Total Domination Versus Domination in Cubic Graphs

A dominating set in a graph G is a set S of vertices of G such that every vertex not in S has a neighbor in S. Further, if every vertex ofG has a neighbor in S, then S is a total dominating set of G. The domination number, γ (G), and total domination number, γt (G), are theminimum cardinalities of a dominating set and total dominating set, respectively, in G. The upper domination number, (G), and the upper total domination number, t (G), are the maximum cardinalities of a minimal dominating set and total dominating set, respectively, in G. It is known that γt (G)/γ (G) ≤ 2 and t (G)/ (G) ≤ 2 for all graphsGwith no isolated vertex. In this paperwe characterize the connected cubic graphs G satisfying γt (G)/γ (G) = 2, and we characterize the connected cubic graphs G satisfying t (G)/ (G) = 2. Research supported in part by the South African National Research Foundation and the University of Johannesburg. B Joanna Cyman joanna.cyman@pg.edu.pl Magda Dettlaff magda.dettlaff1@pg.edu.pl Michael A. Henning mahenning@uj.ac.za Magdalena Lemańska magleman@pg.edu.pl Joanna Raczek joanna.raczek@pg.edu.pl 1 Department of Mathematics, Gdańsk University of Technology, ul. Narutowicza 11/12, 80-952 Gdańsk, Poland 2 Department of Pure and Applied Mathematics, University of Johannesburg, Auckland Park 2006, South Africa


Introduction
In this paper we continue the study of domination and total domination in graphs. Let G = (V, E) be a simple graph. If u, v ∈ V (G), then u dominates v if u and v are adjacent. A dominating set of a graph G is a set S of vertices of G such that every vertex outside S is dominated by at least one vertex in S. A total dominating set, abbreviated TD-set, of G is a set S of vertices of G such that every vertex in V (G) is adjacent to at least one vertex in S. A neighbor of a vertex v in G is a vertex adjacent to v. A vertex totally dominates its neighbors. The domination number of G, denoted γ (G), is the minimum cardinality of a dominating set of G, while the total domination number of G, denoted by γ t (G), is the minimum cardinality of a TD-set of G. A TD-set S in G is minimal if for all vertices v ∈ S, the set S\{v} is not a TD-set of G. The upper total domination number of G, denoted t (G), is the maximum cardinality of a minimal TD-set in G. We refer to a minimum total dominating set of G as a γ t (G)-set. A minimal TD-set of cardinality t (G) we call a t (G)-set.
The literature on this subject of domination and its variations has been surveyed and detailed in the two books by Haynes, Hedetniemi, and Slater [9,10]. For a recent book on total domination in graphs we refer the reader to [15]. A survey of total domination in graphs can also be found in [14]. For a survey of known results on the upper total domination we refer the reader to [16]. Recent papers on total domination in graphs can be found, for example, in [3,6,17,18] and elsewhere.
Let The following property of minimal total dominating sets is established in [5].
A packing in a graph G is a set of vertices whose closed neighborhoods are pairwise disjoint. Thus, if S is a packing in G, then N G [u] ∩ N G [v] = ∅ for all pairs u, v ∈ S; equivalently, the vertices in S are pairwise at distance at least 3 apart in G. A perfect packing in G is a packing whose closed neighborhoods partition V (G).

Domination Versus Total Domination
We shall need the following well known lower bound on the domination number of a graph (see, for example, [9]). For completeness, we provide the elementary proof of this result.
Lemma 2.1 [9] If G is a graph of order n and = (G), then γ (G) ≥ n/( + 1), with equality if and only if every minimum dominating set in G is a perfect packing that contains only vertices of degree .
Proof If D is an arbitrary minimum dominating set of G, then and so γ (G) = |D| ≥ n/( + 1). Further, if γ (G) = n/( + 1), then we must have equality throughout the above inequality chain, implying that d G (v) = for all v ∈ D and that the closed neighborhoods, N G [v], of the vertices v ∈ D are pairwise disjoint. Thus, D is a perfect packing in G that contains only vertices of degree . Conversely, if D is a perfect packing in G that contains only vertices of degree , then every dominating set D of G contains at least one vertex from every closed neighborhood, , for all u ∈ D, and so γ (G) ≥ |D|. However, the set D is a dominating set of G, and so γ (G) ≤ |D|. Consequently, γ (G) = |D| = n/( + 1).
The following relationship between the domination and total domination numbers of a graph with no isolated vertex was first observed by Bollobás and Cockayne [2].
A constructive characterization of trees T satisfying γ t (T ) = 2γ (T ) is given in [12]. However, it remains an open problem to characterize general graphs G satisfying γ t (G) = 2γ (G). We shall need the following lemma.
Proof Let G be a k-regular graph of order n satisfying γ t (G) = 2γ (G). Let D be a minimum dominating set of G, and so |D| = γ (G). For each vertex v ∈ D, let v be an arbitrary neighbor of v, and let Then, X is a TD-set of G, and so γ t (G) ≤ |X |. If v and w are distinct vertices of D, and v = w or v = w , then γ t (G) ≤ |X | < 2|D| = 2γ (G), a contradiction. Therefore, for every pair v and w of vertices in D, the vertices v and w are not adjacent and do not have a neighbor in common; that is, We wish to determine the connected, k-regular graphs achieving equality in the upper bound of Theorem 2.2 for small values of k. If k = 1, then G = K 2 and γ t (G) = 2 and γ (G) = 1, and so γ t (G)/γ (G) = 2. Hence, it is only of interest to consider the cases when k ≥ 2.

