3-Difference Cordial Labeling of Corona Related Graphs

2018·0.14 MB·

by  R. Ponraj

International J.Math. Combin. Vol.2(2018), 122-128
3-Difference Cordial Labeling of Corona Related Graphs
R.Ponraj
DepartmentofMathematics,SriParamakalyaniCollege,Alwarkurichi-627412,India
M.Maria Adaickalam
DepartmentofMathematics
ManonmaniamSundaranarUniversity,Abishekapatti,Tirunelveli-627012,TamilnaduIndia
E-mail: [email protected],[email protected]
Abstract: Let G be a (p,q) graph. Let f :V(G)→{1,2,··· ,k} be a map where k is an
integer 2≤k ≤p. For each edge uv, assign the label |f(u)−f(v)|. f is called k-difference
cordial labeling of G if |v (i)−v (j)| ≤1 and |e (0)−e (1)| ≤1 where v (x) denotes the
f f f f f
numberof verticeslabelled with x,e (1)ande (0) respectivelydenotethenumberofedges
f f
labelled with 1 and not labelled with 1. A graph with a k-difference cordial labeling is
called a k-differencecordial graph. In this paper we investigate 3-difference cordial labeling
behavior of DT ⊙K DT ⊙2K ,DT ⊙K and some more graphs.
n 1 n 1 n 2
KeyWords: Differencecordiallabeling,Smarandachelyk-differencecordiallabeling,path,
complete graph, triangular snake, corona.
AMS(2010): 05C78.
§1. Introduction
All Graphs in this paper are finite ,undirect and simple. Let G , G respectively be (p ,q ),
1 2 1 1
(p ,q )graphs. ThecoronaofG withG ,G G isthegraphobtainedbytakingonecopyof
2 2 1 2 1 2
⊙
G and p copies of G and joining the ith vertex of G with an edge to every vertex in the ith
1 1 2 1
copyofG . Ponrajetal. [3],hasbeenintroducedtheconceptofk-differencecordiallabelingof
2
graphsandstudiedthe3-differencecordiallabelingbehaviorofofsomegraphs. In[4,5,6,7]they
investigate the 3-difference cordial labeling behavior of path, cycle, complete graph, complete
bipartite graph, star, bistar, comb, double comb, quadrilateral snake, C(t), S(K ), S(B )
4 1,n n,n
andcaronaofsomegraphswithdoublealternatetriangularsnakedoublealternatequadrilateral
snake . In this paper we examine the 3-difference cordial labeling behavior of DT K
n 1
⊙
DT 2K ,DT K etc. Terms are not defined here follows from Harary [2].
n 1 n 2
⊙ ⊙
§2. k-Difference Cordial Labeling
Definition 2.1 Let G be a (p,q) graph and let f : V(G) 1,2, ,k be a map. For
→ { ··· }
1ReceivedNovember26,2017,Accepted May29,2018.
3-DifferenceCordialLabelingofCoronaRelatedGraphs 123
each edge uv, assign the label f(u) f(v). f is called a k-difference cordial labeling of G if
| − |
v (i) v (j) 1 and e (0) e (1) 1 where v (x) denotes the number of vertices labelled
f f f f f
| − | ≤ | − |≤
with x, e (1) and e (0) respectively denote the number of edges labelled with 1 and not labelled
f f
with 1. A graph with a k-difference cordial labeling is called a k-difference cordial graph.
On the other hand, if v (i) v (j) 1 or e (0) e (1) 1, such a labeling is called a
f f f f
| − |≥ | − |≥
Smarandachely k-difference cordial labeling of G.
AdoubletriangularsnakeDT consistsoftwotriangularsnakesthathaveacommonpath.
n
That is a double triangularsnakeis obtained froma path u u u by joining u andu to
1 2 n i i+1
···
two new vertices v (1 i n 1) and w (1 i n 1).
i i
≤ ≤ − ≤ ≤ −
First we investigate the 3-difference cordial labeling behavior of DT K .
n 1
⊙
Theorem 2.1 DT K is 3-difference cordial.
n 1
⊙
Proof Let V(DT K ) = V(DT ) x : 1 i n v′,w′ : 1 i n 1 and
n ⊙ 1 n { i ≤ ≤ } { i i ≤ ≤ − }
E(DT K )=E(DT ) u x :1 i n v v′,w w′ :1 i n 1 .
