1 Introduction
A dominating set in a graph , is a set , such that any vertex not in is adjacent to a vertex in . The minimum size of a dominating set is called domination number, denoted by . The Minimum Domination problem involves computing a minimum cardinality dominating set of a graph . The domination number is one of the most studied parameter in the graph theory. A thorough treatment and detailed study on domination can be found in the books [8, 9, 10, 11]. Due to numerous applications in the real world problems, many researchers introduced several variations of domination by imposing one or more additional conditions on dominating set. One of the most important variation of domination is total domination.
In a graph , without isolated vertices, a dominating set is called a total dominating set (TDset in short), if , the graph induced by in has no isolated vertex. The total domination number, denoted by , is the cardinality of a minimum total dominating set of . The Minimum Total Domination problem requires to compute a total dominating set of a graph with no isolated vertex, of size . See [13, 21] for the detailed results on total domination.
Goddard, Henning, and McPillan, introduced a relaxed notion of total domination, called semitotal domination in [7] and further studied in [6, 12, 14, 15, 16, 17, 18, 19, 20, 23, 28, 29, 32], from both algorithmic and combinatorial point of view. In a graph with no isolated vertices, a semitotal dominating set(in short, semiTDset) is a dominating set such that for every vertex , there is another vertex , such that the distance between and is at most two in . The semitotal domination number, denoted by , is the cardinality of a minimum semiTDset of . It follows directly from definitions that every total dominating set is a semitotal dominating set. Hence, for a graph with no isolated vertices, we have the following relation between the three parameters:
Therefore, the semitotal domination number is squeezed between two important parameters, domination number and total domination number. The minimum semitotal domination problem and and its decision version are defined as follows:
Minimum Semitotal Domination problem

Instance: A graph with no isolated vertices.

Solution: A SemiTDset of .

Measure: Cardinality of the set .
Semitotal Domination Decision problem

Instance: A graph and a positive integer .

