# Further results on edge even graceful labeling of the join of two graphs

## Abstract

In this paper, we investigated the edge even graceful labeling property of the join of two graphs. A function f is called an edge even graceful labeling of a graph G=(V(G),E(G)) with p=|V(G)| vertices and q=|E(G)| edges if f:E(G)→{2,4,...,2q} is bijective and the induced function f:V(G) →{0,2,4,,2q−2 }, defined as $$f^{\ast }(x) = ({\sum \nolimits }_{xy \in E(G)} f(xy)~)~\mbox{{mod}}~(2k)$$, where k=max(p,q), is an injective function. Sufficient conditions for the complete bipartite graph Km,n =mK1+nK1 to have an edge even graceful labeling are established. Also, we introduced an edge even graceful labeling of the join of the graph K1 with the star graph K1,n, the wheel graph Wn and the sunflower graph sfn for all $$n \in \mathbb {N}$$. Finally, we proved that the join of the graph $$\overline {K}_{2}~$$ with the star graph K1,n, the wheel graph Wn and the cyclic graph Cn are edge even graceful graphs.

## Introduction

A labeling of a graph is a mapping that carries graph elements (edges or vertices, or both) to positive integers subject to certain constraints. Recently, graph labeling has much attention from different researches in graph theory because it has rigorous applications in many disciplines, e.g., coding theory, X-ray, radar, communication networks, circuit design, astronomy, communication network addressing, and graph decomposition problems. For more interesting applications of graph labeling, see [13].

In graph theory, one can generate many new graphs from given ones by using graph operation. For a graph G, let q=|E(G)| be the cardinality of E(G) and p=|V(G)| be that of V(G). Let G and H be two graphs with no vertex in common. The join of G and H, denoted by G+H, defined to be the graph with vertex set and edge set given as follows: V(G+H)=V(G) V(H),E(G+H)=E(G)E(H){x1x2:x1V(G), x2V(H)}. If G and H are (p1,q1) and (p2,q2) graphs, respectively, then the number of vertices and edges in the join graph are p1+p2 and q1+q2+p1p2 [1].

Elsonbaty and Daoud [4] introduced a new type of labeling of a graph G with p vertices and q edges called an edge even graceful labeling if there is a bijection f from the edges of the graph to the set {2,4,, 2q } such that, when each vertex is assigned the sum of all edges incident to it mod 2k where k=max(p,q), the resulting vertex labels are distinct. The graph that admits edge even graceful labeling is called an edge even graceful graph. They introduced some path and cycle-related graphs which are edge even graceful, then Zeen El Deen [5] studied more graphs having an edge even graceful labeling.

Furthermore, Elsonbaty and Daoud [6] investigate edge even graceful labeling of cylinder grid graphs also, Daoud [7] studied the edge even graceful labeling of Polar grid graphs after that, Zeen El Deen and Omar N. [8] extended the edge even graceful labeling into r- edge even graceful labeling. For a summary of graph labeling, we refer to the dynamic survey by Gallian [9].

It should be noted that the join graph is not necessarily an edge even graceful graph. For example, the wheel graphW3=K1 +C3 is not an edge even graceful graph. In [4], they proved that the fan graphFn=K1 +Pn; n≥2 and Wn=K1 +Cn; n>3 are edge even graceful graphs. Now, we will study the edge even graceful labeling of the join of the graph K1 with the star graphK1,n, the wheel graphWn, and the sunflower graphsfn. Also, we will study the edge even graceful labeling of the complete bipartite graphKm,n =mK1+nK1. Since the double fan graph$$F_{2,n} = \overline {K}_{2}~+P_{n};~ n\geq 2~$$ is an edge even[5], so we will study the edge even graceful labeling of the join of the graph $$\overline {K}_{2}~$$ with the graphs K1,n, Cn and Wn.

## Edge even graceful labeling of the graph Kn,n=nk1+nK1

### Theorem 1

The complete bipartite graph Kn,n=nK1+ nK1 has an edge even graceful labeling when n>1 is an odd number.

### Proof

Let us use the standard notation p=|V(Kn,n)|=2n and q=|E(Kn,n)|=n2. The vertices of Kn,n were divided into two disjoint sets {v1,v2,,vn} {u1,u2,,un} such that every pair of graph vertices in the two sets are adjacent. There are three cases: Case (1) If n≡3 (mod 4),n>3, we define the function f:E(Kn,n)→{2,4,,2n2} as follows:

If i is an odd number, 1≤in

$$~~~~~~~~ f~(u_{i}~v_{j}) = \left \{\begin {array}{ll} ~(i-1)~n~+j+1~~~~ & \text {if}~~ j = ~1,3,\cdots, ~n ; \\[-3pt] ~2n^{2}- [(i-1)n+j]~~~ & \text {if } ~j=2,4,\cdots, ~n-1, \\[-3pt] \par \end {array}\right.$$

and if i is an even number, 2≤in−1

$$~~~~~~~~ f~(u_{i}~v_{j}) = \left \{\begin {array}{ll} ~(i-1)~n~+j+2~~~~ & \text {if}~~ j = ~1,3,\cdots, ~n-2 ; \\[-3pt] \par ~2n^{2}- [(i-1)n+j+1]~~~ & \text {if }~~ ~j=2,4,\cdots, ~n-1 ; \\[-3pt] \par ~2n^{2}- [(i-1)n+1]~~~ & \text {if }~~ ~j=n. \\[-3pt] \end {array}\right.$$

The following matrix X=(aij) shows the methods of labeling, where aij represents the label of the edge uivj.

{\begin{aligned} &\hspace{2.3cm} {v_{1}} \hspace{1.5cm} {v_{2}} \hspace{1.6cm} {v_{3}} \hspace{1.5cm} {v_{4}} \hspace{.7cm} \ldots \hspace{.5cm} {v_{n-2}} \hspace{1.2cm} {v_{n-1}} \hspace{1.5cm} {v_{n}} \\ X&= \begin{array}{c} {u_{1}}\\ {u_{2}}\\ {u_{3}}\\ {u_{4}}\\ \vdots\\ \vdots\\ {u_{n-2}}\\ {u_{n-1}}\\ {u_{n}}\\ \end{array} \left(\begin{array}{cccccccc} \!\!2 & \!\!2n^{2}-2 & \!\!4 & \!\!2n^{2}-4 & \!\!\ldots & \!\!n-1& \!\!2n^{2}-n+1 &n+1 \\ \!\!n+3 & \!\!2n^{2}-n-3 & \!\!n+5 & \!\!2n^{2}-n-5 & \!\!\ldots & \!\!2n& \!\!2n^{2}-2n &2n^{2}-n-1 \\ \!\!2n+2 & \!\!2n^{2}-2n-2 & \!\!2n+4 & \!\!2n^{2}-2n+4 & \!\!\ldots & \!\!3n-1& \!\!2n^{2}-3n+1 &3n+1 \\ \!\!3n+3 & \!\!2n^{2}-3n-3 & \!\!3n+5 & \!\!\!\!2n^{2}-3n-5 & \!\!\ldots & \!\!4n& 2n^{2}-4n &2n^{2}-3n-1\\ \!\!\vdots & \!\!\vdots& \!\!\ldots & \!\!\ldots & \!\!\vdots & \!\!\vdots & \ldots & \ldots \\ \!\! \vdots & \!\!\vdots& \!\!\ldots & \!\!\ldots & \!\!\vdots &\!\!\ddots & \ldots & \ldots \\ \!\! n^{2}-3n+2 & \!\!n^{2}+3n-2 & \!\!n^{2}-3n+4 & \!\!n^{2}+3n-4 & \!\!\ldots & \!\!n^{2}-2n-1& n^{2}+2n+1 &n^{2}-2n+1 \\ \!\!n^{2}-2n+3 & \!\!n^{2}+2n-3 & \!\!n^{2}-2n+5& \!\!n^{2}+2n-5 & \!\!\ldots & \!\!n^{2}-n& n^{2}+n &n^{2}+2n-1 \\ \!\! n^{2}-n+2 & \!\!n^{2}+n-2 &\!\!n^{2}-n+4 & \!\!n^{2}+n-4 & \!\!\ldots & \!\!n^{2}-1 & {2n^{2}}& {n^{2}+1}\\ \end{array}\right) \end{aligned}}

In this case, the equality of the labeling of the two vertices un and vn forces us to change the labels of the two edges unvn−1 and unvn, that is, f(unvn−1)=2n2 and f(unvn) =n2+1. Thus, the induced vertex labels are

1. (i)

The labels of the vertices ui, i=1,2,,n is the sum of rows in the matrix, i.e.,

$$~f^{\ast }(u_{i})= [~\sum _{j=1}^{n-1} ~f (u_{i}v_{j}) ~ ] ~\mbox{{mod}}~(2n^{2}) = f(u_{i} v_{n}),~$$ so the labels of the vertices u1,u2,u3,u4,,un−2,un−1 are n+1, 2n2n−1, 3n+1, 2n2−3n−1,,n2−2n+1,n2+2n−1, respectively, and f(un)=0.

2. (ii)

The labels of the vertices vi is the sum of columns in the matrix and since n≡3 (mod 4) n=3+4k 2q=2n2=32k2+48k+18, then, we have

$$~~~~~~~~~~ f^{\ast }(v_{i}) = \left \{\begin {array}{ll} \frac {2n^{2}+~(2i+3)n-1}{2} ~~~ & \text {if} ~~ i ~=1,3,\cdots, n-2 ~; ~ \\ ~\frac {2n^{2}-(2i+1)n+1~}{2}~~~~~ ~~~ & \text {if}~~ i =2,4,\cdots, n-3.\\ \end {array}\right.$$

Hence, the labels of the vertices v1,v2,v3,v4,,vn−3,vn−2 are $$~ \textstyle \frac {2n^{2}+5n-1}{2}, ~\textstyle ~\frac {2n^{2}-5n+1}{2}, \textstyle ~\frac {2n^{2}+9n-1}{2}, \textstyle ~\frac {2n^{2}-9n+1}{2},\cdots,~ \textstyle \frac {5n+1}{2}, \textstyle ~\frac {4n^{2}-n-1}{2}~$$, respectively.

