Trees with Large Neighborhood Total Domination Number
aa r X i v : . [ m a t h . C O ] A ug Trees with Large Neighborhood Total Domination Number
Michael A. Henning ∗ and Kirsti Wash † Department of Pure and Applied MathematicsUniversity of JohannesburgAuckland Park, 2006, South AfricaEmail: [email protected]: [email protected]
Abstract
In this paper, we continue the study of neighborhood total domination in graphsfirst studied by Arumugam and Sivagnanam [Opuscula Math. 31 (2011), 519–531]. Aneighborhood total dominating set, abbreviated NTD-set, in a graph G is a dominatingset S in G with the property that the subgraph induced by the open neighborhood ofthe set S has no isolated vertex. The neighborhood total domination number, denotedby γ nt ( G ), is the minimum cardinality of a NTD-set of G . Every total dominatingset is a NTD-set, implying that γ ( G ) ≤ γ nt ( G ) ≤ γ t ( G ), where γ ( G ) and γ t ( G ) de-note the domination and total domination numbers of G , respectively. Arumugamand Sivagnanam posed the problem of characterizing the connected graphs G of order n ≥ γ nt ( G ) = ⌈ n/ ⌉ . A partial solution to this problem was presented by Henning and Rad[Discrete Applied Mathematics 161 (2013), 2460–2466] who showed that 5-cycles andsubdivided stars are the only such graphs achieving equality in the bound when n isodd. In this paper, we characterize the extremal trees achieving equality in the boundwhen n is even. As a consequence of this tree characterization, a characterization ofthe connected graphs achieving equality in the bound when n is even can be obtainednoting that every spanning tree of such a graph belongs to our family of extremal trees. Keywords:
Domination; Total domination; Neighborhood total domination.
AMS subject classification: 05C69 ∗ Research supported in part by the South African National Research Foundation and the University ofJohannesburg † Research supported in part by the University of Johannesburg Introduction
In this paper we continue the study of a parameter, called the neighborhood total dom-ination number, that is squeezed between arguably the two most important dominationparameters, namely the domination number and the total domination number. A domi-nating set in a graph G is a set S of vertices of G such that every vertex in V ( G ) \ S isadjacent to at least one vertex in S . The domination number of G , denoted by γ ( G ), isthe minimum cardinality of a dominating set of G . A total dominating set , abbreviated aTD-set, of a graph G with no isolated vertex is a set S of vertices of G such that everyvertex in V ( G ) is adjacent to at least one vertex in S . The total domination number of G , denoted by γ t ( G ), is the minimum cardinality of a TD-set of G . The literature on thesubject of domination parameters in graphs up to the year 1997 has been surveyed anddetailed in the two books [5, 6]. Total domination is now well studied in graph theory. Fora recent book on the topic, see [10]. A survey of total domination in graphs can also befound in [7].Arumugam and Sivagnanam [1] introduced and studied the concept of neighborhood totaldomination in graphs. A neighbor of a vertex v is a vertex different from v that is adjacentto v . The neighborhood of a set S is the set of all neighbors of vertices in S . A neighborhoodtotal dominating set , abbreviated NTD-set, in a graph G is a dominating set S in G withthe property that the subgraph induced by the open neighborhood of the set S has noisolated vertex. The neighborhood total domination number of G , denoted by γ nt ( G ), is theminimum cardinality of a NTD-set of G . A NTD-set of G of cardinality γ nt ( G ) is called a γ nt ( G )- set .Every TD-set is a NTD-set, while every NTD-set is a dominating set. Hence the neigh-borhood total domination number is bounded below by the domination number and aboveby the total domination number as first observed by Arumugam and Sivagnanam in [1]. Observation 1 ([1, 8]) If G is a graph with no isolated vertex, then γ ( G ) ≤ γ nt ( G ) ≤ γ t ( G ) . For notation and graph theory terminology