J. Comp. & Math. Sci. Vol. 1(3), 294-299 (2010).
Strong Domination Critical and Stability in Graphs S.R. RAMACHANDRA and N.D. SONER Department of Studies in Mathematics University of Mysore, Manasagangothri, Mysore-570 006 Karnataka (India) (E-mail : ndsoner@yahoo.co.in) ABSTRACT In general strong domination number s(G) can be made to decrease or increase by removal of vertices from G. In this paper our main objective is the study of this phenomenon. Further the stability of the strong domination number of a graph G is investigated. 2000 Mathematics subject classification : 05C70 Keywords and phrases: graphs, strong domination set.
1. INTRODUCTION Domination alteration sets in graphs was first used by Bauer, Harary, Nieminen and Suffel1. A subset D of V(G) is dominating set in G if every vertex of V(G) D has at least one neighbour in D. Further for a graph G = (V, E) a set DV is a strong dominating set if every vertex v in VD has a neighbour u in D such that the degree of u is not lesser than the degree of v. The minimum cardinality of a strong dominating set of G is the strong domination number s(G). Strong (weak) domination was introduced by Sampathkumar and
Pushpalatha in4. For any graph theoretic parameter, the study of determining the effect of removal of a vertex or an edge from the graph has several important application such as fault tolerance in networks. The terminology of changing and unchanging was first suggested by Harary3. In this paper we initiate the effect of vertices on the strong domination number to study the increasing or decreasing of strong domination number of a graph G, when a vertex is deleted and obtain some new results. Finally we investigate the stability of the strong domination number of a graph. Any undefined term we may
Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)
S.R. Ramachandra et al., J.Comp.&Math.Sci. Vol.1(3), 294-299 (2010) 295 refer to Harary2. Strong domination critical vertices We partition the vertices of G into three sets according to how their removal affects s(G). Let
V V so V s V s for Vso v V : s (G v ) s (G ) V s v V : s (G v) s (G ) V v V : s (G v ) s (G ) s
When the graph under consideration is clear from the context we o s
s
simply write V ,V ,V
s .
Case 1 : Let p0 (mod 3), p3. Let v1, v2,...,vp be the vertices of Pp, then Pp – {v2} consists of an isolated vertex and a path of order p – 2. Thus, s(Pp – {v2}) s(P1) + s(Pp-2)
( p 2) > s (Pp) 3
=1+
Hence s ( Pp ) 1
Case 2 : Let p 1(mod 3), p 5. Let v1, v2,...,vp be the vertices of Pp. Then Pp–{v2,v4, v6}3s(P1)+s(Pp-6)=3
( p 6 + >s(Pp) therefore s ( Pp ) 3 . 3
For example
V s v3 , v 6 V s v5 V7
V1 V3
V4
V6
Case 3 : Let p 2 (mod 3), p 7. The Pp–{v2, v4} consists of two isolated vertices and a path of order p – 4. Hence, s (p –{v2, v4}) = 2s(P1) + s(p – 4)
P 4 > s(Pp) 3
G:
=2 + V8
V2
V5
Figure 1 Proposition 1 : For any path Pp,
s (Pp )
1 2 3
if p 0 (mod 3) p 3 if p 2 (mod 3) p 5 if p 1(mod 3) p 7
Proof : We consider the following three cases.
Thus s ( P p ) 2
Proposition 2 : For any path Pp with P 4,
s (Pp )
1 2
if p 1 (mod 3) if p 2 (mod 3)
3
if p 0(mod 3)
Proof : Case 1 : Let p 1 (mod 3) and v1, v2,...,vp be the vertices of Pp .
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296 S.R. Ramachandra et al., J.Comp.&Math.Sci. Vol.1(3), 294-299 (2010)
p 1 3
Proof : Let Cp = v1, v2,...,vp We consider the following three cases.
Since p 1 (mod 3) then p-1 = 3K
Case 1: Let p 0 (mod 3), p 6. Then removal of the set of vertices {v1, v3, v5} leaves two isolated vertices and a path of order p – 5.
