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Perfectly relating the domination, total domination, and paired domination numbers of a graph
Discrete Mathematics 338 (2015) 1424–1431
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Perfectly relating the domination, total domination, and paired domination numbers of a graph José D. Alvarado a , Simone Dantas a , Dieter Rautenbach b,∗ a b
Instituto de Matemática e Estatística, Universidade Federal Fluminense Niterói, Brazil Institute of Optimization and Operations Research, Ulm University, Ulm, Germany
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Article history: Received 30 May 2014 Received in revised form 6 March 2015 Accepted 8 March 2015
Keywords: Domination Total domination Paired domination Domatic number Total domatic number
abstract The domination number γ (G), the total domination number γt (G), the paired domination number γp (G), the domatic number d(G), and the total domatic number dt (G) of a graph G without isolated vertices are related by trivial inequalities γ (G) ≤ γt (G) ≤ γp (G) ≤ 2γ (G) and dt (G) ≤ d(G). Very little is known about the graphs that satisfy one of these inequalities with equality. We study classes of graphs defined by requiring equality in one of the above inequalities for all induced subgraphs that have no isolated vertices and whose domination number is not too small. Our results are characterizations of several such classes in terms of their minimal forbidden induced subgraphs. Furthermore, we prove some hardness results, which suggest that the extremal graphs for some of the above inequalities do not have a simple structure. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Whenever two graph parameters, say ν and τ , are related by a simple inequality, say ν(G) ≤ τ (G) for every graph G, but the graphs G with ν(G) = τ (G) do not have a simple structure, it makes sense to study the class of the so-called (ν, τ )-perfect graphs, the graphs for which every induced subgraph H satisfies ν(H ) = τ (H ). Famous examples of this approach are the classical perfect graphs [3], where the two parameters are the clique number and the chromatic number, or the domination perfect graphs [22], where the two parameters are the domination number and the independent domination number. In the present paper we consider classes of (ν, τ )-perfect graphs, where ν and τ are well known domination parameters [9]. We consider finite, simple, and undirected graphs and use standard terminology and notation. Let D be a set of vertices of some graph G. The set D is a dominating set of G if every vertex of G that does not belong to D, has a neighbor in D. The set D is a total dominating set of G if it is dominating and the subgraph G[D] of G induced by D does not have any isolated vertices. Finally, the set D is a paired dominating set of G if it is dominating and the graph G[D] has a perfect matching. The domination number γ (G) [9], the total domination number γt (G) [5,14], and the paired domination number γp (G) [10,16] of G are the minimum cardinalities of a dominating, a total dominating, and a paired dominating set of G, respectively. These definitions immediately imply [1,9,10]
γ (G) ≤ γt (G) ≤ γp (G) for every graph G for which γp (G) is defined.
∗
Corresponding author. E-mail addresses: [email protected] (J.D. Alvarado), [email protected] (S. Dantas), [email protected] (D. Rautenbach).
http://dx.doi.org/10.1016/j.disc.2015.03.014 0012-365X/© 2015 Elsevier B.V. All rights reserved.
J.D. Alvarado et al. / Discrete Mathematics 338 (2015) 1424–1431
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Fig. 1. The graphs C5 , P5 , and N.
For a set D of vertices of a graph G and a vertex u in D, let PG (D, u) = {v ∈ V (G) \ D : NG (v) ∩ D = {u}}. The vertices in PG (D, u) are the external private neighbors of u (in G with respect to D). It is known that every graph G without isolated vertices has a minimum dominating set D such that, for every vertex u in D, the set PG (D, u) is non-empty [1]. Therefore, adding to such a dominating set D, for every vertex u in D, an external private neighbor of u, results in a paired dominating set of G, which implies
γp (G) ≤ 2γ (G). Only very little is known about the graphs that satisfy one of the above inequalities with equality. The trees T that satisfy γ (T ) = γt (T ) [8], or γt (T ) = γp (T ) [12,20], or 2γ (T ) = γt (T ) [11], or 2γ (T ) = γp (T ) [13] have been characterized constructively. Examples in [6] show that even very strong structural restrictions do not allow the strengthening of the inequality γt (G) ≤ 2γ (G). The relation between the total domination number and the paired domination number as well as paired dominating sets of a special structure were studied in [2,10,19,18]. The domatic number d(G) of a graph G [5,21] is the maximum integer k such that the vertex set V (G) of G can be partitioned into k dominating sets. The total domatic number dt (G) [4] and the paired domatic number dp (G) of G [10] are defined analogously. The definitions immediately imply [21] dp (G) ≤ dt (G) ≤ d(G) for every graph G without isolated vertices. Furthermore, since the union of at least two disjoint dominating sets is a total dominating set [21], we have dt (G) ≥
d(G)
2
for every graph G without isolated vertices. Having collected the relevant inequalities, we now consider some reasonable classes of ‘‘perfect’’ graphs. Since the total domination number and the paired domination number are only defined for graphs without isolated vertices, it does not make sense to require equalities involving these parameters for all induced subgraphs but only for those that have no isolated vertices. Therefore, if ν and τ are graph parameters with ν(G) ≤ τ (G) for every graph G, then let
G(ν, τ ) = {G : ∀H ⊆ind G : δ(H ) ≥ 1 ⇒ ν(H ) = τ (H )} where H ⊆ind G indicates that H is an induced subgraph of G. Our first results are the following. Note that all proofs are postponed to Section 2. Theorem 1. A graph G belongs to G(γt , γp ) if and only if G is {C5 , P5 , N }-free. Theorem 2. A graph G belongs to G(γt , 2γ ) if and only if G belongs to G(γp , 2γ ) if and only if G is {C4 , P4 }-free. See Fig. 1 for an illustration of the four special graphs in Theorem 1. In general, for a positive integer n, we denote the complete graph, the cycle, and the path of order n by Kn , Cn , and Pn , respectively. Note that K2 is a forbidden induced subgraph for graphs in G(γ , γt ), G(d, dt ), and G(γ , γp ). Therefore, these three classes of graphs consist exactly of all edgeless graphs. In view of this observation, it makes sense to restrict the considered induced subgraphs further. Therefore, for a positive integer k, we consider the class
Gk (ν, τ ) = {G : ∀H ⊆ind G : (δ(H ) ≥ 1 and γ (H ) ≥ k) ⇒ ν(H ) = τ (H )} . Note that G(ν, τ ) coincides with G1 (ν, τ ). For k = 2, we obtain the following, where 2K2 denotes the disjoint union of two copies of K2 . Theorem 3. A graph G belongs to G2 (γ , γt ) if and only if G is {C5 , 2K2 }-free. Theorem 4. A graph G belongs to G2 (d, dt ) if and only if G is {P4 , 2K2 , G1 , G2 }-free. Theorem 5. A graph G belongs to G2 (γ , γp ) if and only if G is {C5 , 2K2 , N }-free. See Fig. 2 for an illustration of the graphs G1 and G2 in Theorem 4.
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Fig. 2. Two graphs G1 and G2 with δ(Gi ) ≥ 1, γ (Gi ) ≥ 2, and d(Gi ) > dt (Gi ) for i ∈ {1, 2}.
For arbitrary values of k, we obtain results subject to girth conditions. In order to phrase our results, we introduce some notation. The order of a graph G is the number of its vertices and is denoted by n(G). If G is a graph, then the graph G ◦ K1 arises from G by adding, for every vertex u of G, a new vertex u′ together with the new edge uu′ . Let Tk be the set of forests of the form ℓ ℓ i=1 (Ti,0 ◦ K1 ) where ℓ is a positive integer, and T1,0 , . . . , Tℓ,0 are non-empty trees with k = i=1 n(Ti,0 ). Let Tk (γ , γt ) be the set containing all forests in Tk for which one of the trees Ti,0 is K1 , that is, Tk (γ , γt ) consists of all forests in Tk that have K2 as a component. Let Tk (γ , γp ) be the set containing all forests T in Tk for which one of the trees Ti,0 has no perfect matching. Our results are as follows. Theorem 6. For some positive integer k, a graph G of girth at least
5k+1
Theorem 7. For some positive integer k, a graph G of girth at least
5k
Theorem 8. For some positive integer k, a graph G of girth at least
5k+1
2
2
2
belongs to Gk (γ , γt ) if and only if G is Tk (γ , γt )-free.
belongs to Gk (d, dt ) if and only if G is Tk -free. belongs to Gk (γ , γp ) if and only if G is Tk (γ , γp )-free.
