Independent Domination in Subcubic Graphs
A. Akbari, S. Akbari, A. Doosthosseini, Z. Hadizadeh, Michael A. Henning, A. Naraghi
IIndependent Domination in Subcubic Graphs A. Akbari, S. Akbari, A. Doosthosseini, Z. Hadizadeh, Michael A. Henning, and A. Naraghi Department of Mathematical SciencesSharif University of TechnologyTehran, Iran Department of Mathematics and Applied MathematicsUniversity of JohannesburgJohannesburg, South Africa
Abstract
A set S of vertices in a graph G is a dominating set if every vertex not in S is adjacentto a vertex in S . If, in addition, S is an independent set, then S is an independent dom-inating set. The independent domination number i ( G ) of G is the minimum cardinalityof an independent dominating set in G . In 2013 Goddard and Henning [Discrete Math313 (2013), 839–854] conjectured that if G is a connected cubic graph of order n , then i ( G ) ≤ n , except if G is the complete bipartite graph K , or the 5-prism C (cid:50) K .Further they construct two infinite families of connected cubic graphs with independentdomination three-eighths their order. They remark that perhaps it is even true thatfor n >
10 these two families are only families for which equality holds. In this paper,we provide a new family of connected cubic graphs G of order n such that i ( G ) = n .We also show that if G is a subcubic graph of order n with no isolated vertex, then i ( G ) ≤ n , and we characterize the graphs achieving equality in this bound. Keywords:
Independent domination; Cubic graph; Subcubic graph
AMS subject classification: E-mail Addresses: [email protected], s [email protected],[email protected], [email protected], [email protected],[email protected] a r X i v : . [ c s . D M ] J a n Introduction
A set S of vertices in a graph G is a dominating set if every vertex not in S is adjacent toa vertex in S . If, in addition, S is an independent set, then S is an independent dominatingset , abbreviated ID-set. The independent domination number , denoted i ( G ), of G is theminimum cardinality of an ID-set in G . The concept of independent domination number ofgraphs is studied extensively in the literature, for example see [1, 2, 3, 4, 5, 7, 8, 9, 10, 11,13, 14, 15, 16, 17, 18, 19, 20]. A survey on independent domination in graphs can be foundin [6].For notation and graph theory terminology we generally follow [12]. The order of agraph G with vertex set V ( G ) and edge set E ( G ) is denoted by n ( G ) = | V ( G ) | and its size by m ( G ) = | E ( G ) | . Two vertices are neighbors in G if they are adjacent. The openneighborhood of a vertex v in G is the set of neighbors of v , denoted N G ( v ). Thus, N G ( v ) = { u ∈ V ( G ) | uv ∈ E ( G ) } . The closed neighborhood of v is the set N G [ v ] = N G ( v ) ∪ { v } .The degree of a vertex v in G is denoted d G ( v ) = | N G ( v ) | . We denote the minimum andmaximum degrees among the vertices of G by δ ( G ) and ∆( G ), respectively. A cubic graph isa graph in which every vertex has degree 3, while a subcubic graph is a graph with maximumdegree at most 3.Further, the subgraph obtained from G by deleting all vertices in S and all edges incidentwith vertices in S is denoted by G − S . If S = { v } , we simply denote G − { v } by G − v .A leaf of a graph G is a vertex of degree 1 in G , while a support vertex of G is a vertexadjacent to a leaf. A star is the graph K ,k , where k ≥
1; that is, a star is a tree withat most one vertex that is not a leaf. A double star is a tree with exactly two (adjacent)non-leaf vertices. Further if one of these vertices is adjacent to r leaves and the other to s leaves, then we denote the double star by S ( r, s ). We denote the path and cycle on n vertices by P n and C n , respectively, and we denote a complete bipartite with partite setsof cardinalities n and m by K n,m . The corona cor( G ) of a graph G , also denoted G ◦ P inthe literature, is the graph obtained from G by adding a pendant edge to each vertex of G .For k ≥ k ] = { , . . . , k } and [ k ] = { , , . . . , k } . As remarked in [6], since every bipartite graph is the union of two independent sets, eachof which dominates the other, we have the following well-known bound on the independentdomination number of a bipartite graph.
