Essential upper bounds on the total domination number

Discrete Applied Mathematics 244 (2018) 103–115

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Essential upper bounds on the total domination number Michael A. Henning Department of Pure and Applied Mathematics, University of Johannesburg, Auckland Park, 2006, South Africa

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Article history: Received 7 January 2017 Received in revised form 15 January 2018 Accepted 7 March 2018 Available online 24 March 2018 Keywords: Total domination Upper bounds

a b s t r a c t A set S of vertices in a graph G is a total dominating set of G if every vertex has a neighbor in S. The total domination number, γt (G), is the minimum cardinality of a total dominating set of G. Let nG and mG denote the number of vertices and edges, respectively, in G. We prove that all the essential upper bounds on the total domination number of a graph G without isolated vertices and isolated edges can be written in the unified form γt (G) ≤ (2anG + 2bmG )/(3a + 2b) for constants b ≥ 0 and a ≥ 23 (1 − b). We also provide unified forms of upper bounds for the total domination number of a graph with minimum degree δ in the cases when δ ∈ {2, 3, 4}. For example, we show that the bound γt (G) ≤ anG + bmG is valid for every graph G with minimum degree δ = 3 if and only if both b ≥ 0 and a ≥ 12 (1 − 3b) hold. © 2018 Elsevier B.V. All rights reserved.

1. Introduction Our aim in this paper is to investigate upper bounds on the total domination number of a graph which does not contain isolated vertices and isolated edges in terms of its order and/or size. We show that all the essential upper bounds on the total domination number of such a graph can be written in a unified form. For notation and graph theory terminology we generally follow [15]. The order of a graph G is denoted by nG = |V (G)| and its size by mG = |E(G)|. We denote the degree of a vertex v in G by dG (v ), or simply d(v ) if the graph G is clear from context. The minimum degree among the vertices of G is denoted by δ (G). For a set S ⊆ V (G), the subgraph induced by S is denoted by G[S ]. Further, the subgraph of G obtained from G by deleting all vertices in S and all edges incident with vertices in S is denoted by G − S. If S = {v}, we simply denote G − {v} by G − v . The distance between two vertices u and v in a connected graph G is the minimum length of a (u, v )-path in G. The diameter, diam(G), of G is the maximum distance among pairs of vertices in G. A dominating set of a graph G is a set S of vertices of G such that every vertex not in S has a neighbor in S, where two vertices are neighbors if they are adjacent. The domination number of G, denoted by γ (G), is the minimum cardinality of a dominating set. The notion of domination and its variations in graphs and has been studied a great deal; a rough estimate says that it occurs in more than 3000 papers to date. A total dominating set of a graph G with no isolated vertex is a set S of vertices such that every vertex in G has a neighbor in S. The total domination number, γt (G), of G is the minimum cardinality of a total dominating set of G. 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 [9]. 2. Known results and motivation A discussion on complexity and algorithmic results on total domination in graphs can be found in [10]. The decision problem to determine the total domination number of a graph is NP-complete (see, for example, [7,17]). Therefore, E-mail address: [email protected]. https://doi.org/10.1016/j.dam.2018.03.008 0166-218X/© 2018 Elsevier B.V. All rights reserved.

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M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

Fig. 1. The 2-corona of a cycle C4 .

Fig. 2. A graph G in the family F .

much interest has focused on determining upper bounds on the total domination number of a graph (see, for example, [1,4–6,8,11–14,16,18–21]). A detailed discussion of such bounds can be found in [11]. We mention here four such results that we will need when proving our main result. 2.1. Minimum degree one The 2-corona of a graph G, denoted G ◦ P2 , is the graph of order 3nG obtained from G by attaching a path of length 2 to each vertex of G so that the resulting paths are vertex–disjoint. The 2-corona of a cycle C4 is illustrated in Fig. 1. Theorem 1 ([2,5]). If G is a connected graph of order n ≥ 3, then γt (G) ≤ some connected graph F .

2 n 3

with equality if and only if G is C3 , C6 or F ◦ P2 for

2 n 3

with equality if and only if every component of G

2.2. Minimum degree two The following result is an immediate consequence of Theorem 1. Theorem 2 ([2,5]). If G is a graph of order n with δ (G) ≥ 2, then γt (G) ≤ is a 3-cycle or a 6-cycle.

In what follows, we adopt the notation from [8]. Let F be the family of all graphs that can be obtained from a connected graph F of order at least 2 as follows: For each vertex v of F , add a 6-cycle and join v either to one vertex of this cycle or to two vertices at distance 2 on this cycle. A graph G in the family F is illustrated in Fig. 2 (here the graph F is a 4-cycle). Theorem 3 ([8]). If G is a connected graph of order n ≥ 11 with δ (G) ≥ 2, then γt (G) ≤ or G is obtained from a 14-cycle by adding certain chords or G ∈ F .

4 n 7

with equality if and only if G = C14

2.3. Minimum degree three In what follows, we adopt the notation from [6]. For k ≥ 1, let Gk be the graph constructed as follows. Consider two copies of the path P2k with respective vertex sequences a1 b1 a2 b2 . . . ak bk and c1 d1 c2 d2 . . . ck dk . Let A = {a1 , a2 , . . . , ak }, B = {b1 , b2 , . . . , bk }, C = {c1 , c2 , . . . , ck }, and D = {d1 , d2 , . . . , dk }. For each i ∈ {1, 2, . . . , k}, join ai to di and bi to ci . To complete the construction of the graph Gk join a1 to c1 and bk to dk . Let Gcubic = {Gk | k ≥ 1}. For k ≥ 2, let Hk be obtained from Gk by deleting the two edges a1 c1 and bk dk and adding the two edges a1 bk and c1 dk . Let Hcubic = {Hk | k ≥ 2}. We note that Gk and Hk are cubic graphs of order 4k. Further, we note that G1 = K4 . The graphs G4 ∈ Gcubic and H4 ∈ Hcubic , for example, are illustrated in Fig. 3(a) and 3(b), respectively. Let GP16 denote the generalized Petersen graph of order 16 shown in Fig. 3(c). Theorem 4 ([1,4,13,21]). If G is a connected graph of order n with δ (G) ≥ 3, then γt (G) ≤ G = GP16 or G ∈ Gcubic ∪ Hcubic .

