Similarity measure of the interval-valued fuzzy numbers and its application in risk analysis in paddy cultivation

Abstract

Generally, in a risk analysis problems, data are collected from experts in terms of linguistic variables, which are then observed as fuzzy numbers. Eventually, the total risk is calculated, and obtained as fuzzy numbers, using the arithmetic operations of fuzzy numbers. Finally, the risk so obtained has to be expressed in terms of linguistic variables, so that it becomes communicable to the general audience. One such tool in fuzzy mathematics is the similarity measure. In this paper, a new method of determining the similarity between two interval-valued fuzzy numbers has been developed. Generally, various parameters were used in deriving a similarity measure. However, the parameters that describe the shape of an interval-valued fuzzy numbers was never a concern in the earlier studies. Hence, in this study, these parameters are incorporated in the similarity measure. Further, a few reasonable properties are being developed, which will be considered as standard in developing a similarity measure. Finally, a risk analysis problem in paddy cultivation is discussed. Using the proposed similarity measure, the final risk is expressed in terms of a linguistic variable. The final risk so obtained for paddy cultivation, under the assumed scenario, is low.

Propositions, Theorems and their proofs

Proposition 1

Consider the interval ${\tilde{a}}=[0,1]$, that is, ${\tilde{a}}=(0,0,1,1;1,1)$ and ${\tilde{b}}$ be any arbitrary fuzzy number such that ${\tilde{b}}\subseteq {\tilde{a}}$. If ${\tilde{b}}$ is obtained by horizontal translation within ${\tilde{a}}$, then the geometric distance between ${\tilde{a}}$ and ${\tilde{b}}$ always remain constant.

Proof

Consider ${\tilde{a}}=(0,0,1,1;1,1)$ and ${\tilde{b}}=(b_1,b_2,b_3,b_4;h_1^{{\tilde{a}}},h_2^{{\tilde{a}}})$ such that ${\tilde{b}} \subseteq {\tilde{a}}$. Let ${\tilde{b}}$ is translated horizontally to obtain ${\tilde{b}}_r$, $r\in I=[0,1]$ such that ${\tilde{b}}_r=(b_1+r,b_2+r,b_3+r,b_4+r;h_1^{{\tilde{a}}},h_2^{{\tilde{a}}})$, then the geometric distance between ${\tilde{a}}$ and ${\tilde{b}}_r$ is $|0-(b_1+r)|+|0-(b_2+r)|+|1-(b_3+r)|+|1-(b_4+r)|$. Equivalently, it can be written as $|0-b_1|+r+|0-b_2|+r+|1-b_3|-r+|1-b_4|-r$, which implies that $|0-b_1|+|0-b_2|+|1-b_3|+|1-b_4|$, which is the geometric distance between ${\tilde{a}}$ and ${\tilde{b}}$. Hence, the proposition. $\square$

Proposition 2

Let ${\tilde{a}}=(a_1,a_2,a_3,a_4;h_1^{{\tilde{a}}},h_2^{{\tilde{a}}})$ and ${\tilde{b}}=(b_1,b_2,b_3,b_4;h_1^{{\tilde{b}}},h_2^{{\tilde{b}}})$ are two arbitrary GFNs, then $\frac{1}{4}\sum \limits _{i=1}^4|a_i-b_i| =1$ if, and only if, either ${\tilde{a}}=(1,1,1,1;h_1^{{\tilde{a}}},h_2^{{\tilde{a}}})$ and ${\tilde{b}}=(0,0,0,0;h_1^{{\tilde{b}}},h_2^{{\tilde{b}}})$ or ${\tilde{a}}=(0,0,0,0;h_1^{{\tilde{a}}},h_2^{{\tilde{a}}})$ and ${\tilde{b}}=(1,1,1,1;h_1^{{\tilde{b}}},h_2^{{\tilde{b}}})$.

Proof

Proof is very trivial. $\square$

Proposition 3

If ${\tilde{a}} \cap {\tilde{b}} \ne \Phi$, then $\frac{1}{4}\sum \limits _{i=1}^4|a_i-b_i|<1$.