Also$$~f^{\ast }(v_{n})= [~\sum _{i=1}^{n} ~f (u_{i}~v_{n}) ~ ] ~\mbox{{mod}}~(2n^{2}) = f(u_{n} ~v_{n}) = ~ n^{2}+1 ~$$ and

$$~~~ f^{\ast }(v_{n-1})= \textstyle [~\frac {-n^{3}+n^{2}+n-1 }{2}~ ] ~\mbox{{mod}}~(2n^{2}) =\textstyle [~\frac {- 2n^{2}+n-1}{2}~ ] ~\mbox{{mod}}~(2n^{2}) ~=~\frac {n^{2}+n-1}{2}$$. Case (2) If n=3, the graph K3,3 is an edge even graceful graph, see the following Fig. 1.Case (3) If n≡1 (mod 4). The labels of the edges incident to the vertices {ui, i=1,2,,n−1 } are similar to the first case, but there are some changes in the label of the edges in the last row in the matrix. The labels of the edges $$u_{n}v_{1},~u_{n}v_{2},~u_{n}v_{3},~u_{n}v_{4},~\cdots,~ u_{n}v_{\frac {n-3}{2}},~u_{n}v_{\frac {n-1}{2}}$$ are given by $$~ (n-1)n+2, ~ 2n^{2}-[(n-1)n+2],~(n-1)n+4,~2n^{2}-[(n-1)n+4], ~\cdots, ~n(n-1)+(\frac {n-1}{2}),~2n^{2}-[n(n-1)+(\frac {n-1}{2}) ]$$ and the edge $$~ u_{n}v_{\frac {n+1}{2}}~$$ label by 2n2. Finally, we swap the direction of labeling to start labels from the end of the row, so the edges $$u_{n}v_{n},~u_{n}v_{n-1},~u_{n}v_{n-2},~u_{n}v_{n-3},~\cdots,~ u_{n}v_{\frac {n+5}{2}},~u_{n}v_{\frac {n+3}{2}}$$ will label by $$(n-1)n+\frac {n+3}{2},~ 2n^{2}-[(n-1)n+\frac {n+3}{2}],~(n-1)n+\frac {n+7}{2}, ~ 2n^{2}-[(n-1)n+\frac {n+7}{2}],~\cdots,~(n+1)(n-1),~2n^{2}-[(n+1)(n-1) ]$$ as shown in the following matrix X=(aij).

{\begin{aligned} &\hspace{2.25cm} {v_{1}}\hspace{1.6cm} {v_{2}} \hspace{1.5cm} {v_{3}} \hspace{.75cm} \ldots \hspace{.35cm} {v_{\frac{n+1}{2}}} \hspace{.3cm} \ldots \hspace{.35cm} {v_{n-2}} \hspace{2cm} {v_{n-1}} \hspace{1.8cm} {v_{n}}\\ X&= \begin{array}{c} {u_{1}}\\ {u_{2}}\\ \vdots\\ \vdots\\ {u_{n-1}}\\ {u_{n}}\\ \end{array} \left(\begin{array}{ccccccccc} \!\!2 & \!\!2n^{2}-2 & \!\!4 & \!\!\ldots& \!\!\frac{n+3}{2} & \!\!\ldots & \!\!n-1& \!\!2n^{2}-n+1 &\!\!n+1 \\ \!\!n+3 & \!\!2n^{2}-(n+3) & \!\!n+5 & \!\!\ldots& \!\!\frac{3n+5}{2} & \!\!\ldots & \!\!2n& \!\!2n^{2}-(2n) 2n^{2}-n-1 \\ \!\!\vdots & \!\!\vdots& \!\!\ldots & \!\!\ddots & \!\!\vdots &\!\!\ddots & \!\!\vdots & \!\!\ldots & \!\!\ldots \\ \!\!\vdots & \!\!\vdots& \!\!\ldots & \!\!\ddots & \!\!\vdots &\!\!\ddots & \!\!\vdots & \!\!\ldots & \!\!\ldots \\ \!\!n^{2}-2n+3 & \!\!n^{2}+2n-3 & \!\!n^{2}-2n+5 & \!\!\ldots& \!\!\frac{(2n-1)n+5}{2} & \!\!\ldots & \!\!(n-1)n& \!\!n^{2}+n &\!\!n^{2}+2n-1 \\ n^{2}-n+2 & \!\!n^{2}+n-2 &\!\!n^{2}-n+4 & \!\!\ldots & \!\!{2n^{2}} & \!\!\ldots & \!\!{n^{2}-(\frac{n-7}{2})} &\!\! {n^{2}+ \frac{n-3}{2}}& \!\!{n^{2}- (\frac{n-3}{2})}\\ \end{array}\right) \end{aligned}}

The algorithm for the matrix X=(aij) of the graph Kn,n when n≡1(mod4) is shown below.

n is odd number. X→ square matrix [n×n]. for i 1 → n, for j 1 → n, if iodd & in−2. if jodd, X(i,j)= (i−1)n+j+1. else if jeven, X(i,j)= 2n2−[(i−1)n+j]. if i ==n, if $$~j~\longrightarrow ~~\mbox{{odd}}~~~~~ ~~\& ~~j\leq ~\frac {n-3}{2},~$$X(i,j)= (n−1)n+j+1. else if $$~~~~~j~\longrightarrow ~~\mbox{{even}}~~~~~ ~~\& ~~j\leq ~\frac {n-1}{2},$$X(i,j)= 2n2−[(i−1)n+j]. else if $$~ ~~~ j~=~\frac {n+1}{2}~,~~$$X(i,j)= 2n2. else if $$~j~\longrightarrow ~~\mbox{{even}}~~~~\& ~~ ~~\frac {n+3}{2} \leq ~j~\leq n-1,$$$$~~~~~~~~~~~~~X(i,j)=~n^{2}-(\frac {n+1}{2})+j.$$ else if $$~j~\longrightarrow ~~\mbox{{odd}}~~~~\& ~~ ~~\& ~~\frac {n+5}{2} \leq ~j~\leq ~n$$,$$~~~~~~~~~~~~~X(i,j)=~n^{2}+(\frac {n+3}{2})-j.~ ~$$

if ieven. if jodd & jn−2, X(i,j)= (i−1)n+j+2. else if jeven, X(i,j)= 2n2−[(i−1)n+j+1]. else if j ==n, X(i,j)= 2n2−[(i−1)n+1].

Thus, the induced vertex labels are

1. (i)

The labels of the vertices ui,i=1,2,,n is the sum of rows in the matrix, so the labels of these vertices are n+1, 2n2n−1, 3n+1, 2n2−3n−1,,n2−2n+1,n2+2n−1,0, respectively.

2. (ii)

The labels of the vertices vi is the sum of columns in the matrix and since n≡1 (mod 4) n=1+4k 2q=2n2=32k2+16k+2, then, we have

$$~~~~~~~~~~ f^{\ast }(v_{i}) = \left \{\begin {array}{ll} \frac {(2i+3)n-1}{2} ~~~ & \text {if} ~~ i ~=1,3,\cdots, \frac {n-3}{2} ~; ~ \\ ~\frac {4n^{2}-(2i+1)n+1~}{2}~~~~~ ~~~ & \text {if}~~ i =2,4,\cdots, \frac {n-1}{2}.\\ \end {array}\right.$$

Hence, the labels of the vertices $$~ v_{1},v_{2},v_{3},v_{4}, \cdots, v_{\frac {n-3}{2}},v_{\frac {n-1}{2}}$$ are

$$~\textstyle \frac {5n-1}{2}, \textstyle ~\frac {4n^{2}-5m+1}{2}, \textstyle ~\frac {9n-1}{2}, \textstyle ~\frac {4n^{2}-9m+1}{2},\cdots,\textstyle \frac {n^{2}+1}{2}, \textstyle \frac {3n^{2}-1}{2},~$$, respectively.

Also,$$~ f^{\ast }(v_{\frac {n+1}{2}})= \textstyle [~\frac {n^{3}-2n^{2}+5n-4}{2}~ ] ~\mbox{{mod}}~(2n^{2})$$

$$~~~~~~~~~~~~~~~~~~~=\textstyle [~\frac {- n^{2}+5n-4}{2}~ ] ~\mbox{{mod}}~(2n^{2})=\textstyle ~\frac {3 n^{2}+5n-4}{2}$$,

$$~f^{\ast }(v_{n})= [~\sum _{i=1}^{n} ~f (u_{i}~v_{n}) ~ ] ~\mbox{{mod}}~(2n^{2}) = f(u_{n} ~v_{n}) = ~\frac { 2n^{2}-n+3}{2}~$$,

$$~~~ f^{\ast }(v_{n-1})= [~\frac {-n^{3}+3n^{2}+2n-4}{2}~ ] ~\mbox{{mod}}~(2n^{2}) ~=~ n^{2}+n-2$$,

$$~~~ f^{\ast }(v_{n-2})= [~\frac {n^{3}+n^{2}-2n+8}{2}~ ] ~\mbox{{mod}}~(2n^{2}) ~=~ n^{2}- n+4$$,

$$~~~ f^{\ast }(v_{n-3})= [~\frac {-n^{3}-n^{2}+6n-12}{2}~ ] ~\mbox{{mod}}~(2n^{2}) ~=~ n^{2}+3n-6$$,

and

$$~~~ f^{\ast }(v_{n-4})= [~\frac {n^{3}+n^{2}-6n+16}{4}~ ] ~\mbox{{mod}}~(2n^{2}) ~=~ n^{2}-3n+8$$.