n⊙ 1 n { i i ≤ ≤S} { i i i i ≤S ≤ − }
Case 1. n is even. S S
First we consider the path vertices u . Assign the label 1 to all the path vertices u
i i
(1 i n). Then assign the label 2 to the path vertices v ,v ,v , and assign the label 1
1 3 5
≤ ≤ ···
to the path vertices v ,v ,v , . Now we consider the vertices w . Assign the label 2 to all
2 4 6 i
···
the vertices w (1 i n 1). Next we move to the vertices v′ and w′. Assign the label 2
i ≤ ≤ − i i
to the vertices v′ for all the values of i=0,1,2,3, and assign the label 1 to the vertices
2i+1 ···
v for i = 1,2,3, . Next we assign the label 1 to the vertex w′ and assign the label 3 to
2i ··· 1
the vertices w′,w′,w′, Finally assign the label 3 to all the vertices of x (1 i n). The
2 3 4 ··· i ≤ ≤
vertex condition and the edge conditions are v (1) = v (2) = 6n−3 and v (3) = 6n−2 and
f f 3 f 3
e (0)=4n 4 and e (1)=4n 3.
f f
− −
Case 2. n is odd.
Assign the label to the path vertices u (1 i n), v (1 i n 1), w (1 i n 1),
i i i
≤ ≤ ≤ ≤ − ≤ ≤ −
v′ (1 i n 1), x (1 i n) as in case 1. Then assign the label 3 to all the vertices
i ≤ ≤ − i ≤ ≤
w′ (1 i n 1). Since e (0) = 4n 3, e (1) = 4n 4 and v (1) = v (3) = 2n 1 and
i ≤ ≤ − f − f − f f −
v (2)=2n 2, DT K is 3-difference cordial.
f n 1
− ⊙ 2
Next investigation about DT 2K .
n 1
⊙
Theorem 2.2 DT 2K is 3-difference cordial.
n 1
⊙
Proof LetV(DT 2K )=V(DT ) x ,y :1 i n v′,v′′,w′,w′′ :1 i n 1
n⊙ 1 n { i i ≤ ≤ } { i i i i ≤ ≤ − }
and E(DT 2K ) = E(DT ) u x ,u y : 1 i n v v′,v v′′,w w′,w w′′ : 1 i
n ⊙ 1 n { i i Si i ≤ ≤ } {Si i i i i i i i ≤ ≤
n 1 .
− } S S
Case 1. n is even.
Consider the path vertices u . Assign the label 1 to the path vertex u . Now we assign
i 1
the labels 1,1,2,2 to the vertices u ,u ,u ,u respectively. Then we assign the labels 1,1,2,2 to
2 3 4 5
124 R.PonrajandM.MariaAdaickalam
the next four vertices u ,u ,u ,u respectively. Proceeding like this we assign the label to the
6 7 8 9
next four verticesand so on. If allthe verticesare labeled then we stopthe process. Otherwise
there are some non labeled vertices are exist. If the number of non labeled vertices are less
than or equal to 3 then assign the labels 1,1,2 to the non labeled vertices. If it is two then
assign the label 1,1 to the non labeled vertices. If only one non labeled vertex is exist then
assign the label 1 only. Next we consider the label v . Assign the label 2 to the vertex v .
i 1
Then we assign the label 2 to the vertices v ,v ,v , and assign the label 3 to the vertices
2 4 6
···
v ,v ,v , . Next we moveto the vertices x and y . Assignthe label 2 to the verticesx and
3 5 7 i i 1
···
x andweassignthe label3tothe verticesy andy . Nowweassignthe label1to thevertices
2 1 2
x and x for all the values of i = 1,2,3, . Then we assign the label 1 to the vertices
4i+1 4i
···
x for i = 0,1,2,3, . Next we assign the label 2 to the vertices x for all the values
4i+3 4i+2
···
of i = 1,2,3, . Now we assign the label 3 to the vertices y for i=0,1,2,3,... For all the
4i+3
···
values of i=1,2,3, . assign the label 3 to the vertices y and y . Then we assign the
4i+1 4i+2
···
label 2 to the vertices y for i = 1,2,3, . Now we consider the vertices v′ and v′′. For all
4i ··· i i
the values of i=1,2,3... assign the label 1 to the vertices v′ ,v′ . Assign the label 1 to the
4i+1 4i+2
verticesv′ for i=1,2,3, . Then we assignthe label2 to the verticesv′ for allthe values
4i ··· 4i+3
of i=0,1,2,3, . Consider the vertices v′′. Assign the label 3 to the path vertex v′′ ,v′′
··· i 4i+1 4i+2
and v′′ for all the values of i = 0,1,2,3, . Next we assign the label 2 to the vertices v′′
4i+3 ··· 4i
for i =1,2,3, . Now we assign the label 3 to the vertices w (1 i n 1). Next we move
i
··· ≤ ≤ −
to the vertices w′ and w′′. Assign the label 1 to all the vertices of w′(1 i n 1) and we
i i i ≤ ≤ −
assignthe label2 toallthe verticesofw′′(1 i n 1). Since v (1)=v (2)=v (3)=3n 2
i ≤ ≤ − f f f −
and e (0)= 11n−10 and e (1)= 11n−8, this labeling is 3-difference cordial labeling.