1 n 2

with equality if and only if

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

(a) G4 .

105

(b) H4 .

(c) GP16 .

Fig. 3. Cubic graphs G4 ∈ Gcubic , H4 ∈ Hcubic and GP16 .

(a) H14 .

¯ 14 . (b) H

Fig. 4. The Heawood graph H14 and its bipartite complement H 14 .

2.4. Minimum degree four The bipartite complement of a bipartite graph G with partite sets X and Y is the bipartite graph with the same vertex set and partite sets as G and where there is an edge between x ∈ X and y ∈ Y in the bipartite complement of G if and only if there is no edge between x ∈ X and y ∈ Y in G. The Heawood graph H14 and its bipartite complement H 14 (or, equivalently, the incidence bipartite graph of the complement of the Fano plane) are shown in Fig. 4(a) and 4(b), respectively. Theorem 5 ([11,20]). If G is a connected graph of order n with δ (G) ≥ 4, then γt (G) ≤ bipartite complement H 14 of the Heawood graph.

3 n 7

with equality if and only if G is a

3. Main result In this paper we prove that all the essential upper bounds on the total domination number of a graph without isolated vertices and isolated edges can be written in a unified form. Thereafter we extend our result to graphs with minimum degree δ in the cases when δ ∈ {2, 3, 4}. More precisely, we shall prove the following unified forms of upper bounds for the total domination number of a graph. Throughout the remaining part of this paper we let nG and mG denote the number of vertices and edges in a graph G, respectively. Theorem 6. For a, b ∈ R, the following holds. (a) The bound γt (G) ≤ anG + bmG is valid for every graph G with δ (G) ≥ 1 and no isolated edge if and only if both b ≥ 0 and a ≥ 23 (1 − b) hold. (b) The bound γt (G) ≤ anG + bmG is valid for every graph G with δ (G) ≥ 2 if and only if both b ≥ 0 and a ≥ 32 − b hold. (c) The bound γt (G) ≤ anG + bmG is valid for every connected graph G with δ (G) ≥ 2 and nG ≥ 11 if and only if both b ≥ 0 and a ≥ 74 − b hold. (d) The bound γt (G) ≤ anG + bmG is valid for every graph G with δ (G) ≥ 3 if and only if both b ≥ 0 and a ≥ 21 (1 − 3b) hold. (e) The bound γt (G) ≤ anG + bmG is valid for every graph G with δ (G) ≥ 4 if and only if both b ≥ 0 and a ≥

3 7

− 2b hold.

We proceed as follows. In Section 3.1, we mention some consequences of our main result, namely Theorem 6. A proof of Theorem 6 is given in Section 4. In Section 5, we characterize the graphs G for which the equality γt (G) = anG + bmG holds in the bounds in Theorem 6. We conclude our discussion in Section 6 with some conjectures that we have yet to settle.

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3.1. Consequences of Theorem 6 We mention here some consequences of Theorem 6. In the special case of Theorem 6(a) when (a, b) = ( 23 , 0), we have the classic upper bound on the total domination number of a graph due to Cockayne, Dawes, and Hedetniemi [5] given in Theorem 1. Thus, Theorem 6(a) gives a new proof of the Cockayne–Dawes–Hedetniemi bound. In the special case of Theorem 6(b) when (a, b) = ( 13 , 31 ), we have that γt (G) ≤ (n + m)/3 for every graph G with minimum degree at least two, a result proven independently by several authors (see [4] for a more general result). In the special case of Theorem 6(c) when (a, b) = ( 47 , 0), we have the upper bound on the total domination number of a graph with minimum degree at least two due to Henning [8] given in Theorem 3. In the special case of Theorem 6(d) when (a, b) = ( 21 , 0), we have the upper bound on the total domination number of a graph with minimum degree at least three due to Archdeacon et al. [1], Chvátal and McDiarmid [4], and Tuza [21] given in Theorem 4. In the special case of Theorem 6(e) when (a, b) = ( 37 , 0), we have the important upper bound on the total domination number of a graph with minimum degree at least four due to Thomassé and Yeo [20] given in Theorem 5. 4. Proof of Theorem 6 In this section, we present a proof of Theorem 6. The ideas in this paper are heavily influenced by a result due to Bujtás, Henning, and Tuza [3] on the domination number of a uniform hypergraph (we do not define these concepts here) with minimum degree at least one. We remark that our proofs of Theorem 6(a) and 6(b) given in Sections 4.1 and 4.2, respectively, are independent of any known results. Our proofs of Theorems 6(c), 6(d) and 6(e) given in Sections 4.3–4.5, respectively, build on known results in [1,4,8,20,21] and extend them to a more general and unified form. 4.1. Proof of Theorem 6(a) In this section, we present a proof of Theorem 6(a). Throughout this section, we let G be the class of all graphs without isolated vertices and isolated edges. We begin with some preliminary lemmas. The first result establishes basic properties of graphs that belong to the family G . Lemma 7. If G ∈ G , then nG ≤

3 m 2 G

and γt (G) ≤ mG . Further, nG =

3 m 2 G

if and only if every component of G is a path P3 .