Proof

From the proposition 2, the result follows immediately. $\square$

Proposition 4

For any two arbitrary fuzzy numbers ${\tilde{a}}$ and ${\tilde{b}}$, $\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})| < 1$.

Proof

As $0 \le \text {ar}({\tilde{a}}), \text {ar}({\tilde{b}}) \le 1$, therefore $\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})| < 1$. $\square$

Proposition 5

Let $L_i^{{\tilde{a}}}$ and $L_i^{{\tilde{b}}}$ are edge-lengths of any two non-empty GFNs ${\tilde{a}}$ and ${\tilde{b}}$, then $\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}}-L_i^{{\tilde{b}}}|<1$.

Proof

As ${\tilde{a}}$ and ${\tilde{b}}$ are non-empty GFNs, therefore $0 < \sum \limits _{i=1}^4L_i^{{\tilde{a}}}, \sum \limits _{i=1}^4L_i^{{\tilde{b}}} \le 4$. Hence, $\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}}-L_i^{{\tilde{b}}}| < 1$. $\square$

Proposition 6

Let $\theta _i^{{\tilde{a}}}$ and $\theta _i^{{\tilde{b}}}$ are angles of any two non-empty GFNs ${\tilde{a}}$ and ${\tilde{b}}$ respectively, then $\prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}},\theta _i^{{\tilde{b}}})}{\max (\theta _i^{{\tilde{a}}},\theta _i^{{\tilde{b}}})} > 0$.

Proof

As ${\tilde{a}}$ and ${\tilde{b}}$ are non-empty GFNs, therefore $\theta _i^{{\tilde{a}}}, \theta _i^{{\tilde{b}}} > 0$. Hence, the result follows immediately. $\square$

Theorem 1

If ${\tilde{a}}$ and ${\tilde{b}}$ are two IVFNs then $S({\tilde{a}},{\tilde{b}})=1$ if, and only if, ${\tilde{a}}$ and ${\tilde{b}}$ are identical.

Proof

Consider two IVFNs ${\tilde{a}}=[{\tilde{a}}^L,{\tilde{a}}^U]=[(a_1^L, a_2^L, a_3^L, a_4^L;h_1^{{\tilde{a}}^L} ,h_2^{{\tilde{a}}^L}),(a_1^U,a_2^U,a_3^U,a_4^U;h_1^{{\tilde{a}}^U},h_2^{{\tilde{a}}^U})]$ and ${\tilde{b}}=[{\tilde{b}}^L,{\tilde{b}}^U]=[(b_1^L, b_2^L, b_3^L, b_4^L;h_1^{{\tilde{b}}^L} ,h_2^{{\tilde{b}}^L}),(b_1^U,b_2^U,b_3^U,b_4^U;h_1^{{\tilde{b}}^U},h_2^{{\tilde{b}}^U})]$. Assume that ${\tilde{a}}$ and ${\tilde{b}}$ are identical; that is, $a_i^L=b_i^L$ and $a_i^U=b_i^U$ for $i=1,2,3,4$; $h_i^{{\tilde{a}}^L}=h_i^{{\tilde{b}}^L}$ and $h_i^{{\tilde{a}}^U}=h_i^{{\tilde{b}}^U}$ for $i=1,2$; $\text {ar}({\tilde{a}}^L)=\text {ar}({\tilde{b}}^L)$ and $\text {ar}({\tilde{a}}^U)=\text {ar}({\tilde{b}}^U)$; $L_i^{{\tilde{a}}^L}=L_i^{{\tilde{b}}^L}$ and $L_i^{{\tilde{a}}^U}=L_i^{{\tilde{b}}^U}$ for $i=1,2,3,4$; $\theta _i^{{\tilde{a}}^L}=\theta _i^{{\tilde{b}}^L}$ and $\theta _i^{{\tilde{a}}^U}=\theta _i^{{\tilde{b}}^U}$ for $i=1,2,3,4$. Then, it follows that

$$\begin{aligned} S\left( {\tilde{a}}^L,{\tilde{b}}^L\right) =&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|a_i^L-b_i^L|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|)\right) \times&\\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{b}}^L)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}&\\ =&(1-0)(1-0)(1-0)(1-0)\times 1 ,&\\ =&1.&\end{aligned}$$

Similarly, it can be obtained that $S\left( {\tilde{a}}^U,{\tilde{b}}^U\right) =1$. Hence $S\left( {\tilde{a}},{\tilde{b}}\right) =1$.

