# A Novel Problem for Solving Permuted Cordial Labeling of Graphs

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## Abstract

**:**

## 1. Introduction

## 2. Terminology and Notation

## 3. Permuted Cordial Labeling of Some Graphs

**Theorem**

**1.**

**Proof.**

**Case (1)**: $n\equiv 0\left(mod3\right)$. The total number of vertices labeled $i$ is denoted by $v\left(i\right)$ is the same as the total number of vertices labeled $f$ and denoted by $v\left(f\right)$ and also equivalent to the total number of vertices labeled $g$ denoted by $v\left(g\right)$ and this number is $r$, i.e., $v\left(i\right)=v\left(f\right)=v\left(g\right)=r$. Obviously, $\left|v\left(x\right)-v\left(y\right)\right|=0,x\ne y$ and $x,y\in \left\{i,f,g\right\}$. Similarly, in this same one can see that the number of edges labeled $i$ denoted by $e\left(i\right)$ the same as the number of edges labeled $f$ denoted by $e\left(f\right)$ and this number is $r$, while the number of edges labeled $g$ denoted by $e\left(g\right)$ and this number is $r-1$. Consequently, $\left|e\left(x\right)-e\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$.

**Case (2)**: $n\equiv 1\left(mod3\right)$. The total number of vertices labeled $g$ is denoted by $v\left(g\right)$ is the same as the total number of vertices labeled $f$ denoted by $v\left(f\right)$ and this number is $r$, while the number of vertices of labeled $i$ denoted by $v\left(i\right)$ and this number is $r+1$. Consequently, $\left|v\left(x\right)-v\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$. Similarly, in this, one can see that the number of edges labeled $i$ denoted by $e\left(i\right)$ is the same as the number of edges labeled $f$ denoted by $e\left(f\right),$ the same as the number of edges labeled $g$ denoted by $e\left(g\right)$ and this number is $r$. Therefore, $\left|e\left(x\right)-e\left(y\right)\right|=0,x\ne y$ and $x,y\in \left\{i,f,g\right\}$.

**Case (3)**: $n\equiv 2\left(mod3\right)$. The total number of vertices labeled $i$ is denoted by $v\left(i\right)$ is the same as the total number of vertices labeled $f$, is denoted by $v\left(f\right)$ and this number is $r+1$, while the total number of vertices labeled $g$ is denoted by $v\left(g\right)$ and this number is $r$. Therefore, $\left|v\left(x\right)-v\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$. Similarly, in this same one can see that the total number of edges labeled $i$ is is denoted by $e\left(i\right)$ is the same as the total number of edges labeled $g$ denoted by $e\left(g\right)$ and this number is $r$, while the number of edges labeled $f$ denoted by $e\left(f\right)$ and this number is $r+1$. Therefore, $\left|e\left(x\right)-e\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$.

**Theorem**

**2.**

**Proof.**

**Theorem**

**3.**

**Proof.**

**Case (1)**: $n\equiv 0\left(mod3\right)$. The total number of vertices labeled $i$ is denoted by $v\left(i\right)$ is the same as the total number of vertices labeled $g$ and denoted by $v\left(g\right)$ and this number is $r$, also the total number of vertices labeled $f$ denoted by $v\left(f\right)$ and this number is $r+1$, i.e., $v\left(i\right)=v\left(g\right)=r,v\left(f\right)=r+1$. Obviously,$\left|v\left(x\right)-v\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$. Similarly, in this same one can see that the number of edges labeled $i$ denoted by $e\left(i\right)$ the same as the number of edges labeled $f$ denoted by $e\left(f\right)$ and also equivalent to the number of edges labeled $g$ denoted by $e\left(g\right)$ and this number is $2r$. Consequently, $\left|e\left(x\right)-e\left(y\right)\right|=0,x\ne y$ and $x,y\in \left\{i,f,g\right\}$.

