Some Information Measures Based on Centroid, Orthocenter, Circumcenter and Incenter Points of Transformed Triangular Fuzzy Numbers and their Applications

Abstract

Cognitive computing has deep extents, which embrace different features of cognition. In the decision-making process, multi-criteria decision making is credited as a cognitive-based human action. However, to treat and unite the information from several resources, the most vital stage is data collection. Intuitionistic fuzzy set (IFS) is one of the most robust and trustworthy tools to accomplish the imprecise information with the help of the membership degrees. In addition to this, an information measure plays an essential role in treating uncertain information to reach the final decision based on the degree of the separation between the pairs of the numbers. Motivated by these, this paper aims to present the novel information measures using four different centers namely centroid, orthocenter, circumcenter and incenter under the IFS environment to address the cognitive-based human decision-making problems. The present work is divided into three folds. The first fold is to propose a technique of transforming intuitionistic fuzzy values into general triangular fuzzy numbers (TFNs). The right-angled and isosceles TFNs are special cases of the proposed transformation technique. The second fold is to develop distance and similarity measures using four different centers namely centroid, orthocenter, circumcenter and incenter of transformed TFNs. The basic axioms of the proposed measures are investigated in detail. The third fold is to justify superiority and validity of the proposed measures. The effectiveness of the developed measures is examined by applying it in clustering as well as the pattern recognition problems, and their results are correlated with some prevailing studies. Additionally, a clustering technique is discussed based on the stated measures to classify the objects. A detailed comparative analysis is done with some of the existing measures and concludes that several existing measures fail to discriminate the results under the different instances such as division by zero problems or counter-intuitive cases while the proposed measure has successfully overcome this drawback.

Appendix

Proof of Theorem 1:

Proof

Let ${\mathcal {P}}=\{\langle x_j,\zeta _{\mathcal {P}}(x_j), \vartheta _{\mathcal {P}}(x_j)\rangle \mid j=1,2, \ldots , n \}$ and ${\mathcal {Q}}=\{\langle x_j,\zeta _{{\mathcal {Q}}}(x_j), \vartheta _{{\mathcal {Q}}}(x_j)\rangle \mid j=1,2, \ldots , n \}$ be two IFSs. Since, $0 \le \zeta _{\mathcal {P}}(x_j),\zeta _{{\mathcal {Q}}}(x_j),\vartheta _{\mathcal {P}}(x_j),\vartheta _{{\mathcal {Q}}}(x_j) \le 1$. Therefore, $-1 \le \zeta _{\mathcal {P}}(x_j)-\zeta _{{\mathcal {Q}}}(x_j) \le 1$ and $-1 \le \vartheta _{\mathcal {P}}(x_j)-\vartheta _{{\mathcal {Q}}}(x_j) \le 1$.

• (P1) For non-negative real number k, we have $-(k+2) \le (k+2)(\zeta _{\mathcal {P}}(x_j)-\zeta _{{\mathcal {Q}}}(x_j)) \le k+2$ and $-(2k+1) \le -(2k+1)(\vartheta _{\mathcal {P}}(x_j)-\vartheta _{{\mathcal {Q}}}(x_j)) \le 2k+1$ which implies that $-(3k+3) \le (k+2)(\zeta _{\mathcal {P}}(x_j)-\zeta _{{\mathcal {Q}}}(x_j))-(2k+1)(\vartheta _{\mathcal {P}}(x_j)-\vartheta _{{\mathcal {Q}}}(x_j)) \le 3k+3$. Thus, $0 \le \frac{\big |(k+2)(\zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j))-(2k+1)(\vartheta _{\mathcal {P}}(x_j)-\vartheta _{\mathcal {Q}}(x_j))\big |}{3(k+1)} \le 1$. Similarly, it can be obtained that, $0 \le \frac{\big |2(\vartheta _{\mathcal {P}}(x_j) - \vartheta _{\mathcal {Q}}(x_j)) - (\zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j))\big |}{3} \le 1$. Hence, using Eq. (4), we obtain that

  $$\begin{aligned} 0\le {\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right)\le & {} \sum \limits _{j=1}^n\psi _j\left[ \left( 1-\frac{h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)}{2}\right) +\left( \frac{h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)}{2}\right) \right] =\sum \limits _{j=1}^n\psi _j=1. \end{aligned}$$

• (P2) For ${\mathcal {P}}={\mathcal {Q}}$, we have $\zeta _{\mathcal {P}}(x_j)=\zeta _{\mathcal {Q}}(x_j)$, $\vartheta _{\mathcal {P}}(x_j)=\vartheta _{\mathcal {Q}}(x_j)$ and $h_{\mathcal {P}}(x_j)=h_{\mathcal {Q}}(x_j)$ $\forall$$j=1,2, \ldots ,n$. Then, from Eq. (4), we get ${\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right) =0$. Conversely, when ${\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right) =0$ then from Eq. (4), we have

  $$\begin{aligned} & \quad \left( \frac{\big |(k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) -(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) \big |}{3(k+1)}\right) \left( 1-\frac{h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)}{2}\right) \\ & + \left( \frac{\big |2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) \big |}{3}\right) \left( \frac{h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)}{2}\right) =0 \end{aligned}$$

  which implies

  $$\begin{aligned}&(k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) -(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) =0 ~~ \text {or} ~~ h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j) = 2 ~~~\\ & \text {and} \ 2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) =0 ~~ \text {or} ~~ h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j) = 0 \end{aligned}$$

  and hence it leads to formulation of three cases which are:

  • Case 1: When $(k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) -(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) = 0$ and $2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) = 0.$ From $2(\vartheta _{\mathcal {P}}(x_j) - \vartheta _{\mathcal {Q}}(x_j))-(\zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)) =0$, we obtain $\zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)=2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right)$. Using this relation in $(k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) -(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) = 0$, we obtain $3\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) = 0$ and it gives $\vartheta _{\mathcal {P}}(x_j) =\vartheta _{\mathcal {Q}}(x_j)$. Further, using $\vartheta _{\mathcal {P}}(x_j) =\vartheta _{\mathcal {Q}}(x_j)$ in the relation $\zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)=2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right)$, we get $\zeta _{\mathcal {P}}(x_j)=\zeta _{\mathcal {Q}}(x_j)$. Hence, ${\mathcal {P}}={\mathcal {Q}}$.

  • Case 2: When $(k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) -(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) = 0$ and $h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)=0$. Since, $h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)=0$ which implies that $\zeta _{\mathcal {P}}(x_j)+\vartheta _{\mathcal {P}}(x_j)=1$ and $\zeta _{\mathcal {Q}}(x_j)+\vartheta _{\mathcal {Q}}(x_j)$ $=1$. Thus, $\zeta _{\mathcal {P}}(x_j) - \zeta _{\mathcal {Q}}(x_j) = -\left( \vartheta _{\mathcal {P}}(x_j) - \vartheta _{\mathcal {Q}}(x_j)\right)$. Using this relation in $(k+2)\left( \zeta _{\mathcal {P}}(x_j)- \zeta _{\mathcal {Q}}(x_j)\right) - (2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) = 0$, we obtain that $-(3k+3)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) =0$ which further gives $\vartheta _{\mathcal {P}}(x_j)=\vartheta _{\mathcal {Q}}(x_j)$. Now, using $\vartheta _{\mathcal {P}}(x_j)=\vartheta _{\mathcal {Q}}(x_j)$ in the relation $\zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)$ $=-\left( \vartheta _{\mathcal {P}}(x_j)-\vartheta _{\mathcal {Q}}(x_j)\right)$, we obtain that $\zeta _{\mathcal {P}}(x_j)=\zeta _{\mathcal {Q}}(x_j)$. Hence, ${\mathcal {P}}={\mathcal {Q}}$.

  • Case 3: When $h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)=2$ and $2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) = 0$. Since, $h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j) = 2$ which implies that $\zeta _{\mathcal {P}}(x_j) + \vartheta _{\mathcal {P}}(x_j) + \zeta _{\mathcal {Q}}(x_j) + \vartheta _{\mathcal {Q}}(x_j)=0$ which is possible only when $\zeta _{\mathcal {P}}(x_j)=\vartheta _{\mathcal {P}}(x_j) = \zeta _{\mathcal {Q}}(x_j) = \vartheta _{\mathcal {Q}}(x_j)=0$. Hence, ${\mathcal {P}}={\mathcal {Q}}$.

  Thus, in all cases, we obtain ${\mathcal {P}}={\mathcal {Q}}$ when ${\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right) =0$.

• (P3) For any two positive real number a, b, we know $\mid a - b\mid = \mid b-a\mid$. Thus, from Eq. (4), we can easily obtain that ${\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right) = {\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {Q}},{\mathcal {P}}\right)$.

• (P4) Let ${\mathcal {R}}=\{\langle x_j,\zeta _{{\mathcal {R}}}(x_j), \vartheta _{{\mathcal {R}}}(x_j)\rangle \mid j=1,2, \ldots , n \}$ be an IFN such that ${\mathcal {P}}\subseteq {\mathcal {Q}}\subseteq {\mathcal {R}}$. Therefore, by using Definition 3, we obtain that $\zeta _{\mathcal {P}}(x_j) \le \zeta _{{\mathcal {Q}}}(x_j) \le \zeta _{{\mathcal {R}}}(x_j)$ and $\vartheta _{\mathcal {P}}(x_j) \ge \vartheta _{{\mathcal {Q}}}(x_j) \ge \vartheta _{{\mathcal {R}}}(x_j)$ which gives that $\zeta _{\mathcal {P}}(x_j)-\zeta _{{\mathcal {Q}}}(x_j) \le 0$ and $\vartheta _{\mathcal {P}}(x_j)-\vartheta _{{\mathcal {Q}}}(x_j) \ge 0$. Therefore, $(k+2)(\zeta _{\mathcal {P}}(x_j)-\zeta _{{\mathcal {Q}}}(x_j))-(2k+1)(\vartheta _{\mathcal {P}}(x_j)-\vartheta _{{\mathcal {Q}}}(x_j)) \le 0$ and hence

  $$\begin{aligned} & {}\frac{\mid (k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) -(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) \mid }{3(k+1)} \\ & = \frac{(2k+1)\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -(k+2)\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) }{3(k+1)}.\end{aligned}$$

