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 K_m,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 K_1,n, the wheel graph W_n and the sunflower graph sf_n for all $n \in \mathbb {N}$. Finally, we proved that the join of the graph $\overline {K}_{2}~$ with the star graph K_1,n, the wheel graph W_n and the cyclic graph C_n 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 [1–3].

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:x₁∈V(G), x₂∈V(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 graphF_n=K₁ +P_n; n≥2 and W_n=K₁ +C_n; 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 graphK_1,n, the wheel graphW_n, and the sunflower graphsf_n. Also, we will study the edge even graceful labeling of the complete bipartite graphK_m,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 K_1,n, C_n and W_n.

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

Theorem 1

The complete bipartite graph K_n,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(K_n,n)|=2n and q=|E(K_n,n)|=n². The vertices of K_n,n were divided into two disjoint sets {v₁,v₂,⋯,v_n} {u₁,u₂,⋯,u_n} 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(K_n,n)→{2,4,⋯,2n²} as follows:

If i is an odd number, 1≤i≤n

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

and if i is an even number, 2≤i≤n−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=(a_ij) shows the methods of labeling, where a_ij represents the label of the edge u_iv_j.

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

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

• (i) The labels of the vertices u_i, 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,⋯,u_n−2,u_n−1 are n+1, 2n²−n−1, 3n+1, 2n²−3n−1,⋯,n²−2n+1,n²+2n−1, respectively, and f^∗(u_n)=0.

• (ii) The labels of the vertices v_i 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,⋯,v_n−3,v_n−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 K_3,3 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 {u_i, 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=(a_ij).

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

Fig. 1

An edge even graceful labeling of the graph K_3,3

The algorithm for the matrix X=(a_ij) of the graph K_n,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 i → odd & i≤ n−2. if j → odd, X(i,j)= (i−1)n+j+1. else if j → even, 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 i → even. if j → odd & j≤ n−2, X(i,j)= (i−1)n+j+2. else if j → even, 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

• (i) The labels of the vertices u_i,i=1,2,⋯,n is the sum of rows in the matrix, so the labels of these vertices are n+1, 2n²−n−1, 3n+1, 2n²−3n−1,⋯,n²−2n+1,n²+2n−1,0, respectively.

• (ii) The labels of the vertices v_i 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 K_n,n=nK₁+ nK₁ has an edge even graceful labeling. □

Illustration: we present an edge even graceful labeling of the graph K_13,13 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 K_m,n=mK₁+ nK₁ has an edge even graceful labeling when m and n are distinct odd numbers, m≠2n−3 and km≠ln, k=1,3,⋯,n−1, l=1,3,⋯,m−1.

Proof

In the graph K_m,n=mK₁+ nK₁, we have p=|V(K_m,n)|=m+n and q=|E(K_m,n)|=mn. Without loss of generality, assume that m > n and the vertices of K_m,n divided into two disjoint sets {v₁,v₂,⋯,v_m} and {u₁,u₂,⋯,u_n}. Put v_i as the columns of the matrix and u_i as the rows. We define the labeling function f:E(K_m,n)→{2,4,⋯,2mn} as follows: first, label the edges incident to the vertex u₁, i.e., u₁v₁, u₁v₂, u₁v₃, u₁v₄,⋯, u₁v_n−2, u₁v_n−1,u₁v_n 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 u₂v_m, u₂v_m−1, u₂v_m−2, u₂v_m−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 u_n will repeat with the labels of the vertex v_m. To avoid this problem we replace the labels of the two edges u_nv_m−1 and u_nv_m, that is f(u_nv_m−1)=2mn and f(u_nv_m) =nm+1. Thus, the induced vertex labels are

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

• (ii) The labels of the vertices v_i,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,⋯,v_m−3,v_m−2 are n+1, 2mn−(n+1), 3n+1, 2mn−(3m+1), ⋯, 2mn−[(m−4)n+1], (m−2)n+1, respectively, and f^∗(v_m−1)=2n−2, f^∗(v_m)=nm+1. □

Illustration: If we take m=15, then we can label the graphs K_15,7, K_15,11, and K_15,13, while we can not find labels of K_15,3, K_15,5, and K_15,9. We present an edge even graceful labeling of the graph K_15,7=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+k_1,n.

Theorem 3

The graph K₁+K_1,n has an edge even graceful labeling.

Proof

Let {v₀,v₁,v₂, ⋯, v_n} be the vertices of the graph K_1,n with central vertex v₀ and {x} be the vertex of K₁ so the edges of K₁+K_1,n are {xv₀, xv_i, v₀v_i,i=1,2,⋯, n }. Here, p=|V(K₁+K_1,n)|=n+2 and q =|E(K₁+K_1,n)|=2n+1. There are two cases:

• when n is even. We define the labeling function f:E(K₁+K_1,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+K_1,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 v_i. Hence, the graph K₁+K_1,n is edge even graceful for all n. □

Illustration: In Fig. 2, we present an edge even graceful labeling of the graphs K₁+ K_1,8 and K₁+ K_1,9