Finally, we establish some hardness results, which yield further motivation to consider the above classes. Our first hardness result implies that even in very restricted classes of graphs, requiring equality of the domination number and the paired domination number does not make the dominating set problem easy. A split graph is a graph whose vertex set can be partitioned into an independent set and a clique. Theorem 9. It is NP-complete to decide for a given split graph G with γ (G) = γt (G) = γp (G) and a given integer k, whether γ (G) ≤ k. The next two results imply that the graphs G with γ (G) = γt (G) or 2γ (G) = γp (G) do most likely not have a simple description. Theorem 10. It is NP-hard to decide for a given graph G, whether γ (G) = γt (G). Theorem 11. It is NP-hard to decide for a given graph G, whether 2γ (G) = γp (G). Section 2 is devoted to the proofs, auxiliary lemmas, corollaries, conjectures, and possible directions for further research. For k ∈ N, let [k] denote the set of the positive integers at most k. 2. Proofs, lemma, corollaries, and conjectures For convenience we restate the results. Theorem 1. A graph G belongs to G(γt , γp ) if and only if G is {C5 , P5 , N }-free. Proof. It is easy to check that the four graphs C5 , P5 , and N are minimal forbidden induced subgraphs for G(γt , γp ), which proves the necessity. In order to prove the sufficiency, we show that a {C5 , P5 , N }-free graph G with δ(G) ≥ 1 satisfies γt (G) = γp (G). For a contradiction, we assume that γt (G) < γp (G). Let Dt be a minimum total dominating set of G. First we assume that some component K of G[Dt ] is not complete. Let P : u0 . . . uℓ be a shortest path of maximum length in K . Since K is not complete, we have ℓ ≥ 2. Furthermore, by the choice of P, no vertex in K − {u0 } or K − {uℓ } is isolated. This implies that PG (Dt , u0 ) contains a vertex v0 and PG (Dt , uℓ ) contains a vertex vℓ . If ℓ = 2, then G[{u0 , u1 , u2 , v0 , v2 }] is either P5 or C5 , which is a contradiction. Hence, we have ℓ ≥ 3, and G[{v0 , u0 , u1 , u2 , u3 }] is P5 , which is a contradiction. Therefore, we may assume that every component of G[Dt ] is complete. Since Dt is not a paired dominating set, some component L of G[Dt ] has odd order at least 3. Let u, v , and w be three vertices from L. Since L is complete, the three sets PG (Dt , u), PG (Dt , v), and PG (Dt , w) are non-empty. If some vertex u′ in PG (Dt , u) is adjacent to all vertices in PG (Dt , v), then D′t = (Dt \ {v}) ∪ {u′ } is a minimum total dominating set of G. Since u′ and w are not adjacent but lie in the same component of G[D′t ], some component of G[D′t ] is not complete, and we can argue as above. Therefore, by symmetry, we may assume that there are vertices v ′ in PG (Dt , v) and w ′ in PG (Dt , w) such that u′ is not adjacent to v ′ or w ′ . If v ′ w ′ ̸∈ E (G), then G[{u, v, w, u′ , v ′ , w ′ }] is N, and if v ′ w ′ ∈ E (G), then G[{u, u′ , v ′ , w ′ }] is 2K2 , which is a contradiction. This completes the proof.
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Theorem 2. A graph G belongs to G(γt , 2γ ) if and only if G belongs to G(γp , 2γ ) if and only if G is {C4 , P4 }-free. Proof. Since γt (G) ≤ γp (G) ≤ 2γ (G) for every graph G without isolated vertices, G(γt , 2γ ) is a subset of G(γp , 2γ ). Since C4 and P4 are minimal forbidden induced subgraphs for G(γp , 2γ ), every graph in G(γp , 2γ ) is {C4 , P4 }-free. In order to complete the proof, it suffices to show that a connected {C4 , P4 }-free graph G with δ(G) ≥ 1 satisfies γt (G) = 2γ (G). For a contradiction, we assume that γt (G) < 2γ (G). Since G is P4 -free, the vertex set of G has a partition into two sets V1 and V2 such that every vertex in V1 is adjacent to every vertex in V2 . Since a set containing one vertex from V1 and one vertex from V2 is a total dominating set of G, we have γt (G) = 2. Since γt (G) < 2γ (G), this implies γ (G) > 1. Hence V1 contains two non-adjacent vertices u1 and v1 , and V2 contains two non-adjacent vertices u2 and v2 . Now G[{u1 , v1 , u2 , v2 }] is C4 , which is a contradiction. This completes the proof. Theorem 3. A graph G belongs to G2 (γ , γt ) if and only if G is {C5 , 2K2 }-free. Proof. It is easy to check that the two graphs C5 and 2K2 are minimal forbidden induced subgraphs for G2 (γ , γt ), which proves the necessity. In order to prove the sufficiency, we show that a {C5 , 2K2 }-free graph G with δ(G) ≥ 1 and γ (G) ≥ 2 satisfies γ (G) = γt (G). For a contradiction, we assume that γ (G) < γt (G). Since G is 2K2 -free and has no isolated vertex, it is connected. We choose a minimum dominating set D of G such that f (D) = u∈D distG (u, D \ {u}) is minimum, where distG (u, D \ {u}) is the minimum number of edges in a path in G between u and a vertex in D \ {u}. Since |D| ≥ 2 and G is connected, we have 0 < f (D) < ∞. Let u ∈ D be such that distG (u, D \ {u}) is maximum and let v ∈ D \ {u} be such that distG (u, v) = distG (u, D \ {u}), that is, G contains a path of length distG (u, v) between u and v . Since D is not a total dominating set of G, we have distG (u, v) ≥ 2. If distG (u, v) ≥ 4, then a vertex at distance 2 from u on a shortest path between u and v has no neighbor in D, which is a contradiction. Hence, we have distG (u, v) ∈ {2, 3}. First, we assume that distG (u, v) = 2. Let x be a common neighbor of u and v . Let D′ = (D \{u})∪{x}. By the choice of D, we obtain that either D′ is no dominating set of G or there is some w in D \{u, v} such that distG (w, D′ \{w}) > distG (w, D \{w}). In both cases, there is a neighbor u′ of u distinct from x such that u′ is not adjacent to v or x. By symmetry, there is a neighbor v ′ of v distinct from x such that v ′ is not adjacent to u or x. If u′ is not adjacent to v ′ , then G[{u, v, u′ , v ′ }] is 2K2 , which is a contradiction. If u′ is adjacent to v ′ , then G[{u, v, x, u′ , v ′ }] is C5 , which is a contradiction. Hence, we have distG (u, v) = 3. Let uxyv be a path in G. Let D′′ = (D \ {u, v}) ∪ {x, y}. By the choice of D, we obtain that either D′′ is no dominating set of G or there is some w in D \ {u, v} such that distG (w, D′′ \ {w}) > distG (w, D \ {w}). In both cases, there is a vertex z such that z is a neighbor of u or v and z is not adjacent to x or y. If z is adjacent to u but not to v , then G[{z , u, y, v}] is 2K2 , which is a contradiction. If z is adjacent to v but not to u, then G[{u, x, v, z }] is 2K2 , which is a contradiction. If z is adjacent to u and v , then G[{u, v, x, y, z }] is C5 , which is a contradiction. This completes the proof. Theorem 4. A graph G belongs to G2 (d, dt ) if and only if G is {P4 , 2K2 , G1 , G2 }-free. Proof. It is easy to check that the four graphs P4 , 2K2 , G1 , and G2 are minimal forbidden induced subgraphs for G2 (d, dt ), which proves the necessity. In order to prove the sufficiency, we show that a {P4 , 2K2 , G1 , G2 }-free graph G with δ(G) ≥ 1 and γ (G) ≥ 2 satisfies d(G) = dt (G). For a contradiction, we assume that d(G) > dt (G). Since the graph G has no isolated vertex and is 2K2 -free, it is connected. Let the positive integer k be maximal such that there is a partition of the vertex set k−1 V (G) of G into k non-empty sets V1 , . . . , Vk−1 , and V≥k such that |V
