Roman Domination of the Comet, Double Comet, and Comb Graphs
AAppeared in Fundamenta Informaticae 178(1) : 1–10 (2021). 1Available at IOS Press through https://doi.org/10.3233/FI-2021-0001
Roman Domination of the Comet, Double Comet, and CombGraphs
Derya DO ˘GAN DURGUN C Department of MathematicsManisa Celal Bayar UniversityMartyr Prof. Dr. Ilhan Varank Campus Manisa, [email protected]
Emre Niyazi TOPRAKKAYA
Department of MathematicsManisa Celal Bayar UniversityMartyr Prof. Dr. Ilhan Varank Campus Manisa, Turkey
Abstract.
One of the well-known measurements of vulnerability in graph theory is domination.There are many kinds of dominating and relative types of sets in graphs. However, we are goingto focus on Roman domination, which is a type of domination that came up with an article by IanStewart [1]. Let G = ( V, E ) be a graph, the function f : V → { , , } satisfying the conditionthat every vertex u for which f ( u ) = 0 is adjacent to at least one vertex v for which f ( v ) = 2 is a Roman dominating function (RDF). The weight of an RDF is the value f ( V ) = (cid:80) u ∈ V f ( u ) ,which equals | V | + 2 | V | , and the minimum weight of an RDF on a graph G is called the Romandomination number of G in [2], denoted by γ R ( G ) . The Roman domination numbers of thecomet, double comet, and comb graphs are given in this paper. Keywords:
Graph Theory, Vulnerability, Domination, Roman Domination. C Corresponding author a r X i v : . [ c s . D M ] F e b D. DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs
1. Introduction
A graph is a pair of sets G = ( V, E ) , where V is the set of vertices and E is the set of edges, formedby pairs of vertices. An area of graph theory that has received attention during recent decades is thatof domination in graphs. A vertex v in a graph G is said to dominate itself and each of its neighbors,that is, v dominates the vertices in its closed neighborhood N [ v ] = { u ∈ V : uv ∈ E } ∪ { v } . Aset S of vertices of G is a dominating set of G if every vertex of G is dominated by at least onevertex of S . Equivalently, a set S of vertices of G is a dominating set if every vertex in V ( G ) − S is adjacent to at least one vertex in S . The minimum cardinality among the dominating sets of G iscalled the domination number of G and is denoted by γ ( G ) . A dominating set of cardinality γ ( G ) isthen referred to as a minimum dominating set.Now consider a military unit. Each military unit from the largest to the smallest has a very clearchain of command. Therefore, there must be free-flowing communication between the commandingechelon and soldiers. For easy commandment, each of the soldiers should be under command of atleast one commander. For example, consider a battalion’s graph model G . Let each person in thebattalion be a vertex. If a soldier and a commander are linked to each other by commandment relation,connect these two vertices with an edge. In the battalions with this characteristic property, a selectedset of soldiers S or a selected set of commanders V ( G ) − S are dominating sets of this battalion’sgraph model.We study a variant of domination, called Roman Domination, of which origin is about the militarystrategy of the Roman Empire in the 4th century, which is why we gave an example about militaryunits before, for making the reader familiar with the subject. A Roman dominating function (RDF)on a graph G = ( V, E ) is a function f : V → { , , } satisfying the condition that every vertex u for which f ( u ) = 0 is adjacent to at least one vertex v for which f ( v ) = 2 . For an RDF f , let V i ( f ) = { v ∈ V ( G ) : f ( v ) = i } . In the context of a fixed RDF, we suppress the argument and simplywrite V , V , and V . Since this partition determines f , we can equivalently write f = ( V , V , V ) [3]. A function f = ( V , V , V ) is a γ R − f unction if it is an RDF and f ( V ) = γ R ( G ) [2].In this paper, the roman domination number of the comet, double comet, and comb graphs aregeneralized and given with their proofs.