Proof. We proceed by induction on nG ≥ 3. If nG = 3, then either G ∼ = P3 , in which case mG = 2, 23 mG = nG and γt (G) = mG , 3 or G ∼ = C3 , in which case mG = 3, 2 mG > nG and γt (G) = mG . This establishes the base case. Let G ∈ G with nG ≥ 4, and assume that the result holds for all graphs F ∈ G with nF < nG . If G is disconnected, then we apply the inductive hypothesis to each component of G to yield the desired result, noting that order and size are additive with respect to vertex–disjoint union. Hence, we may assume that G is connected. Let T be an arbitrary spanning tree of G, and let v be an arbitrary leaf of T . We now consider the graph F = G − v . Since T − v is a spanning tree of F , the graph F is connected. Further, nF = nG − 1 ≥ 3 and mF ≤ mG − 1. In particular, F ∈ G and nF < nG . Applying the inductive hypothesis to the graph F , we note that nG = nF + 1 ≤ 23 mF + 1 ≤ 32 (mG − 1) + 1 < 32 mG . Further, if u is an arbitrary neighbor of v in G, then every total dominating set in F can be extended to a total dominating set of G by adding to it the vertex u if necessary, implying that γt (G) ≤ γt (F ) + 1 ≤ mF + 1 ≤ mG . □ Lemma 8. Let a, b ∈ R. Then, anG + bmG > 0 holds for every G ∈ G if and only if b ≥ 0 and a > − 32 b. Proof. Suppose that anG + bmG > 0 holds for every G ∈ G and for each pair of reals a and b. If b < 0, then there exists a sufficiently large ℓ (simply take ℓ > (b − 2a)/b) such that the complete graph on ℓ vertices gives negative anG + bmG , a contradiction. On the other hand, for all pairs a, b where a ≤ − 32 b, the graph G ∼ = P3 with nG = 3 and mG = 2 shows that anG + bmG is not always positive on G , a contradiction. Hence, both conditions b ≥ 0 and a > − 23 b are necessary. To prove the opposite direction, suppose that a and b are reals satisfying b ≥ 0 and a > − 23 b. We note that 32 a + b > 0. Let G be an arbitrary graph in G . If a > 0, the inequality anG + bmG > 0 trivially holds. Hence we may assume that a ≤ 0, for otherwise the desired result holds. By Lemma 7, nG ≤ 23 mG holds since G ∈ G . This implies that anG + bmG ≥ ( 23 a + b)mG > 0. This establishes the sufficiency. □ In view of Lemma 8, a general way to formulate our problem is as follows. Problem 1. Determine the shape of the surface Γ (x, y, z) which is the subset of 2 D = {(x, y, z) | (y ≥ 0) ∧ (x > − y) } ⊂ R3 3

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

107

defined by the rule

γt (G)

z = sup

xnG + ymG

G∈G

.

In other words, determine z = z(x, y) as a function of x and y. We shall prove the following result. Theorem 9. The surface Γ (x, y, z) is determined by z(x, y) =

2 3x + 2y

.

Proof. Let y ≥ 0 and x > − 23 y be arbitrary real numbers. We observe that the graph G ∼ = P3 with nG = 3 and mG = 2 has γt (G) = 2. Hence a simple general lower bound for z is given by z(x, y) ≥

2 3x + 2y

.

Hence it suffices for us to prove that

γt (G) ≤

2xnG + 2ymG

(1)

3x + 2y

holds for every G ∈ G . For notational simplicity, let

ξ (G) =

2xnG + 2ymG 3x + 2y

denote the right-hand side of Inequality (1). Therefore we wish to show that γt (G) ≤ ξ (G). Suppose that x ≤ 0 (and still x > − 32 y, or, equivalently, 3x + 2y > 0). Let G ∈ G . By Lemma 7, nG ≤ holds. Thus, we obtain the following chain of inequalities:

ξ (G) =

2xnG + 2ymG 3x + 2y

≥

3xmG + 2ymG 3x + 2y

3 m 2 G

and γt (G) ≤ mG

= mG ≥ γt (G),

which establishes the desired inequality. Hence we may assume that x > 0 (and still y ≥ 0). We apply induction on nG + mG to prove that γt (G) ≤ ξ (G) holds for every graph G ∈ G , and for any two reals x and y with x > 0, y ≥ 0. Since nG ≥ 3 and mG ≥ 2, we note that ξ (G) ≥ (6x + 4y)/(3x + 2y) = 2. Thus, the assertion is obvious whenever γt (G) = 2. In particular, this establishes the base case when nG + mG = 5. Let G ∈ G and assume that the result holds for all graphs F ∈ G with nF + mF < nG + mG . If G is a disconnected graph with components G1 , G2 , . . . , Gk , where Gi has order ni and size mi , then

γt (G) =

k ∑ i=1

γt (Gi )

and

ξ (G) =

k ∑

ξ (Gi ).

i=1

Hence both sides of Inequality (1) are additive with respect to vertex–disjoint union. Therefore we may assume without loss of generality that G is connected, for otherwise we apply the result to each component. Suppose that G is not a tree. Let e be a cycle edge of G, and consider the connected graph G − e. Since removing edges does not decrease the total domination number, we have that γt (G) ≤ γt (G − e). By induction, γt (G − e) ≤ ξ (G − e). However, ξ (G − e) ≤ ξ (G). Consequently, γt (G) ≤ ξ (G), yielding the desired bound in Inequality (1). Hence, we may assume that G is a tree. As observed earlier, the assertion is obvious whenever γt (G) = 2. Thus, we may assume that γt (G) ≥ 3, implying that diam(G) ≥ 4. Suppose that G has an edge e such that both components of G − e have more than one vertex and more than one edge. Let G1 and G2 denote the two components of G − e. Applying the induction hypothesis to each component of G − e, we have that γt (G) ≤ γt (G1 ) + γt (G2 ) ≤ ξ (G1 ) + ξ (G2 ) ≤ ξ (G), yielding the desired bound in Inequality (1). Therefore, we may assume that G has no edge e such that both components of G − e have more than one vertex and more than one edge. Let v0 v1 . . . vd denote a longest path in G, and so d = diam(G). If d ≥ 5, then both components of G − v2 v3 have no isolated vertex or isolated edge, a contradiction. Hence, d ≤ 4. By our earlier assumption, d ≥ 4. Consequently, d = 4. If v2 has a neighbor, v2′ say, of degree at least 3, then both components of G − v2 v2′ have no isolated vertex or isolated edge, a contradiction. Hence, every neighbor of v2 that is not a leaf has degree equal to 2. In particular, d(v1 ) = 2 and d(v3 ) = 2. Suppose that v2 has k neighbors of degree 2 and ℓ leaf neighbors. We note that k ≥ 2, ℓ ≥ 0, and d(v2 ) = k + ℓ. Further, we note that G is obtained from a star K1,k+ℓ by subdividing k edges exactly once. Since γt (G) = k + 1, nG = 2k + ℓ + 1 ≥ 2k + 1

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M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

and mG = 2k + ℓ ≥ 2k, we have

ξ (G)

= ≥ =

2xnG + 2ymG 3x + 2y 2x(2k + 1) + 2y(2k) 3x + 2y (3x + 2y)k + (x + 2y)k + 2x 3x + 2y (x + 2y) · 1 + 2x

>

k+

= =

k+1

3x + 2y

γt (G).