not defined herein, we refer the reader to [5].Let G be a graph with vertex set V ( G ) of order n = | V ( G ) | and edge set E ( G ) of size m = | E ( G ) | , and let v be a vertex in V . We denote the degree of v in G by d G ( v ). The minimumdegree among the vertices of G is denoted by δ ( G ). A vertex of degree one is called a leaf and its neighbor a support vertex . We denote the set of leaves in G by L ( G ), and the setof support vertices by S ( G ). A support vertex adjacent to two or more leaves is a strongsupport vertex . For a set S ⊆ V , the subgraph induced by S is denoted by G [ S ]. A 2- packing in G is a set of vertices that are pairwise at distance at least 3 apart in G .A cycle and path on n vertices are denoted by C n and P n , respectively. A star on n ≥ n − K ,n − . A double star is a2ree containing exactly two vertices that are not leaves (which are necessarily adjacent). A subdivided star is a graph obtained from a star on at least two vertices by subdividing eachedge exactly once. The subdivided star obtained from a star K , , for example, is shown inFigure 1. We note that the smallest two subdivided stars are the paths P and P . Let F be the family of all subdivided stars. Let F ∈ F . If F = P , we select a leaf of F and callit the link vertex of F , while if F = P , the link vertex of F is the central vertex of F .Figure 1: A subdivided star.The open neighborhood of v is the set N G ( v ) = { u ∈ V | uv ∈ E } and the closed neigh-borhood of v is N G [ v ] = { v } ∪ N G ( v ). For a set S ⊆ V , its open neighborhood is the set N G ( S ) = S v ∈ S N G ( v ), and its closed neighborhood is the set N G [ S ] = N G ( S ) ∪ S . If thegraph G is clear from the context, we simply write d( v ), N ( v ), N [ v ], N ( S ) and N [ S ] ratherthan d G ( v ), N G ( v ), N G [ v ], N G ( S ) and N G [ S ], respectively. As observed in [8] a NTD-setin G is a set S of vertices such that N [ S ] = V and G [ N ( S )] contains no isolated vertex.A rooted tree distinguishes one vertex r called the root . For each vertex v = r of T , the parent of v is the neighbor of v on the unique ( r, v )-path, while a child of v is any otherneighbor of v . A descendant of v is a vertex u such that the unique ( r, u )-path contains v .Let C ( v ) and D ( v ) denote the set of children and descendants, respectively, of v , and let D [ v ] = D ( v ) ∪ { v } . The maximal subtree at v is the subtree of T induced by D [ v ], and isdenoted by T v . The following upper bound on the neighborhood total domination number of a connectedgraph in terms of its order is established in [8].
Theorem 2 ([8]) If G is a connected graph of order n ≥ , then γ nt ( G ) ≤ ( n + 1) / . In this paper we consider the following problem posed by Arumugam and Sivagnanam [1]to characterize the connected graphs of largest possible neighborhood total dominationnumber.
Problem 1 ([1])
Characterize the connected graphs G of order n for which γ nt ( G ) = ⌈ n/ ⌉ . A partial solution to this problem was presented by Henning and Rad [8] who providedthe following characterization in the case when n is odd.3 heorem 3 ([8]) Let G = C be a connected graph of order n ≥ . If γ nt ( G ) = ( n + 1) / ,then G ∈ F . As first observed in [8], a characterization in the case when n is even and the minimumdegree is at least 2 follows readily from a result on the restrained domination number ofa graph due to Domke, Hattingh, Henning and Markus [4]. Let B , B , . . . , B be the fivegraphs shown in Figure 2. B B B B B Figure 2: The five graphs B , B , . . . , B . Theorem 4 ([8])
