An effective similarity/distance measure between intuitionistic fuzzy sets based on the areas of transformed isosceles right triangle and its applications

Abstract

Since the intuitionistic fuzzy set (IFS) was proposed by Atanassov, many explorations of this particular fuzzy set were conducted. One of the most important areas is the study of similarity and distance between IFSs, which can measure the degree of deviation of objects with uncertain and vague features, and this technique has great value and potential to solve the fuzzy and uncertain problems in the real world. Based on our previous similarity/distance measure model D_JJ (α, β), a new method is proposed for improving the performance of similarity/distance measure model of IFSs, which is derived from the sum of the areas of two triangles constructed by the transformed isosceles triangles of two IFSs. A great effort is made to prove the validity of the proposed method by mathematical derivation. In order to further demonstrate the performance of the proposed method, we apply this method to solve some practical problems such as pattern recognition, medical diagnosis, and cluster analysis. In addition, we also list a series of the existing methods which are used to compare with the proposed method to prove the effectiveness and superiority. The experimental results confirm that the performance of the proposed method exceeds most of the existing methods.

Keywords

Intuitionistic fuzzy set similarity/distance measure transformed isosceles triangle fuzzy number decision-making cluster analysis

1 Introduction

In traditional set theory, there are only two possible relationships (that are belong and not belong to) between each element and a set. A large number of uncertain and fuzzy information exist extensively in the real world causing that the membership between an element and a set is imprecise and vague, so traditional set theory cannot handle these problems effectively [1, 2]. In order to overcome this problem, many theories were explored, and fuzzy set theory is the most representative one [3].

In 1965, Zadeh [4] proposed fuzzy set (FS) theory based on the traditional set theory, this work laid the foundation of the FS theory and fuzzy mathematics. In FS theory, the concept of membership is defined, and a mathematical function is used to represent the degree of membership that an element belongs to a specific fuzzy set. Thus, FS theory can effectively handle the uncertain or vague problems that traditional set theory cannot. However, the concept of membership degree in original FS theory is obviously unable to perfectly solve the uncertain and vague problems in the real world because of the limitation of the membership function. In 1986, Atanassov [5, 6] introduced the concept of IFS and used three indicators including membership, non-membership and hesitation to indicate the relationship between an element and a particular set, which can improve the ability of IFS for describing the fuzzy relationship.

For IFS theory, the research of similarity and distance measure is a hot topic recently [7, 8]. Distance and similarity measure of IFSs is a pair of complementary concept that can be used to measure the degree of deviation and similarity between IFSs [7]. The two techniques have been widely-used in many fields of computer science, such as pattern recognition [7 –10], image processing [11 –13], cluster analysis [13 –17], medical diagnosis [8 , 18], and decision-making [8 , 18–20]. As the core theory of IFSs, similarity and distance measures have been intensively studied by researchers.

In 1997, Chen [21] proposed a distance measure between vague sets and elements, and then this method was applied to the behavior analysis problem by a practical example. However, the method proposed by Chen [21] may produce non-ideal results in some cases which were demonstrated by Hong and Kim in their work published in 1999 [22]. At the same time, Hong and Kim proposed an improved method for measuring the degree of similarity, and then the effectiveness of their method was verified as well. In 2002, Dengfeng and Chuntian [23] gave a definition of the similarity between IFSs, and then proposed several methods of similarity measure for IFSs. In 2003, Mitchell [24] pointed out that there were some counter-intuitive cases of Dengfeng and Chuntian¡¯s measures, a new method was proposed to improve the measure performance, and he also applied it to pattern recognition problem; in the meantime, Liang and Shi [25] summarized several similarity measures and proposed a new similarity measure method for IFSs. Based on Hausdorff distance measure, Hung and Yang [26] proposed a method to measure the degree of similarity between IFSs based on the proof of the properties of similarity measure, and then the effectiveness of this method was proved through some experiments. The similarity measure methods before 2007 were reviewed by Li et al. [2] who made a comprehensive comparison and detailed analysis of these methods. Hung and Yang [27] analyzed several similarity methods between traditional FSs, which inspired two new similarity measure between IFSs, and the new methods were employed to evaluated the studentsąŕ answer-scripts. Ye [28] used the two functions of membership and non-membership to form a vector to represent two elements, and he defined the cosine similarity measure between IFSs and introduced a method to medical diagnosis and pattern recognition.

In addition, Zhang and Yu [29] proposed a distance measure between IFSs and interval-valued fuzzy sets based on the transformed triangle in 2013. Beliakov et al. [30] defined a novel similarity measure that was utilized to the problems of decision making and image segmentation. By introducing the intuitionistic fuzzy number of the centroid point of the right triangle of the IFSs, Chen et al. [8] proposed a method for measuring the relationship between IFSs, this work improved the performance of the distance measure and solved some pattern recognition problems. Jiang et al. [7] proposed a new geometric model based similarity/distance measure method that constructed by a diagonal line and two isosceles triangles transformed by IFSs, and this method was also employed to some practical problems, such as medical diagnosis, cluster analysis, and decision-making. Generally, geometric model-based similarity measure method has favorable interpretability because of the perceptible description.

At present, the most existing methods can be divided into three strategies. (1) Some of these are put forward originally [23 , 28]. (2) Some methods are improved based on the previous models [8 , 26]. (3) The rest are proposed by geometric modeling methods [7 , 30]. Unfortunately, these existing models are not perfect with some unavoidable defects, for example, the intelligibility and unreasonable decision-making. Therefore, we propose a geometric model-based measure model to improve the performance of similarity/distance measure in this work.

In this work, we first introduce the related concepts of IFS theory, and then some existing methods for measuring the similarity/distance between IFSs are reviewed. Second, we present a new similarity measure between IFSs based on the transformation of isosceles triangle fuzzy numbers that is constructed previous work, and a large number of mathematical derivation is used to prove the validity of the proposed method. Some experiments are implemented to compare the performance of the proposed measure with the existing methods.

This paper is structured as follows. In Section 2, the related knowledge of FS and IFS is briefly introduced. Section 3 lists sixteen existing methods for measuring the distance and similarity between IFSs. In Section 4, the proposed model is introduced and verified. Section 5 contains all the experiments and applications. In Section 6, we summarize this work and present our next plan.

2 Basic knowledge related to fuzzy set theory

To help readers understand this paper easily, we introduce some basic concepts related to IFS theory in this section.

For IFSs, Atanassov [5] introduced three intuitionistic fuzzy numbers to indicate the relationship between an element and a special set. IFS α is defined as Definition 2.1.

Definition 2.1 [31, 32] The definition of IFS α in universe of discourse U ={ x₁, x₂, . . . , x_n } is represented as following:

α ={ 〈x_i, μ_α (x_i) , ν_α (x_i) 〉|x_i ∈ U } where μ_α (x_i) represents the membership degree of x_i belonging to IFS α. ν_α (x_i) indicates the non-membership degree of x_i belonging to IFS α. In addition, π_α (x_i) = 1 - μ_α (x_i) - ν_α (x_i), and π_α (x_i) indicates the hesitation degree of x_i belonging to IFS α. μ_α (x_i), ν_α (x_i) and π_α (x_i) are satisfied with 0 ≤ μ_α (x_i) ≤ 1, 0 ≤ ν_α (x_i) ≤ 1, 0 ≤ π_α (x_i) ≤ 1, and μ_α (x_i) + ν_α (x_i) + π_α (x_i) = 1.

In 1975, Zadeh [31] proposed a new extended FS called interval-valued fuzzy set (IvFS) that is defined as the following Definition 2.2:

Definition 2.2 [32] Let U ={ x₁, x₂, . . . , x_n } be an universe of discourse, an IvFS α in U is defined as: $α = {〈 x, F_{α} (x_{i}) 〉 | x_{i} \in U}$ where α: U → ψ ([0, 1]), and ψ ([0, 1]) represents the set of all subintervals in [0,1]. Membership interval function $F_{α} (x_{i}) = [{\bar{F}}_{α} (x_{i}), {\underline{F}}_{α} (x_{i})]$ where ${\bar{F}}_{α} (x_{i})$ denotes the upper membership function and ${\underline{F}}_{α} (x_{i})$ denotes the lower membership function.

There is a conversion relationship between IFS and IvFS as follows. Theorem 2.1 [29] Let the transform T be:

T : IFS (U) → IvFS (U) where IFS (U) and IvFS (U) are respectively denoted by α and α^#, and α^# ={ 〈x_i, μ_α (x_i) 1 - ν_α (x_i) 〉|x_i ∈ U }.

Definition 2.3 [26] The following relationship between IFSs α and β in a given universe of discourse U ={ x₁, x₂, . . . , x_n } is expressed as:

1) α = β ⇔ ∀ x_i ∈ U, μ_α (x_i) = μ_β (x_i) and ν_α (x_i) = ν_β (x_i);

2) α ≤ β ⇔ ∀ x_i ∈ U, μ_α (x_i) ≤ μ_β (x_i) and ν_α (x_i) ≥ ν_β (x_i)

3) α ≥ β ⇔ ∀ x_i ∈ U, μ_α (x_i) ≥ μ_β (x_i) and ν_α (x_i) ≤ ν_β (x_i)

4) α^C ={ 〈x, ν_α (x_i) , μ_α (x_i) 〉|x_i ∈ U }

The distance can be used to represent the deviation between two elements (objects) in the real world. Similarly, the degree of deviation between IFSs can also be measured by distance. The distance between IFSs has the following relationship.

Definition 2.4 [29] Suppose α, β and γ as three IFSs in the universe of discourse U ={ x₁, x₂, . . . , x_n }. There are four properties that the distance measure has to be satisfied.

D1 : 0 ≤ D (α, β) ≤ 1;

D2 : D (α, β) = 0 if and only if α = β;

D3 : D (α, β) = D (β, α), for all α, β ∈ U are established;

D4 : For α, β, γ ∈ U, if α ⊆ β ⊆ γ, then D (α, β) ≤ D (α, γ), D (β, γ) ≤ D (α, γ).

The similarity measure can characterize the degree of similarity between IFSs. Like distance, a reasonable similarity measure must satisfy the following four conditions.

Definition 2.5 [33, 34] Let α, β and γ be three IFSs in the universe of discourse U ={ x₁, x₂, . . . , x_n }, a reasonable similarity measure of IFSs must satisfy the following four properties.

S1 : 0 ≤ S (α, β) ≤ 1;

S2 : S (α, β) = 1 if and only if α = β;

S3 : S (α, β) = S (β, α), for all α, β ∈ U are established;

S4 : For α, β, γ ∈ U, if α ⊆ β ⊆ γ, then S (α, β) ≥ S (α, γ), S (β, γ) ≥ S (α, γ).

Theorem 2.2 [7] If D (α, β) is the degree of deviation between IFSs α and β, the following relationship is satisfied between S (α, β) and D (α, β):

S (α, β) = 1 - D (α, β) or D (α, β) = 1 - S (α, β).

Definition 2.6 [35] Let an association matrix is S = (s_ij) _n×n, if S² = S ∘ S = { max { min { s_ik, s_kj }} } _n×n, S² is the composite matrix of S, where i, j = 1, 2, 3, …, n.

3 Existing distance and similarity measures between IFSs

Since the introduction of IFS by Atanassov, various measure methods have been proposed successively. In this section, sixteen classic similarity/distance measures are listed. α and β denote two IFSs, and the methods are shown as follows.

1) Boran and Akay [10] proposed an IFS similarity measure with two parameters in 2014:

$S_{B} (α, β) = 1 - \sqrt{\frac{\sum_{i = 1}^{n} (\begin{matrix} | \begin{matrix} t (μ_{α} (x_{i}) - μ_{β} (x_{i})) \\ - (ν_{α} (x_{i}) - ν_{β} (x_{i})) \end{matrix} |^{P} \\ + | \begin{matrix} t (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - (μ_{α} (x_{i}) - μ_{β} (x_{i})) \end{matrix} |^{P} \end{matrix})}{2 n {(t + 1)}^{p}} P}$ (1)

where t=2, 3, 4, ⋯, and p=1, 2, 3, ⋯.

2) In 1995, Chen [36] proposed a method S_C (α, β) of similarity measure. The theoretical basis of this method is that IFS and Vague Set are equivalent. $S_{C} (α, β) = 1 - \frac{1}{2 n} [\sum_{i = 1}^{n} | (μ_{α} (x_{i}) - ν_{α} (x_{i})) - (μ_{β} (x_{i}) - ν_{β} (x_{i})) |]$ (2)

3) Chen made a valuable contribution in measuring the distance between IFSs. In 2016, Chen [8] proposed a model to derive an effective similarity measure S_CCL (α, β): $\begin{matrix} S_{CCL} (α, β) = 1 - [\frac{| 2 (μ_{α} (x_{i}) - μ_{β} (x_{i})) - (ν_{α} (x_{i}) - ν_{β} (x_{i})) |}{3} \\ \times (1 - \frac{μ_{β} (x_{i}) - μ_{α} (x_{i}) + ν_{β} (x_{i}) - ν_{α} (x_{i})}{2})] \\ - [\frac{| 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) - (μ_{α} (x_{i}) - μ_{β} (x_{i})) |}{3} \\ \times (\frac{μ_{β} (x_{i}) - μ_{α} (x_{i}) + ν_{β} (x_{i}) - ν_{α} (x_{i})}{2})] \end{matrix}$ (3)

4) The similarity measure S_DC (α, β) proposed by Dengfeng and Chuntian [23] is a highly representative method.

$S_{DC} (α, β) = 1 - \sqrt{\frac{\sum_{i = 1}^{n} {| \begin{matrix} [(μ_{α} (x_{i}) + 1 - ν_{α} (x_{i})) / 2] \\ - [(μ_{β} (x_{i}) + 1 - ν_{β} (x_{i})) / 2] \end{matrix} |}^{P}}{n} p}$ (4)

5) Hong and Kim [22] improved Chen’s method in 1999 to represent a new method S_H (α, β): $S_{H} (α, β) = 1 - \frac{1}{2 n} [\sum_{i = 1}^{n} (| μ_{α} (x_{i}) - μ_{β} (x_{i}) | + | ν_{α} (x_{i}) - ν_{β} (x_{i}) |)]$ (5)

6) Hung and Yang [26] proposed a similarity measure d_HY (α, β) based on Hausdorff distance: $d_{HY} (α, β) = \frac{1}{n} \sum_{i = 1}^{n} H ([μ_{α} (x_{i}), 1 - ν_{α} (x_{i})], [μ_{β} (x_{i}), 1 - ν_{β} (x_{i})])$ (6) where H (α, β) = max{ |a₁ - b₁|, |a₂ - b₂| }.

Based on this distance measure between IFSs, Hung and Yang further proposed a series of corresponding similarity measure methods, as: $S_{HY}^{1} (α, β) = \frac{e^{- d_{HY} (α, β)} - e^{- 1}}{1 - e^{- 1}}$ (7) $S_{HY}^{2} (α, β) = \frac{1 - d_{HY} (α, β)}{1 + d_{HY} (α, β)}$ (8)

7) After a deeply study of L_P measure, Hung [37] proposed a series of similarity measures based on L_P measure in 2007. $S_{Lp}^{l} (α, β) = 1 - \frac{\sum_{i = 1}^{n} {({| g (x_{i}) |}^{P} + {| q (x_{i}) |}^{P})}^{1 / P}}{2^{1 / P} * n}$ (9) $S_{Lp}^{e} (α, β) = \frac{e^{- \frac{1}{n} \sum_{i = 1}^{n} {({| g (x_{i}) |}^{P} + {| q (x_{i}) |}^{P})}^{1 / P}} - e^{- (2^{1 / P})}}{1 - e^{- (2^{1 / P})}}$ (10) $S_{Lp}^{c} (α, β) = \frac{2^{1 / P} - \frac{1}{n} \sum_{i = 1}^{n} {({| g (x_{i}) |}^{P} + {| q (x_{i}) |}^{P})}^{1 / P}}{2^{1 / P} (1 + \frac{1}{n} \sum_{i = 1}^{n} {({| g (x_{i}) |}^{P} + {| q (x_{i}) |}^{P})}^{1 / P})}$ (11) where g (x_i) = μ_α (x_i) - μ_β (x_i), q (x_i) = ν_α (x_i) - ν_β (x_i).