In the general case, we have $$~~ f^{\ast }(v_{n-i}) = \left \{\begin {array}{ll} ~n^{2}+in+2i ~~~ & \text {if }~~ i ~=1,3,\cdots, \frac {n-3}{2} ~; ~ \\ n^{2}-(i-1)n+2i~ ~~~~~ & \text {if}~~ ~ i =2,4,\cdots, \frac {n-4}{2}. \\ \end {array}\right.$$

Here, we notice that $$~ f^{\ast }(v_{n-i}) + f^{\ast }(v_{n-(i+1)}) = 2,~ ~i=1,2,3,\cdots,\frac {n-3}{2}$$

Clearly, all the label of the vertices are even and distinct. Hence, the graph Kn,n=nK1+ nK1 has an edge even graceful labeling. □

Illustration: we present an edge even graceful labeling of the graph K13,13 in the following matrix

\begin{aligned} & \hspace{1.1cm} {v_{1}} \quad {v_{2}} \quad {v_{3}} \quad {v_{4}} \quad {v_{5}} \quad {v_{6}} \quad {v_{7}} \quad {v_{8}} \quad\ {v_{9}} \quad {v_{10}} \quad {v_{11}} \ \ {v_{12}} \ \ \ {v_{13}}\\ &\begin{array}{c} {u_{1}}\\ {u_{2}}\\ {u_{3}}\\ {u_{4}}\\ {u_{5}}\\ {u_{6}}\\ {u_{7}}\\ {u_{8}}\\ {u_{9}}\\ {u_{10}}\\ {u_{11}}\\ {u_{12}}\\ {u_{13}}\\ \end{array} \left(\begin{array}{ccccccccccccc} 2 & 336 & 4 & 334 & 6 & 332 & 8 & 330 & 10 & 328& 12 & 326 & 14 \\ 16 & 322 & 18 & 320 & 20 & 318 & 22 & 316 & 24 & 314& 26 & 312 & 324 \\ 28 & 310 & 30 & 308 & 32 & 306 & 34 & 304 & 36 & 302& 38 & 300 & 40\\ 42 & 296 & 44 & 294 & 46 & 292 & 48 & 290 & 50 & 288& 52 & 286 & 298 \\ 54 & 284 & 56 & 282 & 58 & 280 & 60 & 278 & 62 & 276 & 64 & 244 & 66\\ 68 & 270 & 70 & 268 & 72 & 266 & 74 & 264 & 76 & 262 & 78 & 260 & 272\\ 80 & 258 & 82 & 256 & 84 & 254 & 86 & 252 & 88 & 250& 90 & 248 & 92 \\ 94 & 244 & 96 & 242 & 98 & 240 & 100 & 238 & 102 & 236 & 104 & 234 & 246\\ 106 & 232 & 108 & 230& 110 & 228 & 112 & 226 & 114 & 224 & 116 & 222 & 118 \\ 120 & 218 & 122 & 216 & 124 & 214 & 126 & 212 & 128 & 210 & 130 & 208 & 220\\ 132 & 206 & 134 & 204 & 136 & 202 & 138 & 200 & 140 & 198 & 142 & 196 & 144\\ 146 & 192 & 148 & 190 & 150 & 188 & 152 & 186 & 154 & 184 & 156 & 182 & 194\\ 158 & 180 & 160 & 178 & 162 & 176 & 338 & 170 & 168& 172 & 166 & 174 & 164 \\ \end{array}\right) ~~. \end{aligned}

### Theorem 2

The graph Km,n=mK1+ nK1 has an edge even graceful labeling when m and n are distinct odd numbers, m≠2n−3 and kmln, k=1,3,,n−1, l=1,3,,m−1.

### Proof

In the graph Km,n=mK1+ nK1, we have p=|V(Km,n)|=m+n and q=|E(Km,n)|=mn. Without loss of generality, assume that m > n and the vertices of Km,n divided into two disjoint sets {v1,v2,,vm} and {u1,u2,,un}. Put vi as the columns of the matrix and ui as the rows. We define the labeling function f:E(Km,n)→{2,4,,2mn} as follows: first, label the edges incident to the vertex u1, i.e., u1v1, u1v2, u1v3, u1v4,, u1vn−2, u1vn−1,u1vn by 2, 2mn−2, 4, 2mn−4, , m−1, 2mn−(n−1), n+1, respectively, then reverse the direction of labeling to label the edges incident to the vertex u2 as u2vm, u2vm−1, u2vm−2, u2vm−3, , u2v3, u2v2, u2v1 by 2mn−(n+1), m+3, 2mn−(n+3), n+5,, 2m+4, 2mn−(2m+2),2m+2 and so on.

The following matrix shows the methods of labeling

\begin{aligned} & \hspace{2cm} {v_{1}} \hspace{1.7cm} {v_{2}} \hspace{2cm} {v_{3}} \hspace{2cm} {v_{4}} \hspace{.9cm} \ldots \hspace{.7cm} {v_{m-2}} \hspace{1.7cm} {v_{m-1}} \hspace{1.5cm} {v_{m}} \\ &\begin{array}{c} {u_{1}}\\ {u_{2}}\\ {u_{3}}\\ {u_{4}}\\ \vdots\\ \vdots\\ {u_{n-1}}\\ {u_{n}}\\ \end{array} \left(\begin{array}{cccccccc} 2 & 2mn-2 & 4 & 2mn-4 & \ldots & m-1& 2mn-m+1 &m+1 \\ 2mn-2m & 2m & 2mn-2m+2 & 2mn-2 & \ldots & 2mn-m-3& m+3 & 2mn-m-1 \\ 2m+2 & 2mn-2m-2 & 2m+4 & 2mn -2n-4 & \ldots & 3m-1& 2mn- 3n+1 &3n+1 \\ 4m & 2mn-4m+2 & 4m-2 & 2mn-4m+4 & \ldots & 2mn-3m-3& 3m+3 &2mn-3n-1 \\ \vdots & \vdots& \ldots & \vdots &\ddots & \vdots & \ldots & \ldots \\ \vdots & \vdots& \ldots & \vdots &\ddots & \vdots & \ldots & \ldots \\ mn+m & n m-m & mn+m+2 & n m -m-2 & \ldots & mn+2m-3& n m-2m+3 &mn+2m-1 \\ nm-m+2 & mn+m-2 & nm-m+4 & mn+m-4 & \ldots & mn-1 & {2mn}& {mn+1}\\ \end{array}\right). \end{aligned}

In this case, the label of the vertex un will repeat with the labels of the vertex vm. To avoid this problem we replace the labels of the two edges unvm−1 and unvm, that is f(unvm−1)=2mn and f(unvm) =nm+1. Thus, the induced vertex labels are

1. (i)

The labels of the vertices ui is the sum of rows in the matrix, so the labels of these vertices are m+1, 2mnm−1, 3m+1, 2mn−3m−1,,(n−2)m+1,mn+2n−1, 0, respectively.

2. (ii)

The labels of the vertices vi,i=1,2, ,m is the sum of columns in the matrix, we have

$$~~~~~~~~~~ f^{\ast }(v_{i}) = \left \{\begin {array}{ll} in+1 ~~~ & \text {if }~~ i ~=1,3,,5\cdots, m-2 ~; ~ \\ ~2mn-(i-1)n-1~~~~~~ ~~~ & \text {if}~~ i =2,4,\cdots, m-3. \\ \end {array}\right.$$

Hence, the labels of the vertices v1,v2,v3,v4,,vm−3,vm−2 are n+1, 2mn−(n+1), 3n+1, 2mn−(3m+1), , 2mn−[(m−4)n+1], (m−2)n+1, respectively, and f(vm−1)=2n−2, f(vm)=nm+1. □

Illustration: If we take m=15, then we can label the graphs K15,7, K15,11, and K15,13, while we can not find labels of K15,3, K15,5, and K15,9. We present an edge even graceful labeling of the graph K15,7=15K1+7 K1 in the following matrix

\begin{aligned} &\hspace{1cm} {v_{1}} \hspace{.3cm} {v_{2}} \hspace{.4cm} {v_{3}} \hspace{.4cm} {v_{4}} \hspace{.3cm} {v_{5}} \hspace{.3cm} {v_{6}} \hspace{.3cm} {v_{7}} \hspace{.3cm} {v_{8} } \hspace{.3cm} {v_{9}} \hspace{.3cm} { v_{10}} \hspace{.3cm} {v_{11}} \hspace{.3cm} {v_{12}} \hspace{.3cm} {v_{13}} \hspace{.3cm} {v_{14}} \hspace{.3cm} {v_{15}}\\ &\begin{array}{c} {u_{1}}\\ {u_{2}}\\ {u_{3}}\\ {u_{4}}\\ {u_{5}}\\ {u_{6}}\\ {u_{7}}\\ \end{array} \left(\begin{array}{ccccccccccccccc} 2 & 208 & 4 & 206 & 6 & 204 & 8 & 202 & 10 & 200& 12 & 198 & 14 & 196 & 16 \\ 180 & 30 & 182 & 28 & 184 & 26 & 186 & 24 & 188 & 22& 190 & 20 & 192 & 18 & 194 \\ 32 & 178 & 34 & 176 & 36 & 174 & 38 & 172 & 40 & 170& 42 & 168 & 44 & 166 & 46 \\ 150 & 60 & 152 & 58 & 154 & 56 & 156 & 54 & 158 & 52& 160 & 50 & 162 & 45 & 164 \\ 62 & 148 & 64 & 146 & 66 & 144 & 68 & 142 & 70 & 140 & 72 & 138 & 74 & 136 & 76 \\ 120 & 90 & 122 & 88 & 124& 86 & 126 & 84 & 128 & 82 & 130 & 80 & 132 & 78 & 134 \\ 92 & 118 & 94 & 116 & 96 & 114 & 98 & 112 & 100 & 11 0& 102 & 108 & 104 &210 & 106 \\ \end{array}\right). \end{aligned}

## Edge even graceful labeling of the join graph k1+k1,n.

### Theorem 3

The graph K1+K1,n has an edge even graceful labeling.