Hence, γt (G) ≤ ξ (G), as desired. This completes the proof by induction. □ An equivalent formulation of Theorem 9 gives us the result of Theorem 6(a). Recall its statement. Theorem 6(a). Let a, b ∈ R. Then, the bound γt (G) ≤ anG + bmG is valid for every graph G with no isolated vertex and isolated edge if and only if both 3a + 2b ≥ 2 and b ≥ 0 hold. Proof. Suppose the bound γt (G) ≤ anG + bmG is valid for every G ∈ G . Since every graph has total domination number at least two, anG + bmG ≥ 2, and so, by Lemma 8, b ≥ 0 and a > − 32 b. Further since γt (G)/(anG + bmG ) ≤ 1, Theorem 9 implies that 2/(3a + 2b) = z(a, b) ≤ 1. Thus both 3a + 2b ≥ 2 and b ≥ 0 hold. Conversely, suppose that both 3a + 2b ≥ 2 and b ≥ 0 hold. Then, a > − 23 b and b ≥ 0, and so, by Theorem 9, z(a, b) = 2/(3a + 2b). Thus since 3a + 2b ≥ 2, we have that z(a, b) ≤ 1 and therefore γt (G)/(anG + bmG ) ≤ z(a, b) ≤ 1 holds for every G ∈ G . □ For example, to illustrate Theorem 6(a), let G be a graph without isolated vertices and isolated edges. Taking (a, b) = ( 21 , 14 ), we have that γt (G) ≤ (2n + m)/4. Taking (a, b) = ( 13 , 12 ), we have that γt (G) ≤ (2n + 3m)/6. 4.2. Proof of Theorem 6(b) Let G≥2 be the class of all graphs with minimum degree at least two. We note that each graph G ∈ G≥2 satisfies nG ≤ mG . We shall need the following lemma. Lemma 10. Let a, b ∈ R. Then, anG + bmG > 0 holds for every G ∈ G≥2 if and only if b ≥ 0 and a > −b. Proof. Suppose that anG + bmG > 0 holds for every G ∈ G≥2 and for each pair of reals a and b. If b < 0, then there exists a sufficiently large ℓ such that the complete graph on ℓ vertices gives negative anG + bmG , a contradiction. On the other hand, for all pairs a, b where a ≤ −b, the graph G ∼ = C3 with nG = 3 and mG = 3 shows that anG + bmG is not always positive on G≥2 , a contradiction. Hence, both conditions b ≥ 0 and a > −b are necessary. To prove the opposite direction, suppose that a and b are reals satisfying b ≥ 0 and a > −b. Let G be an arbitrary graph in G≥2 . If a > 0, the inequality anG + bmG > 0 trivially holds. If a ≤ 0, then noting that nG ≤ mG and a + b > 0, we have anG + bmG ≥ (a + b)mG > 0. This establishes the sufficiency. □ In view of Lemma 10, a general way to formulate our problem is as follows. Problem 2. Determine the shape of the surface Θ (x, y, z) which is the subset of D = {(x, y, z) | (y ≥ 0) ∧ (x > −y) } ⊂ R3 defined by the rule z = sup G∈G≥2

γt (G) xnG + ymG

.

We shall prove the following result. Theorem 11. The surface Θ (x, y, z) is determined by z(x, y) =

2 3x + 3y

.

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

109

Proof. Let y ≥ 0 and x > −y be arbitrary real numbers. We observe that the graph G ∼ = C3 with nG = 3 and mG = 3 has γt (G) = 2. Hence a simple general lower bound for z is given by z(x, y) ≥

2 3x + 3y

.

Hence it suffices for us to prove that

γt (G) ≤

2xnG + 2ymG 3x + 3y

holds for every G ∈ G≥2 . However, this is immediate since y ≥ 0 and mG ≥ nG , implying by Theorem 1 that 2 3

(

xnG + ymG

)

x+y

≥

2

(

3

xnG + ynG x+y

) =

2 3

nG ≥ γt (G).

□

An equivalent formulation of Theorem 11 gives us the result of Theorem 6(b). Recall its statement. Theorem 6(b). Let a, b ∈ R. Then, the bound γt (G) ≤ anG + bmG is valid for every graph G with minimum degree at least two if and only if both b ≥ 0 and a ≥ 23 − b hold. Proof. Suppose the bound γt (G) ≤ anG + bmG is valid for every G ∈ G≥2 . Since every graph has total domination number at least two, anG + bmG ≥ 2, and so, by Lemma 10, b ≥ 0 and a > −b. Further since γt (G)/(anG + bmG ) ≤ 1, Theorem 11 implies that 2/(3a + 3b) = z(a, b) ≤ 1. Thus both 3a + 3b ≥ 2 and b ≥ 0 hold. Conversely, suppose that both 3a + 3b ≥ 2 and b ≥ 0 hold. Then, a > −b and b ≥ 0, and so, by Theorem 11, z(a, b) = 2/(3a + 3b). Thus since 3a + 3b ≥ 2, we have that z(a, b) ≤ 1 and therefore γt (G)/(anG + bmG ) ≤ z(a, b) ≤ 1 holds for every G ∈ G≥2 . □ 4.3. Proof of Theorem 6(c) Let F≥2 be the class of all connected graphs with minimum degree at least two and order at least eleven. We note that the class of graphs F≥2 is a proper subfamily of the class of graphs G≥2 . In particular, each graph G ∈ F≥2 satisfies nG ≤ mG . We shall need the following lemma. Lemma 12. Let a, b ∈ R. Then, anG + bmG > 0 holds for every G ∈ F≥2 if and only if b ≥ 0 and a > −b. Proof. Suppose that anG + bmG > 0 holds for every G ∈ F≥2 and for each pair of reals a and b. If b < 0, then there exists a sufficiently large ℓ such that the complete graph on ℓ vertices gives negative anG + bmG , a contradiction. On the other hand, for all pairs a, b where a ≤ −b, the graph G ∼ = C14 with nG = 14 and mG = 14 shows that anG + bmG is not always positive on F≥2 , a contradiction. Hence, both conditions b ≥ 0 and a > −b are necessary. To prove the opposite direction, suppose that a and b are reals satisfying b ≥ 0 and a > −b. Let G be an arbitrary graph in F≥2 . If a > 0, the inequality anG + bmG > 0 trivially holds. If a ≤ 0, then noting that nG ≤ mG and a + b > 0, we have anG + bmG ≥ (a + b)mG > 0. This establishes the sufficiency. □ In view of Lemma 12, a general way to formulate our problem is as follows. A general way to formulate our problem is as follows. Problem 3. Determine the shape of the surface Φ (x, y, z) which is the subset of D = {(x, y, z) | (y ≥ 0) ∧ (x > −y) } ⊂ R3 defined by the rule

γt (G)

z = sup G∈F≥2

xnG + ymG

.

