the distance of from the vertex in the tree. The height of the tree is defined by

We have found a result for the minimum 2-tuple dominating set in terms of the height of interval tree stated as follows.

Lemma 5.2. *Let be the interval tree of the interval graph with height , then
**
where is the set of natural numbers. *

*Proof. *From the definition of interval tree , we know that the vertex 1 of is at level . By the property of interval tree , we know that any vertex at level is not adjacent with a vertex at level and , for every , [8]. Therefore, it is clear that the neighbors of the vertex of level are either at level or at level .

Let such that . As the vertices at level are not adjacent with the vertices at level or at level greater than , two vertices , of must be taken from the level or . For the least possible , we assume that is at level and consequently is either at level or or . If is at level , then possibility of having is either at or at or at which decreases the level from earlier level and hence the number of vertices of may increase. So this last case is excluded from our result as the result demands the lower bound of . Also, in further cases, we neglect such cases for the same reason. Thus, we take at level , at level , at level , at level , at level , and so on. That is, at level and at level , for . So for each there are two vertices from the consecutive levels and .

Now, if , for some , then is the last level, that is, level 0 of .

So,
Thus, there are consecutive levels and hence the least value of is . If , for some , then is not at the last level of . So one vertex is required at level or . In this case, . So there are consecutive levels and hence the least value of is .

Therefore,

Here, we are going to prove a result that removal of a vertex from graph , and have the same minimum 2-tuple dominating set .

Lemma 5.3. *Let and , where and . Then *(i)*the minimum 2-tuple dominating set of and is , if is connected, *(ii)*if is disconnected with components (blocks), say, , then there must exist minimum 2-tuple dominating sets of such that and ’s are pairwise disjoint. *

*Proof. *(i) Suppose is connected. Since, , that is, . By Algorithm 2DIG, at th iteration, say, either *if* or *else-if* condition is satisfied for a vertex of , then , otherwise, . In this case, , at -th iteration, *else* condition is satisfied for the vertex which has no effect on and . Hence, if the vertex is being deleted from the graph , then the new induced subgraph has the same 2-tuple dominating set as .

(ii) Let be disconnected and , where , . Let us decompose into disjoint subsets such that , , where , that is, . As is obtained by Algorithm 2DIG and has no effect on , then has no effect on , and they are also obtained by Algorithm 2DIG. Therefore, are minimum 2-tuple dominating sets of interval graphs , respectively.

The generalized form of the Lemma 5.3 is as follows.

Corollary 5.4. *Let and , where and . Then *(i)*the minimum 2-tuple dominating set of is also , if is connected, *(ii)*if is disconnected with components (blocks), say, , then there must exist minimum 2-tuple dominating sets of such that and ’s are pairwise disjoint. *

*Proof. *(i) By Lemma 5.3, we have seen that the deletion of does not change the minimum 2-tuple dominating set . Let be a graph obtained after the deletion of , so is also the 2-tuple dominating set of . Again, is being deleted from the graph and the graph is obtained. It also has the same 2-tuple dominating set as of . Proceeding in this way, we obtain the graph which has same 2-tuple dominating set as . (ii) The proof of this case follows from (ii) of Lemma 5.3.

In Lemma 5.3 and Corollary 5.4, the graph is a subgraph of the graph induced by whose vertex set is and edge set is the set of those edges of that have both ends in . By keeping the statement of Corollary 5.4 in mind, we define new terms 2*-tuple base graph* and *redundant vertex* as follows.

*Definition 5.5 (2-tuple base graph). * Let a graph be induced subgraph of the graph , where , *. *The graph is called the 2*-*tuple base graph of the graph if the vertex set and edge set , where is the set of leading vertices corresponding to minimum 2*-*tuple dominating set of .

*Note 1. *If , then the graph is the same as and hence the graph is the 2-tuple base graph of the graph itself.

The 2-tuple base graph and its interval representation of the graph of Figure 1 are given in Figures 4 and 5, respectively. Note that, in case of 2-tuple base graph, , but in case of original graph, in general, .

*Definition 5.6 (Redundant vertex). *Let be a given interval graph. A vertex is said to be redundant in , if the minimum 2-tuple dominating set of is same as of*. *

An important conclusion is drawn about 2-tuple base graph as follows.

Lemma 5.7. *Every interval graph has a unique 2-tuple base graph.*

*Proof. *Suppose there exist two distinct 2-tuple base graphs and of the interval graph . Then, either (i) or (ii) . Let be the set of leading vertices corresponding to the minimum 2-tuple dominating set of . Since is the 2-tuple base graph of , and . Again, is the 2-tuple base graph of . Then, and . So and hence , which is a contradiction of our assumption. Therefore, and are same.

Now we define a relation between two interval graphs and it is proved that the relation is an equivalence relation.

Lemma 5.8. *Let be the set of all interval graphs. Let a relation, denoted by , and defined by and have same 2-tuple base graph, for all . Then, the relation is an equivalence relation. *

*Proof. *The relation is an equivalence relation since the following properties hold as well.*Reflexive*

Since every graph has unique 2-tuple base graph, the same graph has the same 2-tuple base graph. Therefore, . *Symmetric*

, for all . Since, means and have the same 2-tuple base graph, then and have the same 2-tuple base graph, that is, . *Transitive*

If and holds for all , then holds. Let us consider , have the same 2-tuple base graph and , have the same 2-tuple base graph . However we know every graph has unique 2-tuple base graph, cannot have distinct 2-tuple base graph and . So and have the 2-tuple base graph same as each of , . So and have the same 2-tuple base graph . So, the transitive property holds for each of .

Since all the properties of equivalence relation hold good in , then the relation defined on is an equivalence relation.

*Definition 5.9 (2-tuple equivalent). *An interval graph is said to be 2*-*tuple equivalent to an interval graph if *, *that is*, *, have the same 2-tuple base graph.

*Definition 5.10 (2-tuple equivalent class). *Let be a set of interval graphs. is said to be 2-tuple equivalent class if any interval graph of is 2-tuple equivalent to each and every graph of *. *

From the above notions, we have an important result about 2-tuple equivalent class.

Lemma 5.11. *Equivalence relation, defined on interval graphs, makes the partition of the set of interval graphs into 2-tuple equivalent classes. *

*Proof. *This result directly follows from the abstract algebra that every equivalence relation defined on a set makes the partition of the set into equivalent classes. Hence, the result follows. Particularly, the partitions can be found by the 2-tuple base graph. That is, we are trying to say that among all interval graphs, for each 2-tuple base graph, there is a 2-tuple equivalent class.

Next, we have an another important result regarding the leading vertex corresponding to 2-tuple dominating set .

Lemma 5.12. *For an interval graph ,
**
where is the number of leading vertices of order 2. *

*Proof. *Let be the number of leading vertices of order 1 and let be the number of leading vertices of order 2. By definition of leading vertex, a leading vertex of order 1 corresponds to a single vertex of and leading vertex of order 2 corresponds to two vertices of . Since there are leading vertices of order 1, then has vertices and also there are leading vertices of order 2, so has vertices. Therefore, . Now, . Thus, .

#### 6. Conclusion

In this paper, we have traced out to find the minimum 2-tuple dominating set on interval graphs. The algorithm we have designed in this paper can be generalized to find minimum -tuple dominating set and -tuple domination number. Further investigations can be done by generalizing our Algorithm 2DIG to find -tuple dominating set of an interval graph. We think it will reduce the next researcher’s labour.

#### Acknowledgment

The authors would like to thank the anonymous referees for valuable comments and also express appreciation of their constructive suggestions to improve the paper.

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