# Defuzzification of Non-Linear Pentagonal Intuitionistic Fuzzy Numbers and Application in the Minimum Spanning Tree Problem

## Abstract

**:**

## 1. Introduction

- The non-linear pentagonal intuitionistic fuzzy number is defined.
- Arithmetic operations of the NLPIFN are formulated using $(\alpha -\beta )$ cuts representation.
- The short-cut formula of the intuitionistic fuzzy weighted averaging based on levels (IF-WABL) value of NLPIFN is derived.
- A solution approach for the intuitionistic fuzzy minimum spanning tree (IF MST) problem based on the IF-WABL value of NLPIFN has been proposed.

## 2. Preliminaries

**Definition**

**1**

**Definition**

**2**

**Definition**

**3**

**Definition**

**4**

**Definition**

**5**

- An intuitionistic fuzzy subset of the real line.
- Normal, that is, there is some ${x}_{0}\in \mathcal{R}$, such that ${\mu}_{\stackrel{~}{A}}\left({x}_{0}\right)=1$ and ${\nu}_{\stackrel{~}{A}}\left({x}_{0}\right)=0$.
- Convex for the membership function ${\mu}_{{A}^{*}}\left(x\right)$, that is, ${\mu}_{\stackrel{~}{A}}\left({\lambda x}_{1}+{(1-\lambda )x}_{2}\right)\ge \mathrm{m}\mathrm{i}\mathrm{n}({\mu}_{\stackrel{~}{A}}\left({x}_{1}\right),{\mu}_{\stackrel{~}{A}}\left({x}_{2}\right))$ for every ${x}_{1},{x}_{2}\in \mathcal{R}$, $\lambda \in \left[\mathrm{0,1}\right]$.
- Concave for the non-membership function ${\nu}_{\stackrel{~}{A}}\left(x\right)$, that is, ${\nu}_{\stackrel{~}{A}}\left({\lambda x}_{1}+{(1-\lambda )x}_{2}\right)\le \mathrm{m}\mathrm{a}\mathrm{x}({\nu}_{\stackrel{~}{A}}\left({x}_{1}\right),{\nu}_{\stackrel{~}{A}}\left({x}_{2}\right))$ for every ${x}_{1},{x}_{2}\in \mathcal{R}$, $\lambda \in \left[\mathrm{0,1}\right]$.

**Definition**

**6**

**Definition**

**7.**

- $\alpha ,\beta \in \left[\mathrm{0,1}\right]$ and $\alpha +\beta \le 1$;
- $\stackrel{~}{{A}_{\alpha}^{\mu}}=\left\{\begin{array}{cc}\left\{x\in R:{\mu}_{\stackrel{~}{A}}\left(x\right)\ge \alpha \right\}& if\alpha \in \left(\mathrm{0,1}\right]\\ cl\left\{x\in R:{\mu}_{\stackrel{~}{A}}\left(x\right)>0\right\}& if\alpha =0\end{array}\right.;$
- $\stackrel{~}{{A}_{\alpha}^{\nu}}=\left\{\begin{array}{cc}\left\{x\in R:{\nu}_{\stackrel{~}{A}}\left(x\right)\le \beta \right\}& if\beta \in \left(\mathrm{0,1}\right]\\ cl\left\{x\in R:{\nu}_{\stackrel{~}{A}}\left(x\right)<1\right\}& if\beta =1\end{array}\right.$.