2-Regular Graphs
We first consider the case when G is a k-regular graph with k = 2. In this case G ∼ = C n and n ≥ 3. It is well known (see, for example, [9,11,15]) that γ t (C n ) = n/2 + n/4 − n/4 and γ (C n ) = n/3 . As an immediate consequence of these results, we note that Thus, we have the following observations. Observation 2.4 If G is a connected 2-regular graph of order n ≥ 3, then if n ≡ 2 (mod 12)

3-Regular Graphs
We next consider the case when k ≥ 3. For this purpose, we shall need a result on the total domination number of a cubic graph. For ≥ 1, let G be the graph constructed in [8] as follows. (This construction is also described in Chapter 5, p. 44, of [15].) Consider two copies of the path P 2 with respective vertex sequences a 1 b 1 a 2 b 2 · · · a b and c 1 d 1 c 2 d 2 · · · c d . For each i ∈ [ ], join a i to d i and b i to c i . To complete the construction of the graph G join a 1 to c 1 and b to d . Let G = {G | ≥ 1}. For ≥ 2, let H be obtained from G by deleting the two edges a 1 c 1 and b d and adding the two edges a 1 b and c 1 d . Let H = {H | ≥ 2}. We note that G and H are cubic graphs of order 4 . Further, we note that G 1 ∼ = K 4 . The graphs G 4 ∈ G and H 4 ∈ H, for example, are illustrated in Fig. 1.
Let H even be the subfamily of H consisting of all graphs H where is even; that is, H even = {H | ≥ 2 is even}.
The following upper bound on the total domination number of a graph with minimum degree at least 3 was established independently by many authors (see, [1,4,19]). Theorem 2.6 [1,4,19] If G is a graph of order n with δ(G) ≥ 3, then γ t (G) ≤ n 2 . The connected graphs that achieve equality in the bound of Theorem 2.6 were characterized in [13].
Theorem 2.7 [13] If G is a connected graph of order n with δ(G) ≥ 3, then γ t (G) ≤ n 2 , with equality if and only if G ∈ G ∪ H or G is the generalized Petersen graph shown in Fig. 2. We are now in a position to characterize the connected, cubic graphs G satisfying