**Case (2)**: $n\equiv 1\left(mod3\right)$. The total number of vertices labeled $g$ is denoted by $v\left(g\right)$ is the same as the total number of vertices labeled $f$ denoted by $v\left(i\right)$ and this number is $r+1$, while the number of vertices of labeled $f$ denoted by $v\left(f\right)$ and this number is $r$. Consequently, $\left|v\left(x\right)-v\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$. Similarly, in this same one can see that the number of edges labeled $i$ denoted by $e\left(i\right)$ is the same as the number of edges labeled $f$ denoted by $e\left(f\right)$ and this number is $2r$, while the number of edges labeled $g$ denoted by $e\left(g\right)$ and this number is $2r+1$. Therefore, $\left|e\left(x\right)-e\left(y\right)\right|\le 1,x\ne y$ and $x,y\in \left\{i,f,g\right\}$.

**Case (3)**: $n\equiv 2\left(mod3\right)$. The total number of vertices labeled $i$ is denoted by $v\left(i\right)$ is the same as the total number of vertices labeled $f$ is denoted by $v\left(f\right)$ and also equivalent to the number of vertices labeled $g$ is denoted by $v\left(g\right)$ and this number is $r+1$. Therefore, $\left|v\left(x\right)-v\left(y\right)\right|=0,x\ne y$ and $x,y\in \left\{i,f,g\right\}$. Similarly, in this same one can see that the number of edges labeled $i$ denoted by $e\left(i\right)$ the same as the number of edges labeled $f$ denoted by $e\left(f\right)$ and also equivalent to the number of edges labeled $g$ denoted by $e\left(g\right)$ and this number is $2r+1$. Consequently, $\left|e\left(x\right)-e\left(y\right)\right|=0,x\ne y$ and $x,y\in \left\{i,f,g\right\}$.

**Theorem**

**4.**

**Proof.**

## 4. The Permuted Cordial Labeling for Union of Paths and Cycles

**Theorem**

**5.**

**Proof.**

**Theorem**

**6.**

**Proof.**

**Theorem**

**7.**

**Proof.**

## 5. Conclusions

## Author Contributions

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

## References

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when $n\equiv 0,1(mod3$) | and when $n\equiv 2(mod3$) |

$h\left({v}_{r}\right)=\left\{\begin{array}{ll}i\hfill & ;0mod3\hfill \\ f\hfill & ;1mod3\hfill \\ g\hfill & ;2mod3\hfill \end{array}\right.$ | $h\left({v}_{r}\right)=\left\{\begin{array}{ll}g\hfill & ;r=0mod3,r\ne n-1\hfill \\ i\hfill & ;r=1mod3,r\ne n\hfill \\ f\hfill & ;r=2mod3\hfill \\ f\hfill & ;r=n-1\hfill \\ g\hfill & ;r=n\hfill \end{array}\right.$ |

${h}^{*}\left(e\right)$ | $.$ | $v\left(i\right)$ | $v\left(f\right)$ | $v\left(g\right)$ |

$v\left(i\right)$ | $i$ | $f$ | $g$ | |

$v\left(f\right)$ | $f$ | $g$ | $i$ | |

$v\left(g\right)$ | $g$ | $i$ | $f$ |

$n$ | $v\left(i\right)$ | $v\left(f\right)$ | $v\left(g\right)$ | $\begin{array}{l}\left|v\left(x\right)-v\left(y\right)\right|x\ne y\hfill \\ \mathrm{and}x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$\begin{array}{l}n\equiv 0mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r\hfill \end{array}$, | $r$ | $r$ | $r$ | $0,0,0$ |

$\begin{array}{l}n\equiv 1mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+1\hfill \end{array},$ | $r+1$ | $r$ | $r$ | $1,1,0$ |

$\begin{array}{l}n\equiv 2mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+2\hfill \end{array},$ | $r$ | $r+1$ | $r+1$ | $1,1,0$ |

$n$ | $e\left(i\right)$ | $e\left(f\right)$ | $e\left(g\right)$ | $\begin{array}{l}\left|e\left(x\right)-e\left(y\right)\right|x\ne y\hfill \\ \mathrm{and}x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$\begin{array}{l}n\equiv 0mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r\hfill \end{array}$, | $r$ | $r$ | $r$ | $0,0,0$ |

$\begin{array}{l}n\equiv 1mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+1\hfill \end{array},$ | $r+1$ | $r$ | $r$ | $1,1,0$ |

$\begin{array}{l}n\equiv 2mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+2\hfill \end{array},$ | $r$ | $r+1$ | $r+1$ | $1,1,0$ |

when $n\equiv 0(mod3$) | when $n\equiv 1(mod3$) | and when $n\equiv 2(mod3$), $n$ odd |