**Definition**

**8.**

**Definition**

**9.**

**Definition**

**10**

## 3. Representations of the NLPIFN and Its Arithmetic Operations

**Definition**

**11.**

**Lemma**

**1.**

**Proof**

**of**

**Lemma**

**1:**

- For ${a}_{1}^{\prime}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{1}^{\prime}\right)=0\mathrm{a}\mathrm{n}\mathrm{d}{\nu}_{\stackrel{~}{A}}\left({a}_{1}^{\prime}\right)=1.0$, so the condition is satisfied.
- For ${a}_{1}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{1}\right)=0$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{1}\right)={k}^{\prime}+(1-k\prime ){\left(\frac{{a}_{2}^{\prime}-{a}_{1}}{{a}_{2}^{\prime}-{a}_{1}^{\prime}}\right)}^{{n}_{1}^{\prime}}$. From Definition 11, we know that $\frac{{a}_{2}^{\prime}-{a}_{1}}{{a}_{2}^{\prime}-{a}_{1}^{\prime}}<1.0$. Based on the value of ${n}_{1}^{\prime}$, we have two possible limit values for ${\nu}_{\stackrel{~}{A}}\left({a}_{1}\right)$: (i) $\underset{{n}_{1}^{\prime}\to {0}^{+}}{\mathrm{lim}}{\nu}_{\stackrel{~}{A}}\left({a}_{1}\right)=1$ and (ii) $\underset{{n}_{1}^{\prime}\to \mathrm{\infty}}{\mathrm{lim}}{\nu}_{\stackrel{~}{A}}\left({a}_{1}\right)={k}^{\prime}$. As a result, the condition is satisfied.
- For ${a}_{2}^{\prime}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{2}^{\prime}\right)=k{\left(\frac{{a}_{2}^{\prime}-{a}_{1}}{{a}_{2}-{a}_{1}}\right)}^{{n}_{1}}$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{2}^{\prime}\right)={k}^{\prime}$. From Definition 11, we know that $\frac{{a}_{2}^{\prime}-{a}_{1}}{{a}_{2}-{a}_{1}}<1.0$. Based on the value of ${n}_{1}$, we have two possible limit values for ${\mu}_{\stackrel{~}{A}}\left({a}_{2}^{\prime}\right)$: (i) $\underset{{n}_{1}\to {0}^{+}}{\mathrm{lim}}{\mu}_{\stackrel{~}{A}}\left({a}_{2}^{\prime}\right)=k$ and (ii) $\underset{{n}_{1}\to \mathrm{\infty}}{\mathrm{lim}}{\mu}_{\stackrel{~}{A}}\left({a}_{2}^{\prime}\right)=0$. As a result, the condition is satisfied.
- For ${a}_{2}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{2}\right)=k$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{2}\right)=k\prime {\left(\frac{{a}_{3}-{a}_{2}}{{a}_{3}-{a}_{2}^{\prime}}\right)}^{{n}_{2}^{\prime}}$. From Definition 11, we know that $\frac{{a}_{3}-{a}_{2}}{{a}_{3}-{a}_{2}^{\prime}}<1.0$. Based on the value of ${n}_{2}^{\prime}$, we have two possible limit values for ${\nu}_{\stackrel{~}{A}}\left({a}_{2}\right)$: (i) $\underset{{n}_{2}^{\prime}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\nu}_{\stackrel{~}{A}}\left({a}_{2}\right)=k\prime $ and (ii) $\underset{{n}_{2}^{\prime}\to \mathrm{\infty}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\nu}_{\stackrel{~}{A}}\left({a}_{2}\right)=0$. As a result, the condition is satisfied.
- For ${a}_{3}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{3}\right)=1.0$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{3}\right)=0$, so the condition is satisfied.
- For ${a}_{4}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{4}\right)=k$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{4}\right)=k\prime {\left(\frac{{a}_{4}-{a}_{3}}{{a}_{4}^{\prime}-{a}_{3}}\right)}^{{m}_{2}^{\prime}}$. From Definition 11, we know that $\frac{{a}_{4}-{a}_{3}}{{a}_{4}^{\prime}-{a}_{3}}<1.0$. Based on the value of ${m}_{2}^{\prime}$, we have two possible limit values for ${\nu}_{\stackrel{~}{A}}\left({a}_{4}\right)$: (i) $\underset{{m}_{2}^{\prime}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\nu}_{\stackrel{~}{A}}\left({a}_{4}\right)=k\prime $ and (ii) $\underset{{m}_{2}^{\prime}\to \mathrm{\infty}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\nu}_{\stackrel{~}{A}}\left({a}_{4}\right)=0$. As a result, the condition is satisfied.
- For ${a}_{4}^{\prime}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{4}^{\prime}\right)=k{\left(\frac{{a}_{5}-{a}_{4}^{\prime}}{{a}_{5}-{a}_{4}}\right)}^{{m}_{1}}$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{4}^{\prime}\right)=k\prime $. From Definition 11, we know that $\frac{{a}_{5}-{a}_{4}^{\prime}}{{a}_{5}-{a}_{4}}<1.0$. Based on the value of ${m}_{1}$, we have two possible limit values for ${\mu}_{\stackrel{~}{A}}\left({a}_{4}^{\prime}\right)$: (i) $\underset{{m}_{1}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\mu}_{\stackrel{~}{A}}\left({a}_{4}^{\prime}\right)=k$ and (ii) $\underset{{m}_{1}\to \mathrm{\infty}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\mu}_{\stackrel{~}{A}}\left({a}_{4}^{\prime}\right)=0$. As a result, the condition is satisfied.
- For ${a}_{5}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{5}\right)=0$ and ${\nu}_{\stackrel{~}{A}}\left({a}_{5}\right)={k}^{\prime}+(1-k\prime ){\left(\frac{x-{a}_{4}^{\prime}}{{a}_{5}^{\prime}-{a}_{4}^{\prime}}\right)}^{{m}_{1}^{\prime}}$. From Definition 11, we know that $\frac{x-{a}_{4}^{\prime}}{{a}_{5}^{\prime}-{a}_{4}^{\prime}}<1.0$. Based on the value of ${m}_{1}^{\prime}$, we have two possible limit values for ${\nu}_{\stackrel{~}{A}}\left({a}_{5}\right)$: (i) $\underset{{m}_{1}^{\prime}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\nu}_{\stackrel{~}{A}}\left({a}_{5}\right)=1$ and (ii) $\underset{{m}_{1}^{\prime}\to \mathrm{\infty}}{\mathrm{lim}}\mathrm{f}\mathrm{o}\mathrm{r}{\nu}_{\stackrel{~}{A}}\left({a}_{5}\right)={k}^{\prime}$. As a result, the condition is satisfied.
- For ${a}_{5}^{\prime}$, ${\mu}_{\stackrel{~}{A}}\left({a}_{5}^{\prime}\right)=0\mathrm{a}\mathrm{n}\mathrm{d}{\nu}_{\stackrel{~}{A}}\left({a}_{5}^{\prime}\right)=1.0$, so the condition is satisfied. □