Let G = C be a connected graph of order n ≥ with δ ( G ) ≥ . If γ nt ( G ) = n/ , then G ∈ { B , B , B , B , B } . T of Trees In this section we define a family of trees T as follows. Let T be an arbitrary tree. Let T be the tree obtained from T by the following operation: for each vertex x ∈ V ( T ), eitheradd a new vertex and an edge joining it to x or add a new path P and an edge joiningits central vertex to x . Let T be the family of all trees T that can be obtained from T byperforming the following operation: • Choose a set of leaves, L , in T , that form a 2-packing (possibly L = ∅ ). For eachvertex v ∈ L , add k ≥ P and join v to exactly one endof each added copy of P . We refer to these k added copies of P as appended P sassociated with x . Figure 3: A tree in the family T .A tree in the family T is illustrated in Figure 3. For ease of reference, we introduce someterminology for a tree T ∈ T . We use the standard notation [ k ] = { , , . . . , k } . First notethat given a tree T ∈ T , a tree T used to construct the tree T may not be unique. That4s, in some cases we may be able to choose two distinct trees T and T ′ such that T isobtained from either T or T ′ by performing different combinations of the above operations.Therefore, we refer to T as an underlying tree of T , and we refer to T as the correspondingbase tree of T .The vertex set V ( T ) of T can be partitioned into sets V , . . . , V ℓ such that each V i contains exactly one vertex of T and T [ V i ] ∈ { P , K , } for each i ∈ [ ℓ ]. We say that T [ V i ]is a P -unit of T if T [ V i ] = P and T [ V i ] is a star-unit otherwise.If x ∈ L belongs to a P -unit of T , then an appended P associated with x we call aType-1 appended P , while if x ∈ L belongs to a star-unit of T , then an appended P associated with x we call a Type-2 appended P .For each vertex v that is the central vertex of a star-unit of T , we denote the two leafneighbors of v in T that do not belong to the underlying tree T by a v and b v . If a v ∈ L ,then a v has appended P ’s in T and, by construction, b v remains a leaf in T (since L isa 2-packing). In this case, we say that v and b v are blocked vertices . Similarly, if b v hasappended P ’s in T , then v and a v are blocked vertices. If a v and b v are both leaves of T ,then only v is a blocked vertex. We shall adopt the convention that if one of a v or b v is ablocked vertex of T , then renaming vertices if necessary, b v is the blocked vertex. By Theorem 2, every connected graph G of order n ≥ γ nt ( G ) ≤ ( n + 1) /
2. If T is a tree of order n ≥ γ nt ( T ) = ( n + 1) /
2, then by Theorem 3, T is a subdivided star.Our aim in this paper is to characterize the trees T of order n ≥ γ nt ( T ) = n/ Theorem 5
Let T be a tree of order n ≥ . If γ nt ( T ) = n/ , then T ∈ T . Proof.
We proceed by induction on the order n ≥ T satisfying γ nt ( T ) = n/ n = 4, then either T = P or T = K , . In both cases, γ nt ( T ) = 2 = n/
2. If T = P (respectively, T = K , ), then T ∈ T with P (respectively, K ) as the unique underlyingtree and the tree T itself as the corresponding base tree. This establishes the base case. Let n ≥ T ′ is a tree of order n ′ where 4 ≤ n ′ < n satisfying γ nt ( T ) = n ′ / T ′ ∈ T . Let T be a tree of order n satisfying γ nt ( T ) = n/
2. Our aim is to show that T ∈ T . For this purpose, we introduce some additional notation.For a subtree T ′ of the tree T that belongs to the family T , we adopt the followingnotation in our proof. Let T ′ be an underlying tree of T ′ with corresponding base tree T ′ .For each vertex x ∈ V ( T ′ ), we let N x = N T ′ ( x ) \ V ( T ′ ), and so N x consists of all neighborsof x in T ′ that do not belong to the base tree T ′ . Further, we let L x consist of all leaves of T ′ at distance 2 from x that do not belong to the base tree T ′ . Necessarily, a vertex in N x is a support vertex of T ′ that belongs to a P appended to x , while a vertex in L x is a leafof T ′ that belongs to a P appended to x . 