Edge even graceful labeling of some graphs

Abstract

Edge even graceful labeling is a new type of labeling since it was introduced in 2017 by Elsonbaty and Daoud (Ars Combinatoria 130:79–96, 2017). In this paper, we obtained an edge even graceful labeling for some path-related graphs like Y- tree, the double star B_n,m, the graph 〈K_1,2n:K_1,2m〉, the graph $~P_{2n-1}\odot \overline { K_{2m}}~$, and double fan graph F_2,n. Also, we showed that some cycle-related graphs like the prism graph $~\prod _{n}~$, the graph $~C_{n}\left (\frac {n}{2}\right)~$, the flag FL_n, the graph K₂⊙C_n, the flower graph FL(n), and the double cycle {C_n,n} are edge even graphs.

Introduction

A graph labeling is an assignment of integers to the edges or vertices, or both, subject to certain condition. The idea of graph labelings was introduced by Rosa in [1]. Following this paper, other studies on different types of labelings (Odd graceful, Chordal graceful, Harmonious, edge odd graceful) introduced by many others [2–4]. A new type of labeling of a graph called an edge even graceful labeling has been introduced by Elsonbaty and Daoud [5]. They introduced some path- and cycle-related graphs which are edge even graceful.

Graph labelings give us useful models for a wide range of applications such as coding theory, X-ray, astronomy, radar, and communication network addressing.

Definition 1

[5] An edge even graceful labeling of a graph G(V(G), E(G)) with p=|V(G)| vertices and q=|E(G)| edges is a bijective mapping f of the edge set E(G) into the set {2,4,6, ⋯,2q } such that the induced mapping f^∗:V(G) →{0,2,4,⋯,2q }, given by: $f^{\ast }(x) = \left ({\sum \nolimits }_{xy \in E(G)} f(xy)~ \right)~mod~(2k)$, is an injective function, where k=max(p,q). The graph that admits an edge even graceful labeling is called an edge even graceful graph.

In Fig. 1, we present an edge even graceful labeling of the Peterson graph and the complete graph K₅.

Fig. 1

An edge even graceful labeling of Peterson graph and the complete graph K₅

Edge even graceful for some path related graphs

A Y- tree is a graph obtained from a path by appending an edge to a vertex of a path adjacent to an end point, and it is denoted by Y_n where n is the number of vertices in the tree.

Lemma 1

The Y-tree Y_n is an edge even graceful graph when n is odd.

Proof

The number of vertices of Y-tree Y_n is n and the number of edges is n−1. Let the vertices and the edges of Y_n be given as in Fig. 2.

Fig. 2

Y_n with ordinary labeling

We define the mapping f:E(Y_n) →{2,4,⋯,2n−2} as follows:

$${\begin{array}{ll} f (e_{i})= 2i & \quad\text{for} \quad 1 \leq i \leq \frac{n+1}{2}\\ f \left(e_{\frac{n+3}{2}}\right)= n+5, \quad & \quad f \left(e_{\frac{n+5}{2}}\right)= n+3\\ f (e_{i})= 2i & \quad\text{for} \quad \frac{n+7}{2} \leq i \leq n-1 \end{array}} $$

Then, the induced vertex labels are

$$\begin{aligned} f^{\ast}(v_{1})&= 2, \qquad f^{\ast}(v_{2})= 2, \quad f^{\ast}(v_{3})=12\\ f^{\ast}(v_{i})&= 4i-2 \qquad \text{for} \quad 4 \leq i \leq \frac{n-1}{2},\\ f^{\ast}\left(v_{\frac{n+1}{2}}\right)&= \left[ f \left(e_{\frac{n-1}{2}}\right) + f \left(e_{\frac{n+1}{2}}\right)\right]~\text{mod}~(2n) =(2n) ~\text{mod}~(2n)~= 0 \end{aligned} $$

Similarly, $~~~f^{\ast }\left (v_{\frac {n+3}{2}}\right)= 6~,~~~~~~~~~~~~~ f^{\ast }\left (v_{\frac {n+5}{2}}\right)= 8~, f^{\ast }\left (v_{\frac {n+7}{2}}\right)= 10 ~$,

$$\begin{array}{*{20}l} f^{\ast}(v_{i})&= (4i-2)~\text{mod}~(2n) ~~~~~ \text{for} ~~~\frac{n+9}{2} \leq i \leq n-1,\\ f^{\ast}(v_{n})&= 2n-2 \end{array} $$

Clearly, all the vertex labels are even and distinct. Thus, Y-tree Y_n is an edge even graceful graph when n is odd. □

Illustration: The edge even labeling of the graph Y₁₃ is shown in Fig. 3.