Theorem 2.8 If G is a connected cubic graph, then γ t (G)/γ (G) ≤ 2 with equality if and only if the following holds.
Proof Let G be a connected, cubic graph of order n. By Theorem 2.2, γ t (G)/γ (G) ≤ 2. Hence it suffices for us to prove that γ t (G)/γ (G) = 2 if and only if (a) or (b) in the statement of the theorem hold. We first prove the sufficiency. Suppose that (a) or (b) in the statement of the theorem holds. If G is the generalized Petersen graph, then γ (G) = 4 and γ t (G) = 8, and so In both cases, D is a dominating set of G, implying that γ (G) This proves the sufficiency.
Next, we prove the necessity. Suppose that G is a connected, cubic graph satisfying It remains for us to show that if G ∈ H, then G ∈ H even . Suppose, to the contrary, Let D be a minimum dominating set in G. Since n = 4 and γ (G) = n/4, we note that |D| = γ (G) = and, by Lemma 2.1, that the set D is a packing in G. In order to dominate the vertex a 1 , the set D contains exactly one of the vertices a 1 , b 1 , b , d 1 .
By symmetry, renaming vertices if necessary, we may assume that either a 1 ∈ D or b 1 ∈ D.
Suppose that b 1 ∈ D. Since D is a packing, this implies that In order to dominate the vertices b and d , we note that either a ∈ D or c ∈ D; that is, D ∩ {a , c } = ∅. In order to dominate the vertex d 1 , we note that c 2 ∈ D. Since D is a packing, this in turn implies that D ∩ V 2 = {c 2 } and D ∩{a 3 , c 3 } = ∅. If = 3, we contradict our earlier observation that D ∩{a , c } = ∅. Therefore, ≥ 5. In order to dominate the vertices a 3 and c 3 , we note that either b 3 In both cases, since D is a packing, we deduce that D ∩ {a 5 , c 5 } = ∅. Continuing in this way, it therefore holds that In particular, noting that is odd, D ∩ {a , c } = ∅. This contradicts our earlier . In order to dominate the vertices a 2 and c 2 , we note that either In order to dominate the vertices a 4 and c 4 , we note that either b 4 In particular, noting that is odd, D ∩ {a , c } = ∅. This contradicts our earlier observation that D ∩ {a , c } = ∅.
Since both cases a 1 ∈ D and b 1 ∈ D produce a contradiction, we deduce that if G ∈ H, then G ∈ H even . This proves the necessity.

Upper Domination Versus Upper Total Domination
Dorbec et al. [7] established the following relationship between the upper domination and upper total domination numbers of a graph with no isolated vertex.
As remarked in [7], it remains an open problem to characterize graphs G satisfying t (G) = 2 (G). We wish to determine the connected, k-regular graphs G achieving equality in the upper bound of Theorem 3.1 for small values of k. If k = 1, then G = K 2 and (G) = 1 and t (G) = 2. Hence, it is only of interest to consider the cases when k ≥ 2.