Proposition 1 If G is a bipartite graph with no isolated vertices of order n , then i ( G ) ≤ n . As noted in [6], the bound in Proposition 1 is sharp as may be seen by taking G = K k,k for any k ≥
1. In particular, if G = K k,k and k ∈ [3], then G is a connected subcubic graphof order n = 2 k such that i ( G ) = n . 2t remains an open problem to determine best possible upper bounds on the independentdomination number of a connected cubic graph in terms of its orders. Lam, Shiu, andSun [14] proved that if G is a connected cubic graph of order n different from K , , then i ( G ) ≤ n , where the graph K , is given in Figure 1(a). This bound is achieved by the5-prism C (cid:50) K which is illustrated in Figure 1(b). (a) K , (b) C (cid:50) K Figure 1: The graphs K , and C (cid:50) K .Goddard and Henning [6] posed the conjecture that the n bound on the independentdomination number can be improved if we forbid the exceptional graphs K , and C (cid:50) K . Conjecture 1 ([6]) If G / ∈ { K , , C (cid:50) K } is a connected, cubic graph of order n , then i ( G ) ≤ n . Dorbec, Henning, Montassier, and Southey [3] proved Conjecture 1 in the case when thereis no subgraph isomorphic to K , . In general, however, Conjecture 1 remains unresolved.Goddard and Henning [6] constructed two infinite families G cubic and H cubic of connectedcubic graphs with independent domination number three-eighths their orders as follows. For k ≥
1, a graph in the family G cubic is constructed by taking two copies of the cycle C k withrespective vertex sequences a b c d . . . a k b k c k d k and w x y z . . . w k x k y k z k , and joining a i to w i , b i to x i , c i to z i , and d i to y i for each i ∈ [ k ]. For (cid:96) ≥
1, a graph in the family H cubic is constructed by taking a copy of a cycle C (cid:96) with vertex sequence a b c . . . a (cid:96) b (cid:96) c (cid:96) , and foreach i ∈ [ (cid:96) ], adding the vertices { w i , x i , y i , z i , z i } , and joining a i to w i , b i to x i , and c i to y i , and further for each j ∈ [2], joining z ji to each of the vertices w i , x i , and y i . Graphs inthe families G cubic and H cubic are illustrated in Figure 2(a) and 2(b), respectively.(a) G (b) H Figure 2: Graphs G ∈ G cubic and H ∈ H cubic of order n with i ( G ) = i ( H ) = n .3 heorem 1 ([6]) If G ∈ G cubic ∪ H cubic has order n , then i ( G ) = n . It is remarked in [6] that “Perhaps even more than Conjecture 1 is true, in that the onlyextremal graphs are those in G cubic ∪ H cubic . We have confirmed by computer search thatthis is true when n ≤ In this paper we have two immediate aims. Our first aim is to provide a new family ofconnected cubic graphs, different from the families G cubic and H cubic , such that every graph G of order n in the family satisfies i ( G ) = n . We shall prove the following result, where F cubic is the family of connected cubic graphs constructed in Section 4. Theorem 2 If G ∈ F cubic has order n , then n ≥ and n ≡ and i ( G ) = n . Our second aim is to provide a tight upper bound on the independent domination numberof a subcubic graph, and to characterize the graphs achieving equality in this bound. Let G ∼ = K , and G ∼ = K , , and let G , G , G be the three graphs shown in Figure 3. (a) G (b) G (c) G Figure 3: The graphs G , G and G We shall prove the following result, a proof of which is presented in Section 4.
Theorem 3 If G is a subcubic graph of order n with no isolated vertex, then i ( G ) ≤ n ,with equality if and only if the following holds. (a) G ∈ { G , G , G , G , G } . (b) n = 2 k for some k ≥ and G = cor( P k ) . (c) n = 2 k for some k ≥ and G = cor( C k ) . Let X and Y be the graphs shown in Figure 4(a) and 4(b), respectively.4 a) X (b) Y Figure 4: The graphs X and Y Let F cubic be the family of graphs constructed as follows. A graph G in the family F cubic is constructed as follows. We start by taking a cycle C : v v . . . v k v on k ≥ k = 2we mean two vertices adjacent with two different edges) vertices and coloring every vertexon the cycle C , red or blue in such a way that the number of red vertices is even. We thenreplace each red vertex on C with a copy of X , and each blue vertex on C with a copy of Y .