COMPLEXITY ISSUES OF PERFECT SECURE DOMINATION IN GRAPHS

Let G = (V, E) be a simple, undirected and connected graph. A dominating set S is called a secure dominating set if for each u ∈ V \ S, there exists v ∈ S such that (u, v) ∈ E and (S \ {v}) ∪ {u} is a dominating set of G. If further the vertex v ∈ S is unique, then S is called a perfect secure dominating set (PSDS). The perfect secure domination number γps(G) is the minimum cardinality of a perfect secure dominating set of G. Given a graph G and a positive integer k, the perfect secure domination (PSDOM) problem is to check whether G has a PSDS of size at most k. In this paper, we prove that PSDOM problem is NP-complete for split graphs, star convex bipartite graphs, comb convex bipartite graphs, planar graphs and dually chordal graphs. We propose a linear time algorithm to solve the PSDOM problem in caterpillar trees and also show that this problem is linear time solvable for bounded tree-width graphs and threshold graphs, a subclass of split graphs. Finally, we show that the domination and perfect secure domination problems are not equivalent in computational complexity aspects. Mathematics Subject Classification. 05C69, 68Q25. Received March 9, 2020. Accepted October 8, 2021.

for each w ∈ N G [v]. A vertex ordering v 1 , v 2 , . . . , v n is a maximum neighbourhood ordering if for each i < n, v i has a maximum neighbour in {v i , v i+1 , . . . , v n } . A graph is dually chordal if it has a maximum neighbourhood ordering. A bipartite graph G = (X, Y, E) is called tree convex if there exists a tree T = (X, F ) such that, for each y in Y , the neighbours of y induce a subtree in T . When T is a star (comb), G is called star (comb) convex bipartite graph [10]. For undefined terminology and notations refer to [18].
A vertex v in G dominates the vertices of its closed neighborhood. A set of vertices S ⊆ V is a dominating set (DS) of G if for every vertex u ∈ V \ S, there exists at least one vertex v ∈ S such that (u, v) ∈ E, i.e., N G [S] = V . A vertex u ∈ V \ S is said to be undominated if N G (u) ∩ S = ∅. The concept of perfect domination (PD) has been introduced by P.M. Weichsel et al. in [17]. A dominating set S of G is called a perfect dominating set (PDS) of G if every vertex in V \ S is adjacent to exactly one vertex in S.
The concept of secure domination has been introduced by E.J. Cockayne et al. in [2] and is studied, for example, in [2,12,13]. A set of vertices S ⊆ V is a secure dominating set (SDS) of G, if S is a dominating set of G and for each u ∈ V \ S, there exists v ∈ S such that (u, v) ∈ E and (S \ {v}) ∪ {u} is a dominating set of G. In this case, we say that u is S-defended by v or v S-defends u.
The concept of perfect secure domination has been introduced by Rashmi et al. in [16]. A set S ⊆ V is said to be a perfect secure dominating set (PSDS) of G, if S is a dominating set of G and for every vertex u ∈ V \ S, there exists a unique vertex v ∈ S such that (u, v) ∈ E and (S \ {v}) ∪ {u} is a dominating set in G. In this case, we say that u is perfectly secure dominated by v or v perfectly secure dominates u. In other words S is not a PSDS of G if any vertex in V \ S is S-defended by more than one vertex in S. The minimum cardinality of a DS, PDS, SDS and PSDS respectively are called domination number γ(G), perfect domination number γ p (G), secure domination number γ s (G) and perfect secure domination number γ ps (G) of G. The MINIMUM PERFECT SECURE DOMINATING SET (MPSDS) problem is to find a perfect secure dominating set of minimum cardinality. The computational complexity of perfect secure domination problem has not been studied in the liturature.
The following results are proved in [16].
if r = s = 2 r + s, otherwise The following observation regarding perfect secure dominating set of a graph will be used throughout this paper.
Proof. Let S be a minimum PSDS of G such that which does not contain a support vertex. Let u be a pendant vertex and v be a support vertex in graph G such that (u, v) ∈ E. Clearly, for the domination property has to be satisfied either u ∈ S or v ∈ S. If v / ∈ S then u ∈ S i.e., v is S-defended by only u. From all such pairs, we can define a minimum PSDS S such that S = (S \ {u}) ∪ {v}.