### Proof

Let {v0,v1,v2, , vn} be the vertices of the graph K1,n with central vertex v0 and {x} be the vertex of K1 so the edges of K1+K1,n are {xv0, xvi, v0vi,i=1,2,, n }. Here, p=|V(K1+K1,n)|=n+2 and q =|E(K1+K1,n)|=2n+1. There are two cases:

• when n is even. We define the labeling function f:E(K1+K1,n)→{2,4,,4n+2} as follows:

f(xv0) =2n,

$$~ f~(x~v_{i}) = \left \{\begin {array}{ll} 2i~~~ & \text {if }~~ i=1,2,\cdots ~, \frac {n}{2} ~; \\ 2n+2i ~~~ & \text {if}~~ i =~\frac {n}{2}+1,\cdots, n ~.~ ~ \\ \end {array}\right.$$

and

$$~f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} 2n+2i ~~~ & \text {if}~~ i=1,2,\cdots ~, \frac {n}{2} ~;~ ~ \\ 4n+2~~~ & \text {if} ~~ i =~\frac {n}{2}+1~; \\ 2i-2~~~ & \text {if} ~~ i =~\frac {n}{2}+2,\cdots, ~ n ~. \\ \end {array}\right.$$

Therefore, the induced vertex labels are

\begin{aligned} f^{\ast}(x)&= [~ f(x~ v_{0})~+\sum_{i=1}^{n} f (x v_{i}~) ~ ] ~{\rm{mod}}~(4n+2)=~ f(x ~v_{0}~)~{\rm{mod}}~(4n+2)=~2n,\\ f^{\ast}(v_{0})&= [~ f(x ~v_{0})~+\sum_{i=1}^{n} f (v_{0}~v_{i}~) ~ ] ~{\rm{mod}}~(4n+2)\\&=~ [f(x ~ v_{0}~)+ f(a_{1})]~{\rm{mod}}~(4n+2)=~0,\\ ~f^{\ast}(v_{i})&= [ f (x v_{i})+ f(v_{0}~v_{i})] ~{\rm{mod}}~(4n+2) \end{aligned}

$$~~~~~~~=\left \{\begin {array}{ll} (2n+4i) ~\mbox{{mod}}~(4n+2)~~~ & \text {if}~~1 \leq i \leq \frac {n }{2}; \\ (2n+ 4i-2) ~\mbox{{mod}}~(4n+2) ~~~ & \text {if} ~~~~~ \frac {n }{2}+2 \leq i < n. \\ \end {array}\right.$$

Hence, the labels of the vertices $$v_{1},v_{2},~ \cdots, v_{\frac {n}{2}-1},v_{\frac {n}{2}}~$$ are 2n+4,2n+8, , 4n−4, 4n, respectively, and the labels of the vertices $$~v_{\frac {n}{2}+1},~ v_{\frac {n}{2}+2},~v_{\frac {n}{2}+3},~ \cdots,~ v_{n-1},~v_{n}$$ are 3n+2, 4,8 ,2n−8, 2n−4, respectively. Finally, $$~ f^{\ast }(v_{\frac {n}{2}+1}) = [ f (x~v_{\frac {n}{2}+1})+ f(v_{0}~v_{\frac {n}{2}+1})] ~\mbox{{mod}}~(4n+2) =~3n+2$$.

• when n is odd. We define the labeling function f:E(K1+K1,n)→{2,4,,4n+2} as follows:

f(xv0) =n+1,

$$f~(x~v_{i}) = \left \{\begin {array}{ll} 2i~~~ ~& \text {if }~~ i=1,2,\cdots ~, \frac {n-1}{2} ; \\ 2n+2i ~~~~~ & \text {if}~~ i =~\frac {n+1}{2},\cdots, n. \\ \end {array}\right.$$

and

$$~f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} n+1+2i ~~~ & \text {if}~~ i=1,2,\cdots ~, \frac {n-1}{2} ~;~ ~ \\ 4n+2~~~ & \text {if} ~~ i =~\frac {n+1}{2}~ ; \\ n-1+2i~~~ & \text {if} ~~ i =~\frac {n+3}{2},\cdots, ~ n ~. \\ \end {array}\right.$$

Considering the vertex labels we find

$$f^{\ast }(v_{0})= [ f(x ~v_{0})~+\sum _{i=1}^{n} f (v_{0}v_{i}~) ] ~\mbox{{mod}}~(4n+2)=~n+1,~~~ f^{\ast }(x)= ~ 0~$$ and

$$~~~~~~~~~~~~~~~~~f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (n+1+4i) ~\mbox{{mod}}~(4n+2)~~~ & \text {if}~~ 1 \leq i \leq \frac {n -1}{2}; \\ (3n-1+ 4i) ~\mbox{{mod}}~(4n+2) ~~~ & \text {if} ~~~~~ \frac {n+3 }{2} \leq i \leq n. \\ \end {array}\right.$$

Then, the labels of the vertices $$v_{1},v_{2},~ \cdots, v_{\frac {n-1}{2}},v_{\frac {n+1}{2}} ~$$ are n+5,n+9, , 3n−1, 3n+1, respectively, and the labels of the vertices $$v_{\frac {n+3}{2}}, v_{\frac {n+5}{2}},~ \cdots,~ v_{n-1},v_{n}$$ are n+3, n+7 ,3n−7, 3n−3, respectively. Finally, $$~f^{\ast }(v_{\frac {n+1}{2}}) = [ f (e_{\frac {n+1}{2}})+ f(a_{\frac {n+}{2}})] ~\mbox{{mod}}~(4n+2) ~ =~3n+1$$.

Overall, all vertex labels are distinct even numbers, also f(x) and f(v0) are different from all the labels of the vertices vi. Hence, the graph K1+K1,n is edge even graceful for all n. □

Illustration: In Fig. 2, we present an edge even graceful labeling of the graphs K1+ K1,8 and K1+ K1,9

## Edge even graceful labeling of the join graph $$\protect \phantom {\dot {i}\!}~K_{1}+ w_{{~}_{n}}$$

### Theorem 4

The join graph K1+ Wn has an edge even graceful labeling for all n.

### Proof

Let {v0,v1,v2, , vn} be the vertices of Wn with central vertex v0 and {x} be the vertex of K1 so the edges of K1+ Wn will be {xv0, xvi, v0vi,vivi+1,i=1,2,, n }. So, p=|V(K1+ Wn)|=n+2 and q=|E(K1+ Wn)|=3n+1. There are five cases:

• For n≡0 (mod 6), orn≡4 (mod 6), we define the labeling function f:E(K1+ Wn)→{2,4,,6n+2} as follows:

f(v0x)=6n+2,

f(v1vn)=3n, f(vivi+1)=n+2ifori=1,2,n−1,

$$f~(x~v_{i}) = \left \{\begin {array}{ll} ~3n+2~~~ & \text {if} ~~ i=1~;~ ~ \\ ~5n-2i+4 ~~~ & \text {if}~~ 2 \leq i \leq n ~.~ ~ \\ \end {array}\right.$$,

and

$$~f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~2i ~~~~~ ~~& \text {if}~~ 1 \leq i \leq \frac {n}{2} ~;~ ~\\ ~4n+2i~~~~~~~ & \text {if} ~~ \frac {n}{2} < i \leq ~~ n~. \\ \end {array}\right.$$

Then, the induced vertex labels are

$$~~ f^{\ast }(v_{0})= [~\sum _{i=1}^{n} f (v_{0}~v_{i}) +f(v_{0}~x)~ ] ~\mbox{{mod}}~(6n+2)=~0~$$ and

$$~~~f^{\ast }(x) = [\sum _{i=1}^{n} f (x~v_{i}) + f(x v_{0})] \mbox{{mod}}~(6n+2)$$

$$~~~~~~~~~~~=[\sum _{i=1}^{n} (3n+2i)~+(6n+2) ] ~\mbox{{mod}}~(6n+2)=(4n^{2}+n)~\mbox{{mod}}(6n+2)$$.