Conversely, let $S({\tilde{a}},{\tilde{b}})=1$, then

$$\begin{aligned}&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|a_i^L-b_i^L|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|)\right) \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{b}}^L)|\right) \times&\\&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}=1 , \end{aligned}$$

and

$$\begin{aligned}&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|a_i^U-b_i^U|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|)\right) \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^U)-\text {ar}({\tilde{b}}^U)|\right) \times&\\&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|U_i^{{\tilde{a}}^U}-U_i^{{\tilde{b}}^U}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}{\max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}=1 , \end{aligned}$$

Thus, it follows that

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|a_i^L-b_i^L|\right) =1,\\ \left( 1-\frac{1}{2}\left( |h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|\right) \right) =1,\\ \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})|\right) =1,\\ \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) =1,\\ \frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}=1 , \end{array}\right. } ~\text {and}~ {\left\{ \begin{array}{ll} \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|a_i^U-b_i^U|\right) =1,\\ \left( 1-\frac{1}{2}\left( |h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|\right) \right) =1,\\ \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})|\right) =1,\\ \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|U_i^{{\tilde{a}}^U}-U_i^{{\tilde{b}}^U}|\right) =1,\\ \frac{\min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}{\max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}=1 , \end{array}\right. } \end{aligned}$$

Eventually, it follows

$$\begin{aligned} {\left\{ \begin{array}{ll} a_i^L=b_i^L ,\\ h_1^{{\tilde{a}}^L}=h_1^{{\tilde{b}}^L} \text {and} h_2^{{\tilde{a}}^L}=h_2^{{\tilde{b}}^L} ,\\ \text {ar}({\tilde{a}})=\text {ar}({\tilde{b}}) \\ L_i^{{\tilde{a}}^L}=L_i^{{\tilde{b}}^L} ,\\ \theta _i^{{\tilde{a}}^L}=\theta _i^{{\tilde{b}}^L}. \end{array}\right. } ~\text {and}~ {\left\{ \begin{array}{ll} a_i^U=b_i^U ,\\ h_1^{{\tilde{a}}^U}=h_1^{{\tilde{b}}^U} \text {and} h_2^{{\tilde{a}}^U}=h_2^{{\tilde{b}}^U} ,\\ \text {ar}({\tilde{a}})=\text {ar}({\tilde{b}}) \\ U_i^{{\tilde{a}}^U}=U_i^{{\tilde{b}}^U} ,\\ \theta _i^{{\tilde{a}}^U}=\theta _i^{{\tilde{b}}^U}. \end{array}\right. } \end{aligned}$$

$\square$

Theorem 2

If ${\tilde{a}}$ and ${\tilde{b}}$ are two IVFNs, then $S({\tilde{a}},{\tilde{b}})=S({\tilde{b}},{\tilde{a}})$.

Proof

Let ${\tilde{a}}=[{\tilde{a}}^L,{\tilde{a}}^U]=[(a_1^L, a_2^L, a_3^L, a_4^L;h_1^{{\tilde{a}}^L} ,h_2^{{\tilde{a}}^L}),(a_1^U,a_2^U,a_3^U,a_4^U;h_1^{{\tilde{a}}^U},h_2^{{\tilde{a}}^U})]$ and ${\tilde{b}}=[{\tilde{b}}^L,{\tilde{b}}^U]=[(b_1^L, b_2^L, b_3^L, b_4^L;h_1^{{\tilde{b}}^L} ,h_2^{{\tilde{b}}^L}),(b_1^U,b_2^U,b_3^U,b_4^U;h_1^{{\tilde{b}}^U},h_2^{{\tilde{b}}^U})]$ be two IVFNs such that $0 \le a_1^L \le a_2^L \le a_3^L \le a_4^L \le 1$, $0 \le a_1^U \le a_2^U \le a_3^U \le a_4^U \le 1$, $0 \le b_1^L \le b_2^L \le b_3^L \le b_4^L \le 1$ and $0 \le b_1^U \le b_2^U \le b_3^U \le b_4^U \le 1$ , then we have

