# Some New Results on Prime Cordial Labeling

Authors: S. K. Vaidya; N. H. Shah
Journal: ISRN Combinatorics
 (2014)

---

## Abstract

A prime cordial labeling of a graphG with the vertex set V
(
G
) is a bijection f
:
V
(
G
)
→
{
1,2
,
3
,
…
,
|
V
(
G
)
|
} such that each edge u
v is assigned the label 1 if g
c
d
(
f
(
u
)
,
f
(
v
)
)
=
1 and 0 if g
c
d
(
f
(
u
)
,
f
(
v
)
)
>
1; then the number of edges labeled with 0 and the number of edges labeled with 1 differ by at most 1. A graph which admits a prime cordial labeling is called a prime cordial graph. In this work we give a method to construct larger prime cordial graph using a given prime cordial graph G. In addition to this we have investigated the prime cordial labeling for double fan and degree splitting graphs of path as well as bistar. Moreover we prove that the graph obtained by duplication of an edge (spoke as well as rim) in wheel W
n admits prime cordial labeling.


---

## Body

## 1. Introduction

We consider a finite, connected, undirected, and simple graphG
=
(
V
(
G
)
,
E
(
G
)
) with p vertices and q edges which is also denoted as G
(
p
,
q
). For standard terminology and notations related to graph theory we follow Balakrishnan and Ranganathan [1] while for any concept related to number theory we refer to Burton [2]. In this section we provide brief summary of definitions and other required information for our investigations.Definition 1.
TheGraph labeling is an assignment of numbers to 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 avertex labeling (edge labeling).
Many labeling schemes have been introduced so far and they are explored as well by many researchers. Graph labelings have enormous applications within mathematics as well as to several areas of computer science and communication networks. Various applications of graph labeling are reported in the work of Yegnanaryanan and Vaidhyanathan [3]. For a dynamic survey on various graph labeling problems along with an extensive bibliography we refer to Gallian [4].Definition 2.
A labelingf
:
V
(
G
)
→
{
0,1
} is calledbinary vertex labeling of G and f
(
v
) is called the label of the vertex v of G under f.Notation 1.
If for an edgee
=
u
v, the induced edge labeling f
*
:
E
(
G
)
→
{
0,1
} is given by f
*
(
e
)
=
|
f
(
u
)
-
f
(
v
)
|. Then
(1)
v
f
(
i
)
=
number
of
vertices
of
G
having
label
i
under
f
e
f
(
i
)
=
number
of
edges
of
G
having
label
i
under
f
*
,
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
where
i
=
0
or
1
.Definition 3.
A binary vertex labelingf of a graph G is called acordial labeling if |
v
f
(
0
)
-
v
f
(
1
)
|
≤
1 and |
e
f
(
0
)
-
e
f
(
1
)
|
≤
1. A graph G iscordial if it admits cordial labeling.
The concept of cordial labeling was introduced by Cahit [5].The notion of prime labeling was originated by Entringer and was introduced by Tout et al. [6].Definition 4.
Aprime labeling of a graph G is an injective function f
:
V
(
G
)
→
{
1,2
,
…
,
|
V
(
G
)
|
} such that for, every pair of adjacent vertices u and v, gcd
(
f
(
u
)
,
f
(
v
)
)
=
1. The graph which admits a prime labeling is called aprime graph.
The concept of prime labeling has attracted many researchers as the study of prime numbers is of great importance because prime numbers are scattered and there are arbitrarily large gaps in the sequence of prime numbers. Vaidya and Prajapati [7, 8] have investigated many results on prime labeling. Same authors [9] have discussed prime labeling in the context of duplication of graph elements. Motivated through the concepts of prime labeling and cordial labeling, a new concept termed as a prime cordial labeling was introduced by Sundaram et al. [10] which contains blend of both the labelings.Definition 5.
Aprime cordial labeling of a graph G with vertex set V
(
G
) is a bijection f
:
V
(
G
)
→
{
1,2
,
3
,
…
,
|
V
(
G
)
|
} and if the induced function f
*
:
E
(
G
)
→
{
0,1
} is defined by
(2)
f
*
(
e
=
u
v
)
=
1
,
if
gcd
(
f
(
u
)
,
f
(
v
)
)
=
1
,
=
0
,
otherwise
,
then the number of edges labeled with 0 and the number of edges labeled with 1 differ by at most 1. A graph which admits prime cordial labeling is called aprime cordial graph.Many graphs are proved to be prime cordial in the work of Sundaram et al. [10]. Prime cordial labeling for some cycle related graphs has been discussed by Vaidya and Vihol [11]. Prime cordial labeling in the context of some graph operations has been discussed by Vaidya and Vihol [12] and Vaidya and Shah [13, 14]. Vaidya and Shah [14] have proved that the wheel graph W
n admits prime cordial labeling for n
≥
8 while same authors in [15] have discussed prime cordial labeling for some wheel related graphs. Babitha and Baskar Babujee [16] have exhibited prime cordial labeling for some cycle related graphs and discussed the duality of prime cordial labeling. The same authors in [17] have derived some characterizations of prime cordial graphs and investigated various methods to construct larger prime cordial graphs using existing prime cordial graphs. We investigate a method different from existing one to construct larger prime cordial graph from an existing prime cordial graph.Definition 6.
Thewheel
W
n is defined to be the join K
1
+
C
n. The vertex corresponding to K
1 is known as apex and vertices corresponding to cycle are known as rim vertices while the edges corresponding to cycle are known as rim edges.Definition 7.