If n≡0 (mod 6) n=6k 2q=6n+2=36k+2, then,

$$\begin{array}{*{20}l} f^{\ast}(x) &= [~4~(6k)^{2}+ (6k) ~]~ {\rm{mod}}~(36 k+2~)~\\&=[ ~4k (36k+2) - (2k)~ ]~ {\rm{mod}}~(36 k+2~)~ \\ &\equiv ~(-2k) ~{\rm{mod}}~(36 k+2~) \equiv (34k+2)~{\rm{mod}}~(36 k+2~) \\&\quad=~\left(\frac{17n+6}{3} \right). ~ \end{array}$$

Similarly, if n≡4 (mod 6) n=6k+4 2q=6n+2=36k+26, then,

$$f^{\ast }(x) = [~4~(6k+4)^{2}+ (6k+4) ~]~ \mbox{{mod}}~(36 k+26~)~= ~\left (\frac {11n+4}{3} \right). ~$$

Also, f(v1)=[f(v1v2)+f(vnv1)+f(xv1)+f(v0v1)] mod (6n+2) = n+4

and f(vi)=[f(vivi+1)+f(vi−1vi)+f(xvi)+f(v0vi)] mod (6n+2)

$$~~~~~~~~~~~~~~ = \left \{\begin {array}{ll} (n+4i) ~\mbox{{mod}}~(6n+2)~~ & \text {if}~~ ~2 \leq i \leq \frac {n }{2}; \\ (5n+4i) ~\mbox{{mod}}~(6n+2) ~~ & \text {if }~~~\frac {n }{2}+1 \leq i \leq n. \\ \end {array}\right.$$

Hence, the labels of the vertices $$v_{1},v_{2},v_{3}, \cdots, v_{\frac {n}{2}}~$$ will be n+4,n+8,n+12,,3n, respectively, and the labels of the vertices $$v_{\frac {n}{2}+1}, v_{\frac {n}{2}+2}, \cdots, v_{n-1},v_{n} ~$$ will be n+2,n+6,,3n−6,3n−2, respectively, which are all even and distinct numbers.

• For n≡2 (mod 6), we define the labeling function f as follows:

f(xv0)=6n+2,

$$~f~(x~v_{i}) =\left \{\begin {array}{ll} ~2i ~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2} ~;~ ~ ~~~~~~\\ ~4n+2i~~~ ~~& \text {if }~~ \frac {n}{2} < i \leq ~~ n~,~~~~~ \\ \end {array}\right.$$

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

and

$$~f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~5n ~~~ & \text {if}~~ i=1 ~;~ ~ ~~~~~\\ ~3n+4-2i~~~ & \text {if} ~~ 2 \leq i \leq ~ n~.~~~~~ \\ \end {array}\right.$$

In view of the above labeling patten and since n≡2 (mod 6) n=6k+2 2q=6n+2=36k+14 then the induced vertex labels are

f(v0)= (2n2+5n−2) mod (6n+2)=[ 2k(36k+14)+(94k+68) ] mod (36k+14)

$$~~~~~~~~\equiv (14k+2) ~\mbox{{mod}}~(36 k+14~) =(\frac {7n-8}{3}). ~$$

By the same way, in the first case, we have

$$~~~~~~~~f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (3n+4i) ~\mbox{{mod}}~(6n+2)~~~ & \text {if}~~ 2 \leq i \leq \frac {n }{2}; \\ (n+4i-2) ~\mbox{{mod}}~(6n+2) ~~~ & \text {if }~~~~~ \frac {n }{2}+1 \leq i < n. \\ \end {array}\right.$$

Finally, f(x)= 0, f(v1)= 3n+4 and f(vn)= n.

Therefore, the labels of the vertices $$v_{1},v_{2},v_{3}, \cdots, v_{\frac {n}{2}}~$$ are 3n+4,3n+8,3n+12,,5n, respectively, and the labels of the vertices $$v_{\frac {n}{2}+1}, v_{\frac {n}{2}+2}, \cdots, v_{n-1},v_{n} ~$$ are 3n+2,3n+6,, 5n−6, n, respectively.

• For n≡3 (mod 6), we define the labeling f:E(K1+ Wn)→{2,4,,6n+2} as follows:

f(xv0)=6n, f(xvi)=3n+3−2ifori=1,2,n,

$$f~(v_{i}~v_{i+1}) = \left \{\begin {array}{ll} ~3n+1+2i~~~~ & \text {if} ~~ i=1,2,\cdots,~n-1 ; \\ ~ 6n+2 ~~~ ~ & \text {if}~~ i = n, \\ \end {array}\right.$$

and

$$f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~2i ~~~ ~~ & \text {if}~~ 1 \leq i \leq \frac {n+1}{2} ;~~ ~~~ \\ ~4n+2i-2~~~~ & \text {if} ~~ \frac {n+3}{2} \leq i \leq n.\\ \end {array}\right.$$

In this labeling, the induced vertex labels are

$$~~~~f^{\ast }(v_{0})= [\sum _{i=1}^{n} f (v_{0}~v_{i}) +f(x~v_{0}) ] ~\mbox{{mod}}~(6n+2)=~0,$$

f(v1)=[f(v1v2)+f(vnv1)+f(v0v1)+f(xv1)] mod (6n+2) = 4,

$$~~f^{\ast }(x)= [~\sum _{i=1}^{n} f (x~v_{i}) ~+ f(v_{0}~x)~] ~\mbox{{mod}}~(6n+2) =~(2n^{2}+2n-2)~\mbox{{mod}}~(6n+2)$$.

Since n≡3 (mod 6) n=6k+3 2q=6n+2=36k+20.

Then, $$~~~f^{\ast }(x) ~= \left (\frac {4n-6}{3}\right). ~$$

And f(vi)=[f(vivi+1)+f(vi−1vi)+f(v0vi)+f(xvi)] mod(6n+2)

$$~~~~~~~~~~= \left \{\begin {array}{ll} (3 n+4i+1) \mbox{{mod}} (6n+2) ~~ & \text {if}~~~~2 \leq i \leq \frac {n+1 }{2} ; \\ (n+4i-3) \mbox{{mod}} (6n+2)~ & \text {if} ~~ ~\frac {n+3 }{2} \leq i \leq n-1. \\ \end {array}\right.$$

Finally, f(vn)=[f(vn−1vn)+f(vnv1)+f(v0vn)+f(xvn)] mod (6n+2)= 6n−2.

Hence, the labels of the vertices $$v_{1},v_{2},v_{3}, \cdots, v_{\frac {n-1}{2}},v_{\frac {n+1}{2}} ~$$ are 4,3n+9,3n+13,,5n−1,5n+3, respectively, and the labels of the vertices $$~v_{\frac {n+3}{2}},v_{\frac {n+5}{2}}, v_{\frac {n+7}{2}}, \cdots, v_{n-1},v_{n}~$$ are 3n+3,3n+7,3n+11,,5n−7,6n−2, respectively. There is no repetition in the vertex label, also f(x) and f(v0) are different from all the labels of the vertices vi.

• For n≡5 (mod 6), n > 5, we define the labeling function f as follows:

f(xv0)=2, f(xvi)=(3n+3)−2ifori=1,2,n,

$$f~(v_{i}~v_{i+1}) = \left \{\begin {array}{ll} ~3n+1 +2i~~~ & \text {if} ~~ i=1,2,\cdots,n-1 ; \\ ~6n+2 ~~~ & \text {if}~~ i = n, \\ \end {array}\right.$$

and

$$f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~ 2i+2~ ~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1}{2} ; \\ ~4n+2i~~~~ & \text {if} ~~ \frac {n+1}{2} \leq i \leq n. \\ \end {array}\right.$$

In view of the above labeling pattern, the induced vertex labels are

Since n≡5 (mod 6) n=6k+5 2q=6n+2=36k+30, then

$$~ f^{\ast }(x) =~(2n^{2}+2n+2)~\mbox{{mod}}~(6n+2)= (\frac {16n+10}{3}) ~ \mbox{{mod}}~(6 n+2~).$$

Also, $$~~~f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (3 n+4i+3) ~\mbox{{mod}}~(6n+2)~~~ & \text {if} ~~ 2 \leq i \leq \frac {n-1 }{2} ; \\ (n+4i-1) ~\mbox{{mod}}~(6n+2)~~~ & \text {if}~~ \frac {n+1 }{2} \leq i < n. \\ \end {array}\right.$$

Finally, f(v0)= 0, f(v1)= 6 and f(vn)= 6n.

Hence, the labels of the vertices $$v_{1},v_{2},v_{3}, \cdots,~v_{\frac {n-3}{2}},~ v_{\frac {n-1}{2}}~~$$ are 6,3n+11,3n+15,,5n−3,5n+1, respectively, and the labels of the vertices $$~ v_{\frac {n+1}{2}},~ v_{\frac {n+3}{2}},~ ~ \cdots, v_{n-1},v_{n} ~~$$ are 3n+1, 3n+5, , 5n−5, 6n, respectively. Clearly, f(x) and f(v0) are different from all the labels of the vertices vi.

When n=5, the graph E(K1+ W5) is an edge even graceful graph but it does not follow this rule, see Fig. 4.

• For n≡1 (mod 6), we defined the labeling function f as follows:

f(v0x)=6n,

f(v1vn)=4n, f(vivi+1)=2n+2ifori=1,2,n−1,

$$f~(x~v_{i}) = \left \{\begin {array}{ll} 2i~~~ & \text {if} ~~ 1 \leq i \leq ~\frac {n+1}{2}; \\ 4n-2+2i ~~~ & \text {if}~~ \frac {n+3}{2} \leq i \leq n, \\ \end {array}\right.$$

and

$$~ f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} 5n+1-2i ~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1}{2} ; \\ 6n+2 ~~~ & \text {if}~~ i = \frac {n+1}{2} ;~~ ~ \\ 3n+3-2i~~~ & \text {if} ~~ \frac {n+3}{2} \leq i \leq n. \\ \end {array}\right.$$

In this case, the induced vertex labels are

f(x)= 0, f(v0)= 6n, f(v1) = 5n+1,

and

$$f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (3 n+4i-3) ~\mbox{{mod}}~(6n+2)~ & \text {if} ~ 2 \leq i \leq \frac {n-1 }{2} \\ (5n+4i-3) ~\mbox{{mod}}~(6n+2)\equiv ~(4i-n-5)~ & \text {if}~ ~\frac {n+3 }{2} \leq i \leq n \\ \end {array}\right.$$

Finally,

\begin{aligned} f^{\ast}(v_{\frac{n+1}{2}}) = \left[ f \left(v_{\frac{n+1}{2}}v_{\frac{n+3}{2}}\right)+ f \left(v_{\frac{n-1}{2}}v_{\frac{n+1}{2}}\right)+ f\left(v_{0}v_{\frac{n+1}{2}}\right)+ f\left(xv_{\frac{n+1}{2}}\right)\right] {\rm{mod}}(6n+2)=~n-1. \end{aligned}

It is clear that for all i{1,2,3,...,n}, the labels of the vertices vi are all distinct, even and different from f(x) and f(v0) which complete the proof. □

Illustration: In Fig. 3, we present an edge even graceful labeling of the graphs K1+ W8 and K1+ W10.

