Research ArticleSome New Results on Prime Cordial Labeling

S K Vaidya1 and N H Shah2

1 Saurashtra University Rajkot Gujarat 360005 India2 Government Polytechnic Rajkot Gujarat 360003 India

Correspondence should be addressed to S K Vaidya samirkvaidyayahoocoin

Received 7 January 2014 Accepted 18 February 2014 Published 23 March 2014

Academic Editors E Manstavicius J Rada and F Rossello

Copyright copy 2014 S K Vaidya and N H ShahThis is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

A prime cordial labeling of a graph119866with the vertex set119881(119866) is a bijection 119891 119881(119866) rarr 1 2 3 |119881(119866)| such that each edge 119906Vis assigned the label 1 if gcd(119891(119906) 119891(V)) = 1 and 0 if gcd(119891(119906) 119891(V)) gt 1 then the number of edges labeled with 0 and the numberof edges labeled with 1 differ by at most 1 A graph which admits a prime cordial labeling is called a prime cordial graph In thiswork we give a method to construct larger prime cordial graph using a given prime cordial graph 119866 In addition to this we haveinvestigated the prime cordial labeling for double fan and degree splitting graphs of path as well as bistar Moreover we prove thatthe graph obtained by duplication of an edge (spoke as well as rim) in wheel119882

119899admits prime cordial labeling

1 Introduction

We consider a finite connected undirected and simplegraph 119866 = (119881(119866) 119864(119866)) with 119901 vertices and 119902 edges whichis also denoted as 119866(119901 119902) For standard terminology andnotations related to graph theory we follow Balakrishnan andRanganathan [1] while for any concept related to numbertheory we refer to Burton [2] In this section we provide briefsummary of definitions and other required information forour investigations

Definition 1 TheGraph labeling is an assignment of numbersto the vertices or edges or both subject to certain condition(s)If the domain of the mapping is the set of vertices (edges)then the labeling is called a vertex labeling (edge labeling)

Many labeling schemes have been introduced so farand they are explored as well by many researchers Graphlabelings have enormous applications within mathematics aswell as to several areas of computer science and communi-cation networks Various applications of graph labeling arereported in the work of Yegnanaryanan and Vaidhyanathan[3] For a dynamic survey on various graph labeling prob-lems along with an extensive bibliography we refer toGallian [4]

Definition 2 A labeling 119891 119881(119866) rarr 0 1 is called binaryvertex labeling of 119866 and 119891(V) is called the label of the vertex Vof 119866 under 119891

Notation 1 If for an edge 119890 = 119906V the induced edge labeling119891lowast

119864(119866) rarr 0 1 is given by 119891lowast

(119890) = |119891(119906) minus 119891(V)| Then

V119891(119894) = number of vertices of 119866 having label 119894 under 119891

119890119891(119894) = number of edges of 119866 having label 119894 under 119891

lowast

where 119894 = 0 or 1

(1)

Definition 3 A binary vertex labeling 119891 of a graph119866 is calleda cordial labeling if |V

119891(0)minus V

119891(1)| le 1 and |119890

119891(0)minus 119890

119891(1)| le 1

A graph 119866 is cordial if it admits cordial labelingThe concept of cordial labeling was introduced by Cahit

[5]

The notion of prime labeling was originated by Entringerand was introduced by Tout et al [6]

Definition 4 A prime labeling of a graph 119866 is an injectivefunction 119891 119881(119866) rarr 1 2 |119881(119866)| such that for every

Hindawi Publishing CorporationISRN CombinatoricsVolume 2014 Article ID 607018 9 pageshttpdxdoiorg1011552014607018

2 ISRN Combinatorics

pair of adjacent vertices 119906 and V gcd(119891(119906) 119891(V)) = 1 Thegraph which admits a prime labeling is called a prime graph

The concept of prime labeling has attracted manyresearchers as the study of prime numbers is of greatimportance because prime numbers are scattered and thereare arbitrarily large gaps in the sequence of prime numbersVaidya and Prajapati [7 8] have investigated many resultson prime labeling Same authors [9] have discussed primelabeling in the context of duplication of graph elementsMotivated through the concepts of prime labeling and cordiallabeling a new concept termed as a prime cordial labelingwas introduced by Sundaram et al [10] which contains blendof both the labelings

Definition 5 A prime cordial labeling of a graph119866with vertexset 119881(119866) is a bijection 119891 119881(119866) rarr 1 2 3 |119881(119866)| and ifthe induced function 119891

lowast

119864(119866) rarr 0 1 is defined by

119891lowast

(119890 = 119906V) = 1 if gcd (119891 (119906) 119891 (V)) = 1

= 0 otherwise(2)

then the number of edges labeled with 0 and the number ofedges labeled with 1 differ by at most 1 A graph which admitsprime cordial labeling is called a prime cordial graph

Many graphs are proved to be prime cordial in the workof Sundaram et al [10] Prime cordial labeling for somecycle related graphs has been discussed by Vaidya and Vihol[11] Prime cordial labeling in the context of some graphoperations has been discussed by Vaidya and Vihol [12] andVaidya and Shah [13 14] Vaidya and Shah [14] have provedthat the wheel graph 119882

119899admits prime cordial labeling for

119899 ge 8 while same authors in [15] have discussed primecordial labeling for some wheel related graphs Babitha andBaskar Babujee [16] have exhibited prime cordial labeling forsome cycle related graphs and discussed the duality of primecordial labeling The same authors in [17] have derived somecharacterizations of prime cordial graphs and investigatedvarious methods to construct larger prime cordial graphsusing existing prime cordial graphs We investigate a methoddifferent from existing one to construct larger prime cordialgraph from an existing prime cordial graph

Definition 6 The wheel 119882119899is defined to be the join 119870

1+ 119862119899

The vertex corresponding to1198701is known as apex and vertices

corresponding to cycle are known as rim vertices while theedges corresponding to cycle are known as rim edges

Definition 7 The bistar 119861119899119899

is a graph obtained by joining thecenter (apex) vertices of two copies of119870

1119899by an edge

Definition 8 The fan 119865119899is the graph obtained by taking 119899 minus 2

concurrent chords in cycle 119862119899+1

The vertex at which all thechords are concurrent is called the apex vertex In otherwords 119865

119899= 119875119899+ 1198701

Definition 9 The double fan DF119899consists of two fan graphs

that have a common path In other words DF119899= 119875119899+ 1198702

Definition 10 The duplication of an edge 119890 = 119906V of graph 119866

produces a new graph 1198661015840 by adding an edge 119890

1015840

= 1199061015840V1015840 such

that119873(1199061015840

) = 119873(119906) cup V1015840 minus V and119873(V1015840) = 119873(V) cup 1199061015840

minus 119906

Definition 11 (see [18]) Let119866 = (119881(119866) 119864(119866)) be a graph with119881 = 119878

1cup 1198782cup 1198783cup sdot sdot sdot 119878

119905cup 119879 where each 119878

119894is a set of vertices

having at least two vertices of the same degree and 119879 =

119881 cup119905

119894=1119878119894 The degree splitting graph of G denoted by 119863119878(119866)

is obtained from 119866 by adding vertices 1199081 1199082 1199083 119908

119905and

joining to each vertex of 119878119894for 1 le 119894 le 119905

2 Main Results

Theorem 12 Let 119866(119901 119902) with 119901 ge 4 be a prime cordial graphand let 119870

2119899be a bipartite graph with bipartition 119881 = 119881

1cup 1198812

with 1198811= V1 V2 and 119881

2= 1199061 1199062 119906

119899 If 119866

1is the graph

obtained by identifying the vertices V1and V2of 1198702119899

with thevertices of 119866 having labels 2 and 4 respectively then 119866

1admits

prime cordial labeling in any of the following cases

(i) 119899 is even and 119866 is of any size 119902

(ii) 119899 119901 and 119902 are odd with 119890119891(0) = lfloor1199022rfloor

(iii) 119899 is odd 119901 is even and 119902 is odd with 119890119891(0) = lceil1199022rceil

Proof Let 119866(119901 119902) be a prime cordial graph and let 1198911be the

prime cordial labeling of 119866 Let 1199081 1199082isin 119881 be the vertices of

119866 such that 1198911(1199081) = 2 and 119891

1(1199082) = 4 Consider the 119870

2119899

with bipartition 119881 = 1198811cup 1198812with 119881

1= V1 V2 and 119881

2=

1199061 1199062 119906

119899 Now identify the vertices V

1to1199081and V2to1199082

and denote the resultant graph as 1198661 Then 119881(119866

1) = 119881(119866) cup

1199061 1199062 119906

119899 and 119864(119866

1) = 119864(119866) cup 119908

1119906119894 11990821199061198941 le 119894 le 119899 so

|119881(1198661)| = 119901 + 119899 and |119864(119866

1)| = 119902 + 2119899 To define 119891 119881(119866

1) rarr

1 2 3 119901 + 119899 we consider the following three casesCase (i) (119899 is even and 119866 is of any size 119902) Since 119866 is aprime cordial graph we assign vertex labels such that119891(119908

119894) =

1198911(119908119894) where 119908

119894isin 119881(119866) = 119881(119866

1) cap 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (3)

Since 119899 is even and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and

1199082are adjacent to each 119906

119894 119894 le 119894 le 119899 And this vertex

assignment generates 119899 edges with label 1 and 119899 edges withlabel 0 Following Table 1 gives edge condition for primecordial labeling for 119866

1under 119891

From Table 1 we have |119890119891(0) minus 119890

119891(1)| le 1

Case (ii) (119899 119901 and 119902 are odd with 119890119891(0) = lfloor1199022rfloor) Here 119901 and

119902 both are odd and 119866 is a prime cordial graph with 119890119891(0) =

lfloor1199022rfloorSince 119866 is a prime cordial graph we keep the vertex label

of all the vertices of119866 in1198661as it isTherefore 119891(119908

119894) = 1198911(119908119894)

where 119908119894isin 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (4)

Since 119899 is odd and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and 119908

2

are adjacent to each 119906119894 119894 le 119894 le 119899 And this vertex assignment

generates 119899+ 1 edges with label 0 and 119899minus 1 edges with label 1

ISRN Combinatorics 3

Table 1

119902 Edge conditions for 119866 Edge conditions for 1198661

Even 119890119891(0) = 119890

119891(1) =

119902

2119890119891(0) = 119890

119891(1) =

119902

2

Odd119890119891(0) minus 1 = 119890

119891(1) = lfloor

119902

2rfloor 119890

119891(0) minus 1 = 119890

119891(1) = lfloor

119902

2rfloor + 119899

119890119891(0) = 119890

119891(1) minus 1 = lfloor

119902

2rfloor 119890

119891(0) = 119890

119891(1) minus 1 = lfloor

119902

2rfloor + 119899

Therefore edge conditions for 1198661under 119891 are 119890

119891(0) =

lfloor1199022rfloor + 119899+ 1 and 119890119891(1) = lceil1199022rceil + 119899minus 1 Therefore 119890

119891(0) minus 1 =

119890119891(1) Hence |119890

119891(0) minus 119890

119891(1)| le 1 for graph 119866

1

Case (iii) (119899 is odd119901 is even and 119902 is oddwith 119890119891(0) = lceil1199022rceil)

Here 119901 is even 119902 is odd and 119866 is a prime cordial graph with119890119891(0) = lceil1199022rceilSince 119866 is a prime cordial graph we keep the vertex label

of all the vertices of119866 in1198661as it isTherefore 119891(119908

119894) = 1198911(119908119894)

where 119908119894isin 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (5)

Since 119899 is odd and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and 119908

2

are adjacent to each 119906119894 119894 le 119894 le 119899 And this vertex assignment

generates 119899minus 1 edges with label 0 and 119899+ 1 edges with label 1Therefore edge conditions for 119866

1under 119891 are 119890

119891(0) =

lceil1199022rceil + 119899 minus 1 and 119890119891(1) = lfloor1199022rfloor + 119899 + 1 Therefore 119890