$$\begin{aligned} S\left( {\tilde{a}}^L,{\tilde{b}}^L\right) =&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|a_i^L-b_i^L|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|)\right) \times&\\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{b}}^L)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})},&\\ =&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|b_i^L-a_i^L|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{b}}^L}-h_1^{{\tilde{a}}^L}|+|h_2^{{\tilde{b}}^L}-h_2^{{\tilde{a}}^L}|)\right) \times&\\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{b}}^L)-\text {ar}({\tilde{a}}^L)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{b}}^L}-L_i^{{\tilde{a}}^L}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{b}}^L},\theta _i^{{\tilde{a}}^L})}{\max (\theta _i^{{\tilde{b}}^L},\theta _i^{{\tilde{a}}^L})},&\\ =&S\left( {\tilde{b}}^L,{\tilde{a}}^L\right) . \end{aligned}$$

Similarly it follows that $S\left( {\tilde{a}}^U,{\tilde{b}}^U\right) = S\left( {\tilde{b}}^U,{\tilde{a}}^U\right)$. Hence, $S({\tilde{a}},{\tilde{b}})=S({\tilde{b}},{\tilde{a}})$. $\square$

Theorem 3

If ${\tilde{a}} \cap {\tilde{b}} \ne \Phi$ then $0<S({\tilde{a}}, {\tilde{b}})$.

Proof

As ${\tilde{a}} \cap {\tilde{b}} \ne \Phi$, therefore ${\tilde{a}}$ and ${\tilde{b}}$ are non-empty GFNs. Hence, from the Propositions 3, 4, 5 and 6 the result follows immediately. $\square$

Theorem 4

If ${\tilde{a}}$ and ${\tilde{b}}$ are two IVFNs, $0 \le S({\tilde{a}},{\tilde{b}}) \le 1$

Proof

Let ${\tilde{a}}=[{\tilde{a}}^L,{\tilde{a}}^U]=[(a_1^L, a_2^L, a_3^L, a_4^L;h_1^{{\tilde{a}}^L} ,h_2^{{\tilde{a}}^L}),(a_1^U,a_2^U,a_3^U,a_4^U;h_1^{{\tilde{a}}^U},h_2^{{\tilde{a}}^U})]$ and ${\tilde{b}}=[{\tilde{b}}^L,{\tilde{b}}^U]=[(b_1^L, b_2^L, b_3^L, b_4^L;h_1^{{\tilde{b}}^L} ,h_2^{{\tilde{b}}^L}),(b_1^U,b_2^U,b_3^U,b_4^U;h_1^{{\tilde{b}}^U},h_2^{{\tilde{b}}^U})]$ be two IVFNs such that $0 \le a_1^L \le a_2^L \le a_3^L \le a_4^L \le 1$, $0 \le a_1^U \le a_2^U \le a_3^U \le a_4^U \le 1$, $0 \le b_1^L \le b_2^L \le b_3^L \le b_4^L \le 1$ and $0 \le b_1^U \le b_2^U \le b_3^U \le b_4^U \le 1$ , then from the Definition 13 , we have

$$\begin{aligned} S({\tilde{a}},{\tilde{b}})=\sqrt{S({\tilde{a}}^L,{\tilde{b}}^L) \times S({\tilde{a}}^U,{\tilde{b}}^U)} \end{aligned}$$

where

$$\begin{aligned} S\left( {\tilde{a}}^L,{\tilde{b}}^L\right) =&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|a_i^L-b_i^L|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|)\right) \times&\\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{b}}^L)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}&\end{aligned}$$

and

$$\begin{aligned} S\left( {\tilde{a}}^U,{\tilde{b}}^U\right) =&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|a_i^U-b_i^U|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|)\right) \times&\\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^U)-\text {ar}({\tilde{b}}^U)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|U_i^{{\tilde{a}}^U}-U_i^{{\tilde{b}}^U}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}{\max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}&\end{aligned}$$