Thebistar
B
n
,
n is a graph obtained by joining the center (apex) vertices of two copies of K
1
,
n by an edge.Definition 8.
Thefan
F
n is the graph obtained by taking n
-
2 concurrent chords in cycle C
n
+
1. The vertex at which all the chords are concurrent is called the apex vertex. In other words, F
n
=
P
n
+
K
1.Definition 9.
Thedouble fan
D
F
n consists of two fan graphs that have a common path. In other words, D
F
n
=
P
n
+
K
2
_
_
_
_.Definition 10.
Theduplication of an edge
e
=
u
v of graph G produces a new graph G
′ by adding an edge e
′
=
u
′
v
′ such that N
(
u
′
)
=
N
(
u
)
∪
{
v
′
}
-
{
v
} and N
(
v
′
)
=
N
(
v
)
∪
{
u
′
}
-
{
u
}.Definition 11 (see [18]).
LetG
=
(
V
(
G
)
,
E
(
G
)
) be a graph with V
=
S
1
∪
S
2
∪
S
3
∪
⋯
S
t
∪
T, where each S
i is a set of vertices having at least two vertices of the same degree and T
=
V
∖
∪
i
=
1
t
S
i. Thedegree splitting graph of G denoted by D
S
(
G
) is obtained from G by adding vertices w
1
,
w
2
,
w
3
,
…
,
w
t and joining to each vertex of S
i for 1
≤
i
≤
t.


## 2. Main Results

Theorem 12.
LetG
(
p
,
q
) with p
≥
4 be a prime cordial graph and let K
2
,
n be a bipartite graph with bipartition V
=
V
1
∪
V
2 with V
1
=
{
v
1
,
v
2
} and V
2
=
{
u
1
,
u
2
,
…
,
u
n
}. If G
1 is the graph obtained by identifying the vertices v
1 and v
2 of K
2
,
n with the vertices of G having labels 2 and 4, respectively, then G
1 admits prime cordial labeling in any of the following cases:(i)
nis even and G is of any size q;
(ii)
n
,
p
,
and
q are odd with e
f
(
0
)
=
⌊
q
/
2
⌋;
(iii)
n is odd, p is even, and q is odd with e
f
(
0
)
=
⌈
q
/
2
⌉.Proof.
LetG
(
p
,
q
) be a prime cordial graph and let f
1 be the prime cordial labeling of G. Let w
1
,
w
2
∈
V be the vertices of G such that f
1
(
w
1
)
=
2 and f
1
(
w
2
)
=
4. Consider the K
2
,
n with bipartition V
=
V
1
∪
V
2 with V
1
=
{
v
1
,
v
2
} and V
2
=
{
u
1
,
u
2
,
…
,
u
n
}. Now identify the vertices v
1 to w
1 and v
2 to w
2 and denote the resultant graph as G
1. Then V
(
G
1
)
=
V
(
G
)
∪
{
u
1
,
u
2
,
…
,
u
n
} and E
(
G
1
)
=
E
(
G
)
∪
{
w
1
u
i
,
w
2
u
i
/
1
≤
i
≤
n
} so |
V
(
G
1
)
|
=
p
+
n and |
E
(
G
1
)
|
=
q
+
2
n. To define f
:
V
(
G
1
)
→
{
1,2
,
3
,
…
,
p
+
n
}, we consider the following three cases.
Case (i) (n is even and G is of any size q). Since G is a prime cordial graph, we assign vertex labels such that f
(
w
i
)
=
f
1
(
w
i
), where w
i
∈
V
(
G
)
=
V
(
G
1
)
∩
V
(
G
) and i
≤
i
≤
p:
(3)
f
(
u
i
)
=
p
+
i
,
i
≤
i
≤
n
.
Since n is even and f
1
(
w
1
)
=
2 and f
1
(
w
2
)
=
4, w
1 and w
2 are adjacent to each u
i, i
≤
i
≤
n. And this vertex assignment generates n edges with label 1 and n edges with label 0. Following Table 1 gives edge condition for prime cordial labeling for G
1 under f.
From Table1, we have |
e
f
(
0
)
-
e
f
(
1
)
|
≤
1.
Case (ii) (n
,
p
,
and
q are odd with e
f
(
0
)
=
⌊
q
/
2
⌋). Here p and q both are odd and G is a prime cordial graph with e
f
(
0
)
=
⌊
q
/
2
⌋.
SinceG is a prime cordial graph, we keep the vertex label of all the vertices of G in G
1 as it is. Therefore f
(
w
i
)
=
f
1
(
w
i
), where w
i
∈
V
(
G
) and i
≤
i
≤
p:
(4)
f
(
u
i
)
=
p
+
i
,
i
≤
i
≤
n
.
Since n is odd and f
1
(
w
1
)
=
2 and f
1
(
w
2
)
=
4, w
1 and w
2 are adjacent to each u
i, i
≤
i
≤
n. And this vertex assignment generates n
+
1 edges with label 0 and n
-
1 edges with label 1.
Therefore edge conditions forG
1 under f are e
f
(
0
)
=
⌊
q
/
2
⌋
+
n
+
1 and e
f
(
1
)
=
⌈
q
/
2
⌉
+
n
-
1. Therefore, e
f
(
0
)
-
1
=
e
f
(
1
). Hence, |
e
f
(
0
)
-
e
f
(
1
)
|
≤
1 for graph G
1.
Case (iii) (n is odd, p is even, and q is odd with e
f
(
0
)
=
⌈
q
/
2
⌉
). Here p is even, q is odd, and G is a prime cordial graph with e
f
(
0
)
=
⌈
q
/
2
⌉.