Fig. 3

The edge even labeling of the graph Y₁₃

Double star is the graph obtained by joining the center of the two stars K_1,n and K_1,m with an edge, denoted by B_n,m. The graph B_n,m has p=n+m+2 and q=n+m+1.

Lemma 2

The double star B_n,m is edge even graceful graph when one (m or n) is an odd number and the other is an even number.

Proof

Without loss of generality, assume that n is odd and m is even. Let the vertex and edge symbols be given as in Fig. 4.

Fig. 4

B_m,n with ordinary labeling

Define the mapping f:E(B_m,n) →{2,4,⋯,2q} as follows:

$$\begin{array}{*{20}l} f (e_{i})&= 2i \qquad \qquad \qquad\text{ for } \quad 1~\leq i \leq\frac{n+1}{2}\\[-2pt] f (E_{i})&= 2p-~2(i +1) \qquad \text{ for } \quad 1~\leq i \leq \frac{n-1}{2}\\[-2pt] f (a_{i})&= m+~2i \qquad \qquad\text{ for } 1~\leq i \leq\frac{m}{2}\\[-2pt] f (b_{i})&= 2p- [ m+~2i ] \qquad \text{ for } 1 ~\leq i \leq \frac{m}{2}\\[-2pt] f (w)&= 2p- 2 \end{array} $$

We realize the following:

$$\begin{array}{*{20}l} f (E_{i})+ f (e_{i+1 })&\equiv 0 ~\text{mod}~(2p) \quad \text{for} ~~~i= 1,2,\cdots, \frac{n-1}{2}\\[-2pt] f (a_{i})+ f (b_{i })&\equiv 0 ~\text{mod}~(2p) \quad \text{for} ~~~j= 1,2,\cdots, \frac{m}{2} \end{array} $$

So, the vertex labels will be

$$\begin{array}{*{20}l} f^{\ast}(u)= [ f (e_{1}) + f (w) ] ~\text{mod}~(2p) =~0 ~ ~ \text{ and }\\[-2pt] f^{\ast}(v)= f(w) = (2p-2)~\text{mod}~(2p)~= 2p-2 \end{array} $$

Also, each pendant vertex takes the labels of its incident edge which are different from the labels of the vertices U and v. □

The graph 〈K_1,n:K_1,m〉 is obtained by joining the center v₁ of the star K_1,n and the center v₂ of the another star K_1,m to a new vertex u, so the number of vertices is p=n+m+3 and the number of edges is q=n+m+2.

Lemma 3

The graph 〈K_1,2n:K_1,2m〉 is an edge even graceful graph.

Proof

Let the vertex and edge symbols be given as in Fig. 5.

Fig. 5

〈K_1,2n:K_1,2m〉 with ordinary labeling

Define the mapping f:E(G) →{2,4,⋯,2p} as follows:

$$\begin{array}{*{20}l} f (e_{1})&= mn~+2 \qquad, \qquad f (e_{2})= 2p-(mn~+2)\\[-2pt] f (A_{i})&= ~2i \qquad \qquad \qquad \text{for} ~~~ i= 1,2,\cdots,n\\[-2pt] f (B_{i})&= 2p- 2i \qquad \qquad \text{for} ~~~ i= 1,2,\cdots,n\\[-2pt] f (C_{i})&= ~2n+~2i \qquad \qquad\text{for} ~~~ i= 1,2,\cdots,m\\[-2pt] f (B_{i})&= 2p- [ 2n+~2i ] \qquad \text{for} ~~~ i= 1,2,\cdots,m \end{array} $$

We can see the following:

$$\begin{array}{*{20}l} f (A_{i})+ f (B_{i })&\equiv 0 ~\text{mod}~(2p) ~~~~~ \text{for} ~~~i= 1,2,\cdots, n\\[-2pt] f (C_{i})+ f (D_{i })&\equiv 0 ~\text{mod}~(2p) ~~~~~ \text{for} ~~~j= 1,2,\cdots, m \end{array} $$

So, the vertex labels will be

$$\begin{array}{*{20}l} f^{\ast}(u)&= [ f (e_{1}) + f (e_{2}) ] ~\text{mod}~(2p) =0,\\[-2pt] f^{\ast}(v_{1})&= f(e_{1}) ~~= mn+~2~ \text{ and }\\[-2pt] f^{\ast}(v_{2})&= f(e_{2}) ~=2p - [ 2n+~2i ] \end{array} $$

which are even and distinct from each pendant vertices. Thus, the graph 〈K_1,2n:K_1,2m〉 is an edge even graceful graph. □

Lemma 4

The corona [6]$~P_{3}\odot \overline { K_{2m}}~$ is an edge even graceful graph.