119891(0) =

119890119891(1) minus 1 Hence |119890

119891(0) minus 119890

119891(1)| le 1 for graph 119866

1

Hence in all the cases discussed above 1198661admits prime

cordial labeling

Illustration 1 Consider the graph 119866 as shown in Figure 1with 119901 = 7 and 119902 = 9 119866 is a prime cordial graph with119890119891(0) = 4 119890

119891(1) = 5 Take 119899 = 3 and construct graph 119866

1

In accordance with Case (ii) of Theorem 12 a prime cordiallabeling of119866

1is as shown in Figure 1 Here 119890

119891(0) = 8 119890

119891(1) =

7

Theorem13 Double fan119863119865119899is a prime cordial graph for 119899 = 8

and 119899 ge 10

Proof Let DF119899be the double fan with apex vertices 119906

1 1199062and

V1 V2 V

119899are vertices common pathThen |119881(DF

119899)| = 119899 +

2 and |119864(DF119899)| = 3119899minus1 To define119891 119881(119866) rarr 1 2 3 119899+

2 we consider the following five casesCase 1 (119899 = 3 to 7 and 119899 = 9) In order to satisfy the edgecondition for prime cordial labeling in DF

3it is essential to

label four edges with label 0 and four edges with label 1 out ofeight edges But all the possible assignments of vertex labelswill give rise to 0 labels for at most one edge and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 6 gt 1 Hence DF

3

is not a prime cordial graphIn order to satisfy the edge condition for prime cordial

labeling in DF4it is essential to label five edges with label

0 and six edges with label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most three edges and 1 label for at least eight edgesThat

is |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence DF

4is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF5it is essential to label seven edges with label 0

and seven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence DF

5is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF6it is essential to label eight edges with label 0

and nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence DF

6is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF7it is essential to label ten edges with label 0

and ten edges with label 1 out of twenty edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most eight edges and 1 label for at least twelve edgesThat is |119890

119891(0) minus 119890

119891(1)| = 4 gt 1 Hence DF

7is not a prime

cordial graphIn order to satisfy the edge condition for prime cordial

labeling in DF9it is essential to label thirteen edges with label

0 and thirteen edges with label 1 out of twenty-six edges Butall the possible assignments of vertex labels will give rise to 0labels for at most twelve edges and 1 label for at least fourteenedges That is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence DF

9is not a

prime cordial graphCase 2 (119899 = 8 10 11 12) For 119899 = 8 119891(119906

1) = 2 119891(119906

2) = 6

and 119891(V1) = 4 119891(V

2) = 8 119891(V

3) = 3 119891(V

4) = 9 119891(V

5) = 10

119891(V6) = 5119891(V

7) = 7 and119891(V

8) = 1Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10 119891(V

7) = 1

119891(V8) = 5 119891(V

9) = 7 and 119891(V

10) = 11 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 11 119891(119906

1) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10 119891(V

7) = 5

119891(V8) = 7 119891(V

9) = 11 119891(V

10) = 13 and 119891(V

11) = 1 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 14 119891(V

7) = 10

119891(V8) = 5 119891(V

9) = 7 119891(V

10) = 11 119891(V

11) = 13 and 119891(V

12) =

1 Then 119890119891(0) = 18 119890

119891(1) = 17

Now for the remaining three cases let

119896 = lfloor119899 + 2

2rfloor 119898 = lfloor

119899 + 2

3rfloor

1199051= lceil

3119899 minus 1

2rceil 119905

2= 3119896 minus 7 + lceil

119898

2rceil

(6)

1199053= largest even number le 119899+2 and 119905

4= largest odd number

le 119899 + 2Case 3 (119905

1= 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (7)

4 ISRN Combinatorics

1

2

3

4

5

6

7

w1

w2

w3

w4w5

w6

w7

G

(a)

1

2

3

4

5

6

7

8

9

10

u1

u2

u3

w1

w2

w3

w4w5

w6

w7

G1

(b)

Figure 1

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 1 1199053 1199053minus 2 1199053minus 4 14 12 10 8 4 3

5 7 9 1199054minus 2 1199054

Case 4 (1199051gt 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (8)

Let 1199055= 1199051minus 1199052 Consider

119891 (V1) = 3

119891 (V2) = 9

119891 (V3) = 15

119891 (V4) = 21

119891 (V1199055

) = 3 (21199055minus 1)

119891 (V1199055+1

) = 119891 (V1199055

) + 6

(9)

Now for remaining vertices V1199055+2

V1199055+3

V119899assign the labels

1 1199053 1199053

minus 2 1199053

minus 4 14 1210 8 4 5 7 all the oddnumbers in ascending orderCase 5 (119905

2gt 1199051) Let 119905

6= 1199052minus 1199051

Sub-Case 1 119899 is even Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (10)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderSub-Case 2 119899 is odd Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (11)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderIn view of the above defined labeling pattern for Cases 3

4 and 5 we have

119890119891(0) = lceil

3119899 minus 1

2rceil 119890

119891(1) = lfloor

3119899 minus 1

2rfloor (12)

Thus we have |119890119891(0) minus 119890

119891(1)| le 1

Hence DF119899is a prime cordial graph for 119899 = 8 and 119899 ge

10

Illustration 2 For the graph DF15 |119881(DF

15)| = 17 and

|119864(DF15)| = 44 In accordance with Theorem 13 we have

119896 = 5 119898 = 5 1199051

= 22 and 1199052

= 20 Here 1199051

gt 1199052so

labeling pattern described in Case 4 will be applicable and1199055= 2 The corresponding prime cordial labeling is shown in

Figure 2 Here 119890119891(0) = 22 = e

119891(1)

Illustration 3 For the graph DF37 |119881(DF

37)| = 39 and

|119864(DF37)| = 110 In accordance with Theorem 13 we have

119896 = 19119898 = 13 1199051= 55 and 119905

2= 57 Here 119905

2gt 1199051and 119899 = 37

so labeling pattern described in Sub-Case 2 of Case 5 will beapplicable and 119905

6= 2 And corresponding labeling pattern is

as below

119891 (1199061) = 2 119891 (119906

2) = 6 (13)

ISRN Combinatorics 5

1

2

3 4 5

6

789 10 1112 131415 16 17

u1

u2

1 23 4 5 6 7 8 9 10 11 12 13 14 15

Figure 2

For the vertices V1 V2 V

37we assign the vertex labels 39

38 37 36 35 34 32 30 28 26 24 22 20 18 16 14 12 10 8 43 5 7 9 11 13 15 17 19 21 23 25 27 29 31 and 33 respectivelywhere 119890

119891(0) = 55 = 119890

119891(1) Therefore DF

37is a prime cordial

graph

Theorem 14 The graph obtained by duplication of an arbi-trary rim edge by an edge in119882

119899is a prime cordial graph where

119899 ge 6

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate therim edge 119890 = V

1V2by an edge 119890

1015840

= 11990611199062and call the resultant

graph as119866Then |119881(119866)| = 119899+3 and |119864(119866)| = 2119899+5 To define119891 119881(119866) rarr 1 2 3 119899+3 we consider the following fourcases

Case 1 (119899 = 3 4 5) For 119899 = 3 to satisfy the edge condition forprime cordial labeling it is essential to label five edges withlabel 0 and six edgeswith label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least seven edgesThatis |119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 3 119866 is not a prime

cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label six edges with label 0 and sevenedges with label 1 out of thirteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost four edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andeight edges with label 1 out of fifteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost six edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graph

Case 2 (119899 = 6 to 10) For 119899 = 6 119891(1199061) = 4 119891(119906

2) = 8 and

119891(V0) = 6 119891(V

1) = 3 119891(V

2) = 9 119891(V

3) = 7 119891(V

4) = 5

119891(V5) = 1 and 119891(V

6) = 2 Then 119890

119891(0) = 8 119890

119891(1) = 9

For 119899 = 7 119891(1199061) = 4 119891(119906

2) = 8 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 9 119891(V

3) = 7 119891(V

4) = 5 119891(V

5) = 10 119891(V

6) = 1 and

119891(V7) = 2 Then 119890

119891(0) = 4 119890

119891(1) = 5

For 119899 = 8 119891(1199061) = 11 119891(119906

2) = 10 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 2 119891(V

3) = 8 119891(V

4) = 4 119891(V

5) = 5 119891(V

6) = 7

119891(V7) = 1 and 119891(V

8) = 9 Then 119890

119891(0) = 10 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 11 119891(119906

2) = 12 and 119891(V

0) = 2 119891(V

1) =

1 119891(V2) = 3 119891(V

3) = 6 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10

119891(V7) = 5119891(V

8) = 7 and119891(V

9) = 9Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 13 119891(119906

2) = 12 and 119891(V

0) = 2

119891(V1) = 9 119891(V

2) = 3 119891(V

3) = 6 119891(V

4) = 4 119891(V

5) = 8

119891(V6) = 10 119891(V

7) = 5 119891(V

8) = 1 119891(V

9) = 7 and 119891(V

10) = 11

Then 119890119891(0) = 12 119890

119891(1) = 13

Case 3 (119899 is even 119899 ge 12) Consider

119891 (V0) = 2

119891 (V1) = 5 119891 (V

2) = 10

119891 (V3) = 4 119891 (V

4) = 8

119891 (V4+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 5

119891 (V1198992

) = 6 119891 (V(1198992)+1

) = 3

119891 (V(1198992)+2

) = 9

119891 (V119899minus1

) = 1 119891 (V119899) = 7

119891 (V(1198992)+2+119894

) = 11 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 4

119891 (1199061) = 119899 + 3 119891 (119906

2) = 119899 + 2

(14)

6 ISRN Combinatorics

In view of the above defined labeling pattern for Case 3 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 4(mod 7) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod 7)

Case 4 (119899 is odd 119899 ge 11) Consider

119891 (V0) = 2

119891 (V1) = 10 119891 (V

2) = 4

119891 (V3) = 8

119891 (V3+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le ((119899 minus 1) 2) minus 4

119891 (V(119899minus1)2

) = 6 119891 (V(119899+1)2

) = 3

119891 (V(119899+3)2

) = 1

119891 (V119899+1minus119894

) = 5 + 2 (119894 minus 1) 1 le 119894 le (119899 minus 3) 2

119891 (1199061) = 119899 + 2 119891 (119906

2) = 119899 + 3

(15)

In view of the above defined labeling pattern for Case 4 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 3(mod5) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod7)

In view of Cases 2 to 4 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence 119866 is a prime cordial graph for 119899 ge 6

Illustration 4 Let 119866 be the graph obtained by duplicationof an arbitrary rim edge by an edge in wheel 119882

13 Then

|119881(119866)| = 16 and |119864(119866)| = 31 In accordance withTheorem 14 Case 4 will be applicable and the correspondingprime cordial labeling is shown in Figure 3 Here 119890

119891(0) = 16

119890119891(1) = 15

Theorem 15 The graph obtained by duplication of an arbi-trary spoke edge by an edge in wheel119882

119899is prime cordial graph

where 119899 = 7 and 119899 ge 9

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate thespoke edge 119890 = V

0V1by an edge 1198901015840 = 119906

11199062and call the resultant

graph 119866 Then |119881(119866)| = 119899 + 3 and |119864(119866)| = 3119899 + 2 To define119891 119881(119866) rarr 1 2 3 119899 + 3 we consider following threecasesCase 1 (119899 = 3 to 6 and 119899 = 8) For 119899 = 3 to satisfy the edgecondition for prime cordial labeling it is essential to label fiveedges with label 0 and six edges with label 1 out of elevenedges But all the possible assignments of vertex labels willgive rise to 0 labels for at most four edges and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for

119899 = 3 119866 is not a prime cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andseven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label eight edges with label 0 and

1

2

3

45

6

7 8

9

10

11

12

13

14

15

16

u1

u2

1

2

3

4

5

6

78

9

10

11

12

13

Figure 3

nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graphFor 119899 = 6 to satisfy the edge condition for prime cordial

labeling it is essential to label ten edges with label 0 and tenedges with label 1 out of twenty edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost eight edges and 1 label for at least twelve edges That is|119890119891(0) minus 119890