SinceG is a prime cordial graph, we keep the vertex label of all the vertices of G in G
1 as it is. Therefore f
(
w
i
)
=
f
1
(
w
i
), where w
i
∈
V
(
G
) and i
≤
i
≤
p:
(5)
f
(
u
i
)
=
p
+
i
,
i
≤
i
≤
n
.
Since n is odd and f
1
(
w
1
)
=
2 and f
1
(
w
2
)
=
4, w
1 and w
2 are adjacent to each u
i, i
≤
i
≤
n. And this vertex assignment generates n
-
1 edges with label 0 and n
+
1 edges with label 1.
Therefore edge conditions forG
1 under f are e
f
(
0
)
=
⌈
q
/
2
⌉
+
n
-
1 and e
f
(
1
)
=
⌊
q
/
2
⌋
+
n
+
1. Therefore, e
f
(
0
)
=
e
f
(
1
)
-
1. Hence, |
e
f
(
0
)
-
e
f
(
1
)
|
≤
1 for graph G
1.
Hence, in all the cases discussed above,G
1 admits prime cordial labeling.Table 1
q
Edge conditions forG
Edge conditions forG
1
Even
e
f
(
0
)
=
e
f
(
1
)
=
q
2
e
f
(
0
)
=
e
f
(
1
)
=
q
2
Odd
e
f
(
0
)
-
1
=
e
f
(
1
)
=
⌊
q
2
⌋
e
f
(
0
)
-
1
=
e
f
(
1
)
=
⌊
q
2
⌋
+
n
e
f
(
0
)
=
e
f
(
1
)
-
1
=
⌊
q
2
⌋
e
f
(
0
)
=
e
f
(
1
)
-
1
=
⌊
q
2
⌋
+
nIllustration  1. Consider the graph G as shown in Figure 1, with p
=
7 and q
=
9. G is a prime cordial graph with e
f
(
0
)
=
4
,
e
f
(
1
)
=
5. Take n
=
3 and construct graph G
1. In accordance with Case (ii) of Theorem 12, a prime cordial labeling of G
1 is as shown in Figure 1. Here e
f
(
0
)
=
8
,
e
f
(
1
)
=
7.(a)
(b)Theorem 13.
Double fanD
F
n is a prime cordial graph for n
=
8 and n
≥
10.Proof.
LetD
F
n be the double fan with apex vertices u
1
,
u
2 and v
1
,
v
2
,
…
,
v
n are vertices common path. Then |
V
(
D
F
n
)
|
=
n
+
2 and |
E
(
D
F
n
)
|
=
3
n
-
1. To define f
:
V
(
G
)
→
{
1,2
,
3
,
…
,
n
+
2
}, we consider the following five cases.
Case
1  (n
=
3 to 7 and n
=
9). In order to satisfy the edge condition for prime cordial labeling in D
F
3 it is essential to label four edges with label 0 and four edges with label 1 out of eight edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most one edge and 1 label for at least seven edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
6
>
1. Hence, D
F
3 is not a prime cordial graph.
In order to satisfy the edge condition for prime cordial labeling inD
F
4 it is essential to label five edges with label 0 and six edges with label 1 out of eleven edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most three edges and 1 label for at least eight edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
5
>
1. Hence, D
F
4 is not a prime cordial graph.
In order to satisfy the edge condition for prime cordial labeling inD
F
5 it is essential to label seven edges with label 0 and seven edges with label 1 out of fourteen edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most four edges and 1 label for at least ten edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
6
>
1. Hence, D
F
5 is not a prime cordial graph.
In order to satisfy the edge condition for prime cordial labeling inD
F
6 it is essential to label eight edges with label 0 and nine edges with label 1 out of seventeen edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most six edges and 1 label for at least eleven edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
5
>
1. Hence, D
F
6 is not a prime cordial graph.
In order to satisfy the edge condition for prime cordial labeling inD
F
7 it is essential to label ten edges with label 0 and ten edges with label 1 out of twenty edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most eight edges and 1 label for at least twelve edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
4
>
1. Hence, D
F
7 is not a prime cordial graph.
In order to satisfy the edge condition for prime cordial labeling inD
F
9 it is essential to label thirteen edges with label 0 and thirteen edges with label 1 out of twenty-six edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most twelve edges and 1 label for at least fourteen edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
2
>
1. Hence, D
F
9 is not a prime cordial graph.
Case
2  (n
=
8,10,11,12). For n
=
8, f
(
u
1
)
=
2, f
(
u
2
)
=
6 and f
(
v
1
)
=
4, f
(
v
2
)
=
8, f
(
v
3
)
=
3, f
(
v
4
)
=
9, f
(
v
5
)
=
10, f
(
v
6
)
=
5, f
(
v
7
)
=
7, and f
(
v
8
)
=
1. Then e
f
(
0
)
=
11, e
f
(
1
)
=
12.
Forn
=
10, f
(
u
1
)
=
2, f
(
u
2
)
=
6 and f
(
v
1
)
=
3, f
(
v
2
)
=
9, f
(
v
3
)
=
12, f
(
v
4
)
=
8
,
f
(
v
5
)
=
4, f
(
v
6
)
=
10, f
(
v
7
)
=
1, f
(
v
8
)
=
5, f
(
v
9
)
=
7, and f
(
v
10
)
=
11. Then e
f
(
0
)
=
15, e
f
(
1
)
=
14.