We shall prove the following result. Theorem 13. The surface Φ (x, y, z) is determined by z(x, y) =

4 7x + 7y

.

Proof. Let y ≥ 0 and x > −y be arbitrary real numbers. We observe that the graph G ∼ = C14 satisfies nG = mG = 14 and γt (G) = 8. Hence a simple general lower bound for z is given by z(x, y) ≥

8 14x + 14y

=

4 7x + 7y

.

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M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

Fig. 5. The hypercube Q3 .

It therefore suffices for us to prove that

γt (G) ≤

4xnG + 4ymG 7x + 7y

holds for every G ∈ F≥2 . However, this is immediate since y ≥ 0 and mG ≥ nG , implying by Theorem 3 that 4 7

(

xnG + ymG

)

x+y

≥

4

(

7

xnG + ynG x+y

) =

4 7

nG ≥ γt (G). □

An equivalent formulation of Theorem 13 gives us the result of Theorem 6(c). Recall its statement. Theorem 6(c). Let a, b ∈ R. Then, the bound γt (G) ≤ anG + bmG is valid for every connected graph G with δ (G) ≥ 2 and nG ≥ 11 if and only if both b ≥ 0 and a ≥ 74 − b hold. Proof. Suppose the bound γt (G) ≤ anG + bmG is valid for every G ∈ F≥2 . Since every graph has total domination number at least two, anG + bmG ≥ 2, and so, by Lemma 12, b ≥ 0 and a > −b. Further since γt (G)/(anG + bmG ) ≤ 1, Theorem 13 implies that 4/(7a + 7b) = z(a, b) ≤ 1. Thus both 7a + 7b ≥ 4 and b ≥ 0 hold. Conversely, suppose that both 7a + 7b ≥ 4 and b ≥ 0 hold. Then, a > −b and b ≥ 0, and so, by Theorem 13, z(a, b) = 4/(7a + 7b). Thus since 7a + 7b ≥ 4, we have that z(a, b) ≤ 1 and therefore γt (G)/(anG + bmG ) ≤ z(a, b) ≤ 1 holds for every G ∈ F≥2 . □ 4.4. Proof of Theorem 6(d) Let G≥3 be the class of all graphs with minimum degree at least three. Each graph G ∈ G≥3 satisfies 3nG ≤ 2mG . We shall need the following lemma. Lemma 14. Let a, b ∈ R. Then, anG + bmG > 0 holds for every G ∈ G≥3 if and only if b ≥ 0 and a > − 23 b. Proof. Suppose that anG + bmG > 0 holds for every G ∈ G≥3 and for each pair of reals a and b. If b < 0, then there exists a sufficiently large ℓ such that the complete graph on ℓ vertices gives negative anG + bmG , a contradiction. On the other hand, for all pairs a, b where a ≤ − 23 b, the hypercube G ∼ = Q3 ∈ G≥3 illustrated in Fig. 5 with nG = 8 and mG = 12 shows that anG + bmG is not always positive on G≥3 , a contradiction. Hence, both conditions b ≥ 0 and a > − 32 b are necessary. To prove the opposite direction, suppose that a and b are reals satisfying b ≥ 0 and a > − 23 b. Let G be an arbitrary graph in G≥3 . If a > 0, the inequality anG + bmG > 0 trivially holds. If a ≤ 0, then noting that nG ≤ anG + bmG ≥ ( 32 a + b)mG > 0. This establishes the sufficiency. □ In view of Lemma 14, a general way to formulate our problem is as follows. Problem 4. Determine the shape of the surface Ψ (x, y, z) which is the subset of 3 D = {(x, y, z) | (y ≥ 0) ∧ (x > − y) } ⊂ R3 2 defined by the rule z = sup G∈G≥3

γt (G) . xnG + ymG

We shall prove the following result. Theorem 15. The surface Ψ (x, y, z) is determined by z(x, y) =

1 2x + 3y

.

2 m 3 G

and 23 a + b > 0, we have

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

111

Proof. Let y ≥ 0 and x > − 23 y be arbitrary real numbers. We observe that the hypercube G ∼ = Q3 ∈ G≥3 illustrated in Fig. 5 with nG = 8 and mG = 12 has γt (G) = 4. Hence a simple general lower bound for z is given by z(x, y) ≥

1 2x + 3y

.

Hence it suffices for us to prove that

γt (G) ≤

xnG + ymG 2x + 3y

holds for every G ∈ G≥3 . However, this is immediate since y ≥ 0 and mG ≥ xnG + ymG 2x + 3y

≥

xnG +

3 ynG 2

2x + 3y

=

1 2

nG ≥ γt (G).

