aa r X i v : . [ m a t h . C O ] O c t On total dominating sets in graphs
Maryam Atapour and Nasrin Soltankhah ∗ Department of MathematicsAlzahra UniversityVanak Square 19834 Tehran, I.R. Iran
Abstract
A set S of vertices in a graph G ( V, E ) is called a dominating setif every vertex v ∈ V is either an element of S or is adjacent to anelement of S . A set S of vertices in a graph G ( V, E ) is called a totaldominating set if every vertex v ∈ V is adjacent to an element of S .The domination number of a graph G denoted by γ ( G ) is the minimumcardinality of a dominating set in G . Respectively the total dominationnumber of a graph G denoted by γ t ( G ) is the minimum cardinality ofa total dominating set in G . An upper bound for γ t ( G ) which hasbeen achieved by Cockayne and et al. in [1] is: for any graph G withno isolated vertex and maximum degree ∆( G ) and n vertices, γ t ( G ) ≤ n − ∆( G ) + 1.Here we characterize bipartite graphs and trees which achieve this upperbound. Further we present some another upper and lower bounds for γ t ( G ). Also, for circular complete graphs, we determine the value of γ t ( G ). total dominating set, total domination number Let G ( V, E ) be a graph. For any vertex x ∈ V , we define the neighborhoodof x , denoted by N ( x ), as the set of all vertices adjacent to x . The closedneighborhood of x , denoted by N [ x ], is the set N ( x ) ∪ { x } . For a set of vertices S , we define N ( S ) as the union of N ( x ) for all x ∈ S , and N [ S ] = N ( S ) ∪ S .The degree of a vertex is the size of its neighborhoods. The maximum degree ∗ Corresponding author: E-mail: [email protected], [email protected]. G is denoted by ∆( G ) and the minimum degree is denoted by δ ( G ).Here n will denote the number of vertices of a graph G . A set S of verticesin a graph G ( V, E ) is called a dominating set if every vertex v ∈ V is eitheran element of S or is adjacent to an element of S . A set S of vertices in agraph G ( V, E ) is called a total dominating set if every vertex v ∈ V is adjacentto an element of S . The domination number of a graph G denoted by γ ( G )is the minimum cardinality of a dominating set in G . Respectively the totaldomination number of a graph G denoted by γ t ( G ) is the minimum cardinalityof a total dominating set in G . clearly γ ( G ) ≤ γ t ( G ), also it has been provedthat γ t ( G ) ≤ γ ( G ).An upper bound for γ t ( G ) has been achieved by Cockayne and et al. in [1] inthe following theorems: THEOREM A
If a graph G has no isolated vertices, then γ t ( G ) ≤ n − ∆( G ) + 1 . THEOREM B If G is a connected graph and ∆( G ) < n − , then γ t ( G ) ≤ n − ∆( G )As a result of the above theorems, if G is a graph with γ t ( G ) = n − ∆( G )+1,then ∆( G ) ≥ n −
1. Hence, if G is a k – regular graph and γ t ( G ) = n − k + 1,then G is K n . As a result of the above theorems, if G is a graph with γ t ( G ) = n − ∆( G ) + 1, then ∆( G ) ≥ n −