Forn
=
11, f
(
u
1
)
=
2, f
(
u
2
)
=
6 and f
(
v
1
)
=
3, f
(
v
2
)
=
9, f
(
v
3
)
=
12, f
(
v
4
)
=
8, f
(
v
5
)
=
4, f
(
v
6
)
=
10, f
(
v
7
)
=
5, f
(
v
8
)
=
7, f
(
v
9
)
=
11, f
(
v
10
)
=
13, and f
(
v
11
)
=
1. Then e
f
(
0
)
=
16, e
f
(
1
)
=
16.
Forn
=
12, f
(
u
1
)
=
2, f
(
u
2
)
=
6 and f
(
v
1
)
=
3, f
(
v
2
)
=
9, f
(
v
3
)
=
12, f
(
v
4
)
=
8, f
(
v
5
)
=
4, f
(
v
6
)
=
14, f
(
v
7
)
=
10, f
(
v
8
)
=
5, f
(
v
9
)
=
7, f
(
v
10
)
=
11, f
(
v
11
)
=
13, and f
(
v
12
)
=
1. Then e
f
(
0
)
=
18, e
f
(
1
)
=
17.
Now for the remaining three cases let(6)
k
=
⌊
n
+
2
2
⌋
,
m
=
⌊
n
+
2
3
⌋
,
t
1
=
⌈
3
n
-
1
2
⌉
,
t
2
=
3
k
-
7
+
⌈
m
2
⌉
,
t
3
= largest even number ≤
n
+
2, and t
4
= largest odd number ≤
n
+
2.
Case
3
(
t
1
=
t
2
). Consider
(7)
f
(
u
1
)
=
2
,
f
(
u
2
)
=
6
.
For the vertices v
1
,
v
2
,
v
3
,
…
,
v
n we assign the vertex labels in the following order: 1, t
3, t
3
-
2, t
3
-
4, …, 14, 12, 10, 8, 4, 3, 5, 7, 9, …, t
4
-
2, t
4.
Case
4
(
t
1
>
t
2
). Consider
(8)
f
(
u
1
)
=
2
,
f
(
u
2
)
=
6
.
Lett
5
=
t
1
-
t
2. Consider
(9)
f
(
v
1
)
=
3
,
f
(
v
2
)
=
9
,
f
(
v
3
)
=
15
,
f
(
v
4
)
=
21
,
⋮
f
(
v
t
5
)
=
3
(
2
t
5
-
1
)
,
f
(
v
t
5
+
1
)
=
f
(
v
t
5
)
+
6
.
Now for remaining vertices v
t
5
+
2
,
v
t
5
+
3
,
…
,
v
n assign the labels 1
,
t
3
,
t
3
-
2
,
t
3
-
4
,
…
,
14,12,10,8
,
4,5
,
7
,
…, all the odd numbers in ascending order.
Case  5
(
t
2
>
t
1
). Let t
6
=
t
2
-
t
1.
Sub-Case
1.
n is even. Consider
(10)
f
(
u
1
)
=
2
,
f
(
u
2
)
=
6
.
For the vertices v
1
,
v
2
,
v
3
,
…
,
v
n we assign the vertex labels in the following order: n
+
2, n
+
1, n, n
-
1, n
-
2, n
-
3
,
n
+
2
-
2
t
6, n
+
2
-
2
(
t
6
+
1
), n
+
2
-
2
(
t
6
+
2
)
,
…
,
10,8
,
4,3
,
5,7
,
… remaining odd numbers in ascending order.
Sub-Case
2.
n is odd. Consider
(11)
f
(
u
1
)
=
2
,
f
(
u
2
)
=
6
.
For the vertices v
1
,
v
2
,
v
3
,
…
,
v
n we assign the vertex labels in the following order: n
+
2, n
+
1, n, n
-
1, n
-
2, n
-
3, n
+
2
-
2
t
6, n
+
2
-
2
(
t
6
+
1
), n
+
2
-
2
(
t
6
+
2
)
,
…
,
10,8
,
4,3
,
5,7
,
…remaining odd numbers in ascending order.
In view of the above defined labeling pattern for Cases 3, 4, and 5, we have(12)
e
f
(
0
)
=
⌈
3
n
-
1
2
⌉
,
e
f
(
1
)
=
⌊
3
n
-
1
2
⌋
.
Thus, we have |
e
f
(
0
)
-
e
f
(
1
)
|
≤
1.
Hence,D
F
n is a prime cordial graph for n
=
8 and n
≥
10.Illustration  2. For the graph D
F
15, |
V
(
D
F
15
)
|
=
17 and |
E
(
D
F
15
)
|
=
44. In accordance with Theorem 13 we have k
=
5, m
=
5, t
1
=
22, and t
2
=
20. Here t
1
>
t
2 so labeling pattern described in Case 4 will be applicable and t
5
=
2. The corresponding prime cordial labeling is shown in Figure 2. Here e
f
(
0
)
=
22
=
e
f
(
1
).Figure 2Illustration  3. For the graph D
F
37, |
V
(
D
F
37
)
|
=
39 and |
E
(
D
F
37
)
|
=
110. In accordance with Theorem 13, we have k
=
19, m
=
13, t
1
=
55, and t
2
=
57. Here t
2
>
t
1 and n
=
37 so labeling pattern described in Sub-Case 2 of Case 5 will be applicable and t
6
=
2. And corresponding labeling pattern is as below:
(13)
f
(
u
1
)
=
2
,
f
(
u
2
)
=
6
.