$h\left(u\right)=i$ $h\left({v}_{r}\right)=\left\{\begin{array}{ll}i\hfill & ;0mod3\hfill \\ f\hfill & ;1mod3\hfill \\ g\hfill & ;2mod3\hfill \end{array}\right.$ | $h\left(u\right)=g$ $h\left({v}_{r}\right)=\left\{\begin{array}{ll}i\hfill & ;0mod3\hfill \\ f\hfill & ;1mod3\hfill \\ g\hfill & ;2mod3\hfill \end{array}\right.$ | $h\left(u\right)=g$ $h\left({v}_{r}\right)=\left\{\begin{array}{ll}i\hfill & ;0mod3\hfill \\ f\hfill & ;1mod3\hfill \\ g\hfill & ;2mod3\hfill \end{array}\right.$ |

when $n\equiv 0(mod3$) | when $n\equiv 1(mod3$) | and when $n\equiv 2(mod3$), $n$ odd |

$h\left(u\right)=i$ $h\left({v}_{r}\right)=\left\{\begin{array}{ll}i\hfill & ;0mod3\hfill \\ f\hfill & ;1mod3\hfill \\ g\hfill & ;2mod3\hfill \end{array}\right.$ | $h\left(u\right)=g$ $h\left({v}_{r}\right)=\left\{\begin{array}{ll}i\hfill & ;0mod3\hfill \\ f\hfill & ;1mod3,\text{}r\ne n\hfill \\ g\hfill & ;2mod3\hfill \\ i\hfill & ;r=n\hfill \end{array}\right.$ | $h\left(u\right)=i$ $h\left({v}_{r}\right)=\left\{\begin{array}{ll}g\hfill & ;r=0mod3,r\ne n-1\hfill \\ i\hfill & ;r=1mod3,r\ne n\hfill \\ f\hfill & ;r=2mod3\hfill \\ f\hfill & ;r=n-1\hfill \\ g\hfill & ;r=n\hfill \end{array}\right.$ |

${h}^{*}\left(e\right)$ | $.$ | $v\left(i\right)$ | $v\left(f\right)$ | $v\left(g\right)$ |

$v\left(i\right)$ | $i$ | $f$ | $g$ | |

$v\left(f\right)$ | $f$ | $g$ | $i$ | |

$v\left(g\right)$ | $g$ | $i$ | $f$ |

$n$ | $v\left(i\right)$ | $v\left(f\right)$ | $v\left(g\right)$ | $\begin{array}{l}\left|v\left(x\right)-v\left(y\right)\right|x\ne y\hfill \\ \mathrm{and}x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$\begin{array}{l}n\equiv 0mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r\hfill \end{array}$, | $r+1$ | $r$ | $r$ | $1,1,0$ |

$\begin{array}{l}n\equiv 1mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+1\hfill \end{array},$ | $r+1$ | $r$ | $r+1$ | $1,0,1$ |

$\begin{array}{l}n\equiv 2mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+2\hfill \end{array},$ $n$ odd | $r+1$ | $r+1$ | $r+1$ | $0,0,0$ |

$n$ | $e\left(i\right)$ | $e\left(f\right)$ | $e\left(g\right)$ | $\begin{array}{l}\left|e\left(x\right)-e\left(y\right)\right|x\ne y\hfill \\ \mathrm{and}x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$\begin{array}{l}n\equiv 0mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r\hfill \end{array}$, | $2r$ | $2r$ | $2r$ | $0,0,0$ |

$\begin{array}{l}n\equiv 1mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+1\hfill \end{array},$ | $2r+1$ | $2r$ | $2r+1$ | $1,0,1$ |

$\begin{array}{l}n\equiv 2mod3\hfill \\ \mathrm{i}.\mathrm{e}.,n=3r+2\hfill \end{array},n$ odd | $2r+2$ | $2r+1$ | $2r+1$ | $1,1,0$ |

$\begin{array}{l}n=3r+j\hfill \\ r\ge 0,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\mathrm{Labeling}\text{}\mathrm{of}\hfill \\ {P}_{n}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ v\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ v\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ v\left(g\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ e\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ e\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ e\left(g\right)\hfill \\ \hfill \end{array}$ |