5et A be the vertex set of the underlying tree T ′ ; that is, A = V ( T ′ ). Let B be theset of vertices in the base tree T ′ that do not belong to the underlying tree T ′ ; that is, B = V ( T ′ ) \ V ( T ′ ). Further, let B be the set of all central vertices of star-units of T ′ . Let C = V ( T ′ ) \ V ( T ′ ) be the set of vertices of T ′ that belong to a Type-1 or Type-2 appended P . We note that ( A, B, C ) is a partition of the vertex set V ( T ′ ), where possibly C = ∅ .Let C (respectively, C ) be the set of all leaves (respectively, support vertices) of T ′ thatdo not belong to the base tree T ′ . We note that ( C , C ) is a partition of the set C . Let D ′ = A ∪ B ∪ C . Then the set D ′ is a NTD-set of T ′ (recall that n ′ ≥ | D ′ | = n ′ / γ nt ( T ′ ), theset D ′ is therefore a γ nt ( T ′ )-set.For each vertex x ∈ A , we let A x be the set of neighbors of x in A that have degree 2 in T ′ and belong to a P -unit in T ′ . Let B x be the set of all vertices in B that are neighborsof vertices in A x and let C x be the set of all vertices of C that are neighbors of vertices in B x . Further, let D x be the set of all vertices of C that are neighbors of vertices in C x . Wenote that each vertex in C x is a support vertex of T ′ that belongs to a Type-1 appended P ,while each vertex in D x is a leaf of T ′ that belongs to a Type-1 appended P . We note that | A x | = | B x | and | C x | = | D x | , although possibly A x = ∅ (in which case B x = ∅ ). Further welet A x be the set of vertices in A x that are support vertices in T ′ and we let B x be the setof leaf-neighbors of vertices in A x . Possibly, A x = ∅ . We note that | A x | = | B x | . If A x = ∅ and A x = A x , then B x is the set B x of leaves of T ′ (and in this case C x = D x = ∅ ).We now return to our proof of Theorem 5. If T is a star, then γ nt ( T ) = 2 < n/
2, acontradiction. If T is a double star, then the two vertices that are not leaves form a NTD-set, implying that γ nt ( T ) = 2 < n/
2, a contradiction. Therefore, diam( T ) ≥
4. Let P be alongest path in T and suppose that P is an ( r, u )-path. Necessarily, r and u are leaves in T .We now root the tree T at the vertex r . Let v be the parent of u , and let w be the parentof v in the rooted tree T . Among all such paths P , we may assume that P is chosen so that d T ( v ) is minimum. Thus if P ′ is an arbitrary longest path in T and P ′ is an ( r ′ , u ′ )-pathwith v ′ the neighbor of u ′ on P ′ , then d T ( v ′ ) ≥ d T ( v ).We proceed further with the following claim. Claim A If d T ( v ) = 2 , then T ∈ T . Proof of Claim A.
Suppose that d T ( v ) = 2. Let T ′ = T − { u, v } have order n ′ , and so n ′ = n − ≥
4. Since n is even, so too is n ′ . By Theorem 2, γ nt ( T ′ ) ≤ n ′ /
2. Let D ∗ be a γ nt ( T ′ )-set. If w ∈ D ∗ , let D = D ∗ ∪ { v } . If w / ∈ D ∗ , let D = D ∗ ∪ { u } . In both cases, theset D is a NTD-set of T , and so n γ nt ( T ) ≤ | D | = | D ∗ | + 1 ≤ n ′ n . Hence we must have equality throughout the above inequality chain. In particular, thisimplies that γ nt ( T ′ ) = n ′ /
2. Applying the inductive hypothesis to the tree T ′ , we have6 ′ ∈ T . Adopting our earlier notation, let D ′ = A ∪ B ∪ C and recall that D ′ is a γ nt ( T ′ )-set. We now consider the parent, w , of the vertex v in the rooted tree T . If w ∈ A , then T ∈ T with T [ A ∪ { v } ] as an underlying tree of T and T [ A ∪ B ∪ { u, v } ] as the correspondingbase tree. If w ∈ B and w is not a blocked vertex, then T ∈ T with T ′ as an underlying treeof T and T ′ as the corresponding base tree. Therefore, we may assume that either w ∈ B and w is a blocked vertex or w ∈ C , for otherwise T ∈ T as desired. We proceed further byconsidering the following three cases. Case 1. w ∈ B and w is a blocked vertex. Thus, w is a blocked vertex contained in astar-unit of T ′ . Let x be the vertex of A that belongs to the star-unit containing w and let y be the central vertex of the star-unit. If w ∈ B (and so, w = y ), then the set( D ′ \ ( A x ∪ { x } )) ∪ ( B x ∪ { v } )is a NTD-set of T of size | D ′ | + | B x | − | A x | = | D ′ | = n ′ / n/ −
1, implying that γ nt ( T ) < n/
2, a contradiction. Hence, w / ∈ B and w is therefore a leaf-neighbor of y inthe star-unit that contains it. Recall that a y and b y denote the two leaf-neighbors of y inthe star-unit that do not belong to A . Since w is a blocked vertex, by convention we have w = b y . We note that at least one Type-2 P is appended to a y in order for b y to be ablocked vertex. If | A | ≥