Complexity results
In this section, we show that the perfect secure domination problem remains NP-complete even when restricted to split, star convex bipartite, comb convex bipartite, planar and dually chordal graphs.

Perfect secure domination in split, star convex and comb convex bipartite graphs
To show that the PSDOM for split, star convex bipartite and comb convex bipartite graphs is NP-complete, we use a well known NP-complete problem, called Exact Cover by 3-Sets (X3C) [6], which is defined as follows. Exact Cover by 3 Sets (X3C) INSTANCE : A finite set X with |X| = 3k and a collection C of 3-element subsets of X. QUESTION : Is there a subcollection C of C such that every element of X appears in exactly one member of C ?
A variant of X3C in which each element appears in at least two subsets has also been proved as NP-complete [7]. Through out this subsection, we use this variant of X3C problem.
Theorem 2.1. PSDOM is NP-complete for split graphs.
Proof. Given a split graph G, a positive integer p and an arbitrary set S ⊆ V , we can check in polynomial time whether S is a PSDS of G of size at most p. Hence PSDOM is in NP.
Let X = {x 1 , x 2 , . . ., x 3k } and C = {c 1 , c 2 , . . ., c t } be an arbitrary instance of X3C, where |X| = 3k, |C| = t and t ≥ 2k. We now construct an instance of PSDOM for split graph from the given instance of X3C as follows. Construct a graph G(V, E) by creating vertices x i for each x i ∈ X, c i and c i for each c i ∈ C. Add edges It is clear by construction that the graph shown in Figure 1 is a split graph since B forms an independent set and A forms a clique. Clearly |V | = 2t + 3k and |E| = 4t + t 2 . Next we show that, X3C has a solution if and only if G has PSDS of size at most k + t.
Suppose C is a solution for X3C with |C | = k.
From the solution of X3C, it follows that S dominates all x i 's and each c i / ∈ S is dominated by c i . Hence, S is a dominating set and it is a SDS since each x i ∈ X is S-defended by c j where (x i , c j ) ∈ E and (c j , c j ) ∈ S, each c i / ∈ S is S-defended by c i . It is also PSDS since each element in V \ S is S-defended by exactly one element of S.
Conversely, suppose that S is a PSDS of G such that |S| ≤ k + t. From Observation 1.4, we can assume that S ∩ C = C. We have the following lemma.   number of c's to S-defend them. But these c's can not S-defend their corresponding c 's. Hence that many c 's must also be part of S then |S| > t + k, a contradiction. Therefore |S ∩ X| = 0.
Since split graphs form a proper subclass of chordal graphs, the following corollary is immediate from Theorem 2.1.