$\begin{array}{l}j=0\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{A}_{0}={A}_{3r}\hfill \\ {A}_{0}^{\prime}={D}_{3r}\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r-1\hfill \end{array}$ | $\begin{array}{l}r-1\hfill \\ r\hfill \end{array}$ |

$j=1$ | ${A}_{1}={D}_{3r}\left(f\right)$ | $r$ | $r+1$ | $r$ | $r$ | $r$ | $r$ |

$j=2$ | ${A}_{2}={A}_{3r}\left(gi\right)$ | $r+1$ | $r$ | $r+1$ | $r$ | $r+1$ | $r$ |

$\begin{array}{l}m=3s+j\hfill \\ s\ge 0,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\mathrm{Labeling}\text{}\mathrm{of}\hfill \\ {P}_{m}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {v}^{\prime}\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {v}^{\prime}\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {v}^{\prime}\left(g\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {e}^{\prime}\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {e}^{\prime}\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {e}^{\prime}\left(g\right)\hfill \\ \hfill \end{array}$ |

$\begin{array}{l}j=0\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{0}={A}_{3s}\hfill \\ {B}_{0}^{\prime}={D}_{3s}\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s-1\hfill \end{array}$ | $\begin{array}{l}s-1\hfill \\ s\hfill \end{array}$ |

$\begin{array}{l}j=1\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{1}={A}_{3s}\left(i\right)\hfill \\ {B}_{1}^{\prime}={A}_{3s}\left(g\right)\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ |

$\begin{array}{l}j=2\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{2}={A}_{3s}\left(gi\right)\hfill \\ {B}_{2}^{\prime}={D}_{3s}\left(fi\right)\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \end{array}$ |

$\begin{array}{l}n=3r+j\hfill \\ r\ge 0,\hfill \\ j=0,1,2\hfill \end{array}$, | $\begin{array}{l}m=3s+j,\hfill \\ s\ge 0,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\hfill \\ {P}_{n}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {P}_{m}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\left|v\left(x\right)-v\left(y\right)\right|\hfill \\ ,x\ne y\mathrm{and}\hfill \\ x,y\in \left\{i,f,g\right\}\hfill \end{array}$ | $\begin{array}{l}\left|e\left(x\right)-e\left(y\right)\right|\hfill \\ ,x\ne y\mathrm{and}\hfill \\ x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$0$ | $0$ | ${A}_{0}$ | ${B}_{0}^{\prime}$ | $0,0,0$ | $1,1,0$ |

$0$ | $1$ | ${A}_{0}^{\prime}$ | ${B}_{1}$ | $1,1,0$ | $1,0,1$ |

$0$ | $2$ | ${A}_{0}^{\prime}$ | ${B}_{2}$ | $1,0,1$ | $0,0,0$ |

$1$ | $1$ | ${A}_{1}$ | ${B}_{1}^{\prime}$ | $1,1,0$ | $0,0,0$ |

$1$ | $2$ | ${A}_{1}$ | ${B}_{2}$ | $0,0,0$ | $1,0,1$ |

$2$ | $2$ | ${A}_{2}$ | ${B}_{2}^{\prime}$ | $1,1,0$ | $1,1,0$ |

$\begin{array}{l}n=3r+j,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\mathrm{Labeling}\text{}\mathrm{of}\hfill \\ {C}_{n}\hfill \end{array}$ | $\begin{array}{l}v\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}v\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}v\left(g\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}e\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}e\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}e\left(g\right)\hfill \\ \hfill \end{array}$ |

$\begin{array}{l}j=0\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{A}_{0}={A}_{3r}\hfill \\ {A}_{0}^{\prime}={D}_{3r}\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ |

$\begin{array}{l}j=1\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{A}_{1}={D}_{3r}\left(f\right)\hfill \\ {A}_{1}^{\prime}={A}_{3r}\left(i\right)\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ |

$\begin{array}{l}j=2\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{A}_{2}={A}_{3r}\left(gi\right)\hfill \\ {A}_{2}^{\prime}={D}_{3s}\left(fi\right)\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ |

$\begin{array}{l}m=3s+j,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\mathrm{Labeling}\text{}\mathrm{of}\hfill \\ {C}_{n}\hfill \end{array}$ | $\begin{array}{l}{v}^{\prime}\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{v}^{\prime}\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{v}^{\prime}\left(g\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{e}^{\prime}\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{e}^{\prime}\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{e}^{\prime}\left(g\right)\hfill \\ \hfill \end{array}$ |

$\begin{array}{l}j=0\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{0}={A}_{3s}\hfill \\ {B}_{0}^{\prime}={D}_{3s}\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ |

$\begin{array}{l}j=1\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{1}={A}_{3s}\left(i\right)\hfill \\ {B}_{1}^{\prime}={A}_{3s}\left(g\right)\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \end{array}$ |

$\begin{array}{l}\hfill \\ j=2\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{2}={A}_{3s}\left(gi\right)\hfill \\ {B}_{2}^{\prime}={D}_{3s}\left(fi\right)\hfill \\ {B}_{2}^{\u2033}={B}_{3s}\left(fg\right)\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \\ s+1\hfill \end{array}$ |

$\begin{array}{l}n=3r+j\hfill \\ r\ge 1,\hfill \\ j=0,1,2\hfill \end{array}$, | $\begin{array}{l}m=3s+j\hfill \\ s\ge 1,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\hfill \\ {C}_{n}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {C}_{m}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\left|v\left(x\right)-v\left(y\right)\right|\hfill \\ x\ne y\mathrm{and}\hfill \\ x,y\in \left\{i,f,g\right\}\hfill \end{array}$ | $\begin{array}{l}\left|e\left(x\right)-e\left(y\right)\right|\hfill \\ x\ne y\mathrm{and}\hfill \\ x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$0$ | $0$ | ${A}_{0}$ | ${B}_{0}$ | $0,0,0$ | $0,0,0$ |

$0$ | $1$ | ${A}_{0}$ | ${B}_{1}$ | $1,1,0$ | $1,1,0$ |

$0$ | $2$ | ${A}_{0}$ | ${B}_{2}$ | $1,0,1$ | $0,1,1$ |

$1$ | $1$ | ${A}_{1}$ | ${B}_{1}$ | $0,1,1$ | $0,1,1$ |

$1$ | $2$ | ${A}_{1}^{\prime}$ | ${B}_{2}^{\u2033}$ | $0,0,0$ | $0,0,0$ |

$2$ | $2$ | ${A}_{2}$ | ${B}_{2}^{\prime}$ | $1,1,0$ | $1,1,0$ |

$\begin{array}{l}n=3r+j\hfill \\ j=0,1,2\hfill \end{array}$, | $\begin{array}{l}\mathrm{Labeling}\text{}\mathrm{of}\hfill \\ {P}_{n}\hfill \end{array}$ | $\begin{array}{l}v\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}v\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}v\left(g\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}e\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}e\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}e\left(g\right)\hfill \\ \hfill \end{array}$ |

$\begin{array}{l}j=0\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{A}_{0}={A}_{3r}\hfill \\ {A}_{0}^{\prime}={B}_{3r}\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r-1\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r-1\hfill \\ r\hfill \end{array}$ |

$j=1$ | ${A}_{1}={A}_{3r}\left(i\right)$ | $r+1$ | $r$ | $r$ | $r$ | $r$ | $r$ |

$\begin{array}{l}j=2\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{A}_{2}={A}_{3r}\left(if\right)\hfill \\ {A}_{2}^{\prime}={A}_{3r}\left(gf\right)\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r\hfill \\ r+1\hfill \end{array}$ | $\begin{array}{l}r+1\hfill \\ r-1\hfill \end{array}$ |

$\begin{array}{l}m=3s+j\hfill \\ j=0,1,2\hfill \end{array}$, | $\begin{array}{l}\mathrm{Labeling}\text{}\mathrm{of}\hfill \\ {C}_{m}\hfill \end{array}$ | $\begin{array}{l}{v}^{\prime}\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{v}^{\prime}\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{v}^{\prime}\left(g\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{e}^{\prime}\left(i\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{e}^{\prime}\left(f\right)\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{e}^{\prime}\left(g\right)\hfill \\ \hfill \end{array}$ |