(In the case k = 2 we only replace each vertex by a copy of Y .) We call each resulting copyof X and Y an X - copy and Y - copy of G , respectively. Let G i be the X -copy or Y -copyassociated with the vertex v i on the cycle C for each i ∈ [ k ]. Thus if v i is colored red, then G i ∼ = X , while if v i is colored blue, then G i ∼ = Y for i ∈ [ k ]. We note that there are an evennumber of X -copies in G . Next we partition these X -copies into pairs. For each resultingpair { X , X } where X i ∼ = X for i ∈ [2], we add two edges as follows: If v i and v i denotethe two (adjacent) vertices of degree 2 in X i for i ∈ [2], then we add the edges v v and v v as illustrated in Figure 5. X X Figure 5: Joining of the pairs X and X We note that each X -copy of G contains a vertex of degree 1 and each Y -copy of G contains two vertices of degree 2. We now complete the construction of the graph G asfollows. Consider the subgraphs G i and G i +1 where addition is taken modulo k and where i ∈ [ k ]. If G i is an X -copy, then let x i denote the vertex of degree 1 in G i , while if G i is a Y -copy, then let y i and y i denote the two vertices of degree 2 in G i . If both G i and G i +1 are X -copies, then add the edge x x . If both G i and G i +1 are Y -copies, then add the edge y i y i +1 . If G i is an X -copy and G i +1 is a Y -copy, then add the edge x y i +1 . If G i is a Y -copyand G i +1 is an X -copy, then add the edge y i x i +1 . We do this for each i ∈ [ k ], and let G denote the resulting graph. An example of a graph G in the family F cubic constructed froma colored 7-cycle (here k = 7) with four red vertices and three blue vertices is illustrated inFigure 6.We are now in a position to prove Theorem 2. Recall its statement. Theorem 2 . If G ∈ F cubic has order n , then n ≥ and n ≡ and i ( G ) = n . G in the family F cubic constructed from a colored 7-cycle Proof. If G ∈ F cubic has order n , then by construction G is obtained from a k -cycle forsome k ≥ X or Y and adding certain edges toproduce a connected cubic graph. Since each copy of X and Y has order 8, we note that n = 8 k . Thus, n ≥
16 and n ≡ i ( G ) = n .Let S be an arbitrary ID-set in G . We show that S contains at least three vertices fromevery X -copy and Y -copy in G . First we consider an X -copy in G , and let the verticesin this X -copy be named as in Figure 7. For notational convenience, we simply call thissubgraph X . We show that | S ∩ V ( X ) | ≥ a a b b b c c c Figure 7: An X -copy in G If { a , a } ∩ S = ∅ , then in order to dominate the vertex b i , we note that either b i ∈ S or c i ∈ S . Thus, |{ b i , c i } ∩ S | = 1 for all i ∈ [3], implying that | S ∩ V ( X ) | ≥
3, asdesired. Hence we may assume that |{ a , a } ∩ S | ≥
1, for otherwise the desired resultfollows. Renaming a and a if necessary, we may further assume that a ∈ S . Since S isan independent set, we note that { b , b , b } ∩ S = ∅ , implying that a ∈ S . We show thatthe set S contains at least one vertex from the set { c , c , c } . Suppose, to the contrary,that { c , c , c } ∩ S = ∅ . By the construction, c i has exactly one neighbor c (cid:48) i in G \ V ( X )for i ∈ [2]. In order to dominate the vertices c and c , our earlier observations imply that c (cid:48) , c (cid:48) ∈ S . However by construction, the vertices c (cid:48) and c (cid:48) are adjacent, implying that theset S contains two adjacent vertices, contradicting the fact that S is an independent set.Hence, { c , c , c } ∩ S (cid:54) = ∅ , implying that | S ∩ V ( X ) | ≥
3, as desired.6ext we consider a Y -copy in G , and let the vertices in this Y -copy be named as inFigure 8. For notational convenience, we simply call this subgraph Y . We show that | S ∩ V ( Y ) | ≥ a a b b b c c c Figure 8: A Y -copy in G If { a , a } ∩ S = ∅ , then as observed earlier, |{ b i , c i } ∩ S | = 1 for all i ∈ [3], implying that | S ∩ V ( X ) | ≥
3, as desired. Hence we may assume that |{ a , a } ∩ S | ≥
1, for otherwisethe desired result follows. Further we may assume that a ∈ S . As observed earlier, thisimplies that { b , b , b } ∩ S = ∅ and a ∈ S . In order to dominate the vertex c , the set S contains at least one of the vertices c , c and c . Thus, |{ c , c , c } ∩ S | ≥
1, and so | S ∩ V ( Y ) | ≥
3, as desired. This completes the proof of Theorem 2. (cid:50)
In this section, we present a proof of Theorem 3. First we prove that the independentdomination number of a subcubic graph with no isolated vertex is at most one-half theorder of the graph.
Theorem 4 If G is a subcubic graph of order n with no isolated vertex, then i ( G ) ≤ n . Proof.
By linearity, the independent domination number of a graph is the sum of theindependent domination numbers of its components. Hence it suffices for us to prove thebound for connected graphs; that is, we prove that if G is a connected subcubic graph oforder n ≥
2, then i ( G ) ≤ n . We proceed by induction on the order n ≥
2. If n = 2, then G = K and i ( G ) = 1 = ×
2. If n = 3, then G = K or G = P , and in both cases, i ( G ) = 1 < ×