f 2 f 2
Case 2. n is odd.
First we consider the path vertices u . Assign the label 1,1,2,2 to the first four path
i
vertices u ,u ,u ,u respectively. Then we assign the labels 1,1,2,2 to the next four vertices
1 2 3 4
u ,u ,u ,u respectively. Continuing like this assign the label to the next four vertices and so
5 6 7 8
on. Ifalltheverticesarelabeledthenwestoptheprocess. Otherwisetherearesomeonlabeled
vertices are exist. If the number of non labeled vertices are less than or equal to 3 then assign
the labels 1,1,2 to the non labeled vertices. If it is 2 assign the labels 1,1 to the non labeled
vertices. If only one non labeled vertex exist then assign the label 1 to that vertex. Consider
the vertices v . Assign the label 2 to the vertices v ,v ,v , and we assign the label 3 to
i 1 3 5
···
the vertices v ,v ,v , . Next we move to the vertices w . Assign the label to the vertices
2 4 6 i
···
w (1 i n ) as in case 1. Now we consider the vertices x and y . Assign the label 2 to
i i i
≤ ≤ −
the vertices x for all the values of i=0,1,2,3, . For all the values of i=0,1,2,3,... assign
4i+1
···
the label 1 to the vertices x and x . Then we assign the label 1 to the vertices x for
4i+2 4i+3 4i
all the values of i = 1,2,3, . Next we assign the label 3 to the vertices y and y for
4i+1 4i+2
···
all the values of i=0,1,2,3, and we assign the label 3 to the vertices y for i=1,2,3, .
4i
··· ···
Then we assign the label 2 to the vertices y for all values i = 0,1,2,3, . Next we move
4i+3
···
to the verticesv′ andv′′. For allthe values of i=0,1,2,3, assignthe label1 to the vertices
i i ···
v′ andv′ . Nowweassignthe label1tothe verticesv′ fori=1,2,3, . Nextweassign
4i+1 4i+3 4i ···
the label 2 to the vertices v′ for i = 01,2,3, . Consider the vertices v′. Assign the label
4i+2 ··· i
3-DifferenceCordialLabelingofCoronaRelatedGraphs 125
3 to the vertices v′′ and v′′ for all the values of i = 0,1,2,3, and we assign the label
4i+1 4i+2 ···
1 to the vertices v for i = 1,2,3, . For the values of i = 0,1,2,3, assign the label 2 to
4i
··· ···
the vertices v . Finally we consider the vertices w′ and w′′. Assign the label to the vertices
4i+3 i i
w′ (1 i n 1) and w′′ (1 i n 1) as in case 1. The vertex and edge condition are
i ≤ ≤ − i ≤ ≤ −
v (1)=v (2)=v (3)=3n 2 and e (0)=e (1)= 11n−9.
f f f − f f 2 2
We now investigate the graph DT K .
n 2
⊙
Theorem 2.3 DT K is 3-difference cordial.
n 2
⊙
Proof Let V(DT K )=V(DT ) x ,y :1 i n v′,v′′,w′,w′′ :1 i n 1
n⊙ 2 n { i i ≤ ≤ } { i i i i ≤ ≤ − }
andE(DT K )=E(DT ) u x ,u y ,x y :1 i n v v′,v v′′,v′v′′,w w′,w w′′,w′w′′ :
n⊙ 2 n { i i i iSi i ≤ ≤ } {Si i i i i i i i i i i i
1 i n 1 .
≤ ≤ − } S S
Case 1. n is even.
Consider the path vertices u . Assign the label 1 to the path vertices u ,u ,u , . Then
i 1 2 3
···
we assign the labels 2 to the vertices v ,v ,v , . Next we assign the labels 3 to the vertices
1 2 3
···
w ,w ,w ,w . Now we consider the vertices v′ and v′′. Assign the label 2 to the vertex v′.