For the vertices v
1
,
v
2
,
…
,
v
37 we assign the vertex labels 39, 38, 37, 36, 35, 34, 32, 30, 28, 26, 24, 22, 20, 18, 16, 14, 12, 10, 8, 4, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, and 33, respectively, where e
f
(
0
)
=
55
=
e
f
(
1
). Therefore D
F
37 is a prime cordial graph.Theorem 14.
The graph obtained by duplication of an arbitrary rim edge by an edge inW
n is a prime cordial graph, where n
≥
6.Proof.
Letv
0 be the apex vertex of W
n and let v
1
,
v
2
,
…
,
v
n be the rim vertices. Without loss of generality we duplicate the rim edge e
=
v
1
v
2 by an edge e
′
=
u
1
u
2 and call the resultant graph as G. Then |
V
(
G
)
|
=
n
+
3 and |
E
(
G
)
|
=
2
n
+
5. To define f
:
V
(
G
)
→
{
1,2
,
3
,
…
,
n
+
3
}, we consider the following four cases.
Case
1
(
n
=
3,4
,
5
). For n
=
3, to satisfy the edge condition for prime cordial labeling, it is essential to label five edges with label 0 and six edges with label 1 out of eleven edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most four edges and 1 label for at least seven edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
3
>
1. Hence, for n
=
3, G is not a prime cordial graph.
Forn
=
4, to satisfy the edge condition for prime cordial labeling, it is essential to label six edges with label 0 and seven edges with label 1 out of thirteen edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most four edges and 1 label for at least nine edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
5
>
1. Hence, for n
=
4, G is not a prime cordial graph.
Forn
=
5, to satisfy the edge condition for prime cordial labeling it is essential to label seven edges with label 0 and eight edges with label 1 out of fifteen edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most six edges and 1 label for at least nine edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
3
>
1. Hence, for n
=
5, G is not a prime cordial graph.
Case
2
(
n
=
6 to 10). For n
=
6, f
(
u
1
)
=
4, f
(
u
2
)
=
8 and f
(
v
0
)
=
6, f
(
v
1
)
=
3, f
(
v
2
)
=
9, f
(
v
3
)
=
7, f
(
v
4
)
=
5, f
(
v
5
)
=
1, and f
(
v
6
)
=
2. Then e
f
(
0
)
=
8, e
f
(
1
)
=
9.
Forn
=
7, f
(
u
1
)
=
4, f
(
u
2
)
=
8 and f
(
v
0
)
=
6, f
(
v
1
)
=
3, f
(
v
2
)
=
9, f
(
v
3
)
=
7, f
(
v
4
)
=
5, f
(
v
5
)
=
10, f
(
v
6
)
=
1, and f
(
v
7
)
=
2. Then e
f
(
0
)
=
4, e
f
(
1
)
=
5.
Forn
=
8, f
(
u
1
)
=
11, f
(
u
2
)
=
10 and f
(
v
0
)
=
6, f
(
v
1
)
=
3, f
(
v
2
)
=
2, f
(
v
3
)
=
8, f
(
v
4
)
=
4, f
(
v
5
)
=
5, f
(
v
6
)
=
7, f
(
v
7
)
=
1, and f
(
v
8
)
=
9. Then e
f
(
0
)
=
10, e
f
(
1
)
=
11.
Forn
=
9, f
(
u
1
)
=
11, f
(
u
2
)
=
12 and f
(
v
0
)
=
2, f
(
v
1
)
=
1, f
(
v
2
)
=
3, f
(
v
3
)
=
6, f
(
v
4
)
=
8, f
(
v
5
)
=
4, f
(
v
6
)
=
10, f
(
v
7
)
=
5, f
(
v
8
)
=
7, and f
(
v
9
)
=
9. Then e
f
(
0
)
=
11, e
f
(
1
)
=
12.
Forn
=
10, f
(
u
1
)
=
13, f
(
u
2
)
=
12 and f
(
v
0
)
=
2, f
(
v
1
)
=
9, f
(
v
2
)
=
3, f
(
v
3
)
=
6, f
(
v
4
)
=
4, f
(
v
5
)
=
8, f
(
v
6
)
=
10, f
(
v
7
)
=
5, f
(
v
8
)
=
1, f
(
v
9
)
=
7, and f
(
v
10
)
=
11. Then e
f
(
0
)
=
12, e
f
(
1
)
=
13.
Case
3 (n is even, n
≥
12
). Consider
(14)
f
(
v
0
)
=
2
,
f
(
v
1
)
=
5
,
f
(
v
2
)
=
10
,
f
(
v
3
)
=
4
,
f
(
v
4
)
=
8
,
f
(
v
4
+
i
)
=
12
+
2
(
i
-
1
)
,
1
≤
i
≤
(
n
/
2
)
-
5
f
(
v
n
/
2
)
=
6
,
f
(
v
(
n
/
2
)
+
1
)
=
3
,
f
(
v
(
n
/
2
)
+
2
)
=
9
,
f
(
v
n
-
1
)
=
1
,
f
(
v
n
)
=
7
,
f
(
v
(
n
/
2
)
+
2
+
i
)
=
11
+
2
(
i
-
1
)
,
1
≤
i
≤
(
n
/
2
)
-
4
f
(
u
1
)
=
n
+
3
,
f
(
u
2
)
=
n
+
2
.
In view of the above defined labeling pattern for Case 3, we have e
f
(
0
)
=
n
+
3 and e
f
(
1
)
=
n
+
2 for n
≡
4
(
mod
⁡
7
) and e
f
(
0
)
=
n
+
2 and e
f
(
1
)
=
n
+
3 for n
≢
4
(
mod
⁡
7
).