3. This establishes the base cases. Let n ≥ G (cid:48) isa connected subcubic graph of order n (cid:48) where 2 ≤ n (cid:48) < n , then i ( G (cid:48) ) ≤ n (cid:48) . Let G bea connected subcubic graph of order n . If G is a bipartite graph, then the desired boundfollows from Proposition 1. Hence we may assume that G contains an odd cycle C .First, assume that there exists a vertex u on the cycle C with a leaf neighbor, say w , and consider the graph H = G − { u, w } . Since the two neighbors of u on the cycle C are connected in H by the path C − u , the graph H is a connected subcubic graph.Since n ≥
4, we note that | V ( H ) | = n − ≥
2. Applying the induction to H , we have i ( H ) ≤ | V ( H ) | = ( n − H can be extended to an ID-set of G by adding to it the vertex w , implying that i ( G ) ≤ i ( H ) + 1 ≤ n . Hence we may assumethat no vertex on the cycle C has a leaf neighbor.7ext, assume there are two consecutive vertices, say u and w , on the cycle C both ofdegree 2 in G . We now consider the graph H = G − { u, w } . We note that the graph H is a connected subcubic graph. Further since n ≥
4, we note that | V ( H ) | = n − ≥ H , we have i ( H ) ≤ | V ( H ) | = ( n − S (cid:48) be a minimumID-set of H , and let u (cid:48) be the neighbor of u different from w and let w (cid:48) be the neighbor of w different from u . (Possibly, u (cid:48) = w (cid:48) .) If u (cid:48) / ∈ S (cid:48) , then let S = S (cid:48) ∪ { u } . If u (cid:48) ∈ S and w (cid:48) / ∈ S , then let S = S (cid:48) ∪ { w } . If u (cid:48) ∈ S and w (cid:48) ∈ S , then let S = S (cid:48) . In all cases, S is anID-set of G , and so i ( G ) ≤ | S | ≤ | S (cid:48) | + 1 = i ( H ) + 1 ≤ n . Hence we may assume that notwo consecutive vertices on the cycle C both have degree 2 in G .Let u and v be two arbitrary consecutive (adjacent) vertices on the cycle C . Suppose thatthere exists a vertex w of degree 2 adjacent to both u and v . In this case, we consider theconnected subcubic graph H = G − { u, v, w } . If | V ( H ) | = 1, then n = 4 and G ∼ = K − e where e is the missing edge of the complete graph K . In this case, i ( G ) = 1 < ×
4. Hencewe may assume that | V ( H ) | ≥
2. Applying the induction to H we have i ( H ) ≤ | V ( H ) | = ( n − H can be extended to an ID-set of G by adding to itthe vertex w , implying that i ( G ) ≤ i ( H ) + 1 < n . Hence we may assume that there is novertex of degree 2 adjacent to both u and v .We now consider the subcubic graph H = G − { u, v } . With our assumptions, we notethat H has at most three components, each of which has order at least 2. Let H , . . . , H t be the components of H , and so t ≤
3. Let S i be a minimum ID-set of H i for i ∈ [ t ]. Bythe inductive hypothesis, | S i | ≤ | V ( H i ) | for i ∈ [ t ]. Let S (cid:48) = t (cid:91) i =1 S i . If u has no neighbor in S (cid:48) , then let S = S (cid:48) ∪ { u } . If u has a neighbor in S (cid:48) and v has noneighbor in S (cid:48) , then let S = S (cid:48) ∪ { v } . If both u and v have a neighbor in S (cid:48) , then let S = S (cid:48) .In all three cases, the set S is an ID-set of G , and so i ( G ) ≤ | S | ≤ | S (cid:48) | +1 ≤ | V ( H ) | +1 = n .This completes the proof of Theorem 4. (cid:50) We are now in a position to present a proof of Theorem 3. Recall its statement.
Theorem 3 . If G is a subcubic graph of order n with no isolated vertex, then i ( G ) ≤ n .Further, if G is connected, then equality holds if and only if the following holds. (a) G ∈ { G , G , G , G , G } . (b) n = 2 k for some k ≥ and G = cor( P k ) . (c) n = 2 k for some k ≥ and G = cor( C k ) . Proof.
The upper bound i ( G ) ≤ n is a restatement of Theorem 4. If G is a connectedsubcubic graph of order n that satisfies (a), (b) or (c) in the statement of the theorem, thenit is a simple exercise to check that i ( G ) = n . Hence it suffices to prove that if G is aconnected subcubic graph of order n ≥ i ( G ) = n , then (a), (b) or (c) in thestatement of the theorem hold. We proceed by induction on the order n ≥
2. We note that n = 2 i ( G ) is even since i ( G ) is an integer. If n = 2, then G = K = cor( P ). Suppose that8 = 4. If ∆( G ) = 3, then i ( G ) = 1 < ×
4, a contradiction. Hence, ∆( G ) = 2, and soeither G = P = cor( P ) or G = K , = G . This establishes the base cases. Let n ≥ G (cid:48) is a connected subcubic graph of even order n (cid:48) where 2 ≤ n (cid:48) < n satisfying i ( G (cid:48) ) = n (cid:48) , then (a), (b) or (c) in the statement of the theorem hold. Let G bea connected subcubic graph of order n satisfying i ( G ) = n . We proceed further with twoclaims. Recall that G = K , . Claim 1
If the graph G contains no support vertex, then G = G . Proof.
