- Email: [email protected]
Domination versus independent domination in graphs of small regularity
Discrete Mathematics xxx (xxxx) xxx
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Domination versus independent domination in graphs of small regularity Ammar Babikir, Michael A. Henning
∗,1
Department of Mathematics and Applied Mathematics, University of Johannesburg, Auckland Park, 2006, South Africa
article
info
Article history: Received 10 May 2019 Received in revised form 31 October 2019 Accepted 3 November 2019 Available online xxxx Keywords: Domination Independent domination Ratio Regular graphs
a b s t r a c t A set S of vertices in a graph G is a dominating set if every vertex not in S is adjacent to a vertex in S. If, in addition, S is an independent set, then S is an independent dominating set. The domination number γ (G) of G is the minimum cardinality of a dominating set in G, while the independent domination number i(G) of G is the minimum cardinality of an independent dominating set in G. It is known (Goddard et al., 2012) that if G is a connected 3-regular graph, then i(G)/γ (G) ≤ 3/2, with equality if and only if G = K3,3 . In this paper, we extend this result to graphs of larger regularity and show that if k ∈ {4, 5, 6} and G is a connected k-regular graph, then i(G)/γ (G) ≤ k/2, with equality if and only if G = Kk,k . © 2019 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we continue the study of domination versus independent domination in regular graphs. A set S of vertices in a graph G is a dominating set if every vertex not in S is adjacent to a vertex in S. If, in addition, S is an independent set, then S is an independent dominating set. The domination number, denoted γ (G), of G is the minimum cardinality of a dominating set of G, and the independent domination number, denoted i(G), of G is the minimum cardinality of an independent dominating set in G. An independent set of vertices in a graph G is a dominating set of G if and only if it is a maximal independent set. Thus, i(G) is equivalently the minimum cardinality of a maximal independent set of vertices in G. A dominating set of cardinality γ (G) is called a γ -set of G, while an independent dominating set of cardinality i(G) is called an i-set of G. A survey on independent domination in graphs can be found in [3]. Independent domination in graphs has been studied, for example, in [2,5–7,9,10]. In this paper, we study the independent domination number versus the domination in regular graphs. A graph G is k-regular if every vertex in G has degree k. Our aim is to provide a best possible upper bound on the ratio of the independent domination number to the domination number in k-regular graphs for small k ∈ {4, 5, 6}. We shall prove the following result. For k ∈ {4, 5, 6} if G is a connected k-regular graph, then
Theorem 1. i(G)
γ (G)
≤
k 2
with equality if and only if G = Kk,k . ∗ Corresponding author. E-mail addresses: [email protected] (A. Babikir), [email protected] (M.A. Henning). 1 Research supported in part by the University of Johannesburg, South Africa. https://doi.org/10.1016/j.disc.2019.111727 0012-365X/© 2019 Elsevier B.V. All rights reserved.
Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
2
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
Fig. 1. The graphs K3,3 and C5 □ K2 .
We proceed as follows. In Section 2, we present the necessary graph theory notation and terminology. In Section 3, we present some known results. Thereafter in Section 4, we present a proof of our main result, namely Theorem 1. 2. Notation and terminology For notation and terminology, we will typically follow [8]. Specifically, let G be a graph with vertex set V (G) and edge set E(G). The order and size of G will be denoted n = |V (G)| and m = |E(G)|, respectively. Two vertices v and w are neighbors in G if they are adjacent; that is, if vw ∈ E(G). The open neighborhood of a vertex v in G is the set of neighbors of v , denoted NG (v ), whereas the closed neighborhood of v is NG [v] = NG (v ) ∪ {v}. The open neighborhood of a set S ⊆ V (G) is the set of all neighbors of vertices in S, denoted NG (S), whereas the closed neighborhood of S is NG [S ] = NG (S) ∪ S. The S-private neighborhood of a vertex v ∈ S is defined by pnG (v, S) = {w ∈ V (G) | NG [w] ∩ S = {v}}. We note that if v ∈ pnG (v, S), then the vertex v is isolated in the subgraph G[S ]. A vertex in pnG (v, S) is called an S-private neighbor of v . The S-external private neighborhood, epnG (v, S), of v is the set of all S-private neighbors of v that do not belong to the set S; that is, epnG (v, S) = pnG (v, S) ∩ (V (G) \ S). A vertex in epnG (v, S) is called an S-external private neighbor of v . If the graph G is clear from the context, we omit the subscript G in the above definitions and write, for example, V , E, N(v ) and N [v] rather than V (G), E(G), NG (v ) and NG [v], respectively. We denote the degree of a vertex v in G by dG (v ) = |NG (v )|. A k-regular graph is a graph in which every vertex has degree k. A 3-regular graph is also called a cubic graph in the literature. Given a graph G, we denote the number of vertices in G of degree i by ni (G). The distance between two vertices u and v in a connected graph G, denoted by dG (u, v ), is the length of a shortest (u, v )-path in G. A component of G is a maximal connected subgraph of G. If H is a graph, then an H-component of G is a component of G isomorphic to H. For a set of vertices S ⊆ V (G), the subgraph induced by S is denoted by G[S ]. We denote by G − S the graph obtained from G by deleting all vertices in S and their incident edges from G. If v ∈ V (G), the graph G −v denotes the graph obtained from G by deleting v (and all edges incident with v ). We denote the path, cycle, and complete graph on n vertices by Pn , Cn , and Kn , respectively. The complete bipartite graph with one partite set of size n and the other of size m is denoted by Kn,m . The 5-prism, C5 □ K2 , is the Cartesian product of a 5-cycle with a copy of K2 . The graphs K3,3 and C5 □ K2 are shown in Figs. 1(a) and 1(b), respectively. For vertex disjoint subsets X and Y of the vertex set of a graph G, we denote the set of edges that join a vertex of X and a vertex of Y by [X , Y ]. Thus, |[X , Y ]| is the number of edges in G with one end in X and the other end in Y . A proper vertex coloring of a graph G is a coloring of the vertices so that every two adjacent vertices are colored differently. A proper vertex coloring whose colors are taken from a set of k colors is called a proper k-coloring. The chromatic number of a graph G, denoted χ (G), is the smallest positive integer k for which G has a proper k-coloring. The independence number, denoted α (G), of G is the maximum cardinality of an independent set in G. An independent set of cardinality α (G) is called an α -set of G. 3. Known results If G is a connected k-regular graph of order n where k = 2, then G is a cycle Cn of order n. In this case, i(G) = γ (G) = ⌈ 31 n⌉ and i(G)/γ (G) = 1. The ratio of the independent domination number to the domination number in a cubic
graph is well studied in the literature. Goddard, Henning, Lyle, and Southey [4] showed that the ratio of the independent domination number and the domination number in a cubic graph cannot be too large. Theorem 2 ([4]). If G is a connected cubic graph, then i(G)/γ (G) ≤
3 , 2
with equality if and only if G = K3,3 .
The result of Theorem 2 was subsequently improved by Southey and Henning [13] who showed that the be strengthened to a 34 -ratio if we forbid K3,3 .
3 -ratio 2
can
Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
Theorem 3 ([13]). If G ̸ = K3,3 is a connected cubic graph, then i(G)/γ (G) ≤
4 , 3
3
with equality if and only if G = C5 □ K2 .
O and West [11] constructed an infinite family of connected cubic graphs G such that i(G)/γ (G) = 5/4. However it remains an open question to determine whether the 43 -ratio in Theorem 3 can be improved to a 54 -ratio if we forbid finitely many graphs. As first observed by Rosenfeld [12], the independence number of a regular graph is at most one-half its order. Theorem 4 ([12]). If G is a regular graph of order n with no isolated vertex, then i(G) ≤ α (G) ≤
1 n. 2
We note that equality in Theorem 4 is only obtainable for graphs with every component a balanced complete bipartite graph, as observed in [4]. For completeness, we present a proof of this fact. Theorem 5.
For k ≥ 1 if G is a k-regular connected graph of order n and i(G) =
1 n, 2
then G = Kk,k .
Proof. Suppose that G is a k-regular connected graph of order n such that i(G) = 12 n. Let X be a maximum independent set of G, and so α (G) = |X |, and let the complement of X be denoted by Y = V (G) \ X . By double counting the edges joining X and its complement Y , we have k|X | = [|X , Y ]| ≤ k|Y | = k(n − |X |), and so 21 n = i(G) ≤ α (G) = |X | ≤ 12 n. Hence we must have equality throughout this inequality chain, implying that |X | = |Y | = 12 n and that each vertex in the complement Y of X has exactly k neighbors in X . In particular, G is a k-regular, bipartite graph with partite sets X and Y . Suppose, to the contrary, that G ̸ = Kk,k . Thus, there exist vertices x ∈ X and y ∈ Y that are not adjacent in G. If {x, y} is a dominating set of G, then n = 2(k + 1) and i(G) = 2 = n/(k + 1) < n/2, a contradiction. Hence, {x, y} is a not a dominating set of G. We now consider the bipartite graph G′ = G − (N [x] ∪ N [y]) of order n′ = n − 2(k + 1). Each vertex in G′ is adjacent in G to at most k − 1 deleted vertices, implying that G′ contains no isolated vertex. Let X ′ = X ∩ V (G′ ). We note that |X ′ | = |X | − (k + 1) = 21 n − k − 1. The set X ′ ∪ {x, y} is an independent dominating set of G, and so i(G) ≤ |X ′ | + 2 = 21 n − k + 1 < 12 n, a contradiction. Hence, G = Kk,k . □ We shall need the following classical result, known as Brooks’s Coloring Theorem. Theorem 6 ([1]). If G is a connected graph with maximum degree ∆, then χ (G) ≤ ∆, except if G is the complete graph K∆+1 or an odd cycle in which case χ (G) = ∆ + 1. Given a (proper) coloring of the vertices of a graph G, the vertices of the same color form a color class. Since every color class is an independent set, we note that if G has order n, then α (G) ≥ n/χ (G). Hence as a consequence of Theorem 6, we have the following lower bound on the independence number of a graph. Corollary 1. If G is a connected graph of order n with maximum degree ∆, then α (G) ≥ n/∆, except if G is a complete graph K∆+1 , in which case α (G) = n/(∆ + 1), or if G is an odd cycle, in which case α (G) = (n − 1)/2 = (n − 1)/∆. 4. Proof of Theorem 1 In this section, we present a proof of our main theorem, namely Theorem 1. Recall its statement. Theorem 1. For k ∈ {4, 5, 6} if G is a connected k-regular graph, then i(G)/γ (G) ≤ k/2, with equality if and only if G = Kk,k . Proof. Among all γ -sets of G, let D be chosen so that G[D] has the fewest edges. Thus, D is a dominating set of G and |D| = γ (G). Further, if D′ is a γ -set of G, then m(G[D]) ≤ m(G[D′ ]). Let F = G[D], and so V (F ) = D. We proceed further with the following series of claims. Claim 1.