Proof

In this graph p=6m+3 and q=6m+2. Let the vertex and edge symbols be given as in Fig. 6.

Fig. 6

$~P_{3}\odot \overline { K_{2m}}~$ with ordinary labeling

We can define the mapping $f: E(P_{3}\odot \overline { K_{2m}}~)~\rightarrow \{2,4,\cdots, 2q \}$ as follows:

$$\begin{array}{*{20}l} f (e_{i})&= 6m+ 2i ~~~~~~~~~~~~~~~~~~~~~~~~~~~ \text{for} ~~~ i= 1,2\\[-2pt] f (A_{ij})&= 2(i-1)m+~2j ~~~~~~~~~~~~~~~~ ~~\text{for} ~~~ i= 1,2,\cdots,m\\[-2pt] f (B_{ij})&= 2p- [ 2(i-1)m+~2j ]~~~~~~~~~ \text{for} ~~~ i= 1,2,\cdots,m \end{array} $$

It is clear that f(A_ij)+f(B_ij)≡0 mod (2p) for j=1,2,⋯,m

So the vertex labels will be,

$$\begin{array}{*{20}l} f^{\ast}(v_{1})&= f(e_{1}) ~= 6m+2~~~~~,~~~~~~~~~f^{\ast}(v_{3})= f(e_{2}) ~=6m+4~~~ \text{ and }\\ f^{\ast}(v_{2})&=[~f(e_{1}) + f(e_{2}) ] ~\text{mod}~(2p) =~ [ (6m+2)+ (6m+4) ]~\text{mod}~(2p)=~0~ \end{array} $$

Therefore, all the vertices are even and distinct which complete the proof. □

Illustration: The edge even labeling of the graph $~P_{3}\odot \overline { K_{6}}~$ is shown in Fig. 7.

Fig. 7

Edge even graceful labeling of the graph $~P_{3}\odot \overline { K_{6}}~$

The generalization of the previous result is presented in the following theory

Theorem 1

The graph $~P_{2n-1}\odot \overline { K_{2m}}~$ is an edge even graceful graph.

Proof

In this graph, p=4nm+2(n−m)−1 and q=4nm+2(n−m)−2. The middle vertex in the path P_2n−1 will be v_n, and we start the labeling from this vertex. Let the vertex and edge symbols be given as in Fig. 8.

Fig. 8

$~P_{2n-1}\odot \overline { K_{2m}}~$ with ordinary labeling

Define the mapping f:E(G) →{2,4,⋯,2q} by the following arrangement

f(e_i)=2ⁱ for i=1,2,⋯,n−1

f(b_i)=2p− 2ⁱ for i=1,2,⋯,n−1

f(A_ij) will take any number from the reminder set of the labeling not contains 2ⁱ nor 2p− 2ⁱ for j=1,2,⋯,m and

f(D_ij)=2p−[f(A_ij)] for j=1,2,⋯,m.

It is clear that [f(A_ij)+f(D_ij)] mod (2p)≡0 mod (2p) for j=1,2,⋯,m.

So, the vertex labels will be

f^∗(v_n)=[f(e₁)+f(b₁)] mod (2p) ≡0 mod (2p),

For any vertex v_k when k<n, let k=n−i and i=1,2,⋯,n−2

$$f^{\ast}(v_{k})= f^{\ast}(v_{n-i}) = \left\{\begin{array}{ll} \left[f(e_{i}) + f(b_{i+1})\right] \ \text{mod}\ (2p)= f(b_{i}) & \text{if}\ i\ \text{is odd} \\ \left[f(b_{i}) + f(e_{i+1})\right] \ \text{mod}\ (2p)= f(e_{i}) & \text{if}\ i\ \text{is even} \end{array}\right. $$

When k>n, let k=n+i and i=1,2,⋯,n−2

$$ f^{\ast}(v_{k})= f^{\ast}(v_{n+i}) = \left\{\begin{array}{ll} \left[f(b_{i})+ f(e_{i+1})\right] \ \text{mod}\ (2p)= f(e_{i}) & \text{if}\ i\ \text{is odd} \\ \left[f(e_{i})+ f(b_{i+1})\right] \ \text{mod}\ (2p)= f(b_{i}) & \text{if}\ i\ \text{is even} \end{array}\right. $$