Then, s (Pp–{v1})= s (P – 1) =
p
< = s (Pp). 3 Thus s ( P p ) 1
Hence s(Cp – {v1, v3, v5}) = 2s(P1) + Case 2 : Let p 2 (mod 3) and v1, v2,...,vp be the vertices of Pp. Then,
p 2 < 3
s(Pp – {v1, v2}) s(Pp-2)
p 3 = s (Pp) Hence s (Pp) = 2
v1,
Case 3 : Let p 0 (mod 3) and v2,...,vp be the vertices Pp. Then,
p 3 s(Pp – {v1, v2, v3}) = s(Pp-3) < 3
Thus s (C p ) 3 .
Case 2 : Let p 1 (mod 3), p1. Then removal of set of vertices {v1,v3, v5, v7, v9} leaves four isolated vertices and a path of order p–9. Hence, s(Cp – {v1, v3, v5, v7, v9}) = 3 s(P1)+s
p 9 > s(Cp) 3
(Pp-9) = 3+
Thus s (C p ) 5
Case 3 : Let p 2 (mod 3), p 14. Then removal of set of vertices {v1,v3,v5,v7} from Cp leaves six isolated vertices and a path of order p–7. Hence,
p 3 = s(Pp) Hence s (Pp)=3
s(Cp–{v1, v2,...,v13})=3s(P1)+s (Pp-7)
Proposition 3 : For any cycle Cp with p 6 vertices.
s (C p )
P 5
s (Pp-5) = 2+ > s(Cp). 3
3 5
if p 0 (mod 3), p 6 if p 1 (mod 3), p 1
7
if p 2(mod 3), p 14
p 7 > s(Cp) 3
= 3+
Thus, s (C p ) 7 .
Proposition 4 : For any cycle Cp with p 6 vertices,
Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)
S.R. Ramachandra et al., J.Comp.&Math.Sci. Vol.1(3), 294-299 (2010) 297
1 if p 1 (mod 3) 2 if p 2 (mod 3) 3 if p 0 (mod 3)
s (C p )
m = n 4. We know that s(Km,n)=2 (if m=n). Let V = V1 V2 be a vertex set of Km,n such that |V1| m, |V2|n. Let u V1 so s (Km,n– u)= m – 1 3 > s (Km,n)=2.
Proof : Proof is similar with path. Proposition 5: Let Wp be a wheel with p 5 vertices, then s (Wp) = 1 Proof : Let Wp be a wheel with p5 vertices, then Wp has a vertex u of degree p – 1 and hence s(Wp)=1. Since Wp–1 is a cycle of length p–1. Hence s (Wp – u)=s (Cp-1)>1= s (Wp) Thus s (W p ) 1 .
Proposition 6 : Let K m,n be a complete bipartite graph, which is neither K2 nor K2,2 then
s
(K m,n )
1 if 1 m n 1 if m n 4 m if m n
Proof : Suppose a complete bipartite graph is K1,n. Then s(K1,n)=1. Let u be a vertex of degree n. Then K1,n – u has atleast two components. Hence s (K1,n – u) 2 > 1 = s (K1,n) Therefore s ( K 1, n ) 1 .
Suppose K m,n is not a star and
Proposition 7 : Let K m,n be a complete bipartite graph which is neither K2 nor K2,2 then
s ( K m, n )
m n 1 if 1 if
mn 3 m n
Proof : Suppose m=n and V=V1 V2 be the vertex set of Km,n such that |V1|=m, |V2|=n, Then, s (Km,n–V1– V2+1)= 1 < 2 = s (Km,n). Hence s ( K m ,n ) m n 1 .
Suppose 3 m<n s(Km,n)=m if m<n. Let u v1 so s(Km,n–u)m–1<m=s(Km,n) Hence s ( K m ,n ) 1
Theorem 1 : For any tree T, V Vs s (T)2 implies s (T) 2.