∆(F ) ≤ k − 2.
Proof. Let v be an arbitrary vertex of D. We show that v has at least two neighbors in the graph G that do not belong to the set D. If all k neighbors of v belong to the set D, then D \ {v} is a dominating set of G, contradicting the minimality of D. Hence, the vertex v has at least one neighbor outside D. Suppose that v has exactly one neighbor v ′ outside D. If v ′ has a neighbor in D different from v , then the set D \ {v} is once again a dominating set of G, contradicting the minimality of D. Hence, v is the only vertex in D that is adjacent to v ′ ; that is, NG (v ′ ) ∩ D = {v}, implying that epn(v, D) = {v ′ }. We now consider the set D′ = (D \ {v}) ∪ {v ′ }. The set D′ is a dominating set of cardinality |D| = γ (G), and is therefore a γ -set of G. However, m(G[D′ ]) = m(G[D]) − k + 1 < m(G[D]), and so the set D′ induces a subgraph with fewer edges than G[D], a contradiction. Hence, v has at least two neighbors outside D. Since v is an arbitrary vertex of D, we deduce that every vertex in D has at least two neighbors outside D, implying by the k-regularity of G that F = G[D] has maximum degree at most k − 2. (□) Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
4
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
By Claim 1, ∆(F ) ≤ k − 2. Let C be the set of all components in F . Let C1 be the set of all K1 -components in F , and let C2 be the set of all K2 -components in F . Let C3 be the set of all components H in F such that ∆(H) ≥ 2 and H ≇ Kk−1 . Let C4 be the set of all Kk−1 -components in F . We call a component of F that belongs to the set Ci a type-i component of F for i ∈ [4]. We note that every component of F is a type-i component for some i ∈ [4]. Thus, (C1 , C2 , C3 , C4 ) is a partition of C . Recall that F = G[D], and so V (F ) = D. Let D1 be a maximum independent set in F , and so |D1 | = α (F ), and let D2 = D \ D1 . Let D = V (G) \ D. For each component H in F , we let Di,H be the set of vertices of Di that belong to the component H, and so Di,H = Di ∩ V (H) for i ∈ [2]. Further, we let ∂2 (H) denote all vertices in D that have a neighbor in the component H that belongs to the set D2,H . Equivalently, ∂2 (H) is the set of all neighbors of vertices of D2,H that belong to D; that is, ∂2 (H) = NG (D2,H ) ∩ D. We call ∂2 (H) the 2-boundary of the component H. More generally, we define the 2-boundary ∂2 (F ) of the graph F = G[D] to be the union of the 2-boundaries of all components of F ; that is,
∂2 (F ) =
⋃
∂2 (H),
H ∈C
where recall that C is the set of all components H in F . Thus, ∂2 (F ) is the set of all vertices outside D that have a neighbor that belongs to the set D2 . Let U1 be the set of vertices not dominated by D1 in G. Since D2 is dominated by D1 , we note that U1 ∩ D = ∅. Further since D = D1 ∪ D2 is a dominating set of G, we note that each vertex in U1 is adjacent to some vertex of D2 . Thus, U1 ⊆ ∂2 (F ). Let I1 be a maximal independent set in G[U1 ] of minimum cardinality, and so I1 is a i-set of G[U1 ]. Equivalently, I1 is an independent dominating set of G[U1 ] of minimum cardinality. Since I1 ⊆ U1 , we have |I1 | ≤ |U1 |. Further, we note that D1 ∪ I1 is an independent dominating set of G. Hence, i(G) ≤ |D1 | + |I1 | ≤ |D1 | + |U1 | ≤ |D1 | + |∂2 (F )|.
(1)
In what follows, we adopt the following notation. If H is a component of F such that ∆(H) ≥ 2, then among all maximum independent sets in H, let IH be chosen so that (1) IH contains the maximum number of vertices of degree 1 in H. (2) Subject to (1), IH contains the maximum number of vertices of degree 2 in H. Claim 2.
If H is a component of F such that ∆(H) ≥ 2, then every vertex of degree 1 in H belongs to the set IH .
Proof. Let v be an arbitrary vertex of degree 1 in H, and let u be its neighbor in H. Since ∆(H) ≥ 2, we note that u has degree at least 2 in H. If v ∈ / IH , then by the maximality of the independent set IH we note that u ∈ IH . However in this case replacing u in IH with the vertex v produces a new α -set of H that contains more vertices of degree 1 in H than does the set IH , a contradiction. Hence, v ∈ IH , implying that every vertex of degree 1 in H belongs to the set IH . (□) If H is a component of F such that ∆(H) ≥ 2, then we let LH be the set of all vertices in H that do not belong to the set IH , and so LH = V (H) \ IH . By Claim 2, every vertex in LH has degree at least 2 in H and is therefore adjacent to at most k − 2 vertices that do no belong to H. We note that since H is a component of F , the neighbors of a vertex in LH that do no belong to H belong to the set V (F ) \ D. We first consider the case when k = 4. Claim 3.
If k = 4, then i(G)/γ (G) ≤ 2, with equality if and only if G = K4,4 .
Proof. Suppose that k = 4. We consider the subgraph F = G[D] induced by the vertices of the γ -set D. By Claim 1, ∆(F ) ≤ 2. Thus, every component of F is a path or a cycle. We proceed further with the following subclaims. Claim 3.1.
If H is a component of F of order nH , then the following hold.
(a) If nH = 1, then |∂2 (H)| < |D2,H | + nH . (b) If nH = 2, then |∂2 (H)| = |D2,H | + nH . (c) If ∆(H) ≥ 2, then nH ≥ 3 and |∂2 (H)| < |D2,H | + nH . Proof. If nH = 1, then H = K1 , implying that D2,H = ∂2 (H) = ∅, and therefore that |D2,H | = 0 and |∂2 (H)| = 0 = |D2,H | + nH − 1 < |D2,H | + nH . If nH = 2, then H = K2 , implying that |D2,H | = 1 and |∂2 (H)| = 3 = |D2,H | + nH . This completes the proof of Part (a) and (b). To prove Part (c), suppose that ∆(H) ≥ 2, implying that nH ≥ 3 and H is a path or a cycle. If H is a path, then α (H) = ⌈ 21 nH ⌉ ≥ 12 nH . If H is a cycle, then α (H) = ⌈ 12 (nH − 1)⌉ ≥ 12 (nH − 1). In both cases, |IH | = α (H) ≥ 21 (nH − 1). By our earlier observations, every vertex in LH has degree at least 2 in H and is therefore Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
5
adjacent to at most two vertices that do no belong to H. Thus since |D2,H | = |LH | and nH ≥ 3, we have
|∂2 (H)| ≤ = = ≤ = <
2|LH | | LH | + | LH | |D2,H | + (nH − |IH |) |D2,H | + nH − 12 (nH − 1) |D2,H | + 21 (nH + 1) |D2,H | + nH .
This completes the proof of Claim 3.1.
(□)
As a consequence of Claim 3.1, we have the following result on the 2-boundary of F . Claim 3.2.
If H is a component of F of order nH , then |∂2 (H)| ≤ |D2,H | + nH , with strict equality if H ̸ = K2 .
Claim 3.3.
|∂2 (F )| ≤ |D2 | + 2|D|, with strict inequality if there is a component in F different from K2 .
Proof. By Claim 3.2, the following holds.
|∂2 (F )| ≤
∑ H ∈C
|∂2 (H)| ≤
∑
(|D2,H | + nH ) = |D2 | + |D|.
H ∈C
Further by Claim 3.2, if there is a component H of F different from K2 , then |∂2 (H)| < |D2,H | + nH , implying that we have strict inequality in the above inequality chain. (□) We now return to the proof of Claim 3. By Eq. (1) and Claim 3.3, we have i(G) ≤ |D1 | + |∂2 (F )| ≤ |D1 | + (|D2 | + |D|) = |D| + |D| = 2|D| = 2γ (G),
(2)
implying that i(G)
γ (G)
≤ 2,
(3)
establishing the desired upper bound. Suppose that i(G)/γ (G) = 2. Then we must have equality throughout the above inequality chains, implying that D1 ∪ I1 is an i-set of G and |I1 | = |U1 | = |∂2 (F )| = |D2 | + |D|. In particular, this implies by Claim 3.2 that every component in F is a K2 -component; that is, F = G[D] = tK2 for some integer t ≥ 1 and |D1 | = |D2 | = 21 |D| = t. Further, I1 = U1 = ∂2 (F ) and NG (D2 ) = D1 ∪ U1 . Therefore, NG (D2 ) is an i-set of G and i(G) = |NG (D2 )| = 4t. In this case, interchanging the roles of D1 and D2 , and defining U2 as the set of vertices not dominated by D2 in G, I2 as a maximal independent set in G[U2 ] of minimum cardinality, and the 1-boundary ∂1 (F ) of F as the set of all vertices outside D that have a neighbor that belongs to the set D1 , we can show using an identical argument that I2 = U2 = ∂1 (F ) and NG (D1 ) = D2 ∪ U2 . Therefore, NG (D1 ) is an i-set of G and i(G) = |NG (D1 )| = 4t. Hence, G is a 4-regular, connected, bipartite graph of order n = 8t with partite sets NG (D1 ) and NG (D2 ) satisfying i(G) = 4t = 21 n. By Theorem 5 this implies that G = K4,4 . This completes the proof of Claim 3. (□) We consider next the case when k = 5. Claim 4.
If k = 5, then i(G)/γ (G) ≤
5 , 2
with equality if and only if G = K5,5 .
Proof. Suppose that k = 5. We consider the subgraph F = G[D] induced by the vertices of the γ -set D. By Claim 1, ∆(F ) ≤ 3. We proceed further with the following subclaims. Claim 4.1. (a) (b) (c) (d)
If If If If
If H is a component of F of order nH , then the following hold.
nH = 1, then |∂2 (H)| = |D2,H | + 32 nH − 32 . nH = 2, then |∂2 (H)| = |D2,H | + 23 nH . H = K4 , then |∂2 (H)| ≤ |D2,H | + 43 nH . ∆(H) ≥ 2 and H ̸= K4 , then |∂2 (H)| ≤ |D2,H | + 43 nH .