**Definition**

**12.**

## 4. Definition of IF-WABL and Formula for NLPIFNs

**Definition**

**13.**

**Definition**

**14.**

**Lemma**

**2.**

**Proof**

**of**

**Lemma**

**2:**

**Theorem**

**1.**

**Proof**

**of**

**Theorem**

**1:**

- $\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\mathsf{\gamma}\stackrel{~}{A}\right)=\mathsf{\gamma}\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\stackrel{~}{A}\right)$, where $\mathsf{\gamma}\ge 0$ and $\mathsf{\gamma}\in \mathrm{R}$.
- $\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\stackrel{~}{A}+\stackrel{~}{B}\right)=$ $\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\stackrel{~}{A}\right)+\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\stackrel{~}{B}\right)$.
- If $\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\stackrel{~}{A}\right)<\mathrm{I}\mathrm{F}-\mathrm{W}\mathrm{A}\mathrm{B}\mathrm{L}\left(\stackrel{~}{B}\right)$, then $\stackrel{~}{A}\stackrel{~}{<}\stackrel{~}{B}$ for IFNs $\stackrel{~}{A}$ and $\stackrel{~}{B}$.

**Theorem**

**2.**

**Proof**

**of**

**Theorem**

**2:**

**Example**

**1.**

## 5. Application of NLPIFNs in the IF Minimum Spanning Tree Problem

_{new}= {x}. Set E

_{new}= $\varnothing $.

_{new}and v is not. If there is more than one arc with the same IF-WABL defuzzified weight, arbitrarily pick one of the arcs.

_{new}and (u, v) to E

_{new}.