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assume that w1 = w2 , in which case G[{u1 , u2 , v1 , v2 , w1 , x}] is either G1 if v1 is not adjacent to v2 or G2 otherwise, which is a contradiction. This completes the proof. Theorem 5. A graph G belongs to G2 (γ , γp ) if and only if G is {C5 , 2K2 , N }-free. Proof. It is easy to check that the four graphs C5 , 2K2 , and N are minimal forbidden induced subgraphs for G2 (γ , γp ), which proves the necessity. In order to prove the sufficiency, we show that a {C5 , 2K2 , N }-free graph G with δ(G) ≥ 1 and γ (G) ≥ 2 satisfies γ (G) = γp (G). For a contradiction, we assume that γ (G) < γp (G). Let Dt be a minimum total dominating set of G. Since G is {C5 , 2K2 }-free, Theorem 3 implies that Dt is also a minimum dominating set of G. Since γ (G) < γp (G), the set Dt is no paired dominating set of G. Furthermore, since every vertex in Dt has a neighbor in Dt , it follows that, for every vertex u in Dt , the set PG (Dt , u) is non-empty. If G[Dt ] is not complete, then it contains an induced path uvw of order 3. Let u′ ∈ PG (Dt , u) and w ′ ∈ PG (Dt , w). Since G is 2K2 -free, the vertices u′ and w ′ are adjacent, and G[{u′ , u, v, w, w ′ }] is C5 , which is a contradiction. Hence, G[Dt ] is complete. Since Dt is not a paired dominating set, the set Dt has odd order at least 3. Let u, v , and w be three vertices in Dt . Let u′ in PG (Dt , u), v ′ in PG (Dt , v), and w ′ in PG (Dt , w). If u′ and v ′ are not adjacent, then, since G is {2K2 , N }-free, the vertex w′ is adjacent to u′ and v ′ , and G[{u′ , u, v, v ′ , w′ }] is C5 , which is a contradiction. Hence, we may assume that for every two distinct vertices x and y in {u, v, w}, every vertex in PG (Dt , x) is adjacent to every vertex in PG (Dt , y). Since (Dt \{v, w})∪{u′ } is not a dominating set of G, there is some vertex x of G with NG (x) ∩ {u, v, w, u′ } = {v, w}. If x is not adjacent to v ′ , then G[{u′ , v ′ , w, x}] is 2K2 , which is a contradiction. Hence, the vertex x is adjacent to v ′ . Now G[{u′ , v ′ , x, w, u}] is C5 , which is a contradiction. This completes the proof. In order to prove Theorems 6, 7, and 8 we need a lemma. An endvertex of a graph is a vertex of degree at most 1. For a graph G, let S (G) be the set of vertices of G that are adjacent to an endvertex of G. The vertices in S (G) are known as support vertices. Lemma 12. Let T be a tree with δ(T ) ≥ 1. (i) If γt (T ) > γ (T ) and T does not have a proper induced subgraph T ′ with • δ(T ′ ) ≥ 1, • γ (T ′ ) ≥ γ (T ), and • γt (T ′ ) > γ (T ′ ), then T = K2 . (ii) If T does not have a proper induced subgraph T ′ with • δ(T ′ ) ≥ 1, and • γ (T ′ ) ≥ γ (T ), then T = T0 ◦ K1 for some tree T0 . (iii) If γp (T ) > γ (T ) and T does not have a proper induced subgraph T ′ with • δ(T ′ ) ≥ 1, • γ (T ′ ) ≥ γ (T ), and • γp (T ′ ) > γ (T ′ ), then T = T0 ◦ K1 for some tree T0 that does not have a perfect matching. Proof. (i) Let T be a tree with the specified properties. Clearly, no vertex in S (T ) is adjacent to more than one endvertex. A vertex of T is special if it is neither an endvertex nor belongs to S (T ). If T has no special vertex, then T = T0 ◦ K1 for some tree T0 . Note that V (T0 ) is a minimum dominating set of T . If γ (T ) ≥ 2, then γ (T ) = γt (T ), which is a contradiction. If γ (T ) = 1, then T is a star, and the properties of T imply that T is K2 . Hence, we may assume that T has a special vertex. We root T in some special vertex and consider a special vertex s of T with largest distance from the root. Let T1 = T − {s}. Clearly, δ(T1 ) ≥ 1. Note that s is not an endvertex of T and that all children of s belong to S (T ). This implies that T has a minimum dominating set D that contains all children of s and does not contain s. Since D is a dominating set of T1 , this implies γ (T1 ) ≤ γ (T ). Conversely, the graph T1 has a minimum dominating set D1 as well as a minimum total dominating set Dt ,1 such that D1 and Dt ,1 contain all children of s. Since D1 is a dominating set of T and Dt ,1 is a total dominating set of T , we obtain γ (T ) ≤ γ (T1 ) and γt (T ) ≤ γt (T1 ). Altogether, it follows that γ (T ) = γ (T1 ) and hence γt (T1 ) ≥ γt (T ) > γ (T ) = γ (T1 ), which is a contradiction. (ii) Let T be a tree with the specified properties. For a contradiction, we assume that there is no tree T0 with T = T0 ◦ K1 . As in the proof of (i), we obtain that no vertex in S (T ) is adjacent to more than one endvertex, and that T has a special vertex. Choosing s exactly as in the proof of (i), we obtain δ(T − {s}) ≥ 1 and γ (T ) = γ (T − {s}) exactly as above, which is a contradiction. (iii) Let T be a tree with the specified properties. For a contradiction, we assume that there is no tree T0 with T = T0 ◦ K1 such that T0 has no perfect matching. As in the proof of (i), we obtain that no vertex in S (T ) is adjacent to more than one endvertex. If T has no special vertex, then T = T0 ◦ K1 for some tree T0 . By assumption, the tree T0 has a perfect matching. This implies that the set V (T0 ) is a minimum dominating set that is a paired dominating set, which is a contradiction. Hence T has a special vertex. Choosing s exactly as in the proof of (i), we obtain δ(T − {s}) ≥ 1 and γ (T ) = γ (T − {s}) exactly as above. Furthermore, the graph T − {s} has a paired dominating set Dp such that Dp contains all of the children of s. Since Dp is a paired dominating set of T , we obtain γp (T ) ≤ γp (T − {s}), which implies a contradiction as in the proof of (i). This completes the proof. In fact, the reverse implications in Lemma 12 also hold.
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Theorem 6. For some positive integer k, a graph G of girth at least
5k+1 2
1429
belongs to Gk (γ , γt ) if and only if G is Tk (γ , γt )-free.
Proof. Let T be a forest in Tk (γ , γt ). Clearly, δ(T ) ≥ 1, γ (T ) = k, and, since T contains K2 as a component, γt (T ) > γ (T ). It is straightforward to check that every proper induced subgraph T ′ of T satisfies either δ(T ) = 0 or γ (T ) < k. This implies that the graphs in Tk (γ , γt ) are minimal forbidden induced subgraphs for Gk (γ , γt ), which proves the necessity. In order to prove the sufficiency, we assume, for a contradiction, that G is a Tk (γ , γt )-free graph of girth at least 5k2+1 with δ(G) ≥ 1,
γ (G) ≥ k, and γt (G) > γ (G) such that G is of minimum order. If k is odd, then every cycle of length at least 5k2+1 contains the induced subgraph K2 ∪ k−2 1 (K2 ◦ K1 ), which belongs to Tk (γ , γt ). If k is even, then every cycle of length at least 5k2+1 contains the induced subgraph 2K2 ∪ k−2 2 (K2 ◦ K1 ), which belongs to Tk (γ , γt ). By the girth condition, this implies that G is a forest. Let T1 , . . . , Tℓ be the components of G. Since γt (G) > γ (G), there is some positive integer r such that γt (Ti ) > γ (Ti ) for i ∈ [r ] and γt (Ti ) = γ (Ti ) for i ∈ [ℓ]\[r ]. If there is some i ∈ [r ] such that Ti ̸= K2 , then Lemma 12(i) implies the existence of a proper induced subgraph Ti′ of Ti with δ(Ti′ ) ≥ 1, γ (Ti′ ) ≥ γ (Ti ), and γt (Ti′ ) > γ (Ti′ ). Now G′ = Ti′ ∪ j∈[ℓ]\{i} Tj is a Tk (γ , γt )-free graph of girth at least 5k2+1 with δ(G′ ) ≥ 1, γ (G′ ) ≥ γ (G) ≥ k, and γt (G′ ) > γ (G′ ) such that G′ has smaller order than G, which contradicts the choice of G. Hence, we have Ti = K2 for i ∈ [r ]. Similarly, by Lemma 12(ii), we have Ti = Ti,0 ◦ K1 for some tree Ti,0 for i ∈ [ℓ] \ [r ]. Altogether, it follows that G belongs to the class Tk′ (γ , γt ) for k′ = γ (G) ≥ k. Note that removing an endvertex from some Ti,0 or removing a K2 component from G, provided that G has more than one K2 component, results in a graph in Tk′ −1 (γ , γt ). Therefore, G contains a graph in Tk (γ , γt ) as an induced subgraph, which is a contradiction. This completes the proof. For k ∈ {3, 4}, Theorem 6 has the following corollaries that rely on slightly weaker girth conditions. Since both corollaries can be proved the same way, we give details only for the first. Corollary 13. A graph G of girth at least 7 belongs to G3 (γ , γt ) if and only if G is {C7 , 3K2 , K2 ∪ P4 }-free. Proof. It is easy to check that C7 is a minimal forbidden induced subgraph for G3 (γ , γt ) and that T3 (γ , γt ) = {3K2 , K2 ∪ P4 }. Since we have already observed that the graphs in T3 (γ , γt ) are minimal forbidden induced for G3 (γ , γt ), this subgraphs proves the necessity. Since a graph of girth at least 7 that is C7 -free, has girth at least 8 = 5·32+1 , the sufficiency follows immediately from Theorem 6. Corollary 14. A graph G of girth at least 10 belongs to G4 (γ , γt ) if and only if G is {C10 , 4K2 , 2K2 ∪ P4 , K2 ∪ (P3 ◦ K1 )}-free. We proceed to the proof of Theorem 7, which is very similar to the proof of Theorem 6. Note that every tree T without isolated vertices satisfies d(T ) = 2 > 1 = dt (T ). Theorem 7. For some positive integer k, a graph G of girth at least
5k 2
belongs to Gk (d, dt ) if and only if G is Tk -free.