This implies,

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|a_i^L-b_i^L|\right) \le 1,\\ \left( 1-\frac{1}{2}\left( |h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|\right) \right) \le 1,\\ \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})|\right) \le 1,\\ \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \le 1,\\ \frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}\le 1 , \end{array}\right. } \text {and} \\ {\left\{ \begin{array}{ll} \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|a_i^U-b_i^U|\right) \le 1,\\ \left( 1-\frac{1}{2}\left( |h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|\right) \right) \le 1,\\ \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})|\right) \le 1,\\ \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|U_i^{{\tilde{a}}^U}-U_i^{{\tilde{b}}^U}|\right) \le 1,\\ \frac{\min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}{\max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}\le 1 , \end{array}\right. } \end{aligned}$$

Therefore $S({\tilde{a}}^L,{\tilde{b}}^L) \le 1$ and $S({\tilde{a}}^U,{\tilde{b}}^U) \le 1$. Further,

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|a_i^L-b_i^L|\right) \ge 0,\\ \left( 1-\frac{1}{2}\left( |h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|\right) \right) \ge 0,\\ \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})|\right) \ge 0,\\ \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \ge 0,\\ \frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}\ge 0 , \end{array}\right. } \text {and} \\ {\left\{ \begin{array}{ll} \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|a_i^U-b_i^U|\right) \ge 0,\\ \left( 1-\frac{1}{2}\left( |h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|\right) \right) \ge 0,\\ \left( 1-\frac{1}{2}|\text {ar}({\tilde{a}})-\text {ar}({\tilde{b}})|\right) \ge 0,\\ \left( 1-\frac{1}{4}\sum \limits _{i=1}^{4}|U_i^{{\tilde{a}}^U}-U_i^{{\tilde{b}}^U}|\right) \ge 0,\\ \frac{\min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}{\max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}\ge 0 , \end{array}\right. } \end{aligned}$$

Therefore $S({\tilde{a}}^L,{\tilde{b}}^L) \ge 0$ and $S({\tilde{a}}^U,{\tilde{b}}^U) \ge 0$. Hence, the property $0 \le S({\tilde{a}},{\tilde{b}}) \le 1$. $\square$

Theorem 5

Let ${\tilde{a}}_1$ and ${\tilde{a}}_2$ are two IVFNs. If ${\tilde{a}}_1 \ne {\tilde{a}}_2$, then $0 \le S({\tilde{a}}_1,{\tilde{a}}_2) < 1$.

Proof

Proof is trivial. $\square$

Theorem 6

Let ${\tilde{a}}_1=\Phi$ and ${\tilde{a}}_2$ be arbitrary normal fuzzy number, then $S({\tilde{a}}_1,{\tilde{a}}_2)=0$

Proof

Proof is trivial. $\square$

Theorem 7

Let ${\tilde{a}}_1=\Phi$ and ${\tilde{a}}_2$ be arbitrary non-empty fuzzy number, then $0 \le S({\tilde{a}}_1,{\tilde{a}}_2) < 1$

Proof

As ${\tilde{a}}_1=\Phi$ and ${\tilde{a}}_2$ is non-empty, therefore they are not identical. Using Theorem 5, it follows that $S({\tilde{a}},{\tilde{b}}) < 1$. Now, from the Theorem 4, the result follows immediately. $\square$

Theorem 8

If ${\tilde{a}}$, ${\tilde{b}}$ and ${\tilde{c}}$ are symmetric IVFNs and ${\tilde{b}}$ is the mirror image of ${\tilde{a}}$ with respect to the axis of symmetric of ${\tilde{c}}$, then $S({\tilde{a}},{\tilde{c}})=S({\tilde{c}},{\tilde{b}})$