Proof. If nH = 1, then H = K1 , implying that D2,H = ∂2 (H) = ∅, and therefore that |D2,H | = 0 and |∂2 (H)| = 0 = |D2,H | + 32 nH − 23 . If nH = 2, then H = K2 , implying that |D2,H | = 1 and |∂2 (H)| = 4 = |D2,H | + 32 nH . If H = K4 , then nH = 4, |D2,H | = 3 and |∂2 (H)| ≤ 3 · 2 = 6 = |D2,H | + 34 nH . This establishes Parts (a), (b) and (c). Suppose that ∆(H) ≥ 2 and H ̸ = K4 . We now consider the maximum independent set IH of H. By Corollary 1, we note that |IH | ≥ 13 nH . By our Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
6
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
earlier observations, every vertex in LH has degree at least 2 in H and is therefore adjacent to at most three vertices that do no belong to H. Thus since |D2,H | = |LH |, we have
|∂2 (H)| ≤ = = ≤ =
3|LH | |LH | + 2|LH | |D2,H | + 2(nH − |IH |) |D2,H | + 2nH − 2 · 31 nH |D2,H | + 34 nH .
This completes the proof of Part (d).
(□)
As a consequence of Claim 4.1, we have the following result on the 2-boundary of F . Claim 4.2.
If H is a component of F of order nH , then |∂2 (H)| ≤ |D2,H | + 23 nH , with strict equality if H ̸ = K2 .
Claim 4.3.
|∂2 (F )| ≤ |D2 | + 23 |D|, with strict inequality if there is a component in F different from K2 .
Proof. By Claim 4.2, the following holds.
|∂2 (F )| ≤
∑ H ∈C
|∂2 (H)| ≤
∑
(|D2,H | +
H ∈C
3 2
nH ) = |D2 | +
3 2
|D|.
Further by Claim 4.2, if there is a component H of F different from K2 , then |∂2 (H)| < |D2,H | + 32 nH , implying that we have strict inequality in the above inequality chain. (□) We now return to the proof of Claim 4. By Eq. (1) and Claim 4.3, we have i(G) ≤ |D1 | + |∂2 (F )| ≤ |D1 | + (|D2 | +
3 2
3
5
2
2
|D|) = |D| + |D| =
|D| =
5 2
γ (G),
(4)
implying that i(G)
γ (G)
≤
5 2
,
(5)
establishing the desired upper bound. Suppose that i(G)/γ (G) = 25 . Then we must have equality throughout the above inequality chains, implying that D1 ∪ I1 is an i-set of G and |I1 | = |U1 | = |∂2 (F )| = |D2 | + |D|. In particular, this implies by Claim 4.2 that every component in F is a K2 -component; that is, F = G[D] = tK2 for some integer t ≥ 1 and |D1 | = |D2 | = 12 |D| = t. Proceeding analogously as in the last paragraph of the proof of Claim 3, the G is a 5-regular, connected, bipartite graph of order n satisfying i(G) = 21 n. By Theorem 5 this implies that G = K5,5 . This completes the proof of Claim 4. (□) We consider finally the case when k = 6. Claim 5.
If k = 6, then i(G)/γ (G) ≤ 3, with equality if and only if G = K6,6 .
Proof. Suppose that k = 6. We consider the subgraph F = G[D] induced by the vertices of the γ -set D. By Claim 1, ∆(F ) ≤ 4. We proceed further with the following subclaims. Claim 5.1.
If H is a component of F of order nH , then the following hold.
(a) If nH = 1, then |∂2 (H)| = |D2,H | + 2nH − 2. (b) If nH = 2, then |∂2 (H)| = |D2,H | + 2nH . (c) If H = K5 , then |∂2 (H)| ≤ |D2,H | + 2nH − 6. Proof. If nH = 1, then H = K1 , implying that D2,H = ∂2 (H) = ∅, and therefore that |D2,H | = 0 and |∂2 (H)| = 0 = |D2,H | + 2nH − 2. If nH = 2, then H = K2 , implying that |D2,H | = 1 and |∂2 (H)| = 5 = 1 + 4 = |D2,H | + 2nH . If H = K5 , then nH = 5, |D2,H | = 4 and |∂2 (H)| ≤ 4 · 2 = 8 = 4 + 4 = |D2,H | + 2nH − 6. (□) Claim 5.2. Let H be a component of F of order nH . If ∆(H) ≥ 2 and H ̸ = K5 , then |∂2 (H)| ≤ |D2,H | + 2nH . Further, if |∂2 (H)| = |D2,H | + 2nH , then H = K3 and every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H . Proof. Suppose that ∆(H) ≥ 2 and H ̸ = K5 . If H is an odd cycle, then nH ≥ 3 and α (H) = 21 (nH − 1) > 14 nH . If H is not an odd cycle, then by Corollary 1, we note that α (H) ≥ 14 nH . In both cases, |IH | = α (H) ≥ 14 nH . By our earlier observations, every vertex in LH has degree at least 2 in H and is therefore adjacent to at most four vertices that do not belong to H. Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
7
Let LH ,2 be the set of vertices in LH of degree 2 in H, and let LH ,≥3 be the set of vertices in LH of degree at least 3 in H. Thus, every vertex in LH ,≥3 has degree 3 or 4 in H. Further, LH = LH ,2 ∪ LH ,≥3 and |D2,H | = |LH | = |LH ,2 | + |LH ,≥3 |. Claim 5.2.1. If LH ,2 = ∅, then |∂2 (H)| < |D2,H | + 2nH . Proof. Suppose that LH ,2 = ∅. Thus, LH = LH ,≥3 , and so every vertex in LH has degree 3 or 4 in H, implying that every vertex in LH has at most three neighbors outside the set D. Thus since |D2,H | = |LH | and |IH | = α (H) ≥ 14 nH , we have
|∂2 (H)| ≤ = = ≤ = = as desired.
3|LH | |D2,H | + 2|LH | |D2,H | + 2(nH − |IH |) |D2,H | + 2nH − 2 · 41 nH |D2,H | + 32 nH |D2,H | + 2nH ,
(□)
By Claim 5.2.1, we may assume that LH ,2 ̸ = ∅, for otherwise the desired result follows. We show now that |∂2 (H)| ≤ |D2,H | + 2nH , with strict inequality if H ̸= K3 . Let Q be the set of all vertices in IH that have a neighbor in LH ,2 , and let q = |Q |. Each vertex in LH ,2 is therefore adjacent to one or two vertices of Q in H. Let L2,1 be the set of vertices in LH ,2 adjacent to exactly one vertex of Q in H, and let L2,2 be the set of vertices in LH ,2 adjacent to two vertices of Q in H. Further, let |L2,1 | = ℓ1 and |L2,2 | = ℓ2 , and so |LH ,2 | = ℓ1 + ℓ2 . Let Qi = {v ∈ Q | dH (v ) = i}
and
qi = |Qi |
for i ∈ [4]. Thus, (Q1 , Q2 , Q3 , Q4 ) is a weak partition of Q (where by a weak partition we mean a partition in which some of the sets may be empty), and q = |Q | = q1 + q2 + q3 + q4 . Let Q≥3 = Q3 ∪ Q4 and let q≥3 = |Q≥3 |. Thus, q≥3 = q3 + q4 and q = q1 + q2 + q≥3 . Claim 5.2.2. Each vertex in LH that has a neighbor in Q≥3 belongs to the set L2,2 . Proof. Let v be a vertex in Q≥3 , implying that v has three or four neighbors in LH . Let w be an arbitrary neighbor of v that belongs to LH . If w ∈ L2,1 , then the vertex w is adjacent to v but to no other vertex of IH in H. In this case replacing v in IH with the vertex w produces a new α -set of H that contains every vertex of degree 1 in H and contains more vertices of degree 2 in H than does the set IH , a contradiction. Hence, w ∈ L2,2 . Thus every neighbor of a vertex of Q≥3 that belongs to the set LH ,2 has two neighbors in Q and therefore belongs to the set L2,2 . (□) Let Q1,j be the set of vertices in Q1 with a neighbor in L2,j for j ∈ [2], and let q1,j = |Q1,j |. Thus, (Q1,1 , Q1,2 ) is a weak partition of Q1 and q1 = q1,1 + q1,2 . For i ∈ [2], let Q2,i be the set of vertices in Q2 with at least one neighbor in L2,i but with no neighbor in L2,3−i . Let Q2,3 be the set of vertices in Q2 with a neighbor in L2,1 and a neighbor in L2,2 . Thus, (Q2,1 , Q2,2 , Q2,3 ) is a weak partition of Q2 . Further, let q2,j = |Q2,j | for j ∈ [3], and so q2 = q2,1 + q2,2 + q2,3 . Claim 5.2.3. The following hold. (a) ℓ1 ≤ q1,1 + 2q2,1 + q2,3 . (b) ℓ2 ≤ 21 q1,2 + q2,2 + 12 q2,3 + 2q≥3 . (c) |LH ,2 | ≤ 2q. Proof. By Claim 5.2.2, no vertex in L2,1 has a neighbor in Q≥3 . Therefore the (unique) neighbor of each vertex L2,1 in Q belongs to the set Q1,1 ∪ Q2,1 ∪ Q2,3 . Counting the edges between L2,1 and Q we therefore have ℓ1 = |[L2,1 , Q ]| = |[L2,1 , Q1,1 ]| + |[L2,1 , Q2,1 ]| + |[L2,1 , Q2,3 ]| ≤ q1,1 + 2q2,1 + q2,3 . This establishes Part (a). Counting the edges between L2,2 and Q , we have 2ℓ2 = |[L2,2 , Q ]| = |[L2,2 , Q1,2 ]| + |[L2,2 , Q2,2 ]| + |[L2,2 , Q2,3 ]| + |[L2,2 , Q≥3 ]| ≤ q1,2 + 2q2,2 + q2,3 + 4q≥3 . This establishes Part (b). By Part (a) and Part (b), we have
|LH ,2 | = ≤ ≤ ≤
ℓ1 + ℓ2
(q1,1 + 21 q1,2 ) + (2q2,1 + q2,2 + 12 q2,3 ) + 2q≥3 q1 + 2q2 + 2q≥3 2q.
This completes the proof of Part (c), and the proof of Claim 5.2.3.
(□)
We now return to the proof of Claim 5.2. Each vertex in the set LH ,2 has four neighbors outside the set D, while each vertex in the set LH ,≥3 has at most three neighbors outside the set D. Recall that D2,H = LH = LH ,2 ∪ LH ,≥3 and Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
8
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
V (H) = IH ∪ LH . By Claim 5.2.3(c), |LH ,2 | ≤ 2q. Since Q ⊆ IH , we note that q = |Q | ≤ |IH |. These observations imply the following inequality chain.
|∂2 (H)| ≤ = = = ≤ ≤ =
4|LH ,2 | + 3|LH ,≥3 | 3(|LH ,2 | + |LH ,≥3 |) + |LH ,2 | 3|LH | + |LH ,2 | |D2,H | + 2|LH | + |LH ,2 | |D2,H | + 2|LH | + 2q |D2,H | + 2|LH | + 2|IH | |D2,H | + 2nH .