2. Preliminary
For t ≥ and r ≥ , the comet graph C t,r with t + r vertices is the graph obtained by identifying oneend of the path P t with the center of the star K ,r [4]. . DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs Figure 1. Comet Graph C t,r For a, b ≥ , n ≥ a + b + 2 by DC ( n, a, b ) we denote a double comet graph , which is a treecomposed of a path containing n − a − b vertices with a pendent vertices attached to one of the endsof the path and b pendent vertices attached to the other end of the path [5]. Figure 2. Double Comet Graph DC ( n, a, b ) A vertex of a graph is said to be pendent if its neighborhood contains exactly one vertex. An edgeof a graph is said to be pendent if one of its vertices is a pendent vertex.The comb graph P n Θ K , is the graph obtained from a path P n by attaching pendent edge at eachvertex of the path and is denoted by P + n [6]. Figure 3. Comb Graph P + n D. DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs
3. Main Results
In this section, we shall give the roman domination numbers of three different graphs. The proof ofthe roman domination number for comb graph, we use v + t for the pendent vertices of v t . Theorem 3.1.
Let G = C t,r be a comet graph where t ≥ and r ≥ . Then the roman dominationnumber of G is equal to γ R ( C t,r ) = (cid:40) t + 1 t ≡ (mod 3) (cid:100) t (cid:101) otherwise (1) Proof:
Roman domination number of the comet graph is considered in three cases. Let f = ( V , V , V ) be a γ R − f unction of G . Case 1: t ≡ (mod 3) In order to dominate u , u , u , . . . , u r , v and v vertices, v vertex should be taken into V set.To dominate v t − , v t − and v t − vertices, v t − vertex should be taken into V set. To dominate v t vertex, v t itself should be taken into V set. For the rest vertices of the graph which are not dominated, v t vertices should be taken into V set which satisfy t ≡ (mod 3). Then for a Roman dominatingfunction (RDF) f , V = { v , v , v , . . . , v t − } and V = { v t } .So that f ( V ) = 1 + 2( t − − + 1) then we get γ R ( C t,r ) ≤ t + 1 .Let f not be a γ R − f unction and by deleting v t vertex from V = { v ∈ V : f ( v ) = 1 } set, let V = ∅ . Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v t ) , obtained function f (cid:48) = ( V ∪ { v t } , ∅ , V ) does not satisfythe condition to be an RDF. According to this γ R ( C t,r ) ≥ t + 1 . For f (cid:48) function to be an RDF, v t vertex should be taken into V set; f (cid:48) = ( V , ∅ , V ∪ { v t } ) . Hence we get f (cid:48) ( V ) = 2 t + 2 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( C t,r ) ≥ t + 1 .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, such as v vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v } , V , V − { v } ) does notsatisfy the condition to be an RDF. According to this γ R ( C t,r ) ≥ t +1 . For f (cid:48)(cid:48) function to be an RDF, v , v , v vertices should be taken into V set; f (cid:48)(cid:48) = ( V −{ v , v } , V ∪{ v , v , v } , V −{ v } ) . Hencewe get f (cid:48)(cid:48) ( V ) = 2 t + 2 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( C t,r ) ≥ t + 1 .Consequently γ R ( C t,r ) = 2 t + 1 . Case 2: t ≡ (mod 3) i) In order to dominate u , u , u , . . . , u r , v and v vertices, v vertex should be taken into V set.To dominate v t − , v t − and v t vertices, v t − vertex should be taken into V set. For the rest vertices ofthe graph which are not dominated, v t vertices should be taken into V set which satisfy t ≡ (mod 3).Because of the v t − vertex is dominated by v t − vertex at the same time, taking v t vertex into V setinstead of v t − vertex does not change the result. Then for an RDF f , V = { v , v , v , . . . , v t − } or V = { v , v , v , . . . , v t } and V = ∅ . . DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs So that f ( V ) = 2( t − − + 1 + 1) then we get γ R ( C t,r ) ≤ (cid:100) t (cid:101) .