Proof

As ${\tilde{a}}$, ${\tilde{b}}$, ${\tilde{c}}$ are symmetric IVFNs, therefore their left height and right heights are equal. As ${\tilde{b}}$ is the image of ${\tilde{a}}$, therefore heights of ${\tilde{a}}^L$ and ${\tilde{a}}^U$ are equal with heights of ${\tilde{b}}^L$ and ${\tilde{b}}^U$ (say $h^L$ and $h^U$) respectively. Let ${\tilde{a}}=[(a_1^L, a_2^L, a_3^L, a_4^L;h^L ,h^L),(a_1^U,a_2^U,a_3^U,a_4^U;h^U,h^U)]$, ${\tilde{b}}=[(b_1^L, b_2^L, b_3^L, b_4^L;h^L ,h^L),(b_1^U,b_2^U,b_3^U,b_4^U;h^U,h^U)]$ and ${\tilde{c}}=[(c_1^L, c_2^L, c_3^L, c_4^L;h^{{\tilde{c}}^L} ,h^{{\tilde{c}}^L}),(c_1^U,c_2^U,c_3^U,c_4^U;h^{{\tilde{c}}^U},h^{{\tilde{c}}^U})]$, as shown in the Fig. 8, where $h^{{\tilde{c}}^L}$ and $h^{{\tilde{c}}^U}$ are the heights of the GFNs ${\tilde{c}}^L$ and ${\tilde{c}}^U$ respectively.

Let $\theta _i^{{\tilde{a}}^L}$, $\theta _i^{{\tilde{a}}^U}$, $\theta _i^{{\tilde{b}}^L}$, $\theta _i^{{\tilde{b}}^U}$, $\theta _i^{{\tilde{c}}^L}$, $\theta _i^{{\tilde{c}}^U}$ are the angles as defined in Definition 12, $L_i^{{\tilde{a}}^L}$, $L_i^{{\tilde{a}}^U}$, $L_i^{{\tilde{b}}^L}$, $L_i^{{\tilde{b}}^U}$, $L_i^{{\tilde{c}}^L}$, $L_i^{{\tilde{c}}^U}$ are the edge-lengths as defined in Definition 11, and $\text {ar}({\tilde{a}}^L)$, $\text {ar}({\tilde{a}}^U)$, $\text {ar}({\tilde{b}}^L)$, $\text {ar}({\tilde{b}}^U)$, $\text {ar}({\tilde{c}}^L)$, $\text {ar}({\tilde{c}}^U)$ are the areas as defined in Definition 10 of ${\tilde{a}}^L$, ${\tilde{a}}^L$, ${\tilde{b}}^L$, ${\tilde{b}}^L$, ${\tilde{c}}^L$, ${\tilde{c}}^L$ respectively. As ${\tilde{c}}$ is symmetric IVFN and ${\tilde{b}}$ is the image of ${\tilde{a}}$ with respect to the symmetric axis of ${\tilde{c}}$, therefore $|a_i^L-c_i^L|=|c_i^L-b_i^L|$, $|a_i^U-c_i^U|=|c_i^U-b_i^U|$, $\theta _i^{{\tilde{a}}^L}=\theta _i^{{\tilde{b}}^L}$, $\theta _i^{{\tilde{a}}^U}=\theta _i^{{\tilde{b}}^U}$, $L_i^{{\tilde{a}}^L}=L_i^{{\tilde{b}}^L}$, $L_i^{{\tilde{a}}^U}=L_i^{{\tilde{b}}^U}$, $\text {ar}({\tilde{a}}^L)=\text {ar}({\tilde{b}}^L)$ and $\text {ar}({\tilde{a}}^U)=\text {ar}({\tilde{b}}^U)$. Thus,