2-Regular Graphs
We first consider the case when k = 2. The following result determines the upper total domination number of a cycle.
Proof We proceed by induction on n ≥ 3. The result is straightforward to check (or use a computer) for small n ≤ 9. Hence we may assume that n ≥ 10. Let G ∼ = C n be the cycle v 1 v 2 · · · v n v 1 . We show firstly that t (C n ) ≥ 2 n 3 + 1 if n ≡ 5 (mod 6) and t (C n ) ≥ 2 n 3 , otherwise. If n ≡ 5 (mod 6), then the set In both cases, the set B is a minimal TD-set of G and |B| = 2 n 3 , and so t (G) ≥ 2 n 3 . We show next that t (C n ) ≤ 2 n 3 + 1 if n ≡ 5 (mod 6) and t (C n ) ≤ 2 n 3 , otherwise. Let S be a t (G)-set. By the minimality of S, no five consecutive vertices of G all belong to S.
Suppose that S contains four consecutive vertices of G. Renaming vertices if necessary, we may assume that {v 3 , v 4 , v 5 , v 6 } ⊆ S. Since no five consecutive vertices all belong to S, we note that v 2 / ∈ S and v 7 / ∈ S. If v 1 ∈ S, then S\{v 3 } is a TD-set of G, contradicting the minimality of S. Hence, v 1 / ∈ S. If v 8 ∈ S, then S\{v 6 } is a TD-set of G, contradicting the minimality of S. Hence, v 8 / ∈ S. This implies that {v 9 , v 10 } ⊂ S. We now consider the cycle G obtained from G by deleting the six vertices v 3 , v 4 , · · · , v 8 and adding the edge v 2 v 9 . Thus, G ∼ = C n where n = n − 6. Let S = S\{v 3 , v 4 , v 5 , v 6 }. Since S is a minimal TD-set of G, the set S is a minimal TD-set of G , implying that |S | ≤ t (G ). If n ≡ 5 (mod 6), then n ≥ 11 and n ≡ 5 (mod 6), and by the inductive hypothesis, Hence, if S contains four consecutive vertices of G, then t (C n ) ≤ 2 n 3 + 1 if n ≡ 5 (mod 6) and t (C n ) ≤ 2 n 3 , otherwise. Hence, we may assume that S contains no four consecutive vertices, for otherwise the desired result follows.
Suppose that S contains three consecutive vertices of G. Renaming vertices if necessary, we may assume that {v 3 , v 4 , v 5 } ⊆ S. Since no four consecutive vertices all belong to S, we note that v 2 / ∈ S and v 6 / ∈ S. By the minimality of S, v 1 / ∈ S and v 7 / ∈ S. This implies that {v n , v n−1 } ⊂ S and {v 8 , v 9 } ⊂ S. We now consider the cycle G obtained from G by deleting the five vertices v 3 , v 4 , . . . , v 7 and adding the edge v 2 v 8 . Thus, G ∼ = C n where n = n − 5. Let S = S\{v 3 , v 4 , v 5 }. Since S is a minimal TD-set of G, the set S is a minimal TD-set of G , implying that t (G) = |S| = |S | + 3 ≤ t (G ) + 3.
If n ≡ 5 (mod 6), then n ≡ 0 (mod 6), and by the inductive hypothesis, t (G) ≤ t (G ) + 3 = 2 n 3 + 3 = 2 n−5 3 + 3 = 2 n 3 + 1. If n ≡ 4 (mod 6), then n ≡ 5 (mod 6), and by the inductive hypothesis, t (G) ≤ t (G ) Suppose that n ≡ 2 (mod 6). We show that t (G) ≤ 2 n 3 . Suppose that v 10 ∈ S. In this case, v 11 / ∈ S and v 12 / ∈ S. This implies that {v 13 , v 14 } ⊂ S. We now consider the cycle G obtained from G by deleting the ten vertices v 3 , v 4 , . . . , v 12 and adding the edge v 2 v 13 . Thus, G ∼ = C n where n = n − 10. We note that n ≡ 4 (mod 6). 9 , v 10 }. Since S is a minimal TD-set of G, the set S is a minimal TD-set of G , implying by the inductive hypothesis that t (G) = |S| = |S | + 6 ≤ t (G ) + 6 ≤ 2 n 3 + 6 = 2 n−10 3 + 6 ≤ 2 n 3 . Hence, we may assume that v 10 / ∈ S, for otherwise t (G) ≤ 2 n 3 as desired. If v 11 / ∈ S, then v 12 ∈ S and v 13 ∈ S. In this case, we consider the cycle G * obtained from G by deleting the four vertices v 6 , v 7 , v 8 , v 9 and adding the edge v 5 v 10 . Thus, G * ∼ = C n * where n * = n − 4. We note that n * ≡ 4 (mod 6). Let S * = S\{v 8 , v 9 }. Since S is a minimal TD-set of G, the set S * is a minimal TD-set of G * , implying by the inductive hypothesis that t (G) = |S| = |S * | + 2 ≤ t (G * ) + 2 ≤ 2 n * 3 + 2 = 2 n−4 3 + 2 ≤ 2 n 3 . Hence, we may assume that v 11 ∈ S, for otherwise the desired result follows, implying that v 12 ∈ S. Using analogous and symmetrical arguments, we may assume that v n−2 / ∈ S and {v n−3 , v n−4 } ⊂ S, for otherwise the desired upper bound follows. If n = 14, then v n−2 = v 12 , contradicting the fact that with our current assumptions v 12 ∈ S and v n−2 / ∈ S. Hence, we note that n ≥ 20. The set Hence, if S contains three consecutive vertices, then we have shown that t (C n ) ≤ 2 n 3 + 1 if n ≡ 5 (mod 6) and t (C n ) ≤ 2 n 3 , otherwise. Hence, we may assume that S contains no three consecutive vertices, for otherwise the desired result follows. With this assumption, t (C n ) ≤ 2 n 3 for all n ≥ 10. This completes the proof of the proposition.
For n ≥ 3, we observe that (C n ) = α(C n ) = n/2 , where α(G) is the size of a maximum independent vertex set of G. We state this result formally as follows.

Observation 3.3 For n ≥ 3, (C n ) = n/2 .
As an immediate consequence of Proposition 3.2 and Observation 3.3, we note that if G ∼ = C n and n ≥ 3, then This yields the following results.

Corollary 3.4 If G is a connected 2-regular graph of order n ≥ 3, then
.

Corollary 3.5
If G is a connected 2-regular graph, then t (G) (G) ≤ 2, with equality if and only if G ∼ = C 3 .

3-Regular Graphs
We next consider the case when k = 3. In order to determine the connected 3-regular (cubic) graphs G satisfying t (G)/ (G) = 2, we analyse the proof of the upper bound of Theorem 3.1 to obtain the structure of these extremal graphs. For this purpose, we shall need the following key lemma in [7]. For completeness, we provide the proof of this lemma as given in [7].