_{new}= V, then STOP, keep E

^{*}= E

_{new}, which describes the IFMST, and compute the total IF-WABL defuzzified values of the NLPIFN weights that belong to the chosen arcs to determine the solution for the IF MSTP. Otherwise, go to Step 2.

**Example**

**2.**

^{*}= {(4,3), (3,2), (3,8), (8,6), (6,5), (6,7), and (7,1)} for the problem. A graphical illustration of the final solution is given in Figure 4, where red lines are used to emphasize the chosen arcs. The total edge weight was equal to 51.099.

## 6. Conclusions

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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Components of the NLPIFN | Type of IFN |
---|---|

${n}_{1}={n}_{2}={m}_{1}={m}_{2}={n}_{1}^{\prime}={n}_{2}^{\prime}={m}_{1}^{\prime}={m}_{2}^{\prime}=1$ | Linear Pentagonal IFN |

$k=1$ , ${k}^{\prime}=0$ | Non-linear Trapezoidal IFN |

$k=1$ , ${k}^{\prime}=0$, ${n}_{1}={m}_{1}={n}_{1}^{\prime}={m}_{1}^{\prime}=1$ | Linear Trapezoidal IFN |

$k=0$ , ${k}^{\prime}=1$ | Non-linear Triangular IFN |

$k=0$ , ${k}^{\prime}=1$ , ${n}_{2}={m}_{2}={n}_{2}^{\prime}={m}_{2}^{\prime}=1$ | Linear Triangular IFN |

$k=1$ , ${k}^{\prime}=0$, ${n}_{1}={m}_{1}={n}_{1}^{\prime}={m}_{1}^{\prime}=\delta $ | Baloui’s Generalized IFN |

$k=0$ , ${k}^{\prime}=1$ , ${n}_{2}={m}_{2}={n}_{2}^{\prime}={m}_{2}^{\prime}=\delta $ | Baloui’s Generalized IFN |

Values of Parameters | ${\mathit{c}}_{\mathit{L}}=0.2$ ${\mathit{c}}_{\mathit{R}}=0.8$ | ${\mathit{c}}_{\mathit{L}}=0.5$ ${\mathit{c}}_{\mathit{R}}=0.6$ | ${\mathit{c}}_{\mathit{L}}=0.6$ ${\mathit{c}}_{\mathit{R}}=0.4$ | ${\mathit{c}}_{\mathit{L}}=0.9$ ${\mathit{c}}_{\mathit{R}}=0.1$ | |
---|---|---|---|---|---|