Case
4 (n is odd, n
≥
11). Consider
(15)
f
(
v
0
)
=
2
,
f
(
v
1
)
=
10
,
f
(
v
2
)
=
4
,
f
(
v
3
)
=
8
,
f
(
v
3
+
i
)
=
12
+
2
(
i
-
1
)
,
1
≤
i
≤
(
(
n
-
1
)
/
2
)
-
4
f
(
v
(
n
-
1
)
/
2
)
=
6
,
f
(
v
(
n
+
1
)
/
2
)
=
3
,
f
(
v
(
n
+
3
)
/
2
)
=
1
,
f
(
v
n
+
1
-
i
)
=
5
+
2
(
i
-
1
)
,
1
≤
i
≤
(
n
-
3
)
/
2
f
(
u
1
)
=
n
+
2
,
f
(
u
2
)
=
n
+
3
.
In view of the above defined labeling pattern for Case  4, we have e
f
(
0
)
=
n
+
3 and e
f
(
1
)
=
n
+
2 for n
≡
3
(
mod
⁡
5
) and e
f
(
0
)
=
n
+
2 and e
f
(
1
)
=
n
+
3 for n
≢
4
(
mod
⁡
7
).
In view of Cases 2 to 4 we have|
e
f
(
0
)
-
e
f
(
1
)
|
≤
1.
Hence,G is a prime cordial graph for n
≥
6.Illustration  4. Let G be the graph obtained by duplication of an arbitrary rim edge by an edge in wheel W
13. Then |
V
(
G
)
|
=
16 and |
E
(
G
)
|
=
31. In accordance with Theorem 14, Case  4 will be applicable and the corresponding prime cordial labeling is shown in Figure 3. Here e
f
(
0
)
=
16, e
f
(
1
)
=
15.Figure 3Theorem 15.
The graph obtained by duplication of an arbitrary spoke edge by an edge in wheelW
n is prime cordial graph, where n
=
7 and n
≥
9.Proof.
Letv
0 be the apex vertex of W
n and let v
1
,
v
2
,
…
,
v
n be the rim vertices. Without loss of generality we duplicate the spoke edge e
=
v
0
v
1 by an edge e
′
=
u
1
u
2 and call the resultant graph G. Then |
V
(
G
)
|
=
n
+
3 and |
E
(
G
)
|
=
3
n
+
2. To define f
:
V
(
G
)
→
{
1,2
,
3
,
…
,
n
+
3
}, we consider following three cases.
Case
1
(
n
=
3 to 6 and n
=
8
). For n
=
3, to satisfy the edge condition for prime cordial labeling it is essential to label five edges with label 0 and six edges with label 1 out of eleven edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most four edges and 1 label for at least seven edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
3
>
1. Hence, for n
=
3, G is not a prime cordial graph.
Forn
=
4, to satisfy the edge condition for prime cordial labeling, it is essential to label seven edges with label 0 and seven edges with label 1 out of fourteen edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most four edges and 1 label for at least ten edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
6
>
1. Hence, for n
=
4, G is not a prime cordial graph.
Forn
=
5, to satisfy the edge condition for prime cordial labeling, it is essential to label eight edges with label 0 and nine edges with label 1 out of seventeen edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most six edges and 1 label for at least eleven edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
5
>
1. Hence, for n
=
5, G is not a prime cordial graph.
Forn
=
6, to satisfy the edge condition for prime cordial labeling, it is essential to label ten edges with label 0 and ten edges with label 1 out of twenty edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most eight edges and 1 label for at least twelve edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
4
>
1. Hence, for n
=
6, G is not a prime cordial graph.
Forn
=
8, to satisfy the edge condition for prime cordial labeling, it is essential to label thirteen edges with label 0 and thirteen edges with label 1 out of twenty-six edges. But all the possible assignments of vertex labels will give rise to 0 labels for at most twelve edges and 1 label for at least fourteen edges. That is, |
e
f
(
0
)
-
e
f
(
1
)
|
=
2
>
1. Hence, for n
=
8, G is not a prime cordial graph.
Case
2
(
n
=
7, n
=
9 to 12, and n
=
14,16,18,20,22
).  For n
=
7, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
9, f
(
v
2
)
=
3, f
(
v
3
)
=
4, f
(
v
4
)
=
8, f
(
v
5
)
=
10, f
(
v
6
)
=
5, and f
(
v
7
)
=
7. Then e
f
(
0
)
=
12, e
f
(
1
)
=
11.
Forn
=
9, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
3, f
(
v
2
)
=
12, f
(
v
3
)
=
4, f
(
v
4
)
=
8, f
(
v
5
)
=
10, f
(
v
6
)
=
5, f
(
v
7
)
=
7, f
(
v
8
)
=
11, and f
(
v
9
)
=
9. Then e
f
(
0
)
=
15, e
f
(
1
)
=
14.
Forn
=
10, f
(
u
1
)
=
2, f
(
u
2
)
=
1and f
(
v
0
)
=
6, f
(
v
1
)
=
3, f
(
v
2
)
=
12, f
(
v
3
)
=
4, f
(
v
4
)
=
8, f
(
v
5
)
=
10, f
(
v
6
)
=
5, f
(
v
7
)
=
7, f
(
v
8
)
=
11, f
(
v
9
)
=
13, and f
(
v
10
)
=
9. Then e
f
(
0
)
=
16, e
f
(
1
)
=
16.