119891(1)| = 4 gt 1 Hence for 119899 = 6 119866 is not a prime

cordial graphFor 119899 = 8 to satisfy the edge condition for prime cordial

labeling it is essential to label thirteen edges with label 0 andthirteen edges with label 1 out of twenty-six edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most twelve edges and 1 label for at least fourteen edgesThat is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence for 119899 = 8 119866 is not a

prime cordial graphCase 2 (119899 = 7 119899 = 9 to 12 and 119899 = 14 16 18 20 22) For119899 = 7 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 9

119891(V2) = 3 119891(V

3) = 4 119891(V

4) = 8 119891(v

5) = 10 119891(V

6) = 5 and

119891(V7) = 7 Then 119890

119891(0) = 12 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 and 119891(V

9) = 9 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 119891(V

9) = 13 and 119891(V

10) = 9 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 11 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9119891(V

8) = 3119891(V

9) = 13119891(V

10) = 11 and119891(V

11) = 7

Then 119890119891(0) = 18 119890

119891(1) = 17

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 ISRN Combinatorics

pair of adjacent vertices 119906 and V gcd(119891(119906) 119891(V)) = 1 Thegraph which admits a prime labeling is called a prime graph

The concept of prime labeling has attracted manyresearchers as the study of prime numbers is of greatimportance because prime numbers are scattered and thereare arbitrarily large gaps in the sequence of prime numbersVaidya and Prajapati [7 8] have investigated many resultson prime labeling Same authors [9] have discussed primelabeling in the context of duplication of graph elementsMotivated through the concepts of prime labeling and cordiallabeling a new concept termed as a prime cordial labelingwas introduced by Sundaram et al [10] which contains blendof both the labelings

Definition 5 A prime cordial labeling of a graph119866with vertexset 119881(119866) is a bijection 119891 119881(119866) rarr 1 2 3 |119881(119866)| and ifthe induced function 119891

lowast

119864(119866) rarr 0 1 is defined by

119891lowast

(119890 = 119906V) = 1 if gcd (119891 (119906) 119891 (V)) = 1

= 0 otherwise(2)

then the number of edges labeled with 0 and the number ofedges labeled with 1 differ by at most 1 A graph which admitsprime cordial labeling is called a prime cordial graph

Many graphs are proved to be prime cordial in the workof Sundaram et al [10] Prime cordial labeling for somecycle related graphs has been discussed by Vaidya and Vihol[11] Prime cordial labeling in the context of some graphoperations has been discussed by Vaidya and Vihol [12] andVaidya and Shah [13 14] Vaidya and Shah [14] have provedthat the wheel graph 119882

119899admits prime cordial labeling for

119899 ge 8 while same authors in [15] have discussed primecordial labeling for some wheel related graphs Babitha andBaskar Babujee [16] have exhibited prime cordial labeling forsome cycle related graphs and discussed the duality of primecordial labeling The same authors in [17] have derived somecharacterizations of prime cordial graphs and investigatedvarious methods to construct larger prime cordial graphsusing existing prime cordial graphs We investigate a methoddifferent from existing one to construct larger prime cordialgraph from an existing prime cordial graph

Definition 6 The wheel 119882119899is defined to be the join 119870

1+ 119862119899

The vertex corresponding to1198701is known as apex and vertices

corresponding to cycle are known as rim vertices while theedges corresponding to cycle are known as rim edges

Definition 7 The bistar 119861119899119899

is a graph obtained by joining thecenter (apex) vertices of two copies of119870

1119899by an edge

Definition 8 The fan 119865119899is the graph obtained by taking 119899 minus 2

concurrent chords in cycle 119862119899+1

The vertex at which all thechords are concurrent is called the apex vertex In otherwords 119865

119899= 119875119899+ 1198701

Definition 9 The double fan DF119899consists of two fan graphs

that have a common path In other words DF119899= 119875119899+ 1198702

Definition 10 The duplication of an edge 119890 = 119906V of graph 119866

produces a new graph 1198661015840 by adding an edge 119890

1015840

= 1199061015840V1015840 such

that119873(1199061015840

) = 119873(119906) cup V1015840 minus V and119873(V1015840) = 119873(V) cup 1199061015840

minus 119906

Definition 11 (see [18]) Let119866 = (119881(119866) 119864(119866)) be a graph with119881 = 119878

1cup 1198782cup 1198783cup sdot sdot sdot 119878

119905cup 119879 where each 119878

119894is a set of vertices

having at least two vertices of the same degree and 119879 =

119881 cup119905

119894=1119878119894 The degree splitting graph of G denoted by 119863119878(119866)

is obtained from 119866 by adding vertices 1199081 1199082 1199083 119908

119905and

joining to each vertex of 119878119894for 1 le 119894 le 119905

2 Main Results

Theorem 12 Let 119866(119901 119902) with 119901 ge 4 be a prime cordial graphand let 119870

2119899be a bipartite graph with bipartition 119881 = 119881

1cup 1198812

with 1198811= V1 V2 and 119881

2= 1199061 1199062 119906

119899 If 119866

1is the graph

obtained by identifying the vertices V1and V2of 1198702119899

with thevertices of 119866 having labels 2 and 4 respectively then 119866

1admits

prime cordial labeling in any of the following cases

(i) 119899 is even and 119866 is of any size 119902

(ii) 119899 119901 and 119902 are odd with 119890119891(0) = lfloor1199022rfloor

(iii) 119899 is odd 119901 is even and 119902 is odd with 119890119891(0) = lceil1199022rceil

Proof Let 119866(119901 119902) be a prime cordial graph and let 1198911be the

prime cordial labeling of 119866 Let 1199081 1199082isin 119881 be the vertices of

119866 such that 1198911(1199081) = 2 and 119891

1(1199082) = 4 Consider the 119870

2119899

with bipartition 119881 = 1198811cup 1198812with 119881

1= V1 V2 and 119881

2=

1199061 1199062 119906

119899 Now identify the vertices V

1to1199081and V2to1199082

and denote the resultant graph as 1198661 Then 119881(119866

1) = 119881(119866) cup

1199061 1199062 119906

119899 and 119864(119866

1) = 119864(119866) cup 119908

1119906119894 11990821199061198941 le 119894 le 119899 so

|119881(1198661)| = 119901 + 119899 and |119864(119866

1)| = 119902 + 2119899 To define 119891 119881(119866

1) rarr

1 2 3 119901 + 119899 we consider the following three casesCase (i) (119899 is even and 119866 is of any size 119902) Since 119866 is aprime cordial graph we assign vertex labels such that119891(119908

119894) =

1198911(119908119894) where 119908

119894isin 119881(119866) = 119881(119866

1) cap 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (3)

Since 119899 is even and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and

1199082are adjacent to each 119906

119894 119894 le 119894 le 119899 And this vertex

assignment generates 119899 edges with label 1 and 119899 edges withlabel 0 Following Table 1 gives edge condition for primecordial labeling for 119866

1under 119891

From Table 1 we have |119890119891(0) minus 119890

119891(1)| le 1

Case (ii) (119899 119901 and 119902 are odd with 119890119891(0) = lfloor1199022rfloor) Here 119901 and

119902 both are odd and 119866 is a prime cordial graph with 119890119891(0) =

lfloor1199022rfloorSince 119866 is a prime cordial graph we keep the vertex label

of all the vertices of119866 in1198661as it isTherefore 119891(119908

119894) = 1198911(119908119894)

where 119908119894isin 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (4)

Since 119899 is odd and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and 119908

2

are adjacent to each 119906119894 119894 le 119894 le 119899 And this vertex assignment

generates 119899+ 1 edges with label 0 and 119899minus 1 edges with label 1

ISRN Combinatorics 3

Table 1

119902 Edge conditions for 119866 Edge conditions for 1198661

Even 119890119891(0) = 119890

119891(1) =

119902

2119890119891(0) = 119890

119891(1) =

119902

2

Odd119890119891(0) minus 1 = 119890

119891(1) = lfloor

119902

2rfloor 119890

119891(0) minus 1 = 119890

119891(1) = lfloor

119902

2rfloor + 119899

119890119891(0) = 119890

119891(1) minus 1 = lfloor

119902

2rfloor 119890

119891(0) = 119890

119891(1) minus 1 = lfloor

119902

2rfloor + 119899

Therefore edge conditions for 1198661under 119891 are 119890

119891(0) =

lfloor1199022rfloor + 119899+ 1 and 119890119891(1) = lceil1199022rceil + 119899minus 1 Therefore 119890

119891(0) minus 1 =

119890119891(1) Hence |119890

119891(0) minus 119890

119891(1)| le 1 for graph 119866

1

Case (iii) (119899 is odd119901 is even and 119902 is oddwith 119890119891(0) = lceil1199022rceil)

Here 119901 is even 119902 is odd and 119866 is a prime cordial graph with119890119891(0) = lceil1199022rceilSince 119866 is a prime cordial graph we keep the vertex label

of all the vertices of119866 in1198661as it isTherefore 119891(119908

119894) = 1198911(119908119894)

where 119908119894isin 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (5)

Since 119899 is odd and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and 119908

2

are adjacent to each 119906119894 119894 le 119894 le 119899 And this vertex assignment

generates 119899minus 1 edges with label 0 and 119899+ 1 edges with label 1Therefore edge conditions for 119866

1under 119891 are 119890

119891(0) =

lceil1199022rceil + 119899 minus 1 and 119890119891(1) = lfloor1199022rfloor + 119899 + 1 Therefore 119890

119891(0) =

119890119891(1) minus 1 Hence |119890

119891(0) minus 119890

119891(1)| le 1 for graph 119866

1

Hence in all the cases discussed above 1198661admits prime

cordial labeling

Illustration 1 Consider the graph 119866 as shown in Figure 1with 119901 = 7 and 119902 = 9 119866 is a prime cordial graph with119890119891(0) = 4 119890

119891(1) = 5 Take 119899 = 3 and construct graph 119866

1

In accordance with Case (ii) of Theorem 12 a prime cordiallabeling of119866

1is as shown in Figure 1 Here 119890

119891(0) = 8 119890

119891(1) =

7

Theorem13 Double fan119863119865119899is a prime cordial graph for 119899 = 8

and 119899 ge 10

Proof Let DF119899be the double fan with apex vertices 119906

1 1199062and

V1 V2 V

119899are vertices common pathThen |119881(DF

119899)| = 119899 +

2 and |119864(DF119899)| = 3119899minus1 To define119891 119881(119866) rarr 1 2 3 119899+

2 we consider the following five casesCase 1 (119899 = 3 to 7 and 119899 = 9) In order to satisfy the edgecondition for prime cordial labeling in DF

3it is essential to

label four edges with label 0 and four edges with label 1 out ofeight edges But all the possible assignments of vertex labelswill give rise to 0 labels for at most one edge and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 6 gt 1 Hence DF

3

is not a prime cordial graphIn order to satisfy the edge condition for prime cordial

labeling in DF4it is essential to label five edges with label

0 and six edges with label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most three edges and 1 label for at least eight edgesThat

is |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence DF

4is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF5it is essential to label seven edges with label 0

and seven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence DF

5is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF6it is essential to label eight edges with label 0

and nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence DF

6is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF7it is essential to label ten edges with label 0

and ten edges with label 1 out of twenty edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most eight edges and 1 label for at least twelve edgesThat is |119890

119891(0) minus 119890

119891(1)| = 4 gt 1 Hence DF

7is not a prime

cordial graphIn order to satisfy the edge condition for prime cordial

labeling in DF9it is essential to label thirteen edges with label

0 and thirteen edges with label 1 out of twenty-six edges Butall the possible assignments of vertex labels will give rise to 0labels for at most twelve edges and 1 label for at least fourteenedges That is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence DF