Assume that the graph G contains no support vertex. Thus, every vertex of G hasdegree at least 2 and degree at most 3. If ∆( G ) = 2, then G is a cycle C n where n ≥
6, andso i ( G ) = i ( C n ) = (cid:100) n (cid:101) < n , a contradiction. Hence, ∆( G ) = 3. Let v be an arbitraryvertex of degree 3 in G and let N G ( v ) = { x, y, z } . We now consider the graph H = G \ N G [ v ].Suppose that H has no isolated vertex. Let H , . . . , H t be the components of H and let S i be a minimum ID-set of H i for i ∈ [ t ]. By Theorem 4, | S i | = i ( H i ) ≤ | V ( H i ) | for i ∈ [ t ].Let S (cid:48) = t (cid:91) i =1 S i and S = S (cid:48) ∪ { v } . The set S is an ID-set of G , implying that i ( G ) ≤ | S | = 1 + | S (cid:48) | = 1 + t (cid:88) i =1 | S i |≤ t (cid:88) i =1 | V ( H i ) | = 1 + 12 | V ( H ) | = 1 + 12 ( n − < n, a contradiction. Hence, H contains at least one isolated vertex. If H contains at leastfour isolated vertices, then since δ ( G ) ≥ H has at least twoneighbors in G belonging to the set { x, y, z } , implying by the Pigeonhole Principle that atleast one of the vertices x , y and z has degree at least 4 in G , a contradiction. Therefore, H contains at most three isolated vertices.We show that H contains at most two isolated vertices. Suppose, to the contrary, that H contains three isolated vertices. Since δ ( G ) ≥ G ) = 3, the graph G is nowdetermined. In this case, n = 7 and this contradicts the fact that n is even. Hence, H contains at most two isolated vertices.Next we show that H contains exactly two isolated vertices. Suppose, to the contrary,that H contains exactly one isolated vertex, say u . In this case, we consider the graph9 (cid:48) = H \ { u } . Since n ≥ H (cid:48) contains no isolated vertex, every component of H (cid:48) has order at least 2. Applying Theorem 4 to H (cid:48) , we have i ( H (cid:48) ) ≤ | V ( H (cid:48) ) | = ( n − n is even, this implies that i ( H (cid:48) ) ≤ ( n − H (cid:48) can be extendedto an ID-set of G by adding to it the vertices u and v , implying that i ( G ) ≤ i ( H (cid:48) ) + 2 < n ,a contradiction. Hence, H contains exactly two isolated vertices.Let u and w be the two isolated vertices of H . Each of u and w has either two or threeneighbors in G that belong to the set { x, y, z } , implying that u and w have at least onecommon neighbor.Suppose that u and w have exactly one common neighbor. Renaming the neighbors of v if necessary, we may assume that N G ( u ) = { x, y } and N G ( w ) = { y, z } . In particular, y isthe common neighbor of u and w . Let H (cid:48) = H − { u, w } . We note that H (cid:48) has no isolatedvertex. Applying Theorem 4, i ( H (cid:48) ) ≤ | V ( H (cid:48) ) | = ( n − S be a minimum ID-set of H (cid:48) . If N ( z ) ∩ S = ∅ , then S ∪ { u, z } is an ID-set of G, a contradiction. If N ( x ) ∩ S = ∅ ,then similarly we get a contradiction. Now, assume that N ( z ) ∩ S (cid:54) = ∅ and N ( x ) ∩ S (cid:54) = ∅ .In this case, S ∪ y is an ID-set of G , a contradiction. Hence, u and w have at least twocommon neighbors.Suppose that u and w have exactly two common neighbors. Renaming neighbors of v ifnecessary, we may assume in this case that { x, y } = N G ( u ) ∩ N G ( w ). By assumption, z isadjacent to at most one of u and w . Renaming u and w , we may assume that z is not adjacentto w . If n = 6, then { w, z } is an ID-set of G , implying that i ( G ) = 2 < ×
6, a contradiction.Hence, n ≥
8. We now consider the connected subcubic graph H (cid:48) = G − { u, v, w, x, y } .Applying Theorem 4 to H (cid:48) , we have i ( H (cid:48) ) ≤ | V ( H (cid:48) ) | = ( n − n is even, thisimplies that i ( H (cid:48) ) ≤ ( n − H (cid:48) can be extended to an ID-set of G by adding to it the vertices x and y , implying that i ( G ) ≤ i ( H (cid:48) ) + 2 < n , a contradiction.Hence, the vertices u and w have three common neighbors. The graph G is now deter-mined, and G = K , = G . This completes the proof of the claim. ( (cid:50) ) By Claim 1, we may assume that the graph G contains at least one support vertex, forotherwise G = G and the desired result follows. Since n ≥
6, we note that every supportvertex of G has at most two leaf neighbors. Recall that G is the double star S (2 ,
2) shownin Figure 3(b).
Claim 2
If the graph G contains a support vertex with two leaf neighbors, then G = G . Proof.