This establishes the desired upper bound |∂2 (H)| ≤ |D2,H | + 2nH . Suppose next that |∂2 (H)| = |D2,H | + 2nH (and still LH ,2 ̸ = ∅). Thus we must have equality throughout the above inequality chain. In particular, |∂2 (H)| = 4|LH ,2 | + 3|LH ,≥3 |, implying that every vertex in LH ,≥3 has exactly three neighbors outside the set D, and therefore every vertex in LH ,≥3 has degree 3 in H. Further, every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H (for otherwise, |∂2 (H)| < 4|LH ,2 | + 3|LH ,≥3 |, a contradiction). We also note that since |∂2 (H)| = |D2,H | + 2nH , we have q = |IH |, and so Q = IH , and |LH ,2 | = 2q. Thus we must have equality throughout the inequalities in the statement and proof of Claim 5.2.3. This implies that q1 = 0 and q2 = q2,1 . Hence, Q1 = ∅ and Q2 = Q2,1 , and if Q2 ̸ = ∅, then |[L2,1 , Q2,1 ]| = 2q2,1 and therefore each vertex in Q2,1 has both its neighbors in H belonging to the set L2,1 . Since Q1,2 = Q2,2 = Q2,3 = ∅, if L2,2 ̸ = ∅, then every vertex of L2,2 has both its neighbors in Q≥3 . Moreover, we must have |[L2,2 , Q≥3 ]| = 4q≥3 , implying that Q≥3 = Q4 and each vertex in Q4 has all four of its neighbors in H belonging to the set L2,2 . These observations imply that Q = Q2,1 ∪ Q4 and LH ,≥3 = ∅, and so LH = LH ,2 . Thus, nH = |LH | + |IH | = |LH ,2 | + |IH | = 2q + q = 3q = 3|IH |. Thus, α (H) = |IH | = 31 nH . Since H is connected, our observations imply that either LH = L2,1 or LH = L2,2 . Suppose that LH = L2,2 . In this case, Q = Q4 and H is a bipartite graph with partite sets IH and LH , where each vertex in IH has four neighbors in LH and each vertex in LH has two neighbors in IH . Thus, α (H) ≥ |LH | = 2|IH |, contradicting the maximality of the independent set IH . Therefore, LH = L2,2 = ∅, and so LH = L2,1 . Thus, Q = Q2,1 and H is a 2-regular graph or, equivalently, H is a cycle, and so α (H) = ⌈ 12 (nH − 1)⌉. However as observed earlier, α (H) = 31 nH . Therefore, nH = 3, and so H is a 3-cycle K3 . This completes the proof of Claim 5.2. (□) As an immediate consequence of Claims 5.1 and 5.2, we have the following result on the 2-boundary of a component in F . Claim 5.3. If H is a component of F of order nH , then |∂2 (H)| ≤ |D2,H | + 2nH . Further, if |∂2 (H)| = |D2,H | + 2nH , then H ∈ {K2 , K3 } and every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H . We are now in a position to establish the following upper bound on the 2-boundary of the graph F . Claim 5.4.
|∂2 (F )| ≤ |D2 | + 2|D|. Further, if |∂2 (F )| = |D2 | + 2|D|, then the following hold.
(a) Every component in H is a K2 -component or a K3 -component. (b) Every vertex in the 2-boundary, ∂2 (F ), of F , has exactly one neighbor in D2 . Proof. By Claim 5.3, the following holds.
|∂2 (F )| ≤
∑
|∂2 (H)| ≤
H ∈C
∑
(|D2,H | + 2nH ) = |D2 | + 2|D|.
H ∈C
Suppose that |∂2 (F )| = |D2 | + 2|D|. In particular, this implies that |∂2 (H)| = |D2,H | + 2nH for every component H of F . Thus by Claim 5.3, such a component H of F is either a K2 -component or a K3 -component. Further by Claim 5.3, every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H . If there is a vertex in the 2-boundary, ∑ ∂2 (F ), of F , that has a neighbor in D2,H ′ and in D2,H ′′ for two distinct components H ′ and H ′′ of F , then |∂2 (F )| < H ∈C |∂2 (H)|, implying that |∂2 (F )| < |D2 | + 2|D|, a contradiction. Hence, every vertex in the 2-boundary, ∂2 (F ), of F , has exactly one neighbor in D2 . Further by Claim 5.3, if there is a component H of F different from K2 or K3 , then |∂2 (H)| < |D2,H | + 2nH , implying that |∂2 (F )| < |D2 | + 2|D|, a contradiction. (□) We now return to the proof of Claim 5. By Eq. (1) and Claim 5.4, we have i(G) ≤ |D1 | + |∂2 (F )| ≤ |D1 | + (|D2 | + 2|D|) = 3|D|.
(6)
Thus noting that |D| = γ (G), we have i(G)
γ (G)
≤
3|D|
|D|
= 3,
(7)
establishing the desired upper bound. Suppose that i(G)/γ (G) = 3. Then we must have equality throughout Eqs. (6) and (7), implying that D1 ∪ I1 is an i-set of G and |I1 | = |U1 | = |∂2 (F )| = |D2 | + 2|D|. In particular, this implies by Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
9
Claim 5.4 that every component in F is a K2 -component or a K3 -component. Thus we can choose the set D1,F to consist of an arbitrary vertex from each component of F . Let the number of K2 -components and K3 -components of F be denoted by k2 and k3 , respectively. By Claim 5.4 and our earlier observations, we note that the 2-boundary ∂2 (F ) is an independent set of cardinality |∂2 (F )| = 5k2 + 8k3 and each vertex in ∂2 (F ) has exactly one neighbor in D2,F . As observed earlier for each K3 -component H of F , we can choose D1,H to consist of any arbitrary vertex from H. Suppose that k3 ≥ 1. Let F ′ be the subgraph of F consisting of all K3 -components of F . Our earlier observations imply that the set, say S ′ , of vertices outside F ′ that have a neighbor in F ′ is an independent set of cardinality |S ′ | = 12k3 and each vertex in S ′ has exactly one neighbor in F ′ . Let v be an arbitrary vertex in the 2-boundary ∂2 (F ′ ). By the 6-regularity of G and our observations to date, the vertex v has exactly one neighbor that belongs to D and has five neighbors outside D. Let u be an arbitrary neighbor of v outside D. Since the vertex u has no neighbor that belongs to F ′ , the vertex u has a neighbor, w say, that belongs to a K2 -component of F . We can therefore choose the set D1,F to contain the neighbor of w in F , or, equivalently, so that w ∈ D2,F . This would imply that both vertices u and v belong to the 2-boundary ∂2 (F ), contradicting the fact that the set ∂2 (F ) is an independent set. Therefore, k3 = 0. Since k3 = 0, we note that k2 ≥ 1 and every component of F is a K2 -component; that is, F = G[D] = k2 K2 for some integer k2 ≥ 1 and |D1 | = |D2 | = 21 |D| = k2 . Further, I1 = U1 = ∂2 (F ) and NG (D2 ) = D1 ∪ U1 . Therefore, NG (D2 ) is an i-set of G and i(G) = |NG (D2 )| = 6k2 . In this case, interchanging the roles of D1 and D2 , and defining U2 as the set of vertices not dominated by D2 in G, I2 as a maximal independent set in G[U2 ] of minimum cardinality, and the 1-boundary ∂1 (F ) of F as the set of all vertices outside D that have a neighbor that belongs to the set D1 , we can show using an identical argument that I2 = U2 = ∂1 (F ) and NG (D1 ) = D2 ∪ U2 . Therefore, NG (D1 ) is an i-set of G and i(G) = |NG (D1 )| = 6k2 . Hence, G is a 6-regular, connected, bipartite graph of order n = 12k2 with partite sets NG (D1 ) and NG (D2 ) satisfying i(G) = 6k2 = 21 n. By Theorem 5 this implies that G = K6,6 . This completes the proof of Claim 5. (□) Theorem 1 now follows from Claims 3–5. □ 5. Closing question We close with the following question that we have yet to settle. Question 1. Is it true that for all integers k ≥ 3 if G is a connected k-regular graph, then i(G)/γ (G) ≤ k/2, with equality if and only if G = Kk,k ? Question 1 is true for k = 3 as shown by Goddard et al. [4]. In this paper, we have shown that Question 1 is true for k ∈ {4, 5, 6}. However, Question 1 remains open for k ≥ 7. We remark that the proof techniques we employ in the proof of Theorem 1 cannot be extended to larger regularity. Hence if Question 1 is true for all k, then new ideas are needed to prove this result for k ≥ 7. Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
R.L. Brooks, On colouring the nodes of a network, Math. Proc. Camb. Phil. Soc. 37 (1941) 194–197. P. Dorbec, M.A. Henning, M. Montassier, J. Southey, Independent domination in cubic graphs, J. Graph Theory 80 (4) (2015) 329–349. W. Goddard, M.A. Henning, Independent domination in graphs: A survey and recent results, Discrete Math. 313 (2013) 839–854. W. Goddard, M.A. Henning, J. Lyle, J. Southey, On the independent domination number of regular graphs, Ann. Combin. 16 (2012) 719–732. J. Haviland, Independent domination in regular graphs, Discrete Math. 143 (1995) 275–280. J. Haviland, Upper bounds for independent domination in regular graphs, Discrete Math. 307 (2007) 2643–2646. M.A. Henning, P.J. Slater, Inequalties relating domination parameters in cubic graphs, Discrete Math. 158 (1996) 87–98. M.A. Henning, A. Yeo, Total Domination in Graphs, in: Springer Monographs in Mathematics, 2013, ISBN: 978-1-4614-6524-9 (Print) 978-1-4614-6525-6 (Online. P.C.B. Lam, W.C. Shiu, L. Sun, On independent domination number of regular graphs, Discrete Math. 202 (1999) 135–144. J. Lyle, A structural approach for independent domination of regular graphs, Graphs Combin. 31 (5) (2015) 1567–1588. Suil. O, D.B. West, Cubic graphs with large ratio of independent domination number to domination number, Graphs Combin. 32 (2) (2016) 773–776. M. Rosenfeld, Independent sets in regular graphs, Israel J. Math. 2 (1964) 262–272. J. Southey, M.A. Henning, Domination versus independent domination in cubic graphs, Discrete Math. 313 (11) (2013) 1212–1220.
Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Domination versus independent domination in graphs of small regularity Ammar Babikir, Michael A. Henning
∗,1
Department of Mathematics and Applied Mathematics, University of Johannesburg, Auckland Park, 2006, South Africa
article
info
Article history: Received 10 May 2019 Received in revised form 31 October 2019 Accepted 3 November 2019 Available online xxxx Keywords: Domination Independent domination Ratio Regular graphs
a b s t r a c t A set S of vertices in a graph G is a dominating set if every vertex not in S is adjacent to a vertex in S. If, in addition, S is an independent set, then S is an independent dominating set. The domination number γ (G) of G is the minimum cardinality of a dominating set in G, while the independent domination number i(G) of G is the minimum cardinality of an independent dominating set in G. It is known (Goddard et al., 2012) that if G is a connected 3-regular graph, then i(G)/γ (G) ≤ 3/2, with equality if and only if G = K3,3 . In this paper, we extend this result to graphs of larger regularity and show that if k ∈ {4, 5, 6} and G is a connected k-regular graph, then i(G)/γ (G) ≤ k/2, with equality if and only if G = Kk,k . © 2019 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we continue the study of domination versus independent domination in regular graphs. A set S of vertices in a graph G is a dominating set if every vertex not in S is adjacent to a vertex in S. If, in addition, S is an independent set, then S is an independent dominating set. The domination number, denoted γ (G), of G is the minimum cardinality of a dominating set of G, and the independent domination number, denoted i(G), of G is the minimum cardinality of an independent dominating set in G. An independent set of vertices in a graph G is a dominating set of G if and only if it is a maximal independent set. Thus, i(G) is equivalently the minimum cardinality of a maximal independent set of vertices in G. A dominating set of cardinality γ (G) is called a γ -set of G, while an independent dominating set of cardinality i(G) is called an i-set of G. A survey on independent domination in graphs can be found in [3]. Independent domination in graphs has been studied, for example, in [2,5–7,9,10]. In this paper, we study the independent domination number versus the domination in regular graphs. A graph G is k-regular if every vertex in G has degree k. Our aim is to provide a best possible upper bound on the ratio of the independent domination number to the domination number in k-regular graphs for small k ∈ {4, 5, 6}. We shall prove the following result. For k ∈ {4, 5, 6} if G is a connected k-regular graph, then
Theorem 1. i(G)
γ (G)
≤
k 2
with equality if and only if G = Kk,k . ∗ Corresponding author. E-mail addresses: [email protected] (A. Babikir), [email protected] (M.A. Henning). 1 Research supported in part by the University of Johannesburg, South Africa. https://doi.org/10.1016/j.disc.2019.111727 0012-365X/© 2019 Elsevier B.V. All rights reserved.
Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
2
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
Fig. 1. The graphs K3,3 and C5 □ K2 .
We proceed as follows. In Section 2, we present the necessary graph theory notation and terminology. In Section 3, we present some known results. Thereafter in Section 4, we present a proof of our main result, namely Theorem 1. 2. Notation and terminology For notation and terminology, we will typically follow [8]. Specifically, let G be a graph with vertex set V (G) and edge set E(G). The order and size of G will be denoted n = |V (G)| and m = |E(G)|, respectively. Two vertices v and w are neighbors in G if they are adjacent; that is, if vw ∈ E(G). The open neighborhood of a vertex v in G is the set of neighbors of v , denoted NG (v ), whereas the closed neighborhood of v is NG [v] = NG (v ) ∪ {v}. The open neighborhood of a set S ⊆ V (G) is the set of all neighbors of vertices in S, denoted NG (S), whereas the closed neighborhood of S is NG [S ] = NG (S) ∪ S. The S-private neighborhood of a vertex v ∈ S is defined by pnG (v, S) = {w ∈ V (G) | NG [w] ∩ S = {v}}. We note that if v ∈ pnG (v, S), then the vertex v is isolated in the subgraph G[S ]. A vertex in pnG (v, S) is called an S-private neighbor of v . The S-external private neighborhood, epnG (v, S), of v is the set of all S-private neighbors of v that do not belong to the set S; that is, epnG (v, S) = pnG (v, S) ∩ (V (G) \ S). A vertex in epnG (v, S) is called an S-external private neighbor of v . If the graph G is clear from the context, we omit the subscript G in the above definitions and write, for example, V , E, N(v ) and N [v] rather than V (G), E(G), NG (v ) and NG [v], respectively. We denote the degree of a vertex v in G by dG (v ) = |NG (v )|. A k-regular graph is a graph in which every vertex has degree k. A 3-regular graph is also called a cubic graph in the literature. Given a graph G, we denote the number of vertices in G of degree i by ni (G). The distance between two vertices u and v in a connected graph G, denoted by dG (u, v ), is the length of a shortest (u, v )-path in G. A component of G is a maximal connected subgraph of G. If H is a graph, then an H-component of G is a component of G isomorphic to H. For a set of vertices S ⊆ V (G), the subgraph induced by S is denoted by G[S ]. We denote by G − S the graph obtained from G by deleting all vertices in S and their incident edges from G. If v ∈ V (G), the graph G −v denotes the graph obtained from G by deleting v (and all edges incident with v ). We denote the path, cycle, and complete graph on n vertices by Pn , Cn , and Kn , respectively. The complete bipartite graph with one partite set of size n and the other of size m is denoted by Kn,m . The 5-prism, C5 □ K2 , is the Cartesian product of a 5-cycle with a copy of K2 . The graphs K3,3 and C5 □ K2 are shown in Figs. 1(a) and 1(b), respectively. For vertex disjoint subsets X and Y of the vertex set of a graph G, we denote the set of edges that join a vertex of X and a vertex of Y by [X , Y ]. Thus, |[X , Y ]| is the number of edges in G with one end in X and the other end in Y . A proper vertex coloring of a graph G is a coloring of the vertices so that every two adjacent vertices are colored differently. A proper vertex coloring whose colors are taken from a set of k colors is called a proper k-coloring. The chromatic number of a graph G, denoted χ (G), is the smallest positive integer k for which G has a proper k-coloring. The independence number, denoted α (G), of G is the maximum cardinality of an independent set in G. An independent set of cardinality α (G) is called an α -set of G. 3. Known results If G is a connected k-regular graph of order n where k = 2, then G is a cycle Cn of order n. In this case, i(G) = γ (G) = ⌈ 31 n⌉ and i(G)/γ (G) = 1. The ratio of the independent domination number to the domination number in a cubic
graph is well studied in the literature. Goddard, Henning, Lyle, and Southey [4] showed that the ratio of the independent domination number and the domination number in a cubic graph cannot be too large. Theorem 2 ([4]). If G is a connected cubic graph, then i(G)/γ (G) ≤
3 , 2
with equality if and only if G = K3,3 .
The result of Theorem 2 was subsequently improved by Southey and Henning [13] who showed that the be strengthened to a 34 -ratio if we forbid K3,3 .
3 -ratio 2
can
Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
Theorem 3 ([13]). If G ̸ = K3,3 is a connected cubic graph, then i(G)/γ (G) ≤
4 , 3
3
with equality if and only if G = C5 □ K2 .
O and West [11] constructed an infinite family of connected cubic graphs G such that i(G)/γ (G) = 5/4. However it remains an open question to determine whether the 43 -ratio in Theorem 3 can be improved to a 54 -ratio if we forbid finitely many graphs. As first observed by Rosenfeld [12], the independence number of a regular graph is at most one-half its order. Theorem 4 ([12]). If G is a regular graph of order n with no isolated vertex, then i(G) ≤ α (G) ≤
1 n. 2
We note that equality in Theorem 4 is only obtainable for graphs with every component a balanced complete bipartite graph, as observed in [4]. For completeness, we present a proof of this fact. Theorem 5.
For k ≥ 1 if G is a k-regular connected graph of order n and i(G) =
1 n, 2
then G = Kk,k .
Proof. Suppose that G is a k-regular connected graph of order n such that i(G) = 12 n. Let X be a maximum independent set of G, and so α (G) = |X |, and let the complement of X be denoted by Y = V (G) \ X . By double counting the edges joining X and its complement Y , we have k|X | = [|X , Y ]| ≤ k|Y | = k(n − |X |), and so 21 n = i(G) ≤ α (G) = |X | ≤ 12 n. Hence we must have equality throughout this inequality chain, implying that |X | = |Y | = 12 n and that each vertex in the complement Y of X has exactly k neighbors in X . In particular, G is a k-regular, bipartite graph with partite sets X and Y . Suppose, to the contrary, that G ̸ = Kk,k . Thus, there exist vertices x ∈ X and y ∈ Y that are not adjacent in G. If {x, y} is a dominating set of G, then n = 2(k + 1) and i(G) = 2 = n/(k + 1) < n/2, a contradiction. Hence, {x, y} is a not a dominating set of G. We now consider the bipartite graph G′ = G − (N [x] ∪ N [y]) of order n′ = n − 2(k + 1). Each vertex in G′ is adjacent in G to at most k − 1 deleted vertices, implying that G′ contains no isolated vertex. Let X ′ = X ∩ V (G′ ). We note that |X ′ | = |X | − (k + 1) = 21 n − k − 1. The set X ′ ∪ {x, y} is an independent dominating set of G, and so i(G) ≤ |X ′ | + 2 = 21 n − k + 1 < 12 n, a contradiction. Hence, G = Kk,k . □ We shall need the following classical result, known as Brooks’s Coloring Theorem. Theorem 6 ([1]). If G is a connected graph with maximum degree ∆, then χ (G) ≤ ∆, except if G is the complete graph K∆+1 or an odd cycle in which case χ (G) = ∆ + 1. Given a (proper) coloring of the vertices of a graph G, the vertices of the same color form a color class. Since every color class is an independent set, we note that if G has order n, then α (G) ≥ n/χ (G). Hence as a consequence of Theorem 6, we have the following lower bound on the independence number of a graph. Corollary 1. If G is a connected graph of order n with maximum degree ∆, then α (G) ≥ n/∆, except if G is a complete graph K∆+1 , in which case α (G) = n/(∆ + 1), or if G is an odd cycle, in which case α (G) = (n − 1)/2 = (n − 1)/∆. 4. Proof of Theorem 1 In this section, we present a proof of our main theorem, namely Theorem 1. Recall its statement. Theorem 1. For k ∈ {4, 5, 6} if G is a connected k-regular graph, then i(G)/γ (G) ≤ k/2, with equality if and only if G = Kk,k . Proof. Among all γ -sets of G, let D be chosen so that G[D] has the fewest edges. Thus, D is a dominating set of G and |D| = γ (G). Further, if D′ is a γ -set of G, then m(G[D]) ≤ m(G[D′ ]). Let F = G[D], and so V (F ) = D. We proceed further with the following series of claims. Claim 1.