3 n , 2 G

implying by Theorem 4 that

□

An equivalent formulation of Theorem 15 gives us the result of Theorem 6(d). Recall its statement. Theorem 6(d). Let a, b ∈ R. Then, the bound γt (G) ≤ anG + bmG is valid for every graph G with minimum degree at least three if and only if both b ≥ 0 and a ≥ 21 (1 − 3b) hold. Proof. Suppose the bound γt (G) ≤ anG + bmG is valid for every G ∈ G≥3 . Since every graph has total domination number at least two, anG + bmG ≥ 2, and so, by Lemma 14, b ≥ 0 and a > − 32 b. Further since γt (G)/(anG + bmG ) ≤ 1, Theorem 15 implies that 1/(2a + 3b) = z(a, b) ≤ 1. Thus both 2a + 3b ≥ 1 and b ≥ 0 hold. Conversely, suppose that both 2a + 3b ≥ 1 and b ≥ 0 hold. Then, a > − 23 b and b ≥ 0, and so, by Theorem 15, z(a, b) = 1/(2a + 3b). Thus since 2a + 3b ≥ 1, we have that z(a, b) ≤ 1 and therefore γt (G)/(anG + bmG ) ≤ z(a, b) ≤ 1 holds for every G ∈ G≥3 . □ 4.5. Proof of Theorem 6(e) Let G≥4 be the class of all graphs with minimum degree at least four. Each graph G ∈ G≥4 satisfies mG ≥ 2nG . We shall need the following lemma. Lemma 16. Let a, b ∈ R. Then, anG + bmG > 0 holds for every G ∈ G≥4 if and only if b ≥ 0 and a > −2b. Proof. Suppose that anG + bmG > 0 holds for every G ∈ G≥4 and for each pair of reals a and b. If b < 0, then there exists a sufficiently large ℓ such that the complete graph on ℓ vertices gives negative anG + bmG , a contradiction. On the other hand, for all pairs a, b where a ≤ −2b, the bipartite complement G ∼ = H 14 ∈ G≥4 of the Heawood graph (or, equivalently, the incidence bipartite graph of the complement of the Fano plane), shown in Fig. 4(b), with nG = 14 and mG = 28 shows that anG + bmG is not always positive on G≥4 , a contradiction. Hence, both conditions b ≥ 0 and a > −2b are necessary. To prove the opposite direction, suppose that a and b are reals satisfying b ≥ 0 and a > −2b. Let G be an arbitrary graph in G≥4 . If a > 0, the inequality anG + bmG > 0 trivially holds. If a ≤ 0, then noting that nG ≤ 12 mG and 12 a + b > 0, we have anG + bmG ≥ ( 21 a + b)mG > 0. This establishes the sufficiency. □ In view of Lemma 16, a general way to formulate our problem is as follows. Problem 5. Determine the shape of the surface Υ (x, y, z) which is the subset of D = {(x, y, z) | (y ≥ 0) ∧ (x > −2y) } ⊂ R3 defined by the rule z = sup G∈G≥4

γt (G) xnG + ymG

.

We shall prove the following result. Theorem 17. The surface Υ (x, y, z) is determined by z(x, y) =

3 7x + 14y

.

Proof. Let y ≥ 0 and x > −2y be arbitrary real numbers. We observe that the bipartite complement G ∼ = H 14 ∈ G≥4 of the Heawood graph, illustrated in Fig. 4, with nG = 14 and mG = 28 has γt (G) = 6. Hence a simple general lower bound for z is given by z(x, y) ≥

3 7x + 14y

.

112

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

Hence it suffices for us to prove that

γt (G) ≤

3

(

xnG + ymG

7

)

x + 2y

holds for every G ∈ G≥4 . However, this is immediate since y ≥ 0 and mG ≥ 2nG , implying by Theorem 5 that 3 7

(

xnG + ymG x + 2y

) ≥

3 7

(

xnG + 2ynG

) =

x + 2y

3 7

nG ≥ γt (G).

□

An equivalent formulation of Theorem 17 gives us the result of Theorem 6(e). Recall its statement. Theorem 6(e). Let a, b ∈ R. Then, the bound γt (G) ≤ anG + bmG is valid for every graph G with minimum degree at least four if and only if both b ≥ 0 and a ≥ 37 − 2b hold. Proof. Suppose the bound γt (G) ≤ anG + bmG is valid for every G ∈ G≥4 . Since every graph has total domination number at least two, anG + bmG ≥ 2, and so, by Lemma 16, b ≥ 0 and a > −2b. Further since γt (G)/(anG + bmG ) ≤ 1, Theorem 17 implies that 3/(7a + 14b) = z(a, b) ≤ 1. Thus both 7a + 14b ≥ 3 and b ≥ 0 hold. Conversely, suppose that both 7a + 14b ≥ 3 and b ≥ 0 hold. Then, a > −2b and b ≥ 0, and so, by Theorem 17, z(a, b) = 3/(7a + 14b). Thus since 7a + 14b ≥ 3, we have that z(a, b) ≤ 1 and therefore γt (G)/(anG + bmG ) ≤ z(a, b) ≤ 1 holds for every G ∈ G≥4 . □

5. Graphs achieving equality in Theorem 6 In this section, we characterize the graphs G for which the equality γt (G) = anG + bmG holds in the bounds in Theorem 6. Recall that G is the class of all graphs without isolated vertices and isolated edges. Further recall that G≥2 , G≥3 and G≥4 are the class of all graphs with minimum degree at least two, three and four, respectively. More generally for δ ≥ 1, let G≥δ be the class of all graphs with minimum degree at least δ . Let F≥δ be the subclass of the class G≥δ such that every graph G ∈ F≥δ satisfies the following properties. (A) γt (G) ≤ cδ nG , where cδ is a constant depending only on δ = δ (G). (B) For a, b ∈ R, the following holds. The bound γt (G) ≤ anG + bmG is valid if and only if both b ≥ 0 and a ≥ cδ − 21 bδ hold. We proceed further with the following lemma. Lemma 18. For δ ≥ 1, if G ∈ F≥δ satisfies γt (G) = anG + bmG , then γt (G) = cδ nG and either b = 0 and a = cδ or b > 0, G is δ -regular and a = cδ − 21 bδ . Proof. Let δ ≥ 1 and let G ∈ F≥δ , and suppose that γt (G) = anG + bmG . Thus, by definition of the family F≥δ and by simple algebra, cδ nG ≥ γt (G) = anG + bmG = cδ nG +

( a − (cδ −

1 2

)

bδ ) nG + b(mG −

1 2

nG δ ).

(2)

Clearly, mG − nG δ/2 ≥ 0 with equality if and only if G is δ -regular. Since b ≥ 0 and a ≥ cδ − bδ/2 hold, we have anG + bmG ≥ cδ nG with equality if and only if either b = 0 and a = cδ or G is δ -regular, b > 0 and a = cδ − bδ/2. Thus, by Inequality (2), cδ nG ≥ γt (G) = anG + bmG ≥ cδ nG .