1 2 3 4 i i 1
Then we assign the label 1 to the vertices v′,v′,v′,v′, . Now we assign the label 3 to the
2 3 4 6 ···
vertices v′′,v′′,v′′,v′′, . Next we move to the vertices w′ and w′′. Assign the label 1 to the
1 2 3 4 ··· i i
vertex w′. Then we assign the label 1 to the vertices w′,w′,w′, and assign the label 2 to
1 2 4 6 ···
the vertices w′,w′,w′, . Assign the label 2 to the vertices w′′,w′′,w′′,w′′, . Finally we
3 5 7 ··· 1 2 3 4 ···
move to the vertices x and y . Assign the label 1 to the vertices x ,x ,x , and we assign
i i 1 3 5
···
the label2 to the verticesx ,x ,x , then weassignthe label3to the verticesy ,y ,y , .
2 4 6 1 2 3
··· ···
Clearly in this case the vertex and edge condition is given in v (1) = v (2) = v (3) = 3n 2
f f f
−
and e (0)=7n 5 and e (1)=7n 6.
f f
− −
Case 2. n is odd.
Assignthe labelto the verticesu (1 i n), v (1 i n 1)andw (1 i n 1)as
i i i
≤ ≤ ≤ ≤ − ≤ ≤ −
in case 1. Consider the vertices v′ and v′′. Assign the label 1 to the vertices v′,v′,v′,v′, .
i i 1 2 3 4 ···
Then assign the label to the vertices v′′ (1 i n 1) as in case 1. Now we move to the
i ≤ ≤ −
vertices w′ and w′′. Assign the label 1 to the vertices w′,w′,w′, and we assign the label 3
i i 1 3 5 ···
to the vertices w′,w′,w′, . Next we assignthe label to the verticesw′′ (1 i n 1) as in
2 4 6 ··· i ≤ ≤ −
case 1. Now we consider the vertices x and y . Assignthe label 2 to the vertices x ,x ,x ,
i i 1 3 5
···
andweassignthe label1to the verticesx ,x ,x , .Thenwe assignthe labelto the vertices
2 4 6
···
y (1 i n) as in case 1. Since v (1) = v (2) = v (3) = 3n 2 and e (0) = 7n 6 and
i f f f f
≤ ≤ − −
e (1)=7n 5, this labeling is 3-difference cordial labeling.
f
− 2
AdoublequadrilateralsnakeDQ consistsoftwoquadrilateralsnakesthathaveacommon
n
path. Let V(DQ ) = u : 1 i n v ,w ,xi,y : 1 i n 1 and E(DQ ) =
n i i i i n
{ ≤ ≤ } { ≤ ≤ − }
u u ,v w ,x y ,w u , y u :1 i n 1 .
{ i i+1 i i i i i i+1 i i+1 ≤ ≤S− }
Now we investigate the graphs DQ K ,DQ 2K and DQ K .
n 1 n 1 n 2
⊙ ⊙ ⊙
Theorem 2.4 DQ K is 3-difference cordial.
n 1
⊙
126 R.PonrajandM.MariaAdaickalam
Proof Let V(DQ K ) = V(DQ ) u′ : 1 i n v′,w′,x′,y′ : 1 i n 1
n ⊙ 1 n { i ≤ ≤ } { i i i i ≤ ≤ − }
and E(DQ K ) = E(DQ ) u u′ : 1 i n u v′,w w′,x x′,y ,y′ : 1 i n 1 .
n⊙ 1 n { i i ≤S ≤ } { i i iSi i i i i ≤ ≤ − }
Assign the label 1 to the path vertex u . Next we assign the labels 1,1,2 to the vertices
S 1 S
u ,u ,u respectively. Then weassignthe labels 1,1,2to the nextthree pathverticesu ,u ,u
2 3 4 5 6 7
respectively. Proceedinglike this we assignthe labelto the next three verticesand so on. If all
theverticesarelabeledthenwestoptheprocess. Otherwisetherearesomenonlabeledvertices
areexist. Ifthe numberofnonlabeledverticesarelessthanorequalto2thenassignthelabels
1,1tothe nonlabeledvertices. Ifonlyonenonlabeledvertexexistthenassignthe label1only.