Proof. Let T be a forest in Tk . Clearly, δ(T ) ≥ 1, γ (T ) = k, and d(T ) = 2 > 1 = dt (T ). Again every proper induced subgraph T ′ of T satisfies either δ(T ) = 0 or γ (T ) < k. This implies that the graphs in Tk are minimal forbidden induced subgraphs for Gk (d, dt ), which proves the necessity. In order to prove the sufficiency, we assume, for a contradiction, that G is a Tk -free graph of girth at least 5k with δ(G) ≥ 1, γ (G) ≥ k, and d(G) > dt (G) such that G is of minimum order. If k 2
5k contains the induced subgraph K2 ∪ k−2 1 (K2 ◦ K1 ), which belongs to Tk . If k is 5k2 even, then every cycle of length at least 2 contains the induced subgraph 2k (K2 ◦ K1 ), which belongs to Tk . By the girth condition, this implies that G is a forest. Let T1 , . . . , Tℓ be the components of G. If there is some i ∈ [ℓ] such that there is no tree Ti,0 with Ti = Ti,0 ◦ K1 , then Lemma 12(ii) implies the existence of a proper induced subgraph Ti′ of Ti with δ(Ti′ ) ≥ 1 and γ (Ti′ ) ≥ γ (Ti ). Now G′ = Ti′ ∪ j∈[ℓ]\{i} Tj is a Tk -free graph of girth at least 5k with δ(G′ ) ≥ 1 and γ (G′ ) ≥ γ (G) ≥ k. Since 2 ′ ′ ′ ′ G is a forest without isolated vertices, we have d(G ) > dt (G ). Since G has smaller order than G, we obtain a contradiction to the choice of G. Hence, we have Ti = Ti,0 ◦ K1 for some tree Ti,0 for i ∈ [ℓ]. This implies that G belongs to the class Tk′ for k′ = γ (G) ≥ k. Again it follows that G contains a graph in Tk as an induced subgraph, which is a contradiction. This is odd, then every cycle of length at least
completes the proof.
The proof of Theorem 8 is again very similar to the proof of Theorem 6 and uses Lemma 12(iii) and (ii) instead of Lemma 12(i) and (ii). We leave the details to the reader. We proceed to the hardness results. The proof of our next result relies on a simple modification of the NP-completeness proofs given by Laskar and Pfaff [15] and McRae [17]. Theorem 9. It is NP-complete to decide for a given split graph G with γ (G) = γt (G) = γp (G) and a given integer k, whether γ (G) ≤ k. Proof. Clearly, the considered decision problem is in NP. We describe a reduction from the well known NP-complete problem Exact Cover By 3-Sets, that is, given an instance X of the decision problem Exact Cover By 3-Sets, we will construct a split graph G with γ (G) = γt (G) = γp (G) and an integer k such that the order of G is polynomially bounded in the encoding length of X and X is a ‘‘yes’’-instance if and only if γ (G) ≤ k. Therefore, let (X0 , C0 ) be an instance of Exact Cover By 3-Sets, that is, X0 is a set of order divisible by 3, C0 is a set of 3-element subsets of X0 such that X0 = C ∈C0 C , and (X0 , C0 ) is a ‘‘yes’’-instance if and only if there is a subset D0 of C0 such that X0 = C and 3 | D | = | X | , that is, every element 0 0 C ∈D0
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Fig. 3. The graph G(xi ) created for the variable xi .
of X0 is contained in exactly one of the sets in D0 . Let the instance (X , C ) of Exact Cover By 3-Sets arise from (X0 , C0 ) by adding to X , for every element x of X0 , two distinct elements x and x′ , and by adding to C , for every element {x, y, z } of C0 , the two sets {x, y, z } and {x′ , y′ , z ′ }. Let X = {x1 , . . . , x3n } and C = {C1 , . . . , Cm }. Clearly, the two instances (X0 , C0 ) and (X , C ) are equivalent. We construct a split graph G with γ (G) = γt (G) = γp (G) whose order is polynomially bounded in terms of n and m as well as an integer k such that there is a subset D of C with |D| = n and X = C ∈D C if and only if γ (G) ≤ k. For every element xi of X , we create a vertex xi . For every element Cj of C , we create a vertex Cj . We add edges such that {C1 , . . . , Cm } is a clique. For every C in C with elements x, y, and z, we add the three edges Cx, Cy, and Cz. Let k = n. This completes the construction of the split graph G and the integer k. Since for every xi in X and every Cj in C with xi ∈ Cj , we have NG [xi ] ⊆ NG [Cj ], the graph G has a minimum dominating set D that is a subset of {C1 , . . . , Cm }. Since every vertex Cj has exactly three neighbors in X and n ≥ 2, we have |D| ≥ |X |/3 = n. By the way in which (X , C ) is obtained from (X0 , C0 ), it follows easily that |D| is even. Since {C1 , . . . , Cm } is a clique, the set D is also a paired dominating set, which implies γ (G) = γt (G) = γp (G). If γ (G) ≤ n, then γ (G) = n and D is subset of C with |D| = n and X = C ∈D C . Conversely, if such a subset D exists, then it yields a dominating set of G of order n, which implies γ (G) ≤ n. This completes the proof. Theorem 10. It is NP-hard to decide for a given graph G, whether γ (G) = γt (G). Proof. We describe a reduction from 3Sat. Therefore, let f be a 3Sat instance with clauses C1 , . . . , Cm over the Boolean variables x1 , . . . , xn . We construct a graph G whose order is polynomially bounded in terms of n and m such that f is satisfiable if and only if γ (G) = γt (G). For every variable xi , we create a copy G(xi ) of the graph shown in Fig. 3 and denote its vertices as indicated in the figure. For every clause Cj with literals x, y, and z, we create a copy G(Cj ) of K2 and denote one of its two vertices by uj . Furthermore, we create the three edges uj x, uj y, and uj z. If, for example, C1 = x1 ∨¯x2 ∨¯x3 , then these are the three edges u1 x1 , u1 x¯ 2 , and u1 x¯ 3 . This completes the construction of G. Clearly, the order of G is 5n + 2m. Since {xi : i ∈ [n]} ∪ {yi : i ∈ [n]} ∪ {uj : j ∈ [m]} is a dominating set of G, we have γ (G) ≤ 2n + m. Since every dominating set of G contains at least two vertices from each G(xi ) and at least one vertex from each G(Cj ), we have γ (G) = 2n + m and γt (G) ≥ 2n + m. Furthermore, if γt (G) = 2n + m, then G has a total dominating set Dt that contains exactly two vertices from each G(xi ) and exactly one vertex from each G(Cj ). Note that this implies that Dt ∩ V (G(xi )) ∈ {{xi , yi }, {¯xi , yi }} and Dt ∩ V (G(Cj )) = {uj }. Therefore, if γt (G) = γ (G) = 2n + m, then the intersection of a minimum total dominating set with the graphs G(xi ) indicates a satisfying truth assignment for f . Conversely, if f is satisfiable, we consider a satisfying truth assignment for f . Now {xi : i ∈ [n] and xi is set to true} ∪ {¯xi : i ∈ [n] and xi is set to false} ∪ {yi : i ∈ [n]} ∪ {uj : j ∈ [m]} is a total dominating set of G of order 2n + m, which implies γt (G) = γ (G). This completes the proof. Theorem 11. It is NP-hard to decide for a given graph G, whether 2γ (G) = γp (G). Proof. We describe a reduction from 3Sat. Therefore, let f be a 3Sat instance with clauses C1 , . . . , Cm over the Boolean variables x1 , . . . , xn . We construct a graph G whose order is polynomially bounded in terms of n and m such that f is satisfiable if and only if 2γ (G) = γp (G). For every variable xi , we create a copy G(xi ) of K3 and denote two vertices of G(xi ) by xi and x¯ i . For every clause Cj with literals x, y, and z, we create a vertex Cj and the three edges Cj x, Cj y, and Cj z. This completes the construction of G. Clearly, the order of G is 3n + m. Since {xi : i ∈ [n]} ∪ {¯xi : i ∈ [n]} is a paired dominating set of G, we have γp (G) ≤ 2n. Since every dominating set of G contains a vertex from each G(xi ), it follows that γ (G) ≥ n. For a minimum paired dominating set of G, this easily implies that γp (G) ≥ 2n, that is, γp (G) = 2n. If γ (G) = n, then G has a minimum dominating set D that contains exactly one vertex from each G(xi ). Therefore, if γ (G) = n, then a minimum dominating set of G indicates a satisfying truth assignment for f . Conversely, if f is satisfiable, we consider a satisfying truth assignment for f . Now {xi : i ∈ [n] and xi is set to true} ∪ {¯xi : i ∈ [n] and xi is set to false} is a dominating set of G of order n, which implies γ (G) = n. This completes the proof. We pose two conjectures concerning hardness results. Conjecture 15. It is NP-hard to decide for a given graph G whether γ (G) = γp (G). Conjecture 16. It is NP-hard to decide for a given graph G whether 2γ (G) = γt (G). It seems a challenging problem to prove variants of Theorems 6, 7, and 8 without a girth condition. Furthermore, one can study classes derived from the inequality dt (G) ≥ context.