  Similarly, it can be obtained that

  $$\begin{aligned} &\frac{\big |2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) \big |}{3} \\ &\quad =\frac{2\left( \vartheta _{\mathcal {P}}(x_j) -\vartheta _{\mathcal {Q}}(x_j)\right) -\left( \zeta _{\mathcal {P}}(x_j)-\zeta _{\mathcal {Q}}(x_j)\right) }{3}. \end{aligned}$$

  Thus, based on these equations, we have

  $$\begin{aligned} &{\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {R}}\right) -{\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right) \\ & = {} \sum \limits _{j=1}^n\psi _j \begin{bmatrix} \left( \frac{(2k+1)\left( \vartheta _{\mathcal {Q}}(x_j) -\vartheta _{\mathcal {R}}(x_j)\right) -(k+2)\left( \zeta _{\mathcal {Q}}(x_j)-\zeta _{\mathcal {R}}(x_j)\right) }{3(k+1)}\right) \\ +\left( \frac{h_{\mathcal {R}}(x_j)-h_{\mathcal {P}}(x_j)}{3(k+1)}\right) \left( \frac{h_{\mathcal {P}}(x_j)+h_{\mathcal {R}}(x_j)}{2}\right) \\ +\left( \frac{h_{\mathcal {P}}(x_j)-h_{\mathcal {Q}}(x_j)}{3(k+1)}\right) \left( \frac{h_{\mathcal {P}}(x_j)+h_{\mathcal {Q}}(x_j)}{2}\right) \end{bmatrix} \\ &= {} \sum \limits _{j=1}^n\psi _j \begin{bmatrix} \left( \frac{(2k+1)\left( \vartheta _{\mathcal {Q}}(x_j) -\vartheta _{\mathcal {R}}(x_j)\right) -(k+2)\left( \zeta _{\mathcal {Q}}(x_j)-\zeta _{\mathcal {R}}(x_j)\right) }{3(k+1)}\right) \\ + \frac{\left( \left( \vartheta _{\mathcal {Q}}(x_j) -\vartheta _{\mathcal {R}}(x_j)\right) +\left( \zeta _{\mathcal {Q}}(x_j) -\zeta _{\mathcal {R}}(x_j)\right) \right) \left( h_{\mathcal {R}}(x_j)+h_{\mathcal {Q}}(x_j)\right) }{6(k+1)} \end{bmatrix}\end{aligned}$$
  $$\begin{aligned}= & {} \sum \limits _{j=1}^n\psi _j \left[ \begin{aligned}&\left( \frac{2k+1}{3(k+1)}+\frac{h_{\mathcal {R}}(x_j)+h_{\mathcal {Q}}(x_j)}{6(k+1)}\right) \left( \vartheta _{\mathcal {Q}}(x_j) -\vartheta _{\mathcal {R}}(x_j)\right) \\&\qquad -\left( \frac{k+2}{3(k+1)}-\frac{h_{\mathcal {R}}(x_j)+h_{\mathcal {Q}}(x_j)}{6(k+1)}\right) \left( \zeta _{\mathcal {Q}}(x_j) -\zeta _{\mathcal {R}}(x_j)\right) \end{aligned} \right] \\= & {} \sum \limits _{j=1}^n\psi _j \left[ \begin{aligned}&\left( \frac{2k+1}{3(k+1)}+\frac{h_{\mathcal {R}}(x_j)+h_{\mathcal {Q}}(x_j)}{6(k+1)}\right) \left( \vartheta _{\mathcal {Q}}(x_j) -\vartheta _{\mathcal {R}}(x_j)\right) \\ & \quad +\left( \frac{4+\zeta _{\mathcal {Q}}(x_j) +\zeta _{\mathcal {R}}(x_j)+\vartheta _{\mathcal {Q}}(x_j) +\vartheta _{\mathcal {R}}(x_j)}{6(k+1)}\right) \\ & \left( \zeta _{\mathcal {R}}(x_j) -\zeta _{\mathcal {Q}}(x_j)\right) \end{aligned} \right] \\ \ge & {} \ 0 \end{aligned}$$

  Thus, ${\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {R}}\right) \ge {\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {Q}}\right)$. Similarly, it can be obtained that ${\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {P}},{\mathcal {R}}\right) \ge {\mathcal {D}}_{\mathcal {C}}\left( {\mathcal {Q}},{\mathcal {R}}\right)$. Hence, ’${\mathcal {D}}_{\mathcal {C}}$’ is a valid distance measure.