8) Afterwards, Hung [27] proposed a series of methods based on the model proposed by Pappis [38]. The specific method is as follows: $S_{H}^{1} (α, β) = \frac{1}{n} \sum_{i = 1}^{n} \frac{min (μ_{α} (x_{i}), μ_{β} (x_{i})) + min (ν_{α} (x_{i}), ν_{β} (x_{i}))}{max (μ_{α} (x_{i}), μ_{β} (x_{i})) + max (ν_{α} (x_{i}), ν_{β} (x_{i}))}$ (12) $S_{H}^{2} (α, β) = \frac{1}{n} \sum_{i = 1}^{n} (1 - \frac{1}{2} (| μ_{α} (x_{i}) - μ_{β} (x_{i}) | + | ν_{α} (x_{i}) - ν_{β} (x_{i}) |))$ (13) $S_{H}^{pk 1} (α, β) = \frac{\sum_{i = 1}^{n} min (μ_{α} (x_{i}), μ_{β} (x_{i})) + min (ν_{α} (x_{i}), ν_{β} (x_{i}))}{\sum_{i = 1}^{n} max (μ_{α} (x_{i}), μ_{β} (x_{i})) + max (ν_{α} (x_{i}), ν_{β} (x_{i}))}$ (14) $S_{H}^{pk 2} (α, β) = 1 - \frac{\sum_{i = 1}^{n} (| μ_{α} (x_{i}) - μ_{β} (x_{i}) | + | ν_{α} (x_{i}) - ν_{β} (x_{i}) |)}{\sum_{i = 1}^{n} (| μ_{α} (x_{i}) + μ_{β} (x_{i}) | + | ν_{α} (x_{i}) + ν_{β} (x_{i}) |)}$ (15)

9) In 2018, Hwang [13] proposed a new similarity measure S_HW (α, β) using the Jaccard-factor: $S_{HW} (α, β) = \frac{1}{3 n} \sum_{i = 1}^{n} (\begin{matrix} g (μ_{α} (x_{i}), μ_{β} (x_{i})) \\ + g (1 - ν_{α} (x_{i}), 1 - ν_{β} (x_{i})) \\ + g (\frac{1}{2} (1 + μ_{α} (x_{i}) - ν_{α} (x_{i})), \\ \frac{1}{2} (1 + μ_{β} (x_{i}) - ν_{β} (x_{i}))) \end{matrix})$ (16) where $g (a, b) = {\begin{matrix} 1, if a = b = 0 \\ \frac{a \times b}{a^{2} + b^{2} - a \times b}, if a > 0 and b > 0 \end{matrix}$ .

10) In 2019, Jiang [7] derived a new distance measure D_JJ (α, β) based on a diagonal line and the intersection of isosceles triangles transformed IFSs. $D_{JJ} (α, β) = \frac{1}{2 n} \sum_{i = 1}^{n} [ω (x_{i}) + ζ (x_{i})]$ (17) where $ω (x_{i}) = | \frac{\begin{matrix} 2 (μ_{α} (x_{i}) π_{β} (x_{i}) - μ_{β} (x_{i}) π_{α} (x_{i})) \\ - 4 (μ_{α} (x_{i}) - μ_{α} (x_{i})) \end{matrix}}{4 - π_{α} (x_{i}) π_{β} (x_{i})} |$ , and $ζ (x_{i}) = | \frac{\begin{matrix} 4 (ν_{α} (x_{i}) - ν_{β} (x_{i})) + 2 [\begin{matrix} ν_{α} (x_{i}) π_{β} (x_{i}) \\ - ν_{β} (x_{i}) π_{α} (x_{i}) \end{matrix}] \\ + 2 [(π_{α} (x_{i})) - (π_{β} (x_{i}))] \end{matrix}}{4 - (π_{α} (x_{i})) (π_{β} (x_{i}))} |$

11) Liang [25] proposed three similarity measures $S_{LS}^{1} (α, β)$ , $S_{LS}^{2} (α, β)$ , $S_{LS}^{3} (α, β)$ in 2003:

$S_{LS}^{1} (α, β) = 1 - \sqrt{\frac{\sum_{i = 1}^{n} {| \begin{matrix} (| μ_{α} (x_{i}) - μ_{β} (x_{i}) | / 2) \\ - (| 1 - ν_{α} (x_{i}) - (1 - ν_{β} (x_{i})) | / 2) \end{matrix} |}^{P}}{n} p}$ (18) $S_{LS}^{2} (A, B) = 1 - \sqrt{\frac{\sum_{i = 1}^{n} {| \begin{matrix} (| φ_{α 1} (x_{i}) - φ_{β 1} (x_{i}) | / 2) \\ - (| φ_{α 2} (x_{i}) - φ_{β 2} (x_{i}) | / 2) \end{matrix} |}^{P}}{n} p}$ (19)

where φ_α1 (x_i) = (μ_α (x_i) - φ_α (x_i))/2, φ_β1 (x_i) = (μ_β (x_i) - φ_β (x_i))/2, φ_α2 (x_i) = (φ_α (x_i) +1 - ν_α (x_i))/2, φ_β2 (x_i) = (φ_β (x_i) +1 - ν_β (x_i))/2, φ_α (x_i) = (μ_α (x_i) +1 - ν_α (x_i))/2, φ_β (x_i) = (μ_β (x_i) +1 - ν_β (x_i))/2. $S_{LS}^{3} (α, β) = 1 - \sqrt{\sum_{i = 1}^{n} w_{i} {(\sum_{j = 1}^{3} ω_{j} φ_{j} (x_{i}))}^{P} p}$ (20) where 0 ≤ w_i ≤ 1, $\sum_{j = 1}^{3} ω_{j} = 1$ , φ₁ (x_i) = (|μ_A (x_i) - μ_B (x_i) |/2) + (|1 - ν_A (x_i) - (1 - ν_B (x_i)) |/2), φ₂ (x_i) = [(μ_A (x_i) +1 - ν_A (x_i))/2] + [(μ_B (x_i) +1 - ν_B (x_i))/ 2], $φ_{3} (x_{i}) = max (\frac{1 - μ_{α} (x_{i}) - ν_{α} (x_{i})}{2}, \frac{1 - μ_{β} (x_{i}) - ν_{β} (x_{i})}{2})$

$- min (\frac{1 - μ_{α} (x_{i}) - ν_{α} (x_{i})}{2}, \frac{1 - μ_{β} (x_{i}) - ν_{β} (x_{i})}{2})$

12) Mitchell [24] pointed out that Dengfeng and Chuntian’s [23] similarity measure had some unavoidable problems, so he proposed a new method S_M (α, β) that was expected to overcome the shortcomings of S_DC (α, β).

$S_{M} (α, β) = 1 - \frac{1}{2} (\sqrt{\frac{\sum_{i = 1}^{n} {| g (x_{i}) |}^{p}}{n} p} + \sqrt{\frac{\sum_{i = 1}^{n} {| q (x_{i}) |}^{p}}{n} p})$ (21)

where g (x_i) = μ_α (x_i) - μ_β (x_i), q (x_i) = ν_α (x_i) - ν_β (x_i)

13) Song [39] proposed a similarity measure S_S (α, β).

$S_{S} (α, β) = \frac{1}{2 n} \sum_{i = 1}^{n} [\begin{matrix} \sqrt{μ_{α} (x_{i}) μ_{β} (x_{i})} + 2 \sqrt{ν_{α} (x_{i}) ν_{β} (x_{i})} \\ + \sqrt{(1 - ν_{A} (x_{i})) (1 - ν_{B} (x_{i}))} \\ + \sqrt{g (x_{i}) q (x_{i})} \end{matrix}]$ (22)

where g (x_i) =1 - μ_α (x_i) - ν_α (x_i), q (x_i) =1 - μ_β (x_i) - ν_β (x_i)

In 2019, Song proposed [20] an improved measure of similarity S_SN (α, β) after he presented the shortcomings of the previous method S_S (α, β). $S_{SN} (α, β) = \frac{1}{2 n} \sum_{i = 1}^{n} [\begin{matrix} 2 \sqrt{μ_{α} (x_{i}) μ_{β} (x_{i})} \\ + 2 \sqrt{ν_{α} (x_{i}) ν_{β} (x_{i})} \\ + \sqrt{(1 - ν_{A} (x_{i})) (1 - ν_{B} (x_{i}))} \\ + \sqrt{(1 - μ_{A} (x_{i})) (1 - ν_{B} (x_{i}))} \\ + \sqrt{g (x_{i}) q (x_{i})} \end{matrix}]$ (23) where g (x_i) =1 - μ_α (x_i) - ν_α (x_i), q (x_i) =1 - μ_β (x_i) - ν_β (x_i)

14) In 2005, Wang and Liu [33, 40] proposed two methods, one was the similarity measure S_L (α, β) and the other was the distance measure D_W (α, β). $D_{W} (α, β) = \sum_{i = 1}^{n} [\frac{φ (x_{i}) + φ (x_{i})}{4} + \frac{max (φ (x_{i}), φ (x_{i}))}{2}]$ (24)

$S_{L} (α, β) = 1 - \sqrt{\frac{\sum_{i = 1}^{n} φ^{2} (x_{i}) + φ^{2} (x_{i}) + {[2 + g (x_{i}) + q (x_{i})]}^{2}}{2 n}}$ (25)

where φ (x_i) = |μ_α (x_i) - μ_β (x_i) |, φ (x_i) = |ν_α (x_i) - ν_β (x_i) |, g (x_i) = μ_β (x_i) - μ_α (x_i), q (x_i) = ν_β (x_i) - ν_α (x_i)

15) Ye [28] explored similarity measure from a new perspective, and proposed a similarity measure S_Y (α, β) based on the cosine theorem. $S_{Y} (α, β) = \frac{1}{n} \sum_{i = 1}^{n} \frac{μ_{α} (x_{i}) μ_{β} (x_{i}) + ν_{α} (x_{i}) ν_{β} (x_{i})}{\sqrt{({μ_{α}}^{2} (x_{i}) + {ν_{α}}^{2} (x_{i})) ({μ_{β}}^{2} (x_{i}) + {ν_{β}}^{2} (x_{i}))}}$ (26)

16) Vlachos [11] proposed a distance measure method D_V (α, β) using cross-entropy, as: $\begin{matrix} \begin{matrix} D_{V} (α, β) = \sum_{i = 1}^{n} [μ_{A} (x_{i}) ln \frac{2 μ_{α} (x_{i})}{μ_{α} (x_{i}) + μ_{β} (x_{i})} \\ + ν_{α} (x_{i}) ln \frac{2 ν_{α} (x_{i})}{ν_{α} (x_{i}) + ν_{β} (x_{i})}] \\ + \sum_{i = 1}^{n} [μ_{β} (x_{i}) ln \frac{2 μ_{β} (x_{i})}{μ_{β} (x_{i}) + μ_{α} (x_{i})} + ν_{β} (x_{i}) ln \frac{2 ν_{β} (x_{i})}{ν_{β} (x_{i}) + ν_{α} (x_{i})}] \end{matrix} \end{matrix}$ (27)

4 The proposed distance and similarity measure for IFSs

The existing measure methods have some defects in some cases, which may lead to unsatisfactory results in its applications. To improve the performance of similarity/distance measure, we therefore propose an effective distance/similarity measure method of IFSs based on the area of the transformed isosceles right triangle fuzzy number, which is modified on the basis of our previous work [7].

Firstly, we transform IFSs α and β into IvFSs which are expressed as:

For convenience, the transformed intuitionistic fuzzy number (μ_α (x_i) , 1 - ν_α (x_i)) and (μ_β (x_i) , 1 - ν_β (x_i) is expressed by (Z₁, Z₂) and (H₁, H₂), and 0 ≤ H₁ ≤ H₂ ≤ 1, 0 ≤ Z₁ ≤ Z₂ ≤ 1. As shown in Fig. 1, triangle H₁H₂H₃ and triangle Z₁Z₂Z₃ are two isosceles triangles obtained by the transformation of IFSs α and β, respectively. H₃ and Z₃ are the vertices of the two isosceles triangles.

Fig. 1

The isosceles right triangles mentioned in the proposed method

The intersection points of two isosceles triangles transformed by IFSs α and β are defined as P₁ and P₂. Lines $\bar{P_{1} Q_{1}}$ and $\bar{P_{2} Q_{2}}$ are parallel to the Z-axis and intersects the diagonal line at point Q₁ and Q₂. Lines $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ are perpendicular to the diagonal line, and the vertical points are O₁ and O₂ respectively.

$\bar{P_{1} Q_{1}}$ and $\bar{P_{2} Q_{2}}$ parallel to the Z-axis, and the angle between diagonal line and the Z-axis is 45°, so ∠P₁Q₁O₁ = ∠ P₂Q₂O₂ = 45° and ∠O₁P₁Q₁ = ∠ O₂P₂Q₂ = 180° - 45° - 90° = 45°. Accordingly, triangle P₁O₁Q₁ and triangle P₂O₂Q₂ are isosceles right triangles, so $\bar{Q_{1} O_{1}} = \bar{P_{1} O_{1}}, \bar{Q_{2} O_{2}} = \bar{P_{2} O_{2}}$ .

The area of triangle P₁O₁Q₁ is: $S_{1} = \frac{1}{2} \times \bar{Q_{1} O_{1}} \times \bar{P_{1} O_{1}} = \frac{1}{2} \times {(\bar{P_{1} O_{1}})}^{2}$ (28)

The area of triangle P₂O₂Q₂ is: $S_{2} = \frac{1}{2} \times \bar{Q_{2} O_{2}} \times \bar{P_{2} O_{2}} = \frac{1}{2} \times {(\bar{P_{2} O_{2}})}^{2}$ (29)

Point P₁ is the intersection point of line $\bar{Z_{1} Z_{3}}$ and line $\bar{H_{1} H_{3}}$ . The coordinates of point P₁ can be calculated as follows.

The equation of line $\bar{Z_{1} Z_{3}}$ is: $y_{\bar{Z_{1} Z_{3}}} = \frac{Z_{2} - Z_{1}}{2} x + Z_{1}$ (30)

The equation of line $\bar{H_{1} H_{3}}$ is: $y_{\bar{H_{1} H_{3}}} = \frac{2}{H_{2} - H_{1}} x - \frac{2 H_{1}}{H_{2} - H_{1}}$ (31)

According to Eq. (30) = Eq. (31), the coordinate of point P₁ on the X-axis is: $x_{P_{1}} = \frac{4 H_{1} + 2 Z_{1} H_{2} - 2 Z_{1} H_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}}$

The coordinate of point P₁ on Y-axis is: $y_{P_{1}} = \frac{Z_{2} - Z_{1}}{2} x + Z_{1} = \frac{Z_{2} - Z_{1}}{2} \times \frac{4 H_{1} + 2 Z_{1} H_{2} - 2 Z_{1} H_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1} + Z_{1}} = \frac{2 H_{1} Z_{2} - 2 Z_{1} H_{1} + 4 Z_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}}$

Point P₂ is the intersection of line Z₂Z₃ and line H₂H₃. The coordinates of point P₂ can be calculated as follows.