d(G) 2
, and one might consider the paired domatic number [10] in this
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Acknowledgments José Alvarado and Simone Dantas were partially supported by FAPERJ, CNPq, and CAPES. Dieter Rautenbach was supported by CAPES and DAAD (process number BEX 11620/13-7). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22]
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Perfectly relating the domination, total domination, and paired domination numbers of a graph José D. Alvarado a , Simone Dantas a , Dieter Rautenbach b,∗ a b
Instituto de Matemática e Estatística, Universidade Federal Fluminense Niterói, Brazil Institute of Optimization and Operations Research, Ulm University, Ulm, Germany
article
info
Article history: Received 30 May 2014 Received in revised form 6 March 2015 Accepted 8 March 2015
Keywords: Domination Total domination Paired domination Domatic number Total domatic number
abstract The domination number γ (G), the total domination number γt (G), the paired domination number γp (G), the domatic number d(G), and the total domatic number dt (G) of a graph G without isolated vertices are related by trivial inequalities γ (G) ≤ γt (G) ≤ γp (G) ≤ 2γ (G) and dt (G) ≤ d(G). Very little is known about the graphs that satisfy one of these inequalities with equality. We study classes of graphs defined by requiring equality in one of the above inequalities for all induced subgraphs that have no isolated vertices and whose domination number is not too small. Our results are characterizations of several such classes in terms of their minimal forbidden induced subgraphs. Furthermore, we prove some hardness results, which suggest that the extremal graphs for some of the above inequalities do not have a simple structure. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Whenever two graph parameters, say ν and τ , are related by a simple inequality, say ν(G) ≤ τ (G) for every graph G, but the graphs G with ν(G) = τ (G) do not have a simple structure, it makes sense to study the class of the so-called (ν, τ )-perfect graphs, the graphs for which every induced subgraph H satisfies ν(H ) = τ (H ). Famous examples of this approach are the classical perfect graphs [3], where the two parameters are the clique number and the chromatic number, or the domination perfect graphs [22], where the two parameters are the domination number and the independent domination number. In the present paper we consider classes of (ν, τ )-perfect graphs, where ν and τ are well known domination parameters [9]. We consider finite, simple, and undirected graphs and use standard terminology and notation. Let D be a set of vertices of some graph G. The set D is a dominating set of G if every vertex of G that does not belong to D, has a neighbor in D. The set D is a total dominating set of G if it is dominating and the subgraph G[D] of G induced by D does not have any isolated vertices. Finally, the set D is a paired dominating set of G if it is dominating and the graph G[D] has a perfect matching. The domination number γ (G) [9], the total domination number γt (G) [5,14], and the paired domination number γp (G) [10,16] of G are the minimum cardinalities of a dominating, a total dominating, and a paired dominating set of G, respectively. These definitions immediately imply [1,9,10]
γ (G) ≤ γt (G) ≤ γp (G) for every graph G for which γp (G) is defined.
∗
Corresponding author. E-mail addresses: [email protected] (J.D. Alvarado), [email protected] (S. Dantas), [email protected] (D. Rautenbach).
http://dx.doi.org/10.1016/j.disc.2015.03.014 0012-365X/© 2015 Elsevier B.V. All rights reserved.
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Fig. 1. The graphs C5 , P5 , and N.
For a set D of vertices of a graph G and a vertex u in D, let PG (D, u) = {v ∈ V (G) \ D : NG (v) ∩ D = {u}}. The vertices in PG (D, u) are the external private neighbors of u (in G with respect to D). It is known that every graph G without isolated vertices has a minimum dominating set D such that, for every vertex u in D, the set PG (D, u) is non-empty [1]. Therefore, adding to such a dominating set D, for every vertex u in D, an external private neighbor of u, results in a paired dominating set of G, which implies
γp (G) ≤ 2γ (G). Only very little is known about the graphs that satisfy one of the above inequalities with equality. The trees T that satisfy γ (T ) = γt (T ) [8], or γt (T ) = γp (T ) [12,20], or 2γ (T ) = γt (T ) [11], or 2γ (T ) = γp (T ) [13] have been characterized constructively. Examples in [6] show that even very strong structural restrictions do not allow the strengthening of the inequality γt (G) ≤ 2γ (G). The relation between the total domination number and the paired domination number as well as paired dominating sets of a special structure were studied in [2,10,19,18]. The domatic number d(G) of a graph G [5,21] is the maximum integer k such that the vertex set V (G) of G can be partitioned into k dominating sets. The total domatic number dt (G) [4] and the paired domatic number dp (G) of G [10] are defined analogously. The definitions immediately imply [21] dp (G) ≤ dt (G) ≤ d(G) for every graph G without isolated vertices. Furthermore, since the union of at least two disjoint dominating sets is a total dominating set [21], we have dt (G) ≥
d(G)
2
for every graph G without isolated vertices. Having collected the relevant inequalities, we now consider some reasonable classes of ‘‘perfect’’ graphs. Since the total domination number and the paired domination number are only defined for graphs without isolated vertices, it does not make sense to require equalities involving these parameters for all induced subgraphs but only for those that have no isolated vertices. Therefore, if ν and τ are graph parameters with ν(G) ≤ τ (G) for every graph G, then let
G(ν, τ ) = {G : ∀H ⊆ind G : δ(H ) ≥ 1 ⇒ ν(H ) = τ (H )} where H ⊆ind G indicates that H is an induced subgraph of G. Our first results are the following. Note that all proofs are postponed to Section 2. Theorem 1. A graph G belongs to G(γt , γp ) if and only if G is {C5 , P5 , N }-free. Theorem 2. A graph G belongs to G(γt , 2γ ) if and only if G belongs to G(γp , 2γ ) if and only if G is {C4 , P4 }-free. See Fig. 1 for an illustration of the four special graphs in Theorem 1. In general, for a positive integer n, we denote the complete graph, the cycle, and the path of order n by Kn , Cn , and Pn , respectively. Note that K2 is a forbidden induced subgraph for graphs in G(γ , γt ), G(d, dt ), and G(γ , γp ). Therefore, these three classes of graphs consist exactly of all edgeless graphs. In view of this observation, it makes sense to restrict the considered induced subgraphs further. Therefore, for a positive integer k, we consider the class
Gk (ν, τ ) = {G : ∀H ⊆ind G : (δ(H ) ≥ 1 and γ (H ) ≥ k) ⇒ ν(H ) = τ (H )} . Note that G(ν, τ ) coincides with G1 (ν, τ ). For k = 2, we obtain the following, where 2K2 denotes the disjoint union of two copies of K2 . Theorem 3. A graph G belongs to G2 (γ , γt ) if and only if G is {C5 , 2K2 }-free. Theorem 4. A graph G belongs to G2 (d, dt ) if and only if G is {P4 , 2K2 , G1 , G2 }-free. Theorem 5. A graph G belongs to G2 (γ , γp ) if and only if G is {C5 , 2K2 , N }-free. See Fig. 2 for an illustration of the graphs G1 and G2 in Theorem 4.
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Fig. 2. Two graphs G1 and G2 with δ(Gi ) ≥ 1, γ (Gi ) ≥ 2, and d(Gi ) > dt (Gi ) for i ∈ {1, 2}.
For arbitrary values of k, we obtain results subject to girth conditions. In order to phrase our results, we introduce some notation. The order of a graph G is the number of its vertices and is denoted by n(G). If G is a graph, then the graph G ◦ K1 arises from G by adding, for every vertex u of G, a new vertex u′ together with the new edge uu′ . Let Tk be the set of forests of the form ℓ ℓ i=1 (Ti,0 ◦ K1 ) where ℓ is a positive integer, and T1,0 , . . . , Tℓ,0 are non-empty trees with k = i=1 n(Ti,0 ). Let Tk (γ , γt ) be the set containing all forests in Tk for which one of the trees Ti,0 is K1 , that is, Tk (γ , γt ) consists of all forests in Tk that have K2 as a component. Let Tk (γ , γp ) be the set containing all forests T in Tk for which one of the trees Ti,0 has no perfect matching. Our results are as follows. Theorem 6. For some positive integer k, a graph G of girth at least
5k+1
Theorem 7. For some positive integer k, a graph G of girth at least
5k
Theorem 8. For some positive integer k, a graph G of girth at least
5k+1
2
2
2
belongs to Gk (γ , γt ) if and only if G is Tk (γ , γt )-free.
belongs to Gk (d, dt ) if and only if G is Tk -free. belongs to Gk (γ , γp ) if and only if G is Tk (γ , γp )-free.