Suppose that G contains a support vertex v with two leaf neighbors, say u and w .Let x be the third neighbor of v . Since n ≥
6, we note that d G ( x ) ≥
2. We show that x isa support vertex. Suppose, to the contrary, that x is not a support vertex. In this case, weconsider the subcubic graph H = G − N G [ v ] = G − { u, v, w, x } . Since x is not a supportvertex in G , every component of H has order at least 2. Applying Theorem 4 to H , wehave i ( H ) ≤ | V ( H ) | = ( n − H can be extended to an ID-setof G by adding to it the vertex v , implying that i ( G ) ≤ i ( H ) + 1 < n , a contradiction.Hence, x is a support vertex. 10ext, we show that x has two leaf neighbors. Suppose, to the contrary, that x has exactlyone leaf neighbor, say y . Since n ≥
6, we note that in this case the vertex x has degree 3.We consider the connected subcubic graph H = G − { u, v, w, x, y } . We note that H hasorder at least 2. Applying Theorem 4 to H , we have i ( H ) ≤ | V ( H ) | = ( n − n is even, this implies that i ( H ) ≤ ( n − H can be extended to anID-set of G by adding to it the vertices v and y , implying that i ( G ) ≤ i ( H ) + 2 < n , acontradiction. Hence, x has exactly two leaf neighbors; that is, G = G . ( (cid:50) ) By Claim 2, we may assume that every support of G has exactly one leaf neighbor, forotherwise G = G and the desired result follows. Among all support vertices of G , let v bechosen so that the following holds, where u is the leaf neighbor of v .(1) The degree, d G ( v ), of v is a minimum.(2) Subject to (1), the number of components of G − { u, v } is a minimum.We note that either d G ( v ) = 2 or d G ( v ) = 3. Further, we note that either G − { u, v } isconnected or has two components. Let H = G − { u, v } . Each component of H contains aneighbor of v . Since u is the only leaf neighbor of v , the graph H has no isolated vertex,and so each component of H has order at least 2. Applying Theorem 4 to H , we have i ( H ) ≤ | V ( H ) | = ( n − H can be extended to an ID-set of G by adding to it the vertex u , implying that n = i ( G ) ≤ i ( H ) + 1 ≤ n . Hence we musthave equality throughout this inequality chain, implying that i ( H ) = | V ( H ) | and thatevery component H (cid:48) of H satisfies i ( H (cid:48) ) = | V ( H (cid:48) ) | . Applying the inductive hypothesis toeach component H (cid:48) of H , the component H (cid:48) satisfies (a), (b) or (c) in the statement of thetheorem. Recall that G and G are the graphs shown in Figure 3(a) and 3(b), respectively. Claim 3
The graph H is connected. Proof.
Suppose, to the contrary, that H is disconnected. Thus, H has two components,say H and H . In particular, this implies that d G ( v ) = 3. Let v i be the neighbor of v thatbelongs to H i for i ∈ [2]. Let H i have order n i for i ∈ [2]. As observed earlier, n i ≥ i ( H i ) = n i for i ∈ [2]. Further, H i satisfies (a), (b) or (c) in the statement of the theoremfor i ∈ [2]. Claim 3.1 H i / ∈ { G , G , G , G , G } for i ∈ [2] . Proof.
Suppose, to the contrary, that H ∈ { G , G , G , G , G } . We note that H (cid:54) = G = K , since the vertex v has degree at most 2 in H . Thus, H = G , in which case n = 4,or H ∈ { G , G , G } , in which case n = 6. Further we note that H − v is a connectedgraph of odd order n − ≥
3, implying by Theorem 4 that i ( H − v ) ≤ ( n − S be a minimum ID-set of H . We note that the set S contains no neighbor of v . Wenow consider the connected subcubic graph G (cid:48) = G − ( V ( H ) \ { v } ). Let G (cid:48) has order n (cid:48) .Since n (cid:48) = n − n + 1 is odd, Theorem 4 implies that i ( G (cid:48) ) ≤ ( n (cid:48) −
1) = ( n − n ). If S (cid:48) is an ID-set of G (cid:48) of minimum cardinality, then S (cid:48) ∪ S is an ID-set of G , implying that i ( G ) ≤ | S | + | S (cid:48) | ≤
12 ( n −
2) + 12 ( n − n ) < n,
11 contradiction. ( (cid:50) ) By Claim 3.1 and our earlier observations, H i satisfies (b) or (c) in the statement of thetheorem for i ∈ [2]. Thus, n i = 2 k i and H i = cor( P k i ) for some k i ≥ H i = cor( C k i )for some k i ≥ i ∈ [2]. Recall that among all support vertices of G , the vertex v was chosen to have minimum degree. This implies that if H i = cor( P k i ), then k i ≥ i ∈ [2], for otherwise if H i = cor( P ) = P , then the vertex v i would be a supportvertex of G of degree 2, a contradiction. In particular, we note that n i ≥ i ∈ [2]. If H = cor( P k ) for some k ≥
2, then at least one of the two support vertices of degree 2 in H is a support vertex of degree 2 in G , contradicting our choice of the support vertex v .If H = cor( C k ) for some k ≥
3, then at least one support vertex (of degree 3) in H is asupport vertex in G . However, the removal of such a support vertex and its leaf neighbor in G produces a connected graph, once again contradicting our choice of the support vertex v .This completes the proof of Claim 3. ( (cid:50) ) By Claim 3, the graph H is connected. Let H have order n (cid:48) , and so n (cid:48) = n −
2. Asobserved earlier, H satisfies (a), (b) or (c) in the statement of the theorem. Thus, H ∈{ G , G , G , G , G } or n (cid:48) = 2 k (cid:48) for some k (cid:48) ≥ H = cor( P k (cid:48) ) or n (cid:48) = 2 k (cid:48) for some k (cid:48) ≥ H = cor( C k (cid:48) ). Claim 4 If H ∈ { G , G , G , G , G } , then G = G . Proof.