$t=0.3$ | $l=0.45$ | 11.19250 | 9.51232 | 8.95226 | 7.27208 |

$l=0.75$ | 11.12610 | 9.38482 | 8.80438 | 7.06306 | |

$l=2.0$ | 10.89610 | 9.04328 | 8.42566 | 6.57280 | |

$l=3.7$ | 10.71070 | 8.82510 | 8.19659 | 6.31105 | |

$t=0.6$ | $l=0.45$ | 11.12430 | 9.50938 | 8.97107 | 7.35615 |

$l=0.75$ | 11.05800 | 9.38189 | 8.82320 | 7.14714 | |

$l=2.0$ | 10.82790 | 9.04033 | 8.44447 | 6.65686 | |

$l=3.7$ | 10.64250 | 8.82218 | 8.21541 | 6.39513 | |

$t=3.0$ | $l=0.45$ | 10.92930 | 9.53205 | 9.06630 | 7.66906 |

$l=0.75$ | 10.86290 | 9.40455 | 8.91842 | 7.46004 | |

$l=2.0$ | 10.63290 | 9.06300 | 8.53969 | 6.96977 | |

$l=3.7$ | 10.44740 | 8.84484 | 8.31064 | 6.70803 | |

$t=4.2$ | $l=0.45$ | 10.90790 | 9.54557 | 9.09146 | 7.72912 |

$l=0.75$ | 10.84160 | 9.41807 | 8.94358 | 7.52011 | |

$l=2.0$ | 10.61155 | 9.07652 | 8.56485 | 7.02983 | |

$l=3.7$ | 10.42610 | 8.85836 | 8.33580 | 6.76810 |

Weight | ${\mathit{a}}_{1}^{\prime}$ | ${\mathit{a}}_{1}$ | ${\mathit{a}}_{2}^{\prime}$ | ${\mathit{a}}_{2}$ | ${\mathit{a}}_{3}$ | ${\mathit{a}}_{4}$ | ${\mathit{a}}_{4}^{\prime}$ | ${\mathit{a}}_{5}$ | ${\mathit{a}}_{5}^{\prime}$ | IF-WABL |
---|---|---|---|---|---|---|---|---|---|---|

$\stackrel{~}{{A}_{1}}$ | 10.6 | 11.47 | 11.82 | 12.12 | 12.62 | 12.66 | 12.85 | 13.77 | 13.94 | 12.761 |

$\stackrel{~}{{A}_{2}}$ | 2.07 | 3.07 | 3.99 | 4.34 | 4.44 | 5.06 | 5.11 | 5.96 | 6.38 | 4.914 |

$\stackrel{~}{{A}_{3}}$ | 4.91 | 5.52 | 6.10 | 6.31 | 6.72 | 6.89 | 7.72 | 8.21 | 8.52 | 7.151 |

$\stackrel{~}{{A}_{4}}$ | 5.38 | 6.15 | 6.40 | 6.79 | 7.20 | 7.68 | 8.24 | 8.33 | 8.71 | 7.568 |

$\stackrel{~}{{A}_{5}}$ | 6.75 | 7.30 | 7.34 | 7.46 | 7.88 | 8.83 | 9.18 | 9.82 | 9.98 | 8.685 |

$\stackrel{~}{{A}_{6}}$ | 8.97 | 9.32 | 10.13 | 10.40 | 10.51 | 11.17 | 11.72 | 11.77 | 12.70 | 11.238 |

$\stackrel{~}{{A}_{7}}$ | 2.93 | 3.83 | 3.88 | 4.36 | 4.49 | 4.51 | 5.27 | 5.81 | 6.68 | 5.081 |

$\stackrel{~}{{A}_{8}}$ | 10.85 | 11.74 | 12.16 | 12.89 | 13.22 | 13.71 | 14.44 | 15.27 | 16.14 | 14.034 |

$\stackrel{~}{{A}_{9}}$ | 8.04 | 8.72 | 9.61 | 9.64 | 10.41 | 10.52 | 10.96 | 11.04 | 11.47 | 10.401 |

$\stackrel{~}{{A}_{10}}$ | 5.50 | 5.80 | 6.54 | 6.59 | 6.88 | 7.16 | 7.94 | 8.39 | 8.43 | 7.327 |

$\stackrel{~}{{A}_{11}}$ | 6.10 | 6.55 | 7.42 | 7.77 | 8.17 | 8.58 | 9.29 | 9.82 | 10.00 | 8.616 |

$\stackrel{~}{{A}_{12}}$ | 11.89 | 12.41 | 13.27 | 13.73 | 14.32 | 14.79 | 15.57 | 15.83 | 16.04 | 14.662 |

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**MDPI and ACS Style**

Mert, A.
Defuzzification of Non-Linear Pentagonal Intuitionistic Fuzzy Numbers and Application in the Minimum Spanning Tree Problem. *Symmetry* **2023**, *15*, 1853.
https://doi.org/10.3390/sym15101853

**AMA Style**

Mert A.
Defuzzification of Non-Linear Pentagonal Intuitionistic Fuzzy Numbers and Application in the Minimum Spanning Tree Problem. *Symmetry*. 2023; 15(10):1853.
https://doi.org/10.3390/sym15101853

**Chicago/Turabian Style**

Mert, Ali.
2023. "Defuzzification of Non-Linear Pentagonal Intuitionistic Fuzzy Numbers and Application in the Minimum Spanning Tree Problem" *Symmetry* 15, no. 10: 1853.
https://doi.org/10.3390/sym15101853