Further results on equitable near proper coloring of derived graph families

Proof. Consider $V(P_n)=\{v_1,v_2,\dots ,v_n\}$ and $E(P_n)=\{e_1,e_2,\dots ,e_{n-1}\}$. Then, $V((M(P_n))=V(P_n)\cup E(P_n)$. Thus, $M(P_n)$ contains $2n-1$ vertices, and observe that ${\chi }_e(M(P_n))=3$. In an ENP $k$-coloring, only one case $k=2$ is to be considered.

It is possible to partition the vertex set into two subsets, $V_1=\{v_1,v_2,\dots ,v_n\}$ and $$\begin{gathered} V_2=\{e_1,e_2,\dots , \\ {e}_{n-1}\} \end{gathered}$$. Now, assign the vertices in $V_1$ with the color $c_1$ and $V_2$ with the color $c_2$. Since $e_i$ $;$ $1\le i\le n-1$ forms a path of order $n-2$, $b_{{\chi }_e}^k(M(P_n))=n-2$, as $n-2$ monochromatic edges are formed among the vertices $e_i$.

Another possible way of an ENP $k$-coloring is defined as follows: $$c(v_i)=\{\begin{array}{ll}{c}_1,& \text{if}i\equiv 1,2\mathrm{m}\mathrm{o}\mathrm{d}4;\\ c_2,& \text{if}i\equiv 0,3\mathrm{m}\mathrm{o}\mathrm{d}4.\end{array}$$

This coloring results in $n-2$ monochromatic edges of the form $v_i{e}_i$ $;$ $1\le i\le n-1$ or $e_i{v}_j$ $;$ $1\le i\le n-1$ $,$ $1\le j\le n$. It also yields in $n-2$ monochromatic edges, and thus $b_{{\chi }_e}^k(M(P_n))=n-2$. 

Proof. Let $V((T(P_n))=\{v_1,v_2,\dots ,v_n,e_1,e_2,\dots ,e_{n-1}\}$, and in an ENP $k$-coloring, the below situations are investigated.

Case 1: When $n$ is even, we partition the vertex set $V(T(P_n))$ as follows:

$V_1=\{v_1,e_1,v_3,e_3,\dots ,v_{n-1}{e}_{n-1}\}$ and $V_2=\{v_2,e_2,v_4,e_4,\dots ,v_{n-2},e_{n-2},v_n\}$.

Case 2: When $n$ is odd, $V(T(P_n))$ is partitioned as below:

$V_1=\{v_1,e_1,v_3,e_3,\dots ,v_{n-2}{e}_{n-2},v_n\}$ and $V_2=\{v_2,e_2,v_4,e_4,\dots ,v_{n-1},e_{n-1}\}$.

In both cases, it is evident that $||V_1|-|V_2||=1$. Now, assigning the vertices of $V_1$ with the color $c_1$ and $V_2$ with the color $c_2$, we observe that every edge $v_i{e}_i$ $;$ $1\le i\le n-1$ is a monochromatic edge, and hence the equitable defective number is $n-1$. 

Proof. Let $V(C_n)=\{v_1,v_2,\cdots ,v_n\}$ and $E(C_n)=\{e_1,e_2,\cdots ,e_n\}$. Now,
$V(M(C_n))=V(C_n)\cup E(C_n)$, and note that ${\chi }_e(M(C_n))=3$, and thus the ENP $k$-coloring considers only one case $k=2$. In the first possible ENP $k$-coloring, assign all of the vertices $v_i$ and $e_i$ with the colors $c_1$ and $c_2$ respectively, and it can be observed that every edge between the vertices $e_i$ is a monochromatic edge, and thus the equitable defective number is $n$.

The second possible ENP $k$-coloring may be performed as in Theorem 2.1, and this coloring also produces $n$ monochromatic edges. Additionally, there are $n$ monochromatic edges of the form $v_i{e}_i$ $;$ $1\le i\le n$ for even $n$; however, for odd $n$, there are $n-1$ monochromatic edges of the form $v_i{e}_i$ $;$ $1\le i\le n-1$ and one $e_{n-1}{e}_n$ monochromatic edge. Thus, $b_{{\chi }_e}^2(M(C_n))=n$. 