Suppose that H ∈ { G , G , G , G , G } . We consider each possibility in turn.Suppose that H = G = C , and so n = 6. Let H be the cycle C : w w w w w , where vw is an edge of G . If vw is not an edge of G , then { v, w } is an ID-set of G , and so i ( G ) = 2 < ×
6, a contradiction. Hence, vw is an edge of G , implying that G = G .We note that the vertex v has one or two neighbors in H , and each neighbor of v in H has degree at most 2 in H , implying that H (cid:54) = G ∼ = K , .Suppose that H = G , and so n = 8. Let a and a be the two vertices of H with threecommon neighbors, say b , b and b , where b has degree 3 in H . Let w be the leaf neighborof b in H . If vw ∈ E ( G ), then let S = { a , a , v } . If vw / ∈ E ( G ) and v is adjacent to both b and b , then let S = { b , v } . If vw / ∈ E ( G ) and v is adjacent to exactly one of b and b ,say to b , then let S = { b , b , v } . In all three cases, the set S is an ID-set of G and | S | ≤ i ( G ) ≤ < ×
8, a contradiction.Suppose that H = G , and so n = 8. Let x and y be the two central vertices of thedouble star H . By our earlier assumptions, every support vertex of G has exactly one leafneighbor. Hence, the vertex v is adjacent in G to a leaf neighbor in H of x and a leafneighbor in H of y . Thus if x (cid:48) be the leaf neighbor of x in H that is not adjacent to v in G , then the set { v, x (cid:48) , y } is an ID-set of G , and so i ( G ) ≤ < ×
8, a contradiction.Suppose that H = G , and so n = 8. Thus, H is obtained from a path a a a a a a byadding the edge a a . Suppose that v is adjacent to a or a , say to a . If v is not adjacentto a , then let S = { v, a , a } . If v is adjacent to a , then let S = { v, a } . In both cases,12he set S is an ID-set of G and | S | ≤
3. Thus, i ( w G ) ≤ < ×
8, a contradiction. Hence, v is adjacent to neither a nor a . Thus, the only possible neighbors of v in H are a or a .By symmetry, we may assume that va ∈ E ( G ). Thus, { v, a , a } is an ID-set of G , and so i ( G ) ≤ < ×
8, a contradiction. This completes the proof of Claim 4. ( (cid:50) ) Let n (cid:48) = | V ( H ) | , and so n (cid:48) = n −
2. Recall that n ≥
6, and so n (cid:48) ≥
4. By Claim 4, wemay assume that
H / ∈ { G , G , G , G , G } , for otherwise G = G , and the desired resultfollows. Hence n (cid:48) = 2 k (cid:48) and H = cor( P k (cid:48) ) for some k (cid:48) ≥ H = cor( C k (cid:48) ) for some k (cid:48) ≥ Claim 5 H = cor( P k (cid:48) ) for some k (cid:48) ≥ . Proof.
Suppose that H = cor( C k (cid:48) ) for some k (cid:48) ≥
3. Thus, n (cid:48) = 2 k (cid:48) and n = 2 k (cid:48) + 2. Let H be the corona of the cycle C : v v . . . v k (cid:48) v , and let u i be the resulting leaf neighbor of v i in H for i ∈ [ k (cid:48) ]. Since G is a subcubic graph, we note that the only possible neighborsof v that belong to H are the leaves of H . We now consider the connected subcubicgraph G (cid:48) = G − N G [ v ] of order at least 2. Applying Theorem 4 to the graph G (cid:48) , we have i ( G (cid:48) ) ≤ | V ( G (cid:48) ) | ≤ ( n − n is even, this implies that i ( G (cid:48) ) ≤ ( n − G (cid:48) can be extended to an ID-set of G by adding to it the vertex v , implying that i ( G ) ≤ i ( G (cid:48) ) + 1 < n , a contradiction. ( (cid:50) ) By Claim 5, H = cor( P k (cid:48) ) for some k (cid:48) ≥
2. Thus, n (cid:48) = 2 k (cid:48) and n = 2 k (cid:48) + 2. Let H bethe corona of the path P : v v . . . v k (cid:48) , and let u i be the resulting leaf neighbor of v i in H for i ∈ [ k (cid:48) ]. Since G is a subcubic graph, we note that the only possible neighbors of v thatbelong to H are the vertices u i for i ∈ [ k (cid:48) ] or the vertices v and v k (cid:48) of degree 2 in H . Claim 6 If d G ( v ) = 2 , then G = cor( P k ) where k = k (cid:48) + 1 . Proof.