(3)

Hence, we must have equality throughout Inequality Chain (3), implying that γt (G) = cδ nG and anG + bmG = cδ nG . As observed earlier, if anG + bmG = cδ nG , then either b = 0 and a = cδ or G is δ -regular, b > 0 and a = cδ − bδ/2. □ We are now in a position to prove the following results. Theorem 19. Equality γt (G) = anG + bmG holds in Theorem 6(b) if and only if the following hold. (a) Every component of G is a 3-cycle or a 6-cycle. (b) (a, b) = ( 32 − b, b) where b ≥ 0. Proof. Suppose that the equality γt (G) = anG + bmG holds in Theorem 6(b). We note that in this case G ∈ G≥2 , and both b ≥ 0 and a ≥ 32 − b hold. By Theorem 1, γt (G) ≤ 2nG /3. Thus, G ∈ F≥2 with c2 = 2/3 and γt (G) = anG + bmG . By Lemma 18,

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

113

we have γt (G) = 2nG /3. Further either b = 0 and a = 2/3 or b > 0, G is 2-regular and a = 2/3 − b. Since γt (G) = 2nG /3, Theorem 1 implies that every component of G is a 3-cycle or a 6-cycle. Thus, nG = mG and

γt (G) =

2 3

( nG =

2 3

) − b nG + bmG

for all b ≥ 0. Hence, (a, b) = ( 23 − b, b) where b ≥ 0. Conversely, if (a) and (b) hold in the statement of the theorem, then γt (G) = anG + bmG . □ Theorem 20. Equality γt (G) = anG + bmG holds in Theorem 6(d) if and only if the following hold. (a) Every component of G is the generalized Petersen graph GP16 or belongs to the family Gcubic ∪ Hcubic . (b) (a, b) = ( 12 − 32 b, b) where b ≥ 0. Proof. Suppose that the equality γt (G) = anG + bmG holds in Theorem 6(d). We note that in this case G ∈ G≥3 , and both b ≥ 0 and a ≥ (1 − 3b)/2 hold. By Theorem 4, γt (G) ≤ nG /2. Thus, G ∈ F≥3 with c3 = 1/2 and γt (G) = anG + bmG . By Lemma 18, we have γt (G) = nG /2. Further either b = 0 and a = 1/2 or b > 0, G is 3-regular and a = (1 − 3b)/2. Since γt (G) = nG /2, Theorem 4 implies that every component of G is the generalized Petersen graph GP16 or belongs to the family Gcubic ∪ Hcubic . In particular, we note that G is a cubic graph and mG = 3nG /2. Thus,

γt (G) =

1 2

( nG =

1 2

) (1 − 3b) nG + bmG

for all b ≥ 0. Hence, (a, b) = ( 12 − 23 b, b) where b ≥ 0. Conversely, if (a) and (b) hold in the statement of the theorem, then γt (G) = anG + bmG . □ Theorem 21. Equality γt (G) = anG + bmG holds in Theorem 6(e) if and only if the following hold. (a) G is a bipartite complement H 14 of the Heawood graph. (b) (a, b) = ( 37 − 2b, b) where b ≥ 0. Proof. Suppose that the equality γt (G) = anG + bmG holds in Theorem 6(e). We note that in this case G ∈ G≥4 , and both b ≥ 0 and a ≥ 73 − 2b hold. By Theorem 5, we have γt (G) ≤ 3nG /7. Thus, G ∈ F≥4 with c3 = 3/7 and γt (G) = anG + bmG . By Lemma 18, we have γt (G) = 3nG /7. Further either b = 0 and a = 3/7 or b > 0, G is 4-regular and a = 73 − 2b. Since γt (G) = 3nG /7, Theorem 5 implies that G is a bipartite complement H 14 of the Heawood graph. In particular, we note that G is a 4-regular graph and mG = 2nG . Thus,

γt (G) =

3 7

( nG =

3 7

) − 2b nG + bmG

for all b ≥ 0. Hence, (a, b) = ( 37 − 2b, b) where b ≥ 0. Conversely, if (a) and (b) hold in the statement of the theorem, then γt (G) = anG + bmG . □ It remains for us to characterize the graphs G for which the equality γt (G) = anG +bmG holds in the bounds in Theorem 6(a) and 6(c). Theorem 22. Equality γt (G) = anG + bmG holds in Theorem 6(a) if and only if the following hold. (a) (a, b) = ( 23 , 0) and every component of G is C3 , C6 or F ◦ P2 for some connected graph F . (b) (a, b) = ( 23 (1 − b), b) where b > 0 and every component of G is a path P3 . Proof. Suppose that the equality γt (G) = anG + bmG holds in Theorem 6(a). We note that in this case G ∈ G , and both b ≥ 0 and a ≥ 32 (1 − b) hold. By Theorem 1, γt (G) ≤ 2nG /3. Thus, 2 3

nG ≥ γt (G) = anG + bmG =

2 3

( nG +

a−

2 3

) (1 − b) nG + b(mG −

2 3

nG ) .

(4)

By Lemma 7, mG − 23 nG ≥ 0 with equality if and only if every component of G is a path P3 . Since b ≥ 0 and a ≥ hold, we have anG + bmG ≥ path P3 and a = 2 3

2 (1 3

2 n 3 G

with equality if and only if either b = 0 and a =

2 3

2 (1 3

− b)

or b > 0, every component of G is a

− b). Thus, by Inequality (6),

nG ≥ γt (G) = anG + bmG ≥

2 3

nG .

(5)

114

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115

Hence, we must have equality throughout Inequality Chain (7), implying that γt (G) = 32 nG and anG + bmG = 32 nG . Since γt (G) = 32 nG , Theorem 1 implies that every component of G is C3 , C6 or F ◦ P2 for some connected graph F . As observed earlier, if anG + bmG = 23 nG , then either b = 0 and a = 23 or b > 0, every component of G is a path P3 and a ≥ 32 (1 − b). □ The total domination number of a cycle Cn is easy to compute (see, for example, [13]): For n ≥ 3, γt (Cn ) = ⌊n/2⌋ + ⌈n/4⌉ − ⌊n/4⌋. We are now in a position to characterize the graphs G for which the equality γt (G) = anG + bmG holds in the bound in Theorem 6(c). Theorem 23. Equality γt (G) = anG + bmG holds in Theorem 6(c) if and only if the following hold. (a) (a, b) = ( 74 , 0) and G is a 14-cycle or obtained from a 14-cycle by adding certain chords or belongs to the family F . (b) (a, b) = ( 74 − b, b) where b > 0 and G is a 14-cycle. Proof. Suppose that the equality γt (G) = anG + bmG holds in Theorem 6(c). We note that in this case G ∈ F≥2 , and both b ≥ 0 and a ≥ 74 − b hold. By Theorem 3, γt (G) ≤ 4nG /7. Thus, 4 7

nG ≥ γt (G) = anG + bmG =

4 7

( nG +

a−(

4 7

) − b) nG + b(mG − nG ).