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, such as v vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48) = ( V ∪ { v } , ∅ , V − { v } ) doesnot satisfy the condition to be an RDF. According to this γ R ( C t,r ) ≥ (cid:100) t (cid:101) . For f (cid:48) function to be anRDF, v , v , v vertices should be taken into V set; f (cid:48) = ( V − { v , v } , V ∪ { v , v , v } , V − { v } ) .Hence we get f (cid:48) ( V ) = 2 (cid:100) t (cid:101) + 1 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) .In this case γ R ( C t,r ) ≥ (cid:100) t (cid:101) . ii) In order to dominate u , u , u , . . . , u r , v and v vertices, v vertex should be taken into V set. To dominate v t − , v t − and v t − vertices, v t − vertex should be taken into V set. To dominate v t and v t − vertices, v t and v t − themselves should be taken into V set. For the rest vertices of thegraph which are not dominated, v t vertices should be taken into V set which satisfy t ≡ (mod 3).Then V = { v , v , v , . . . , v t − } and V = { v t − , v t } .Therefore f ( V ) = 2( t − − + 1) + 2 then we get γ R ( C t,r ) ≤ (cid:100) t (cid:101) .Let f not be a γ R − f unction and delete any vertex from V set, then the result will be the sameas above. So that any vertex of V set, such as v t vertex, should be deleted. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v t ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v t } , V − { v t } , V ) does not satisfy the condition to bean RDF. According to this γ R ( C t,r ) ≥ (cid:100) t (cid:101) . For f (cid:48)(cid:48) function to be an RDF, v t vertex should be takeninto V set; f (cid:48)(cid:48) = ( V , V − { v t } , V ∪ { v t } ) . Hence we get f (cid:48)(cid:48) ( V ) = 2 (cid:100) t (cid:101) + 1 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( C t,r ) ≥ (cid:100) t (cid:101) .Consequently γ R ( C t,r ) = 2 (cid:100) t (cid:101) . Case 3: t ≡ (mod 3) In order to dominate u , u , u , . . . , u r , v and v vertices, v vertex should be taken into V set. Todominate v t − , v t − and v t vertices, v t − vertex should be taken into V set. For the rest vertices ofthe graph which are not dominated, v t vertices should be taken into V set which satisfy t ≡ (mod 3).Then V = { v , v , v , . . . , v t − } and V = ∅ .So that f ( V ) = 2( t − − + 1) then we get γ R ( C t,r ) ≤ (cid:100) t (cid:101) .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, such as v vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48) = ( V ∪ { v } , ∅ , V − { v } ) does notsatisfy the condition to be an RDF. According to this γ R ( C t,r ) ≥ (cid:100) t (cid:101) . For f (cid:48) function to be an RDF, v , v , v vertices should be taken into V set; f (cid:48) = ( V −{ v , v } , V ∪{ v , v , v } , V −{ v } ) . Hencewe get f (cid:48) ( V ) = 2 (cid:100) t (cid:101) + 1 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( C t,r ) ≥ (cid:100) t (cid:101) .Consequently γ R ( C t,r ) = 2 (cid:100) t (cid:101) . (cid:117)(cid:116) Theorem 3.2.
For p = n − a − b and p (cid:54) = 2 , let G = DC ( n, a, b ) be a double comet graph.The romandomination number of G is equal to γ R ( DC ( n, a, b )) = p + 1) p ≡ (mod 3) (cid:100) p (cid:101) p ≡ (mod 3) (cid:100) p (cid:101) + 1 p ≡ (mod 3) (2) D. DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs
Proof:
