###
**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 *v*_{1}*v*_{2}⋯*v*_{n}*v*_{1}. Without loss of generality, we assume that the chord joins *v*_{1} with any one *v*_{i}, where 3≤*i*≤*n*−1. This graph is denoted by *C*_{n}(*i*).

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**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*_{i}*v*_{i+1}) for *i*≤*i*≤*n*−1 and the chord \(e_{0}=(v_{1}v_{\frac {n}{2}}) \) connecting the vertex *v*_{1} with \(v_{\frac {n}{2}}\) as in Fig. 12.

Here, *p*=*n* and *q*=*n*+1, so 2*k*=2*q*=2*n*+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,⋯, 2*n*−2, 2, 4,⋯, 2*n* 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 *F**L*_{n} is obtained by joining one vertex of *C*_{n} to an extra vertex called the root, in this graph *p*=*q*=*n*+1.

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**Lemma 6**

The flag graph *F**L*_{n} is edge even graceful graph when *n* is even.

###
*Proof*

Let {*v*_{1},*v*_{2},⋯,*v*_{n}} be the vertices of the cycle *C*_{n} and the edges are *e*_{i}=(*v*_{i}*v*_{i+1}) *f**o**r* 1≤*i*≤*n* and the edge *e*=(*v*_{1}*v*_{0}) connecting the vertex *v*_{1} with *v*_{0} as in Fig. 13.

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*^{∗}(*v*_{0})=*f*(*e*_{0})=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,⋯, 2*n*−2, 0, 4,⋯, 2*n* respectively. □

###
**Lemma 7**

The graph *K*_{2}⊙*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 *K*_{2}⊙*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*=2*n* and *q*=2*n*+1, so 2*k*=4*n*+2.

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(2*q*)=(4*n*+2) mod(4*n*+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*(*e*_{1})+*f*(*e*_{n})+*f*(*e*_{n+1}) ] mod(4*n*+2)= *f*(*e*_{1}) mod(4*n*+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 *K*_{2}⊙*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 {*v*_{1}} 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*=2*m*(*n*−1)+1.

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**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.

Define the mapping *f*:*E*(*G*) →{2,4,⋯,2*q*} 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 (2*q*)≡0 mod (2*q*) for *j*=1,2,⋯,*m*−1

Also, [*f*(*E*_{i−1})+*f*(*e*_{i})] mod (2*q*)≡0 mod (2*q*) *f**o**r* *i*=2,3,⋯,*n*

So, verifying the vertex labels, we get that,

*f*^{∗}(*v*_{1})=[*f*(*E*_{1})+*f*(*E*_{n})] mod (2*q*)=(2+2*q*) mod (2*q*)=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 *v*_{1},*v*_{2},⋯,*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 \( ~ \{C_{6}-\{v_{1} \} \} \odot \overline {K_{3}}~ \).

###
**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*=2*n*. Let the vertex and edge symbols be given as in Fig. 17.

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

*f*(*e*_{i})=2*i* for *i*=1,2,⋯,*n*. So, the vertex labels will be

*f*^{∗}(*v*_{1})=[*f*(*e*_{1})+*f*(*e*_{n}) +*f*(*e*_{n+1}) +*f*(*e*_{2n})]mod (4*n*)=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,⋯,4*n*−4 □

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

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**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 *v*_{1},*v*_{2},⋯,*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.

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 *v*_{1},*v*_{2},⋯,*v*_{n} will be 4*n*+4, 4*n*+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,⋯,2*n*, 2*n*+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 \( ~\prod _{6} \).

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*=2*n*+1 and *q*=4*n*. 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

{ *e*_{1},*e*_{2},*e*_{3},⋯,*e*_{3n},*E*_{1},*E*_{2},*E*_{3},⋯,*E*_{n} } be the edges of FL(*n*) as in Fig. 20.

First, define the mapping *f*:*E*(*F**l*(*n*))→{2,4,⋯,8*n* } 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).