The equation of line Z₂Z₃ is: $y_{\bar{Z_{2} Z_{3}}} = - \frac{Z_{2} - Z_{1}}{2} x + Z_{2}$ (32) The equation of line H₂H₃ is: $y_{\bar{H_{2} H_{3}}} = - \frac{2}{H_{2} - H_{1}} x + \frac{2 H_{2}}{H_{2} - H_{1}}$ (33)

According to Eq. (32) = Eq. (33), the coordinate of point P₂ on X-axis is:

$x_{P_{2}} = \frac{4 H_{2} - 2 H_{2} Z_{2} + 2 H_{1} Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}}$

The coordinate of point P₂ on Y-axis is:

$\begin{matrix} y_{P_{2}} = - \frac{Z_{2} - Z_{1}}{2} x + Z_{2} \\ = - \frac{Z_{2} - Z_{1}}{2} \times \frac{4 H_{2} - 2 H_{2} Z_{2} + 2 H_{1} Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} + Z_{2} \\ = \frac{2 Z_{1} H_{2} - 2 Z_{2} H_{2} + 4 Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} \end{matrix}$

The length of $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ are calculated as follows.

The equation for the diagonal is: x - y = 0. According to the distance equation $d = \frac{| A x_{1} + B y_{1} + C |}{\sqrt{A^{2} + B^{2}}}$ from point (x₁, y₁) to Ax + By + C = 0, so we can obtain:

$\begin{matrix} \bar{P_{1} O_{1}} = \frac{1}{\sqrt{2}} \times | \begin{matrix} \frac{4 H_{1} + 2 Z_{1} H_{2} - 2 Z_{1} H_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} \\ - \frac{2 H_{1} Z_{2} - 2 Z_{1} H_{1} + 4 Z_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} \end{matrix} | \\ = \sqrt{2} \times | \frac{2 H_{1} + Z_{1} H_{2} - Z_{2} H_{1} - 2 Z_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} | \end{matrix}$ ,

$\begin{matrix} \bar{P_{2} O_{2}} = \frac{1}{\sqrt{2}} \times | \begin{matrix} \frac{4 H_{2} - 2 H_{2} Z_{2} + 2 H_{1} Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} \\ - \frac{2 Z_{1} H_{2} - 2 Z_{2} H_{2} + 4 Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} \end{matrix} | \\ = \sqrt{2} \times | \frac{2 H_{2} + Z_{2} H_{1} - H_{2} Z_{1} - 2 Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}} | \end{matrix}$ .

The proposed distance measure is denoted by D (α, β) that is the distance between IFSs α and β. D (α, β) can be calculated as:

$\begin{matrix} D (α, β) = 2 \times [\frac{1}{2} \times {(\bar{P_{1} O_{1}})}^{2} + \frac{1}{2} \times {(\bar{P_{2} O_{2}})}^{2}] \\ = {(\bar{P_{1} O_{1}})}^{2} + {(\bar{P_{2} O_{2}})}^{2} \end{matrix}$ .

So, the distance D (α, β) between α and β can be expressed as:

$D (α, β) = 2 \times [\begin{matrix} {(\frac{2 H_{1} + Z_{1} H_{2} - Z_{2} H_{1} - 2 Z_{1}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}})}^{2} \\ + {(\frac{2 H_{2} + Z_{2} H_{1} - H_{2} Z_{1} - 2 Z_{2}}{4 - H_{2} Z_{2} + H_{2} Z_{1} + H_{1} Z_{2} - H_{1} Z_{1}})}^{2} \end{matrix}]$ $= 2 \times [\begin{matrix} {(\frac{- μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i})}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))})}^{2} \\ + {(\frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))})}^{2} \end{matrix}]$ (34)

Because of S (α, β) = 1 - D (α, β), the corresponding similarity measure S (α, β) is:

$S (α, β) = 1 - 2 [\begin{matrix} {(\frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) \\ + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}})}^{2} \\ + {(\frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) \\ + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}})}^{2} \end{matrix}]$ (35)

Because of the complementary characteristic of distance measure and similarity measure, the distance measure method must satisfy properties (D1-D4) while the similarity measure method satisfies properties (S1-S4). Therefore, we only need to prove that the similarity measure method satisfies properties (S1-S4).

The proof as follows:

Property (S1) : 0≤ S (α, β) ≤1 ;

Proof.

(a) Because of $S (α, β) = 1 - [{(\bar{P_{1} O_{1}})}^{2} + {(\bar{P_{2} O_{2}})}^{2}]$ , the value of S (α, β) is the smallest when $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ take the maximum values, which can be represented as:

$S_{min} (α, β) = 1 - [{({\bar{P_{1} O_{1}}}_{max})}^{2} + {({\bar{P_{2} O_{2}}}_{max})}^{2}]$ .

According to Fig 1 , $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ are perpendicular to the diagonal line, $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ take the maximum value when μ_α (x_i) = 1 and ν_α (x_i) = 0, μ_β (x_i) = 0 and ν_β (x_i) = 1 or μ_α (x_i) = 0 and ν_α (x_i) = 1, μ_β (x_i) = 1 and ν_β (x_i) = 0 .

The maximum values of $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ are ${\bar{P_{1} O_{1}}}_{max} = {\bar{P_{2} O_{2}}}_{max} = \frac{\sqrt{2}}{2}$ . So we can obtain:

$S_{min} (α, β) = 1 - [{({\bar{P_{1} O_{1}}}_{max})}^{2} + {({\bar{P_{2} O_{2}}}_{max})}^{2}]$ = $1 - [{(\frac{\sqrt{2}}{2})}^{2} + {(\frac{\sqrt{2}}{2})}^{2}] = 1 - (\frac{1}{2} + \frac{1}{2}) = 0$ .

(b) When $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ take the minimum value, $S_{max} (α, β) = 1 - [{({\bar{P_{1} O_{1}}}_{min})}^{2} + {({\bar{P_{2} O_{2}}}_{min})}^{2}]$ .

When μ_α (x_i) = μ_β (x_i) andν_α (x_i) = ν_β (x_i) , we can find that P₁ and O₁ become a coincide point, P₂ and O₂ become a coincide point. So $\bar{P_{1} O_{1}}$ and $\bar{P_{2} O_{2}}$ take the minimum values as ${\bar{P_{1} O_{1}}}_{min} = {\bar{P_{2} O_{2}}}_{min} = 0$ , we can further get: $S_{max} (α, β) = 1 - [{({\bar{P_{1} O_{1}}}_{min})}^{2} + {({\bar{P_{2} O_{2}}}_{min})}^{2}] = 1 - (0 + 0) = 1$

The above analysis can get this conclusion:

0 ≤ S (α, β) ≤1.

The proof is completed.

Property: (S2) : S (α, β) =1 if and only if α = β ;

Proof.

(a) S (α, β) =1 ⇒ α = β;

According to $S (α, β) = 1 - [{(\bar{P_{1} O_{1}})}^{2} + {(\bar{P_{2} O_{2}})}^{2}]$ , if and only if ${(\bar{P_{1} O_{1}})}^{2} + {(\bar{P_{2} O_{2}})}^{2} = 0$ , we can get: S (α, β) =1. So $\bar{P_{1} O_{1}} = 0$ and $\bar{P_{2} O_{2}} = 0$ , and we can get equation set, as: ${\begin{matrix} | \frac{- μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i})}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))} | = 0 \\ | \frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) \\ - ν_{β} (x_{i})) - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))} | = 0 \end{matrix}$ (36)

Because 0 ≤ 1 - μ_α (x_i) - ν_α (x_i) ≤1, 0 ≤ 1 - μ_β (x_i) - ν_β (x_i) ≤1, and 4 - (1 - μ_α (x_i) - ν_α (x_i)) (1 - μ_β (x_i) - ν_β (x_i)) > 0, we only need to prove the next equation set:

${(\begin{matrix} ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) \\ + μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) = 0 \end{matrix})$ (37) $(\begin{matrix} μ_{β} (x_{i}) - μ_{α} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) = 0 \end{matrix})$ (38)

Take Eq. (37) into Eq. (38), we can get μ_β (x_i) - μ_α (x_i) = ν_β (x_i) - ν_α (x_i) which is then took into Eq. (37), and an item μ_β (x_i) ν_β (x_i) is also took into Eq. (37). Then, we can get follows:

$\begin{matrix} Eq . (37) = - μ_{α} (x_{i}) + μ_{β} (x_{i}) \\ + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \\ = - μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{β} (x_{i}) ν_{β} (x_{i}) \\ + μ_{β} (x_{i}) ν_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \\ = [ν_{β} (x_{i}) - ν_{α} (x_{i})] - μ_{β} (x_{i}) [ν_{β} (x_{i}) - ν_{α} (x_{i})] \\ + ν_{β} (x_{i}) [ν_{β} (x_{i}) - ν_{α} (x_{i})] \\ = [ν_{β} (x_{i}) - ν_{α} (x_{i})] [1 - μ_{β} (x_{i}) + ν_{β} (x_{i})] = 0 \end{matrix}$ .

Thus, we can infer that ν_β (x_i) - ν_α (x_i) =0 or 1 - μ_β (x_i) + ν_β (x_i) =0.

For the first case: ν_β (x_i) - ν_α (x_i) =0 ⇒ ν_β (x_i) = ν_α (x_i), and then we take it into Eq. (37) and Eq. (38) to get μ_α (x_i) = μ_β (x_i). By this way, we can obtain α = β.

For the second case: 1 - μ_β (x_i) + ν_β (x_i) =0 ⇒ μ_β (x_i) - ν_β (x_i) =1. Because 0 ≤ μ_β (x_i) ≤1 and 0 ≤ ν_β (x_i) ≤1, we can infer that μ_β (x_i) - ν_β (x_i) =1 is established if and only if μ_β (x_i) =1 and ν_β (x_i) =0. We take μ_β (x_i) =1 and ν_β (x_i) =0 into Eq. (38) to get μ_α (x_i) =1 and ν_α (x_i) =0. We can obtain α = β.

To sum up, we can get: μ_α (x_i) = μ_β (x_i) , ν_α (x_i) = ν_β (x_i) ⇒ α = β.

(b) α = β ⇒ S (α, β) = 1;

We have known that α = β is established, so μ_α (x_i) = μ_β (x_i) and ν_α (x_i) = ν_β (x_i) can be inferred. Take μ_β (x_i) = μ_α (x_i) and ν_β (x_i) = ν_α (x_i) into S (α, β), we can obtain:

$\begin{matrix} S (α, β) \\ = 1 - 2 [\begin{matrix} {(\frac{\begin{matrix} - μ_{α} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) \\ + μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}})}^{2} \\ + {(\frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) \\ - ν_{β} (x_{i})) - ν_{α} (x_{i}) μ_{β} (x_{i}) \\ + μ_{α} (x_{i}) ν_{β} (x_{i}) + 2 (ν_{α} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}})}^{2} \end{matrix}] \\ = 1 - 2 (0 + 0) = 1 \end{matrix}$ .

Hence, we can prove that S (α, β) = 1 if and only if α = β.

The proof is completed.

Property: (S3) : S (α, β) = S (β, α) ;

Proof.

According the feature of square operation and Eq. (35), we can get:

$\begin{matrix} {(\frac{- μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i})}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))})}^{2} \\ = {(\frac{μ_{α} (x_{i}) - μ_{β} (x_{i}) - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i})}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))})}^{2} \end{matrix}$ and,

${(\frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))})}^{2}$

$= {(\frac{\begin{matrix} μ_{α} (x_{i}) - μ_{β} (x_{i}) - 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))})}^{2}$

Hence, we can get: $\begin{matrix} S (α, β) = 1 - 2 [\begin{matrix} {(\frac{\begin{matrix} - μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \\ + μ_{α} (x_{i}) - ν_{α} (x_{i}) μ_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \\ (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \end{matrix}})}^{2} \\ + {(\frac{\begin{matrix} μ_{α} (x_{i}) + 2 (ν_{β} (x_{i}) - ν_{α} (x_{i})) \\ - μ_{α} (x_{i}) ν_{β} (x_{i}) - μ_{β} (x_{i}) \\ + ν_{α} (x_{i}) μ_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \\ (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \end{matrix}})}^{2} \end{matrix}] \\ = S (β, α) \end{matrix}$ .

To summarize the above, S (α, β) = S (β, α).

The proof is completed.

Property: (S4) : If α ⊆ β ⊆ γ, then S (α, β) ≥ S (α, γ) and S (β, γ) ≥ S (α, γ).

Proof.

Let $S_{1} (α, β) = \frac{\begin{matrix} μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) \\ - μ_{α} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}}$ , and, $S_{2} (α, β) = \frac{\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}}{\begin{matrix} 4 - (1 - μ_{β} (x_{i}) - ν_{α} (x_{i})) \\ (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}}$ .

Hence, we can prove that S (α, β) = 1 if and only if α = β.

The proof is completed.

Property: (S3) : S (α, β) = S (β, α) ;

Proof.

So, we can rewrite Eq. (35) as S (α, β) =1 - 2 [(S₁ (α, β)) ² + (S₂ (α, β)) ²], and we can further get: S (α, γ) =1 - 2 [(S₁ (α, γ)) ² + (S₂ (α, γ)) ²]. If S₁ (α, β) ≤ S₁ (α, γ), and S₂ (α, β) ≤ S₂ (α, γ), we can get S (α, β) ≥ S (α, γ). Therefore, we divide S (α, β) ≥ S (α, γ) into two steps to prove it, and the proof as follows:

(a) First step: we should prove S₁ (α, β) ≤ S₁ (α, γ). So we can get:

$\begin{matrix} S_{1} (α, γ) - S_{1} (α, β) \\ = (\begin{matrix} \frac{- μ_{α} (x_{i}) + μ_{γ} (x_{i}) + ν_{α} (x_{i}) μ_{γ} (x_{i}) - μ_{α} (x_{i}) ν_{γ} (x_{i})}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))} \\ - \frac{- μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i})}{4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))} \end{matrix}) \end{matrix}$

$\begin{matrix} = \frac{{\begin{matrix} (\begin{matrix} (\begin{matrix} μ_{γ} (x_{i}) + ν_{α} (x_{i}) μ_{γ} (x_{i}) \\ - μ_{α} (x_{i}) ν_{γ} (x_{i}) - μ_{α} (x_{i}) \end{matrix}) \\ [\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}] \end{matrix}) \\ - (\begin{matrix} [\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i})) \end{matrix}] \\ (\begin{matrix} μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) \\ - μ_{α} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}) \end{matrix}) \end{matrix}}}{(\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \end{matrix})} \end{matrix}$ .

Because 0 ≤ 1 - μ_α (x_i) - ν_α (x_i) ≤1, 0 ≤ 1 - μ_β (x_i) - ν_β (x_i) ≤1, 0 ≤ 1 - μ_γ (x_i) - ν_γ (x_i) ≤1, 4 - (1 - μ_α (x_i) - ν_α (x_i)) (1 - μ_β (x_i) - ν_β (x_i)) ≥ 0, and 4 - (1 - μ_α (x_i) - ν_α (x_i)) (1 - μ_γ (x_i) - ν_γ (x_i)) ≥ 0. We can get:

$(\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \end{matrix}) \geq 0$ .

We only need to prove that the numerator part of S₁ (α, γ) - S₁ (α, β) is larger than 0, as:

$\begin{matrix} (\begin{matrix} [- μ_{α} (x_{i}) + μ_{γ} (x_{i}) + ν_{α} (x_{i}) μ_{γ} (x_{i}) - μ_{α} (x_{i}) ν_{γ} (x_{i})] \\ \times [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \end{matrix}) - \\ (\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ \times [- μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i})] \end{matrix}) \\ \geq 0 \end{matrix}$ (39)

For convenience, we divide Eq. (39) into two parts as:

$\begin{matrix} [- μ_{α} (x_{i}) + μ_{γ} (x_{i}) + ν_{α} (x_{i}) μ_{γ} (x_{i}) - μ_{α} (x_{i}) ν_{γ} (x_{i})] \\ \times [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \end{matrix}$ (40)

$\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ \times [- μ_{α} (x_{i}) + μ_{β} (x_{i}) + ν_{α} (x_{i}) μ_{β} (x_{i}) - μ_{α} (x_{i}) ν_{β} (x_{i})] \end{matrix}$ (41) then we can get: $\begin{matrix} Eq . (40) - Eq . (41) = \\ [(1 + μ_{α} (x_{i})) (3 + μ_{α} (x_{i})) (μ_{γ} (x_{i}) - μ_{β} (x_{i}))] \\ + 2 ν_{α} (x_{i}) (1 + μ_{α} (x_{i}) + ν_{α} (x_{i})) (ν_{β} (x_{i}) - ν_{γ} (x_{i})) \\ + (μ_{β} (x_{i}) ν_{γ} (x_{i}) - ν_{β} (x_{i}) μ_{γ} (x_{i})) [{(μ_{α} (x_{i}) + ν_{α} (x_{i}))}^{2} - 1] \\ + [{(μ_{α} (x_{i}) - 1)}^{2} - 1] (μ_{β} (x_{i}) - μ_{γ} (x_{i})) \end{matrix}$ .