Proof. Note that ${\chi }_e(T(C_n))=3$ when $n\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}3$ and $4$ otherwise. Hence, the following cases need to be addressed here.

Case 1: Assume that $k=2$ and $n$ is even. We partition the vertex set so that $$\begin{gathered} V_1=\{v_1,e_1,v_3,e_3,\dots \\ {v}_{n-1},e_{n-1}\} \end{gathered}$$ and $V_2=\{v_2,e_2,v_4,e_4,\dots v_n,e_n\}$. Now, $|V_1|=|V_2|$, and every $v_i{e}_i$ edge, where $1\le i\le n$, is a monochromatic edge. Thus, the equitable defective number is $n$.

Case 2: Assume that $k=2$ and $n$ is odd. Let us partition the vertex set such that $V_1=\{v_1,e_1,v_3,e_3,\dots v_{n-2},e_{n-2},v_n\}$ and $V_2=\{v_2,e_2,v_4,e_4,\dots v_{n-1},e_{n-1},e_n\}$. Here, $|V_1|=|V_2|$ and $n-1$ monochromatic edges are obtained connecting $v_i$ and $e_i$ $;$ $1\le i\le n-1$. Furthermore, another $v_1{v}_n$ and $e_{n-1}{e}_n$ monochromatic edges are obtained. Thus, the equitable defective number, in this case, is $n+1$ (see Figure 4 for illustration).

Case 3: When $k=3$ and $n\not\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}3$, assign the vertices with the three available colors in a cyclic order by considering the equitability condition. Here, the equitable defective number is $1$. 

Proof. Consider $V(L(S{l}_n))=\{v_1,v_2,\dots ,v_n,u_1,u_2,\dots ,u_n\}$, where $v_i$ induces a cycle of order $n$, and $u_i$ are the vertices that are adjacent with both $v_i$ and $v_{i+1}$. The equitable coloring of $L(S{l}_n)$ contains $3$ colors (see [7]), and thus, in an ENP $k$-coloring, only one case as $k=2$ is considered. Let $c_1$ and $c_2$ be the two available colors, and the following cases need to be considered.

Case 1: Let $n$ be even. Then, the $v_i$ $;$ $1\le i\le n$ vertices can be properly colored with the two available colors. When the remaining vertices are assigned the same colors equitably, $n$ monochromatic edges are obtained between $v_i$ and $u_i$ $;$ $1\le i\le n$.

Case 2: Let $n$ be odd and follow the same coloring procedure as in Case 1. That is, assign $v_1$ with the color $c_1$, $v_2$ with the color $c_2$, and so on. Since $n$ is odd, both $v_1$ and $v_n$ receive the same color, and thus there is a $v_1{v}_n$ monochromatic edge on the rim. Now, the vertex adjacent to both $v_1$ and $v_n$ receives an alternate color from the color assigned to both of them. Further, the remaining vertices can be assigned in an equitable manner. Thus, $n-1$ monochromatic edges are obtained between the vertices $v_i$ and $u_i$, and the equitable defective number is $n$.

Combining the above two cases, $b_{{\chi }_e}^2(L(S{l}_n))=n$. 

Proof. Consider $V(M(S{l}_n))=\{v_1,v_2,\dots ,v_n,u_1,u_2,\dots ,u_n,v_1^{\prime },v_2^{\prime },\dots ,v_n^{\prime },u_1^{\prime },u_2^{\prime },\dots ,u_n^{\prime }\}$. Here, the vertices $v_1,u_1,v_2,u_2,\dots ,v_n,u_n$ forms an inner $2n$ – cycle termed as the rim vertices of $M(S{l}_n)$, and each $v_i^{\prime }$; $1\le i\le n$ is connected to the corresponding $v_i$; $1\le i\le n$ forms an outer $n$ – cycle. Again, $u_i^{\prime };$ $1\le i\le n$ are the pendant vertices connected to the corresponding $v_i^{\prime }$. Recall that ${\chi }_e(M(S{l}_n))=4$ (see [7]), and in an ENP $k$-coloring, two cases are to be considered as $k=2$ and $k=3$.

Case 1: Assume $k=2$ and we partition the vertex set as follows.