2, then the set( D ′ \ ( A x ∪ D x ∪ { x } )) ∪ ( B x ∪ C x ∪ { u } )is a NTD-set of T of size | D ′ | + ( | B x | − | A x | ) + ( | C x | − | D x | ) = | D ′ | , a contradiction. Hence, A consists only of the vertex x . Thus, T ∈ T with T [ { v, w } ] as an underlying tree of T and T [ { u, v, w, x, y, a y } ] as the corresponding base tree. Case 2. w ∈ C and w belongs to a Type- appended P . Suppose firstly that w is a leafof T ′ , and so w ∈ C . Let x be the neighbor of w that belongs to C , let y be the neighborof x that belongs to B and let z be the neighbor of y that belongs to A . If the vertex z hasa neighbor in A that is not a support vertex of degree 2 in T ′ , then the set( D ′ \ ( A z ∪ { w, z } )) ∪ ( B z ∪ { u, x } )is a NTD-set of T of size | D ′ | + | B z | − | A z | = | D ′ | , a contradiction. Hence, A = { z } orevery neighbor of z that belongs to A is a support vertex of degree 2 in T ′ . In this case, A = A z ∪ { z } , C = N y and C = L y . Thus, T ∈ T with T [ N y ∪ { y } ] as an underlying treeof T and T [ C ∪ { y, z } ] as the corresponding base tree.Suppose secondly that w is a support vertex of T ′ , and so w ∈ C . Let x be the leaf-neighbor of w in T ′ . Let y be the neighbor of w that belongs to B and let z be the neighborof y that belongs to A . If the vertex z has a neighbor in A that is not a support vertex ofdegree 2 in T ′ , then the set ( D ′ \ ( A z ∪ { x, z } )) ∪ ( B z ∪ { v, w } )is a NTD-set of T of size | D ′ | + | B z | − | A z | = | D ′ | , a contradiction. Hence, A = { z } orevery neighbor of z that belongs to A is a support vertex of degree 2 in T ′ . In this case,7 = A z ∪ { z } , C = N y and C = L y . Thus, T ∈ T with T [ N y ∪ { v, y } ] as an underlyingtree of T and T [ C ∪ { u, v, y, z } ] as the corresponding base tree. Case 3. w ∈ C and w belongs to a Type- appended P . Suppose firstly that w is a leaf of T ′ , and so w ∈ C . Let x be the neighbor of w that belongs to C . Let a y be the neighborof x that belongs to B and let y be the central vertex of the star-unit that contains a y . Wenote that b y is a leaf in T ′ . Let z be the neighbor of y that belongs to A . If the vertex z has a neighbor in A that is not a support vertex of degree 2 in T ′ , then the set( D ′ \ ( A z ∪ { w, y, z } )) ∪ ( B z ∪ { u, x, b y } )is a NTD-set of T of size | D ′ | , a contradiction. Hence, A = { z } or every neighbor of z thatbelongs to A is a support vertex of degree 2 in T ′ . In this case, A = A z ∪ { z } and C = N a y .Thus, T ∈ T with T [ C ∪ { a y } ] as an underlying tree of T and T [ C ∪ { y, a y , b y , z } ] as thecorresponding base tree.Suppose secondly that w is a support vertex of T ′ , and so w ∈ C . Let x be the leaf-neighbor of w in T ′ . Let a y be the neighbor of w that belongs to B and let y be the centralvertex of the star-unit that contains a y . Let z be the neighbor of y that belongs to A . Ifthe vertex z has a neighbor in A that is not a support vertex of degree 2 in T ′ , then the set( D ′ \ ( A z ∪ { x, y, z } )) ∪ ( B z ∪ { v, w, b y } )is a NTD-set of T of size | D ′ | , a contradiction. Hence, A = { z } or every neighbor of z thatbelongs to A is a support vertex of degree 2 in T ′ . In this case, A = A z ∪ { z } and C = N a y .Thus, T ∈ T with T [ C ∪ { v, a y } ] as an underlying tree of T and T [ C ∪ { u, v, y, a y , b y , z } ] asthe corresponding base tree. In all three cases above, we have that T ∈ T . This completesthe proof of Claim A. ( ✷ ) By Claim A, we may assume that d T ( v ) ≥
3, for otherwise T ∈ T as desired. By ourchoice of the path P , every child of w that is not a leaf has degree at least as large as d T ( v ).Let x be the parent of w in T . Since diam( T ) ≥
4, we note that x = r , and so d T ( x ) ≥ w have ℓ ≥ k ≥ W be theset consisting of the vertex w and its k children that are support vertices. Then, | W | = k + 1and, as observed earlier, every vertex in W \ { w } has degree at least d T ( v ) ≥
3. Let thesubtree, T w , of T rooted at w have order n w , and so n w ≥ k + ℓ + 1. Let T ′ = T − V ( T w )be the tree obtained from T by deleting the vertices in the subtree T w of T rooted at w .Let T ′ have order n ′ . Then, n ′ ≥ n ′ = n − n w . Claim B If n ′ = 2 , then T ∈ T . Proof of Claim B.