Because α ⊆ β ⊆ γ that is 0 ≤ μ_α (x_i) ≤ μ_β (x_i) ≤ μ_γ (x_i) ≤1 and 0 ≤ ν_γ (x_i) ≤ ν_β (x_i) ≤ ν_α (x_i) ≤1, then we can get:

${\begin{matrix} {\begin{matrix} (1 + μ_{α} (x_{i})) (3 + μ_{α} (x_{i})) \geq 0 \\ μ_{γ} (x_{i}) - μ_{β} (x_{i}) \geq 0 \end{matrix} \\ \Rightarrow (\begin{matrix} (1 + μ_{α} (x_{i})) (3 + μ_{α} (x_{i})) \\ (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \end{matrix}) \geq 0; \\ {\begin{matrix} 2 ν_{α} (x_{i}) (1 + μ_{α} (x_{i}) + ν_{α} (x_{i})) \geq 0 \\ ν_{β} (x_{i}) - ν_{γ} (x_{i}) \geq 0 \end{matrix} \\ \Rightarrow (\begin{matrix} 2 ν_{α} (x_{i}) (1 + μ_{α} (x_{i}) + ν_{α} (x_{i})) \\ \times (ν_{β} (x_{i}) - ν_{γ} (x_{i})) \end{matrix}) \geq 0; \\ {\begin{matrix} μ_{β} (x_{i}) ν_{γ} (x_{i}) - ν_{β} (x_{i}) μ_{γ} (x_{i}) \leq 0 \\ {(μ_{α} (x_{i}) + ν_{α} (x_{i}))}^{2} - 1 \leq 0 \end{matrix} \\ \Rightarrow (\begin{matrix} (μ_{β} (x_{i}) ν_{γ} (x_{i}) - ν_{β} (x_{i}) μ_{γ} (x_{i})) \\ \times [{(μ_{α} (x_{i}) + ν_{α} (x_{i}))}^{2} - 1] \end{matrix}) \geq 0; \\ {\begin{matrix} {(μ_{α} (x_{i}) - 1)}^{2} - 1 \leq 0 \\ μ_{β} (x_{i}) - μ_{γ} (x_{i}) \leq 0 \end{matrix} \\ \Rightarrow (\begin{matrix} [{(μ_{α} (x_{i}) - 1)}^{2} - 1] \\ \times [μ_{β} (x_{i}) - μ_{γ} (x_{i})] \end{matrix}) \geq 0 . \end{matrix}$

Thus, we can get that Eq. (40)- Eq. (41) ≥ 0 is always satisfied.

From the above analysis, we can know that the numerator of S₁ (α, γ) - S₁ (α, β) is always larger than 0, which means S₁ (α, β) ≤ S₁ (α, γ).

(b) Second step: we proof S₂ (α, β) ≤ S₂ (α, γ). So we can get: $\begin{array}{l} S_{2} (α, γ) - S_{2} (α, β) = \frac{(\begin{matrix} - μ_{α} (x_{i}) + μ_{γ} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{γ} (x_{i})) \\ - ν_{α} (x_{i}) μ_{γ} (x_{i}) + μ_{α} (x_{i}) ν_{γ} (x_{i}) \end{matrix})}{(\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i})) \end{matrix})} \\ - \frac{(\begin{matrix} - μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix})}{(\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix})} \end{array}$

$= \frac{{\begin{matrix} [\begin{matrix} μ_{γ} (x_{i}) - μ_{α} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{γ} (x_{i})) \\ - ν_{α} (x_{i}) μ_{γ} (x_{i}) + μ_{α} (x_{i}) ν_{γ} (x_{i}) \end{matrix}] \\ \times [\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix}] \\ - [\begin{matrix} 4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i})) \end{matrix}] \\ \times [\begin{matrix} μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}] \end{matrix}}}{\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ \times [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \end{matrix}}$ .

From the proof of the first step, we know that:

$\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ \times [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \geq 0 \end{matrix}$ .

We only need to prove that the numerator of S₂ (α, γ) - S₂ (α, β) is larger than 0, as: $\begin{matrix} (\begin{matrix} μ_{γ} (x_{i}) - μ_{α} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{γ} (x_{i})) \\ - ν_{α} (x_{i}) μ_{γ} (x_{i}) + μ_{α} (x_{i}) ν_{γ} (x_{i}) \end{matrix}) \times \\ (4 - (\begin{matrix} (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{β} (x_{i}) - ν_{β} (x_{i})) \end{matrix})) \\ - (4 - (\begin{matrix} (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) \\ \times (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i})) \end{matrix})) \times \\ (\begin{matrix} μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}) \geq 0 \end{matrix}$ (42)

For convenience, we divided Eq. (42) into two parts, as: $\begin{matrix} [\begin{matrix} μ_{γ} (x_{i}) - μ_{α} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{γ} (x_{i})) \\ - ν_{α} (x_{i}) μ_{γ} (x_{i}) + μ_{α} (x_{i}) ν_{γ} (x_{i}) \end{matrix}] \\ \times [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{β} (x_{i}) - ν_{β} (x_{i}))] \end{matrix}$ (43) $\begin{matrix} [4 - (1 - μ_{α} (x_{i}) - ν_{α} (x_{i})) (1 - μ_{γ} (x_{i}) - ν_{γ} (x_{i}))] \\ \times [\begin{matrix} μ_{α} (x_{i}) + μ_{β} (x_{i}) + 2 (ν_{α} (x_{i}) - ν_{β} (x_{i})) \\ - ν_{α} (x_{i}) μ_{β} (x_{i}) + μ_{α} (x_{i}) ν_{β} (x_{i}) \end{matrix}] \end{matrix}$ (44)

Then, we can get:

$\begin{matrix} Eq . (43) - Eq . (44) \\ = [3 + 2 μ_{α} (x_{i}) + {ν_{α}}^{2} (x_{i})] \times (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \\ + 2 \times (3 + 2 ν_{α} (x_{i})) (ν_{β} (x_{i}) - μ_{γ} (x_{i})) \\ + [3 + {(μ_{α} (x_{i}) + ν_{α} (x_{i}))}^{2}] \times \\ [ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i})] \\ + 2 \times {μ_{α}}^{2} (x_{i}) (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \\ + ν_{α} (x_{i}) [\begin{matrix} 2 \times (μ_{γ} (x_{i}) - μ_{β} (x_{i})) + \\ 4 \times (ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i})) \\ + ν_{α} (x_{i}) (ν_{β} (x_{i}) - ν_{γ} (x_{i})) \end{matrix}] \end{matrix}$ .

$+ μ_{α} (x_{i}) [\begin{matrix} 2 ν_{α} (x_{i}) (μ_{γ} (x_{i}) - μ_{β} (x_{i})) + \\ 4 \times (ν_{β} (x_{i}) - ν_{γ} (x_{i})) + \\ 4 \times (ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i})) \end{matrix}]$

Because α ⊆ β ⊆ γ (that is 0 ≤ μ_α (x_i) ≤ μ_β (x_i) ≤ μ_γ (x_i) ≤1 and 0 ≤ ν_γ (x_i) ≤ ν_β (x_i) ≤ ν_α (x_i) ≤1), we can get:

${\begin{matrix} 3 + 2 μ_{α} (x_{i}) + {ν_{α}}^{2} (x_{i}) > 0 and μ_{γ} (x_{i}) - μ_{β} (x_{i}) \geq 0 \\ \Rightarrow [3 + 2 μ_{α} (x_{i}) + {ν_{α}}^{2} (x_{i})] (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \geq 0; \\ 2 \times (3 + 2 ν_{α} (x_{i})) > 0 and ν_{β} (x_{i}) - ν_{γ} (x_{i}) \geq 0 \\ \Rightarrow 2 \times (3 + 2 ν_{α} (x_{i})) (ν_{β} (x_{i}) - ν_{γ} (x_{i})) \geq 0; \\ {\begin{matrix} 3 + {(μ_{α} (x_{i}) + ν_{α} (x_{i}))}^{2} > 0 \\ ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i}) \geq 0 \end{matrix} \\ \Rightarrow (\begin{matrix} (3 + {(μ_{α} (x_{i}) + ν_{α} (x_{i}))}^{2}) \times \\ (ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i})) \end{matrix}) \geq 0; \\ {\begin{matrix} μ_{γ} (x_{i}) - μ_{β} (x_{i}) \geq 0, ν_{β} (x_{i}) - ν_{γ} (x_{i}) \geq 0 \\ ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i}) \geq 0 \end{matrix} \\ \Rightarrow [\begin{matrix} 2 {μ_{α}}^{2} (x_{i}) (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \\ + ν_{α} (x_{i}) (\begin{matrix} 2 (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \\ + ν_{α} (x_{i}) (ν_{β} (x_{i}) - ν_{γ} (x_{i})) \\ + 4 (\begin{matrix} ν_{β} (x_{i}) μ_{γ} (x_{i}) \\ - μ_{β} (x_{i}) ν_{γ} (x_{i}) \end{matrix}) \end{matrix}) \end{matrix}] \geq 0; \\ {\begin{matrix} 2 \times ν_{α} (x_{i}) (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \geq 0 \\ ν_{β} (x_{i}) - ν_{γ} (x_{i}) \geq 0 \\ ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i}) \geq 0 \end{matrix} \\ \Rightarrow μ_{α} (x_{i}) [\begin{matrix} 2 ν_{α} (x_{i}) (μ_{γ} (x_{i}) - μ_{β} (x_{i})) \\ + 4 (ν_{β} (x_{i}) - ν_{γ} (x_{i})) \\ + 4 (ν_{β} (x_{i}) μ_{γ} (x_{i}) - μ_{β} (x_{i}) ν_{γ} (x_{i})) \end{matrix}] \geq 0 . \end{matrix}$

Hence, Eq. (43)- Eq. (44) ≥ 0 is always satisfied.

From the above analysis, we can know that the numerator of S₂ (α, γ) - S₂ (α, β) is always larger than 0, which means that S₁ (α, β) ≤ S₁ (α, γ).

Summarize the above analysis, we can get S (α, β) ≥ S (α, γ). Correspondingly, S (β, γ) ≥ S (α, γ) can be proved in the same way.

Overall, we can draw a conclusion that if α ⊆ β ⊆ γ, then S (α, β) ≥ S (α, γ) and S (β, γ) ≥ S (α, γ).

The proof is completed.

5 Numerical experiments and analysis

Numerical experiment is the most basic task of verifying the effectiveness of similarity/distance measure method. In this section, two sets of numerical experiments are performed, and two results are shown in Table 1 and Table 2. Each numerical experiment contains six pairs of IFSs α and β. If the similarity of the six pairs of IFSs are different, we can infer that the measure method is reasonable in these cases; otherwise, the method is unreasonable. Some classical similarity/distance measure methods from the Section 2 are conducted to compare with the proposed method; and a parameter p is required in S_DC, $S_{LS}^{1}$ , $S_{LS}^{2}$ , $S_{LS}^{3}$ , S_M, and S_B, and p = 1; in S_B, the value of parameter p is 2; the value of parameter ω_i is 1/3 in $S_{LS}^{3}$ .

Table 1
The first set of numerical results of the proposed method compared with other methods

1 2 3 4 5 6

Case α={<x,0.3,0.3> } β={<x,0.4,0.4> } α={<x,0.3,0.4> } β={<x,0.4,0.3> } α={<x,1.0,0.0> } β={<x,0.0,0.0> }

α={<x,0.5,0.5> } β={<x,0.0,0.0> } α={<x,0.4,0.2> } β={<x,0.5,0.3> } α={<x,0.4,0.2> } β={<x,0.5,0.2> }

S _B 0.9667 0.9000 0.5000 0.8333 0.9667 0.9500

S _C 1 0.9000 0.5000 1 1 0.9500

S _CCL 0.9667 0.9000 0.5000 0.8333 0.9667 0.9450

S _DC 1 0.9000 0.5000 1 1 0.9500

S _HK 0.9000 0.9000 0.5000 0.5000 0.9000 0.9500

d _HY 0.9000 0.9000 0 0.5000 0.9000 0.9000

$S_{HY}^{1}$ 0.8495 0.8495 0 0.3775 0.8495 0.8495

$S_{HY}^{2}$ 0.8182 0.8182 0 0.3333 0.8182 0.8182

$S_{LP}^{l}$ 0.9000 0.9000 0.5000 0.5000 0.9000 0.9500

$S_{LP}^{e}$ 0.7904 0.7904 0.2689 0.2689 0.7904 0.8899

$S_{LP}^{c}$ 0.7500 0.7500 0.2500 0.2500 0.7500 0.8636

$S_{H}^{l}$ 0.7500 0.7500 0 0 0.7500 0.8571

$S_{H}^{2}$ 0.9000 0.9000 0.5000 0.5000 0.9000 0.9500

$S_{H}^{pk 1}$ 0.7500 0.7500 0 0 0.7500 0.8571

$S_{H}^{pk 2}$ 0.8571 0.8571 0 0 0.8571 0.9231

S _JJ 0.9490 0.8977 0.5000 0.7500 0.9592 0.9450

$S_{LS}^{1}$ 0.9000 0.9000 0.5000 0.5000 0.9000 0.9500

$S_{LS}^{2}$ 0.9500 0.9000 0.5000 0.7500 0.9500 0.9500

$S_{LS}^{3}$ 0.9333 0.9333 0.5000 0.6667 0.9333 0.9500

S _M 0.9000 0.9000 0.5000 0.5000 0.9000 0.9500

D _W 0.9000 0.9000 0.2500 0.5000 0.9000 0.9250

S _L 0.9000 0.9000 0.2929 0.5000 0.9000 0.9293

S _Y 1.0000 0.9600 N/A N/A 0.9971 0.9965

D _V 0.9857 0.9857 N/A N/A 0.9844 0.9944

S _HJF 0.9974 0.9893 0.7500 0.9375 0.9983 0.9972

	1	2	3	4	5	6
Case	α={<x,0.3,0.3> }	β={<x,0.4,0.4> }	α={<x,0.3,0.4> }	β={<x,0.4,0.3> }	α={<x,1.0,0.0> }	β={<x,0.0,0.0> }
	α={<x,0.5,0.5> }	β={<x,0.0,0.0> }	α={<x,0.4,0.2> }	β={<x,0.5,0.3> }	α={<x,0.4,0.2> }	β={<x,0.5,0.2> }
S _B	0.9667	0.9000	0.5000	0.8333	0.9667	0.9500
S _C	1	0.9000	0.5000	1	1	0.9500
S _CCL	0.9667	0.9000	0.5000	0.8333	0.9667	0.9450
S _DC	1	0.9000	0.5000	1	1	0.9500
S _HK	0.9000	0.9000	0.5000	0.5000	0.9000	0.9500
d _HY	0.9000	0.9000	0	0.5000	0.9000	0.9000
$S_{HY}^{1}$	0.8495	0.8495	0	0.3775	0.8495	0.8495
$S_{HY}^{2}$	0.8182	0.8182	0	0.3333	0.8182	0.8182
$S_{LP}^{l}$	0.9000	0.9000	0.5000	0.5000	0.9000	0.9500
$S_{LP}^{e}$	0.7904	0.7904	0.2689	0.2689	0.7904	0.8899
$S_{LP}^{c}$	0.7500	0.7500	0.2500	0.2500	0.7500	0.8636
$S_{H}^{l}$	0.7500	0.7500	0	0	0.7500	0.8571
$S_{H}^{2}$	0.9000	0.9000	0.5000	0.5000	0.9000	0.9500
$S_{H}^{pk 1}$	0.7500	0.7500	0	0	0.7500	0.8571
$S_{H}^{pk 2}$	0.8571	0.8571	0	0	0.8571	0.9231
S _JJ	0.9490	0.8977	0.5000	0.7500	0.9592	0.9450
$S_{LS}^{1}$	0.9000	0.9000	0.5000	0.5000	0.9000	0.9500
$S_{LS}^{2}$	0.9500	0.9000	0.5000	0.7500	0.9500	0.9500
$S_{LS}^{3}$	0.9333	0.9333	0.5000	0.6667	0.9333	0.9500
S _M	0.9000	0.9000	0.5000	0.5000	0.9000	0.9500
D _W	0.9000	0.9000	0.2500	0.5000	0.9000	0.9250
S _L	0.9000	0.9000	0.2929	0.5000	0.9000	0.9293
S _Y	1.0000	0.9600	N/A	N/A	0.9971	0.9965
D _V	0.9857	0.9857	N/A	N/A	0.9844	0.9944
S _HJF	0.9974	0.9893	0.7500	0.9375	0.9983	0.9972