$j=0$ | ${B}_{0}={A}_{3s}$ | $s$ | $s$ | $s$ | $s$ | $s$ | $s$ |

$\begin{array}{l}\hfill \\ j=1\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{1}={A}_{3s}\left(i\right)\hfill \\ {B}_{1}^{\prime}={A}_{3s}\left(g\right)\hfill \\ {B}_{1}^{*}={D}_{3s}\left(g\right)\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \\ s-1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \\ s+1\hfill \end{array}$ |

$\begin{array}{l}\hfill \\ j=2\hfill \\ \hfill \end{array}$ | $\begin{array}{l}{B}_{2}={B}_{3s}\left(gi\right)\hfill \\ {B}_{2}^{\prime}={B}_{3s}\left(fg\right)\hfill \\ {B}_{2}^{*}={A}_{3s}\left(gi\right)\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \\ s\hfill \end{array}$ | $\begin{array}{l}s+1\hfill \\ s+1\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s\hfill \\ s+1\hfill \\ s+1\hfill \end{array}$ | $\begin{array}{l}s+2\hfill \\ s+1\hfill \\ s\hfill \end{array}$ |

$\begin{array}{l}n=3r+j\hfill \\ r\ge 0,\hfill \\ j=0,1,2\hfill \end{array}$, | $\begin{array}{l}m=3s+j,\hfill \\ s\ge 1,\hfill \\ j=0,1,2\hfill \end{array}$ | $\begin{array}{l}\hfill \\ {P}_{n}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\hfill \\ {C}_{m}\hfill \\ \hfill \end{array}$ | $\begin{array}{l}\left|v\left(x\right)-v\left(y\right)\right|\hfill \\ ,x\ne y\mathrm{and}\hfill \\ x,y\in \left\{i,f,g\right\}\hfill \end{array}$ | $\begin{array}{l}\left|e\left(x\right)-e\left(y\right)\right|\hfill \\ ,x\ne y\mathrm{and}\hfill \\ x,y\in \left\{i,f,g\right\}\hfill \end{array}$ |

$0$ | $0$ | ${A}_{0}$ | ${B}_{0}$ | $0,0,0$ | $0,1,1$ |

$0$ | $1$ | ${A}_{0}^{\prime}$ | ${B}_{1}$ | $1,1,0$ | $0,0,0$ |

$0$ | $2$ | ${A}_{0}$ | ${B}_{2}$ | $1,0,1$ | $0,1,1$ |

$1$ | $0$ | ${A}_{1}$ | ${B}_{0}$ | $1,1,0$ | $0,0,0$ |

$1$ | $1$ | ${A}_{1}$ | ${B}_{1}^{\prime}$ | $1,0,1$ | $1,0,1$ |

$1$ | $2$ | ${A}_{1}$ | ${B}_{2}^{\prime}$ | $0,0,0$ | $1,1,0$ |

$2$ | $0$ | ${A}_{2}$ | ${B}_{0}$ | $0,1,1$ | $0,1,1$ |

$2$ | $1$ | ${A}_{2}$ | ${B}_{1}^{\prime}$ | $0,0,0$ | $1,1,0$ |

$2$ | $2$ | ${A}_{2}$ | ${B}_{2}^{*}$ | $1,1,0$ | $0,0,0$ |

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## Share and Cite

**MDPI and ACS Style**

ELrokh, A.; Al-Shamiri, M.M.A.; Almazah, M.M.A.; El-hay, A.A.
A Novel Problem for Solving Permuted Cordial Labeling of Graphs. *Symmetry* **2023**, *15*, 825.
https://doi.org/10.3390/sym15040825

**AMA Style**

ELrokh A, Al-Shamiri MMA, Almazah MMA, El-hay AA.
A Novel Problem for Solving Permuted Cordial Labeling of Graphs. *Symmetry*. 2023; 15(4):825.
https://doi.org/10.3390/sym15040825

**Chicago/Turabian Style**

ELrokh, Ashraf, Mohammed M. Ali Al-Shamiri, Mohammed M. A. Almazah, and Atef Abd El-hay.
2023. "A Novel Problem for Solving Permuted Cordial Labeling of Graphs" *Symmetry* 15, no. 4: 825.
https://doi.org/10.3390/sym15040825