Roman domination number of the double comet graph is considered in three cases. Let f = ( V , V , V ) be a γ R − f unction of G . Case 1: p ≡ (mod 3) In order to dominate u , u , . . . , u a and k , k vertices, k vertex should be taken into V set andsimilarly to dominate v , v , . . . , v b and k p , k p − vertices, k p vertex should be taken into V set. Forthe rest vertices of the graph which are not dominated k t vertices should be taken into V set whichsatisfy t ≡ (mod 3) Then V = { k , k , . . . , k p − , k p } and V = ∅ .So that f ( V ) = 2( p − − + 1 + 1) then we get γ R ( DC ( n, a, b )) ≤ p + 1) .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, such as k vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( k ) , obtained function f (cid:48) = ( V ∪ { k } , ∅ , V − { k } ) does notsatisfy the condition to be an RDF. According to this γ R ( DC ( n, a, b )) ≥ p + 1) . For f (cid:48) function tobe an RDF, k , k , k vertices should be taken into V set; f (cid:48) = ( V − { k , k } , V ∪ { k , k , k } , V −{ k } ) . Hence we get f (cid:48) ( V ) = 2( p + 1) + 1 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( DC ( n, a, b )) ≥ p + 1) .Consequently γ R ( DC ( n, a, b )) = 2( p + 1) . Case 2: p ≡ (mod 3) In order to dominate u , u , . . . , u a and k , k vertices, k vertex should be taken into V set andsimilarly to dominate v , v , . . . , v b and k p , k p − vertices, k p vertex should be taken into V set. Forthe rest vertices of the graph which are not dominated k t vertices should be taken into V set whichsatisfy t ≡ (mod 3) Then V = { k , k , . . . , k p − , k p } and V = ∅ .So that f ( V ) = 2( p − − + 1 + 1) then we get γ R ( DC ( n, a, b )) ≤ (cid:100) p (cid:101) .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, such as k vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( k ) , obtained function f (cid:48) = ( V ∪ { k } , ∅ , V − { k } ) does notsatisfy the condition to be an RDF. According to this γ R ( DC ( n, a, b )) ≥ (cid:100) p (cid:101) . For f (cid:48) function to bean RDF, k , k , k vertices should be taken into V set; f (cid:48) = ( V − { k , k } , V ∪ { k , k , k } , V −{ k } ) . Hence we get f (cid:48) ( V ) = 2 (cid:100) p (cid:101) + 1 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( DC ( n, a, b )) ≥ (cid:100) p (cid:101) .Consequently γ R ( DC ( n, a, b )) = 2 (cid:100) p (cid:101) . Case 3: p ≡ (mod 3) In order to dominate u , u , . . . , u a and k , k vertices, k vertex should be taken into V set andsimilarly to dominate v , v , . . . , v b and k p , k p − vertices, k p vertex should be taken into V set. Todominate k p − vertex, k p − vertex itself should be taken into V set. For the rest vertices of the graphwhich are not dominated k t vertices should be taken into V set which satisfy t ≡ (mod 3) Then V = { k , k , . . . , k p − , k p } and V = { k p − } .So that f ( V ) = 2( p − − + 1 + 1) + 1 then we get γ R ( DC ( n, a, b )) ≤ (cid:100) p (cid:101) + 1 . . DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs Let f not be a γ R − f unction and by deleting k p − vertex from V = { v ∈ V : f ( v ) = 1 } set,let V = ∅ . Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( k p − ) , obtained function f (cid:48) = ( V ∪ { k p − } , ∅ , V ) does notsatisfy the condition to be an RDF. According to this γ R ( DC ( n, a, b )) ≥ (cid:100) p (cid:101) + 1 . For f (cid:48) function tobe an RDF, k p − vertex should be taken into V set; f (cid:48) = ( V , ∅ , V ∪ { k p − } ) . Hence we get f (cid:48) ( V ) =2 (cid:100) p (cid:101) + 2 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( DC ( n, a, b )) ≥ (cid:100) p (cid:101) + 1 .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, such as k vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( k ) , obtained function f (cid:48)(cid:48) = ( V ∪ { k } , V , V − { k } ) does notsatisfy the condition to be an RDF. According to this γ R ( DC ( n, a, b )) ≥ (cid:100) p (cid:101) + 1 . For f (cid:48)(cid:48) function tobe an RDF, k , k , k vertices should be taken into V set; f (cid:48)(cid:48) = ( V − { k , k } , V ∪ { k , k , k } , V −{ k } ) . Hence we get f (cid:48)(cid:48) ( V ) = 2 (cid:100) p (cid:101) + 2 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( DC ( n, a, b )) ≥ (cid:100) p (cid:101) + 1 .Consequently γ R ( DC ( n, a, b )) = 2 (cid:100) p (cid:101) + 1 . (cid:117)(cid:116) Theorem 3.3.