Question: Does there exist a SemiTDset in such that ?
The Semitotal Domination Decision problem in NPcomplete [7] for general graphs. The problem remains NPcomplete, even when restricted to chordal graphs, chordal bipartite graphs, and planar graphs [20]. On positive side, we have polynomialtime algorithms to compute a minimum cardinality semiTDset in trees [7], interval graphs [20, 28] and block graphs [19]. Galby et al. [6] proved that, a minimum semiTDset can be computed in polynomialtime in bounded MIMwidth graphs, which includes many important graph classes. The complexity status of the problem in some well known graph classes is shown in Fig. 1. In the figure, P stands for polynomialtime and NPC stands for NPcomplete. The complexity status of the problem in graph classes with question mark is still unknown.
Henning and Pandey studied the approximation hardness of Minimum Semitotal Domination problem [20]. They proved that the problem can not be approximated within ln for any , unless . On other side, they proved that the Minimum Semitotal Domination problem is in the class logAPX. They also proved that the problem is APXcomplete for the bipartite graphs with maximum degree .
Although, we have noticed that the semitotal domination number is squeezed between domination number and total domination number. But, Minimum Semitotal Domination problem and Minimum Total Domination problem differs in complexity, see [20]. Indeed, the Minimum Total Domination problem is polynomialtime solvable in chordal bipartite graphs but decision version of Minimum Semitotal Domination is NPcomplete for chordal bipartite graphs. Further, Galby et. al. proved that it is NPhard to decide , even when is a planar graph with degree at most , see [6].
As the MIMwidth of strongly chordal graphs is unbounded, the complexity of the problem was left open in strongly chordal graphs by Galby et al. [6]. Henning and Pandey in [20], also asked to find the complexity status of the problem in strongly chordal graphs. In this paper, we prove that the Minimum Semitotal Domination problem can be solved in lineartime in strongly chordal graphs.
The further structure of the paper is as follows. In Section 2, we discuss some notations and definitions. In Section 3, we discuss strongly chordal graphs and their properties. In Section 4, we design a lineartime algorithm to compute a semiTDset in strongly chordal graphs. Finally, Section 5, concludes the paper.
2 Preliminaries
Let be a simple graph, where and . Two distinct vertices , said to be adjacent if . For a vertex , the set denotes the open neighbourhood of in and the set denotes the closed neighbourhood of in . A path , is a sequence of distinct vertices, such that , where and . Such a path, is called, a path between and . We denote . The length of the path is . The distance between two distinct vertices , denoted by , is the length of the shortest path between and in . Further, we call , a distance two neighbour of , if .
A path with an additional condition that, is known as a cycle on vertices, denoted by . In a cycle , where , a chord is an edge joining two nonconsecutive vertices of . A graph is called chordal, if any cycle of length at least in , has a chord.
Let , and be any ordering of the vertex set . For a vertex in the ordering , we define the sets and . Further, we define, and .
For other notations and graph theoretic terminology, we refer [21]. In this paper, we consider only simple and connected graphs with at least vertices. Also, for a positive integer , we use the standard notation, .
3 Strongly Chordal Graphs
Strongly chordal graphs is an important subclass of chordal graphs introduced by several researchers in the literature [3, 5, 22]. Strongly chordal graphs includes interval graphs, block graphs, directed path graphs, and trees as subclass. Many variations of domination are polynomialtime solvable on strongly chordal graphs, see [2, 3, 4, 24, 25, 31]. There are many equivalent definitions of strongly chordal graphs. We follow, the definition given in [5].
Let be a graph. A vertex is called simple if the vertices in the closed neighbourhood of can be ordered, where , such that for . A graph is strongly chordal if every induced subgraph of has a simple vertex. An ordering of vertices of is called strong elimination ordering(SEO) if implies for .
Many algorithms are studied to recognise a strongly chordal graph, . In [1, 22], the authors designed an time algorithm to recognise a strongly chordal graph. In [26], a time algorithm is given which later, improved to time algorithm in [27]. Spinrad in [30], gave an time algorithm to recognise a strongly chordal graph. The same algorithm also computes, a strong elimination ordering, if the graph is strongly chordal. The graph in Fig. 2, is a strongly chordal graph with strong elimination ordering .
Given a strongly chordal graph and a strong elimination ordering of the vertex set , we have the following observation.
Observation 3.1.
If such that , then . Further, if then .
Let be a strongly chordal graph and be its SEO. For a vertex , denotes the highest index neighbour of according to SEO, where . In particular . Our algorithm is an iterative algorithm which process the vertices as they appear in SEO. We use the following labels on the vertices during the execution of the algorithm to construct a minimum semiTDset of .
Further, represents the set of neighbours of such that and one of the neighbour of is already dominated. Formally, and there is a vertex such that . We note that, if then . Also by Observation 3.1, . Hence, for a vertex , we have . In our algorithm, we also use two special types of operation on a particular vertex , MARK and UNMARK which are defined as follows: if then the operation MARK updates for all such that . While in operation UNMARK we update for all . Before designing the algorithm, we first prove the following results.
Lemma 3.1.
For a vertex , let such that then .
Proof : Since , using the property of SEO. Now, consider a vertex such that and . Let be a shortest path between and . An illustration of possible positions of is given in Fig. 3. As is the highest index neighbour of , we have . Indeed, , as if then using property of SEO we have , a contradiction. Hence, we have , implying that . Consequently, we have . Hence, the result follows. ∎
Lemma 3.2.
If such that then .
Proof : Let in the SEO. Clearly, and using property of SEO, . Let be an arbitrary vertex. If , then as . Now suppose, and be a shortest path between and . Clearly . Since , we have . Hence, . Therefore, the lemma follows. ∎
Lemma 3.3.
If such that and then .
Proof : Using property of SEO, we note that . Therefore, for any vertex , we have . Now consider a vertex such that and . Let be a shortest path between and . If , then using the property of SEO, we have , a contradiction. Hence, . Now, if then using property of SEO, , implying that . Further, if then . This implies that and hence, . Consequently, we have . Therefore the lemma follows. ∎
Lemma 3.4.
If for a vertex , and then .
Proof : The proof directly follows from the property of strong elimination ordering. For completeness, suppose . We note that . Hence, if , Lemma follows. Now, assume that where . Let be a shortest path of length two in . As , if then using property of SEO, we have , a contradiction. Hence, . Further, using fact that, , we have . Consequently, we have . ∎
Lemma 3.5.
If for a vertex , we have then .
Proof : On contrary, suppose but . Now since the graph is connected there exists a path joining the vertices and . Suppose, is a shortest such path. Note that the path must contain a vertex such that and a vertex such that . Let be the highest index vertex and be the least index in the path such that . Since, is a shortest path, we have where . An illustration is given in Fig. 4. As , therefore, using the property of SEO, we have , a contradiction on choice of . Hence, the result follows. ∎
Lemma 3.6.
Let be a connected strongly chordal graph and be an SEO of . For a vertex , let . Suppose there is a path from to a vertex such that . If has a neighbour such that then . Specifically, .
Proof : The proof directly follows using property of strong elimination ordering. Hence, omitted. ∎
4 Algorithm for Semitotal Domination in Strongly Chordal Graphs
In this section, we propose, a lineartime algorithm to compute a minimum semiTDset in strongly chordal graphs. But, before designing the algorithm first, we discuss the idea of the algorithm.
Outline of the Algorithm
Let be a strongly chordal graph and be a strong elimination ordering of the vertex set of . In our algorithm, we process the vertices iteratively as they appear in and in each iteration we maintain a set containing the selected vertices. In the iteration we will process the vertex . We ensure that the vertices having index at most are dominated by at least one vertex of . Now in iteration, we update the set in the following way:

If the vertex is not dominated, and is not the last vertex, then we look for a vertex such that, and there exists a vertex in which is at distance at most two from .

If such a vertex exists, we include it in . We maintain the information that all the neighbours of are dominated now. Also, as there is already a vertex in which is at distance at most from , we maintain the information that a distance two neighbour of is already selected. In the same iteration, we also check if there is a vertex , such that distance two neighbour of is not selected till iteration. If such a vertex exists, and , then for all such vertices , we update the information that a distance two neighbour of is selected, by updating and UNMARKoperation.

Otherwise, we include in to dominate and updated the information that all the neighbours of are dominated now. Further, using MARK operation, we also maintain the information that we need to select a distance two neighbour of , in one of the further iterations.


If the vertex is not dominated, and is the last vertex, that is, , then we include in the . We update the status of as dominated. Also, we may note that, all neighbours of are already dominated. Hence, there is a vertex in which is at distance at most from . Therefore, we updated .

If is already dominated and , then we do not need to update .

If is already dominated and , that is, is selected in and a distance two neighbour of is still need to be selected then we updated as follows:

If is less than the index of , we simply update .

If and then as is the last vertex, we include any neighbour of in and update .

If and then we include the highest index neighbour, of in . We update the information that all the neighbour of are dominated now. We also update the information that a distance two neighbour of is selected by updating and UNMARK operation. For all vertices, , such that distance two neighbour of is not selected till iteration and , we update that a distance two neighbour of is selected, by updating and UNMARKoperation.

Algorithm
Now, we design a lineartime algorithm, Algorithm 1, to compute a minimum semiTDset in a strongly chordal graph, given its strong elimination ordering.
Now to show the correctness of Algorithm 1 and to prove that given a strongly chordal graph and its strong elimination ordering , it computes a minimum semiTDset of , we prove the following lemmas (Lemma 4.1 to Lemma 4.6).
Lemma 4.1.
At the beginning of the iteration, the following statements are true:

for all .

for all .