The pendant vertices v₁ and v_2n−1 of the path P_2n−1 will take the labels of its pendant edges of P_2n−1,i.e.,

If n is even, then f^∗(v₁)=f(e_n−1), and f^∗(v_2n−1)=f(b_n−1)

If n is odd, then f^∗(v₁)=f(b_n−1), and f^∗(v_2n−1)=f(e_n−1)

Then, the labels of the vertices of the path P_2n−1 takes the labels of the edges of the path P_2n−1, and each pendant vertex takes the labels of its incident edge. Then, there are no repeated vertex labels, which complete the proof. □

Illustration: The graph $~P_{11}\odot \overline { K_{4}}~$ labeled according to Theorem 1 is presented in Fig. 9.

Fig. 9

Edge even graceful labeling of the graph $~P_{11}\odot \overline { K_{4}}~$

A double fan graph F_2,n is defined as the graph join $\overline {K_{2}} + P_{n}~$ where $\overline {K_{2}}$ is the empty graph on two vertices and P_n be a path of length n.

Theorem 2

The double fan graph F_2,n is an edge even graceful labeling when n is even.

Proof

In the graph F_2,n we have p=n+2 and q=3n−1. Let the graph F_2,n be given as indicated in Fig. 10.

Fig. 10

F_2,n with ordinary labeling

Define the edge labeling function f:E(F_2,n)→ {2,4,⋯, 6n−2} as follows:

f(a_i)=2i; i=1,2,⋯n

f(b_i)=2q−2i=6n−2(i+1); i=1,2,⋯n

$$f~(e_{i}) = \left\{\begin{array}{lll} 2n+2i ~~~ & \text{if} & 1 \leq i < \frac{n}{2}\\ 6n-2~~~ & \text{if} & i=\frac{n}{2}\\ 2n+2(i-1)~~~ & \text{if} & \frac{n}{2} < i \leq n-1\\ \end{array}\right. $$

Hence, the induced vertex labels are

$~ f^{\ast }(u) = \left (\sum _{i=1}^{n}(f(a_{i}))\right)~\text {mod}(6n-2)\,=\, \left (\sum _{i=1}^{n}(2i)\right)~\text {mod}(6n-2) =(n^{2}+n)~\text {mod}(6n-2)~~$

$~ f^{\ast }(v) = \left (\sum _{i=1}^{n}(f(b_{i})) \right)~\text {mod}(6n-2)= \left (\sum _{i=1}^{n}(2q-2i)\right)~\text {mod}(6n-2) =\left (- n^{2}-n\right)~\text {mod}(6n-2)$

$f^{\ast }(v_{i}) \,=\, \left [\sum _{i=1}^{n}(f(a_{i})\,+\,f(b_{i})\,+\,f (e_{i}) \,+\,f(e_{i-1}))\right ]~\text {mod}(6n\,-\,2)~ \,=\,(4n\,+\,4i\,-\,2)~\text {mod}(6n\,-\,2)~,~~2 \leq i \leq \frac {n}{2}- 1$

$f^{\ast }(v_{i}) = \left [\sum _{i=1}^{n}(f(a_{i})+f(b_{i})+f (e_{i}) +f(e_{i-1}))\right ]~\text {mod}(6n-2) =(4n+4i-6)~\text {mod}(6n-2),~~\frac {n}{2}+2 \leq i \leq n- 1$

Since [f(a_i)+f(b_i)] mod(6n−2)=0, we see that

f^∗(v₁)= f(e₁)=2n+2,

f^∗(v_n)= f(e_n−1)=4n−4,

$~~~~~ f^{\ast }(v_{\frac {n}{2}})=~f (e_{\frac {n}{2}-1})= 3n-2 ~$ and

$~~~~f^{\ast }(v_{\frac {n}{2}+1})=~f (e_{\frac {n}{2}})= 3n ~$

Thus, the set of vertices $v_{1}, v_{2},v_{3},\cdots ~,v_{\frac {n}{2}-1},v_{\frac {n}{2} },v_{\frac {n}{2}+1},v_{\frac {n}{2}+2},v_{\frac {n}{2}+3},\cdots ~,v_{n-2},v_{n-1},v_{n}$ are labeled by 2n+2, 4n+6, 4n+10,⋯,6n−6, 3n−2, 3n, 4,8, ⋯,2n−12, 2n−8, 4n−4 respectively.