9is not a

prime cordial graphCase 2 (119899 = 8 10 11 12) For 119899 = 8 119891(119906

1) = 2 119891(119906

2) = 6

and 119891(V1) = 4 119891(V

2) = 8 119891(V

3) = 3 119891(V

4) = 9 119891(V

5) = 10

119891(V6) = 5119891(V

7) = 7 and119891(V

8) = 1Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10 119891(V

7) = 1

119891(V8) = 5 119891(V

9) = 7 and 119891(V

10) = 11 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 11 119891(119906

1) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10 119891(V

7) = 5

119891(V8) = 7 119891(V

9) = 11 119891(V

10) = 13 and 119891(V

11) = 1 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 14 119891(V

7) = 10

119891(V8) = 5 119891(V

9) = 7 119891(V

10) = 11 119891(V

11) = 13 and 119891(V

12) =

1 Then 119890119891(0) = 18 119890

119891(1) = 17

Now for the remaining three cases let

119896 = lfloor119899 + 2

2rfloor 119898 = lfloor

119899 + 2

3rfloor

1199051= lceil

3119899 minus 1

2rceil 119905

2= 3119896 minus 7 + lceil

119898

2rceil

(6)

1199053= largest even number le 119899+2 and 119905

4= largest odd number

le 119899 + 2Case 3 (119905

1= 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (7)

4 ISRN Combinatorics

1

2

3

4

5

6

7

w1

w2

w3

w4w5

w6

w7

G

(a)

1

2

3

4

5

6

7

8

9

10

u1

u2

u3

w1

w2

w3

w4w5

w6

w7

G1

(b)

Figure 1

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 1 1199053 1199053minus 2 1199053minus 4 14 12 10 8 4 3

5 7 9 1199054minus 2 1199054

Case 4 (1199051gt 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (8)

Let 1199055= 1199051minus 1199052 Consider

119891 (V1) = 3

119891 (V2) = 9

119891 (V3) = 15

119891 (V4) = 21

119891 (V1199055

) = 3 (21199055minus 1)

119891 (V1199055+1

) = 119891 (V1199055

) + 6

(9)

Now for remaining vertices V1199055+2

V1199055+3

V119899assign the labels

1 1199053 1199053

minus 2 1199053

minus 4 14 1210 8 4 5 7 all the oddnumbers in ascending orderCase 5 (119905

2gt 1199051) Let 119905

6= 1199052minus 1199051

Sub-Case 1 119899 is even Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (10)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderSub-Case 2 119899 is odd Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (11)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderIn view of the above defined labeling pattern for Cases 3

4 and 5 we have

119890119891(0) = lceil

3119899 minus 1

2rceil 119890

119891(1) = lfloor

3119899 minus 1

2rfloor (12)

Thus we have |119890119891(0) minus 119890

119891(1)| le 1

Hence DF119899is a prime cordial graph for 119899 = 8 and 119899 ge

10

Illustration 2 For the graph DF15 |119881(DF

15)| = 17 and

|119864(DF15)| = 44 In accordance with Theorem 13 we have

119896 = 5 119898 = 5 1199051

= 22 and 1199052

= 20 Here 1199051

gt 1199052so

labeling pattern described in Case 4 will be applicable and1199055= 2 The corresponding prime cordial labeling is shown in

Figure 2 Here 119890119891(0) = 22 = e

119891(1)

Illustration 3 For the graph DF37 |119881(DF

37)| = 39 and

|119864(DF37)| = 110 In accordance with Theorem 13 we have

119896 = 19119898 = 13 1199051= 55 and 119905

2= 57 Here 119905

2gt 1199051and 119899 = 37

so labeling pattern described in Sub-Case 2 of Case 5 will beapplicable and 119905

6= 2 And corresponding labeling pattern is

as below

119891 (1199061) = 2 119891 (119906

2) = 6 (13)

ISRN Combinatorics 5

1

2

3 4 5

6

789 10 1112 131415 16 17

u1

u2

1 23 4 5 6 7 8 9 10 11 12 13 14 15

Figure 2

For the vertices V1 V2 V

37we assign the vertex labels 39

38 37 36 35 34 32 30 28 26 24 22 20 18 16 14 12 10 8 43 5 7 9 11 13 15 17 19 21 23 25 27 29 31 and 33 respectivelywhere 119890

119891(0) = 55 = 119890

119891(1) Therefore DF

37is a prime cordial

graph

Theorem 14 The graph obtained by duplication of an arbi-trary rim edge by an edge in119882

119899is a prime cordial graph where

119899 ge 6

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate therim edge 119890 = V

1V2by an edge 119890

1015840

= 11990611199062and call the resultant

graph as119866Then |119881(119866)| = 119899+3 and |119864(119866)| = 2119899+5 To define119891 119881(119866) rarr 1 2 3 119899+3 we consider the following fourcases

Case 1 (119899 = 3 4 5) For 119899 = 3 to satisfy the edge condition forprime cordial labeling it is essential to label five edges withlabel 0 and six edgeswith label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least seven edgesThatis |119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 3 119866 is not a prime

cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label six edges with label 0 and sevenedges with label 1 out of thirteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost four edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andeight edges with label 1 out of fifteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost six edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graph

Case 2 (119899 = 6 to 10) For 119899 = 6 119891(1199061) = 4 119891(119906

2) = 8 and

119891(V0) = 6 119891(V

1) = 3 119891(V

2) = 9 119891(V

3) = 7 119891(V

4) = 5

119891(V5) = 1 and 119891(V

6) = 2 Then 119890

119891(0) = 8 119890

119891(1) = 9

For 119899 = 7 119891(1199061) = 4 119891(119906

2) = 8 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 9 119891(V

3) = 7 119891(V

4) = 5 119891(V

5) = 10 119891(V

6) = 1 and

119891(V7) = 2 Then 119890

119891(0) = 4 119890

119891(1) = 5

For 119899 = 8 119891(1199061) = 11 119891(119906

2) = 10 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 2 119891(V

3) = 8 119891(V

4) = 4 119891(V

5) = 5 119891(V

6) = 7

119891(V7) = 1 and 119891(V

8) = 9 Then 119890

119891(0) = 10 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 11 119891(119906

2) = 12 and 119891(V

0) = 2 119891(V

1) =

1 119891(V2) = 3 119891(V

3) = 6 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10

119891(V7) = 5119891(V

8) = 7 and119891(V

9) = 9Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 13 119891(119906

2) = 12 and 119891(V

0) = 2

119891(V1) = 9 119891(V

2) = 3 119891(V

3) = 6 119891(V

4) = 4 119891(V

5) = 8

119891(V6) = 10 119891(V

7) = 5 119891(V

8) = 1 119891(V

9) = 7 and 119891(V

10) = 11

Then 119890119891(0) = 12 119890

119891(1) = 13

Case 3 (119899 is even 119899 ge 12) Consider

119891 (V0) = 2

119891 (V1) = 5 119891 (V

2) = 10

119891 (V3) = 4 119891 (V

4) = 8

119891 (V4+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 5

119891 (V1198992

) = 6 119891 (V(1198992)+1

) = 3

119891 (V(1198992)+2

) = 9

119891 (V119899minus1

) = 1 119891 (V119899) = 7

119891 (V(1198992)+2+119894

) = 11 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 4

119891 (1199061) = 119899 + 3 119891 (119906

2) = 119899 + 2

(14)

6 ISRN Combinatorics

In view of the above defined labeling pattern for Case 3 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 4(mod 7) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod 7)

Case 4 (119899 is odd 119899 ge 11) Consider

119891 (V0) = 2

119891 (V1) = 10 119891 (V

2) = 4

119891 (V3) = 8

119891 (V3+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le ((119899 minus 1) 2) minus 4

119891 (V(119899minus1)2

) = 6 119891 (V(119899+1)2

) = 3

119891 (V(119899+3)2

) = 1

119891 (V119899+1minus119894

) = 5 + 2 (119894 minus 1) 1 le 119894 le (119899 minus 3) 2

119891 (1199061) = 119899 + 2 119891 (119906

2) = 119899 + 3

(15)

In view of the above defined labeling pattern for Case 4 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 3(mod5) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod7)

In view of Cases 2 to 4 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence 119866 is a prime cordial graph for 119899 ge 6

Illustration 4 Let 119866 be the graph obtained by duplicationof an arbitrary rim edge by an edge in wheel 119882

13 Then

|119881(119866)| = 16 and |119864(119866)| = 31 In accordance withTheorem 14 Case 4 will be applicable and the correspondingprime cordial labeling is shown in Figure 3 Here 119890

119891(0) = 16

119890119891(1) = 15

Theorem 15 The graph obtained by duplication of an arbi-trary spoke edge by an edge in wheel119882

119899is prime cordial graph

where 119899 = 7 and 119899 ge 9

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate thespoke edge 119890 = V

0V1by an edge 1198901015840 = 119906

11199062and call the resultant

graph 119866 Then |119881(119866)| = 119899 + 3 and |119864(119866)| = 3119899 + 2 To define119891 119881(119866) rarr 1 2 3 119899 + 3 we consider following threecasesCase 1 (119899 = 3 to 6 and 119899 = 8) For 119899 = 3 to satisfy the edgecondition for prime cordial labeling it is essential to label fiveedges with label 0 and six edges with label 1 out of elevenedges But all the possible assignments of vertex labels willgive rise to 0 labels for at most four edges and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for

119899 = 3 119866 is not a prime cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andseven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label eight edges with label 0 and

1

2

3

45

6

7 8

9

10

11

12

13

14

15

16

u1

u2

1

2

3

4

5

6

78

9

10

11

12

13

Figure 3

nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graphFor 119899 = 6 to satisfy the edge condition for prime cordial

labeling it is essential to label ten edges with label 0 and tenedges with label 1 out of twenty edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost eight edges and 1 label for at least twelve edges That is|119890119891(0) minus 119890

119891(1)| = 4 gt 1 Hence for 119899 = 6 119866 is not a prime

cordial graphFor 119899 = 8 to satisfy the edge condition for prime cordial

labeling it is essential to label thirteen edges with label 0 andthirteen edges with label 1 out of twenty-six edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most twelve edges and 1 label for at least fourteen edgesThat is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence for 119899 = 8 119866 is not a

prime cordial graphCase 2 (119899 = 7 119899 = 9 to 12 and 119899 = 14 16 18 20 22) For119899 = 7 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 9

119891(V2) = 3 119891(V

3) = 4 119891(V

4) = 8 119891(v

5) = 10 119891(V

6) = 5 and

119891(V7) = 7 Then 119890

119891(0) = 12 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 and 119891(V

9) = 9 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 119891(V

9) = 13 and 119891(V

10) = 9 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 11 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9119891(V

8) = 3119891(V

9) = 13119891(V

10) = 11 and119891(V

11) = 7

Then 119890119891(0) = 18 119890

119891(1) = 17

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Combinatorics 3

Table 1

119902 Edge conditions for 119866 Edge conditions for 1198661

Even 119890119891(0) = 119890

119891(1) =

119902

2119890119891(0) = 119890

119891(1) =

119902

2

Odd119890119891(0) minus 1 = 119890

119891(1) = lfloor

119902

2rfloor 119890

119891(0) minus 1 = 119890

119891(1) = lfloor

119902

2rfloor + 119899

119890119891(0) = 119890

119891(1) minus 1 = lfloor

119902

2rfloor 119890

119891(0) = 119890

119891(1) minus 1 = lfloor

119902

2rfloor + 119899

Therefore edge conditions for 1198661under 119891 are 119890

119891(0) =

lfloor1199022rfloor + 119899+ 1 and 119890119891(1) = lceil1199022rceil + 119899minus 1 Therefore 119890

119891(0) minus 1 =

119890119891(1) Hence |119890

119891(0) minus 119890

119891(1)| le 1 for graph 119866

1

Case (iii) (119899 is odd119901 is even and 119902 is oddwith 119890119891(0) = lceil1199022rceil)

Here 119901 is even 119902 is odd and 119866 is a prime cordial graph with119890119891(0) = lceil1199022rceilSince 119866 is a prime cordial graph we keep the vertex label

of all the vertices of119866 in1198661as it isTherefore 119891(119908

119894) = 1198911(119908119894)

where 119908119894isin 119881(119866) and 119894 le 119894 le 119901

119891 (119906119894) = 119901 + 119894 119894 le 119894 le 119899 (5)