Finally, we establish some hardness results, which yield further motivation to consider the above classes. Our first hardness result implies that even in very restricted classes of graphs, requiring equality of the domination number and the paired domination number does not make the dominating set problem easy. A split graph is a graph whose vertex set can be partitioned into an independent set and a clique. Theorem 9. It is NP-complete to decide for a given split graph G with γ (G) = γt (G) = γp (G) and a given integer k, whether γ (G) ≤ k. The next two results imply that the graphs G with γ (G) = γt (G) or 2γ (G) = γp (G) do most likely not have a simple description. Theorem 10. It is NP-hard to decide for a given graph G, whether γ (G) = γt (G). Theorem 11. It is NP-hard to decide for a given graph G, whether 2γ (G) = γp (G). Section 2 is devoted to the proofs, auxiliary lemmas, corollaries, conjectures, and possible directions for further research. For k ∈ N, let [k] denote the set of the positive integers at most k. 2. Proofs, lemma, corollaries, and conjectures For convenience we restate the results. Theorem 1. A graph G belongs to G(γt , γp ) if and only if G is {C5 , P5 , N }-free. Proof. It is easy to check that the four graphs C5 , P5 , and N are minimal forbidden induced subgraphs for G(γt , γp ), which proves the necessity. In order to prove the sufficiency, we show that a {C5 , P5 , N }-free graph G with δ(G) ≥ 1 satisfies γt (G) = γp (G). For a contradiction, we assume that γt (G) < γp (G). Let Dt be a minimum total dominating set of G. First we assume that some component K of G[Dt ] is not complete. Let P : u0 . . . uℓ be a shortest path of maximum length in K . Since K is not complete, we have ℓ ≥ 2. Furthermore, by the choice of P, no vertex in K − {u0 } or K − {uℓ } is isolated. This implies that PG (Dt , u0 ) contains a vertex v0 and PG (Dt , uℓ ) contains a vertex vℓ . If ℓ = 2, then G[{u0 , u1 , u2 , v0 , v2 }] is either P5 or C5 , which is a contradiction. Hence, we have ℓ ≥ 3, and G[{v0 , u0 , u1 , u2 , u3 }] is P5 , which is a contradiction. Therefore, we may assume that every component of G[Dt ] is complete. Since Dt is not a paired dominating set, some component L of G[Dt ] has odd order at least 3. Let u, v , and w be three vertices from L. Since L is complete, the three sets PG (Dt , u), PG (Dt , v), and PG (Dt , w) are non-empty. If some vertex u′ in PG (Dt , u) is adjacent to all vertices in PG (Dt , v), then D′t = (Dt \ {v}) ∪ {u′ } is a minimum total dominating set of G. Since u′ and w are not adjacent but lie in the same component of G[D′t ], some component of G[D′t ] is not complete, and we can argue as above. Therefore, by symmetry, we may assume that there are vertices v ′ in PG (Dt , v) and w ′ in PG (Dt , w) such that u′ is not adjacent to v ′ or w ′ . If v ′ w ′ ̸∈ E (G), then G[{u, v, w, u′ , v ′ , w ′ }] is N, and if v ′ w ′ ∈ E (G), then G[{u, u′ , v ′ , w ′ }] is 2K2 , which is a contradiction. This completes the proof.
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Theorem 2. A graph G belongs to G(γt , 2γ ) if and only if G belongs to G(γp , 2γ ) if and only if G is {C4 , P4 }-free. Proof. Since γt (G) ≤ γp (G) ≤ 2γ (G) for every graph G without isolated vertices, G(γt , 2γ ) is a subset of G(γp , 2γ ). Since C4 and P4 are minimal forbidden induced subgraphs for G(γp , 2γ ), every graph in G(γp , 2γ ) is {C4 , P4 }-free. In order to complete the proof, it suffices to show that a connected {C4 , P4 }-free graph G with δ(G) ≥ 1 satisfies γt (G) = 2γ (G). For a contradiction, we assume that γt (G) < 2γ (G). Since G is P4 -free, the vertex set of G has a partition into two sets V1 and V2 such that every vertex in V1 is adjacent to every vertex in V2 . Since a set containing one vertex from V1 and one vertex from V2 is a total dominating set of G, we have γt (G) = 2. Since γt (G) < 2γ (G), this implies γ (G) > 1. Hence V1 contains two non-adjacent vertices u1 and v1 , and V2 contains two non-adjacent vertices u2 and v2 . Now G[{u1 , v1 , u2 , v2 }] is C4 , which is a contradiction. This completes the proof. Theorem 3. A graph G belongs to G2 (γ , γt ) if and only if G is {C5 , 2K2 }-free. Proof. It is easy to check that the two graphs C5 and 2K2 are minimal forbidden induced subgraphs for G2 (γ , γt ), which proves the necessity. In order to prove the sufficiency, we show that a {C5 , 2K2 }-free graph G with δ(G) ≥ 1 and γ (G) ≥ 2 satisfies γ (G) = γt (G). For a contradiction, we assume that γ (G) < γt (G). Since G is 2K2 -free and has no isolated vertex, it is connected. We choose a minimum dominating set D of G such that f (D) = u∈D distG (u, D \ {u}) is minimum, where distG (u, D \ {u}) is the minimum number of edges in a path in G between u and a vertex in D \ {u}. Since |D| ≥ 2 and G is connected, we have 0 < f (D) < ∞. Let u ∈ D be such that distG (u, D \ {u}) is maximum and let v ∈ D \ {u} be such that distG (u, v) = distG (u, D \ {u}), that is, G contains a path of length distG (u, v) between u and v . Since D is not a total dominating set of G, we have distG (u, v) ≥ 2. If distG (u, v) ≥ 4, then a vertex at distance 2 from u on a shortest path between u and v has no neighbor in D, which is a contradiction. Hence, we have distG (u, v) ∈ {2, 3}. First, we assume that distG (u, v) = 2. Let x be a common neighbor of u and v . Let D′ = (D \{u})∪{x}. By the choice of D, we obtain that either D′ is no dominating set of G or there is some w in D \{u, v} such that distG (w, D′ \{w}) > distG (w, D \{w}). In both cases, there is a neighbor u′ of u distinct from x such that u′ is not adjacent to v or x. By symmetry, there is a neighbor v ′ of v distinct from x such that v ′ is not adjacent to u or x. If u′ is not adjacent to v ′ , then G[{u, v, u′ , v ′ }] is 2K2 , which is a contradiction. If u′ is adjacent to v ′ , then G[{u, v, x, u′ , v ′ }] is C5 , which is a contradiction. Hence, we have distG (u, v) = 3. Let uxyv be a path in G. Let D′′ = (D \ {u, v}) ∪ {x, y}. By the choice of D, we obtain that either D′′ is no dominating set of G or there is some w in D \ {u, v} such that distG (w, D′′ \ {w}) > distG (w, D \ {w}). In both cases, there is a vertex z such that z is a neighbor of u or v and z is not adjacent to x or y. If z is adjacent to u but not to v , then G[{z , u, y, v}] is 2K2 , which is a contradiction. If z is adjacent to v but not to u, then G[{u, x, v, z }] is 2K2 , which is a contradiction. If z is adjacent to u and v , then G[{u, v, x, y, z }] is C5 , which is a contradiction. This completes the proof. Theorem 4. A graph G belongs to G2 (d, dt ) if and only if G is {P4 , 2K2 , G1 , G2 }-free. Proof. It is easy to check that the four graphs P4 , 2K2 , G1 , and G2 are minimal forbidden induced subgraphs for G2 (d, dt ), which proves the necessity. In order to prove the sufficiency, we show that a {P4 , 2K2 , G1 , G2 }-free graph G with δ(G) ≥ 1 and γ (G) ≥ 2 satisfies d(G) = dt (G). For a contradiction, we assume that d(G) > dt (G). Since the graph G has no isolated vertex and is 2K2 -free, it is connected. Let the positive integer k be maximal such that there is a partition of the vertex set k−1 V (G) of G into k non-empty sets V1 , . . . , Vk−1 , and V≥k such that |V