Illustration: In Fig. 4, we present an edge even graceful labeling of the graphs K1+ W9, K1+ W11, K1+ W5 and K1+ W7.

## Edge even graceful labeling of the join graph K1+sfn

The sunflower graph, sfn, is defined as a graph obtained by starting with an n-cycle Cn with a consecutive vertices v1,v2,,vn and creating new vertices u1,u2,,un, with ui connected to vi and vi+1. The graph sfn has number of vertices p=2n and a number of edges q=3n.

### Theorem 5

The join graph K1+sfn has an edge even graceful labeling for all n.

### Proof

Let { x } be the vertex of K1 and the edges of the graph K1+ sfn will be

{ xvi, xui, vivi+1, viui, vi+1ui, i=1,2,,n }. Let us use the standard notation p=|V(K1+ sfn)|=2n+1 and q =|E(K1+ sfn)|=5n.

We define the labeling function f:E(K1+ sfn)→{2,4,,10n} as follows:

fori=1, 2, , n, f(xvi)=2i, f(xui)=10 n−2i,

fori=1, 2, , n, f(viui)=4n+2i, f(uivi+1)=2n+2i,

and

$$~~f~(v_{1}~v_{n}) =6n+2,~~~~~~~~ f~(v_{i}~v_{i+1}) = \left \{\begin {array}{ll} ~10~n~~~~~ & \text {if}~~ i = ~1~; ~ \\ ~8n+2-2i~~~ & \text {if} ~~ ~i=2,3,\cdots, ~n -1. \\ \end {array}\right.$$

Considering the given vertex labels, the induced vertex labels are

$$~~~~f^{\ast }(x) = [ \sum _{i=1}^{n} f (x~v_{i}) + \sum _{i=1}^{n} f (x~u_{i})] ~\mbox{{mod}}~(10~n) ~= 0$$,

f(v1)=4n+6, f(v2)=4n+8,

f(vi)=2n+2i+4 fori=3, 4, , n,

and

f(ui)=6n+2ifori=1,2, , n.

Thus, the labels of the vertices v3, v4, , vn are 2n+10, 2n+12, , 4n+4, respectively, and the labels of the vertices u1, u2, , un are 6n+2, 6n+4, , 8n, respectively. Clearly, all the labels are even and distinct. Thus, the graph K1+ sfn is an edge even graceful labeling. □

Illustration: In Fig. 5, we present an edge even graceful labeling of the graph K1+ sf8.

## Edge even graceful labeling of the join graph $$~\overline {K}_{2}+K_{_{1,n}}$$

### Theorem 6

The graph $$~\overline {K}_{2} + ~K_{1,n}$$ has an edge even graceful labeling for all n.

### Proof

Let {v0,v1,v2, , vn} be the vertices of the graph K1,n with central vertex v0 and {x, y} be the vertices of $$~\overline {K}_{2}$$ so the edges of $$\overline {K}_{2} + K_{1,n}$$ are {xv0, xvi, v0vi,yv0,yvi, i=1,2,, n }. Let us use the standard notation $$p= | V (~\overline {K}_{2} + ~K_{1,n})| = n+3 ~$$ and $$~ q~= | E (~\overline {K}_{1} + ~K_{1,n})| =3n+2~$$. There are two cases:

• when n is even. We define the labeling function $$~f:E(\overline {K}_{2} + ~K_{1,n})\longrightarrow \{2,4,\cdots,6n+4\}$$ as follows:

$$~~~~~~~~~~~~~f(x ~v_{i}~)= \left \{\begin {array}{ll} 6n+4~~~ & \text {if} ~~ i=~0 ~; \\ 2i ~~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2} ; \\ 4n+2+2i ~~~~ & \text {if}~~ \frac {n}{2}+1 \leq i \leq ~~ n, \\ \end {array}\right.$$

$$~~~~~~~~ ~~~~~f(y ~v_{i}~)= \left \{\begin {array}{ll} ~ 3n+4~~~~~ & \text {if} ~~ i=~0 ~; ~~\\ ~ n+2i ~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2} ;~~~~ \\ ~ 3n+2+2i~~~~~~~ & \text {if} ~~ \frac {n}{2}+1 \leq i \leq ~~ n ~,~ \\ \end {array}\right.$$

and

$$~~~~~~~~~~~~ f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~2n+2i~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2}+1 ; \\ ~2n+2+2i~~ & \text {if} ~~ ~\frac {n}{2}+2 \leq i \leq n. \\ \end {array}\right.$$

Considering the given vertex labels, the induced vertex labels are

$$f^{\ast }(x)= [~ f(x~ v_{0})~+\sum _{i=1}^{n} f (x~ v_{i}~) ~ ] ~\mbox{{mod}}~(6n+4)=~ f(x ~v_{0}~)~\mbox{{mod}}~(6n+4)=~0$$,

$$f^{\ast }(y)= [ f(y v_{0})+\sum _{i=1}^{n} f (y v_{i}) ] \mbox{{mod}}(6n+4)= f(y v_{0})\mbox{{mod}}~(6n+4)=3n+4$$,

$$~~~f^{\ast }(v_{0}) = [ \sum _{i=1}^{n} f (v_{0}~v_{i}) + f(x~v_{0})+f(y~v_{0})] ~\mbox{{mod}}~(6n+4) ~= 3n+2$$,

f(vi)=[f(v0vi)+f(xvi)+f(yvi)] mod (6n+4)

$$~~~~~~~~~= \left \{\begin {array}{ll} (3 n+6i) ~\mbox{{mod}}~(6n+4)~~~ & \text {if}~~ ~2 \leq i \leq \frac {n }{2}; \\ (3n+2+6i) ~\mbox{{mod}}~(6n+4) ~~~ & \text {if} ~~~~~\frac {n }{2}+2 \leq i \leq n. \\ \end {array}\right.$$

and $$~~~ f^{\ast }(v_{\frac {n}{2}+1}) = \left [ f \left (v_{0}v_{\frac {n}{2}+1}\right)+ f\left (x~v_{\frac {n}{2}+1}\right)+f\left (y~v_{\frac {n}{2}+1}\right)\right ] ~\mbox{{mod}}~(6n+4) ~ =~2$$.

Thus, the labels of the vertices $$v_{1},~v_{2},~ \cdots, ~ v_{\frac {n}{2}-1}, ~ v_{\frac {n}{2}}~~$$ are 3n+6, 3n+12, , 6n−6, 6n, respectively, and the labels of the vertices $$~v_{\frac {n}{2}+1},~ v_{\frac {n}{2}+2},~ v_{\frac {n}{2}+3},~ \cdots, ~v_{n-1},~v_{n} ~~$$ are 2, 10, 16, , 3n−8, 3n−2, respectively. Clearly, f(x), f(y) and f(v0) are different from all the labels of the vertices vi.

• when n is odd. We introduce two different labeling

Method 1: We define the labeling function f as follows:

$$~~~~~~~~~~~~~~~f~(v_{0}~v_{i}) =\left \{\begin {array}{ll} ~2n+2+2i~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1}{2} ; \\ ~5n+1-2i~~~ & \text {if} ~~ ~\frac {n+1}{2} \leq i \leq ~n-1 ; \\ ~6n+2~~~ & \text {if} ~~ ~ i =~n~,~~ \\ \end {array}\right.$$,

$$~~~~~~~~~~~~~~~~f(x ~v_{i}~)= \left \{\begin {array}{ll} ~4~~~ & \text {if} ~~ i=~0 ~; \\ 4+2i ~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1}{2} ~;~\\ 4n+2i ~~~~~ & \text {if}~~ \frac {n+1}{2} \leq i \leq ~~ n, ~~ \\ \end {array}\right.$$

and

$$~~ ~~~~~~~~~~~~~f(y ~v_{i}~)= \left \{\begin {array}{ll} ~ 2~~~~~ & \text {if} ~~ i=~0 ~; ~~\\ ~ 5n+1-2i ~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1}{2} ; \\ ~ 3n+3-2i~~~~~~~ & \text {if} ~~ \frac {n+1}{2} \leq i \leq ~~ n-1 ~;~ ~\\ ~ 6n+4~~~~~ & \text {if} ~~ i=~n.~ ~~\\ \end {array}\right.$$

Hence, the induced vertex labels are

\begin{aligned} f^{\ast}(x)&= \left[ f(x~ v_{0})~+\sum_{i=1}^{n} f (x~ v_{i}) \right] {\rm{mod}}~(6n+4)= f(x ~v_{0}~)~{\rm{mod}}~(6n+4)=~0,\\ ~~~f^{\ast}(y)&= \left[ f(y ~v_{0})~+\sum_{i=1}^{n} f (y ~v_{i}~) ~ \right] ~{\rm{mod}}~(6n+4)=~ f(y ~ v_{0}~)~{\rm{mod}}~(6n+4)=2,\\ f^{\ast}(v_{0}) &= \left[ \sum_{i=1}^{n} f (v_{0}v_{i}) + f(xv_{0})+f(yv_{0})\right] {\rm{mod}}(6n+4) = ~4, \end{aligned}

$$\hspace {-6pt} f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (n+3+2i) ~\mbox{{mod}}~(6n+4)~~~ & \text {if}~~1 \leq i \leq \frac {n-1 }{2}; \\ (6n-2i) ~\mbox{{mod}}~(6n+4) ~~~ & \text {if} ~~~~~ \frac {n+1 }{2} \leq i \leq n-1~. \\ \end {array}\right.$$

and f(vn)=[f(v0vn)+f(xvn)+f(yvn)] mod (6n+4) = 6n−2.