Clearly, f^∗(u) and f^∗(v) are different from all the labels of the vertices. Hence F_2,n is an edge even graceful when n is even. □

Illustration: The double fan F_2,10 labeled according to Theorem 2 is presented in Fig. 11.

Fig. 11

Edge even graceful labeling of the double fan F_2,10

Edge even graceful for some cycle related graphs

Definition 2

For n≥4, a cycle (of order n) with one chord is a simple graph obtained from an n-cycle by adding a chord. Let the n-cycle be v₁v₂⋯v_nv₁. Without loss of generality, we assume that the chord joins v₁ with any one v_i, where 3≤i≤n−1. This graph is denoted by C_n(i).

Lemma 5

The graph $~C_{n}\left (\frac {n}{2}\right)~$ is an edge even graceful graph if n is even.

Proof

Let $\{ v_{1}, v_{2},\cdots, v_{\frac {n}{2}-1},v_{\frac {n}{2}},v_{\frac {n}{2}+1}, \cdots,v_{n} \}$ be the vertices of the graph $C_{n}\left (\frac {n}{2}\right)$, and the edges are e_i=(v_iv_i+1) for i≤i≤n−1 and the chord $e_{0}=(v_{1}v_{\frac {n}{2}})$ connecting the vertex v₁ with $v_{\frac {n}{2}}$ as in Fig. 12.

Fig. 12

$~C_{n}\left (\frac {n}{2}\right)~$ with ordinary labeling

Here, p=n and q=n+1, so 2k=2q=2n+2; first, we label the edges as follows:

$$f(e_{i})= 2i+2 ; ~~~~i=0,1,2,\cdots n $$

Then, the induced vertex labels are as follows:

$$\begin{array}{*{20}l} f^{\ast}(v_{1})&= \left[ f(e_{0}) +f(e_{1})+f(e_{n})\right]\text{mod}(2n+2)=(2+4+2n+2)\text{mod}(2n+2)= 6,\\ f^{\ast}(v_{\frac{n}{2}})&= \left[ f(e_{0}+f(e_{\frac{n}{2}})+f(e_{\frac{n}{2}-1})\right]\text{mod}(2n+2)=~(~2n+4)~\text{mod}(2n+2)~= 2 \end{array} $$

for any other vertex $~~~v_{i}, ~~~i\neq 1,\frac {n}{2}$

$$\begin{array}{*{20}l} f^{\ast}(v_{i})=~ f(e_{i})~+~f(e_{i-1})~=~(~4n+4)~\text{mod}(2n+2)~= (4i+2)\text{mod}(2n+2) \end{array} $$

Hence, the labels of the vertices $v_{0}, v_{1},v_{2}, \cdots,v_{\frac {n}{2}-1},v_{\frac {n}{2}},v_{\frac {n}{2}+1}, \cdots v_{n}$ are 6, 10, 14,⋯, 2n−2, 2, 4,⋯, 2n respectively, which are even and distinct. So, the graph $~C_{n}\left (\frac {n}{2}\right)~$ is an edge even graceful graph if n is even. □

Definition 3

Let C_n denote the cycle of length n. The flag FL_n is obtained by joining one vertex of C_n to an extra vertex called the root, in this graph p=q=n+1.

Lemma 6

The flag graph FL_n is edge even graceful graph when n is even.

Proof

Let {v₁,v₂,⋯,v_n} be the vertices of the cycle C_n and the edges are e_i=(v_iv_i+1) for 1≤i≤n and the edge e=(v₁v₀) connecting the vertex v₁ with v₀ as in Fig. 13.

Fig. 13

Fl_n with ordinary labeling

First, we label the edges as follows:

$$f(e_{i})= 2i+2 ; ~~~~i=0,1,2,\cdots n $$

Then, the induced vertex labels are as follows f^∗(v₀)=f(e₀)=2,

$$\begin{array}{*{20}l} f^{\ast}(v_{1})&= \left[ f(e_{0}) +f(e_{1})+f(e_{n})\right]\text{mod}(2n+2)=(~2+4+2n+2)~\text{mod}(2n+2)~= 6\\ f^{\ast}(v_{i})&= \left[f(e_{i})~+~f(e_{i-1})\right]~\text{mod}(2n+2) =~ (4i+2)\text{mod}(2n+2)~~~~i=2,3,\cdots, n \end{array} $$

Hence, the labels of the vertices $v_{0}, v_{1},v_{2}, \cdots,v_{\frac {n}{2}-1},v_{\frac {n}{2}},v_{\frac {n}{2}+1}, \cdots v_{n}$ will be 2, 6, 10, 14,⋯, 2n−2, 0, 4,⋯, 2n respectively. □

Lemma 7

The graph K₂⊙C_n is edge even graceful graph when n is odd.