If for , then there exists a neighbour of such that and .
Proof : The proof directly follows from Algorithm 1. ∎
For each , let be the set of selected vertices after the iteration of the algorithm. Indeed, is the set of vertices selected by the algorithm after all the vertices are processed. In particular, suppose . In order to prove the correctness of the algorithm first we will show that is a semiTDset of and then we will show that is contained in a minimum semiTDset of for each .
Lemma 4.2.
is a semiTDset of strongly chordal graph .
Proof : In the iteration of the algorithm, we process the vertex . First we check whether the vertex is dominated or not using the vertices selected till iteration. Indeed, in the iteration, represents that the vertex is not dominated by the current set and hence, we pick a neighbour of to dominate . As we visit every vertex in some iteration, after the iteration, all vertices will be dominated by set . Further in any iteration, if we pick any vertex such that no vertex is already selected such as then we perform MARK operation. Let . According to the algorithm, we select a vertex such in some iteration where and update . Hence, after the iteration we note that or for all . Hence is a semiTDset of . ∎
Using Lemma 4.2, we note that the set is a semiTDset of the strongly chordal graph . Now we claim that for each , there is a minimum semiTDset of containing . We will prove this using induction on . If , is contained in any minimum semiTDset of . Now we assume that there is minimum semiTDset of containing . In the iteration, depending upon the several cases of the algorithm we prove the following lemmas to show that there is a minimum semiTDset of containing .
Lemma 4.3.
If , , and then there exists a minimum semiTDset of such that .
Proof : Using induction hypothesis, we have a minimum semiTDset of containing . Clearly, if then is the required minimum semiTDset. Hence, we assume that . Let . Since , . Let and be the minimum index vertices in dominating and respectively. We note that . Now we prove the result by considering the following two cases:
Case 1: If
In this case is the least index vertex in dominating both and . As and , therefore . Indeed as if then , a contradiction. Further, if then using Lemma 4.1, we have . Since and with hence, , a contradiction as . Therefore . As and , using the property of SEO, we have and hence using Lemma 4.1, we note that is a dominating set of such that . Now in order to prove that is a semiTDset, we need to prove the following claim.
Claim 4.1.
For all , there exists a vertex such that and there exists a vertex .
Proof : From previous arguments we note that . Also . Let be an arbitrary vertex. If , that is, there is a vertex other than in such that , then the result follows. Hence, we assume that .
If then we note that as , and the result follows. Assume that . Let be any shortest path between and in . If then using the property SEO, we have , implying that , hence the result follows.
Now suppose, . We have . Therefore, if , then using property of SEO, , contradicting the assumption that . Hence, we have . Further, using Lemma 3.4, we note that if has a neighbour such that , then . Hence, the result follows. Suppose, has no neighbour such that . In this case, we show that either the vertex already has a distance two neighbour in or we can remove to get a semiTDset of smaller cardinality.
Here, first we show that there exists a vertex such that and . We have and since , . We note that, as if then , a contradiction. Hence, we have a vertex such that that is, can not be dominated by . Hence, there exists a vertex such that . Therefore, we have a vertex such that .
Now, suppose . Using Lemma 4.1 and the fact that has no neighbour such that , the set is a semiTDset of such that , a contradiction as is a minimum semiTDset of . If and then by Lemma 4.1, there is a neighbour of having index greater than which is marked for , a contradiction as has no neighbour having index greater than . If and , then there exists a vertex such that . Also we have a vertex such that and hence, the claim follows. ∎
Consequently, in this case the result follows. Now we consider the other case.
Case 2: If
Since we have , . If , then using property of SEO, we note that . Also if , then implying that . Hence, the set is a dominating set of . Also as is dominated by in and hence, for we have such that . Consequently, to prove is a semiTDset of we need to prove the following claim.
Claim 4.2.
For any vertex , there exists a vertex such that .
Proof : Let be an arbitrary vertex. Note that if there exists a vertex then the result follows. Hence we assume that . If , then we can give similar arguments as we gave in Claim 4.1, to show that there exists a vertex such that
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