1. Hence, if G is a k – regular graph and γ t ( G ) = n − k + 1, then G is K n . Total domination and upper bounds on thetotal domination number in graphs were intensively investigated, see e. g. ([3], [4]).Here we characterize bipartite graphs and trees which achieve the upper boundin Theorem A. Further we present some another upper and lower bounds for γ t ( G ). Also, for circular complete graphs, we determine the value of γ t ( G ).It is easy to prove that for n ≥ γ t ( C n ) = γ t ( P n ) = n if n ≡ γ t ( C n ) = γ t ( P n ) = ⌊ n ⌋ + 1 otherwise.for the definitions and notations not defined here we refer the reader totexts, such as [2]. γ t ( G ) In this section we introduce some other upper bounds for γ t ( G ). Theorem 2.1
Let G be a connected graph, then γ t ( G ) ≥ ⌈ n ∆( G ) ⌉ . Proof:
Let S ⊆ V ( G ) be a total dominating set in G . Every vertex in S dominates at most ∆( G ) − V ( G ) − S and dominate at least oneof the vertices in S . Hence, | S | (∆( G ) −
1) + | S | ≥ n . Since, S is an arbitrarytotal dominating set, then γ t ( G ) ≥ ⌈ n ∆( G ) ⌉ .If G = K n , G = C n , or G = P n then γ t ( G ) = ⌈ n ∆( G ) ⌉ . so the above bound issharp. Theorem 2.2
Let G be a graph with diam ( G ) = 2 then, γ t ( G ) ≤ δ ( G ) + 1 . Proof:
Let x ∈ V ( G ) and deg( x ) = δ ( G ). Since, diam( G ) = 2, then N ( x ) isa dominating set for G .Now S = N ( x ) ∪ { x } is a total dominating set for G and | S | = δ ( G ) + 1.Hence, γ t ( G ) ≤ δ ( G ) + 1.As we know, γ t ( C ) = 3 and also δ ( C ) = 2, diam ( C ) = 2 then γ t ( C ) = δ ( C ) + 1. Hence, the above bound is sharp. Theorem 2.3 If G is a connected graph with the girth of length g ( G ) ≥ and δ ( G ) ≥ , then γ t ( G ) ≤ n − ⌈ g ( G )2 ⌉ + 1 . Proof:
Let G be a connected graph with g ( G ) ≥ C be a cycle oflength g ( G ). Remove C from G to form a graph G ′ . Suppose an arbitraryvertex v ∈ V ( G ′ ), since δ ( G ) ≥
2, then v has at least two neighbors say x and y . Let x, y ∈ C . If d ( x, y ) ≥
3, then replacing the path from x to y on C withthe path x, v, y reduces the girth of G , a contradiction. If d ( x, y ) ≤
2, then x, y, v are on either C or C in G , contradicting the hypothesis that g ( G ) ≥ G ′ has two or more neighbors on C . Since δ ( G ) ≥
2, thegraph G ′ has minimum degree at least δ ( G ) − ≥
1. Then G ′ has no isolatedvertex. Now let S ′ be a γ t –set for C . Then S = S ′ ∪ V ( G ′ ) is a total dominatingset for G . Hence, γ t ( G ) ≤ n − ⌈ g ( G )2 ⌉ + 1(note that γ t ( C ) ≤ ⌊ g ( G )2 ⌋ + 1) . γ t ( G ) = n − ∆( G ) + 1 In this section we charactrize the bipartite graphs achieving the upper boundin the theorem A.
Theorem 3.4
Let G be a bipartite graph with no isolated vertices. Then γ t ( G ) = n − ∆( G ) + 1 if and only if G is a graph in form of K ,t S rK for r ≥ . Proof: If G is K ,t ∪ rK ( r ≥ γ t ( G ) = n − ∆( G ) + 1. Now let G bea bipartite graph with partitions A S B and x ∈ A where deg( x ) = ∆( G ) = t .We continue our proof in four stages: S tage 1: We claim that for every vertex y ∈ A − { x } , N ( y ) − N ( x ) = ∅ . If itis not true, there exists a vertex in A − { x } , say y , such that N ( y ) ⊆ N ( x ).So let u ∈ N ( y ), the set S = V − ( N ( x ) ∪ { y } ) S { u } is a total dominating setand | S | = n − ∆( G ), a contradiction. So we have n ≥ | A | + ∆( G ) − S tage 2: For every vertex y ∈ A , let u y ∈ N ( y ). Clearly the set S = A ∪ ( ∪ y ∈ A { u y } ) is a total dominating set for G and | S | ≤ | A | , so γ t ( G ) ≤ | A | .Now let y ∈ A − { x } such that | N ( y ) − N ( x ) | ≥