$$\begin{aligned} {\left\{ \begin{array}{ll} |a_i^L-c_i^L|=|c_i^L-b_i^L| ,\\ |\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{c}}^L)|=|\text {ar}({\tilde{c}}^L)-\text {ar}({\tilde{b}}^L)|, \\ |L_i^{{\tilde{a}}^L}-L_i^{{\tilde{c}}^L}|=|L_i^{{\tilde{c}}^L}-L_i^{{\tilde{b}}^L}|, \\ \min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{c}}^L})=\min (\theta _i^{{\tilde{c}}^L},\theta _i^{{\tilde{b}}^L}), \\ \max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{c}}^L})=\max (\theta _i^{{\tilde{c}}^L},\theta _i^{{\tilde{b}}^L}), \end{array}\right. } ~\text {and}\\ {\left\{ \begin{array}{ll} |a_i^U-c_i^U|=|c_i^U-b_i^U| ,\\ |\text {ar}({\tilde{a}}^U)-\text {ar}({\tilde{c}}^U)|=|\text {ar}({\tilde{c}}^U)-\text {ar}({\tilde{b}}^U)|, \\ |U_i^{{\tilde{a}}^U}-U_i^{{\tilde{c}}^U}|=|U_i^{{\tilde{c}}^U}-U_i^{{\tilde{b}}^U}|, \\ \min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{c}}^U})=\min (\theta _i^{{\tilde{c}}^U},\theta _i^{{\tilde{b}}^U}), \\ \max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{c}}^U})=\max (\theta _i^{{\tilde{c}}^U},\theta _i^{{\tilde{b}}^U}), \end{array}\right. } \end{aligned}$$

Hence from the definition 13 , the result follow immediately. $\square$

Fig. 8

${\tilde{a}}$, ${\tilde{b}}$ and ${\tilde{c}}$ are symmetric IVFNs and ${\tilde{b}}$ is the mirror image of ${\tilde{a}}$ with respect to symmetric axis of ${\tilde{c}}$

Theorem 9

Consider ${\tilde{a}}=[0,1]$, ${\tilde{b}}_r \subseteq {\tilde{a}}$ and ${\tilde{b}}_r$ translate in ${\tilde{a}}$, then $0 \le S({\tilde{a}},{\tilde{b}}_r)=S({\tilde{a}},{\tilde{b}}_s) \le 1$, $r,s\in I=[0,1]$, .

Proof

Consider ${\tilde{a}}=[(0,0,1,1;1,1),(0,0,1,1;1,1)]$ and ${\tilde{b}}_r, {\tilde{b}}_s\subseteq {\tilde{a}}$. Then from the Proposition 1, geometric distance among ${\tilde{a}}^L$ and ${\tilde{b}}_r^L$, ${\tilde{a}}^U$ and ${\tilde{b}}_r^U$, ${\tilde{a}}^L$ and ${\tilde{b}}_s^L$, ${\tilde{a}}^U$ and ${\tilde{b}}_s^U$ are always remain constant. Again, under horizontal translation, height, edge-lengths, area and angles of ${\tilde{b}}_r^L, {\tilde{b}}_r^U, {\tilde{b}}_s^L, {\tilde{b}}_s^U$ remain invariant. Hence, the result follows immediately. $\square$

Corollary 1

If ${\tilde{a}}=[(0,0,0,0;1,1),(0,0,0,0;1,1)]$ and ${\tilde{b}}=[(1,1,1,1;1,1),(1,1,1,1;1,1)]$, then $S({\tilde{a}},{\tilde{b}})=0$

Proof

The proof is trivial. $S({\tilde{a}},{\tilde{b}})=0$ that means ${\tilde{a}}$ and ${\tilde{b}}$ are completely dissimilar. $\square$

Corollary 2

If ${\tilde{a}}=[(m,m,m,m;1,1),(m,m,m,m;1,1)]$ and ${\tilde{b}}=[(n,n,n,n;1,1),(n,n,n,n;1,1)]$, then $S({\tilde{a}},{\tilde{b}})= 1-|m-n|$.