Since 119899 is odd and 1198911(1199081) = 2 and 119891

1(1199082) = 4 119908

1and 119908

2

are adjacent to each 119906119894 119894 le 119894 le 119899 And this vertex assignment

generates 119899minus 1 edges with label 0 and 119899+ 1 edges with label 1Therefore edge conditions for 119866

1under 119891 are 119890

119891(0) =

lceil1199022rceil + 119899 minus 1 and 119890119891(1) = lfloor1199022rfloor + 119899 + 1 Therefore 119890

119891(0) =

119890119891(1) minus 1 Hence |119890

119891(0) minus 119890

119891(1)| le 1 for graph 119866

1

Hence in all the cases discussed above 1198661admits prime

cordial labeling

Illustration 1 Consider the graph 119866 as shown in Figure 1with 119901 = 7 and 119902 = 9 119866 is a prime cordial graph with119890119891(0) = 4 119890

119891(1) = 5 Take 119899 = 3 and construct graph 119866

1

In accordance with Case (ii) of Theorem 12 a prime cordiallabeling of119866

1is as shown in Figure 1 Here 119890

119891(0) = 8 119890

119891(1) =

7

Theorem13 Double fan119863119865119899is a prime cordial graph for 119899 = 8

and 119899 ge 10

Proof Let DF119899be the double fan with apex vertices 119906

1 1199062and

V1 V2 V

119899are vertices common pathThen |119881(DF

119899)| = 119899 +

2 and |119864(DF119899)| = 3119899minus1 To define119891 119881(119866) rarr 1 2 3 119899+

2 we consider the following five casesCase 1 (119899 = 3 to 7 and 119899 = 9) In order to satisfy the edgecondition for prime cordial labeling in DF

3it is essential to

label four edges with label 0 and four edges with label 1 out ofeight edges But all the possible assignments of vertex labelswill give rise to 0 labels for at most one edge and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 6 gt 1 Hence DF

3

is not a prime cordial graphIn order to satisfy the edge condition for prime cordial

labeling in DF4it is essential to label five edges with label

0 and six edges with label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most three edges and 1 label for at least eight edgesThat

is |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence DF

4is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF5it is essential to label seven edges with label 0

and seven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence DF

5is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF6it is essential to label eight edges with label 0

and nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence DF

6is not a prime cordial

graphIn order to satisfy the edge condition for prime cordial

labeling in DF7it is essential to label ten edges with label 0

and ten edges with label 1 out of twenty edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most eight edges and 1 label for at least twelve edgesThat is |119890

119891(0) minus 119890

119891(1)| = 4 gt 1 Hence DF

7is not a prime

cordial graphIn order to satisfy the edge condition for prime cordial

labeling in DF9it is essential to label thirteen edges with label

0 and thirteen edges with label 1 out of twenty-six edges Butall the possible assignments of vertex labels will give rise to 0labels for at most twelve edges and 1 label for at least fourteenedges That is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence DF

9is not a

prime cordial graphCase 2 (119899 = 8 10 11 12) For 119899 = 8 119891(119906

1) = 2 119891(119906

2) = 6

and 119891(V1) = 4 119891(V

2) = 8 119891(V

3) = 3 119891(V

4) = 9 119891(V

5) = 10

119891(V6) = 5119891(V

7) = 7 and119891(V

8) = 1Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10 119891(V

7) = 1

119891(V8) = 5 119891(V

9) = 7 and 119891(V

10) = 11 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 11 119891(119906

1) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10 119891(V

7) = 5

119891(V8) = 7 119891(V

9) = 11 119891(V

10) = 13 and 119891(V

11) = 1 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 6 and 119891(V

1) = 3 119891(V

2) =

9 119891(V3) = 12 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 14 119891(V

7) = 10

119891(V8) = 5 119891(V

9) = 7 119891(V

10) = 11 119891(V

11) = 13 and 119891(V

12) =

1 Then 119890119891(0) = 18 119890

119891(1) = 17

Now for the remaining three cases let

119896 = lfloor119899 + 2

2rfloor 119898 = lfloor

119899 + 2

3rfloor

1199051= lceil

3119899 minus 1

2rceil 119905

2= 3119896 minus 7 + lceil

119898

2rceil

(6)

1199053= largest even number le 119899+2 and 119905

4= largest odd number

le 119899 + 2Case 3 (119905

1= 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (7)

4 ISRN Combinatorics

1

2

3

4

5

6

7

w1

w2

w3

w4w5

w6

w7

G

(a)

1

2

3

4

5

6

7

8

9

10

u1

u2

u3

w1

w2

w3

w4w5

w6

w7

G1

(b)

Figure 1

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 1 1199053 1199053minus 2 1199053minus 4 14 12 10 8 4 3

5 7 9 1199054minus 2 1199054

Case 4 (1199051gt 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (8)

Let 1199055= 1199051minus 1199052 Consider

119891 (V1) = 3

119891 (V2) = 9

119891 (V3) = 15

119891 (V4) = 21

119891 (V1199055

) = 3 (21199055minus 1)

119891 (V1199055+1

) = 119891 (V1199055

) + 6

(9)

Now for remaining vertices V1199055+2

V1199055+3

V119899assign the labels

1 1199053 1199053

minus 2 1199053

minus 4 14 1210 8 4 5 7 all the oddnumbers in ascending orderCase 5 (119905

2gt 1199051) Let 119905

6= 1199052minus 1199051

Sub-Case 1 119899 is even Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (10)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderSub-Case 2 119899 is odd Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (11)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderIn view of the above defined labeling pattern for Cases 3

4 and 5 we have

119890119891(0) = lceil

3119899 minus 1

2rceil 119890

119891(1) = lfloor

3119899 minus 1

2rfloor (12)

Thus we have |119890119891(0) minus 119890

119891(1)| le 1

Hence DF119899is a prime cordial graph for 119899 = 8 and 119899 ge

10

Illustration 2 For the graph DF15 |119881(DF

15)| = 17 and

|119864(DF15)| = 44 In accordance with Theorem 13 we have

119896 = 5 119898 = 5 1199051

= 22 and 1199052

= 20 Here 1199051

gt 1199052so

labeling pattern described in Case 4 will be applicable and1199055= 2 The corresponding prime cordial labeling is shown in

Figure 2 Here 119890119891(0) = 22 = e

119891(1)

Illustration 3 For the graph DF37 |119881(DF

37)| = 39 and

|119864(DF37)| = 110 In accordance with Theorem 13 we have

119896 = 19119898 = 13 1199051= 55 and 119905

2= 57 Here 119905

2gt 1199051and 119899 = 37

so labeling pattern described in Sub-Case 2 of Case 5 will beapplicable and 119905

6= 2 And corresponding labeling pattern is

as below

119891 (1199061) = 2 119891 (119906

2) = 6 (13)

ISRN Combinatorics 5

1

2

3 4 5

6

789 10 1112 131415 16 17

u1

u2

1 23 4 5 6 7 8 9 10 11 12 13 14 15

Figure 2

For the vertices V1 V2 V

37we assign the vertex labels 39

38 37 36 35 34 32 30 28 26 24 22 20 18 16 14 12 10 8 43 5 7 9 11 13 15 17 19 21 23 25 27 29 31 and 33 respectivelywhere 119890

119891(0) = 55 = 119890

119891(1) Therefore DF

37is a prime cordial

graph

Theorem 14 The graph obtained by duplication of an arbi-trary rim edge by an edge in119882

119899is a prime cordial graph where

119899 ge 6

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate therim edge 119890 = V

1V2by an edge 119890

1015840

= 11990611199062and call the resultant

graph as119866Then |119881(119866)| = 119899+3 and |119864(119866)| = 2119899+5 To define119891 119881(119866) rarr 1 2 3 119899+3 we consider the following fourcases

Case 1 (119899 = 3 4 5) For 119899 = 3 to satisfy the edge condition forprime cordial labeling it is essential to label five edges withlabel 0 and six edgeswith label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least seven edgesThatis |119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 3 119866 is not a prime

cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label six edges with label 0 and sevenedges with label 1 out of thirteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost four edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andeight edges with label 1 out of fifteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost six edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graph

Case 2 (119899 = 6 to 10) For 119899 = 6 119891(1199061) = 4 119891(119906

2) = 8 and

119891(V0) = 6 119891(V

1) = 3 119891(V

2) = 9 119891(V

3) = 7 119891(V

4) = 5

119891(V5) = 1 and 119891(V

6) = 2 Then 119890

119891(0) = 8 119890

119891(1) = 9

For 119899 = 7 119891(1199061) = 4 119891(119906

2) = 8 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 9 119891(V

3) = 7 119891(V

4) = 5 119891(V

5) = 10 119891(V

6) = 1 and

119891(V7) = 2 Then 119890

119891(0) = 4 119890

119891(1) = 5

For 119899 = 8 119891(1199061) = 11 119891(119906

2) = 10 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 2 119891(V

3) = 8 119891(V

4) = 4 119891(V

5) = 5 119891(V

6) = 7

119891(V7) = 1 and 119891(V

8) = 9 Then 119890

119891(0) = 10 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 11 119891(119906

2) = 12 and 119891(V

0) = 2 119891(V

1) =

1 119891(V2) = 3 119891(V

3) = 6 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10

119891(V7) = 5119891(V

8) = 7 and119891(V

9) = 9Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 13 119891(119906

2) = 12 and 119891(V

0) = 2

119891(V1) = 9 119891(V

2) = 3 119891(V

3) = 6 119891(V

4) = 4 119891(V

5) = 8

119891(V6) = 10 119891(V

7) = 5 119891(V

8) = 1 119891(V

9) = 7 and 119891(V

10) = 11

Then 119890119891(0) = 12 119890

119891(1) = 13

Case 3 (119899 is even 119899 ge 12) Consider

119891 (V0) = 2

119891 (V1) = 5 119891 (V

2) = 10

119891 (V3) = 4 119891 (V

4) = 8

119891 (V4+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 5

119891 (V1198992

) = 6 119891 (V(1198992)+1

) = 3

119891 (V(1198992)+2

) = 9

119891 (V119899minus1

) = 1 119891 (V119899) = 7

119891 (V(1198992)+2+119894

) = 11 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 4

119891 (1199061) = 119899 + 3 119891 (119906

2) = 119899 + 2

(14)

6 ISRN Combinatorics

In view of the above defined labeling pattern for Case 3 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 4(mod 7) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod 7)

Case 4 (119899 is odd 119899 ge 11) Consider

119891 (V0) = 2

119891 (V1) = 10 119891 (V

2) = 4

119891 (V3) = 8

119891 (V3+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le ((119899 minus 1) 2) minus 4

119891 (V(119899minus1)2

) = 6 119891 (V(119899+1)2

) = 3

119891 (V(119899+3)2

) = 1

119891 (V119899+1minus119894

) = 5 + 2 (119894 minus 1) 1 le 119894 le (119899 minus 3) 2

119891 (1199061) = 119899 + 2 119891 (119906

2) = 119899 + 3

(15)

In view of the above defined labeling pattern for Case 4 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 3(mod5) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod7)

In view of Cases 2 to 4 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence 119866 is a prime cordial graph for 119899 ge 6

Illustration 4 Let 119866 be the graph obtained by duplicationof an arbitrary rim edge by an edge in wheel 119882

13 Then

|119881(119866)| = 16 and |119864(119866)| = 31 In accordance withTheorem 14 Case 4 will be applicable and the correspondingprime cordial labeling is shown in Figure 3 Here 119890

119891(0) = 16

119890119891(1) = 15

Theorem 15 The graph obtained by duplication of an arbi-trary spoke edge by an edge in wheel119882

119899is prime cordial graph

where 119899 = 7 and 119899 ge 9

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate thespoke edge 119890 = V

0V1by an edge 1198901015840 = 119906

11199062and call the resultant

graph 119866 Then |119881(119866)| = 119899 + 3 and |119864(119866)| = 3119899 + 2 To define119891 119881(119866) rarr 1 2 3 119899 + 3 we consider following threecasesCase 1 (119899 = 3 to 6 and 119899 = 8) For 119899 = 3 to satisfy the edgecondition for prime cordial labeling it is essential to label fiveedges with label 0 and six edges with label 1 out of elevenedges But all the possible assignments of vertex labels willgive rise to 0 labels for at most four edges and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for