Proof. Clearly, PSDOM is a member of NP. Let X = {x 1 , x 2 , . . ., x 3k } and C = {c 1 , c 2 , . . . , c t } be an arbitrary instance of X3C, where |X| = 3k, |C| = t and t ≥ 2k. We now construct an instance of PSDOM for star convex bipartite graph from the given instance of X3C as follows. Construct a graph G(V, E) by creating vertices x i for each x i ∈ X, c i for each c i ∈ C and also create vertices a, b, c, d and e. Add edges (x i , a) for each x i ∈ X, (a, b), (a, c), (a, d) and (d, e). Next add edges (c i , c), The set A induces an independent set, but can be associated with a star with vertex a as central vertex and the neighbors of each element in B induce a subtree of star. Therefore G is a star convex bipartite graph and can be constructed from the given instance of X3C in polynomial time. The graph constructed and its associated star is shown in Figure 2. Next we show that, X3C has a solution if and only if G has PSDS of size at most k + 3.
Conversely, suppose that S is a PSDS of G such that |S| ≤ k + 3 and we have the following lemmas.  Proof. The proof is obtained with the similar arguments as in Lemma 2.5. From Observation 1.4, we can assume that a, d ∈ S. It is clear that c ∈ S since d can not S-defend the elements of C.
Proof. The proof is same as in Lemma 2.2.
Proof. Clearly, PSDOM is a member of NP. Let X = {x 1 , x 2 , . . ., x 3k } and C = {c 1 , c 2 , . . ., c t } be an arbitrary instance of X3C, where |X| = 3k, |C| = t and t ≥ 2k. We now construct an instance of PSDOM for comb convex bipartite graph from the given instance of X3C as follows. Construct a graph G(V, E) by creating vertices x i for each x i ∈ X, c i , c i , c i and c i for each c i ∈ C. Create four more vertices c a , c a , c a and c b if t is odd. Add edges (c i , c i ), (c i , c i ) for each c i ∈ C and (c j , x i ) if x i ∈ c j . Next add edges by joining each c j to every x i and also add (c i , c j ), (c i , c j ), where 1 ≤ i < t, i mod 2 = 1 and j = i + 1. If t is odd then add edges (c a , c t ), (c a , c a ), (c a , c b ), (c a , c t ), (c a , c a ) and (c a , c b ). Assume The set A induces a comb with elements {c i : 1 ≤ i ≤ t} ∪ {c a } as backbone and {c i : 1 ≤ i ≤ t} ∪ {c a } as teeth and the neighbors of each element in B induce a subtree of the comb. Therefore G is a comb convex bipartite graph and can be constructed from the given instance of X3C in polynomial time. The graph constructed and its associated comb is shown in Figure 3. Next we show that, X3C has a solution if and only if G has PSDS of size at most 2t The set S is also PSDS since each element in V \ S is S-defended by exactly one element of S.
Conversely, suppose that S is a PSDS of G such that |S| ≤ 2t + k + 2(t − 2 t 2 ). From Observation 1.4, each c i ∈ S. These make at least t vertices. Let S i = {c i , c i−1 , c i−1 , c i } where 1 ≤ i ≤ t and i mod 2 = 0. Clearly, each S i is a cycle graph C 4 . If S contains more than two vertices from a C 4 then each x i is S-defended by more than one vertex of the C 4 . By Theorem 1.2, S contains atleast two vertices from each C 4 . If t is odd then by contradiction it can be easily verified that |S ∩ {c t , c t , c a , c a , c a , c b }| = 3. The total are at least 2t + 2(t − 2 t 2 ) vertices and the following lemmas holds. Lemma 2.9. |S ∩ X| = 0.
Proof. The proof is same as in Lemma 2.2.
The following result is immediate from Theorems 2.4 and 2.8.