2. Hence, we have: n ≥ | A | + ∆( G ) ⇒ γ t ( G ) + ∆( G ) − ≥ | A | + ∆( G ) ⇒ γ t ( G ) ≥ | A | + 1,a contradiction. Hence, for every vertex y ∈ A − { x } , | N ( y ) − N ( x ) | = 1. S tage 3: Let y ∈ A − { x } and N ( y ) ∩ N ( x ) = ∅ . Let u ∈ N ( y ) ∩ N ( x ). Now, S = ( V − N ( x ) ∪ { y } ) ∪ { u } is a total dominating set and | S | = n − ∆( G ).So, γ t ( G ) ≤ n − ∆( G ), a contradiction. S tage 4: Let y, z ∈ A − { x } and N ( y ) ∩ N ( z ) = ∅ . Now S = ( V − ( { z } ∪ N ( x ))) ∪ { u } , where u ∈ N ( x ), is a total dominating set and | S | = n − ∆( G ).So, γ t ( G ) ≤ n − ∆( G ), a contradiction. Hence, G is a graph in form of K ,t ∪ rK . COROLLARY 3.1
Let T is a Tree. Then γ t ( T ) = n − ∆( T ) + 1 if and onlyif T is a star. If n and d are positive integers with n ≥ d , then circular complete graph K n,d is the graph with vertex set { v , v , . . . , v n − } in which v i is adjacent to v j if andonly if d ≤ | i − j | ≤ n − d . In this section we determine the total dominationof circular complete graphs. It is easy to see that K n, is the complete graph K n and K n, is a circle on n vertices, therefore we assume that d ≥ Theorem 4.5
For n ≥ d − and d ≥ , γ t ( K n,d ) = 2 . Proof:
Clearly, γ t ( K n,d ) ≥
2. Let S = { v , v d − } . We will show that S isa total dominating set for K n,d . Since n ≥ d − d − ≤ d , then2 d − ≤ n − d . Also 2 d − ≥ d since d ≥
3. Thus d ≤ d − ≤ n − d and v v d − ∈ E ( K n,d ). By definition of K n,d , v is adjacent to each of the vertices v d , v d +1 , . . . , v n − d .Now for each 1 ≤ i ≤ d − n − d + i − (2 d −
1) = n − d + i + 1 ≥ d − − d + i + 1 ≥ d and n − d + i − (2 d −
1) = n − d + i + 1 ≤ n − d + d = n − d < n − d. Thus v d − is adjacent to each of the vertices v n − d +1 , . . . , v n − . On the otherhand, for each 1 ≤ i ≤ d − d − − i ≤ d − ≤ d − ≤ n − d and 2 d − − i ≥ d − − d + 1 = d. Hence v d − is adjacent to each of the vertices v , v , . . . , v d − and so S is atotal dominating set for K n,d and γ t ( K n,d ) = 2. Theorem 4.6
For d ≤ n ≤ d − and d ≥ , γ t ( K n,d ) = 3 . Proof:
Let S = { v , v d , v d − } . We prove that S is a γ t ( K n,d )- set. Since d ≤ d − ≤ n − d , G [ S ] contains no isolated vertices. Clearly v and v d are adjacent to each of the vertices v d , v d +1 , . . . , v n − d and v d , v d +1 , . . . , v n − d respectively. For 1 ≤ i ≤ d − d − − i ≤ d − − d + 1 = d and 2 d − − i ≤ d − ≤ d ≤ n − d Thus v d − is adjacent to each of the vertices v , v , . . . , v d − . Hence S is atotal dominating set for K n,d and so γ t ( K n,d ) ≤
3. Now we prove that thereis no total dominating set for K n,d of size 2. Let S ′ = { u, v } be a γ t ( K n,d )-set. Without loss of generality, let u = v and v = v j . Clearly d ≤ j ≤ n − d .Since v v n − d +1 / ∈ E ( K n,d ), d ≤ n − d + 1 − j ≤ n − d and so 1 ≤ j ≤ d + 1.Thus j = d or j = d + 1. In both cases, S ′ is not a total dominating set since v , v , . . . , v d − are not dominated by S ′ a contradiction. This completes theproof. References [1]
E.J. Cockayne, R.M. Dawes, and S.T. Hedetniemi , Total domina-tions in graphs , Networks, (1980), 211–219.[2] T.W. Haynes and S.T. Hedetniemi , Funamentals domination ingraphs , Marcel Dekker, New York, 1998.[3]
M.A. Henning and A. Yeo , A new upper bound on the total dominationnumber of a graph , Electronic J. of Combinatorics, (2007), P.C.B. Lam and B. Wei , On the total domination number of graphs ,Utilitas Math.,72