Fig. 2

An edge even graceful labeling of K₁+K_1,8, and K₁+K_1,9

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

Theorem 4

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

Proof

Let {v₀,v₁,v₂, ⋯, v_n} be the vertices of W_n with central vertex v₀ and {x} be the vertex of K₁ so the edges of K₁+ W_n will be {xv₀, xv_i, v₀v_i,v_iv_i+1,i=1,2,⋯, n }. So, p=|V(K₁+ W_n)|=n+2 and q=|E(K₁+ W_n)|=3n+1. There are five cases:

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

  f(v₀x)=6n+2,

  f(v₁v_n)=3n, f(v_iv_i+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(v_nv₁)+f(xv₁)+f(v₀v₁)] mod (6n+2) = n+4

  and f^∗(v_i)=[f(v_iv_i+1)+f(v_i−1v_i)+f(xv_i)+f(v₀v_i)] 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^∗(v_n)= 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+ W_n)→{2,4,⋯,6n+2} as follows:

  f(xv₀)=6n, f(xv_i)=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(v_nv₁)+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^∗(v_i)=[f(v_iv_i+1)+f(v_i−1v_i)+f(v₀v_i)+f(xv_i)] 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^∗(v_n)=[f(v_n−1v_n)+f(v_nv₁)+f(v₀v_n)+f(xv_n)] 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 v_i.

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

  f(xv₀)=2, f(xv_i)=(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^∗(v_n)= 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 v_i.

  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(v₁v_n)=4n, f(v_iv_i+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 v_i 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.

Fig. 3

An edge even graceful labeling of the graphs K₁+ W₈ and K₁+ W₁₀

Fig. 4

An edge even graceful labeling of the graphs K₁+ W₉, K₁+ W₁₁, K₁+ W₅ and K₁+ W₇

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+sf_n

The sunflower graph, sf_n, is defined as a graph obtained by starting with an n-cycle C_n with a consecutive vertices v₁,v₂,⋯,v_n and creating new vertices u₁,u₂,⋯,u_n, with u_i connected to v_i and v_i+1. The graph sf_n has number of vertices p=2n and a number of edges q=3n.

Theorem 5

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

Proof

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

{ xv_i, xu_i, v_iv_i+1, v_iu_i, v_i+1u_i, i=1,2,⋯,n }. Let us use the standard notation p=|V(K₁+ sf_n)|=2n+1 and q =|E(K₁+ sf_n)|=5n.

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

fori=1, 2, ⋯, n, f(xv_i)=2i, f(xu_i)=10 n−2i,

fori=1, 2, ⋯, n, f(v_iu_i)=4n+2i, f(u_iv_i+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^∗(v_i)=2n+2i+4 fori=3, 4, ⋯, n,

and

f^∗(u_i)=6n+2ifori=1,2, ⋯, n.

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

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

Fig. 5

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, ⋯, v_n} be the vertices of the graph K_1,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, xv_i, v₀v_i,yv₀,yv_i, 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^∗(v_i)=[f(v₀v_i)+f(xv_i)+f(yv_i)] 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 v_i.

• 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^∗(v_n)=[f(v₀v_n)+f(xv_n)+f(yv_n)] 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 v_i. 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 v_i. □

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

Fig. 6

An edge even graceful labeling of the graphs $\overline {K}_{2} + ~K_{1,10}$ 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, ⋯, v_n} be the vertices of the wheel graph W_n 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(v₁v_n)=8n+4, f(v_iv_i+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^∗(v_n)= 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^∗(v_i)=[f(v_iv_i+1)+f(v_i−1v_i)+f(v₀v_i)+f(xv_i)+f(yv_i)] 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 v_i.

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

  f(v₁v_n)=5n+1, f(v_iv_i+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^∗(v_n)=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 v_i.

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 v_i. 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 $~ \overline {K}_{2} + ~W_{11}$.

Fig. 7

An edge even graceful labeling of the graphs $\overline {K}_{2} + ~W_{10}$ 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, ⋯, v_n} be the vertices of the graph C_n so the edges are { xv_i, yv_i, v_iv_i+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(xv_i)=2ifori=1,2,⋯n,

  f(yv_i)=6n−2ifori=1,2,⋯n,

  and

  f(v₁v_n)=6n, f(v_iv_i+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(v_nv₁)+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^∗(v_n)=[f(v_nv₁)+f(v_n−1v_n)+f(xv_n)+f(yv_n)] 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 v_i.

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

  f(yv_i)=2n+2ifori=1,2,⋯n,

  f(v₁v_n)=4n+2, f(v_iv_i+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^∗(v_i)=(4n−4i+10) mod (6n) for 2≤i≤n.

  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 v_i. 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(v₁v_n)=4n, f(v_iv_i+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(v_nv₁)] 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 v_i. 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. □

Fig. 8

An edge even graceful labeling of the double cone $~\overline {K}_{2} + ~C_{2}$

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 $\overline {K}_{2} + ~C_{12}$

Fig. 9

An edge even graceful labeling of the double cones $\overline {K}_{2} + ~C_{7}$, $\overline {K}_{2} + ~C_{9}$, $\overline {K}_{2} + ~C_{11}$, and $\overline {K}_{2} + ~C_{12}$