Perfect secure domination in planar graphs
To show that the PSDOM is NP-complete for planar graphs, we use a well known NP-complete problem, called Planar Exact Cover by 3-Sets (Planar X3C) [14], which is defined as follows. Planar Exact Cover by 3 Sets (Planar X3C) INSTANCE : A finite set X with |X| = 3k and a collection C of 3-element subsets of X such that (i) every element of X occurs in at most three subsets and such that (ii) the induced graph is planar. (This induced graph H is formed by creating vertices x i for each x i ∈ X, c i for each c i ∈ C and by adding edges (c j , x i ) if x i ∈ c j .) QUESTION : Is there a subcollection C of C such that every element of X appears in exactly one member of C ? Theorem 2.11. PSDOM is NP-complete for planar graphs.
Proof. The proof is same as in Theorem 2.1, in which the vertices of C does not form a clique.

Perfect secure domination in dually chordal graphs
To prove the NP-completeness of the PSDOM for dually chordal graphs, we consider the following perfect domination decision problem which has been proved as NP-complete [3]. Proof. Clearly, PSDOM is a member of NP. We now construct an instance of PSDOM for dually chordal graph from the given instance of PDOM as follows. Given an instance Since G admits a maximum neighbourhood ordering {v 1 , v 2 , . . . , v n , x 1 , x 2 , x 3 }, it is a dually chordal graph and the construction of G can be accomplished in polynomial time.
Next we show that G has a perfect dominating set of size at most k if and only if G has a PSDS of size k + 1. Suppose D be a perfect dominating set in G of size at most k. Then the set D ∪ {x 1 } is a PSDS of G of size at most k + 1 since x 1 cannot S-defend the vertices of V (G).
On the other hand, let S be a PSDS of G with |S| ≤ k + 1. We have the following lemma.
is not a PSDS of G since x 1 would be S-defended by more than one vertex. Therefore, our assumption leads to contradiction. Hence the lemma.
Therefore, S ∩ V is a perfect dominating set of G of size at most k.

Caterpillar tree
A caterpillar is a tree with the property that the removal of its pendant vertices and incident edges results in a path, which we call the central path. In this section we propose a linear algorithm to compute perfect secure domination number of a caterpillar graph. Let P be the set of pendant vertices and Q be the set of support vertices. Let N p (u) = {v : (u, v) ∈ T and v ∈ P } be the pendant neighbors of a support vertex u. Also let N p [u] = N p (u) ∩ {u}. Our algorithm is given in Algorithm 1.  Let S = ∅.

5:
for each vertex u ∈ Q do 6:  Correctness of the Algorithm 1 : It is easy to verify that, the vertices selected in steps 5 to 7 perfectly secure dominate all the vertices in u∈Q N p [u] and the vertices selected in steps 9 to 11 perfectly secure dominate the remaining vertices of T i.e., V (T ) \ ∪ u∈Q N p [u]. Hence the set S returned in step 12 of Algorithm 1 is a PSDS of T . From Lemma 3.1, it also follows that |S| = γ ps (T ) i.e., S is a minimum PSDS of T .

Threshold graphs
Threshold graphs have been studied with the following definition [11]. Although several characterizations defined for threshold graphs, we use the following characterization of threshold graphs given in [11] in solving the PSDOM problem.
A graph G is a threshold graph if and only if it is a split graph and, for split partition (C, I) of V where C is a clique and I is an independent set, there is an ordering {c 1 , c 2 , . . . , c n } of vertices of C such that , and there is an ordering {i 1 , i 2 , . . . , i m } of the vertices of I such that Theorem 4.2. Let G(V, E) be a threshold graph with split partition (C, I). Then Proof. If G ∼ = K 1 then γ ps (G) = 1. Otherwise, let S be a minimum PSDS of G. If |C| = 1 i.e., the graph G is in the form K 1,p then, by Observation 1.4 and Theorem 1.3, S contains p vertices of G. The following lemmas are used to complete the proof when |C| > 1.  Our assumption leads to contradiction. Hence |S| ≥ n + m. Therefore |S| = n + m.
If the threshold graph G is disconnected with k connected components G 1 , G 2 , . . . , G k then it is easy to verify that γ ps (G) = k i=1 γ ps (G i ). Now, the following result is immediate from Theorem 4.2. Theorem 4.5. PSDOM problem can be solvable in linear time for threshold graphs.
Proof. Since the split partition can be obtained in linear time [11], the result follows.