Now we consider the vertices v and w . Assign the label 2 to the vertices v and v for
i i 3i+1 3i+2
all the values of i=0,1,2,3... For all the vales of i=1,2,3, assign the label 1 to the vertices
···
v . Then we assign the label 3 to the vertices w (1 i n). Next we move to the vertices x
3i i i
≤ ≤
and y . Assign the labels 2,3 to the vertices x and y respectively. Then we assign the label 2
i 1 1
to the vertices x ,x ,x , . Now we assign the label 1 to the vertices x ,x ,x , and the
2 5 8 3 6 9
··· ···
verticesx ,x ,x , .Assignthelabel3totheverticesy ,y ,y , .Weconsiderthevertices
4 7 10 1 2 3
··· ···
u′. Assignthelabels2,3totheverticesu′ andu′ respectively. Nowweassignthelabel1tothe
i 1 2
vertices u′,u′,u′, and we assign the label 3 to the vertices u′,u′,u′ , . Then we assign
3 6 9 ··· 4 7 10 ···
the label 2 to the vertices u′,u′,u′ , . Next we move to the vertices v′ and w′. Assign the
5 8 11 ··· i i
the label 3 to the vertex w′. Now assign the label 1 to all the vertices of v′ (1 i n 1)
1 i ≤ ≤ −
and we assign the label 2 to the vertices w′,w′,w′, . We consider the vertices x′ and y′.
2 3 4 ··· i i
Assign the label 2,1 to the vertices x′ and y′ respectively. Also we assign the label 2 to the
1 1
verticesx′,x′,x′, andthe verticesx′,x′,x′, .Thenwe assignthe label1to the vertices
2 5 8 ··· 3 6 9 ···
x′,x′,x′ , . Next we assign the label 3 to the vertices y′,y′,y′,... The vertex condition is
4 7 10 ··· 2 3 4
e (0)=6n 5 and e (1)=6n 6. Also the edge condition is given in Table 1 following.
f f
− − 2
Nature of n v (1) v (2) v (3)
f f f
n 0 (mod 3) 10n−9 10n−6 10n−9
≡ 3 3 3
n 1 (mod 3) 10n−10 10n−7 10n−7
≡ 3 3 3
n 2 (mod 3) 10n−8 10n−8 10n−8
≡ 3 3 3
Table 1
Theorem 2.5 DQ 2K is 3-difference cordial.
n 1
⊙
Proof Let V(DQ 2K )=V(DQ ) u′,u′′ :1 i n v′,v′′,w′,w′′,x′,x′′,y′,y′′ :
n⊙ 1 n { i i ≤ ≤ } { i i i i i i i i
1 i n 1 andE(DQ 2K )=E(DQ ) u u′,u u′′ :1 i n v v′,v v′′,w w′,w w′′,
≤ ≤ − } n⊙ 1 nS { i i i i ≤ S≤ } { i i i i i i i i
x x′,x x′′,y y′′,y y′′ :1 i n 1 . Firstweconsiderthepathverticesu . Assignthe label1
i i i i i i i i ≤ ≤ − } S S i
to the path vertices u ,u ,u , and we assign the label 2 to the path vertices u ,u ,u , .
1 3 5 2 4 6
··· ···
Clearly the lastvertex u receivedthe label 2 or 1 accordingas n 0mod 2 or n 1 (mod 2).
n
≡ ≡
Next we move to the vertices v and w . Assign the label 1 to all the vertices of v (1 i n)
i i i
≤ ≤
and we assign the label 3 to the vertices w ,w ,w ,... Then we assign the label to the vertices
1 2 3
x (1 i n 1) is same as assign the label to the vertices v (1 i n 1) and we
i i
≤ ≤ − ≤ ≤ −
assign the label to the vertices y (1 i n 1) is same as assign the label to the vertices w
i i
≤ ≤ −
(1 i n 1). Next we move to the vertices u′ and v′′. Assign the label 2 to the vertices
≤ ≤ − i i
3-DifferenceCordialLabelingofCoronaRelatedGraphs 127
v′,v′,v′, thenweassignthelabel3tothevertexv′′. Assignthelabel3totheverticesv′′ for
1 2 3 ··· 1 2i
all the values of i=1,2,3, and we assignthe label 2 to the vertices v′′ for i=1,2,3, .
··· 2i+1 ···
Next we consider the vertices w′ and w′′. Assign the label 1 to the vertices w′,w′,w′, and
i i 1 2 3 ···
we assign the label 3 to the vertices w′′,w′′,w′′, . Next we move to the vertices x′ and x′′.