∆(F ) ≤ k − 2.
Proof. Let v be an arbitrary vertex of D. We show that v has at least two neighbors in the graph G that do not belong to the set D. If all k neighbors of v belong to the set D, then D \ {v} is a dominating set of G, contradicting the minimality of D. Hence, the vertex v has at least one neighbor outside D. Suppose that v has exactly one neighbor v ′ outside D. If v ′ has a neighbor in D different from v , then the set D \ {v} is once again a dominating set of G, contradicting the minimality of D. Hence, v is the only vertex in D that is adjacent to v ′ ; that is, NG (v ′ ) ∩ D = {v}, implying that epn(v, D) = {v ′ }. We now consider the set D′ = (D \ {v}) ∪ {v ′ }. The set D′ is a dominating set of cardinality |D| = γ (G), and is therefore a γ -set of G. However, m(G[D′ ]) = m(G[D]) − k + 1 < m(G[D]), and so the set D′ induces a subgraph with fewer edges than G[D], a contradiction. Hence, v has at least two neighbors outside D. Since v is an arbitrary vertex of D, we deduce that every vertex in D has at least two neighbors outside D, implying by the k-regularity of G that F = G[D] has maximum degree at most k − 2. (□) Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
4
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
By Claim 1, ∆(F ) ≤ k − 2. Let C be the set of all components in F . Let C1 be the set of all K1 -components in F , and let C2 be the set of all K2 -components in F . Let C3 be the set of all components H in F such that ∆(H) ≥ 2 and H ≇ Kk−1 . Let C4 be the set of all Kk−1 -components in F . We call a component of F that belongs to the set Ci a type-i component of F for i ∈ [4]. We note that every component of F is a type-i component for some i ∈ [4]. Thus, (C1 , C2 , C3 , C4 ) is a partition of C . Recall that F = G[D], and so V (F ) = D. Let D1 be a maximum independent set in F , and so |D1 | = α (F ), and let D2 = D \ D1 . Let D = V (G) \ D. For each component H in F , we let Di,H be the set of vertices of Di that belong to the component H, and so Di,H = Di ∩ V (H) for i ∈ [2]. Further, we let ∂2 (H) denote all vertices in D that have a neighbor in the component H that belongs to the set D2,H . Equivalently, ∂2 (H) is the set of all neighbors of vertices of D2,H that belong to D; that is, ∂2 (H) = NG (D2,H ) ∩ D. We call ∂2 (H) the 2-boundary of the component H. More generally, we define the 2-boundary ∂2 (F ) of the graph F = G[D] to be the union of the 2-boundaries of all components of F ; that is,
∂2 (F ) =
⋃
∂2 (H),
H ∈C
where recall that C is the set of all components H in F . Thus, ∂2 (F ) is the set of all vertices outside D that have a neighbor that belongs to the set D2 . Let U1 be the set of vertices not dominated by D1 in G. Since D2 is dominated by D1 , we note that U1 ∩ D = ∅. Further since D = D1 ∪ D2 is a dominating set of G, we note that each vertex in U1 is adjacent to some vertex of D2 . Thus, U1 ⊆ ∂2 (F ). Let I1 be a maximal independent set in G[U1 ] of minimum cardinality, and so I1 is a i-set of G[U1 ]. Equivalently, I1 is an independent dominating set of G[U1 ] of minimum cardinality. Since I1 ⊆ U1 , we have |I1 | ≤ |U1 |. Further, we note that D1 ∪ I1 is an independent dominating set of G. Hence, i(G) ≤ |D1 | + |I1 | ≤ |D1 | + |U1 | ≤ |D1 | + |∂2 (F )|.
(1)
In what follows, we adopt the following notation. If H is a component of F such that ∆(H) ≥ 2, then among all maximum independent sets in H, let IH be chosen so that (1) IH contains the maximum number of vertices of degree 1 in H. (2) Subject to (1), IH contains the maximum number of vertices of degree 2 in H. Claim 2.
If H is a component of F such that ∆(H) ≥ 2, then every vertex of degree 1 in H belongs to the set IH .
Proof. Let v be an arbitrary vertex of degree 1 in H, and let u be its neighbor in H. Since ∆(H) ≥ 2, we note that u has degree at least 2 in H. If v ∈ / IH , then by the maximality of the independent set IH we note that u ∈ IH . However in this case replacing u in IH with the vertex v produces a new α -set of H that contains more vertices of degree 1 in H than does the set IH , a contradiction. Hence, v ∈ IH , implying that every vertex of degree 1 in H belongs to the set IH . (□) If H is a component of F such that ∆(H) ≥ 2, then we let LH be the set of all vertices in H that do not belong to the set IH , and so LH = V (H) \ IH . By Claim 2, every vertex in LH has degree at least 2 in H and is therefore adjacent to at most k − 2 vertices that do no belong to H. We note that since H is a component of F , the neighbors of a vertex in LH that do no belong to H belong to the set V (F ) \ D. We first consider the case when k = 4. Claim 3.
If k = 4, then i(G)/γ (G) ≤ 2, with equality if and only if G = K4,4 .
Proof. Suppose that k = 4. We consider the subgraph F = G[D] induced by the vertices of the γ -set D. By Claim 1, ∆(F ) ≤ 2. Thus, every component of F is a path or a cycle. We proceed further with the following subclaims. Claim 3.1.
If H is a component of F of order nH , then the following hold.
(a) If nH = 1, then |∂2 (H)| < |D2,H | + nH . (b) If nH = 2, then |∂2 (H)| = |D2,H | + nH . (c) If ∆(H) ≥ 2, then nH ≥ 3 and |∂2 (H)| < |D2,H | + nH . Proof. If nH = 1, then H = K1 , implying that D2,H = ∂2 (H) = ∅, and therefore that |D2,H | = 0 and |∂2 (H)| = 0 = |D2,H | + nH − 1 < |D2,H | + nH . If nH = 2, then H = K2 , implying that |D2,H | = 1 and |∂2 (H)| = 3 = |D2,H | + nH . This completes the proof of Part (a) and (b). To prove Part (c), suppose that ∆(H) ≥ 2, implying that nH ≥ 3 and H is a path or a cycle. If H is a path, then α (H) = ⌈ 21 nH ⌉ ≥ 12 nH . If H is a cycle, then α (H) = ⌈ 12 (nH − 1)⌉ ≥ 12 (nH − 1). In both cases, |IH | = α (H) ≥ 21 (nH − 1). By our earlier observations, every vertex in LH has degree at least 2 in H and is therefore Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
5
adjacent to at most two vertices that do no belong to H. Thus since |D2,H | = |LH | and nH ≥ 3, we have
|∂2 (H)| ≤ = = ≤ = <
2|LH | | LH | + | LH | |D2,H | + (nH − |IH |) |D2,H | + nH − 12 (nH − 1) |D2,H | + 21 (nH + 1) |D2,H | + nH .
This completes the proof of Claim 3.1.
(□)
As a consequence of Claim 3.1, we have the following result on the 2-boundary of F . Claim 3.2.
If H is a component of F of order nH , then |∂2 (H)| ≤ |D2,H | + nH , with strict equality if H ̸ = K2 .
Claim 3.3.
|∂2 (F )| ≤ |D2 | + 2|D|, with strict inequality if there is a component in F different from K2 .
Proof. By Claim 3.2, the following holds.
|∂2 (F )| ≤
∑ H ∈C
|∂2 (H)| ≤
∑
(|D2,H | + nH ) = |D2 | + |D|.
H ∈C
Further by Claim 3.2, if there is a component H of F different from K2 , then |∂2 (H)| < |D2,H | + nH , implying that we have strict inequality in the above inequality chain. (□) We now return to the proof of Claim 3. By Eq. (1) and Claim 3.3, we have i(G) ≤ |D1 | + |∂2 (F )| ≤ |D1 | + (|D2 | + |D|) = |D| + |D| = 2|D| = 2γ (G),
(2)
implying that i(G)
γ (G)
≤ 2,
(3)
establishing the desired upper bound. Suppose that i(G)/γ (G) = 2. Then we must have equality throughout the above inequality chains, implying that D1 ∪ I1 is an i-set of G and |I1 | = |U1 | = |∂2 (F )| = |D2 | + |D|. In particular, this implies by Claim 3.2 that every component in F is a K2 -component; that is, F = G[D] = tK2 for some integer t ≥ 1 and |D1 | = |D2 | = 21 |D| = t. Further, I1 = U1 = ∂2 (F ) and NG (D2 ) = D1 ∪ U1 . Therefore, NG (D2 ) is an i-set of G and i(G) = |NG (D2 )| = 4t. In this case, interchanging the roles of D1 and D2 , and defining U2 as the set of vertices not dominated by D2 in G, I2 as a maximal independent set in G[U2 ] of minimum cardinality, and the 1-boundary ∂1 (F ) of F as the set of all vertices outside D that have a neighbor that belongs to the set D1 , we can show using an identical argument that I2 = U2 = ∂1 (F ) and NG (D1 ) = D2 ∪ U2 . Therefore, NG (D1 ) is an i-set of G and i(G) = |NG (D1 )| = 4t. Hence, G is a 4-regular, connected, bipartite graph of order n = 8t with partite sets NG (D1 ) and NG (D2 ) satisfying i(G) = 4t = 21 n. By Theorem 5 this implies that G = K4,4 . This completes the proof of Claim 3. (□) We consider next the case when k = 5. Claim 4.
If k = 5, then i(G)/γ (G) ≤
5 , 2
with equality if and only if G = K5,5 .
Proof. Suppose that k = 5. We consider the subgraph F = G[D] induced by the vertices of the γ -set D. By Claim 1, ∆(F ) ≤ 3. We proceed further with the following subclaims. Claim 4.1. (a) (b) (c) (d)
If If If If
If H is a component of F of order nH , then the following hold.
nH = 1, then |∂2 (H)| = |D2,H | + 32 nH − 32 . nH = 2, then |∂2 (H)| = |D2,H | + 23 nH . H = K4 , then |∂2 (H)| ≤ |D2,H | + 43 nH . ∆(H) ≥ 2 and H ̸= K4 , then |∂2 (H)| ≤ |D2,H | + 43 nH .