Proof

Let $\left \{ v_{1}, v_{2},\cdots, v_{n},v^{'}_{1},v^{'}_{2}, \cdots,v^{'}_{n} \right \}$ be the vertices of the graph K₂⊙C_n and the edges are $\left \{e_{1}, e_{2},\cdots, e_{n},~e^{'}_{1},e^{'}_{2}, \cdots,e^{'}_{n} \right \}$ as shown in Fig. 14. Here, p=2n and q=2n+1, so 2k=4n+2.

Fig. 14

The graphs K₂⊙C_n with ordinary labeling

First, we label the edges as follows:

$$\begin{array}{*{20}l} f(e_{i})&= 2i ; ~~~~~~~~~~~~~~~~~~~~~~~1\leq~i~\leq n+1\\ f(e^{\prime}_{i})&= 2n+2(i+1) ; ~~~~~~1\leq~i~\leq n \end{array} $$

We can see that [f(e_n)+f(e_n+1)]mod(2q)=(4n+2) mod(4n+2)≡0

Then, the induced vertex labels are as follows

$$\begin{array}{*{20}l} f^{\ast}(v_{i}) &= [ f(e_{i}) +f(e_{i+1})]~\text{mod}(4n+2)~~ ~~~ ~~~1\leq i \leq n-1 \\ &= [~6+4(i-1)]~\text{mod}(4n+2),~~~~~~1\leq i \leq n-1 \end{array} $$

$f^{\ast }(v^{\prime }_{i})= \left [ f(e^{\prime }_{i}) +f(e^{\prime }_{i+1})\right ]~\text {mod}(4n+2)=[8+4~(i-1)]~\text {mod}(4n+2)~,~~1 \leq i < n$

f^∗(v_n)= [f(e₁)+f(e_n)+f(e_n+1) ] mod(4n+2)= f(e₁) mod(4n+2) ≡2

$~~~f^{\ast }(v^{\prime }_{n})=~ \left [ f(e^{\prime }_{1}) +f(e_{n+1})~+~f\left (e^{\prime }_{n}\right)\right ]~\text {mod}(4n+2) ~\equiv ~4$

Clearly, the vertex labels are all even and distinct. Hence, the graph K₂⊙C_n is edge even graceful for odd n. □

Let C_n denote the cycle of length n. Then, the corona of all vertices of C_n except one vertex {v₁} with the complement graph $\overline {K_{2m-1}}$ is denoted by $~ \{C_{n}-\{v_{1} \} \} \odot \overline {K_{2m-1}}~$, in this graph p=q=2m(n−1)+1.

Lemma 8

The graph $\{C_{n}-\{v_{1} \} \} \odot \overline {K_{2m-1}}~$ is an edge even graceful graph.

Proof

Let the vertex and edge symbols be given as in Fig. 15.

Fig. 15

$~ \{C_{n}-\{v_{1} \} \} \odot \overline {K_{2m-1}}~$ with ordinary labeling

Define the mapping f:E(G) →{2,4,⋯,2q} as follows:

$$\begin{array}{*{20}l} f (E_{i})&= 2(i-1)m+2 ~~~~~ ~~~~~~~~~~~~~~ \text{for} ~~~i= 1,2,\cdots, n-1\\ f(E_{n}) &=2q\\ f (e_{i+1})&= 2q- [2(i-1)m+2 ] ~~~~~~~~~ ~ \text{for} ~~~ i= 1,2,\cdots, n-1\\ f (A_{ij})&= 2(i-1)m+2j+2 ~~~~~ ~~~~~~~~~ \text{for} ~~~~ j= 1,2,\cdots, m-1\\ f (B_{ij})&= 2q - [2(i-1)m+2j+2 ]~~~~~ \text{for} ~~~j= 1,2,\cdots, m-1 \end{array} $$

We realize the following:

[f(A_ij)+f(B_ij)] mod (2q)≡0 mod (2q) for j=1,2,⋯,m−1

Also, [f(E_i−1)+f(e_i)] mod (2q)≡0 mod (2q) for i=2,3,⋯,n

So, verifying the vertex labels, we get that,

f^∗(v₁)=[f(E₁)+f(E_n)] mod (2q)=(2+2q) mod (2q)=2,

$$\begin{array}{*{20}l} {} f^{\ast}(v_{i}) &\,=\, \left[~ \sum_{j=1}^{m-1}f (A_{ij})\,+\, \sum_{j=1}^{m-1} f (B_{ij}) \,+\, f(E_{i}) +f(E_{i-1})~+ f(e_{i})~\right]~ \text{mod}~(2q)~~ ~~~i= 2,3,\cdots, n \\ &= f(E_{i})~\text{mod}~(2q)~=2(i-1)m+2~\text{mod}~(2q),~~ ~~~i= 2,3,\cdots, n \end{array} $$

Hence, the labels of the vertices v₁,v₂,⋯,v_n takes the label of the edges of the cycles and each of the pendant vertices takes the label of its edge, so they are all even and different numbers. □

Illustration: In Fig. 16, we present an edge even graceful labeling of the graph $~ \{C_{6}-\{v_{1} \} \} \odot \overline {K_{3}}~$.

Fig. 16

$~ \{C_{6}-\{v_{1} \} \} \odot \overline {K_{3}}~$

Lemma 9

The double cycle graph {C_n,n} is an edge even graceful graph when n is odd.

Proof

Here, p=n and q=2n. Let the vertex and edge symbols be given as in Fig. 17.

Fig. 17

{C_n,n} with ordinary labeling

Define the mapping f:E(G) →{2,4,⋯,4n} by

f(e_i)=2i for i=1,2,⋯,n. So, the vertex labels will be

f^∗(v₁)=[f(e₁)+f(e_n) +f(e_n+1) +f(e_2n)]mod (4n)=4

$$\begin{array}{*{20}l} f^{\ast}(v_{i}) &= \left[f(e_{i}) +f(e_{i-1})~+ f(e_{i+n})~+ f(e_{i+n-1})\right] \text{mod}~(4n)~~ ~~~i= 2, {,} \cdots, n \\ &= (8i-4)~\text{mod}~(4n)~~ ~~~i= 2,3, \cdots, n \end{array} $$

Hence, the labels of the vertices $v_{1},v_{2},v_{3}, \cdots, v_{\frac {n+1}{2}}, v_{\frac {n+3}{2}}, \cdots, v_{n}$ will be 4,12,20,⋯,0,8,⋯,4n−4 □

The prism graph $\prod _{n}$ is the cartesian product C_n□K₂ of a cycle C_n by an edge K₂, and an n-prism graph has p=2n vertices and q=3n edges.

Theorem 3

The prism graph $~\prod _{n}~$ is edge even graceful graph.

Proof

In the prism graph $\prod _{n}$ we have two copies of the cycle C_n, let the vertices in one copy be v₁,v₂,⋯,v_n and the vertices on the other copy be $~v^{\prime }_{1}, v^{\prime }_{2}, \cdots, v^{\prime }_{n}~$. In $~\prod _{n}~$, the edges will be

$v_{i}v_{i+1}, ~~ ~~~v^{\prime }_{i}v^{\prime }_{i+1},~~~ \text {and} ~~~v_{i}v^{\prime }_{i}$. Let the vertex and edge symbols be given as in Fig. 18.

Fig. 18

$~\prod _{n}~$ with ordinary labeling

Define the mapping $f: E\left (\prod _{n}\right)~\rightarrow \{2,4,\cdots, 6n \}$ by

$$\begin{array}{*{20}l} f (e_{i})&= 2i ~~~~~~~~~ ~~~~~~~~~~~~~~\text{for} ~~~ i= 1,2,\cdots, n\\ f (e^{\prime}_{i})&= 4n +2i ~~~~~~~~~~~~~~ \text{for} ~~~ i= 1,2,\cdots, n\\ f (E_{i})&= 2n +2i ~~~~~~~~~~~~~~~~~~ \text{for} ~~~ i= 1,2,\cdots, n \end{array} $$

So, the vertex labels will be

$$\begin{array}{*{20}l} f^{\ast}(v_{1}) &\,=\, \left[f(e_{1}) +f(e_{n})+ f(E_{1})\right] \text{mod}~(6n)=4n+4\\ f^{\ast}(v_{i}) &\,=\, \left[f(e_{i}) +f(e_{i-1})+ f(E_{n-i+2})\right] \text{mod}~(6n)= (2i+4n \,+\,2)~\text{mod}~(6n) ~~i= 2,3, \cdots, n \end{array} $$

Hence, the labels of the vertices v₁,v₂,⋯,v_n will be 4n+4, 4n+6,⋯,0, 2 respectively.