Suppose that n ′ = 2. Then, n ≥ k + ℓ + 3 and the set W ∪ { x } is aNTD-set of T ′ , and so n/ γ nt ( T ) ≤ | W | + 1 = k + 2 ≤ k + ( k + ℓ + 3) / ≤ n/
2. Hencewe must have equality throughout this inequality chain, implying that k = 1, ℓ = 0, n w = 4and n = 6. Thus, T ∈ T with T [ { w, x } ] as the underlying tree of T and T itself as thecorresponding base tree. ✷
8y Claim B, we may assume that n ′ ≥
3, for otherwise T ∈ T as desired. ApplyingTheorem 2 and Theorem 3 to the tree T ′ , we have that γ nt ( T ′ ) ≤ ( n ′ + 1) /
2, with equalityif and only if T ′ is a subdivided star. Claim C T ′ ∈ T , d T ( w ) = 2 , and d T ( v ) = 3 . Proof of Claim C.
We show firstly that γ nt ( T ′ ) ≤ n ′ /
2. Suppose, to the contrary, that γ nt ( T ′ ) = ( n ′ + 1) / T ′ is a subdivided star. Let y be the link vertex of T ′ , and let Y and Y be the set of vertices at distance 1 and 2, respectively, from y in T ′ . Select anarbitrary vertex y ∈ Y and let y be the common neighbor of y and y , and so yy y is apath in T ′ . Renaming vertices if necessary, we may assume that x ∈ { y, y , y } . If x = y , let Y = Y . If x = y , let Y = ( Y \ { y } ) ∪ { x } . If x = y , let Y = ( Y \ { y } ) ∪ { y } . In all threecases, | Y | = ( n ′ − / W ∪ Y is a NTD-set of T . Recall that n w ≥ k + ℓ + 1and n ′ = n − n w . Hence, γ nt ( T ) ≤ | Y | + | W | = n ′ −
12 + k + 1 ≤ n − k − ℓ −
22 + k + 1= n − k − ℓ ≤ n − , a contradiction. Therefore, γ nt ( T ′ ) ≤ n ′ /
2. Every γ nt ( T ′ )-set can be extended to a NTD-setof T by adding to it the set W . Hence, n γ nt ( T ) ≤ γ nt ( T ′ ) + | W | ≤ n ′ k + 1 ≤ n − k − ℓ + 12 ≤ n . Consequently, we must have equality throughout this inequality chain, implying that k = 1, ℓ = 0, d T ( w ) = 2, d T ( v ) = 3, and γ nt ( T ′ ) = n ′ /
2. Applying the inductive hypothesis to thetree T ′ of (even) order n ′ ≥
4, we deduce that T ′ ∈ T . ✷ By Claim C, d T ( w ) = 2 and d T ( v ) = 3. Thus, N ( w ) = { v, x } . Let u and u be thetwo children of v where u = u . By Claim C, T ′ ∈ T . Adopting our earlier notation, let T ′ have order n ′ . In this case, n ′ = n −
4. Further, let D ′ = A ∪ B ∪ C and recall that D ′ is a γ nt ( T ′ )-set. We now consider the parent, x , of the vertex w in the rooted tree T .If x ∈ A , then T ∈ T with T [ A ∪ { w } ] as an underlying tree of T and T [ A ∪ B ∪ N [ v ]] asthe corresponding base tree. Hence we may assume that x ∈ B ∪ C , for otherwise T ∈ T as desired. Claim D If x ∈ B , then T ∈ T . Proof of Claim D.