Table 2

The second set of numerical results of the proposed method compared with other methods

	1	2	3	4	5	6
Case	α={<x,0.5,0.5> }	β={<x,0.0,0.0> }	α={<x,0.6,0.4> }	β={<x,0.0,0.0> }	α={<x,0.00,0.87> }	β={<x,0.28,0.55> }
	α={<x,0.60,0.27> }	β={<x,0.28,0.55> }	α={<x,0.125,0.075> }	β={<x,0.175,0.025> }	α={<x,0.5,0.45> }	β={<x,0.5,0.40> }
S _B	0.8333	0.8333	0.7000	0.7000	0.9500	0.9500
S _C	1	0.9000	0.7000	0.7000.	0.9500	0.9500
S _CCL	0.8333	0.8333	0.7047	0.6953	0.8825	0.5000
S _DC	1	0.9000	0.7000	0.7000	0.9500	0.9500
S _HK	0.5000	0.5000	0.7000	0.7000	0.9500	0.9500
d _HY	0.5000	0.4000	0.6800	0.6800	0.9500	0.9500
$S_{HY}^{1}$	0.3775	0.2862	0.5668	0.5668	0.9229	0.9229
$S_{HY}^{2}$	0.3333	0.2500	0.5152	0.3333	0.8182	0.8182
$S_{LP}^{l}$	0.5000	0.5000	0.7000	0.7000	0.9000	0.9500
$S_{LP}^{e}$	0.2500	0.2500	0.4375	0.4375	0.7500	0.8636
$S_{LP}^{c}$	0.2689	0.2689	0.4782	0.4782	0.7904	0.8899
$S_{H}^{l}$	1	1	0.5217	0.5217	0.4000	0.1000
$S_{H}^{2}$	0.5000	0.5000	0.3000	0.3000	0.0500	0.0500
$S_{H}^{pk 1}$	1	1	0.5217	0.5217	0.4000	0.1000
$S_{H}^{pk 2}$	1	1	0.3529	0.3529	0.2500	0.0526
S _JJ	0.7500	0.8000	0.6983	0.6983	0. 9405	0.9500
$S_{LS}^{1}$	0.5000	0.5000	0.7000	0.7000	0.9500	0.9500
$S_{LS}^{2}$	0.7500	0.7500	0.7000	0.7000	0.9500	0.9500
$S_{LS}^{3}$	0.6667	0.6333	0.7933	0.7933	0.9667	0.9667
S _M	0.5000	0.5000	0.7000	0.7000	0.9500	0.9500
D _W	0.5000	0.5500	0.3100	0.3100	0.0500	0.0500
S _L	0.1340	0.1282	0.6980	0.6980	0.9500	0.9500
S _Y	N/A	N/A	0.8912	0.7794	0.9216	0.9946
D _V	N/A	N/A	N/A	0.8917	0.9827	0.9973
S _HJF	0.9375	0.9500	0.9075	0.9088	0.9959	0.9975

The result of the first experiment is shown in Table 1, the data were also used in [8 , 41]. Most results of contrast methods are unreasonable, two of the twenty-five IFS pairs are reasonable, and one of them is the proposed method (expressed as S_HJF), and another is S_JJ. The unreasonable methods are S_C, S_HK, S_DC, $S_{LS}^{1}$ , $S_{LS}^{2}$ , $S_{LS}^{3}$ , S_M, d_HY, $S_{HY}^{1}$ , $S_{HY}^{2}$ , $S_{Lp}^{l}$ , $S_{Lp}^{e}$ , $S_{Lp}^{c}$ , D_W, S_L, $S_{H}^{1}$ , $S_{H}^{2}$ , $S_{H}^{pk 1}$ , $S_{H}^{pk 2}$ , S_B, and S_CCL because the same similarity are appeared in different IFS pairs. Besides, D_V and S_Y do not produce similarity results because denominator is zero in these two methods(this situation is represented in the table as N/A).

The similarity of the second experiment are shown in Table 2, the data were also used in [41]. Although S_JJ obtained reasonable results in the first numerical experiment, the results produced by S_JJ are unreasonable in the second experiment. Some unreasonable results are produced in S_Y and D_V because the denominator is zero(this situation is represented in the table as N/A). Expect the proposed method, other comparison methods are all unreasonable because the same similarity are appeared in different IFSspairs.

Overall, none of the comparison methods satisfy both the two numerical experiments. Fortunately, the results which are produced by the proposed method are reasonable in both numerical experiments, which also confirms that our proposed method is more effective than other methods.

6 Applications

In this section, we introduce three kinds of applications, such as pattern recognition, medical diagnosis, and cluster analysis, to test the performance of the proposed method. In this section, the parameter p is set as 1 in S_DC, $S_{LS}^{1}$ , $S_{LS}^{2}$ , $S_{LS}^{3}$ , S_M, and S_B; the parameter t is set as 2 in S_B; ω_i is set as 1/3in $S_{LS}^{3}$ .

6.1 Applications for pattern recognition

We tend to classify two most similar object into one class in the real world. The similarity measure method can be used to measure the degree of similarity between IFSs from the real world, such as pattern recognition. In this sub-section, we give four pattern recognition problems. Several different patterns are given in each experiment, and then the degree of similarity between a test pattern and the given patterns are calculated. There is a basic rule: the test pattern should be classified into the class (one of the given pattern) that is most similar with the test pattern. However, if the similarity between a test pattern and the given patterns are all same, it will be a failed case, which indicates that the decision result of the method is unreasonable.

Example 1. Given three patterns S₁, S₂, and S₃, which are represented by three IFSs α₁, α₂ and α₃ in the universe of discourse U ={ x₁, x₂, x₃ } and they were used in [25][34]. These given three patterns are expressed as:

α₁ ={ 〈x₁, 0.1, 0.1〉, 〈x₂, 0.5, 0.1〉, 〈x₃, 0.1, 0.9〉 }

α₂ ={ 〈x₁, 0.5, 0.5〉, 〈x₂, 0.7, 0.3〉, 〈x₃, 0.0, 0.8〉 }

α₃ ={ 〈x₁, 0.7, 0.2〉, 〈x₂, 0.1, 0.8〉, 〈x₃, 0.4, 0.4〉 }

The test pattern is denoted by β in U ={ x₁, x₂, x₃ }, as:

β ={ 〈x₁, 0.4, 0.4〉, 〈x₂, 0.6, 0.2〉, 〈x₃, 0.0, 0.8〉 }

The results of different measure methods are shown as Table 3. Obviously, there are two similarity measure methods, as S_C and S_DC cannot classify the unknown pattern into one given class because S₁ (α₁, β) = S₂ (α₂, β) and S₁ (α₁, β) = S₂ (α₂, β) > S₃ (α₃, β). D_V cannot classify the test pattern because the problem of dividing by zero in this method(this situation is represented in the table as N/A). However, other methods exclusively classify β into S2, and the result is the same as the proposed method.

Table 3
The first experiment of different similarity measure methods on pattern recognition problem

Measures S₁ (α₁, β) S₂ (α₂, β) S₃ (α₃, β) Classification result

S _B 0.9444 0.9778 0.6000 S ₂

S _C 1 1 0.6000 Cannot be determined

S _CCL 0.9444 0.9989 0.6709 S ₂

S _DC 1 1 0.6000 Cannot be determined

S _HK 0.8333 0.9333 0.6000 S ₂

d _HY 0.8333 0.9333 0.5667 S ₂

$S_{HY}^{1}$ 0.7571 0.8980 0.4437 S ₂

$S_{HY}^{2}$ 0.7143 0.8750 0.3954 S ₂

$S_{LP}^{l}$ 0.8333 0.9333 0.6000 S ₂

$S_{LP}^{e}$ 0.6863 0.8602 0.3788 S ₂

$S_{LP}^{c}$ 0.6458 0.8333 0.3492 S ₂

$S_{H}^{l}$ 0.6000 0.8667 0.3644 S ₂

$S_{H}^{2}$ 0.8333 0.9333 0.6000 S ₂

$S_{H}^{pk 1}$ 0.6000 0.8667 0.3644 S ₂

$S_{H}^{pk 2}$ 0.7153 0.9259 0.5196 S ₂

S _JJ 0.9077 0.9733 0.5973 S ₂

$S_{LS}^{1}$ 0.8333 0.9333 0.6000 S ₂

$S_{LS}^{2}$ 0.9167 0.9667 0.6000 S ₂

$S_{LS}^{3}$ 0.8889 0.9556 0.7222 S ₂

S _M 0.8333 0.9333 0.6000 S ₂

S _S 0.9225 0.9626 0.8655 S ₂

D _W 0.8333 0.9333 0.5833 S ₂

S _L 0.6683 0.8586 0.3957 S ₂

S _Y 0.9954 0.9986 0.6709 S ₂

D _V N/A N/A N/A Cannot be determined

S _HJF 0.9840 0.9989 0.8205 S ₂

Measures	S₁ (α₁, β)	S₂ (α₂, β)	S₃ (α₃, β)	Classification result
S _B	0.9444	0.9778	0.6000	S ₂
S _C	1	1	0.6000	Cannot be determined
S _CCL	0.9444	0.9989	0.6709	S ₂
S _DC	1	1	0.6000	Cannot be determined
S _HK	0.8333	0.9333	0.6000	S ₂
d _HY	0.8333	0.9333	0.5667	S ₂
$S_{HY}^{1}$	0.7571	0.8980	0.4437	S ₂
$S_{HY}^{2}$	0.7143	0.8750	0.3954	S ₂
$S_{LP}^{l}$	0.8333	0.9333	0.6000	S ₂
$S_{LP}^{e}$	0.6863	0.8602	0.3788	S ₂
$S_{LP}^{c}$	0.6458	0.8333	0.3492	S ₂
$S_{H}^{l}$	0.6000	0.8667	0.3644	S ₂
$S_{H}^{2}$	0.8333	0.9333	0.6000	S ₂
$S_{H}^{pk 1}$	0.6000	0.8667	0.3644	S ₂
$S_{H}^{pk 2}$	0.7153	0.9259	0.5196	S ₂
S _JJ	0.9077	0.9733	0.5973	S ₂
$S_{LS}^{1}$	0.8333	0.9333	0.6000	S ₂
$S_{LS}^{2}$	0.9167	0.9667	0.6000	S ₂
$S_{LS}^{3}$	0.8889	0.9556	0.7222	S ₂
S _M	0.8333	0.9333	0.6000	S ₂
S _S	0.9225	0.9626	0.8655	S ₂
D _W	0.8333	0.9333	0.5833	S ₂
S _L	0.6683	0.8586	0.3957	S ₂
S _Y	0.9954	0.9986	0.6709	S ₂
D _V	N/A	N/A	N/A	Cannot be determined
S _HJF	0.9840	0.9989	0.8205	S ₂

Example 2. Given three patterns S₁, S₂, and S₃, which are represented by three IFSs α₁, α₂ and α₃ in the universe of discourse U ={ x₁, x₂, x₃, x₄ } and they were used in [8][10][41]. These given three patterns are expressed as:

α₁=< x₁,0.5,0.3>,< x₂,0.7,0.0>,< x₃,0.4,0.5>,< x₄,0.7,0.3>

α₂=< x₁,0.5,0.2>,< x₂,0.6,0.1>,< x₃,0.2,0.7>,< x₄,0.7,0.3>

α₃=< x₁,0.5,0.4>,< x₂,0.7,0.1>,< x₃,0.4,0.6>,< x₄,0.7,0.2>

The test pattern is denoted by β in U ={ x₁, x₂, x₃ }, as:

β ={ 〈x₁, 0.4, 0.3〉, 〈x₂, 0.7, 0.1〉

〈x₃, 0.3, 0.6〉, 〈x₄, 0.7, 0.3〉 }

The results of different similarity measure methods are shown in Table 4. S_HK, $S_{LS}^{1}$ , $S_{LS}^{3}$ , S_M, $S_{Lp}^{l}$ , $S_{Lp}^{e}$ , $S_{Lp}^{c}$ , D_W and $S_{H}^{2}$ cannot make a decision because S₁ (α₁, β) = S₃ (α₃, β) > S₂ (α₂, β). d_HY, $S_{HY}^{1}$ , and $S_{HY}^{2}$ cannot make a decision because S₁ (α₁, β) = S₂ (α₂, β) = S₃ (α₃, β). S_L cannot make a decision because S₁ (α₁, β) = S₂ (α₂, β) > S₃ (α₃, β). D_V cannot classify test pattern because the problem of dividing by zero in this method(this situation is represented in the table as N/A). Fortunately, other methods including S_C, S_DC, $S_{LS}^{2}$ , $S_{H}^{1}$ , $S_{H}^{pk 1}$ , $S_{H}^{pk 2}$ , S_Y, S_B, S_S, S_CCL, S_JJ, and the proposed method can make a decision, but most of these methods classify β into S₃, expect S_S classifies β into S₁.