(6)

Since G ∈ F≥2 , we note that nG ≤ mG with equality if and only if G is a cycle. Since b ≥ 0 and a ≥ have anG + bmG ≥ Inequality (6), 4 7

4 n 7 G

with equality if and only if either b = 0 and a =

nG ≥ γt (G) = anG + bmG ≥

4 7

4 7

or b > 0, G is a cycle and a =

nG .

4 7 4 7

− b hold, we − b. Thus, by (7)

Hence, we must have equality throughout Inequality Chain (7), implying that γt (G) = 47 nG and anG + bmG = 74 nG . Since γt (G) = 74 nG , Theorem 3 implies that G is a 14-cycle or obtained from a 14-cycle by adding certain chords or belongs to the family F . As observed earlier, if anG + bmG = 47 nG , then either b = 0 and a = 47 or b > 0, G is a cycle and a = 74 − b. If G is a cycle, then G = Cn where n = nG ≥ 11, implying that 4n/7 = γt (G) = ⌊n/2⌋ + ⌈n/4⌉ − ⌊n/4⌋. This in turn implies that n = 14, and so if b > 0, then G is a 14-cycle and a = 47 − b. □ 6. Closing comments and conjectures Our main contributions in this paper are to prove that all the essential upper bounds on the total domination number of a graph without isolated vertices and isolated edges can be written in a unified form, and to extend this result to graphs with minimum degree δ for small values of δ . These unified forms of upper bounds for the total domination number of a graph are summarized in Theorem 6. In Section 5, we characterize the graphs G for which the equality γt (G) = anG + bmG holds in the bounds in Theorem 6. We close with the following conjecture that we have yet to settle. Conjecture 1. Let a, b ∈ R. Then, the bound γt (G) ≤ anG + bmG is valid for every graph G with minimum degree at least five if 4 − 52 b hold. and only if both b ≥ 0 and a ≥ 11 4 In the special case when (a, b) = ( 11 , 0), Conjecture 1 simplifies to the following conjecture by Thomassé and Yeo [20].

Conjecture 2 (Thomassé – Yeo Conjecture). If G is a graph of order n with δ (G) ≥ 5, then γt (G) ≤

4 n. 11

Acknowledgments Research supported in part by the South African National Research Foundation and the University of Johannesburg. The author express his sincere thanks to the anonymous referees for their meticulous and thorough reading of the paper that greatly improved its exposition and depth. References [1] D. Archdeacon, J. Ellis-Monaghan, D. Fischer, D. Froncek, P.C.B. Lam, S. Seager, B. Wei, R. Yuster, Some remarks on domination, J. Graph Theory 46 (2004) 207–210. [2] R.C. Brigham, J.R. Carrington, R.P. Vitray, Connected graphs with maximum total domination number, J. Combin. Comput. Combin. Math. 34 (2000) 81–96. [3] Cs. Bujtás, M.A. Henning, Zs. Tuza, Transversals and domination in uniform hypergraphs, European J. Combin. 33 (2012) 62–71. [4] V. Chvátal, C. McDiarmid, Small transversals in hypergraphs, Combinatorica 12 (1992) 19–26. [5] E.J. Cockayne, R.M. Dawes, S.T. Hedetniemi, Total domination in graphs, Networks 10 (1980) 211–219. [6] O. Favaron, M.A. Henning, C.M. Mynhardt, J. Puech, Total domination in graphs with minimum degree three, J. Graph Theory 34 (1) (2000) 9–19.

M.A. Henning / Discrete Applied Mathematics 244 (2018) 103–115 [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21]

115

M.R. Garey, D.S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completenes, W. H. Freeman, San Francisco, 1979. M.A. Henning, Graphs with large total domination number, J. Graph Theory 35 (1) (2000) 21–45. M.A. Henning, Recent results on total domination in graphs: A survey, Discrete Math. 309 (2009) 32–63. M.A. Henning, A. Yeo, Complexity and algorthmic results. Chapter 3 in [15], pp. 19–29. M.A. Henning, A. Yeo, Total domination and minimum degree. Chapter 5 in [15], pp. 39–54. M.A. Henning, A. Yeo, A new upper bound on the total domination number of a graph, Electron. J. Combin. 14 (2007) #R65. M.A. Henning, A. Yeo, Hypergraphs with large transversal number and with edge sizes at least three, J. Graph Theory 59 (2008) 326–348. M.A. Henning, A. Yeo, Total domination in 2-connected graphs and in graphs with no induced 6-cycles, J. Graph Theory 60 (2009) 55–79. M.A. Henning, A. Yeo, Total Domination in Graphs, in: Springer Monographs in Mathematics, 2013, ISBN: 978-1-4614-6524-9 (Print) 978-1-46146525-6 (Online). P.C.B. Lam, B. Wei, On the total domination number of graphs, Util. Math. 72 (2007) 223–240. R.C. Laskar, J. Pfaff, S.M. Hedetniemi, S.T. Hedetniemi, On the algorithmic complexity of total domination, SIAM J. Algebr. Discrete Methods 5 (1984) 420–425. S.O., D.B. West, Balloons, cut-edges, matchings, and total domination in regular graphs of odd degree, J. Graph Theory 64 (2010) 116–131. L. Sun, An upper bound for the total domination number, J. Beijing Inst. Tech. 4 (1995) 111–114. S. Thomassé, A. Yeo, Total domination of graphs and small transversals of hypergraphs, Combinatorica 27 (2007) 473–487. Zs. Tuza, Covering all cliques of a graph, Discrete Math. 86 (1990) 117–126.