$V_1=\{u_1,u_2,\dots ,u_n,u_1^{\prime },u_2^{\prime },\dots ,u_n^{\prime }\}$ and $V_2=\{v_1,v_2,\dots ,v_n,v_1^{\prime },v_2^{\prime },\dots ,v_n^{\prime }\}$. Assign the vertices in $V_1$ with the color $c_1$ and $V_2$ with the color $c_2$. Now, $|V_1|=|V_2|$, and $n$ monochromatic edges generated among the vertices $u_i$. Additionally, the vertices $v_i$ and $v_i^{\prime }$ have $n$ monochromatic edges connecting them, where $1\le i\le n$. Thus, the equitable defective number is $2n$.

Case 2: Let $k=3$. Among the $4n$ vertices of $M(S{l}_n)$, $2n$ vertices are on the rim. Assign all the rim vertices with the three available colors, say $c_1$, $c_2$, and $c_3$, in a cyclic order, and consider the following three subcases.

Subcase 2.1: When $n\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}3$, $2n\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}3$, and assign the rim vertices with the three colors in a cyclic order such that each color repeats exactly $\frac{2n}{3}$ times without creating any monochromatic edges. Now, assign the vertices $v_i^{\prime }$ with the same color received by the vertices $v_i$, and $n$ monochromatic edges are obtained in between the vertices $v_i$ and $v_i^{\prime }$ $;$ $1\le i\le n$. Further, the vertices $u_i^{\prime }$ can be equitably assigned these three colors without creating monochromatic edges. Thus, there are $n$ monochromatic edges.

Subcase 2.2: When $n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}3$, $2n\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}3$, follow the same coloring procedure for the rim vertices $v_i$ and $u_i$ where $1\le i\le n$ creates a $u_1{u}_n$ monochromatic edge. Since $2n\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}3$, to attain the equitability, color $c_3$ can be assigned to $v_1^{\prime }$. The remaining $v_i^{\prime }$; ($2\le i\le n$) can receive the same color received by $v_i$ to obtain $n-1$ monochromatic edges of the form $v_i{v}_i^{\prime }$ ($2\le i\le n$). Further, each $u_i^{\prime }$; ($1\le i\le n$) can be received any of the three colors equitably with the condition that $v_i^{\prime }$ and the corresponding $u_i^{\prime }$ receive different colors leading to the equitable defective number as $n$.

Subcase 2.3: When $n\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}3$, $2n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}3$, and the rim vertices follow the same coloring pattern as in the previous subcases leads to a $v_1{u}_n$ monochromatic edge. Further, the remaining vertices $v_i^{\prime }$ and $u_i^{\prime }$ can receive any of the three colors by considering the equitability condition, creates $n-1$ monochromatic edges as in Subcase 2.2. Hence, the equitable defective number in this case is $n$.

Thus, combining the above three subcases, $b_{{\chi }_e}^3(M(S{l}_n))=n$. 

Proof. Consider the vertex sets and the adjacencies as in Theorem 2.6. Observe that ${\chi }_e(T(S{l}_n))=4$ (see [7]); thus, in an ENP $k$-coloring, the following cases need to be considered.

Case 1: When $k=2$, repeat the same coloring procedure as in Case-1 of Theorem 2.6, and we obtain the equitable defective number as $3n$.

Case 2: When $k=3$, follow the same coloring pattern as in Case-2 of Theorem 2.6 by considering the equitability condition and maximising the properness of the coloring, there are the following observations.

Subcase 2.1: When $n\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}k$, $n$ monochromatic edges are obtained of the form $v_i{v}_i^{\prime }$ $;$ $1\le i\le n$ as in Theorem 2.6.

Subcase 2.2: When $n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}k$, two monochromatic edges are obtained, one among the vertices $v_i$ and one among the vertices $u_i$. Apart from that, the vertices $v_i$ and $v_i^{\prime }$ $;$ $2\le i\le n-1$ have $n-1$ monochromatic edges connecting them. Thus, the equitable defective number is $n+1$. (See Figure $\text{???}$ for reference.)

Subcase 2.3: When $n\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}k$, we get the same $n-1$ monochromatic edges as in Subcase 2.2 and an additional monochromatic edge $v_1{u}_n$. Thus, the equitable defective number is $n$. (refer to Figure 7 for illustration.) 