Lemma 3.6 [7] Every t (G)-set contains as a subset a minimal dominating set S such that
Proof Let D be an arbitrary t (G)-set, and let Then, D = A ∪ B ∪ C. As shown in [7], if C = ∅, then G[C] = |C| 2 K 2 and each vertex of C has degree 1 in G[D]. Following the notation introduced in [7], we call two adjacent vertices in G[C] partners in C and we let (X, Y ) be the partite sets in the graph G [C]. Thus, each vertex in X (resp., in Y ) is adjacent in G[D] only to its partner in Y (resp., in X ). For each x ∈ X , let y x be the partner of x in C. As shown in [7], for each vertex v ∈ B, we have pn(v, (1) Following the notation introduced in the proof of Theorem 3.1 given in [7], we let As observed in [7], the set S = A ∪ B Y ∪ X is a dominating set of G, although not necessarily minimal. For each vertex x ∈ X we systematically delete x from S if epn(x, S) = ∅ at each stage in the resulting set S, and let X * be the resulting subset of vertices of X that belong to the set S upon the completion of this process. Thus, S = A ∪ B Y ∪ X * and, by construction, |epn(v, S)| ≥ 1 for each v ∈ S. As observed in [7], the set S is a minimal dominating set of G, and so (G) ≥ |S|. For every vertex x ∈ X , the set S contains at least one of x and its partner in C, noting that if the partner y x ∈ Y of x in C is not in S, then y x ∈ epn(x, S), and so x is not deleted from S. Thus, |S ∩ C| ≥ |X | = 1 2 |C|. As shown earlier, |A| ≥ |B|. Hence, or equivalently, t (G) ≤ 2 (G). As observed earlier, S ⊂ D and S is a minimal dominating set of G. Further, |S| ≥ 1 2 t (G) and |epn(v, S)| ≥ 1 for each v ∈ S.
We note that Theorem 3.1 follows as an immediate consequence of Lemma 3.6. We proceed further with the following lemma which gives structural properties of graphs G that achieve equality in the bound of Theorem 3.1. We remark that the following properties of graphs achieving equality in the bound of Theorem 3.1 is implicitly implied in [7], albeit without proof. We provide here a complete proof since we will need these properties in order to characterize the cubic graphs achieving equality in the bound of Theorem 3.1.