119899 = 3 119866 is not a prime cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andseven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label eight edges with label 0 and

1

2

3

45

6

7 8

9

10

11

12

13

14

15

16

u1

u2

1

2

3

4

5

6

78

9

10

11

12

13

Figure 3

nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graphFor 119899 = 6 to satisfy the edge condition for prime cordial

labeling it is essential to label ten edges with label 0 and tenedges with label 1 out of twenty edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost eight edges and 1 label for at least twelve edges That is|119890119891(0) minus 119890

119891(1)| = 4 gt 1 Hence for 119899 = 6 119866 is not a prime

cordial graphFor 119899 = 8 to satisfy the edge condition for prime cordial

labeling it is essential to label thirteen edges with label 0 andthirteen edges with label 1 out of twenty-six edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most twelve edges and 1 label for at least fourteen edgesThat is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence for 119899 = 8 119866 is not a

prime cordial graphCase 2 (119899 = 7 119899 = 9 to 12 and 119899 = 14 16 18 20 22) For119899 = 7 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 9

119891(V2) = 3 119891(V

3) = 4 119891(V

4) = 8 119891(v

5) = 10 119891(V

6) = 5 and

119891(V7) = 7 Then 119890

119891(0) = 12 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 and 119891(V

9) = 9 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 119891(V

9) = 13 and 119891(V

10) = 9 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 11 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9119891(V

8) = 3119891(V

9) = 13119891(V

10) = 11 and119891(V

11) = 7

Then 119890119891(0) = 18 119890

119891(1) = 17

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 ISRN Combinatorics

1

2

3

4

5

6

7

w1

w2

w3

w4w5

w6

w7

G

(a)

1

2

3

4

5

6

7

8

9

10

u1

u2

u3

w1

w2

w3

w4w5

w6

w7

G1

(b)

Figure 1

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 1 1199053 1199053minus 2 1199053minus 4 14 12 10 8 4 3

5 7 9 1199054minus 2 1199054

Case 4 (1199051gt 1199052) Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (8)

Let 1199055= 1199051minus 1199052 Consider

119891 (V1) = 3

119891 (V2) = 9

119891 (V3) = 15

119891 (V4) = 21

119891 (V1199055

) = 3 (21199055minus 1)

119891 (V1199055+1

) = 119891 (V1199055

) + 6

(9)

Now for remaining vertices V1199055+2

V1199055+3

V119899assign the labels

1 1199053 1199053

minus 2 1199053

minus 4 14 1210 8 4 5 7 all the oddnumbers in ascending orderCase 5 (119905

2gt 1199051) Let 119905

6= 1199052minus 1199051

Sub-Case 1 119899 is even Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (10)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderSub-Case 2 119899 is odd Consider

119891 (1199061) = 2 119891 (119906

2) = 6 (11)

For the vertices V1 V2 V3 V

119899we assign the vertex labels in

the following order 119899+2 119899+1 119899 119899minus1 119899minus2 119899minus3 119899+2minus21199056

119899+2minus2(1199056+1) 119899+2minus2(119905

6+2) 10 8 4 3 5 7 remaining

odd numbers in ascending orderIn view of the above defined labeling pattern for Cases 3

4 and 5 we have

119890119891(0) = lceil

3119899 minus 1

2rceil 119890

119891(1) = lfloor

3119899 minus 1

2rfloor (12)

Thus we have |119890119891(0) minus 119890

119891(1)| le 1

Hence DF119899is a prime cordial graph for 119899 = 8 and 119899 ge

10

Illustration 2 For the graph DF15 |119881(DF

15)| = 17 and

|119864(DF15)| = 44 In accordance with Theorem 13 we have

119896 = 5 119898 = 5 1199051

= 22 and 1199052

= 20 Here 1199051

gt 1199052so

labeling pattern described in Case 4 will be applicable and1199055= 2 The corresponding prime cordial labeling is shown in

Figure 2 Here 119890119891(0) = 22 = e

119891(1)

Illustration 3 For the graph DF37 |119881(DF

37)| = 39 and

|119864(DF37)| = 110 In accordance with Theorem 13 we have

119896 = 19119898 = 13 1199051= 55 and 119905

2= 57 Here 119905

2gt 1199051and 119899 = 37

so labeling pattern described in Sub-Case 2 of Case 5 will beapplicable and 119905

6= 2 And corresponding labeling pattern is

as below

119891 (1199061) = 2 119891 (119906

2) = 6 (13)

ISRN Combinatorics 5

1

2

3 4 5

6

789 10 1112 131415 16 17

u1

u2

1 23 4 5 6 7 8 9 10 11 12 13 14 15

Figure 2

For the vertices V1 V2 V

37we assign the vertex labels 39

38 37 36 35 34 32 30 28 26 24 22 20 18 16 14 12 10 8 43 5 7 9 11 13 15 17 19 21 23 25 27 29 31 and 33 respectivelywhere 119890

119891(0) = 55 = 119890

119891(1) Therefore DF

37is a prime cordial

graph

Theorem 14 The graph obtained by duplication of an arbi-trary rim edge by an edge in119882

119899is a prime cordial graph where

119899 ge 6

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate therim edge 119890 = V

1V2by an edge 119890

1015840

= 11990611199062and call the resultant

graph as119866Then |119881(119866)| = 119899+3 and |119864(119866)| = 2119899+5 To define119891 119881(119866) rarr 1 2 3 119899+3 we consider the following fourcases

Case 1 (119899 = 3 4 5) For 119899 = 3 to satisfy the edge condition forprime cordial labeling it is essential to label five edges withlabel 0 and six edgeswith label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least seven edgesThatis |119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 3 119866 is not a prime

cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label six edges with label 0 and sevenedges with label 1 out of thirteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost four edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andeight edges with label 1 out of fifteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost six edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graph

Case 2 (119899 = 6 to 10) For 119899 = 6 119891(1199061) = 4 119891(119906

2) = 8 and

119891(V0) = 6 119891(V

1) = 3 119891(V

2) = 9 119891(V

3) = 7 119891(V

4) = 5

119891(V5) = 1 and 119891(V

6) = 2 Then 119890

119891(0) = 8 119890

119891(1) = 9

For 119899 = 7 119891(1199061) = 4 119891(119906

2) = 8 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 9 119891(V

3) = 7 119891(V

4) = 5 119891(V

5) = 10 119891(V

6) = 1 and

119891(V7) = 2 Then 119890

119891(0) = 4 119890

119891(1) = 5

For 119899 = 8 119891(1199061) = 11 119891(119906

2) = 10 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 2 119891(V

3) = 8 119891(V

4) = 4 119891(V

5) = 5 119891(V

6) = 7

119891(V7) = 1 and 119891(V

8) = 9 Then 119890

119891(0) = 10 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 11 119891(119906

2) = 12 and 119891(V

0) = 2 119891(V

1) =

1 119891(V2) = 3 119891(V

3) = 6 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10

119891(V7) = 5119891(V

8) = 7 and119891(V

9) = 9Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 13 119891(119906

2) = 12 and 119891(V

0) = 2

119891(V1) = 9 119891(V

2) = 3 119891(V

3) = 6 119891(V

4) = 4 119891(V

5) = 8

119891(V6) = 10 119891(V

7) = 5 119891(V

8) = 1 119891(V

9) = 7 and 119891(V

10) = 11

Then 119890119891(0) = 12 119890

119891(1) = 13

Case 3 (119899 is even 119899 ge 12) Consider

119891 (V0) = 2

119891 (V1) = 5 119891 (V

2) = 10

119891 (V3) = 4 119891 (V

4) = 8

119891 (V4+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 5

119891 (V1198992

) = 6 119891 (V(1198992)+1

) = 3

119891 (V(1198992)+2

) = 9

119891 (V119899minus1

) = 1 119891 (V119899) = 7

119891 (V(1198992)+2+119894

) = 11 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 4

119891 (1199061) = 119899 + 3 119891 (119906

2) = 119899 + 2

(14)

6 ISRN Combinatorics

In view of the above defined labeling pattern for Case 3 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 4(mod 7) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod 7)

Case 4 (119899 is odd 119899 ge 11) Consider

119891 (V0) = 2

119891 (V1) = 10 119891 (V

2) = 4

119891 (V3) = 8

119891 (V3+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le ((119899 minus 1) 2) minus 4

119891 (V(119899minus1)2

) = 6 119891 (V(119899+1)2

) = 3

119891 (V(119899+3)2

) = 1

119891 (V119899+1minus119894

) = 5 + 2 (119894 minus 1) 1 le 119894 le (119899 minus 3) 2

119891 (1199061) = 119899 + 2 119891 (119906

2) = 119899 + 3

(15)

In view of the above defined labeling pattern for Case 4 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 3(mod5) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod7)

In view of Cases 2 to 4 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence 119866 is a prime cordial graph for 119899 ge 6

Illustration 4 Let 119866 be the graph obtained by duplicationof an arbitrary rim edge by an edge in wheel 119882

13 Then

|119881(119866)| = 16 and |119864(119866)| = 31 In accordance withTheorem 14 Case 4 will be applicable and the correspondingprime cordial labeling is shown in Figure 3 Here 119890

119891(0) = 16

119890119891(1) = 15

Theorem 15 The graph obtained by duplication of an arbi-trary spoke edge by an edge in wheel119882

119899is prime cordial graph

where 119899 = 7 and 119899 ge 9

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate thespoke edge 119890 = V

0V1by an edge 1198901015840 = 119906

11199062and call the resultant

graph 119866 Then |119881(119866)| = 119899 + 3 and |119864(119866)| = 3119899 + 2 To define119891 119881(119866) rarr 1 2 3 119899 + 3 we consider following threecasesCase 1 (119899 = 3 to 6 and 119899 = 8) For 119899 = 3 to satisfy the edgecondition for prime cordial labeling it is essential to label fiveedges with label 0 and six edges with label 1 out of elevenedges But all the possible assignments of vertex labels willgive rise to 0 labels for at most four edges and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for

119899 = 3 119866 is not a prime cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andseven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label eight edges with label 0 and

1

2

3

45

6

7 8

9

10

11

12

13

14

15

16

u1

u2

1

2

3

4

5

6

78

9

10

11

12

13

Figure 3

nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graphFor 119899 = 6 to satisfy the edge condition for prime cordial

labeling it is essential to label ten edges with label 0 and tenedges with label 1 out of twenty edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost eight edges and 1 label for at least twelve edges That is|119890119891(0) minus 119890

119891(1)| = 4 gt 1 Hence for 119899 = 6 119866 is not a prime

cordial graphFor 119899 = 8 to satisfy the edge condition for prime cordial

labeling it is essential to label thirteen edges with label 0 andthirteen edges with label 1 out of twenty-six edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most twelve edges and 1 label for at least fourteen edgesThat is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence for 119899 = 8 119866 is not a

prime cordial graphCase 2 (119899 = 7 119899 = 9 to 12 and 119899 = 14 16 18 20 22) For119899 = 7 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 9

119891(V2) = 3 119891(V

3) = 4 119891(V

4) = 8 119891(v

5) = 10 119891(V

6) = 5 and

119891(V7) = 7 Then 119890

119891(0) = 12 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 and 119891(V

9) = 9 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 119891(V

9) = 13 and 119891(V

10) = 9 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 11 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9119891(V

8) = 3119891(V

9) = 13119891(V

10) = 11 and119891(V

11) = 7

Then 119890119891(0) = 18 119890

119891(1) = 17

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Combinatorics 5

1

2

3 4 5

6

789 10 1112 131415 16 17

u1

u2

1 23 4 5 6 7 8 9 10 11 12 13 14 15

Figure 2

For the vertices V1 V2 V

37we assign the vertex labels 39

38 37 36 35 34 32 30 28 26 24 22 20 18 16 14 12 10 8 43 5 7 9 11 13 15 17 19 21 23 25 27 29 31 and 33 respectivelywhere 119890