Proof. Consider $V(K_{1,n})=\{v,v_1,v_2,\dots ,v_n\}$ and $E(K_{1,n})=\{e_1,e_2,\dots ,e_n\}$. Consider $u_1,u_2,\dots ,u_n$ as the corresponding vertex set of $E(K_{1,n})$. These edges form a clique of order $n$ in $M(K_{1,n})$, and hence ${\chi }_e(M(K_{1,n}))=n+1$ as the vertices $v,u_1,u_2,\dots ,u_n$ induce a complete graph of order $n+1$. Thus, in an ENP $k$-coloring, $2\le k\le n$ and the following cases need to be considered.

Case 1: When $k\ge 2$ and $k\ne n$, assign the $k$ available colors, say $c_1,c_2,\dots ,c_k$ alternately to the vertices $u_i$ $;$ $1\le i\le n$. Now, color the $v_i$ vertices so that each $u_i$ and the corresponding $v_i$ receive different colors in an equitable manner. Also, assign the vertex $v$ with any available colors such that $v$ receives the least used color assigned to the vertices $u_i$. Since, $\{v,u_1,u_2,\dots ,u_n\}$ induces a clique of order $n+1$, the equitable defective number in this case is $b_{{\chi }_e}^k(K_{n+1})$.

Case 2: When $k=n$, assign the vertices $u_i$ using the $k$ colors, and the corresponding vertices $v_i$ can be assigned using the same colors such that both $u_i$ and the corresponding $v_i$ receive different colors, and as a result of this assignment no monochromatic edges obtained. Further, assign the vertex $v$ with any available colors; only one $v{u}_i$ monochromatic edge is obtained, and thus the equitable defective number is $1$. 

Proof. Consider the vertices as in Theorem 2.9. Note that ${\chi }_e(T(K_{1,n}))=n+1$ as $\{v,u_1,u_2,\dots ,u_n\}$ induces a clique of order $n+1$. The vertices $u_1,u_2,\dots ,u_n$ induce a clique of order $n$, and let $v_1,v_2,\dots ,v_n$ be the vertices that are adjacent to $u_1,u_2,\dots ,u_n$, respectively. In $T(K_{1,n})$, the vertex $v$ is adjacent to all the vertices $u_i$ and $v_i$ ($1\le i\le n$). The ENP $k$-coloring of complete graphs is discussed in Theorem 2.8. Hence, the following cases need to be addressed here.

Case 1: When $k=2$, assign the two available colors, say $c_1$ and $c_2$, for every alternate vertex of the complete graph. Further, if $c(u_i)=c_1$ (or $c_2$), then assign $c(v_i)=c_2$ (or $c_1$). Now, $c(v)=c_1$ or $c_2$, and based on this assignment, the following observations are to be noted.

When $n$ is even, $\frac{n}{2}$ monochromatic edges of the form $v{u}_i$ and $\frac{n}{2}$ monochromatic edges of the form $v{v}_i$ are obtained. When $n$ is odd, we get $\lfloor \frac{n}{2}\rfloor$ monochromatic edges of the form $v{u}_i$ and $\lceil \frac{n}{2}\rceil$ monochromatic edges of the form $v{v}_i$. Consequently, in both cases, the equitable defective number is $b_{{\chi }_e}^k(K_n)+n$.

Case 2: When $k\ge 3$ and $k\ne n$, we can assign all the vertices with the $k$ available colors in an equitable manner as follows. When the vertex set is partitioned into $k$ color classes, each color class consists of either $\lfloor \frac{2n+1}{k}\rfloor$ vertices or $\lceil \frac{2n+1}{k}\rceil$ vertices. If the vertex $v$ is placed in the color class with minimum cardinality $\lfloor \frac{2n+1}{k}\rfloor$, we obtain $\lfloor \frac{2n+1}{k}\rfloor -1$ monochromatic edges. Hence, together with the monochromatic edges from $K_n$, the equitable defective number is $b_{{\chi }_e}^k(K_n)+\lfloor \frac{2n+1}{k}\rfloor -1$.

Case 3: When $k=n$, we can properly color the complete graph of order $n$ with the $k$ available colors, and the vertices $v_1,v_2,\dots ,v_n$ can also be properly colored with these $k$ colors. Further, assign the vertex $v$ with any of the $k$ available colors, from which we obtain two monochromatic edges. 