1 2 3 ··· i i
Assign the label 1 to all the vertices of x′ (1 i n 1) and we assign the label 2 to all
i ≤ ≤ −
the vertices of x′′ (1 i n 1). Now we assign the label 2 to the vertices y′,y′,y′,
i ≤ ≤ − 1 2 3 ···
and we assign the label 3 to the vertices y′′,y′′,y′′, . Finally we move to the vertices u′ and
1 2 3 ··· i
u′′. Assign the label 2 to the vertices u′,u′,u′, and we assign the label 1 to the vertices
i 1 3 5 ···
u′,u′,u′, . Next we assign the label 2 to the vertices u′′,u′′,u′′, and we assign the label
2 4 6 ··· 1 3 5 ···
3 to the vertices u′′,u′′,u′′, . The vertex condition is v (1)=v (2)= v (3)= 15n−12. Also
2 4 6 ··· f f f 3
the edge condition is given in Table 2.
2
Values of n e (0) e (1)
f f
n 0 (mod 2) 17n−16 17n−14
≡ 2 2
n 1 (mod 2) 17n−15 17n−15
≡ 2 2
Table 2
Theorem 2.6 DQ K is 3-difference cordial.
n 2
⊙
Proof Let V(DQ K ) = V(DQ ) u′,u′′ : 1 i n v′,v′′,w′, w′′,x′,x′′,y′,y′′ :
n⊙ 2 n { i i ≤ ≤ } { i i i i i i i i
1 i n 1 andE(DQ K )=E(DQ ) u u′,u u′′,u′u′′ :1 i n v v′,v v′′,v′v′′,
≤ ≤ − } n⊙ 2 Sn { i i i i i i S ≤ ≤ } { i i i i i i
w w′,w w′′,w′w′′,x x′, x x′′,x′x′′,y y′′,y y′′,y′y′′ :1 i n 1 . First we consider the path
i i i i i i i i i i i i i i i i Si i ≤ ≤ − } S
vertices u . Assign the label 1 to the vertex u . Then we assign the label 1 to the vertices
i 1
u ,u ,u , and we assign the label 2 to the path vertices u ,u ,u , . Note that in this
2 4 6 1 3 5
··· ···
case the last vertex u received the label 1or 2 according as n 0 (mod 2) or n 1 (mod 2).
n
≡ ≡
Nextwemovetotheverticesv andw . Assignthelabel2tothevertexv . Thenweassignthe
i i 1
label 3 to all the vertices of w (1 i n 1). Assign the label 1 to the vertices v ,v ,v ,...
i 2 3 4
≤ ≤ −
We consider the vertices x and y . Assign the label to the vertices x (1 i n 1) is same
i i i
≤ ≤ −
as assign the label to the vertices v (1 i n 1) and assign the label to the vertices y
i i
≤ ≤ −
(1 i n 1) is same as assign the label to the vertices w (1 i n 1). Next we move
i
≤ ≤ − ≤ ≤ −
to the vertices v′ and v′′. Assign the label 2 to the vertices v′,v′,v′, and assign the label
i i 1 2 3 ···
3 to the vertices v′′,v′′,v′′, . Consider the vertices x′ and x′′. Assign the label 1 to all the
1 2 3 ··· i i
vertices of x′ (1 i n 1). Assign the label 2 to the vertex x′′. Then we assign the label 3
i ≤ ≤ − 1
tothe verticesx′′,x′′,x′′, .Nowweassignthe label2to allthe verticesofw′ (1 i n 1)
2 3 4 ··· i ≤ ≤ −
and assignthe label 3 to all the vertices of w′′ (1 i n 1). Now we move to the vertices y′
i ≤ ≤ − i
andy′′. Assignthe label1 tothe verticesy′,y′,y′, andweassignthe label2to the vertices
i 1 2 3 ···
y′′,y′′,y′′, . Next we move to the vertices u′ and u′′. Assign the label 1,3 to the vertices u′
1 2 3 ··· i i 1
and u′′ respectively. Assign the label 1 to the vertices u′ for all the values of i = 1,2,3,
1 2i ···
and assign the label 2 to the vertices u for i=1,2,3, then we assign the label 2 to the
2i+1
···
vertices u′′,u′′,u′′, . The vertex and edge conditions are
2 3 4 ···
15n 12
v (1)=v (2)=v (3)= −
f f f
3
128 R.PonrajandM.MariaAdaickalam
and
e (0)=11n 9, e (1)=11n 10.
f f
− − 2
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