119891(0) = 55 = 119890

119891(1) Therefore DF

37is a prime cordial

graph

Theorem 14 The graph obtained by duplication of an arbi-trary rim edge by an edge in119882

119899is a prime cordial graph where

119899 ge 6

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate therim edge 119890 = V

1V2by an edge 119890

1015840

= 11990611199062and call the resultant

graph as119866Then |119881(119866)| = 119899+3 and |119864(119866)| = 2119899+5 To define119891 119881(119866) rarr 1 2 3 119899+3 we consider the following fourcases

Case 1 (119899 = 3 4 5) For 119899 = 3 to satisfy the edge condition forprime cordial labeling it is essential to label five edges withlabel 0 and six edgeswith label 1 out of eleven edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least seven edgesThatis |119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 3 119866 is not a prime

cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label six edges with label 0 and sevenedges with label 1 out of thirteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost four edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andeight edges with label 1 out of fifteen edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost six edges and 1 label for at least nine edges That is|119890119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graph

Case 2 (119899 = 6 to 10) For 119899 = 6 119891(1199061) = 4 119891(119906

2) = 8 and

119891(V0) = 6 119891(V

1) = 3 119891(V

2) = 9 119891(V

3) = 7 119891(V

4) = 5

119891(V5) = 1 and 119891(V

6) = 2 Then 119890

119891(0) = 8 119890

119891(1) = 9

For 119899 = 7 119891(1199061) = 4 119891(119906

2) = 8 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 9 119891(V

3) = 7 119891(V

4) = 5 119891(V

5) = 10 119891(V

6) = 1 and

119891(V7) = 2 Then 119890

119891(0) = 4 119890

119891(1) = 5

For 119899 = 8 119891(1199061) = 11 119891(119906

2) = 10 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 2 119891(V

3) = 8 119891(V

4) = 4 119891(V

5) = 5 119891(V

6) = 7

119891(V7) = 1 and 119891(V

8) = 9 Then 119890

119891(0) = 10 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 11 119891(119906

2) = 12 and 119891(V

0) = 2 119891(V

1) =

1 119891(V2) = 3 119891(V

3) = 6 119891(V

4) = 8 119891(V

5) = 4 119891(V

6) = 10

119891(V7) = 5119891(V

8) = 7 and119891(V

9) = 9Then 119890

119891(0) = 11 119890

119891(1) =

12For 119899 = 10 119891(119906

1) = 13 119891(119906

2) = 12 and 119891(V

0) = 2

119891(V1) = 9 119891(V

2) = 3 119891(V

3) = 6 119891(V

4) = 4 119891(V

5) = 8

119891(V6) = 10 119891(V

7) = 5 119891(V

8) = 1 119891(V

9) = 7 and 119891(V

10) = 11

Then 119890119891(0) = 12 119890

119891(1) = 13

Case 3 (119899 is even 119899 ge 12) Consider

119891 (V0) = 2

119891 (V1) = 5 119891 (V

2) = 10

119891 (V3) = 4 119891 (V

4) = 8

119891 (V4+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 5

119891 (V1198992

) = 6 119891 (V(1198992)+1

) = 3

119891 (V(1198992)+2

) = 9

119891 (V119899minus1

) = 1 119891 (V119899) = 7

119891 (V(1198992)+2+119894

) = 11 + 2 (119894 minus 1) 1 le 119894 le (1198992) minus 4

119891 (1199061) = 119899 + 3 119891 (119906

2) = 119899 + 2

(14)

6 ISRN Combinatorics

In view of the above defined labeling pattern for Case 3 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 4(mod 7) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod 7)

Case 4 (119899 is odd 119899 ge 11) Consider

119891 (V0) = 2

119891 (V1) = 10 119891 (V

2) = 4

119891 (V3) = 8

119891 (V3+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le ((119899 minus 1) 2) minus 4

119891 (V(119899minus1)2

) = 6 119891 (V(119899+1)2

) = 3

119891 (V(119899+3)2

) = 1

119891 (V119899+1minus119894

) = 5 + 2 (119894 minus 1) 1 le 119894 le (119899 minus 3) 2

119891 (1199061) = 119899 + 2 119891 (119906

2) = 119899 + 3

(15)

In view of the above defined labeling pattern for Case 4 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 3(mod5) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod7)

In view of Cases 2 to 4 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence 119866 is a prime cordial graph for 119899 ge 6

Illustration 4 Let 119866 be the graph obtained by duplicationof an arbitrary rim edge by an edge in wheel 119882

13 Then

|119881(119866)| = 16 and |119864(119866)| = 31 In accordance withTheorem 14 Case 4 will be applicable and the correspondingprime cordial labeling is shown in Figure 3 Here 119890

119891(0) = 16

119890119891(1) = 15

Theorem 15 The graph obtained by duplication of an arbi-trary spoke edge by an edge in wheel119882

119899is prime cordial graph

where 119899 = 7 and 119899 ge 9

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate thespoke edge 119890 = V

0V1by an edge 1198901015840 = 119906

11199062and call the resultant

graph 119866 Then |119881(119866)| = 119899 + 3 and |119864(119866)| = 3119899 + 2 To define119891 119881(119866) rarr 1 2 3 119899 + 3 we consider following threecasesCase 1 (119899 = 3 to 6 and 119899 = 8) For 119899 = 3 to satisfy the edgecondition for prime cordial labeling it is essential to label fiveedges with label 0 and six edges with label 1 out of elevenedges But all the possible assignments of vertex labels willgive rise to 0 labels for at most four edges and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for

119899 = 3 119866 is not a prime cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andseven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label eight edges with label 0 and

1

2

3

45

6

7 8

9

10

11

12

13

14

15

16

u1

u2

1

2

3

4

5

6

78

9

10

11

12

13

Figure 3

nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graphFor 119899 = 6 to satisfy the edge condition for prime cordial

labeling it is essential to label ten edges with label 0 and tenedges with label 1 out of twenty edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost eight edges and 1 label for at least twelve edges That is|119890119891(0) minus 119890

119891(1)| = 4 gt 1 Hence for 119899 = 6 119866 is not a prime

cordial graphFor 119899 = 8 to satisfy the edge condition for prime cordial

labeling it is essential to label thirteen edges with label 0 andthirteen edges with label 1 out of twenty-six edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most twelve edges and 1 label for at least fourteen edgesThat is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence for 119899 = 8 119866 is not a

prime cordial graphCase 2 (119899 = 7 119899 = 9 to 12 and 119899 = 14 16 18 20 22) For119899 = 7 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 9

119891(V2) = 3 119891(V

3) = 4 119891(V

4) = 8 119891(v

5) = 10 119891(V

6) = 5 and

119891(V7) = 7 Then 119890

119891(0) = 12 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 and 119891(V

9) = 9 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 119891(V

9) = 13 and 119891(V

10) = 9 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 11 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9119891(V

8) = 3119891(V

9) = 13119891(V

10) = 11 and119891(V

11) = 7

Then 119890119891(0) = 18 119890

119891(1) = 17

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 ISRN Combinatorics

In view of the above defined labeling pattern for Case 3 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 4(mod 7) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod 7)

Case 4 (119899 is odd 119899 ge 11) Consider

119891 (V0) = 2

119891 (V1) = 10 119891 (V

2) = 4

119891 (V3) = 8

119891 (V3+119894

) = 12 + 2 (119894 minus 1) 1 le 119894 le ((119899 minus 1) 2) minus 4

119891 (V(119899minus1)2

) = 6 119891 (V(119899+1)2

) = 3

119891 (V(119899+3)2

) = 1

119891 (V119899+1minus119894

) = 5 + 2 (119894 minus 1) 1 le 119894 le (119899 minus 3) 2

119891 (1199061) = 119899 + 2 119891 (119906

2) = 119899 + 3

(15)

In view of the above defined labeling pattern for Case 4 wehave 119890

119891(0) = 119899 + 3 and 119890

119891(1) = 119899 + 2 for 119899 equiv 3(mod5) and

119890119891(0) = 119899 + 2 and 119890

119891(1) = 119899 + 3 for 119899 equiv 4(mod7)

In view of Cases 2 to 4 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence 119866 is a prime cordial graph for 119899 ge 6

Illustration 4 Let 119866 be the graph obtained by duplicationof an arbitrary rim edge by an edge in wheel 119882

13 Then

|119881(119866)| = 16 and |119864(119866)| = 31 In accordance withTheorem 14 Case 4 will be applicable and the correspondingprime cordial labeling is shown in Figure 3 Here 119890

119891(0) = 16

119890119891(1) = 15

Theorem 15 The graph obtained by duplication of an arbi-trary spoke edge by an edge in wheel119882

119899is prime cordial graph

where 119899 = 7 and 119899 ge 9

Proof Let V0be the apex vertex of119882

119899and let V

1 V2 V

119899be

the rim vertices Without loss of generality we duplicate thespoke edge 119890 = V

0V1by an edge 1198901015840 = 119906

11199062and call the resultant

graph 119866 Then |119881(119866)| = 119899 + 3 and |119864(119866)| = 3119899 + 2 To define119891 119881(119866) rarr 1 2 3 119899 + 3 we consider following threecasesCase 1 (119899 = 3 to 6 and 119899 = 8) For 119899 = 3 to satisfy the edgecondition for prime cordial labeling it is essential to label fiveedges with label 0 and six edges with label 1 out of elevenedges But all the possible assignments of vertex labels willgive rise to 0 labels for at most four edges and 1 label for atleast seven edges That is |119890

119891(0) minus 119890

119891(1)| = 3 gt 1 Hence for

119899 = 3 119866 is not a prime cordial graphFor 119899 = 4 to satisfy the edge condition for prime cordial

labeling it is essential to label seven edges with label 0 andseven edges with label 1 out of fourteen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most four edges and 1 label for at least ten edges Thatis |119890119891(0) minus 119890

119891(1)| = 6 gt 1 Hence for 119899 = 4 119866 is not a prime

cordial graphFor 119899 = 5 to satisfy the edge condition for prime cordial

labeling it is essential to label eight edges with label 0 and

1

2

3

45

6

7 8

9

10

11

12

13

14

15

16

u1

u2

1

2

3

4

5

6

78

9

10

11

12

13

Figure 3

nine edges with label 1 out of seventeen edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most six edges and 1 label for at least eleven edges Thatis |119890119891(0) minus 119890

119891(1)| = 5 gt 1 Hence for 119899 = 5 119866 is not a prime

cordial graphFor 119899 = 6 to satisfy the edge condition for prime cordial

labeling it is essential to label ten edges with label 0 and tenedges with label 1 out of twenty edges But all the possibleassignments of vertex labels will give rise to 0 labels for atmost eight edges and 1 label for at least twelve edges That is|119890119891(0) minus 119890

119891(1)| = 4 gt 1 Hence for 119899 = 6 119866 is not a prime

cordial graphFor 119899 = 8 to satisfy the edge condition for prime cordial

labeling it is essential to label thirteen edges with label 0 andthirteen edges with label 1 out of twenty-six edges But all thepossible assignments of vertex labels will give rise to 0 labelsfor at most twelve edges and 1 label for at least fourteen edgesThat is |119890

119891(0) minus 119890

119891(1)| = 2 gt 1 Hence for 119899 = 8 119866 is not a

prime cordial graphCase 2 (119899 = 7 119899 = 9 to 12 and 119899 = 14 16 18 20 22) For119899 = 7 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 9

119891(V2) = 3 119891(V

3) = 4 119891(V

4) = 8 119891(v

5) = 10 119891(V

6) = 5 and

119891(V7) = 7 Then 119890

119891(0) = 12 119890

119891(1) = 11

For 119899 = 9 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) = 3

119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 and 119891(V

9) = 9 Then 119890

119891(0) = 15

119890119891(1) = 14For 119899 = 10 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

3 119891(V2) = 12 119891(V

3) = 4 119891(V

4) = 8 119891(V

5) = 10 119891(V

6) = 5

119891(V7) = 7 119891(V

8) = 11 119891(V

9) = 13 and 119891(V

10) = 9 Then

119890119891(0) = 16 119890

119891(1) = 16

For 119899 = 11 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9119891(V

8) = 3119891(V

9) = 13119891(V

10) = 11 and119891(V

11) = 7

Then 119890119891(0) = 18 119890

119891(1) = 17

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Combinatorics 7

For 119899 = 12 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 14 119891(V