Forn
=
11, f
(
u
1
)
=
2, f
(
u
2
)
=
1and f
(
v
0
)
=
6, f
(
v
1
)
=
5, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
14, f
(
v
6
)
=
12, f
(
v
7
)
=
9, f
(
v
8
)
=
3, f
(
v
9
)
=
13, f
(
v
10
)
=
11, and f
(
v
11
)
=
7. Then e
f
(
0
)
=
18, e
f
(
1
)
=
17.
Forn
=
12, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
5, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
14, f
(
v
6
)
=
12, f
(
v
7
)
=
9, f
(
v
8
)
=
3, f
(
v
9
)
=
13, f
(
v
10
)
=
15, f
(
v
11
)
=
11, and f
(
v
12
)
=
7. Then e
f
(
0
)
=
19, e
f
(
1
)
=
19.
Forn
=
14, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
5, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
12, f
(
v
6
)
=
14, f
(
v
7
)
=
16, f
(
v
8
)
=
3, f
(
v
9
)
=
9, f
(
v
10
)
=
15, f
(
v
11
)
=
17, f
(
v
12
)
=
13, f
(
v
13
)
=
11, and f
(
v
14
)
=
7. Then e
f
(
0
)
=
22, e
f
(
1
)
=
22.
Forn
=
16, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
19, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
12, f
(
v
6
)
=
14, f
(
v
7
)
=
16, f
(
v
8
)
=
18, f
(
v
9
)
=
3, f
(
v
10
)
=
9, f
(
v
11
)
=
5, f
(
v
12
)
=
7, f
(
v
13
)
=
11, f
(
v
14
)
=
13, f
(
v
15
)
=
15, and f
(
v
16
)
=
17. Then e
f
(
0
)
=
25, e
f
(
1
)
=
25.
Forn
=
18, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
21, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
12, f
(
v
6
)
=
14, f
(
v
7
)
=
16, f
(
v
8
)
=
18,f
(
v
9
)
=
20, f
(
v
10
)
=
3, f
(
v
11
)
=
9, f
(
v
12
)
=
5, f
(
v
13
)
=
7, f
(
v
14
)
=
11, f
(
v
15
)
=
13, f
(
v
16
)
=
15, f
(
v
17
)
=
17, and f
(
v
18
)
=
19. Then e
f
(
0
)
=
28, e
f
(
1
)
=
28.
Forn
=
20, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
23, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
12, f
(
v
6
)
=
14, f
(
v
7
)
=
16, f
(
v
8
)
=
18, f
(
v
9
)
=
20, f
(
v
10
)
=
22, f
(
v
11
)
=
3, f
(
v
12
)
=
9, f
(
v
13
)
=
5, f
(
v
14
)
=
7, f
(
v
15
)
=
11, f
(
v
16
)
=
13, f
(
v
17
)
=
15, f
(
v
18
)
=
17, f
(
v
19
)
=
19, and f
(
v
20
)
=
21. Then e
f
(
0
)
=
31, e
f
(
1
)
=
31.
Forn
=
22, f
(
u
1
)
=
2, f
(
u
2
)
=
1 and f
(
v
0
)
=
6, f
(
v
1
)
=
25, f
(
v
2
)
=
4, f
(
v
3
)
=
8, f
(
v
4
)
=
10, f
(
v
5
)
=
12, f
(
v
6
)
=
14, f
(
v
7
)
=
16, f
(
v
8
)
=
18, f
(
v
9
)
=
20, f
(
v
10
)
=
22, f
(
v
11
)
=
24, f
(
v
12
)
=
3, f
(
v
13
)
=
5, f
(
v
14
)
=
7, f
(
v
15
)
=
9, f
(
v
16
)
=
11, f
(
v
17
)
=
13, f
(
v
18
)
=
15, f
(
v
19
)
=
17, f
(
v
20
)
=
19, f
(
v
21
)
=
21, and f
(
v
22
)
=
23. Then e
f
(
0
)
=
34, e
f
(
1
)
=
34.
For the next case lett
1
=
⌊
(
n
+
3
)
/
2
⌋, m
=
⌊
(
n
+
3
)
/
3
⌋, t
2
=
⌈
m
/
2
⌉, k
1
=
⌊
(
3
n
+
2
)
/
2
⌋, k
2
=
2
t
1
+
t
2
-
4,
(16)
t
=
k
1
-
k
2
,
t
3
=
{
t
-
2
;
n
=
13,15,17
,
t
-
1
;
n
=
19,21
,
n
≥
23
.
Case 3  (t
1
-
t
≥
3 (n
=
13,15,17,19,21 and n
≥
23)). Consider
(17)
f
(
u
1
)
=
2
,
f
(
u
2
)
=
5
,
f
(
v
0
)
=
6
,
f
(
v
1
)
=
3
,
f
(
v
2
)
=
4
,
f
(
v
n
)
=
1
,
f
(
v
2
+
i
)
=
8
+
2
(
i
-
1
)
,
1
≤
i
≤
t
f
(
v
t
+
3
)
=
{
9
,
if
t
≡
4
(
mod 7
)
7
,
otherwise
f
(
v
t
+
4
)
=
{
7
,
if
t
≡
4
(
mod 7
)
9
,
otherwise
f
(
v
t
+
4
+
i
)
=
9
+
2
i
,
1
≤
i
≤
t
3
.
For 2
t
+
3
<
n
-
1,
(18)
f
(
v
2
t
+
3
+
i
)
=
f
(
v
2
t
+
3
)
+
i
,
1
≤
i
≤
(
n
-
1
)
-
(
2
t
+
3
)
.