Proof. Let $V(G_{1,n})=\{v,v_1,v_2,\dots ,v_{2n}\}$ and $$\begin{gathered} E(G_{1,n})=\{e_1,e_2,\dots ,e_n,e_1^{\prime },e_2^{\prime }\dots , \\ {e}_{2n}^{\prime }\} \end{gathered}$$ such that $e_i=v{v}_{2i-1}$ $;$ $1\le i\le n$, $e_i^{\prime }=v_i{v}_{i+1}$ $;$ $1\le i\le 2n-1$, and $e_{2n}^{\prime }=v_{2n}{v}_1$. Consequently, $V(L(G_{1,n}))=\{e_1,e_2,\dots ,e_n,e_1^{\prime },e_2^{\prime }\dots ,e_{2n}^{\prime }\}$. Since $e_i$ $;$ $1\le i\le n$ form a clique of order $n$ in $L(G_{1,n})$, ${\chi }_e(L(G_{1,n}))=n$. Subsequently, in an ENP $k$-coloring, consider $2\le k\le n-1$ and address the following cases.

Case 1: Assume $k=2$ and assign the vertices $e_i$ with the two available colors, say $c_1$ and $c_2$ alternately. This assignment yields the same amount of monochromatic edges as in the equitable defective number of a complete graph, that is stated in Theorem 2.8. Further, assign the vertices $e_j^{\prime }$ with the same available colors in a cyclic order satisfying the equitability condition, and we obtain $n$ monochromatic edges in between the vertices $e_i$ and $e_j^{\prime }$ $;$ $1\le i,j^{\prime }\le n$. Thus, $b_{{\chi }_e}^k(K_n)+n$ gives the equitable defective number in this case.

Case 2: When $k\ge 3$, assign the vertices $e_i$ with the $k$ available colors, $b_{{\chi }_e}^k(K_n)$ monochromatic edges are obtained. By the definition of $L(G_{1,n})$, each $e_i$ is adjacent to two distinct vertices $e_j^{\prime }$, which forms a $K_3$ that can be colored properly using the three available colors such that each vertex of $K_3$ receives different color. This assignment of colors does not create any additional monochromatic edges; therefore, the equitable defective number is $b_{{\chi }_e}^k(K_n)$. 

Proof. Consider $V(G_{1,n})=\{v,v_1,v_2,\dots ,v_{2n}\}$. Let $E(G_{1,n})=\{e_1,e_2,\dots ,e_{2n}\}\cup \{e_i^{\prime },e_2^{\prime },\dots ,e_n^{\prime }\}$ such that $e_i=v_i{v}_{i+1}$ $;$ $1\le i\le 2n-1$, $e_{2n}=v_{2n}{v}_1$ and $e_i^{\prime }=v{v}_{2i-1}$ $;$ $1\le i\le n$. Consequently, $$\begin{gathered} V(M(G_{1,n}))=\{v,v_1,v_2,\dots ,v_{2n},e_1,e_2,\dots ,e_{2n},e_1^{\prime },e_2^{\prime }, \\ {e}_3^{\prime },\dots ,e_n^{\prime }\} \end{gathered}$$. It is evident that the edges $v{e}_i^{\prime }$ $;$ $1\le i\le n$, along with the vertex $v$ form a clique of order $n+1$, and thus ${\chi }_e(M(G_{1,n}))=n+1$ (see [6]). In an ENP $k$-coloring, consider $2\le k\le n$ and address the following cases.

Case 1: Let $k=2$ and consider the subcases as given below.

Subcase 1.1: When $n$ is even, partition the vertex set into two subsets as follows.

Let $V_1=\{v_i:1\le i\le 2n-3;i\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{v_j:2\le j\le 2n-2;j\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_k^{\prime }:1\le k\le n-1;k\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2\}$ $\cup$ $\{e_l:3\le l\le 2n-1;l\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_m:4\le m\le 2n;m\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}4\}$.

$V_2=\{v_i:3\le i\le 2n-1;i\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{v_j:4\le j\le 2n;j\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_k^{\prime }:2\le k\le n;k\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}2\}$ $\cup$ $\{e_l:1\le l\le 2n-3;l\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_m:2\le m\le 2n-2;m\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}4\}$.