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assume that w1 = w2 , in which case G[{u1 , u2 , v1 , v2 , w1 , x}] is either G1 if v1 is not adjacent to v2 or G2 otherwise, which is a contradiction. This completes the proof. Theorem 5. A graph G belongs to G2 (γ , γp ) if and only if G is {C5 , 2K2 , N }-free. Proof. It is easy to check that the four graphs C5 , 2K2 , and N are minimal forbidden induced subgraphs for G2 (γ , γp ), which proves the necessity. In order to prove the sufficiency, we show that a {C5 , 2K2 , N }-free graph G with δ(G) ≥ 1 and γ (G) ≥ 2 satisfies γ (G) = γp (G). For a contradiction, we assume that γ (G) < γp (G). Let Dt be a minimum total dominating set of G. Since G is {C5 , 2K2 }-free, Theorem 3 implies that Dt is also a minimum dominating set of G. Since γ (G) < γp (G), the set Dt is no paired dominating set of G. Furthermore, since every vertex in Dt has a neighbor in Dt , it follows that, for every vertex u in Dt , the set PG (Dt , u) is non-empty. If G[Dt ] is not complete, then it contains an induced path uvw of order 3. Let u′ ∈ PG (Dt , u) and w ′ ∈ PG (Dt , w). Since G is 2K2 -free, the vertices u′ and w ′ are adjacent, and G[{u′ , u, v, w, w ′ }] is C5 , which is a contradiction. Hence, G[Dt ] is complete. Since Dt is not a paired dominating set, the set Dt has odd order at least 3. Let u, v , and w be three vertices in Dt . Let u′ in PG (Dt , u), v ′ in PG (Dt , v), and w ′ in PG (Dt , w). If u′ and v ′ are not adjacent, then, since G is {2K2 , N }-free, the vertex w′ is adjacent to u′ and v ′ , and G[{u′ , u, v, v ′ , w′ }] is C5 , which is a contradiction. Hence, we may assume that for every two distinct vertices x and y in {u, v, w}, every vertex in PG (Dt , x) is adjacent to every vertex in PG (Dt , y). Since (Dt \{v, w})∪{u′ } is not a dominating set of G, there is some vertex x of G with NG (x) ∩ {u, v, w, u′ } = {v, w}. If x is not adjacent to v ′ , then G[{u′ , v ′ , w, x}] is 2K2 , which is a contradiction. Hence, the vertex x is adjacent to v ′ . Now G[{u′ , v ′ , x, w, u}] is C5 , which is a contradiction. This completes the proof. In order to prove Theorems 6, 7, and 8 we need a lemma. An endvertex of a graph is a vertex of degree at most 1. For a graph G, let S (G) be the set of vertices of G that are adjacent to an endvertex of G. The vertices in S (G) are known as support vertices. Lemma 12. Let T be a tree with δ(T ) ≥ 1. (i) If γt (T ) > γ (T ) and T does not have a proper induced subgraph T ′ with • δ(T ′ ) ≥ 1, • γ (T ′ ) ≥ γ (T ), and • γt (T ′ ) > γ (T ′ ), then T = K2 . (ii) If T does not have a proper induced subgraph T ′ with • δ(T ′ ) ≥ 1, and • γ (T ′ ) ≥ γ (T ), then T = T0 ◦ K1 for some tree T0 . (iii) If γp (T ) > γ (T ) and T does not have a proper induced subgraph T ′ with • δ(T ′ ) ≥ 1, • γ (T ′ ) ≥ γ (T ), and • γp (T ′ ) > γ (T ′ ), then T = T0 ◦ K1 for some tree T0 that does not have a perfect matching. Proof. (i) Let T be a tree with the specified properties. Clearly, no vertex in S (T ) is adjacent to more than one endvertex. A vertex of T is special if it is neither an endvertex nor belongs to S (T ). If T has no special vertex, then T = T0 ◦ K1 for some tree T0 . Note that V (T0 ) is a minimum dominating set of T . If γ (T ) ≥ 2, then γ (T ) = γt (T ), which is a contradiction. If γ (T ) = 1, then T is a star, and the properties of T imply that T is K2 . Hence, we may assume that T has a special vertex. We root T in some special vertex and consider a special vertex s of T with largest distance from the root. Let T1 = T − {s}. Clearly, δ(T1 ) ≥ 1. Note that s is not an endvertex of T and that all children of s belong to S (T ). This implies that T has a minimum dominating set D that contains all children of s and does not contain s. Since D is a dominating set of T1 , this implies γ (T1 ) ≤ γ (T ). Conversely, the graph T1 has a minimum dominating set D1 as well as a minimum total dominating set Dt ,1 such that D1 and Dt ,1 contain all children of s. Since D1 is a dominating set of T and Dt ,1 is a total dominating set of T , we obtain γ (T ) ≤ γ (T1 ) and γt (T ) ≤ γt (T1 ). Altogether, it follows that γ (T ) = γ (T1 ) and hence γt (T1 ) ≥ γt (T ) > γ (T ) = γ (T1 ), which is a contradiction. (ii) Let T be a tree with the specified properties. For a contradiction, we assume that there is no tree T0 with T = T0 ◦ K1 . As in the proof of (i), we obtain that no vertex in S (T ) is adjacent to more than one endvertex, and that T has a special vertex. Choosing s exactly as in the proof of (i), we obtain δ(T − {s}) ≥ 1 and γ (T ) = γ (T − {s}) exactly as above, which is a contradiction. (iii) Let T be a tree with the specified properties. For a contradiction, we assume that there is no tree T0 with T = T0 ◦ K1 such that T0 has no perfect matching. As in the proof of (i), we obtain that no vertex in S (T ) is adjacent to more than one endvertex. If T has no special vertex, then T = T0 ◦ K1 for some tree T0 . By assumption, the tree T0 has a perfect matching. This implies that the set V (T0 ) is a minimum dominating set that is a paired dominating set, which is a contradiction. Hence T has a special vertex. Choosing s exactly as in the proof of (i), we obtain δ(T − {s}) ≥ 1 and γ (T ) = γ (T − {s}) exactly as above. Furthermore, the graph T − {s} has a paired dominating set Dp such that Dp contains all of the children of s. Since Dp is a paired dominating set of T , we obtain γp (T ) ≤ γp (T − {s}), which implies a contradiction as in the proof of (i). This completes the proof. In fact, the reverse implications in Lemma 12 also hold.
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Theorem 6. For some positive integer k, a graph G of girth at least
5k+1 2
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belongs to Gk (γ , γt ) if and only if G is Tk (γ , γt )-free.
Proof. Let T be a forest in Tk (γ , γt ). Clearly, δ(T ) ≥ 1, γ (T ) = k, and, since T contains K2 as a component, γt (T ) > γ (T ). It is straightforward to check that every proper induced subgraph T ′ of T satisfies either δ(T ) = 0 or γ (T ) < k. This implies that the graphs in Tk (γ , γt ) are minimal forbidden induced subgraphs for Gk (γ , γt ), which proves the necessity. In order to prove the sufficiency, we assume, for a contradiction, that G is a Tk (γ , γt )-free graph of girth at least 5k2+1 with δ(G) ≥ 1,
γ (G) ≥ k, and γt (G) > γ (G) such that G is of minimum order. If k is odd, then every cycle of length at least 5k2+1 contains the induced subgraph K2 ∪ k−2 1 (K2 ◦ K1 ), which belongs to Tk (γ , γt ). If k is even, then every cycle of length at least 5k2+1 contains the induced subgraph 2K2 ∪ k−2 2 (K2 ◦ K1 ), which belongs to Tk (γ , γt ). By the girth condition, this implies that G is a forest. Let T1 , . . . , Tℓ be the components of G. Since γt (G) > γ (G), there is some positive integer r such that γt (Ti ) > γ (Ti ) for i ∈ [r ] and γt (Ti ) = γ (Ti ) for i ∈ [ℓ]\[r ]. If there is some i ∈ [r ] such that Ti ̸= K2 , then Lemma 12(i) implies the existence of a proper induced subgraph Ti′ of Ti with δ(Ti′ ) ≥ 1, γ (Ti′ ) ≥ γ (Ti ), and γt (Ti′ ) > γ (Ti′ ). Now G′ = Ti′ ∪ j∈[ℓ]\{i} Tj is a Tk (γ , γt )-free graph of girth at least 5k2+1 with δ(G′ ) ≥ 1, γ (G′ ) ≥ γ (G) ≥ k, and γt (G′ ) > γ (G′ ) such that G′ has smaller order than G, which contradicts the choice of G. Hence, we have Ti = K2 for i ∈ [r ]. Similarly, by Lemma 12(ii), we have Ti = Ti,0 ◦ K1 for some tree Ti,0 for i ∈ [ℓ] \ [r ]. Altogether, it follows that G belongs to the class Tk′ (γ , γt ) for k′ = γ (G) ≥ k. Note that removing an endvertex from some Ti,0 or removing a K2 component from G, provided that G has more than one K2 component, results in a graph in Tk′ −1 (γ , γt ). Therefore, G contains a graph in Tk (γ , γt ) as an induced subgraph, which is a contradiction. This completes the proof. For k ∈ {3, 4}, Theorem 6 has the following corollaries that rely on slightly weaker girth conditions. Since both corollaries can be proved the same way, we give details only for the first. Corollary 13. A graph G of girth at least 7 belongs to G3 (γ , γt ) if and only if G is {C7 , 3K2 , K2 ∪ P4 }-free. Proof. It is easy to check that C7 is a minimal forbidden induced subgraph for G3 (γ , γt ) and that T3 (γ , γt ) = {3K2 , K2 ∪ P4 }. Since we have already observed that the graphs in T3 (γ , γt ) are minimal forbidden induced for G3 (γ , γt ), this subgraphs proves the necessity. Since a graph of girth at least 7 that is C7 -free, has girth at least 8 = 5·32+1 , the sufficiency follows immediately from Theorem 6. Corollary 14. A graph G of girth at least 10 belongs to G4 (γ , γt ) if and only if G is {C10 , 4K2 , 2K2 ∪ P4 , K2 ∪ (P3 ◦ K1 )}-free. We proceed to the proof of Theorem 7, which is very similar to the proof of Theorem 6. Note that every tree T without isolated vertices satisfies d(T ) = 2 > 1 = dt (T ). Theorem 7. For some positive integer k, a graph G of girth at least
5k 2
belongs to Gk (d, dt ) if and only if G is Tk -free.