6) = 12

119891(V7) = 9 119891(V

8) = 3 119891(V

9) = 13 119891(V

10) = 15 119891(V

11) = 11

and 119891(V12) = 7 Then 119890

119891(0) = 19 119890

119891(1) = 19

For 119899 = 14 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

5 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 3 119891(V

9) = 9 119891(V

10) = 15 119891(V

11) = 17

119891(V12) = 13 119891(V

13) = 11 and 119891(V

14) = 7 Then 119890

119891(0) = 22

119890119891(1) = 22For 119899 = 16 119891(119906

1) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

19 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 3 119891(V

10) = 9 119891(V

11) =

5 119891(V12) = 7 119891(V

13) = 11 119891(V

14) = 13 119891(V

15) = 15 and

119891(V16) = 17 Then 119890

119891(0) = 25 119890

119891(1) = 25

For 119899 = 18 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

21 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18119891(V

9) = 20 119891(V

10) = 3 119891(V

11) = 9

119891(V12) = 5 119891(V

13) = 7 119891(V

14) = 11 119891(V

15) = 13 119891(V

16) = 15

119891(V17) = 17 and 119891(V

18) = 19 Then 119890

119891(0) = 28 119890

119891(1) = 28

For 119899 = 20 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

23 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 3

119891(V12) = 9 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 11 119891(V

16) = 13

119891(V17) = 15 119891(V

18) = 17 119891(V

19) = 19 and 119891(V

20) = 21

Then 119890119891(0) = 31 119890

119891(1) = 31

For 119899 = 22 119891(1199061) = 2 119891(119906

2) = 1 and 119891(V

0) = 6 119891(V

1) =

25 119891(V2) = 4 119891(V

3) = 8 119891(V

4) = 10 119891(V

5) = 12 119891(V

6) = 14

119891(V7) = 16 119891(V

8) = 18 119891(V

9) = 20 119891(V

10) = 22 119891(V

11) = 24

119891(V12) = 3 119891(V

13) = 5 119891(V

14) = 7 119891(V

15) = 9 119891(V

16) = 11

119891(V17) = 13 119891(V

18) = 15 119891(V

19) = 17 119891(V

20) = 19 119891(V

21) =

21 and 119891(V22) = 23 Then 119890

119891(0) = 34 119890

119891(1) = 34

For the next case let 1199051

= lfloor(119899 + 3)2rfloor 119898 = lfloor(119899 + 3)3rfloor1199052= lceil1198982rceil 119896

1= lfloor(3119899 + 2)2rfloor 119896

2= 21199051+ 1199052minus 4

119905 = 1198961minus 1198962 1199053=

119905 minus 2 119899 = 13 15 17

119905 minus 1 119899 = 19 21 119899 ge 23(16)

Case 3 (1199051minus119905 ge 3(119899 = 13 15 17 19 21 and 119899 ge 23)) Consider

119891 (1199061) = 2 119891 (119906

2) = 5

119891 (V0) = 6

119891 (V1) = 3 119891 (V

2) = 4

119891 (V119899) = 1

119891 (V2+119894

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119905

119891 (V119905+3

) = 9 if 119905 equiv 4 (mod 7)7 otherwise

119891 (V119905+4

) = 7 if 119905 equiv 4 (mod 7)9 otherwise

119891 (V119905+4+119894

) = 9 + 2119894 1 le 119894 le 1199053

(17)

For 2119905 + 3 lt 119899 minus 1

119891 (V2119905+3+119894

) = 119891 (V2119905+3

) + 119894 1 le 119894 le (119899 minus 1) minus (2119905 + 3)

(18)

In view of the above defined labeling pattern for Case 3we have 119890

119891(0) = lfloor(3119899 + 2)2rfloor and 119890

119891(1) = lceil(3119899 + 2)2rceil

Thus for Cases 2 and 3 we have |119890119891(0) minus 119890

119891(1)| le 1

Hence119866 is a prime cordial graph for 119899 = 7 and 119899 ge 9

Illustration 5 Let 1198661be the graph obtained by duplication of

arbitrary spoke edge by an edge of wheel11988223 Then |119881(119866)| =

26 and |119864(119866)| = 71 In accordance with Theorem 15 we have1199051

= 13 119898 = 8 1199052

= 4 1198961

= 35 1198962

= 26 and 119905 = 9Here 119905

1minus 119905 = 4 gt 3 so labeling pattern described in Case 3

will be applicable The corresponding prime cordial labelingis shown in Figure 4 It is easy to visualise that 119890

119891(0) = 35

119890119891(1) = 36

Theorem 16 DS(119875119899) is a prime cordial graph

Proof Consider 119875119899with 119881(119875

119899) = V

119894 1 le 119894 le 119899 Here

119881(119875119899) = 119881

1cup 1198812 where 119881

1= V119894 2 le 119894 le 119899 minus 1 and

1198812= V1 V119899 Now in order to obtain DS(119875

119899) from 119866 we add

1199081 1199082corresponding to 119881

1 1198812 Then |119881(DS(119875

119899))| = 119899 + 2

and 119864(DS(119875119899)) = V

11199082 V21199082 cup 119908

1V119894 2 le 119894 le 119899 minus 1 cup

119864(119875119899) so |119864(DS(119875

119899)| = 2119899 minus 1 We define vertex labeling

119891 119881(DS(119875119899)) rarr 1 2 119899 + 2 as follows

Let 1199011be the highest prime number lt 119899+2 and 119896 = lfloor(119899+

2)2rfloor Consider

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (V1) = 1

119891 (V119899) = 9 119891 (V

119899minus1) = 1199011

(19)

For 0 le 119894 lt 119896 minus 1

119891 (V2+119894

) = (119899 + 2) minus 2119894 119899 119894119904 even(119899 + 1) minus 2119894 119899 119894119904 odd

(20)

And for vertices V119896+2

V119896+3

V119899minus2

we assign distinct oddnumbers (lt 119899 + 2) in ascending order starting from 5

In view of the above defined labeling pattern if 119899 is evennumber then 119890

119891(0) = 119899 119890

119891(1) = 119899minus1 otherwise 119890

119891(0) = 119899minus1

119890119891(1) = 119899Thus |119890

119891(0) minus 119890

119891(1)| le 1

Hence DS(119875119899) is a prime cordial graph

Illustration 6 Prime cordial labeling of the graph DS(1198757) is

shown in Figure 5

Theorem 17 DS(119861119899119899

) is a prime cordial graph

Proof Consider 119861119899119899

with 119881(119861119899119899

) = 119906 V 119906119894 V119894 1 le 119894 le 119899

where 119906119894 V119894are pendant vertices Here 119881(119861

119899119899) = 119881

1cup 1198812

where 1198811= 119906119894 V119894 1 le 119894 le 119899 and 119881

2= 119906 V Now in order

to obtain DS(119861119899119899

) from 119866 we add 1199081 1199082corresponding

to 1198811 1198812 Then |119881(DS(119861

119899119899)| = 2119899 + 4 and 119864(DS(119861

119899119899)) =

119906V 1199061199082 V1199082 cup 119906119906

119894 VV119894 1199081119906119894 1199081V119894

1 le 119894 le 119899

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 ISRN Combinatorics

1 3

2

4

8

10

12

14

16

18

20

22

24

5

79

11

13

15

17

19

21

6

23

25

26

u1

u2

1

2

3

4

5

6

7

8

9

10

1112

13

14

15

16

17

18

19

20

21

22

23

0

Figure 4

1 2 3 4 5 6 7

w1

w2

1

3

4

2

568 7 9

Figure 5

so |119864(DS(119861119899119899

)| = 4119899 + 3 We define vertex labeling 119891

119881(119863119878(119861119899119899

)) rarr 1 2 2119899 + 4 as follows

119891 (119906) = 6 119891 (V) = 1

119891 (1199081) = 2 119891 (119908

2) = 3

119891 (1199061) = 4

119891 (119906119894+1

) = 8 + 2 (119894 minus 1) 1 le 119894 le 119899 minus 1

119891 (V119894) = 5 + 2 (119894 minus 1) 1 le 119894 le 119899

(21)

w1

w2

3

2

12

1

1011

134 14 5

6

8 79

u1

u

u2u3

u4

u5 12

3

4

5

Figure 6

In view of the above defined labeling pattern we have 119890119891(0) =

2119899 + 1 119890119891(1) = 2119899 + 2

Thus |119890119891(0) minus 119890

119891(1)| le 1

Hence DS(119861119899119899

) is a prime cordial graph

Illustration 7 Prime cordial labeling of the graph DS(11986155

) isshown in Figure 6

3 Conclusion

A new approach for constructing larger prime cordial graphfrom the existing prime cordial graph is investigated

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Combinatorics 9

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authorsrsquo thanks are due to the anonymous refereesfor careful reading and constructive suggestions for theimprovement in the first draft of this paper

References

[1] R Balakrishnan and K Ranganathan A Textbook of GraphTheory Springer New York NY USA 2nd edition 2012

[2] D M Burton Elementary Number Theory Brown Publishers2nd edition 1999

[3] V Yegnanarayanan and P Vaidhyanathan ldquoSome interestingapplications of graph labellingsrdquo Journal of Mathematical andComputational Science vol 2 no 5 pp 1522ndash1531 2012

[4] J A Gallian ldquoA dynamic survey of graph labelingrdquo TheElectronic Journal of Combinatorics vol 16 article DS6 2013

[5] I Cahit ldquoCordial graphs a weaker version of graceful andharmonious graphsrdquo Ars Combinatoria vol 23 pp 201ndash2071987

[6] A Tout A N Dabboucy and K Howalla ldquoPrime labeling ofgraphsrdquo National Academy Science Letters vol 11 pp 365ndash3681982

[7] S K Vaidya and U M Prajapati ldquoSome new results on primegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 3 pp99ndash104 2012

[8] S K Vaidya and U M Prajapati ldquoSome switching invariantprime graphsrdquoOpen Journal of Discrete Mathematics vol 2 no1 pp 17ndash20 2012

[9] S K Vaidya and U M Prajapati ldquoPrime labeling in the contextof duplication of graph elementsrdquo International Journal ofMathematics and Soft Computing vol 3 no 1 pp 13ndash20 2013

[10] M Sundaram R Ponraj and S Somasundram ldquoPrime cordiallabelling of graphsrdquo Journal of the IndianAcademy ofMathemat-ics vol 27 no 2 pp 373ndash390 2005

[11] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somecycle related graphsrdquo International Journal of Open Problems inComputer Science and Mathematics vol 3 no 5 pp 223ndash2322010

[12] S K Vaidya and P L Vihol ldquoPrime cordial labeling for somegraphsrdquoModern Applied Science vol 4 no 8 pp 119ndash126 2010

[13] S K Vaidya and N H Shah ldquoSome new families of primecordial graphsrdquo Journal of Mathematics Research vol 3 no 4pp 21ndash30 2011

[14] S K Vaidya and N H Shah ldquoPrime cordial labeling of somegraphsrdquo Open Journal of Discrete Mathematics vol 2 no 1 pp11ndash16 2012

[15] S K Vaidya and N H Shah ldquoPrime cordial labeling of somewheel related graphsrdquoMalaya Journal of Matematik vol 4 no1 pp 148ndash156 2013

[16] S Babitha and J Baskar Babujee ldquoPrime cordial labeling anddualityrdquoAdvances andApplications inDiscreteMathematics vol11 no 1 pp 79ndash102 2013

[17] S Babitha and J Baskar Babujee ldquoPrime cordial labeling ongraphsrdquo International Journal of Mathematical Sciences vol 7no 1 pp 43ndash48 2013

[18] P Selvaraju P Balaganesan J Renuka and V Balaj ldquoDegreesplitting graph on graceful felicitous and elegant labelingrdquoInternational Journal of Mathematical Combinatorics vol 2 pp96ndash102 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of