Table 4

The second experiment of different similarity measure methods on pattern recognition problem

Measures	S₁ (α₁, β)	S₂ (α₂, β)	S₃ (α₃, β)	Classification result
S _B	0.9500	0.9375	0.9545	S ₃
S _C	0.9500	0.9375	0.9750	S ₃
S _CCL	0.9500	0.9354	0.9667	S ₃
S _DC	0.9500	0.9375	0.9750	S ₃
S _HK	0.9500	0.9375	0.9500	Cannot be determined
d _HY	0.9250	0.9250	0.9250	Cannot be determined
$S_{HY}^{1}$	0.8871	0.8871	0.8871	Cannot be determined
$S_{HY}^{2}$	0.8636	0.8636	0.8636	Cannot be determined
$S_{LP}^{l}$	0.9500	0.9375	0.9500	Cannot be determined
$S_{LP}^{e}$	0.8926	0.8677	0.8926	Cannot be determined
$S_{LP}^{c}$	0.8693	0.8409	0.8693	Cannot be determined
$S_{H}^{l}$	0.8875	0.8562	0.8944	S ₃
$S_{H}^{2}$	0.9500	0.9375	0.9500	Cannot be determined
$S_{H}^{pk 1}$	0.8875	0.8562	0.8944	S ₃
$S_{H}^{pk 2}$	0.9389	0.9198	0.9424	S ₃
S _JJ	0.9496	0.9367	0.9637	S ₃
$S_{LS}^{1}$	0.9500	0.9375	0.9500	Cannot be determined
$S_{LS}^{2}$	0.9500	0.9375	0.9625	S ₃
$S_{LS}^{3}$	0.9583	0.9542	0.9583	Cannot be determined
S _S	0.9844	0.9957	0.9807	S ₁
S _M	0.9500	0.9375	0.9500	Cannot be determined
D _W	0.9375	0.9312	0.9375	Cannot be determined
S _L	0.9250	0.9250	0.9067	Cannot be determined
S _Y	0.9906	0.9871	0.9959	S ₃
D _V	N/A	0.9916	0.9925	Cannot be determined
S _HJF	0.9949	0.9942	0.9976	S ₃

Example 3. Given three patterns S₁, S₂, and S₃, which are represented by three IFSs α₁, α₂ and α₃ in the universe of discourse U ={ x₁, x₂, x₃ } and they were used in [41, 42]. These given three patterns are expressed as:

α₁ ={ 〈x₁, 0.34, 0.34〉, 〈x₂, 0.19, 0.48〉, 〈x₃, 0.02, 0.12〉 }

α₂ ={ 〈x₁, 0.35, 0.33〉, 〈x₂, 0.20, 0.47〉, 〈x₃, 0.0, 0.14〉 }

α₃ ={ 〈x₁, 0.33, 0.35〉, 〈x₂, 0.21, 0.46〉, 〈x₃, 0.01, 0.13〉 }

The test pattern is denoted by β in U ={ x₁, x₂, x₃ }, as:

β ={ 〈x₁, 0.37, 0.31〉, 〈x₂, 0.23, 0.44〉,

〈x₃, 0.04, 0.10〉 }

The results of different similarity measure methods are shown in Table 5. There are only seven methods that can classify the unknown pattern, including the proposed method, $S_{H}^{l}$ , $S_{H}^{pk 1}$ , $S_{H}^{pk 2}$ , S_Y, S_S, S_JJ, and all these methods classify the unknown pattern β into S₁. S_C, S_HK, S_DC, $S_{LS}^{1}$ , $S_{LS}^{2}$ , $S_{LS}^{3}$ , S_M, d_HY, $S_{HY}^{1}$ , $S_{HY}^{2}$ , $S_{LP}^{l}$ , $S_{LP}^{e}$ , $S_{LP}^{c}$ , D_W, S_L, $S_{H}^{2}$ , S_B, S_CCL cannot make a decision because S₁ (α₁, β) = S₂ (α₂, β) = S₃ (α₃, β). D_V cannot classify the test pattern because the problem of dividing by zero in this method(this situation is represented in the table as N/A).

Table 5

The third experiment of different similarity measure methods on pattern recognition problem

Measures	S₁ (α₁, β)	S₂ (α₂, β)	S₃ (α₃, β)	Classification result
S _B	0.9700	0.9700	0.9700	Cannot be determined
S _C	0.9700	0.9700	0.9700	Cannot be determined
S _CCL	0.9700	0.9700	0.9700	Cannot be determined
S _DC	0.9700	0.9700	0.9700	Cannot be determined
S _HK	0.9700	0.9700	0.9700	Cannot be determined
d _HY	0.9700	0.9700	0.9700	Cannot be determined
$S_{HY}^{1}$	0.9533	0.9533	0.9533	Cannot be determined
$S_{HY}^{2}$	0.9418	0.9418	0.9418	Cannot be determined
$S_{LP}^{l}$	0.9700	0.9700	0.9700	Cannot be determined
$S_{LP}^{e}$	0.9328	0.9328	0.9328	Cannot be determined
$S_{LP}^{c}$	0.9154	0.9154	0.9154	Cannot be determined
$S_{H}^{l}$	0.8509	0.8042	0.8260	S ₁
$S_{H}^{2}$	0.9700	0.9700	0.9700	Cannot be determined
$S_{H}^{pk 1}$	0.8509	0.8042	0.8260	S ₁
$S_{H}^{pk 2}$	0.9178	0.8800	0.8990	S ₁
S _JJ	0.9679	0.9665	0.9672	S ₁
$S_{LS}^{1}$	0.9700	0.9700	0.9700	Cannot be determined
$S_{LS}^{2}$	0.9700	0.9700	0.9700	Cannot be determined
$S_{LS}^{3}$	0.9800	0.9800	0.9800	Cannot be determined
S _S	0.9990	0.9958	0.9984	S ₁
S _M	0.9700	0.9700	0.9700	Cannot be determined
D _W	0.9700	0.9700	0.9700	Cannot be determined
S _L	0.9700	0.9700	0.9700	Cannot be determined
S _Y	0.9892	0.9745	0.9819	S ₁
D _V	0.9972	N/A	0.9951	Cannot be determined
S _HJF	0.9989	0.9986	0.9988	S ₁

Example 4. Given three patterns S₁, S₂, and S₃, which are represented by three IFSs α₁, α₂ and α₃ in the universe of discourse U ={ x₁, x₂, x₃, x₄, x₅, x₆ } and they were used in [43]. These given three patterns are expressed as:

α₁=< x₁,0.94,0.00>,< x₂,0.88,0.00>,< x₃,0.82,0.00>,

< x₄,0.78,0.02>,< x₅,0.75,0.05>,< x₆,0.72,0.08>

α₂=< x₁,0.86,0.07>,< x₂,0.92,0.04>,< x₃,0.98,0.01>,

< x₄,0.98,0.00>,< x₅,0.95,0.00>,< x₆,0.92,0.00>

α₃=< x₁,0.66,0.14>,< x₂,0.72,0.08>,< x₃,0.78,0.02>,

< x₄,0.84,0.00>,< x₅,0.90,0.00>,< x₆,0.96,0.00>

The test pattern is denoted by β in U = x₁,x₂,x₃,x₄,

x₅,{x₆, as:

β=< x₁,0.53,0.27>,< x₂,0.56,0.24>,< x₃,0.59,0.21>,

< x₄,0.64,0.18>,< x₅,0.70,0.15>,< x₆,0.76,0.12>

The results of different similarity measure methods are shown in Table 6. There is only one method (D_V) cannot classify the unknown pattern because the problem of dividing by zero(this situation is represented in the table as N/A). Other methods (including the proposed method) can make a decision that classifies the unknown pattern β into S₃.

Table 6

The fourth experiment of different similarity measure methods on pattern recognition problem

Measures	S₁ (α₁, β)	S₂ (α₂, β)	S₃ (α₃, β)	Classification result
S _B	0.8181	0.7600	0.8325	S ₃
S _C	0.8225	0.7600	0.8325	S ₃
S _CCL	0.8169	0.7432	0.8292	S ₃
S _DC	0.8225	0.7600	0.8325	S ₃
S _HK	0.8158	0.7600	0.8325	S ₃
d _HY	0.7900	0.6950	0.8200	S ₃
$S_{HY}^{1}$	0.7105	0.5877	0.7399	S ₃
$S_{HY}^{2}$	0.6708	0.5383	0.6958	S ₃
$S_{LP}^{l}$	0.8158	0.7600	0.8325	S ₃
$S_{LP}^{e}$	0.6620	0.5634	0.6715	S ₃
$S_{LP}^{c}$	0.6239	0.5192	0.6247	S ₃
$S_{H}^{l}$	0.6568	0.5785	0.6668	S ₃
$S_{H}^{2}$	0.8158	0.7600	0.8325	S ₃
$S_{H}^{pk 1}$	0.6568	0.5785	0.0668	S ₃
$S_{H}^{pk 2}$	0.7801	0.7299	0.7995	S ₃
S _JJ	0.8202	0.7596	0.8313	S ₃
$S_{LS}^{1}$	0.8158	0.7600	0.8325	S ₃
$S_{LS}^{2}$	0.8192	0.7600	0.8325	S ₃
$S_{LS}^{3}$	0.8686	0.8183	0.8842	S ₃
S _S	0.9204	0.9154	0.9431	S ₃
S _M	0.8158	0.7600	0.8325	S ₃
D _W	0.8029	0.7275	0.8263	S ₃
S _L	0.8087	0.7346	0.8307	S ₃
S _Y	0.9517	0.9557	0.9669	S ₃
D _V	N/A	N/A	N/A	Cannot be determined
S _HJF	0.9527	0.9400	0.9709	S ₃

In this sub-section, the effectiveness of the proposed method is proved because obtains more reasonable classification results compared with most of the existing methods.

6.2 Applications for medical diagnosis

Medical diagnosis is one of the most common problems in the real world, the idea of classifying patients is similar to that of pattern recognition. Hence, we apply the idea of pattern recognition to medical diagnosis. After analyzing the fuzzy numbers of an unknown patient, the obtained similarity can be used to diagnose the unknown patient according to the given patterns of disease. In this sub-section, the value of the special parameter in comparison methods is the same as that of the sub-section 6.1.

Example 5. Suppose a set of diagnosis is α =

{α₁, α₂, α₃, α₄, α₅}, the five elements are represented by the IFSs S₁, S₂, S₃, S₄, and S₅ in the universe of discourse U ={ x₁, x₂, x₃, x₄, x₅ } and they were used in [30,43,44 , 30,43,44], where x₁ means Temperature, x₂ means Headache, x₃ means StomachPain, x₄ means Cough, and x₅ means ChestPain. The experimental results are shown in Table 7. The IFSs S₁, S₂, S₃, S₄, and S₅ are expressed as follows:

Table 7
The result of the comparison between the proposed method and the existing method

Measures S₁ (α₁, β) S₂ (α₂, β) S₃ (α₃, β) S₄ (α₃, β) S₅ (α₃, β) Result

S _B 0.8191 0.8236 0.8000 0.5492 0.5000 S ₂

S _C 0.8500 0.8400 0.8000 0.5600 0.5000 S ₁

S _CCL 0.8417 0.8360 0.8067 0.5593 0.5047 S ₁

S _DC 0.8500 0.8400 0.8000 0.5600 0.5000 S ₁

d _HY 0.7600 0.7800 0.7200 0.4800 0.4400 S ₂

$S_{HY}^{1}$ 0.6738 0.7006 0.6389 0.3893 0.3629 S ₂

$S_{HY}^{2}$ 0.6320 0.6606 0.6047 0.3569 0.3412 S ₂

$S_{LP}^{l}$ 0.8100 0.8200 0.8000 0.5400 0.5000 S ₂

$S_{LP}^{e}$ 0.6524 0.6770 0.6643 0.3529 0.3415 S ₂

$S_{LP}^{c}$ 0.6136 0.6387 0.6379 0.3319 0.3347 S ₂

$S_{H}^{l}$ 0.6077 0.6566 0.6427 0.3134 0.3107 S ₂

$S_{H}^{2}$ 0.8100 0.8200 0.8000 0.5400 0.5000 S ₂

$S_{H}^{pk 1}$ 0.6077 0.6566 0.6427 0.3134 0.3107 S ₂

$S_{H}^{pk 2}$ 0.7370 0.7698 0.7408 0.4338 0.3917 S ₂

S _JJ 0.8314 0.8350 0.7979 0.5413 0.4968 S ₂

$S_{LS}^{1}$ 0.8100 0.8200 0.8000 0.5400 0.5000 S ₂

$S_{LS}^{2}$ 0.8400 0.8300 0.8000 0.5500 0.5000 S ₁

$S_{LS}^{3}$ 0.7533 0.7600 0.7400 0.6600 0.6467 S ₂

S _S 0.9347 0.9228 0.9223 0.7673 0.7490 S ₁

S _M 0.8100 0.8200 0.8000 0.5400 0.5000 S ₂

D _W 0.7850 0.8000 0.7600 0.5100 0.4700 S ₂

S _L 0.7416 0.7791 0.7412 0.5120 0.4823 S ₂

S _Y 0.9046 0.8602 0.8510 0.5033 0.4542 S ₁

D _V N/A N/A 0.8987 N/A N/A Not

S _HJF 0.9623 0.9671 0.9206 0.7305 0.6827 S ₂

Measures	S₁ (α₁, β)	S₂ (α₂, β)	S₃ (α₃, β)	S₄ (α₃, β)	S₅ (α₃, β)	Result
S _B	0.8191	0.8236	0.8000	0.5492	0.5000	S ₂
S _C	0.8500	0.8400	0.8000	0.5600	0.5000	S ₁
S _CCL	0.8417	0.8360	0.8067	0.5593	0.5047	S ₁
S _DC	0.8500	0.8400	0.8000	0.5600	0.5000	S ₁
d _HY	0.7600	0.7800	0.7200	0.4800	0.4400	S ₂
$S_{HY}^{1}$	0.6738	0.7006	0.6389	0.3893	0.3629	S ₂
$S_{HY}^{2}$	0.6320	0.6606	0.6047	0.3569	0.3412	S ₂
$S_{LP}^{l}$	0.8100	0.8200	0.8000	0.5400	0.5000	S ₂
$S_{LP}^{e}$	0.6524	0.6770	0.6643	0.3529	0.3415	S ₂
$S_{LP}^{c}$	0.6136	0.6387	0.6379	0.3319	0.3347	S ₂
$S_{H}^{l}$	0.6077	0.6566	0.6427	0.3134	0.3107	S ₂
$S_{H}^{2}$	0.8100	0.8200	0.8000	0.5400	0.5000	S ₂
$S_{H}^{pk 1}$	0.6077	0.6566	0.6427	0.3134	0.3107	S ₂
$S_{H}^{pk 2}$	0.7370	0.7698	0.7408	0.4338	0.3917	S ₂
S _JJ	0.8314	0.8350	0.7979	0.5413	0.4968	S ₂
$S_{LS}^{1}$	0.8100	0.8200	0.8000	0.5400	0.5000	S ₂
$S_{LS}^{2}$	0.8400	0.8300	0.8000	0.5500	0.5000	S ₁
$S_{LS}^{3}$	0.7533	0.7600	0.7400	0.6600	0.6467	S ₂
S _S	0.9347	0.9228	0.9223	0.7673	0.7490	S ₁
S _M	0.8100	0.8200	0.8000	0.5400	0.5000	S ₂
D _W	0.7850	0.8000	0.7600	0.5100	0.4700	S ₂
S _L	0.7416	0.7791	0.7412	0.5120	0.4823	S ₂
S _Y	0.9046	0.8602	0.8510	0.5033	0.4542	S ₁
D _V	N/A	N/A	0.8987	N/A	N/A	Not
S _HJF	0.9623	0.9671	0.9206	0.7305	0.6827	S ₂

S₁(Viral fever)=< x₁,0.4,0.0>,< x₂,0.3,0.5>,

< x₃,0.1,0.7>,< x₄,0.4,0.3>,< x₅,0.1,0.7>

S₂(Malaria)=< x₁,0.7,0.0>,< x₂,0.2,0.6>,

< x₃,0.0,0.9>,< x₄,0.7,0.0>,< x₅,0.1,0.8>

S₃(Typhoid)=< x₁,0.3,0.03,< x₂,0.6,0.1>,

< x₃,0.2,0.7>,< x₄,0.2,0.6>,< x₅,0.1,0.9>

S₄(Stomach problem)=< x₁,0.1,0.7>,< x₂,0.2,0.4>,

< x₃,0.8,0.0>,< x₄,0.2,0.7>,< x₅,0.2,0.7>

S₅(Chest problem)=< x₁,0.1,0.8>,< x₂,0.0,0.8>,

< x₃,0.2,0.8>,< x₄,0.2,0.8>,< x₅,0.8,0.1>

Suppose the diagnosis of a patient is unknown, but the patient’s symptoms are expressed by IFS β in the universe of discourse U ={ x₁, x₂, x₃, x₄, x₅ } as follows:

β(Patient)=< x₁,0.8,0.1>,< x₂,0.6,0.1>,

< x₃,0.2,0.8>,< x₄,0.6,0.1>,< x₅,0.1,0.6>

In this experiment, twenty-four methods are selected to compare with the proposed method. As shown in Table 7, D_V cannot make a decision because the problem of dividing by zero(this situation is represented in the table as N/A). The classification results that obtained by S_C, S_DC, $S_{LS}^{2}$ , S_Y, S_S, and S_CCL classify the unknown patient β into S₁; however, other methods classify the unknown patient β into S₂, and these methods are $S_{LS}^{1}$ , $S_{LS}^{3}$ , S_M, d_HY, $S_{HY}^{1}$ , $S_{HY}^{2}$ , $S_{LP}^{l}$ , $S_{LP}^{e}$ , $S_{LP}^{c}$ , S_L, D_W, $S_{H}^{l}$ , $S_{H}^{2}$ , $S_{H}^{pk 1}$ , $S_{H}^{pk 2}$ , S_B, S_JJ and the proposed method.