Also, $f^{\ast }(v^{\prime }_{1}) = \left [~ f(e^{\prime }_{1}) +f(e^{\prime }_{n})~+ f(E_{1})\right ] \text {mod}~(6n)= 12n+ 4~ \text {mod}~(6n)= 4$

$$\begin{array}{*{20}l} f^{\ast}(v^{\prime}_{i}) &= \left[~ f(e^{\prime}_{i}) +f(e^{\prime}_{i-1})~+ f\right] \text{mod}~(6n)~~ ~~~i= 2, {,} \cdots, n \\ &= (2i+12n +2)~\text{mod}~(6n)~= 2i+2~ ~~~i= 2,3, \cdots, n \end{array} $$

Hence, the labels of the vertices $v^{\prime }_{1},v^{\prime }_{2}, \cdots, v^{\prime }_{n}$ are 4, 6,8,⋯,2n, 2n+2 respectively. Overall, the vertices are even and different. Thus, the prism graph $~\prod _{n}~$ is an edge even graceful graph. □

Illustration: In Fig. 19, we present an edge even graceful labeling of of prism graphs $~\prod _{5}~$ and $~\prod _{6}$.

Fig. 19

An edge even graceful labeling of prism graphs $~\prod _{5}~$ and $~\prod _{6}$

The flower graph FL(n) (n≥3) is the graph obtained from a helm H_n by joining each pendant vertex to the center of the helm.

Theorem 4

The flower graph FL(n) (n≥4) is an edge even graceful graph.

Proof

In the flower graph FL(n) (n≥4), we have p=2n+1 and q=4n. Let $\{~v_{0},v_{1}, v_{2}, \cdots, v_{n}~,v^{\prime }_{1}, v^{\prime }_{2}, \cdots, v^{\prime }_{n}\}$ be the vertices of FL(n) and

{ e₁,e₂,e₃,⋯,e_3n,E₁,E₂,E₃,⋯,E_n } be the edges of FL(n) as in Fig. 20.

Fig. 20

The flower graph FL(n) with ordinary labeling

First, define the mapping f:E(Fl(n))→{2,4,⋯,8n } as the following:

$$f (E_{i})= 8n-2i ~~~~~~~~~ \text{for} ~~~ i= 1,2,\cdots, n ~~~ \text{and} $$

$$f~(e_{i}) = \left\{\begin{array}{ll} 2i ~~~ & \text{if~~ \(1 \leq i \leq 3n-1 ~~ ~ \) }\\ 8n~~~ & \text{if ~~\(i = 3 n~~ ~ \)} \\ \end{array}\right. $$

Then, the induced vertex labels are

$$\begin{array}{*{20}l} f^{\ast}(v_{0})&= \left[ \sum_{i=1}^{n}(f(e_{i})+ f(E_{i})\right]~\text{mod}(8n)=~0\\ f^{\ast}(v_{i})&= \left[f(e_{i})+ f(e_{n+i})\right]~\text{mod}(8n)=~2n+4i~,~~~~ i= 1,2,\cdots, n\\ f^{\ast}(v^{\prime}_{1})&= \left[f(e_{3n})+ f(e_{2n+1})+f(e_{n+1}) + f(E_{1})\right]~\text{mod}(8n)=~6n+2\\ f^{\ast}(v^{\prime}_{2})&= \left[f(e_{3n})+ f(e_{3n-1})+f(e_{n+2}) + f(E_{2})\right]~\text{mod}(8n)=~8n-2 \end{array} $$

$$\begin{array}{*{20}l} f^{\ast}(v^{\prime}_{i}) &= \left[ f(e_{3n-i+1})+ f(e_{3n-i+2})+f(e_{n+i}) + f(E_{i})\right]~\text{mod}(8n),~~ ~~~ ~~~3\leq i \leq n \\ &= \left[6 (n+1)-4i\right]~\text{mod}(8n) ~~~~~~3\leq i \leq n \end{array} $$

Overall, all the vertex labels are even and distinct which complete the proof. □

Illustration: In Fig. 21, we present an edge even graceful labeling of of the flower graphs FL(6)and FL(7).

Fig. 21

An edge even graceful labeling of The flower graphs FL(6) and FL(7)