Proof. Let T be a forest in Tk . Clearly, δ(T ) ≥ 1, γ (T ) = k, and d(T ) = 2 > 1 = dt (T ). Again every proper induced subgraph T ′ of T satisfies either δ(T ) = 0 or γ (T ) < k. This implies that the graphs in Tk are minimal forbidden induced subgraphs for Gk (d, dt ), which proves the necessity. In order to prove the sufficiency, we assume, for a contradiction, that G is a Tk -free graph of girth at least 5k with δ(G) ≥ 1, γ (G) ≥ k, and d(G) > dt (G) such that G is of minimum order. If k 2
5k contains the induced subgraph K2 ∪ k−2 1 (K2 ◦ K1 ), which belongs to Tk . If k is 5k2 even, then every cycle of length at least 2 contains the induced subgraph 2k (K2 ◦ K1 ), which belongs to Tk . By the girth condition, this implies that G is a forest. Let T1 , . . . , Tℓ be the components of G. If there is some i ∈ [ℓ] such that there is no tree Ti,0 with Ti = Ti,0 ◦ K1 , then Lemma 12(ii) implies the existence of a proper induced subgraph Ti′ of Ti with δ(Ti′ ) ≥ 1 and γ (Ti′ ) ≥ γ (Ti ). Now G′ = Ti′ ∪ j∈[ℓ]\{i} Tj is a Tk -free graph of girth at least 5k with δ(G′ ) ≥ 1 and γ (G′ ) ≥ γ (G) ≥ k. Since 2 ′ ′ ′ ′ G is a forest without isolated vertices, we have d(G ) > dt (G ). Since G has smaller order than G, we obtain a contradiction to the choice of G. Hence, we have Ti = Ti,0 ◦ K1 for some tree Ti,0 for i ∈ [ℓ]. This implies that G belongs to the class Tk′ for k′ = γ (G) ≥ k. Again it follows that G contains a graph in Tk as an induced subgraph, which is a contradiction. This is odd, then every cycle of length at least
completes the proof.
The proof of Theorem 8 is again very similar to the proof of Theorem 6 and uses Lemma 12(iii) and (ii) instead of Lemma 12(i) and (ii). We leave the details to the reader. We proceed to the hardness results. The proof of our next result relies on a simple modification of the NP-completeness proofs given by Laskar and Pfaff [15] and McRae [17]. Theorem 9. It is NP-complete to decide for a given split graph G with γ (G) = γt (G) = γp (G) and a given integer k, whether γ (G) ≤ k. Proof. Clearly, the considered decision problem is in NP. We describe a reduction from the well known NP-complete problem Exact Cover By 3-Sets, that is, given an instance X of the decision problem Exact Cover By 3-Sets, we will construct a split graph G with γ (G) = γt (G) = γp (G) and an integer k such that the order of G is polynomially bounded in the encoding length of X and X is a ‘‘yes’’-instance if and only if γ (G) ≤ k. Therefore, let (X0 , C0 ) be an instance of Exact Cover By 3-Sets, that is, X0 is a set of order divisible by 3, C0 is a set of 3-element subsets of X0 such that X0 = C ∈C0 C , and (X0 , C0 ) is a ‘‘yes’’-instance if and only if there is a subset D0 of C0 such that X0 = C and 3 | D | = | X | , that is, every element 0 0 C ∈D0
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Fig. 3. The graph G(xi ) created for the variable xi .
of X0 is contained in exactly one of the sets in D0 . Let the instance (X , C ) of Exact Cover By 3-Sets arise from (X0 , C0 ) by adding to X , for every element x of X0 , two distinct elements x and x′ , and by adding to C , for every element {x, y, z } of C0 , the two sets {x, y, z } and {x′ , y′ , z ′ }. Let X = {x1 , . . . , x3n } and C = {C1 , . . . , Cm }. Clearly, the two instances (X0 , C0 ) and (X , C ) are equivalent. We construct a split graph G with γ (G) = γt (G) = γp (G) whose order is polynomially bounded in terms of n and m as well as an integer k such that there is a subset D of C with |D| = n and X = C ∈D C if and only if γ (G) ≤ k. For every element xi of X , we create a vertex xi . For every element Cj of C , we create a vertex Cj . We add edges such that {C1 , . . . , Cm } is a clique. For every C in C with elements x, y, and z, we add the three edges Cx, Cy, and Cz. Let k = n. This completes the construction of the split graph G and the integer k. Since for every xi in X and every Cj in C with xi ∈ Cj , we have NG [xi ] ⊆ NG [Cj ], the graph G has a minimum dominating set D that is a subset of {C1 , . . . , Cm }. Since every vertex Cj has exactly three neighbors in X and n ≥ 2, we have |D| ≥ |X |/3 = n. By the way in which (X , C ) is obtained from (X0 , C0 ), it follows easily that |D| is even. Since {C1 , . . . , Cm } is a clique, the set D is also a paired dominating set, which implies γ (G) = γt (G) = γp (G). If γ (G) ≤ n, then γ (G) = n and D is subset of C with |D| = n and X = C ∈D C . Conversely, if such a subset D exists, then it yields a dominating set of G of order n, which implies γ (G) ≤ n. This completes the proof. Theorem 10. It is NP-hard to decide for a given graph G, whether γ (G) = γt (G). Proof. We describe a reduction from 3Sat. Therefore, let f be a 3Sat instance with clauses C1 , . . . , Cm over the Boolean variables x1 , . . . , xn . We construct a graph G whose order is polynomially bounded in terms of n and m such that f is satisfiable if and only if γ (G) = γt (G). For every variable xi , we create a copy G(xi ) of the graph shown in Fig. 3 and denote its vertices as indicated in the figure. For every clause Cj with literals x, y, and z, we create a copy G(Cj ) of K2 and denote one of its two vertices by uj . Furthermore, we create the three edges uj x, uj y, and uj z. If, for example, C1 = x1 ∨¯x2 ∨¯x3 , then these are the three edges u1 x1 , u1 x¯ 2 , and u1 x¯ 3 . This completes the construction of G. Clearly, the order of G is 5n + 2m. Since {xi : i ∈ [n]} ∪ {yi : i ∈ [n]} ∪ {uj : j ∈ [m]} is a dominating set of G, we have γ (G) ≤ 2n + m. Since every dominating set of G contains at least two vertices from each G(xi ) and at least one vertex from each G(Cj ), we have γ (G) = 2n + m and γt (G) ≥ 2n + m. Furthermore, if γt (G) = 2n + m, then G has a total dominating set Dt that contains exactly two vertices from each G(xi ) and exactly one vertex from each G(Cj ). Note that this implies that Dt ∩ V (G(xi )) ∈ {{xi , yi }, {¯xi , yi }} and Dt ∩ V (G(Cj )) = {uj }. Therefore, if γt (G) = γ (G) = 2n + m, then the intersection of a minimum total dominating set with the graphs G(xi ) indicates a satisfying truth assignment for f . Conversely, if f is satisfiable, we consider a satisfying truth assignment for f . Now {xi : i ∈ [n] and xi is set to true} ∪ {¯xi : i ∈ [n] and xi is set to false} ∪ {yi : i ∈ [n]} ∪ {uj : j ∈ [m]} is a total dominating set of G of order 2n + m, which implies γt (G) = γ (G). This completes the proof. Theorem 11. It is NP-hard to decide for a given graph G, whether 2γ (G) = γp (G). Proof. We describe a reduction from 3Sat. Therefore, let f be a 3Sat instance with clauses C1 , . . . , Cm over the Boolean variables x1 , . . . , xn . We construct a graph G whose order is polynomially bounded in terms of n and m such that f is satisfiable if and only if 2γ (G) = γp (G). For every variable xi , we create a copy G(xi ) of K3 and denote two vertices of G(xi ) by xi and x¯ i . For every clause Cj with literals x, y, and z, we create a vertex Cj and the three edges Cj x, Cj y, and Cj z. This completes the construction of G. Clearly, the order of G is 3n + m. Since {xi : i ∈ [n]} ∪ {¯xi : i ∈ [n]} is a paired dominating set of G, we have γp (G) ≤ 2n. Since every dominating set of G contains a vertex from each G(xi ), it follows that γ (G) ≥ n. For a minimum paired dominating set of G, this easily implies that γp (G) ≥ 2n, that is, γp (G) = 2n. If γ (G) = n, then G has a minimum dominating set D that contains exactly one vertex from each G(xi ). Therefore, if γ (G) = n, then a minimum dominating set of G indicates a satisfying truth assignment for f . Conversely, if f is satisfiable, we consider a satisfying truth assignment for f . Now {xi : i ∈ [n] and xi is set to true} ∪ {¯xi : i ∈ [n] and xi is set to false} is a dominating set of G of order n, which implies γ (G) = n. This completes the proof. We pose two conjectures concerning hardness results. Conjecture 15. It is NP-hard to decide for a given graph G whether γ (G) = γp (G). Conjecture 16. It is NP-hard to decide for a given graph G whether 2γ (G) = γt (G). It seems a challenging problem to prove variants of Theorems 6, 7, and 8 without a girth condition. Furthermore, one can study classes derived from the inequality dt (G) ≥ context.
d(G) 2
, and one might consider the paired domatic number [10] in this
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Acknowledgments José Alvarado and Simone Dantas were partially supported by FAPERJ, CNPq, and CAPES. Dieter Rautenbach was supported by CAPES and DAAD (process number BEX 11620/13-7). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22]
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