Example 6. Suppose a set of diagnoses is α =

{α₁, α₂, α₃, α₄, α₅}, the four elements are represented by the IFSs S₁, S₂, S₃, and S₄ in the universe of discourse U ={ x₁, x₂, x₃, x₄, x₅ } and they were used in [7] where x₁ means characterofstool, x₂ means bellyache, x₃ means ictussileus, x₄ means chronicsileus, and x₅ means anemia. IFSs S₁, S₂, S₃, and S₄ are represented as:

S₁(metastasis)=< x₁,0.4,0.4>,< x₂,0.3,0.3>,

< x₃,0.5,0.1>,< x₄,0.5,0.2>,< x₅,0.6,0.2>

S₂(recurrence)=< x₁,0.2,0.6>,< x₂,0.3,0.5>,

< x₃,0.2,0.3>,< x₄,0.7,0.1>,< x₅,0.8,0.0>

S₃(bad)=< x₁,0.1,0.9,< x₂,0.0,0.1>,

< x₃,0.2,0.7>,< x₄,0.1,0.8>,< x₅,0.2,0.8>

S₄(well)=< x₁,0.8,0.2>,< x₂,0.9,0.0>,

< x₃,1.0,0.0>,< x₄,0.7,0.2>,< x₅,0.6,0.4>

Suppose the diagnosis of a patient is unknown, but the patient’s symptoms are expressed as IFS β in the universe of discourse U ={ x₁, x₂, x₃, x₄, x₅ } as follows:

β(Patient)=< x₁,0.3,0.5>,< x₂,0.4,0.4>,

< x₃,0.6,0.2>,< x₄,0.5,0.1>,< x₅,0.9,0.0>

The experimental results of different methods are shown in Table 8. As it can be seen from Table 8, S_HK, $S_{LS}^{1}$ , S_M, $S_{LP}^{l}$ , $S_{LP}^{e}$ , $S_{LP}^{c}$ , and $S_{H}^{2}$ cannot make a reasonable decision because S₁ (α₁, β) = linebreakS₂ (α₂, β) > S₃ (α₃, β) and S₁ (α₁, β) = S₂ (α₂, β) > S₄ (α₄, β), these methods cannot be used to make a diagnosis in this experiment. The reason why D_V cannot make a diagnosis is that the denominator is zero(this situation is represented in the table as N/A). Other methods (including the proposed method) can make a diagnosis and classify the patientas S₁.

Table 8

The result of the comparison between the proposed method and the existing method

Measures	S₁ (α₁, β)	S₂ (α₂, β)	S₃ (α₃, β)	S₄ (α₃, β)	Result
S _B	0.9029	0.8800	0.5800	0.6755	S ₁
S _C	0.9200	0.8800	0.5800	0.6900	S ₁
S _CCL	0.9053	0.8720	0.5833	0.6750	S ₁
S _DC	0.9200	0.8800	0.5800	0.6900	S ₁
S _HK	0.8800	0.8800	0.5200	0.6700	Cannot
d _HY	0.8600	0.8200	0.4400	0.6000	S ₁
$S_{HY}^{1}$	0.7976	0.7480	0.3334	0.4850	S ₁
$S_{HY}^{2}$	0.7622	0.7100	0.2956	0.4381	S ₁
$S_{LP}^{l}$	0.8800	0.8800	0.5200	0.6700	Cannot
$S_{LP}^{e}$	0.7612	0.7612	0.3079	0.4546	Cannot
$S_{LP}^{c}$	0.7227	0.7227	0.2847	0.4193	Cannot
$S_{H}^{l}$	0.7361	0.7278	0.2408	0.4641	S ₁
$S_{H}^{2}$	0.8800	0.8800	0.5200	0.6700	Cannot
$S_{H}^{pk 1}$	0.7361	0.7278	0.2408	0.4641	S ₁
$S_{H}^{pk 2}$	0.8436	0.8327	0.3673	0.6249	S ₁
S _HW	0.7515	0.8344	0.4289	0.2633	S ₁
S _JJ	0.9029	0.8779	0.5528	0.6894	S ₁
$S_{LS}^{1}$	0.8800	0.8800	0.5200	0.6700	Cannot
$S_{LS}^{2}$	0.9000	0.8800	0.5550	0.6850	S ₁
$S_{LS}^{3}$	0.9133	0.9000	0.6533	0.7567	S ₁
S _S	0.9669	0.9841	0.7309	0.8372	S ₁
S _M	0.8800	0.8800	0.5200	0.6700	Cannot
D _W	0.8700	0.8500	0.4800	0.6350	S ₁
S _L	0.8378	0.8279	0.4448	0.6269	S ₁
S _Y	0.9789	0.9468	0.5479	0.8383	S ₁
D _V	N/A	N/A	N/A	N/A	Cannot
S _HJF	0.9839	0.9800	0.7524	0.8836	S ₁

The proposed method obtains reasonable results that are analyzed in these experiments. However, some of comparison methods cannot make a decision in above experiments, which reveal the proposed method has competitiveness and potential to medical diagnosis.

6.3 Applications for cluster analysis

Fuzzy theory-based clustering methods is suitable for handling the uncertain and vague problem in the real world. In 1969, Ruspini [18] introduced the concept of fuzzy partition and then applied it for handling cluster analysis problems. In 2008, the clustering method based on intuitionistic fuzzy was proposed by Xu [15]. We employed the proposed similarity measure method to produce an association matrix according to the work from [15], so the proposed method can be used for dealing with cluster problems. In this sub-section, two clustering experiments are used to test the performance of the similarity measure-based clustering method.

Example 7. Given a car dataset that has ten kinds of cars from the market in Guangzhou, China [15]. These ten kinds of cars are expressed as: A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈, A₉, and A₁₀ that are shown in Table 9. Each car has six attributes, as x₁: fuel economy, x₂: aerodynamic degree, x₃: price, x₄: comfort, x₅: design, and x₆: safety. Each attribute is represented by set of intuitionistic fuzzy numbers 〈μ_α (x_i) v_α (x_i) 〉.

Table 9
The car data set

x ₁ x ₂ x ₃ x ₄ x ₅ x ₆

〈μ_α (x₁) , v_α (x₁) 〉〈μ_α (x₂) , v_α (x₂) 〉〈μ_α (x₃) , v_α (x₃) 〉〈μ_α (x₄) , v_α (x₄) 〉〈μ_α (x₅) , v_α (x₅) 〉〈μ_α (x₆) , v_α (x₆) 〉

A ₁ <0.30,0.40> <0.20, 0.70> <0.40,0.50> <0.80, 0.10> <0.40, 0.50> <0.20, 0.70>

A ₂ <0.40,0.30> <0.50, 0.10> <0.60,0.20> <0.20, 0.70> <0.30, 0.60> <0.70, 0.20>

A ₃ <0.40,0.20> <0.60, 0.10> <0.80, 0.10> <0.20, 0.60> <0.30, 0.70> <0.50, 0.20>

A ₄ <0.30,0.40> <0.90, 0.00> <0.80, 0.10> <0.70, 0.10> <0.10, 0.80> <0.20, 0.80>

A ₅ <0.80,0.10> <0.70, 0.20> <0.70, 0.00> <0.40, 0.10> <0.80, 0.20> <0.40, 0.60>

A ₆ <0.40,0.30> <0.30, 0.50> <0.20, 0.60> <0.70, 0.10> <0.50, 0.40> <0.30, 0.60>

A ₇ <0.60,0.40> <0.40, 0.20> <0.70, 0.20> <0.30, 0.60> <0.30, 0.70> <0.60, 0.10>

A ₈ <0.90,0.10> <0.70, 0.20 <0.70, 0.10> <0.40, 0.50> <0.40, 0.50> <0.80, 0.00>

A ₉ <0.40,0.40> <1.00, 0.00> <0.90, 0.10> <0.60, 0.20> <0.20, 0.70> <0.10, 0.80>

A ₁₀ <0.90,0.10> <0.80, 0.00> <0.60, 0.30> <0.50, 0.20> <0.80, 0.10> <0.60, 0.40>

	x ₁	x ₂	x ₃	x ₄	x ₅	x ₆
	〈μ_α (x₁) , v_α (x₁) 〉	〈μ_α (x₂) , v_α (x₂) 〉	〈μ_α (x₃) , v_α (x₃) 〉	〈μ_α (x₄) , v_α (x₄) 〉	〈μ_α (x₅) , v_α (x₅) 〉	〈μ_α (x₆) , v_α (x₆) 〉
A ₁	<0.30,0.40>	<0.20, 0.70>	<0.40,0.50>	<0.80, 0.10>	<0.40, 0.50>	<0.20, 0.70>
A ₂	<0.40,0.30>	<0.50, 0.10>	<0.60,0.20>	<0.20, 0.70>	<0.30, 0.60>	<0.70, 0.20>
A ₃	<0.40,0.20>	<0.60, 0.10>	<0.80, 0.10>	<0.20, 0.60>	<0.30, 0.70>	<0.50, 0.20>
A ₄	<0.30,0.40>	<0.90, 0.00>	<0.80, 0.10>	<0.70, 0.10>	<0.10, 0.80>	<0.20, 0.80>
A ₅	<0.80,0.10>	<0.70, 0.20>	<0.70, 0.00>	<0.40, 0.10>	<0.80, 0.20>	<0.40, 0.60>
A ₆	<0.40,0.30>	<0.30, 0.50>	<0.20, 0.60>	<0.70, 0.10>	<0.50, 0.40>	<0.30, 0.60>
A ₇	<0.60,0.40>	<0.40, 0.20>	<0.70, 0.20>	<0.30, 0.60>	<0.30, 0.70>	<0.60, 0.10>
A ₈	<0.90,0.10>	<0.70, 0.20	<0.70, 0.10>	<0.40, 0.50>	<0.40, 0.50>	<0.80, 0.00>
A ₉	<0.40,0.40>	<1.00, 0.00>	<0.90, 0.10>	<0.60, 0.20>	<0.20, 0.70>	<0.10, 0.80>
A ₁₀	<0.90,0.10>	<0.80, 0.00>	<0.60, 0.30>	<0.50, 0.20>	<0.80, 0.10>	<0.60, 0.40>

Firstly, according to the intuitionistic fuzzy numbers in Table 9, a similarity matrix S₍₁₎ is obtained by the calculation of Eq. (35).

Secondly, according to the operation method in Definition 2.6, we can get the intuitionistic fuzzy synthetic matrix $S_{(1)}^{2}$ of the similar matrix S as:

$\begin{matrix} S_{(1)} = \\ (\begin{matrix} 1 & 0 . 857209 & 0 . 839350 & 0 . 887748 & 0 . 870987 & 0 . 985949 & 0 . 874329 & 0 . 831257 & 0 . 870946 & 0 . 869733 \\ 0 . 857209 & 1 & 0 . 989929 & 0 . 886779 & 0 . 897207 & 0 . 856192 & 0 . 994531 & 0 . 966235 & 0 . 899027 & 0 . 902147 \\ 0 . 839350 & 0 . 989929 & 1 & 0 . 915177 & 0 . 915831 & 0 . 823741 & 0 . 991858 & 0 . 970943 & 0 . 930726 & 0 . 899590 \\ 0 . 887748 & 0 . 886779 & 0 . 915177 & 1 & 0 . 899362 & 0 . 849401 & 0 . 903786 & 0 . 876398 & 0 . 994562 & 0 . 862552 \\ 0 . 870987 & 0 . 897207 & 0 . 915831 & 0 . 899362 & 1 & 0 . 863369 & 0 . 911527 & 0 . 943871 & 0 . 921765 & 0 . 979855 \\ 0 . 985949 & 0 . 856192 & 0 . 823741 & 0 . 849401 & 0 . 863369 & 1 & 0 . 864807 & 0 . 834342 & 0 . 835752 & 0 . 894173 \\ 0 . 874329 & 0 . 994531 & 0 . 991858 & 0 . 903786 & 0 . 911527 & 0 . 864807 & 1 & 0 . 975384 & 0 . 911803 & 0 . 906323 \\ 0 . 831257 & 0 . 966235 & 0 . 970943 & 0 . 876398 & 0 . 943871 & 0 . 834342 & 0 . 975384 & 1 & 0 . 890247 & 0 . 948360 \\ 0 . 870946 & 0 . 899027 & 0 . 930726 & 0 . 994562 & 0 . 921765 & 0 . 835752 & 0 . 911803 & 0 . 890247 & 1 & 0 . 880995 \\ 0 . 869733 & 0 . 902147 & 0 . 899590 & 0 . 862552 & 0 . 979855 & 0 . 894173 & 0 . 906323 & 0 . 948360 & 0 . 880995 & 1 \end{matrix}) \end{matrix}$

$\begin{matrix} S_{(1)}^{2} = S_{(1)} \circ S_{(1)} = \\ (\begin{matrix} 1 & 0 . 886779 & 0 . 887748 & 0 . 887748 & 0 . 887748 & 0 . 985949 & 0 . 887748 & 0 . 876398 & 0 . 887748 & 0 . 894173 \\ 0 . 886779 & 1 & 0 . 991858 & 0 . 915177 & 0 . 943871 & 0 . 894173 & 0 . 994531 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 887748 & 0 . 991858 & 1 & 0 . 930726 & 0 . 943871 & 0 . 894173 & 0 . 991858 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 887748 & 0 . 915177 & 0 . 930726 & 1 & 0 . 921765 & 0 . 887748 & 0 . 915177 & 0 . 915177 & 0 . 994562 & 0 . 903786 \\ 0 . 887748 & 0 . 943871 & 0 . 943871 & 0 . 921765 & 1 & 0 . 894173 & 0 . 9438715 & 0 . 948360 & 0 . 921765 & 0 . 979855 \\ 0 . 985949 & 0 . 894173 & 0 . 894173 & 0 . 887748 & 0 . 894173 & 1 & 0 . 894173 & 0 . 894173 & 0 . 8809948 & 0 . 894173 \\ 0 . 887748 & 0 . 994531 & 0 . 991858 & 0 . 915177 & 0 . 943871 & 0 . 894173 & 1 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 876398 & 0 . 975384 & 0 . 975384 & 0 . 915177 & 0 . 948360 & 0 . 894173 & 0 . 975384 & 1 & 0 . 930726 & 0 . 948360 \\ 0 . 887748 & 0 . 930726 & 0 . 930726 & 0 . 994531 & 0 . 921765 & 0 . 880995 & 0 . 930726 & 0 . 930726 & 1 & 0 . 921765 \\ 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 903786 & 0 . 979855 & 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 921765 & 1 \end{matrix}) \end{matrix}$

$\begin{matrix} S_{(1)}^{4} = S_{(1)}^{2} \circ S_{(1)}^{2} = \\ (\begin{matrix} 1 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 985949 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 \\ 0 . 894173 & 1 & 0 . 991858 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 0 . 994531 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 991858 & 1 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 0 . 991858 & 0 . 9753836 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 930726 & 0 . 930726 & 1 & 0 . 930726 & 0 . 894173 & 0 . 930726 & 0 . 930726 & 0 . 994562 & 0 . 930726 \\ 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 1 & 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 0 . 979855 \\ 0 . 985949 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 1 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 \\ 0 . 894173 & 0 . 994531 & 0 . 991858 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 1 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 975384 & 0 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 0 . 975384 & 1 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 930726 & 0 . 930726 & 0 . 994562 & 0 . 930726 & 0 . 894173 & 0 . 930726 & 0 . 930726 & 1 & 0 . 930726 \\ 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 0 . 979855 & 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 1 \end{matrix}) \end{matrix}$