Proof

For crisp-valued IVFN numbers ${\tilde{a}}$ and ${\tilde{b}}$ heights difference, areas difference and edge-lengths difference are zero. That is, $|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|=0$, $|h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|=0$, $\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{b}}^L)=0$, $\text {ar}({\tilde{a}}^U)-\text {ar}({\tilde{b}}^U)=0$, $\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|=0$ and $\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^U}-L_i^{{\tilde{b}}^U}|=0$. Also $\theta _i^{{\tilde{a}}^L}=\theta _i^{{\tilde{b}}^L}=90^{\circ }$ and $\theta _i^{{\tilde{a}}^U}=\theta _i^{{\tilde{b}}^U}=90^{\circ }$ ($i=1,2,3,4$); that is, $\prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{b}}^L},\theta _i^{{\tilde{a}}^L})}{\max (\theta _i^{{\tilde{b}}^L},\theta _i^{{\tilde{a}}^L})}=1$ and $\prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{b}}^U},\theta _i^{{\tilde{a}}^U})}{\max (\theta _i^{{\tilde{b}}^U},\theta _i^{{\tilde{a}}^U})}=1$. Hence, by the definition of proposed similarity measure

$$\begin{aligned} S({\tilde{a}}^L,{\tilde{b}}^L)=&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|m_i-n_i|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|)\right) \times \\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^L)-\text {ar}({\tilde{b}}^L)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|L_i^{{\tilde{a}}^L}-L_i^{{\tilde{b}}^L}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})}{\max (\theta _i^{{\tilde{a}}^L},\theta _i^{{\tilde{b}}^L})},\\ =&(1-|m-n|)\times 1 \times 1 \times 1 \times 1 , \\ =&(1-|m-n|). \end{aligned}$$

and

$$\begin{aligned} S({\tilde{a}}^U,{\tilde{b}}^U)=&\left( 1-\frac{1}{4}\sum \limits _{i=1}^4|m_i-n_i|\right) \left( 1-\frac{1}{2}(|h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|)\right) \times \\&\left( 1-\frac{1}{2}|\text {ar}({\tilde{a}}^U)-\text {ar}({\tilde{b}}^U)|\right) \left( 1-\frac{1}{4}\sum \limits _{i=1}^4|U_i^{{\tilde{a}}^U}-U_i^{{\tilde{b}}^U}|\right) \prod _{i=1}^4\frac{\min (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})}{\max (\theta _i^{{\tilde{a}}^U},\theta _i^{{\tilde{b}}^U})},\\ =&(1-|m-n|)\times 1 \times 1 \times 1 \times 1 , \\ =&(1-|m-n|). \end{aligned}$$

Thus, $S({\tilde{a}},{\tilde{b}})= (1-|m-n|)$. $\square$

Corollary 3

If ${\tilde{a}}=[(m_1,m_2,m_3,m_4;1.0,1.0),(m_1,m_2,m_3,m_4;1.0,1.0)]$ and ${\tilde{b}}=[(n_1,n_2,n_3,n_4;0.0,0.0),(n_1,n_2,n_3,n_4;0.0,0.0)]$, then $S({\tilde{a}},{\tilde{b}})=0$.

Proof

Here $h_1^{{\tilde{a}}^L}=h_2^{{\tilde{a}}^L}=1$, $h_1^{{\tilde{a}}^U}=h_2^{{\tilde{a}}^U}=1$, $h_1^{{\tilde{b}}^L}=h_2^{{\tilde{b}}^L}=0$, $h_1^{{\tilde{b}}^U}=h_2^{{\tilde{b}}^U}=0$. As, $|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|=|h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|=1$ and $|h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|=|h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|=1$, so $1-\frac{1}{2}(|h_1^{{\tilde{a}}^L}-h_1^{{\tilde{b}}^L}|+ |h_2^{{\tilde{a}}^L}-h_2^{{\tilde{b}}^L}|)=0$ and $1-\frac{1}{2}(|h_1^{{\tilde{a}}^U}-h_1^{{\tilde{b}}^U}|+ |h_2^{{\tilde{a}}^U}-h_2^{{\tilde{b}}^U}|)=0$, Hence, using the proposed similarity measure 13, it follows that $S({\tilde{a}},{\tilde{b}})=0$. $\square$