Proof. If nH = 1, then H = K1 , implying that D2,H = ∂2 (H) = ∅, and therefore that |D2,H | = 0 and |∂2 (H)| = 0 = |D2,H | + 32 nH − 23 . If nH = 2, then H = K2 , implying that |D2,H | = 1 and |∂2 (H)| = 4 = |D2,H | + 32 nH . If H = K4 , then nH = 4, |D2,H | = 3 and |∂2 (H)| ≤ 3 · 2 = 6 = |D2,H | + 34 nH . This establishes Parts (a), (b) and (c). Suppose that ∆(H) ≥ 2 and H ̸ = K4 . We now consider the maximum independent set IH of H. By Corollary 1, we note that |IH | ≥ 13 nH . By our Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
6
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
earlier observations, every vertex in LH has degree at least 2 in H and is therefore adjacent to at most three vertices that do no belong to H. Thus since |D2,H | = |LH |, we have
|∂2 (H)| ≤ = = ≤ =
3|LH | |LH | + 2|LH | |D2,H | + 2(nH − |IH |) |D2,H | + 2nH − 2 · 31 nH |D2,H | + 34 nH .
This completes the proof of Part (d).
(□)
As a consequence of Claim 4.1, we have the following result on the 2-boundary of F . Claim 4.2.
If H is a component of F of order nH , then |∂2 (H)| ≤ |D2,H | + 23 nH , with strict equality if H ̸ = K2 .
Claim 4.3.
|∂2 (F )| ≤ |D2 | + 23 |D|, with strict inequality if there is a component in F different from K2 .
Proof. By Claim 4.2, the following holds.
|∂2 (F )| ≤
∑ H ∈C
|∂2 (H)| ≤
∑
(|D2,H | +
H ∈C
3 2
nH ) = |D2 | +
3 2
|D|.
Further by Claim 4.2, if there is a component H of F different from K2 , then |∂2 (H)| < |D2,H | + 32 nH , implying that we have strict inequality in the above inequality chain. (□) We now return to the proof of Claim 4. By Eq. (1) and Claim 4.3, we have i(G) ≤ |D1 | + |∂2 (F )| ≤ |D1 | + (|D2 | +
3 2
3
5
2
2
|D|) = |D| + |D| =
|D| =
5 2
γ (G),
(4)
implying that i(G)
γ (G)
≤
5 2
,
(5)
establishing the desired upper bound. Suppose that i(G)/γ (G) = 25 . Then we must have equality throughout the above inequality chains, implying that D1 ∪ I1 is an i-set of G and |I1 | = |U1 | = |∂2 (F )| = |D2 | + |D|. In particular, this implies by Claim 4.2 that every component in F is a K2 -component; that is, F = G[D] = tK2 for some integer t ≥ 1 and |D1 | = |D2 | = 12 |D| = t. Proceeding analogously as in the last paragraph of the proof of Claim 3, the G is a 5-regular, connected, bipartite graph of order n satisfying i(G) = 21 n. By Theorem 5 this implies that G = K5,5 . This completes the proof of Claim 4. (□) We consider finally the case when k = 6. Claim 5.
If k = 6, then i(G)/γ (G) ≤ 3, with equality if and only if G = K6,6 .
Proof. Suppose that k = 6. We consider the subgraph F = G[D] induced by the vertices of the γ -set D. By Claim 1, ∆(F ) ≤ 4. We proceed further with the following subclaims. Claim 5.1.
If H is a component of F of order nH , then the following hold.
(a) If nH = 1, then |∂2 (H)| = |D2,H | + 2nH − 2. (b) If nH = 2, then |∂2 (H)| = |D2,H | + 2nH . (c) If H = K5 , then |∂2 (H)| ≤ |D2,H | + 2nH − 6. Proof. If nH = 1, then H = K1 , implying that D2,H = ∂2 (H) = ∅, and therefore that |D2,H | = 0 and |∂2 (H)| = 0 = |D2,H | + 2nH − 2. If nH = 2, then H = K2 , implying that |D2,H | = 1 and |∂2 (H)| = 5 = 1 + 4 = |D2,H | + 2nH . If H = K5 , then nH = 5, |D2,H | = 4 and |∂2 (H)| ≤ 4 · 2 = 8 = 4 + 4 = |D2,H | + 2nH − 6. (□) Claim 5.2. Let H be a component of F of order nH . If ∆(H) ≥ 2 and H ̸ = K5 , then |∂2 (H)| ≤ |D2,H | + 2nH . Further, if |∂2 (H)| = |D2,H | + 2nH , then H = K3 and every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H . Proof. Suppose that ∆(H) ≥ 2 and H ̸ = K5 . If H is an odd cycle, then nH ≥ 3 and α (H) = 21 (nH − 1) > 14 nH . If H is not an odd cycle, then by Corollary 1, we note that α (H) ≥ 14 nH . In both cases, |IH | = α (H) ≥ 14 nH . By our earlier observations, every vertex in LH has degree at least 2 in H and is therefore adjacent to at most four vertices that do not belong to H. Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
7
Let LH ,2 be the set of vertices in LH of degree 2 in H, and let LH ,≥3 be the set of vertices in LH of degree at least 3 in H. Thus, every vertex in LH ,≥3 has degree 3 or 4 in H. Further, LH = LH ,2 ∪ LH ,≥3 and |D2,H | = |LH | = |LH ,2 | + |LH ,≥3 |. Claim 5.2.1. If LH ,2 = ∅, then |∂2 (H)| < |D2,H | + 2nH . Proof. Suppose that LH ,2 = ∅. Thus, LH = LH ,≥3 , and so every vertex in LH has degree 3 or 4 in H, implying that every vertex in LH has at most three neighbors outside the set D. Thus since |D2,H | = |LH | and |IH | = α (H) ≥ 14 nH , we have
|∂2 (H)| ≤ = = ≤ = = as desired.
3|LH | |D2,H | + 2|LH | |D2,H | + 2(nH − |IH |) |D2,H | + 2nH − 2 · 41 nH |D2,H | + 32 nH |D2,H | + 2nH ,
(□)
By Claim 5.2.1, we may assume that LH ,2 ̸ = ∅, for otherwise the desired result follows. We show now that |∂2 (H)| ≤ |D2,H | + 2nH , with strict inequality if H ̸= K3 . Let Q be the set of all vertices in IH that have a neighbor in LH ,2 , and let q = |Q |. Each vertex in LH ,2 is therefore adjacent to one or two vertices of Q in H. Let L2,1 be the set of vertices in LH ,2 adjacent to exactly one vertex of Q in H, and let L2,2 be the set of vertices in LH ,2 adjacent to two vertices of Q in H. Further, let |L2,1 | = ℓ1 and |L2,2 | = ℓ2 , and so |LH ,2 | = ℓ1 + ℓ2 . Let Qi = {v ∈ Q | dH (v ) = i}
and
qi = |Qi |
for i ∈ [4]. Thus, (Q1 , Q2 , Q3 , Q4 ) is a weak partition of Q (where by a weak partition we mean a partition in which some of the sets may be empty), and q = |Q | = q1 + q2 + q3 + q4 . Let Q≥3 = Q3 ∪ Q4 and let q≥3 = |Q≥3 |. Thus, q≥3 = q3 + q4 and q = q1 + q2 + q≥3 . Claim 5.2.2. Each vertex in LH that has a neighbor in Q≥3 belongs to the set L2,2 . Proof. Let v be a vertex in Q≥3 , implying that v has three or four neighbors in LH . Let w be an arbitrary neighbor of v that belongs to LH . If w ∈ L2,1 , then the vertex w is adjacent to v but to no other vertex of IH in H. In this case replacing v in IH with the vertex w produces a new α -set of H that contains every vertex of degree 1 in H and contains more vertices of degree 2 in H than does the set IH , a contradiction. Hence, w ∈ L2,2 . Thus every neighbor of a vertex of Q≥3 that belongs to the set LH ,2 has two neighbors in Q and therefore belongs to the set L2,2 . (□) Let Q1,j be the set of vertices in Q1 with a neighbor in L2,j for j ∈ [2], and let q1,j = |Q1,j |. Thus, (Q1,1 , Q1,2 ) is a weak partition of Q1 and q1 = q1,1 + q1,2 . For i ∈ [2], let Q2,i be the set of vertices in Q2 with at least one neighbor in L2,i but with no neighbor in L2,3−i . Let Q2,3 be the set of vertices in Q2 with a neighbor in L2,1 and a neighbor in L2,2 . Thus, (Q2,1 , Q2,2 , Q2,3 ) is a weak partition of Q2 . Further, let q2,j = |Q2,j | for j ∈ [3], and so q2 = q2,1 + q2,2 + q2,3 . Claim 5.2.3. The following hold. (a) ℓ1 ≤ q1,1 + 2q2,1 + q2,3 . (b) ℓ2 ≤ 21 q1,2 + q2,2 + 12 q2,3 + 2q≥3 . (c) |LH ,2 | ≤ 2q. Proof. By Claim 5.2.2, no vertex in L2,1 has a neighbor in Q≥3 . Therefore the (unique) neighbor of each vertex L2,1 in Q belongs to the set Q1,1 ∪ Q2,1 ∪ Q2,3 . Counting the edges between L2,1 and Q we therefore have ℓ1 = |[L2,1 , Q ]| = |[L2,1 , Q1,1 ]| + |[L2,1 , Q2,1 ]| + |[L2,1 , Q2,3 ]| ≤ q1,1 + 2q2,1 + q2,3 . This establishes Part (a). Counting the edges between L2,2 and Q , we have 2ℓ2 = |[L2,2 , Q ]| = |[L2,2 , Q1,2 ]| + |[L2,2 , Q2,2 ]| + |[L2,2 , Q2,3 ]| + |[L2,2 , Q≥3 ]| ≤ q1,2 + 2q2,2 + q2,3 + 4q≥3 . This establishes Part (b). By Part (a) and Part (b), we have
|LH ,2 | = ≤ ≤ ≤
ℓ1 + ℓ2
(q1,1 + 21 q1,2 ) + (2q2,1 + q2,2 + 12 q2,3 ) + 2q≥3 q1 + 2q2 + 2q≥3 2q.
This completes the proof of Part (c), and the proof of Claim 5.2.3.
(□)
We now return to the proof of Claim 5.2. Each vertex in the set LH ,2 has four neighbors outside the set D, while each vertex in the set LH ,≥3 has at most three neighbors outside the set D. Recall that D2,H = LH = LH ,2 ∪ LH ,≥3 and Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
8
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
V (H) = IH ∪ LH . By Claim 5.2.3(c), |LH ,2 | ≤ 2q. Since Q ⊆ IH , we note that q = |Q | ≤ |IH |. These observations imply the following inequality chain.
|∂2 (H)| ≤ = = = ≤ ≤ =
4|LH ,2 | + 3|LH ,≥3 | 3(|LH ,2 | + |LH ,≥3 |) + |LH ,2 | 3|LH | + |LH ,2 | |D2,H | + 2|LH | + |LH ,2 | |D2,H | + 2|LH | + 2q |D2,H | + 2|LH | + 2|IH | |D2,H | + 2nH .