Bounded tree-width graphs
Let G be a graph, T be a tree and v be a family of vertex sets V t ⊆ V (G) indexed by the vertices t of T . The pair (T, v ) is called a tree-decomposition of G if it satisfies the following three conditions: (i) V (G) = t∈V (T ) V t , (ii) for every edge e ∈ E(G) there exists a t ∈ V (T ) such that both ends of e lie in V t , (iii) V t1 ∩ V t3 ⊆ V t2 whenever t 1 , t 2 , t 3 ∈ V (T ) and t 2 is on the path in T from t 1 to t 3 . The width of (T, v ) is the number max{|V t | − 1 : t ∈ T }, and the tree-width tw(G) of G is the minimum width of any tree-decomposition of G. By Courcelle's Thoerem, it is well known that every graph problem that can be described by counting monadic second-order logic (CMSOL) can be solved in linear-time in graphs of bounded tree-width, given a tree decomposition as input [4]. We show that PSDOM problem can be expressed in CMSOL. Proof. First, we present the CMSOL formula which expresses that the graph G has a dominating set S of size at most k.
where adj(p, q) is the binary adjacency relation which holds if and only if, p, q are two adjacent vertices of G. Dominating(S) ensures that for every vertex p ∈ V , either p ∈ S or p is adjacent to a vertex in S and the cardinality of S is at most k. Now, by using the above CMSOL formula we can express PSDOM in CMSOL formula as follows.
The above statement states that if an element x / ∈ S is adjacent to two or more elements of S then it should be S-defended by exactly any one adjacent element of S.
Therefore, PSDOM can be expressed in CMSOL.
Now, the following result is immediate from Theorems 5.1 and 5.2.
Theorem 5.3. PSDOM problem can be solvable in linear time for bounded tree-width graphs.

Complexity difference in domination and perfect secure domination
Although perfect secure domination is one of the several variants of domination problem, these two differ in computational complexity. In particular, there exist graph classes for which the decision version of the first problem is polynomial-time solvable whereas the decision version of the second problem is NP-complete and vice versa. Similar study has been made between domination and other domination parameters in [8,15].
The DOMINATION problem is polynomial-time solvable for dually chordal graphs [1,5], but the PSDOM problem is NP-complete for this class of graphs which is proved in Section 2.3. Now, we construct a new class of graphs in which the MINIMUM PERFECT SECURE DOMINATION problem can be solved trivially, whereas the decision version of the DOMINATION problem is NP-complete, which is defined as follows.

Add edges
General GP graph construction is shown in Figure 4.
It can be observed that S is a PSDS of G of size 2n and hence γ ps (G ) ≤ 2n.
Let S be any PSDS of G . Clearly, |S ∩ {b i , c i : 1 ≤ i ≤ n}| ≥ n and these vertices can not S-defend any other vertex in V . Therefore, either v i or a i , for each i, where 1 ≤ i ≤ n must be included in every PSDS of G . Hence γ ps (G ) ≥ 2n. This completes the proof of the theorem. Proof. Suppose D be dominating set of G of size at most k, then it is clear that D ∪ {b i : 1 ≤ i ≤ n} is a dominating set of G of size at most n + k.
Conversely, suppose D is a dominating set of G of size at most n + k. Then at least one vertex from each of the vertices b i and c i must be included in D . These are at least n vertices. Let A * = {v i : a i ∈ D }. Clearly, V ∩ (A * ∪ D ) is a dominating set of G of size at most k. Hence the lemma.
The following result is well known for the DOMINATION DECISION problem. From Theorem 6.4 and Lemma 6.3, it follows that DOMINATION DECISION problem is NP-hard. Hence the following theorem.
Theorem 6.5. The DOMINATION DECISION problem is NP-complete for GP graphs.

Conclusion
In this paper, we have shown that PSDOM problem is NP-complete for split graphs, star convex bipartite graphs, comb convex bipartite graphs, planar graphs and dually chordal graphs. We also proved that this problem is linear time solvable for threshold graphs, caterpillar trees and bounded tree-width graphs. Further, it is interesting to investigate the algorithmatic complexity of PSDOM for other subclasses of chordal graphs namely, strongly chordal graphs, doubly chordal graphs, block graphs, etc. We remark, however, that the two problems, domination and perfect secure domination are not equivalent in computational complexity aspects. A good example is when the input graph is a GP graph, the domination problem is known to be NP-complete whereas the PSDOM problem is trivially solvable. Thus, there is a scope to study each of these problems on its own for particular graph classes.