Proof : Let v be a vertex of T which is adjacent to an end-vertex u of T and V Vs . If s(T–v) < s(T) we e proved. If not since we know s (T–v) s (T). It follows that s(T–v) = s(T). However T v = T{u} where T is a subtree of T and hence s(Tv) s(T)+l. But then, s(T–u–v)
= s(T)<s(T–v)=s(T)and so s (T) 2.
Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)
298 S.R. Ramachandra et al., J.Comp.&Math.Sci. Vol.1(3), 294-299 (2010) Theorem 2 : For any graph G, s(G) s(G–v) for all vV if and only if V = Vs or V= Vs
with n 2 and V Vs , there exists a vertex v V, such that s(T–v) = s(T).
Proof : Obviously, if V = Vs or V= Vs , s (G) s (G–v) for all v V.. Assume that s(G) s(G–v) for all vV. Then Vs and Vs partition V..
Proof : Clearly the result is true if T=K2. Assume that T has at least one vertex v that is adjacent to two (or more) endvertices u1, u2. Then v is in every s-set for T and s (T–u1)=s(T). If not {since V Vs then there existss
Theorem 3 : A vertex v of V(G) is in Vso if and only if G=
Theorem 4 : For any three T
t
Ki
i2
where t 2 and K i, i=2,3, …, t are complete graphs. Proof : Let S be an s – set of G. Clearly |S|=t–1. Since there is no path between Ki, i=2,3,…,t. Let V= V2 V3, ..., Vt where Vi are vertex sets of Ki respectively. Suppose vVr, 2 r t. As Kr–v is a complete graph of order at least r–1 and s (Kr–v) = s (Kr) then S will still s–set of G if such a vertex is removed thus for any vV. o s .
We have v V
o s
Conversely, let V(G) = V
u Vs or Vso , if u Vso we proved. Assu ssume u Vs }, then v is adjacent to one end vertex u and deg (v) = 2. Suppose N[v]D = {V} and let T=T–v–u. For every graph G, if deg u=1, then s(G–u) s(G) Hence s(T) s (T–u) s(T). However, s(T) s(T) –1, if s = s (T) –1, then s (T) = s (T–v). Otherwise s (T)= s (T–u). If N[v] D{v}. Since V Vs then there existss V Vs or
Vso . If v Vso we have done..
If not v Vs so we know that deg v2 (G)
and S be an s – set of G. Note that every vertex of degree at least one must be in S. Also note that all vertices of V–S must be in <N[S]>. Further more there can be no path in G, connecting two vertices of S. A graph with these properties is only union of complete graphs of order at least two.
and there exists at least one vertex u V that is adjacent with v such that s (T–v) = s (T). Theorem 5 : A vertex v of V(G) t
is in Vs if and only if G =
K r ,r .
r 2
Proof : Let S be an s – set of G, clearly |S|=2(t–3). Since there is
Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)
S.R. Ramachandra et al., J.Comp.&Math.Sci. Vol.1(3), 294-299 (2010) 299 no path between Ki,i, i=4,5,…,t. Let V = V4 V5 ,…, Vt. Where Vi are vertex sets of Ki,i respectively. Suppose v Vr, 4 r t. As Kr,r–v is a Complete bipartite graph of order at least r–1 and s (Kr,r–v) = r–1> 2 = s (Kr,r) then S will still s–set of G such a vertex is removed therefore for any vV. We
have v Vs . Conversely, let V(G)= Vs (G) and S be an s-set of G. Note that every vertex of maximum degree must be in S. Also note that all vertices of V–S must be in N[S] . Further more there can be no path in G connecting
two vertices of S. A graph with these properties is only union of Complete bipartite graph of order at least 4. References 1. D. Bauer, F. Harary, J. Nieminen and C.L. Suffel, Domination alteration sets in graphs, Discrete Math. 47, 153-161 (1983). 2. F. Harary, Graph Theory, Addison– Wesley Reading, MA. (1969). 3. F. Harary, Changing and unchanging invariants for graphs, Bull. Malaysian Math. Soc. 5, 73-78 (1982). 4. E. Sampathkumar and L. Pushpalatha, Strong (Weak) domination and domination balance in graph, Discrete Math., 161, 235-242 (1996).
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