In view of the above defined labeling pattern for Case 3, we havee
f
(
0
)
=
⌊
(
3
n
+
2
)
/
2
⌋ and e
f
(
1
)
=
⌈
(
3
n
+
2
)
/
2
⌉.
Thus for Cases  2 and  3 we have|
e
f
(
0
)
-
e
f
(
1
)
|
≤
1.
Hence,G is a prime cordial graph for n
=
7 and n
≥
9.Illustration  5. Let G
1 be the graph obtained by duplication of arbitrary spoke edge by an edge of wheel W
23. Then |
V
(
G
)
|
=
26 and |
E
(
G
)
|
=
71. In accordance with Theorem 15 we have t
1
=
13, m
=
8, t
2
=
4, k
1
=
35, k
2
=
26, and t
=
9. Here t
1
-
t
=
4
>
3 so labeling pattern described in Case 3 will be applicable. The corresponding prime cordial labeling is shown in Figure 4. It is easy to visualise that e
f
(
0
)
=
35, e
f
(
1
)
=
36.Figure 4Theorem 16.
DS(
P
n
) is a prime cordial graph.Proof.
ConsiderP
n with V
(
P
n
)
=
{
v
i
:
1
≤
i
≤
n
}. Here V
(
P
n
)
=
V
1
∪
V
2, where V
1
=
{
v
i
:
2
≤
i
≤
n
-
1
} and V
2
=
{
v
1
,
v
n
}. Now in order to obtain DS
(
P
n
) from G, we add w
1
,
w
2 corresponding to V
1
,
V
2. Then |
V
(
DS
(
P
n
)
)
|
=
n
+
2 and E
(
DS
(
P
n
)
)
=
{
v
1
w
2
,
v
2
w
2
}
∪
{
w
1
v
i
:
2
≤
i
≤
n
-
1
}
∪
E
(
P
n
) so |
E
(
DS
(
P
n
)
|
=
2
n
-
1. We define vertex labeling f
:
V
(
DS
(
P
n
)
)
→
{
1,2
,
…
,
n
+
2
} as follows.
Letp
1 be the highest prime number <
n
+
2 and k
=
⌊
(
n
+
2
)
/
2
⌋. Consider
(19)
f
(
w
1
)
=
2
,
f
(
w
2
)
=
3
,
f
(
v
1
)
=
1
,
f
(
v
n
)
=
9
,
f
(
v
n
-
1
)
=
p
1
.
For 0
≤
i
<
k
-
1,
(20)
f
(
v
2
+
i
)
=
{
(
n
+
2
)
-
2
i
;
n
i
s
even
(
n
+
1
)
-
2
i
;
n
i
s
odd
.
And for vertices v
k
+
2
,
v
k
+
3
,
…
,
v
n
-
2 we assign distinct odd numbers (<
n
+
2) in ascending order starting from 5.
In view of the above defined labeling pattern, ifn is even number, then e
f
(
0
)
=
n, e
f
(
1
)
=
n
-
1; otherwise e
f
(
0
)
=
n
-
1, e
f
(
1
)
=
n.
Thus,|
e
f
(
0
)
-
e
f
(
1
)
|
≤
1.
Hence,DS
(
P
n
) is a prime cordial graph.Illustration  6. Prime cordial labeling of the graph DS
(
P
7
) is shown in Figure 5.Figure 5Theorem 17.
DS(
B
n
,
n
) is a prime cordial graph.Proof.
ConsiderB
n
,
n with V
(
B
n
,
n
)
=
{
u
,
v
,
u
i
,
v
i
:
1
≤
i
≤
n
}, where u
i
,
v
i are pendant vertices. Here V
(
B
n
,
n
)
=
V
1
∪
V
2, where V
1
=
{
u
i
,
v
i
:
1
≤
i
≤
n
} and V
2
=
{
u
,
v
}. Now in order to obtain DS
(
B
n
,
n
) from G, we add w
1
,
w
2 corresponding to V
1
,
V
2. Then |
V
(
DS
(
B
n
,
n
)
|
=
2
n
+
4 and E
(
DS
(
B
n
,
n
)
)
=
{
u
v
,
u
w
2
,
v
w
2
}
∪
{
u
u
i
,
v
v
i
,
w
1
u
i
,
w
1
v
i
:
1
≤
i
≤
n
} so |
E
(
D
S
(
B
n
,
n
)
|
=
4
n
+
3. We define vertex labeling f
:
V
(
DS
(
B
n
,
n
)
)
→
{
1,2
,
…
,
2
n
+
4
} as follows:
(21)
f
(
u
)
=
6
,
f
(
v
)
=
1
,
f
(
w
1
)
=
2
,
f
(
w
2
)
=
3
,
f
(
u
1
)
=
4
,
f
(
u
i
+
1
)
=
8
+
2
(
i
-
1
)
,
1
≤
i
≤
n
-
1
f
(
v
i
)
=
5
+
2
(
i
-
1
)
,
1
≤
i
≤
n
.
In view of the above defined labeling pattern we have e
f
(
0
)
=
2
n
+
1, e
f
(
1
)
=
2
n
+
2.
Thus,|
e
f
(
0
)
-
e
f
(
1
)
|
≤
1.
Hence,DS
(
B
n
,
n
) is a prime cordial graph.Illustration  7. Prime cordial labeling of the graph DS
(
B
5,5
) is shown in Figure 6.Figure 6


## 3. Conclusion

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



---

*Source: 607018-2014-03-23.xml*