Correction: Cabrera Martínez et al. On the Secure Total Domination Number of Graphs. Symmetry 2019, 11, 1165

Abel Cabrera Martínez; Luis P. Montejano; Juan A. Rodríguez-Velázquez

doi:10.3390/sym13091668

,

and

Departament d’Enginyeria Informàtica i Matemàtiques, Universitat Rovira i Virgili, Av. Països Catalans 26, 43007 Tarragona, Spain

^*

Author to whom correspondence should be addressed.

Symmetry2021, 13(9), 1668;https://doi.org/10.3390/sym13091668

This article belongs to the Special Issue Theoretical Computer Science and Discrete Mathematics

Version Notes

Order Reprints

The authors wish to make the following corrections on paper [1]:

(1): Eliminate Lemma 1 because we have found that this lemma is not correct.
(2): Theorem 3 states that for any graph G with no isolated vertex,

$γ_{s t} (G) \leq α (G) + γ (G) .$

The result is correct, but the proof uses Lemma 1. For this reason, we propose the following alternative proof for Theorem 3.

Proof.

Let D be a

γ (G)

-set. Let I be an

α (G)

-set such that

| D \cap I |

is at its maximum among all

α (G)

-sets. Notice that for any

x \in D \cap I

,

e p n (x, D \cup I) \cup i p n (x, D \cup I) \subseteq e p n (x, I) .

(1)

We next define a set

S \subseteq V (G)

of minimum cardinality among the sets satisfying the following properties.

(a)

D \cup I \subseteq S

.

(b)

For every vertex

x \in D \cap I

,

(b1): if $e p n (x, D \cup I) \neq ⌀$ , then $S \cap e p n (x, D \cup I) \neq ⌀$ ;
(b2): if $e p n (x, D \cup I) = ⌀$ , $i p n (x, D \cup I) \neq ⌀$ and $e p n (x, I) \ i p n (x, D \cup I) \neq ⌀$ , then either $e p n (x, I) \ D = ⌀$ or $S \cap e p n (x, I) \ D \neq ⌀$ ;
(b3): if $e p n (x, D \cup I) = ⌀$ and $e p n (x, I) = i p n (x, D \cup I) \neq ⌀$ , then $S \cap N (e p n (x, I)) \ {x} \neq ⌀$ ;
(b4): if $e p n (x, D \cup I) = i p n (x, D \cup I) = ⌀$ , then $N (x) \ (D \cup I) = ⌀$ or $S \cap N (x) \ (D \cup I) \neq ⌀$ .

Since D and I are dominating sets, from (a) and (b) we conclude that S is a TDS. From now on, let

v \in V (G) \ S

. Observe that there exists a vertex

u \in N (v) \cap I \subseteq N (v) \cap S

, as

I \subseteq S

is an

α (G)

-set. To conclude that S is a STDS, we only need to prove that

S^{'} = (S \ {u}) \cup {v}

is a TDS of G.

First, notice that every vertex in

V (G) \ N (u)

is dominated by some vertex in

S^{'}

, because S is a TDS of G. Let

w \in N (u)

. Now, we differentiate two cases with respect to vertex u.

Case 1. $u \in I \ D$ . If $w \notin D$ , then there exists some vertex in $D \subseteq S^{'}$ which dominates w, as D is a dominating set. Suppose that $w \in D$ . If $w \in i p n (u, D \cup I)$ , then $I^{'} = (I \cup {w}) \ {u}$ is an $α (G)$ -set such that $| D \cap I^{'} | > | D \cap I |$ , which is a contradiction. Hence, $w \notin i p n (u, D \cup I)$ , which implies that there exists some vertex in $(D \cup I) \ {u} \subseteq S^{'}$ which dominates w.
Case 2. $u \in I \cap D$ . We first suppose that $w \notin D$ . If $w \notin e p n (u, D \cup I)$ , then w is dominated by some vertex in $(D \cup I) \ {u} \subseteq S^{'}$ . If $w \in e p n (u, D \cup I)$ , then by $(b 1)$ and the fact that in this case all vertices in $e p n (u, D \cup I)$ form a clique, w is dominated by some vertex in $S \ {u} \subseteq S^{'}$ . From now on, suppose that $w \in D$ . If $w \notin i p n (u, D \cup I)$ , then there exists some vertex in $(D \cup I) \ {u} \subseteq S^{'}$ which dominates w. Finally, we consider the case in that $w \in i p n (u, D \cup I)$ .

We claim that

i p n (u, D \cup I) = {w}

. In order to prove this claim, suppose that there exists

w^{'} \in i p n (u, D \cup I) \ {w}

. Notice that

w^{'} \in D

. By (1) and the fact that all vertices in

e p n (u, I)

form a clique, we prove that

w w^{'} \in E (G)

, and so

w \notin i p n (u, D \cup I)

, which is a contradiction. Therefore,

i p n (u, D \cup I) = {w}

and, as a result,

e p n (u, D \cup I) \cup {w} \subseteq e p n (u, I) .

(2)

In order to conclude the proof, we consider the following subcases.

Subcase 2.1.

e p n (u, D \cup I) \neq ⌀

. By (2),

(b 1)

, and the fact that all vertices in

e p n (u, I)

form a clique, we conclude that w is adjacent to some vertex in

S \ {u} \subseteq S^{'}

, as desired.

Subcase 2.2.

e p n (u, D \cup I) = ⌀

and

e p n (u, I) \ {w} \neq ⌀

. By (2),

(b 2)

, and the fact that all vertices in

e p n (u, I)

form a clique, we show that w is dominated by some vertex in

S \ {u} \subseteq S^{'}

, as desired.

Subcase 2.3.

e p n (u, D \cup I) = ⌀

and

e p n (u, I) = {w}

. In this case, by

(b 3)

we deduce that w is dominated by some vertex in

S \ {u} \subseteq S^{'}

, as desired.

According to the two cases above, we can conclude that

S^{'}

is a TDS of G, and so S is a STDS of G. Now, by the the minimality of

| S |

, we show that

| S | \leq | D \cup I | + | D \cap I | = | D | + | I |

. Therefore,

γ_{s t} (G) \leq | S | \leq | I | + | D | = α (G) + γ (G)

, which completes the proof. □

The authors would like to apologize for any inconvenience caused to the readers by these changes. The changes do not affect the scientific results.

Reference

Cabrera Martínez, A.; Montejano, L.P.; Rodríguez-Velázquez, J.A. On the secure total domination number of graphs. Symmetry 2019, 11, 1165. [Google Scholar] [CrossRef] [Green Version]

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Correction: Cabrera Martínez et al. On the Secure Total Domination Number of Graphs. Symmetry 2019, 11, 1165

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