Suppose that d G ( v ) = 2. Let w be the neighbor of v different from u . If w = u i forsome i ∈ [ k (cid:48) ], then we consider the connected subcubic graph G (cid:48) = G − { u, v, w } of order atleast 3. Applying Theorem 4 to the graph G (cid:48) , we have i ( G (cid:48) ) ≤ | V ( G (cid:48) ) | = ( n − n is even, this implies that i ( G (cid:48) ) ≤ ( n − G (cid:48) can be extendedto an ID-set of G by adding to it the vertex v , implying that i ( G ) ≤ i ( G (cid:48) ) + 1 < n , acontradiction. Hence, either w = v or w = v k (cid:48) . In both cases, G = cor( P k ) where k = k (cid:48) +1,as desired. ( (cid:50) ) By Claim 6, we may assume that d G ( v ) = 3. Hence, the vertex v has two neighbors in H , say w and x . If both neighbors w and x are leaves in H , then we consider the connectedsubcubic graph G (cid:48) = G − { u, v, w, x } of order at least 2. Applying Theorem 4 to the graph G (cid:48) , we have i ( G (cid:48) ) ≤ | V ( G (cid:48) ) | = ( n − G (cid:48) can be extended toan ID-set of G by adding to it the vertex v , implying that i ( G ) ≤ i ( G (cid:48) ) + 1 < n , acontradiction. Hence, renaming w and x if necessary, we may assume that w = v .If x = v k (cid:48) , then G = cor( C k ) where k = k (cid:48) + 1, and the desired result follows. Hence, wemay assume that x is a leaf of H . 13f x = u , then again we consider the connected subcubic graph G (cid:48) = G − { u, v, w, x } ,and as before obtain the contradiction i ( G ) ≤ i ( G (cid:48) ) + |{ v }| < n . Hence, x = u i for some i ∈ [ k (cid:48) ] \ { } . Suppose that k (cid:48) ≥
3. In this case, we consider the connected subcubic graph G (cid:48) = G − { u, v, w, x, u } of order at least 3. Applying Theorem 4 to the graph G (cid:48) , wehave i ( G (cid:48) ) ≤ | V ( G (cid:48) ) | = ( n − n is even, this implies that i ( G (cid:48) ) ≤ ( n − G (cid:48) can be extended to an ID-set of G by adding to it the vertices u and v , implying that i ( G ) ≤ i ( G (cid:48) ) + 2 < n , a contradiction. Hence, k (cid:48) = 2, implying that G = G , and the desired result follows. This completes the proof of Theorem 3. ( (cid:50) ) References [1] G. Abrishami, and M.A. Henning, Independent domination in subcubic graphs of girthat least six,
Discrete Math. (2018), 155–164.[2] C. Brause and M.A. Henning, Independent domination in bipartite cubic graphs,
Graphs Combin (4) (2019), 881–919.[3] P. Dorbec, M.A. Henning, M. Montassier, and J. Southey, Independent domination incubic graphs, J. Graph Theory (4) (2015), 329–349.[4] O. Favaron, A bound on the independent domination number of a tree, Vishwa Inter-nat. J. Graph Theory (1992), 19–27.[5] M. Furuya, K. Ozeki, and A. Sasaki, On the ratio of the domination number and theindependent domination number in graphs, Discrete Appl. Math. (2014), 157–159.[6] W. Goddard and M.A. Henning, Independent domination in graphs: A survey andrecent results,
Discrete Math . (2013), 839–854.[7] W. Goddard, M.A. Henning, J. Lyle, and J. Southey, On the independent dominationnumber of regular graphs, Annals Combin . (2012), 719–732.[8] W. Goddard and J. Lyle, Independent dominating sets in triangle-free graphs, J. Comb.Optim. (1) (2012), 9–20.[9] J. Haviland, Independent domination in regular graphs, Discrete Math. (1995),275–280.[10] J. Haviland, Upper bounds for independent domination in regular graphs,
DiscreteMath. (2007), 2643–2646.[11] M.A. Henning, C. L¨owenstein, and D. Rautenbach, Independent domination in subcu-bic bipartite graphs of girth at least six,
Discrete Appl. Math. (2014), 399–403.[12] M.A. Henning and A. Yeo,
Total Domination in Graphs (Springer Monographs inMathematics)
Graphs Combin. (3) (1993), 235–237.[14] P.C.B. Lam, W.C. Shiu, and L. Sun, On independent domination number of regulargraphs, Discrete Math. (1999), 135–144.[15] J. Lyle, A note on independent sets in graphs with large minimum degree and smallcliques,
Electr. J. Comb. (2) P2.38 (2014).[16] J. Lyle, A structural approach for independent domination of regular graphs, GraphsCombin. (5) (2015), 1567–1588.[17] S. O and D.B. West, Cubic graphs with large ratio of independent domination numberto domination number, Graphs Combin. (2) (2016), 773–776.[18] N.J. Rad and L. Volkmann, A note on the independent domination number in graphs, Discrete Appl. Math. (2013) 3087–3089.[19] J. Southey and M.A. Henning, Domination versus independent domination in cubicgraphs,
Discrete Math. (11) (2013), 1212–1220.[20] S. Wang and B. Wei, A note on the independent domination number versus the domi-nation number in bipartite graphs,
Czechoslovak Math. J.67