This establishes the desired upper bound |∂2 (H)| ≤ |D2,H | + 2nH . Suppose next that |∂2 (H)| = |D2,H | + 2nH (and still LH ,2 ̸ = ∅). Thus we must have equality throughout the above inequality chain. In particular, |∂2 (H)| = 4|LH ,2 | + 3|LH ,≥3 |, implying that every vertex in LH ,≥3 has exactly three neighbors outside the set D, and therefore every vertex in LH ,≥3 has degree 3 in H. Further, every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H (for otherwise, |∂2 (H)| < 4|LH ,2 | + 3|LH ,≥3 |, a contradiction). We also note that since |∂2 (H)| = |D2,H | + 2nH , we have q = |IH |, and so Q = IH , and |LH ,2 | = 2q. Thus we must have equality throughout the inequalities in the statement and proof of Claim 5.2.3. This implies that q1 = 0 and q2 = q2,1 . Hence, Q1 = ∅ and Q2 = Q2,1 , and if Q2 ̸ = ∅, then |[L2,1 , Q2,1 ]| = 2q2,1 and therefore each vertex in Q2,1 has both its neighbors in H belonging to the set L2,1 . Since Q1,2 = Q2,2 = Q2,3 = ∅, if L2,2 ̸ = ∅, then every vertex of L2,2 has both its neighbors in Q≥3 . Moreover, we must have |[L2,2 , Q≥3 ]| = 4q≥3 , implying that Q≥3 = Q4 and each vertex in Q4 has all four of its neighbors in H belonging to the set L2,2 . These observations imply that Q = Q2,1 ∪ Q4 and LH ,≥3 = ∅, and so LH = LH ,2 . Thus, nH = |LH | + |IH | = |LH ,2 | + |IH | = 2q + q = 3q = 3|IH |. Thus, α (H) = |IH | = 31 nH . Since H is connected, our observations imply that either LH = L2,1 or LH = L2,2 . Suppose that LH = L2,2 . In this case, Q = Q4 and H is a bipartite graph with partite sets IH and LH , where each vertex in IH has four neighbors in LH and each vertex in LH has two neighbors in IH . Thus, α (H) ≥ |LH | = 2|IH |, contradicting the maximality of the independent set IH . Therefore, LH = L2,2 = ∅, and so LH = L2,1 . Thus, Q = Q2,1 and H is a 2-regular graph or, equivalently, H is a cycle, and so α (H) = ⌈ 12 (nH − 1)⌉. However as observed earlier, α (H) = 31 nH . Therefore, nH = 3, and so H is a 3-cycle K3 . This completes the proof of Claim 5.2. (□) As an immediate consequence of Claims 5.1 and 5.2, we have the following result on the 2-boundary of a component in F . Claim 5.3. If H is a component of F of order nH , then |∂2 (H)| ≤ |D2,H | + 2nH . Further, if |∂2 (H)| = |D2,H | + 2nH , then H ∈ {K2 , K3 } and every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H . We are now in a position to establish the following upper bound on the 2-boundary of the graph F . Claim 5.4.
|∂2 (F )| ≤ |D2 | + 2|D|. Further, if |∂2 (F )| = |D2 | + 2|D|, then the following hold.
(a) Every component in H is a K2 -component or a K3 -component. (b) Every vertex in the 2-boundary, ∂2 (F ), of F , has exactly one neighbor in D2 . Proof. By Claim 5.3, the following holds.
|∂2 (F )| ≤
∑
|∂2 (H)| ≤
H ∈C
∑
(|D2,H | + 2nH ) = |D2 | + 2|D|.
H ∈C
Suppose that |∂2 (F )| = |D2 | + 2|D|. In particular, this implies that |∂2 (H)| = |D2,H | + 2nH for every component H of F . Thus by Claim 5.3, such a component H of F is either a K2 -component or a K3 -component. Further by Claim 5.3, every vertex in the 2-boundary, ∂2 (H), of H, has exactly one neighbor in D2,H . If there is a vertex in the 2-boundary, ∑ ∂2 (F ), of F , that has a neighbor in D2,H ′ and in D2,H ′′ for two distinct components H ′ and H ′′ of F , then |∂2 (F )| < H ∈C |∂2 (H)|, implying that |∂2 (F )| < |D2 | + 2|D|, a contradiction. Hence, every vertex in the 2-boundary, ∂2 (F ), of F , has exactly one neighbor in D2 . Further by Claim 5.3, if there is a component H of F different from K2 or K3 , then |∂2 (H)| < |D2,H | + 2nH , implying that |∂2 (F )| < |D2 | + 2|D|, a contradiction. (□) We now return to the proof of Claim 5. By Eq. (1) and Claim 5.4, we have i(G) ≤ |D1 | + |∂2 (F )| ≤ |D1 | + (|D2 | + 2|D|) = 3|D|.
(6)
Thus noting that |D| = γ (G), we have i(G)
γ (G)
≤
3|D|
|D|
= 3,
(7)
establishing the desired upper bound. Suppose that i(G)/γ (G) = 3. Then we must have equality throughout Eqs. (6) and (7), implying that D1 ∪ I1 is an i-set of G and |I1 | = |U1 | = |∂2 (F )| = |D2 | + 2|D|. In particular, this implies by Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.
A. Babikir and M.A. Henning / Discrete Mathematics xxx (xxxx) xxx
9
Claim 5.4 that every component in F is a K2 -component or a K3 -component. Thus we can choose the set D1,F to consist of an arbitrary vertex from each component of F . Let the number of K2 -components and K3 -components of F be denoted by k2 and k3 , respectively. By Claim 5.4 and our earlier observations, we note that the 2-boundary ∂2 (F ) is an independent set of cardinality |∂2 (F )| = 5k2 + 8k3 and each vertex in ∂2 (F ) has exactly one neighbor in D2,F . As observed earlier for each K3 -component H of F , we can choose D1,H to consist of any arbitrary vertex from H. Suppose that k3 ≥ 1. Let F ′ be the subgraph of F consisting of all K3 -components of F . Our earlier observations imply that the set, say S ′ , of vertices outside F ′ that have a neighbor in F ′ is an independent set of cardinality |S ′ | = 12k3 and each vertex in S ′ has exactly one neighbor in F ′ . Let v be an arbitrary vertex in the 2-boundary ∂2 (F ′ ). By the 6-regularity of G and our observations to date, the vertex v has exactly one neighbor that belongs to D and has five neighbors outside D. Let u be an arbitrary neighbor of v outside D. Since the vertex u has no neighbor that belongs to F ′ , the vertex u has a neighbor, w say, that belongs to a K2 -component of F . We can therefore choose the set D1,F to contain the neighbor of w in F , or, equivalently, so that w ∈ D2,F . This would imply that both vertices u and v belong to the 2-boundary ∂2 (F ), contradicting the fact that the set ∂2 (F ) is an independent set. Therefore, k3 = 0. Since k3 = 0, we note that k2 ≥ 1 and every component of F is a K2 -component; that is, F = G[D] = k2 K2 for some integer k2 ≥ 1 and |D1 | = |D2 | = 21 |D| = k2 . Further, I1 = U1 = ∂2 (F ) and NG (D2 ) = D1 ∪ U1 . Therefore, NG (D2 ) is an i-set of G and i(G) = |NG (D2 )| = 6k2 . In this case, interchanging the roles of D1 and D2 , and defining U2 as the set of vertices not dominated by D2 in G, I2 as a maximal independent set in G[U2 ] of minimum cardinality, and the 1-boundary ∂1 (F ) of F as the set of all vertices outside D that have a neighbor that belongs to the set D1 , we can show using an identical argument that I2 = U2 = ∂1 (F ) and NG (D1 ) = D2 ∪ U2 . Therefore, NG (D1 ) is an i-set of G and i(G) = |NG (D1 )| = 6k2 . Hence, G is a 6-regular, connected, bipartite graph of order n = 12k2 with partite sets NG (D1 ) and NG (D2 ) satisfying i(G) = 6k2 = 21 n. By Theorem 5 this implies that G = K6,6 . This completes the proof of Claim 5. (□) Theorem 1 now follows from Claims 3–5. □ 5. Closing question We close with the following question that we have yet to settle. Question 1. Is it true that for all integers k ≥ 3 if G is a connected k-regular graph, then i(G)/γ (G) ≤ k/2, with equality if and only if G = Kk,k ? Question 1 is true for k = 3 as shown by Goddard et al. [4]. In this paper, we have shown that Question 1 is true for k ∈ {4, 5, 6}. However, Question 1 remains open for k ≥ 7. We remark that the proof techniques we employ in the proof of Theorem 1 cannot be extended to larger regularity. Hence if Question 1 is true for all k, then new ideas are needed to prove this result for k ≥ 7. Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
R.L. Brooks, On colouring the nodes of a network, Math. Proc. Camb. Phil. Soc. 37 (1941) 194–197. P. Dorbec, M.A. Henning, M. Montassier, J. Southey, Independent domination in cubic graphs, J. Graph Theory 80 (4) (2015) 329–349. W. Goddard, M.A. Henning, Independent domination in graphs: A survey and recent results, Discrete Math. 313 (2013) 839–854. W. Goddard, M.A. Henning, J. Lyle, J. Southey, On the independent domination number of regular graphs, Ann. Combin. 16 (2012) 719–732. J. Haviland, Independent domination in regular graphs, Discrete Math. 143 (1995) 275–280. J. Haviland, Upper bounds for independent domination in regular graphs, Discrete Math. 307 (2007) 2643–2646. M.A. Henning, P.J. Slater, Inequalties relating domination parameters in cubic graphs, Discrete Math. 158 (1996) 87–98. M.A. Henning, A. Yeo, Total Domination in Graphs, in: Springer Monographs in Mathematics, 2013, ISBN: 978-1-4614-6524-9 (Print) 978-1-4614-6525-6 (Online. P.C.B. Lam, W.C. Shiu, L. Sun, On independent domination number of regular graphs, Discrete Math. 202 (1999) 135–144. J. Lyle, A structural approach for independent domination of regular graphs, Graphs Combin. 31 (5) (2015) 1567–1588. Suil. O, D.B. West, Cubic graphs with large ratio of independent domination number to domination number, Graphs Combin. 32 (2) (2016) 773–776. M. Rosenfeld, Independent sets in regular graphs, Israel J. Math. 2 (1964) 262–272. J. Southey, M.A. Henning, Domination versus independent domination in cubic graphs, Discrete Math. 313 (11) (2013) 1212–1220.
Please cite this article as: A. Babikir and M.A. Henning, Domination versus independent domination in graphs of small regularity, Discrete Mathematics (2019) 111727, https://doi.org/10.1016/j.disc.2019.111727.