Then the labels of the vertices $$v_{1},~v_{2},~ \cdots, ~ v_{\frac {n-3}{2}}, ~ v_{\frac {n-1}{2}}~~$$ are n+5, n+7, , 2n, 2n+2, respectively, and the labels of the vertices $$~v_{\frac {n+1}{2}},~ v_{\frac {n+3}{2}},~~ \cdots, v_{n-2},~v_{n-1},~v_{n}$$ are 5n−1, 5n−3, , 4n+4, 4n+2, 6n−2, respectively. Clearly f(x), f(y) and f(v0) are different from all the labels of the vertices vi. Thus the graph $$~\overline {K}_{2} + ~K_{1,n}$$ has an edge even graceful labeling for all n.

Method 2: We can find another labeling when n is an odd number, by redefining the labeling function as follows:

$$~~~~~~~~~~~~~~~f(x ~v_{i}~)=\left \{\begin {array}{ll} ~5n+3~~~ & \text {if} ~~ i=~0 ~; \\ 2i ~~~ & \text {if}~~ 1 \leq i \leq \frac {n+1}{2} ~;~ \\ 4n+2i+2 ~~~ & \text {if}~~ \frac {n+3}{2} \leq i \leq ~~ n, ~ \\ \end {array}\right.$$

$$~~~~~~~~~~~~~~~f(y ~v_{i}~)= \left \{\begin {array}{ll} ~ ~ 6n+4~~~~~~ & \text {if} ~~ i=~0 ~; ~~\\ ~ n+1+2i ~~~~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n+1}{2} ~;~~ \\ ~ 3n+1+2i~~~~~~~~~~ & \text {if} ~~ \frac {n+3}{2} \leq i \leq ~~ n ~,~~~~ ~\\ \end {array}\right.$$

and

$$~~~~~~~~~~~~~~~ f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~2n+2+2i~~~~ & \text {if}~~ ~~ 1 \leq i \leq \frac {n+1}{2} ;\\ ~5n+5-2i~~~~ & \text {if} ~~~~ ~\frac {n+3}{2} \leq i \leq ~n. \\ \end {array}\right.$$

Then, by the same way, we can calculate f(x),f(y) and f(v0) and prove that they are different from all the labels of the vertices vi. □

Illustration: In Fig. 6, we present an edge even graceful labeling of $$\overline {K}_{2} + ~K_{1,10}$$ and $$~ \overline {K}_{2} + ~K_{1,11}$$.

## Edge even graceful labeling of the join graph $$~\overline {K}_{2}+ w_{_{n}}$$

### Theorem 7

The graph $$~\overline {K}_{2} + ~W_{n}$$ has an edge even graceful labeling for all n.

### Proof

Let {v0, v1,v2, , vn} be the vertices of the wheel graph Wn with central vertex v0 and {x, y} be the vertices of the graph $$~\overline {K}_{2},~$$ so $$~ E(\overline {K}_{2} + ~W_{n})= \{ x~v_{0},~ x~v_{i},~y~v_{0},y~v_{i},v_{0}~v_{i}~,v_{i}~v_{i+1},~~i=1,2, \cdots,~n~\}$$. In this graph, $$p = | V (~\overline {K}_{2} + ~w_{n})| = n+3$$ and $$q~= | E (~\overline {K}_{2} + ~W_{n})| =4n+2~$$. There are two cases:

• when n is even, we define the labeling $$f:E(\overline {K}_{2} + ~W_{n})\longrightarrow \{2,4,\cdots,8n+4\}$$ as follows:

f(v1vn)=8n+4, f(vivi+1)=3n+2i+2 fori=1,2,n−1,

$$f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~2i +2~~~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2} ; \\ ~6n+2i~~~~~~~ & \text {if }~~ \frac {n}{2} < i \leq ~~ n~, \\ \end {array}\right.$$

$$f(x ~v_{i}~)= \left \{\begin {array}{ll} ~2~~~ & \text {if }~~ i=~0 ~; \\ ~n+2+2i ~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2} ; \\ ~5n+2i ~~~ & \text {if}~~ \frac {n}{2} < i \leq ~~ n, \\ \end {array}\right.$$,

and

$$~~ ~f(y ~v_{i}~)= \left \{\begin {array}{ll} ~8n~~~ & \text {if} ~~ i=~0 ~; \\ ~2n+2+2i ~~~ & \text {if}~~ 1 \leq i \leq \frac {n}{2} ;~ \\ ~4n+2i ~~~ & \text {if}~~ \frac {n}{2} < i \leq ~~ n.~ \\ \end {array}\right.$$

In view of the above labeling pattern we have

f(v1)= 6n+16, f(vn)= 2n−12,

$$f^{\ast }(x)= \left [~ f(x~ v_{0})~+\sum _{i=1}^{n} f (x~ v_{i}~) ~ \right ] ~\mbox{{mod}}~(8n+4)=~ f(x ~v_{0}~)~\mbox{{mod}}~(8n+4)=~2$$,

$$f^{\ast }(y)= \left [ f(y v_{0})+\sum _{i=1}^{n} f (y v_{i}) \right ] ~\mbox{{mod}}(8n+4)=~f(y v_{0})~\mbox{{mod}}(8n+4)=8n+2$$,

$$f^{\ast }(v_{0}) = \left [ \sum _{i=1}^{n} f (v_{0}~v_{i}) + f(x~v_{0})+f(y~v_{0})\right ] ~\mbox{{mod}}~(8n+4) ~ =~0$$,

and f(vi)=[f(vivi+1)+f(vi−1vi)+f(v0vi)+f(xvi)+f(yvi)] mod (8n+4).

Therefore, $$~~~~~f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (n+4+10i) ~\mbox{{mod}}~(8n+4)~~~ & \text {if}~~ ~2 \leq i \leq \frac {n }{2}; \\ (5n-6+10i) ~\mbox{{mod}}~(8n+4) ~~~ & \text {if} ~~~~~ \frac {n }{2}+1 \leq i < n. \\ \end {array}\right.$$

Hence, the labels of the vertices $$v_{2},v_{3}, v_{4},\cdots, v_{\frac {n}{2}}~$$ are n+24,n+34,n+44,,6n+4, respectively, and the labels of the vertices $$v_{\frac {n}{2}+1}, v_{\frac {n}{2}+2}, \cdots, v_{n-1},v_{n}$$ are 2n,2n+10,,7n−20, respectively, which are even and distinct numbers. Clearly, f(x) and f(v0) are even and different from all the labels of the vertices vi.

• when n is odd, we define the labeling function f as follows:

f(v1vn)=5n+1, f(vivi+1)=5n−2i+1 fori=1,2,n−1,

$$~~~~~~~~~~~~~~~ f~(v_{0}~v_{i}) = \left \{\begin {array}{ll} ~n+2i +1~~~ & \text {if}~~ 1 \leq i \leq \frac {n+1}{2}; ~~ \\ ~5n-1+2i~~~ & \text {if} ~~ \frac {n+3}{2} \leq i \leq ~~ n, \\ \end {array}\right.$$

$$~~~~~~~~~~~~~~~f(x ~v_{i}~)= \left \{\begin {array}{ll} n+1~~~ & \text {if} ~~ i=~0 ~; \\ 2i ~~~~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1}{2} ; \\ 6n+2+2i ~~~~~~ & \text {if}~~ \frac {n+1}{2} \leq i \leq ~~ n,\\ \end {array}\right.$$

and

$$~~ ~~~~~~~~~~~~~f(y ~v_{i}~)=\left \{\begin {array}{ll} ~ 7n+1~~~~~ & \text {if} ~~ i=~0 ~; ~~\\ ~ 2n+2+2i ~~ & \text {if}~ 1 \leq i \leq \frac {n-1}{2} ; \\ ~ 8n+4~~ & \text {if} ~ i=~\frac {n+1}{2} ~;~~ \\ ~4n+2i ~~ & \text {if}~ \frac {n+3}{2} \leq i \leq ~~ n. \\ \end {array}\right.$$

In view of the above labeling pattern and by the same way in Case (1), we have $$~f^{\ast }(x)= 0,~ ~f^{\ast }(y)=7n+1,~~~f^{\ast }(v_{0}) = n+1,~ ~f^{\ast }(v_{\frac {n+1}{2}}) = n-1~$$ and f(vn)=5n−7.

Finally,$$~~~~~f^{\ast }(v_{i}) = \left \{\begin {array}{ll} (5n+3+2i) ~\mbox{{mod}}~(8n+4)~~~ & \text {if}~~ 1 \leq i \leq \frac {n-1 }{2}; \\ (n-7+2i) ~\mbox{{mod}}~(8n+4) ~~~ & \text {if} ~~~~ \frac {n+3 }{2} \leq i \leq n-1.\\ \end {array}\right.$$

Hence, the labels of the vertices $$~v_{1}, ~v_{2},~v_{3},~ \cdots, ~ v_{\frac {n-3}{2}},~v_{\frac {n-1}{2}}~$$ will be 5n+5, 5n+7,5n+9 , 6n, 6n+2, respectively. Also, the labels of the vertices $$~v_{\frac {n+3}{2}},~v_{\frac {n+5}{2}},~ \cdots, ~ v_{n-2},~v_{n-1}$$ will be 2n−4, 2n−2, , 3n−11, 3n−9, respectively. It is clear that f(x),f(y) and f(v0) are even and different from all the labels of the vertices vi.