Suppose that x ∈ B . We consider two subclaims.9 laim D.1 If x belongs to a P -unit in T ′ , then T ∈ T . Proof of Claim D.1
Suppose that x belongs to a P -unit in T ′ . Let y be the neighborof x that belongs to A . If the vertex y has a neighbor in A that is not a support vertex ofdegree 2 in T ′ , then the set ( D ′ \ ( A y ∪ { y } )) ∪ ( B y ∪ { v, w } )is a NTD-set of T of size | D ′ | + 1 + ( | B y | − | A y | ) = | D ′ | + 1 = n ′ / n/ −
1, acontradiction. Hence, A = { y } or every neighbor of y that belongs to A is a support vertexof degree 2 in T ′ . In this case, A = A y ∪ { y } , C = N x and C = L x . Thus, T ∈ T with T [ N x ∪ { w, x } ] as an underlying tree of T and T [ C ∪ { u , u , v, w, x, y } ] as the correspondingbase tree. ( ✷ ) Claim D.2 If x belongs to a star-unit in T ′ , then T ∈ T . Proof of Claim D.2.
Suppose that x belongs to a star-unit in T ′ . Let z be the vertex of A that belongs to the star-unit containing x and let y be the central vertex of the star-unit.If x ∈ B (and so, x = y ), then the set( D ′ \ ( A z ∪ { z } )) ∪ ( B z ∪ { v, w } )is a NTD-set of T of size | D ′ | +1 = n/ −
1, a contradiction. Hence, x / ∈ B and x is thereforea leaf-neighbor in its star-unit. Recall that a y and b y denote the two leaf-neighbors of y inthe star-unit that do not belong to A . By convention the vertex b y is a leaf in T ′ and thevertex a y has ℓ ≥ P ’s appended to it.Suppose x = a y . If the vertex z has a neighbor in A that is not a support vertex ofdegree 2 in T ′ , then the set( D ′ \ ( A z ∪ { y, z } )) ∪ ( B z ∪ { b y , v, w } )is a NTD-set of T of size | D ′ | + 1, a contradiction. Hence, A = { z } or every neighborof z that belongs to A is a support vertex of degree 2 in T ′ . In this case A = A z ∪ { z } , C = N a y and C = L a y . Thus, T ∈ T with T [ C ∪ { w, a y } ] as an underlying tree of T and T [ C ∪ { a y , b y , u , u , v, w, y, z } ] as the corresponding base tree.Suppose that x = b y . If a y is a leaf of T ′ , then renaming a y and b y , we may assume that x = a y . In this case, we have shown that T ∈ T . Hence we may assume that a y has at leastone Type-2 P appended to it. If | A | ≥
2, then the set( D ′ \ ( A z ∪ D z ∪ { z } )) ∪ ( B z ∪ C z ∪ { v, w } )is a NTD-set of T of size | D ′ | + 1, a contradiction. Hence, A = { z } . Thus, T ∈ T with T [ { b y , w } ] as an underlying tree of T and T [ { a y , b y , u , u , v, w, y, z } ] as the correspondingbase tree. This completes the proof of Claim D.2. ( ✷ ) Claim D now follows from Claim D.1 and Claim D.2. ✷ By Claim D, we may assume that x ∈ C , for otherwise T ∈ T as desired.10 laim E If x belongs to a Type- appended P , then T ∈ T . Proof of Claim E.