Now, $|V_1|=|V_2|=\lfloor \frac{5n+1}{2}\rfloor$. Hence, the central vertex $v$ can be placed into any one of the color classes, resulting in $||V_1|-|V_2||=1$. Further, assign the vertices in the set $V_1$ with the color $c_1$ and $V_2$ with the color $c_2$. As a result of this assignment of colors, we obtain $n$ monochromatic edges of the form $e_i^{\prime }{v}_j$ $;$ $1\le i\le n$ and $1\le j\le 2n-1;j\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2$. Also, the vertices $v_i$ and $e_i$ have $n-1$ monochromatic edges connecting them, of the form $v_i{e}_{i+1}$, where $2\le i\le 2n-2$ $;$ $i\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}2$ and there is a monochromatic edge $e_1{v}_{2n}$. Further, $n$ monochromatic edges are obtained in between the vertices $e_i$ of the form $e_i{e}_{i+1}$, where $1\le i\le 2n-1$ $;$ $\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2$. Furthermore, together with the monochromatic edges from the complete graph that is investigated in Theorem 2.8, the equitable defective number is $b_{{\chi }_e}^k(K_{n+1})+3n$.

Subcase 1.2: When $n$ is odd, let us partition the vertex set as follows.

Let $V_1=\{v_i:1\le i\le 2n-1;i\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{v_j:2\le j\le 2n;j\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_k^{\prime }:1\le k\le n;k\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2\}$ $\cup$ $\{e_l:3\le l\le 2n-3;l\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_m:4\le m\le 2n-2;m\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}4\}$.

$V_2=\{v\}$ $\cup$ $\{v_i:3\le i\le 2n-3;i\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{v_j:4\le j\le 2n-2;j\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_k^{\prime }:2\le k\le n-1;k\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}2\}$ $\cup$ $\{e_l:1\le l\le 2n-1;l\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\}$ $\cup$ $\{e_m:2\le m\le 2n;m\equiv 2\mathrm{m}\mathrm{o}\mathrm{d}4\}$.

Assign the vertices in $V_1$ with the color $c_1$ and $V_2$ with the color $c_2$. Now, $|V_1|=|V_2|=\frac{5n+1}{2}$ and the vertices $v_i$ and $e_i$ have $n-1$ monochromatic edges connecting them, of the form $v_i{e}_i$, where $2\le i\le 2n-2$ $;$ $i\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}2$. Moreover, the vertices $e_i^{\prime }$ and $v_j$ have $n$ monochromatic edges connecting them of the form $e_i^{\prime }{v}_j$ $;$ $1\le i\le n$ and $1\le j\le 2n-1$ $;$ $j\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2$. Furthermore, a monochromatic edge $e_1{e}_{2n}$ and $n$ monochromatic edges among the vertices $e_i$ of the form $e_i{e}_{i+1}$ are generated, where $1\le i\le 2n-1$ $;$ $i\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2$. Now, along with the monochromatic edges obtained in the ENP $k$-coloring of complete graphs, the equitable defective number is $b_{{\chi }_e}^k(K_{n+1})+3n$.

Case 2: When $k=3$, the following subcases are to be addressed.

Subcase 2.1: Let $n\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}3$. Assign the $4n$ vertices on the outer rim ($v_i$ $;$ $1\le i\le 2n$ and $e_i$ $;$ $1\le i\le 2n$) using the three available colors in a cyclic order. Further, color the inner rim vertices ($e_i^{\prime }$ $;$ $1\le i\le n$) in the same cyclic order using the three colors such that both $e_1^{\prime }$ and $e_1$ receive the same color. Then, assign $v$ with any one of the three colors. It can be observed that along with the monochromatic edges obtained from the complete graph, the vertices $e_i^{\prime }$ and $v_i$ generate $n$ monochromatic edges of the form $e_i^{\prime }{v}_j$ $;$ $1\le i\le n$ and $1\le j\le 2n$ $;$ $j\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}2$. Thus, the equitable defective number is $b_{{\chi }_e}^k(K_{n+1})+n$ (see Figure 11 for illustration).