Obviously, the vertex labels are all even and distinct. Also f(x) and f(y) are even and different from all the labels of the vertices vi. Thus, the graph $$~ \overline {K}_{2} + ~W_{n}$$ is an edge even graceful graph for all n. □

Illustration: In Fig. 7, we present an edge even graceful labeling of $$\overline {K}_{2} + ~W_{10}$$ and $$~ \overline {K}_{2} + ~W_{11}$$.

## Edge even graceful labeling of the double cone $$~\overline {K}_{2} + ~C_{n}$$

### Theorem 8

The double cone $$\overline {K}_{2} + C_{n}$$ has an edge even graceful labeling for all n.

### Proof

Let {x, y} be the vertices of $$~\overline {K}_{2}$$ and {v1,v2, , vn} be the vertices of the graph Cn so the edges are { xvi, yvi, vivi+1, i=1,2,, n }. Let us use the standard notation $$p= | V (~\overline {K}_{2} + ~C_{n})| = n+2 ~$$ and $$~ q~= | E (~\overline {K}_{2} + ~C_{n})| =3n$$. There are three cases:

• When n≡1 (mod 6) orn≡3 (mod 6). We define the labeling $$f:E(\overline {K}_{2} + ~C_{n})\longrightarrow \{2,4,\cdots,6n\}$$ as follows:

f(xvi)=2ifori=1,2,n,

f(yvi)=6n−2ifori=1,2,n,

and

f(v1vn)=6n, f(vivi+1)=2n+2ifori=1,2,n−1,

The induced vertex labels are

$$f^{\ast }(x)= \left [\sum _{i=1}^{n} f (xv_{i}) \right ]~ \mbox{{mod}}(6n)=\left [\sum _{i=1}^{n} (2i) \right ]~ \mbox{{mod}}(6n)=(n^{2}+n)\mbox{{mod}}(6n)$$,

If n≡1 (mod 6) 2q=6n=36k+6, then f(x)≡[ 12k+2] mod (36k+6)= 2n.

If n≡3 (mod 6) 2q=6n=36k+18, then f(x)= ≡[ 24k+12] mod (36k+12)= 4n.

Similarly, $$~~f^{\ast }(y)= [~\sum _{i=1}^{n} f (yv_{i}) ~ ] ~\mbox{{mod}}~(~6n)=\left [~\sum _{i=1}^{n} (6n-2i) \right ]~ \mbox{{mod}}~(6n)$$.

$$\therefore ~~~ f^{\ast }(y)=(5 n^{2}-n)~\mbox{{mod}}(6n) =\left \{\begin {array}{ll} 4n ~~~ & \text {if}~~ n \equiv 1~(\mbox{{mod}}~ 6) ;\\ 2n~~~ & \text {if} ~~ n \equiv 3~(\mbox{{mod}}~ 6)~. ~\\ \end {array}\right.$$

f(v1)=[f(v1v2)+f(vnv1)+f(xv1)+f(yv1)] mod (6n) =2n+2,

\begin{aligned} f^{\ast}(v_{i}) &= [ f (v_{i}~v_{i+1})+ f (v_{i-1}v_{i})+ f(x~v_{i})+ f(y~v_{i})] ~{\rm{mod}}~(6n) \\ &=[~ f (v_{i}~v_{i+1})+ f (v_{i-1}v_{i})] ~{\rm{mod}}~(6n)\\ &=(4n+4i-2) ~{\rm{mod}} \quad(6n)~, ~~~2 \leq i \leq n-1, \end{aligned}

and

f(vn)=[f(vnv1)+f(vn−1vn)+f(xvn)+f(yvn)] mod (6n)= 4n−2.

Hence, the labels of the vertices $$~v_{1},~v_{2},~v_{3},~ \cdots,~ v_{\frac {n-1}{2}}$$ will be 2n+2, 4n+6, 4n+10, , 6n−4, respectively, and the labels of the vertices $$~v_{\frac {n+1}{2}}, ~v_{\frac {n+3}{2}}, ~ \cdots,~ v_{n-1},~v_{n}~$$ will be 0,4,,4n−6,4n−2, respectively. Clearly, f(x) and f(y) are different from all the labels of the vertices vi.

• When n≡5 (mod 6), we define the labeling f as follows:

f(yvi)=2n+2ifori=1,2,n,

f(v1vn)=4n+2, f(vivi+1)=6n−2i+2 fori=1,2,n−1,

and

$$f~(x~v_{i}) = \left \{\begin {array}{ll} 2 ~~~ & \text {if}~~ i=1~~ ~\\ 2n-2i+4~~~ & \text {if} ~~ i=2,3~,\cdots n ~~ \\ \end {array}\right.$$

Since n≡5 (mod 6) n=6k+5 2q=6n=36k+30.

Then the induced vertex labels are

$$~f^{\ast }(y) = \left [~ \sum _{i=1}^{n} (2n+2i) ~\right ] ~\mbox{{mod}}~(6n)=~(3 n^{2}+n)~\mbox{{mod}}~(6n)$$

$$~~~~~~~\equiv \left [~24(\frac {n-5}{6})+ 20 ~\right ]~ \mbox{{mod}}~(6 n~)= ~4n$$,

f(x)=(n2+n) mod (6n)≡0, f(v1)= 6 and

f(vi)=(4n−4i+10) mod (6n) for 2≤in.

Hence the labels of the vertices $$v_{1},v_{2},v_{3}, \cdots, v_{\frac {n-1}{2}},v_{\frac {n+1}{2}}, v_{\frac {n+3}{2}}, \cdots, v_{n-1},v_{n} ~~$$ are 6, 4n+2, 4n−2, , 2n+14, 2n+8, 2n+4, , 14, 10, respectively. Clearly, f(x) and f(y) are even and different from all the labels of the vertices vi. Thus, the graph $$\overline {K}_{2} + ~C_{n}$$ is an edge even graceful graph.

• When n is even, n≥4, we define the labeling function f as follows:

f(v1vn)=4n, f(vivi+1)=2n+2ifori=1,2,n−1,

$$f~(x~v_{i}) = \left \{\begin {array}{ll} 2i ~~~ & \text {if}~~ i=1,2,\cdots,~\frac {n}{2}; \\ 4n +2(i-1)~~~ & \text {if} ~~ \frac {n}{2}+1 \leq i \leq ~n, \\ \end {array}\right.$$

and

$$~f~(y~v_{i}) = \left \{\begin {array}{ll} 2n-2(i-1) ~~~ & \text {if}~~ 1 \leq i\leq \frac {n}{2};\\ 6n ~~~ & \text {if }~~ i ~= \frac {n}{2}+1; \\ 6n -2(i-1)~~~ & \text {if} ~~ \frac {n}{2}+2 \leq i \leq ~n. \\ \end {array}\right.$$

Thus, the induced vertex labels are

$$~~~~~~~~~f^{\ast }(x)= \left [~\sum _{i=1}^{n} ~f (x~v_{i}) ~ \right ] ~\mbox{{mod}}~(6n) = 0$$,

$$~~~~~~~~~ f^{\ast }(y)= \left [~\sum _{i=1}^{n} ~f (y~v_{i}) ~ \right ] ~\mbox{{mod}}~(6n)= \left [~f(y~v_{1})+ f(y~v_{\frac {n}{2}}) \right ] ~\mbox{{mod}}~(6n) = ~2n$$,

f(v1)=[f(xv1)+f(yv1)+f(v1v2)+f(vnv1)] mod (6n)=2n+4,

and

$$~~~~~~~~~~ f^{\ast }(v_{i}) =\left \{\begin {array}{ll} ~4n-4i+8~~~~~~ ~~~ & \text {if}~~ ~2 \leq i \leq \frac {n}{2};\\ 5n+2 ~~~ & \text {if }~~ i ~= \frac {n}{2}+1~; ~ \\ 6n -4i+6~~~ & \text {if} ~~\frac {n}{2}+2 \leq i \leq ~n~. ~\\ \end {array}\right.$$

Hence, the labels of the vertices $$~ v_{1},v_{2},v_{3}, \cdots, v_{\frac {n}{2}-1},v_{\frac {n}{2}}$$ are 2n+4, 4n, 4n−4,,2n+12,2n+8, respectively, and the labels of the vertices $$~~v_{\frac {n}{2}+1},~v_{\frac {n}{2}+2},~v_{\frac {n}{2}+3},~ \cdots, ~ v_{n-1},~v_{n}~$$ are 5n+2, 4n−2, 4n−6, ,2n+10,2n+6, respectively. Obviously the vertex labels are all even and distinct. Also, f(x) and f(y) are even and different from all the labels of the vertices vi. Thus, the double cone $$~\overline {K}_{2} + ~C_{n}$$ is an edge even graceful labeling when n is even.

∙ If n=2, the double cone $$~\overline {K}_{2} + ~C_{2}$$ has an edge even graceful labeling, see the following Fig. 8. □

Illustration: In Fig. 9, we present an edge even graceful labeling of $$\overline {K}_{2} + ~C_{7}, \overline {K}_{2} + ~C_{9}, \overline {K}_{2} + ~C_{11}$$ and $$\overline {K}_{2} + ~C_{12}$$

Not applicable.

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## Acknowledgements

The authors wish to express their thanks to the reviewers for the insightful comments and suggestions which have improved aspects of the work.

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Zeen El Deen, M.R., Omar, N.A. Further results on edge even graceful labeling of the join of two graphs. J Egypt Math Soc 28, 21 (2020). https://doi.org/10.1186/s42787-020-00077-5