Suppose that x belongs to a Type-1 appended P . Suppose firstly that x is a leaf of T ′ , and so x ∈ C . Let y be the neighbor of x that belongs to C , let y bethe neighbor of y that belongs to B and let z be the neighbor of y that belongs to A . If y has degree at least 3 in T ′ , then | N y | ≥ D ′ \ ( A z ∪ L y ∪ { z } )) ∪ ( B z ∪ ( N y \ { y } ) ∪ { v, w, y } )is a NTD-set of T of size | D ′ | + ( | B z |− | A z | )+ ( | N y |− | L y | )+ 3 − | D ′ | + 1, a contradiction.Hence, y has degree 2 in T ′ . If | A | ≥
2, then the set( D ′ \ ( A z ∪ D z ∪ { z } )) ∪ ( B x ∪ C x ∪ { v, w, y } )is a NTD-set of T of size | D ′ | + ( | B x |− | A x | )+ ( | C x |− | D x | )+ 3 − | D ′ | + 1, a contradiction.Hence, A consists only of the vertex z . Therefore, T is obtained from the path u vwxy yz by adding a new vertex u and the edge u v . Thus, T ∈ T with T [ { w, x } ] as an underlyingtree of T and T [ { u , u , v, w, x, y } ] as the corresponding base tree.Suppose secondly that x is a support vertex of T ′ , and so x ∈ C . Let x be the leaf-neighbor of x in T ′ . Let y be the neighbor of x that belongs to B and let z be the neighborof y that belongs to A . If the vertex z has a neighbor in A that is not a support vertex ofdegree 2 in T ′ , then the set( D ′ \ ( A z ∪ { x , z } )) ∪ ( B z ∪ { v, w, x } )is a NTD-set of T of size | D ′ | + 1, a contradiction. Hence, A = { z } or every neighbor of z that belongs to A is a support vertex of degree 2 in T ′ . In this case, A = A z ∪ { z } , C = N y and C = L y . Thus, T ∈ T with T [ N y ∪ { w, y } ] as an underlying tree of T and T [ C ∪ { u , u , v, w, y, z } ] as the corresponding base tree. ✷ By Claim E, we may assume that x belongs to a Type-2 appended P , for otherwise T ∈ T as desired. Claim F
The vertex x is a support vertex of T ′ . Proof of Claim F.
Suppose to the contrary that x is a leaf of T ′ , and so x ∈ C . Let x be the neighbor of x that belongs to C . Let a y be the neighbor of x that belongs to B and let y be the central vertex of the star-unit that contains a y . We note that b y is a leafin T ′ . Let z be the neighbor of y that belongs to A . Then the set( D ′ \ ( A z ∪ { x, z } )) ∪ ( B z ∪ { v, w, a y } )is a NTD-set of T of size | D ′ | + 1, a contradiction. ✷ We now return to the proof of Theorem 5 one last time. By Claim F, the vertex x isa support vertex of T ′ , and so x ∈ C . Let x be the leaf-neighbor of x in T ′ . Let a y be11he neighbor of x that belongs to B and let y be the central vertex of the star-unit thatcontains a y . Let z be the neighbor of y that belongs to A . If the vertex z has a neighbor in A that is not a support vertex of degree 2 in T ′ , then the set( D ′ \ ( A z ∪ { x , y, z } )) ∪ ( B z ∪ { v, w, x, b y } )is a NTD-set of T of size | D ′ | + 1, a contradiction. Hence, A = { z } or every neighborof z that belongs to A is a support vertex of degree 2 in T ′ . In this case, A = A z ∪{ z } and C = N a y . Thus, T ∈ T with T [ C ∪ { w, a y } ] as an underlying tree of T and T [ C ∪ { u , u , v, w, y, a y , b y , z } ] as the corresponding base tree. This completes the proof ofTheorem 5. ✷ As a consequence of our tree characterization provided in Theorem 5, we remark that acomplete solution to the Arumugam-Sivagnanam Problem 1 can be obtained as follows.Let G be a connected graph of (even) order n ≥ γ nt ( G ) = n/ T of G . By Theorem 2, γ nt ( T ) ≤ n/
2. Every NTD-set of T is anNTD-set of G , implying that n/ γ nt ( G ) ≤ γ nt ( T ) ≤ n/
2. Consequently, γ nt ( T ) = n/ T ∈ T . This is true for every spanning tree T of the graph G . Anexhaustive case analysis of the allowable edges that can be added to trees in the family T without lowering their neighborhood total domination number produces the connectedgraphs G of order n ≥ γ nt ( G ) = n/
2. Since our detailed case analysis of theresulting such graphs exceeds the length of the current paper, we omit the details herewhich can be found in [9].
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