Let G = P + n be a comb graph. The roman domination number of G is equal to γ R ( P + n ) = n n ≡ (mod 3) (cid:98) n (cid:99) + 2 n ≡ (mod 3) (cid:100) n (cid:101) − n ≡ (mod 3) (3) Proof:
Roman domination number of the comb graph is considered in three cases. Let f = ( V , V , V ) be a γ R − f unction of G . Case 1: n ≡ (mod 3) In order to dominate v t − , v t , v t +1 and v + t vertices, v t vertices should be taken into V set which satisfy t ≡ (mod 3). For the rest vertices of the graph which are not dominated, v + t − and v + t +1 verticesshould be taken into V set which satisfy t ≡ (mod 3). Then V = { v , v , v , . . . , v n − , v n − } and V = { v +0 , v +2 , v +3 , v +5 , . . . , v + n − , v + n − } .So that f ( V ) = ( n − − + 1 + n − − + 1) + 2( n − − + 1) then we get γ R ( P + n ) ≤ n .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 1 } set, such as v +0 vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v +0 ) , obtained function f (cid:48) = ( V ∪ { v +0 } , V − { v +0 } , V ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ n . For f (cid:48) function to bean RDF, v +0 vertex should be taken into V set; f (cid:48) = ( V , V − { v +0 } , V ∪ { v +0 } ) . Hence we get f (cid:48) ( V ) = 4 n + 1 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ n .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, suchas v vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v } , V , V − { v } ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ n . For f (cid:48)(cid:48) function tobe an RDF, v , v , v and v +1 vertices should be taken into V set; f (cid:48)(cid:48) = ( V − { v , v , v +1 } , V ∪{ v , v , v , v +1 } , V − { v } ) . Hence we get f (cid:48)(cid:48) ( V ) = 4 n + 2 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ n . D. DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs
Consequently γ R ( P + n ) = 4 n . Case 2: n ≡ (mod 3) i) In order to dominate v t − , v t , v t +1 and v + t vertices, v t vertices should be taken into V set whichsatisfy t ≡ (mod 3) . To dominate v n − and v + n − vertices, v n − or v + n − vertex should be taken into V set. For the rest vertices of the graph which are not dominated, v + t − and v + t +1 vertices shouldbe taken into V set which satisfy t ≡ (mod 3) . Then V = { v , v , v , . . . , v n − , v n − , v n − } or V = { v , v , v , . . . , v n − , v n − , v + n − } and V = { v +0 , v +2 , v +3 , v +5 , . . . , v + n − , v + n − } .Thus, f ( V ) = ( n − − + 1 + n − − + 1) + 2( n − − + 1 + 1) then we get γ R ( P + n ) ≤ (cid:98) n (cid:99) + 2 .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 1 } set, such as v +0 vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v +0 ) , obtained function f (cid:48) = ( V ∪ { v +0 } , V − { v +0 } , V ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 . For f (cid:48) functionto be an RDF, v +0 vertex should be taken into V set; f (cid:48) = ( V , V − { v +0 } , V ∪ { v +0 } ) . Hence we get f (cid:48) ( V ) = 4 (cid:98) n (cid:99) + 3 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, suchas v vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v } , V , V − { v } ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 . For f (cid:48)(cid:48) functionto be an RDF, v , v , v and v +1 vertices should be taken into V set; f (cid:48)(cid:48) = ( V − { v , v , v +1 } , V ∪{ v , v , v , v +1 } , V − { v } ) . Hence we get f (cid:48)(cid:48) ( V ) = 4 (cid:98) n (cid:99) + 4 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 . ii) In order to dominate v t − , v t , v t +1 and v + t vertices, v t vertices should be taken into V setwhich satisfy t ≡ (mod 3) . To dominate v n − and v + n − vertices, v n − and v + n − vertices should betaken into V set. For the rest vertices of the graph which are not dominated, v + t − and v + t +1 verticesshould be taken into V set which satisfy t ≡ (mod 3) . Then V = { v , v , v , . . . , v n − , v n − } and V = { v +0 , v +2 , v +3 , v +5 , . . . , v + n − , v + n − , v + n − , v n − } .So that f ( V ) = ( n − − + 1 + n − − + 1 + 2) + 2( n − − + 1) then we get γ R ( P + n ) ≤ (cid:98) n (cid:99) + 2 .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 1 } set, such as v +0 vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v +0 ) , obtained function f (cid:48) = ( V ∪ { v +0 } , V − { v +0 } , V ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 . For f (cid:48) functionto be an RDF, v +0 vertex should be taken into V set; f (cid:48) = ( V , V − { v +0 } , V ∪ { v +0 } ) . Hence we get f (cid:48) ( V ) = 4 (cid:98) n (cid:99) + 3 . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, suchas v vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v } , V , V − { v } ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 . For f (cid:48)(cid:48) functionto be an RDF, v , v , v and v +1 vertices should be taken into V set; f (cid:48)(cid:48) = ( V − { v , v , v +1 } , V ∪{ v , v , v , v +1 } , V − { v } ) . Hence we get f (cid:48)(cid:48) ( V ) = 4 (cid:98) n (cid:99) + 4 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:98) n (cid:99) + 2 .Consequently, we could say that γ R ( P + n ) = 4 (cid:98) n (cid:99) + 2 . . DO ˘GAN DURGUN, E.N. TOPRAKKAYA / Roman Dom. of the Comet, D. Comet, and Comb Graphs Case 3: n ≡ (mod 3) i) In order to dominate v t − , v t , v t +1 and v + t vertices, v t vertices should be taken into V set whichsatisfy t ≡ (mod 3) . For the rest vertices of the graph which are not dominated, v + t − and v + t +1 ver-tices should be taken into V set which satisfy t ≡ (mod 3) . Then V = { v , v , v , . . . , v n − , v n − } and V = { v +0 , v +2 , v +3 , v +5 , . . . , v + n − , v + n − } .So that f ( V ) = ( n − − + 1 + n − − + 1) + 2( n − − + 1) then we get γ R ( P + n ) ≤ (cid:100) n (cid:101) − .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 1 } set, such as v +0 vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v +0 ) , obtained function f (cid:48) = ( V ∪ { v +0 } , V − { v +0 } , V ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:100) n (cid:101) − . For f (cid:48) functionto be an RDF, v +0 vertex should be taken into V set; f (cid:48) = ( V , V − { v +0 } , V ∪ { v +0 } ) . Hence we get f (cid:48) ( V ) = 4 (cid:100) n (cid:101) . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:100) n (cid:101) − .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, suchas v vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v } , V , V − { v } ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:100) n (cid:101) − . For f (cid:48)(cid:48) functionto be an RDF, v , v , v and v +1 vertices should be taken into V set; f (cid:48)(cid:48) = ( V − { v , v , v +1 } , V ∪{ v , v , v , v +1 } , V − { v } ) . Hence we get f (cid:48)(cid:48) ( V ) = 4 (cid:100) n (cid:101) + 1 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:100) n (cid:101) − . ii) To dominate v n − , v n − and v + n − vertices, v n − vertex should be taken into V set. To dominate v + n − vertex, v + n − vertex itself should be taken into V set. For the rest vertices of the graph whichare not dominated, in order to dominate v t − , v t , v t +1 and v + t vertices, v t vertices should be takeninto V set , and v + t − and v + t +1 vertices should be taken into V set which satisfy t ≡ (mod 3) . Then V = { v , v , v , . . . , v n − , v n − } and V = { v +0 , v +2 , v +3 , v +5 , . . . , v + n − , v + n − } .Therefore f ( V ) = ( n − − +1+ n − − +1+1)+2( n − − +1+1) then we get γ R ( P + n ) ≤ (cid:100) n (cid:101)− .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 1 } set, such as v +0 vertex. Since f (cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v +0 ) , obtained function f (cid:48) = ( V ∪ { v +0 } , V − { v +0 } , V ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:100) n (cid:101) − . For f (cid:48) functionto be an RDF, v +0 vertex should be taken into V set; f (cid:48) = ( V , V − { v +0 } , V ∪ { v +0 } ) . Hence we get f (cid:48) ( V ) = 4 (cid:100) n (cid:101) . Since f ( V ) < f (cid:48) ( V ) that f (cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:100) n (cid:101) − .Let f not be a γ R − f unction and delete any vertex from V = { v ∈ V : f ( v ) = 2 } set, suchas v vertex. Since f (cid:48)(cid:48) ( v ) (cid:54) = 2 for ∀ v ∈ N ( v ) , obtained function f (cid:48)(cid:48) = ( V ∪ { v } , V , V − { v } ) does not satisfy the condition to be an RDF. According to this γ R ( P + n ) ≥ (cid:100) n (cid:101) − . For f (cid:48)(cid:48) functionto be an RDF, v , v , v and v +1 vertices should be taken into V set; f (cid:48)(cid:48) = ( V − { v , v , v +1 } , V ∪{ v , v , v , v +1 } , V − { v } ) . Hence we get f (cid:48)(cid:48) ( V ) = 4 (cid:100) n (cid:101) + 1 . Since f ( V ) < f (cid:48)(cid:48) ( V ) that f (cid:48)(cid:48) ( V ) (cid:54) = γ R ( G ) . In this case γ R ( P + n ) ≥ (cid:100) n (cid:101) − .Consequently, we had γ R ( P + n ) = 4 (cid:100) n (cid:101) − . (cid:117)(cid:116) References [1] Stewart I. Defend the Roman empire!
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