(a) (G) = α(G). (b) Every t (G)-set induces a subgraph that consists of disjoint copies of K 2 .
(c) Every vertex that does not belong to an arbitrary t (G)-set is contained in a common triangle with two vertices of the set.
Proof Suppose that G is a graph with no isolated vertex satisfying (G) = 1 2 t (G). Let D be an arbitrary t (G)-set. Following exactly the notation introduced in the proof of Lemma 3.6, we must have equality throughout inequality chain (2) in this proof. As observed in the proof of Lemma 3.6, for every vertex x ∈ X , the set S contains at least one of x and its partner in C, and so |S ∩ C| ≥ |C|/2. However, |S ∩ C| = |C|/2, implying that the set S contains exactly one of x and its partner y x in C. Renaming vertices of C if necessary, we may assume that S ∩ C = X . Since A is an independent set and there is no edge between A and C, the set S = A ∪ X is an independent set, and so α(G) ≥ |S| = |A| + |C|/2. Since every maximum independent set in a graph is a minimal dominating set in the graph, we note that |S| = (G) ≥ α(G) = |S|. Consequently, the set S = A ∪ X is a maximum independent set in G, and so (G) = α(G) = |S|. This proves Part (a).
Suppose that A = ∅ and consider an arbitrary vertex u ∈ A. Let v be the neighbor of u in G [D], and so v ∈ B and pn(v, D) = {u}. Further, let w ∈ epn(u, S). The vertex w exists since S ⊆ D and u ∈ A implies that u has an external private neighbor with respect to D. Let |C| = 2c and recall that G[C] = cK 2 and each vertex of C has degree 1 in G [D]. In particular, we note that v is adjacent to no vertex of A ∪ C different from u. Thus, the set (S\{u}) ∪ {v, w} is an independent set in G, implying that α(G) ≥ |S| + 1, a contradiction. Hence, A = ∅, which in turn implies that B = ∅. To prove Part (c), let v be an arbitrary vertex in V (G)\D. As observed above, D = C and G[C] = cK 2 . We show that v is contained in a common triangle with two vertices of C. Suppose, to the contrary, that v does not belong to a common triangle with two vertices of C. If v is adjacent to both x and its partner y x for some vertex x ∈ C, then vx y x v is a common triangle containing v and two vertices of C, a contradiction. Hence, v is adjacent to at most one of x and its partner y x for every vertex x ∈ C. Let C be a subset of C constructed as follows. Initially, let C = ∅. For each pair x and y x where x ∈ C (and x y x is an edge), select one vertex that is not adjacent to v and add it to the set C . Thus, |C | = c and C is an independent set. By construction, no vertex of C is adjacent to v. Hence, C ∪ {v} is an independent set on G, and so α(G) ≥ |C | + 1 = c + 1 > α(G), a contradiction. Therefore, the vertex v is adjacent to both x and its partner y x in C for at least one vertex x ∈ C.
We are now in a position to characterize the connected cubic graphs achieving equality in Theorem 3.1.
. In particular, we note that D = D ≥1 . Suppose that u 1 and u 2 are not adjacent. Thus, c ≥ 2. If for every i ∈ [c]\{1}, there exists a vertex in {x i , y i } that is adjacent to neither u 1 nor u 2 , then there exists a subset D ≥2 of D ≥2 such that |D ≥2 | = c − 1 and D ≥2 ∪ {u 1 , u 2 } is an independent set, implying that α(G) ≥ |D ≥2 | + 2 = c + 1, a contradiction. Hence for some i ∈ [c]\{1}, the vertex u 1 is adjacent to one of x i and y i , say y i , and the vertex u 2 is adjacent to x i . Renaming the sets V 2 , . . . , V c if necessary, we may assume that i = 2. Thus, u 1 is adjacent to y 2 and u 2 is adjacent to x 2 . Let u 3 be the neighbor of x 2 different from u 2 and y 2 . We note that u 3 ∈ V i for some i ∈ [c]\{1}. Suppose that u 3 ∈ V 2 . Thus, the vertex u 3 is adjacent to both x 2 and y 2 , and is adjacent to at most one vertex from D ≥3 . Hence there exists a subset D ≥3 of D ≥3 such that |D ≥3 | = c − 2 and D ≥3 ∪ {u 1 , u 2 , u 3 } is an independent set, implying that α(G) ≥ |D ≥3 | + 3 = c + 1, a contradiction. Therefore, u 3 / ∈ V 2 , and so u 3 ∈ V i for some i ∈ [c]\{1, 2}. Renaming the sets V 3 , . . . , V c if necessary, we may assume that u 3 ∈ V 3 . Thus, u 3 is adjacent to both x 3 and y 3 .
Let u 4 be the neighbor of x 3 different from u 3 and y 3 . We note that u 4 ∈ V i for some i ∈ [c]\{1, 2}. Suppose that u 4 ∈ V 3 . Thus, the vertex u 4 is adjacent to both x 3 and y 3 , and is adjacent to at most one vertex from D ≥4 . Hence there exists a subset D ≥4 of D ≥4 such that |D ≥4 | = c − 3 and D ≥4 ∪ {u 1 , u 2 , u 3 , u 4 } is an independent set, implying that α(G) ≥ |D ≥4 | + 4 = c + 1, a contradiction. Therefore, u 4 / ∈ V 3 , and so u 4 ∈ V i for some i ∈ [c]\{1, 2, 3}. Renaming the sets V 4 , . . . , V c if necessary, we may assume that u 4 ∈ V 4 . Thus, u 4 is adjacent to both x 4 and y 4 . The graph shown in Fig. 3 is therefore a subgraph of G, where the darkened vertices belong to the set D.

Closing Questions and Problems
We pose the following two questions that we have yet to settle. We have shown that Question 2 is true when k = 2 (see Corollary 3.5) and when k = 3 (see Theorem 3.8). However, we have yet to answer Question 2 when k ≥ 4.
In summary, our focus in this paper is to characterize the connected cubic graphs G satisfying γ t (G)/γ (G) = 2, and to characterize the connected cubic graphs G satisfying t (G)/ (G) = 2. These characterizations are presented in Theorem 2.8 and Theorem 3.8, respectively. It would be interesting to extend these characterizations to connected subcubic graphs. We remark that there are many families of subcubic graphs G that are not cubic graphs satisfying γ t (G)/γ (G) = 2. For example, the 2-corona of a path or a cycle is such a family of subcubic graphs, where we recall that the 2-corona H • P 2 of a graph H is the graph of order 3|V (H )| obtained from H by attaching a path of length 2 to each vertex of H so that the resulting paths are vertex-disjoint. We close with the following two open problems.
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