$\begin{matrix} S_{(1)}^{8} = S_{(1)}^{4} \circ S_{(1)}^{4} = \\ (\begin{matrix} 1 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 985949 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 \\ 0 . 894173 & 1 & 0 . 991858 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 0 . 994531 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 991858 & 1 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 0 . 991858 & 0 . 9753836 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 930726 & 0 . 930726 & 1 & 0 . 930726 & 0 . 894173 & 0 . 930726 & 0 . 930726 & 0 . 994562 & 0 . 930726 \\ 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 1 & 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 0 . 979855 \\ 0 . 985949 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 1 & 0 . 894173 & 0 . 894173 & 0 . 894173 & 0 . 894173 \\ 0 . 894173 & 0 . 994531 & 0 . 991858 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 1 & 0 . 975384 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 975384 & 0 & 0 . 930726 & 0 . 948360 & 0 . 894173 & 0 . 975384 & 1 & 0 . 930726 & 0 . 948360 \\ 0 . 894173 & 0 . 930726 & 0 . 930726 & 0 . 994562 & 0 . 930726 & 0 . 894173 & 0 . 930726 & 0 . 930726 & 1 & 0 . 930726 \\ 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 0 . 979855 & 0 . 894173 & 0 . 948360 & 0 . 948360 & 0 . 930726 & 1 \end{matrix}) \end{matrix}$

$S_{(2)} = (\begin{matrix} 1 & 0 . 9998 & 0 . 9984 & 0 . 9964 & 0 . 9856 & 0 . 4816 & 0 . 3991 & 0 . 3026 & 0 . 3434 & 0 . 4444 & 0 . 8278 & 0 . 7838 & 0 . 7837 & 0 . 8019 & 0 . 8277 \\ 0 . 9998 & 1 & 0 . 9970 & 0 . 9944 & 0 . 9818 & 0 . 4598 & 0 . 3759 & 0 . 2775 & 0 . 3193 & 0 . 4224 & 0 . 8151 & 0 . 7696 & 0 . 7696 & 0 . 7884 & 0 . 8151 \\ 0 . 9984 & 0 . 9970 & 1 & 0 . 9996 & 0 . 9936 & 0 . 5376 & 0 . 4596 & 0 . 3678 & 0 . 4066 & 0 . 5024 & 0 . 8594 & 0 . 8194 & 0 . 8193 & 0 . 8359 & 0 . 8593 \\ 0 . 9964 & 0 . 9944 & 0 . 9996 & 1 & 0 . 9964 & 0 . 5644 & 0 . 4886 & 0 . 3993 & 0 . 4370 & 0 . 5302 & 0 . 8740 & 0 . 8360 & 0 . 8359 & 0 . 8517 & 0 . 8739 \\ 0 . 9856 & 0 . 9818 & 0 . 9936 & 0 . 9964 & 1 & 0 . 6400 & 0 . 5708 & 0 . 4887 & 0 . 5234 & 0 . 6087 & 0 . 9130 & 0 . 8810 & 0 . 8809 & 0 . 8943 & 0 . 9129 \\ 0 . 4816 & 0 . 4598 & 0 . 5376 & 0 . 5644 & 0 . 6400 & 1 & 0 . 9962 & 0 . 9866 & 0 . 9888 & 0 . 9935 & 0 . 9070 & 0 . 9350 & 0 . 9335 & 0 . 9229 & 0 . 9061 \\ 0 . 3991 & 0 . 3759 & 0 . 4596 & 0 . 4886 & 0 . 5708 & 0 . 9962 & 1 & 0 . 9960 & 0 . 9979 & 0 . 9963 & 0 . 8700 & 0 . 9034 & 0 . 9031 & 0 . 8902 & 0 . 8698 \\ 0 . 3026 & 0 . 2775 & 0 . 3678 & 0 . 3993 & 0 . 4887 & 0 . 986 & 0 . 9960 & 1 & 0 . 9968 & 0 . 9862 & 0 . 8235 & 0 . 8630 & 0 . 8616 & 0 . 8463 & 0 . 8226 \\ 0 . 3434 & 0 . 3193 & 0 . 4066 & 0 . 4370 & 0 . 5234 & 0 . 9888 & 0 . 9979 & 0 . 9968 & 1 & 0 . 9950 & 0 . 8426 & 0 . 8793 & 0 . 8801 & 0 . 8657 & 0 . 8430 \\ 0 . 4444 & 0 . 4224 & 0 . 5024 & 0 . 5302 & 0 . 6087 & 0 . 9935 & 0 . 9963 & 0 . 9862 & 0 . 9950 & 1 & 0 . 8886 & 0 . 9186 & 0 . 9206 & 0 . 9090 & 0 . 8899 \\ 0 . 8278 & 0 . 8151 & 0 . 8594 & 0 . 8740 & 0 . 9130 & 0 . 9070 & 0 . 8700 & 0 . 8235 & 0 . 8426 & 0 . 8886 & 1 & 0 . 9975 & 0 . 9969 & 0 . 9985 & 0 . 9996 \\ 0 . 7838 & 0 . 7696 & 0 . 8194 & 0 . 8360 & 0 . 8810 & 0 . 9350 & 0 . 9034 & 0 . 8630 & 0 . 8793 & 0 . 9186 & 0 . 9975 & 1 & 0 . 9992 & 0 . 9988 & 0 . 9970 \\ 0 . 7837 & 0 . 7696 & 0 . 8193 & 0 . 8359 & 0 . 8809 & 0 . 9335 & 0 . 9031 & 0 . 8616 & 0 . 8801 & 0 . 9206 & 0 . 9969 & 0 . 9992 & 1 & 0 . 9996 & 0 . 9975 \\ 0 . 8019 & 0 . 7884 & 0 . 8359 & 0 . 8517 & 0 . 8943 & 0 . 9229 & 0 . 8902 & 0 . 8463 & 0 . 8657 & 0 . 9090 & 0 . 9985 & 0 . 9988 & 0 . 9996 & 1 & 0 . 9991 \\ 0 . 8277 & 0 . 8151 & 0 . 8593 & 0 . 8739 & 0 . 9129 & 0 . 9061 & 0 . 8698 & 0 . 8226 & 0 . 8430 & 0 . 8899 & 0 . 9996 & 0 . 9970 & 0 . 9979 & 0 . 9991 & 1 \end{matrix})$

Obviously, $S_{(1)}^{2}$ ≠ S₍₁₎ , thus we repeat to the operation in Definition 2.6 to get $S_{(1)}^{4}$ .

By comparing $S_{(1)}^{4}$ and $S_{(1)}^{2}$ , we can know that $S_{(1)}^{4}$ ≠ $S_{(1)}^{2}$ . Hence, the operation is carried out again to get $S_{(1)}^{8}$ .

Thirdly, we get $S_{(1)}^{8} = S_{(1)}^{4}$ , so $S_{(1)}^{4}$ is an intuitionistic fuzzy equivalence matrix. From the study of Xu [15], we can know that the settings of confidence level λ can determine the final clustering result. Therefore, different confidence level λ will result in different clustering results. In this sub-section, all the possible confidence levels λ are considered for clustering analysis. The specific classification results obtained by the proposed method are shown in Table 10.

Table 10

The clustering results of the proposed method

No.	Confidence level	Clustering result
1	0< λ< =0.89417	{A₁,A₂,A₃,A₄,A₅,A₆,A₇,A₈,A₉,A₁₀}
2	0. 89417< λ< =0.93073	{A₁,A₆}{A₂,A₃,A₄,A₅,A₇,A₈,A₉,A₁₀}
3	0. 93073< λ< =0.94836	{A₁,A₆},{A₄,A₉},{A₂,A₃,A₅,A₇,A₈,A₁₀}
4	0. 94835< λ< =0.97538	{A₁,A₆},{A₂,A₃,A₇},{A₄,A₉},{A₈},{A₅,A₁₀}
5	0. 97538< λ< =0.97985	{A₁,A₆},{A₄,A₉},{A₂,A₃,A₇,A₈},{A₅,A₁₀}
6	0. 97985< λ< =0.98595	{A₁,A₆},{A₄,A₉},{A₂,A₃,A₇},{A₈},{A₅,A₁₀}
7	0. 98595< λ< =0.99186	{A₁},{A₆},{A₄,A₉},{A₂,A₃,A₇},{A₈},{A₅,A₁₀}
8	0. 99186< λ< =0.99453	{A₁},{A₆},{A₄,A₉},{A₂,A₇},{A₃},{A₈},{A₅},{A₁₀}
9	0. 99453< λ< =0.99456	{A₁},{A₆},{A₄,A₉},{A₂},{A₇},{A₃},{A₈},{A₅},{A₁₀}
10	0. 99456< λ< =1.00	{A₁},{A₂},{A₃},{A4},{A₅},{A₆},{A₇},{A₈},{A9},{A₁₀}

According to different confidence levels, a variety of clustering results are obtained. In Table 10, the confidence level of No. 2, 3, 5, 6, 7, and 8 respectively classify these cars into two, three, four, five, seven, and eight categories, which are the same as that of [7]. In addition, the clustering result of No. 2 classifies the data set into two categories: {A₁, A₆} and {A₂, A₃, A₄, A₅, A₇, A₈, A₉, A₁₀}, and the same results can be found in [9, 15]. The clustering result of three categories: {A₁, A₆}, {A₄, A₉}, and {A₂, A₃, A₅, A₇, A₈, A₁₀} can be also found in [9][13]. The results of the four categories: {A₁, A₆}, {A₄, A₉}, {A₂, A₃, A₇, A₈} and {A₅, A₁₀} can be verified by [9, 10]. Overall, the proposed method can be used for clustering.

Example 8. There are other fifteen patterns for cluster analysis, which are from [9]. These patterns are represented by intuitionistic fuzzy number as follows.

A₁ ={ x, (0.910, 0.080) } A₂ ={ x, (0.930, 0.070) } A₃ ={ x, (0.870, 0.120) } A₄ ={ x, (0.850, 0.140) } A₅ ={ x, (0.790, 0.200) } A₆ ={ x, (0.190, 0.800) } A₇ ={ x, (0.100, 0.820) } A₈ ={ x, (0.450, 0.550) } A₉ ={ x, (0.030, 0.820) } A₁₀ ={ x, (0.070, 0.730) } A₁₁ ={ x, (0.500, 0.500) } A₁₂ ={ x, (0.910, 0.080) } A₁₃ ={ x, (0.400, 0.500) } A₁₄ ={ x, (0.420, 0.480) } A₁₅ ={ x, (0.460, 0.460) }

Similar to Example 7, the similarity matrix of 15 IFS patterns is got, and then we obtained that $S_{(2)}^{8} = S_{(2)}^{16}$ . The similarity matrix and equivalent similarity matrix are shown as follows. The clustering results of the proposed method are shown in Table 11.

Table 11

The clustering results of the proposed method

No.	Confidence level	Clustering result
1	0< λ< =0.9130	{A₁,A₂,A₃,A₄,A₅,A₆,A₇,A₈,A₉,A₁₀,A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
2	0.9130< λ< =0.9350	{A₁,A₂,A₃,A₄,A₅},{A₆,A₇,A₈,A₉,A₁₀,A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
3	0.9350< λ< =0.9962	{A₁,A₂,A₃,A₄,A₅},{A₆,A₇,A₈,A₉,A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
4	0.9962< λ< =0.9963	{A₁,A₂,A₃,A₄,A₅},{A₆},{A₇,A₈,A₉,A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
5	0.9963< λ< =0.9964	{A₁,A₂,A₃,A₄,A₅},{A₆},{A₇,A₈,A₉},{A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
6	0.9964< λ< =0.9967	{A₁,A₂,A₃,A₄},{A₅},{A₆},{A₇,A₈,A₉},{A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
7	0.9967< λ< =0.9979	{A₁,A₂,A₃,A₄},{A₅},{A₆},{A₇, A₉}, {A₈},{A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
8	0.9979< λ< =0.9984	{A₁,A₂,A₃,A₄},{A₅},{A₆},{A₇},{A₉},{A₈},{A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
9	0.9984< λ< =0.9991	{A₁,A₂},{A₃,A₄},{A₅},{A₆},{A₇},{A₉},{A₈},{A₁₀},{A₁₁,A₁₂,A₁₃,A₁₄,A₁₅}
10	0.9991< λ< =0.9992	{A₁,A₂},{A₃,A₄},{A₅},{A₆},{A₇},{A₉},{A₈},{A₁₀},{A₁₁,A₁₅},{A₁₂,A₁₃,A₁₄}
11	0.9992< λ< =0.9995	{A₁,A₂},{A₃,A₄},{A₅},{A₆},{A₇},{A₉},{A₈},{A₁₀},{A₁₁,A₁₅},{A₁₂},{A₁₃,A₁₄}
12	0.9995< λ< =0.9996	{A₁,A₂},{A₃,A₄},{A₅},{A₆},{A₇},{A₉},{A₈},{A₁₀},{A₁₁,A₁₅},{A₁₂},{A₁₃},{A₁₄}
13	0.9996< λ< =0.9998	{A₁,A₂},{A₃},{A₄},{A₅},{A₆},{A₇},{A₉},{A₈},{A₁₀},{A₁₁},{A₁₅},{A₁₂},{A₁₃},{A₁₄}
14	0.9998< λ< =1.00	{A₁},{A₂},{A₃},{A4},{A₅},{A₆},{A₇},{A₈},{A₉},{A₁₀},{A₁₁},{A₁₂},{A₁₃},{A₁₄},{A₁₅}

$\begin{matrix} S_{(2)}^{16} = S_{(2)}^{8} \circ S_{(2)}^{8} = \\ (\begin{matrix} 1 & 0 . 9998 & 0 . 9984 & 0 . 9984 & 0 . 9964 & 0 . 9130 & 0 . 9130 \\ 0 . 9998 & 1 & 0 . 9984 & 0 . 9984 & 0 . 9964 & 0 . 9130 & 0 . 9130 \\ 0 . 9984 & 0 . 9984 & 1 & 0 . 9996 & 0 . 9964 & 0 . 9130 & 0 . 9130 \\ 0 . 9984 & 0 . 9984 & 0 . 9996 & 1 & 0 . 9964 & 0 . 9130 & 0 . 9130 \\ 0 . 9964 & 0 . 9964 & 0 . 9964 & 0 . 9964 & 1 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 1 & 0 . 9962 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9962 & 1 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9962 & 0 . 9968 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9962 & 0 . 9979 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9962 & 0 . 9963 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9350 & 0 . 9350 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9350 & 0 . 9350 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9350 & 0 . 9350 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9350 & 0 . 9350 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9350 & 0 . 9350 \end{matrix} \begin{matrix} 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9962 & 0 . 9962 & 0 . 9962 & 0 . 9350 \\ 0 . 9968 & 0 . 9980 & 0 . 9963 & 0 . 9350 \\ 1 & 0 . 9968 & 0 . 9963 & 0 . 9350 \\ 0 . 9968 & 1 & 0 . 9963 & 0 . 9350 \\ 0 . 9963 & 0 . 9963 & 1 & 0 . 9350 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 1 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9991 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9991 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9991 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9996 \end{matrix} \begin{matrix} 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9130 & 0 . 9130 & 0 . 9130 & 0 . 9130 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9350 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9350 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9350 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9350 \\ 0 . 9350 & 0 . 9350 & 0 . 9350 & 0 . 9350 \\ 0 . 9991 & 0 . 9991 & 0 . 9991 & 0 . 9996 \\ 1 & 0 . 9992 & 0 . 9992 & 0 . 9991 \\ 0 . 9992 & 1 & 0 . 9995 & 0 . 9991 \\ 0 . 9992 & 0 . 9995 & 1 & 0 . 9991 \\ 0 . 9991 & 0 . 9991 & 0 . 9991 & 1 \end{matrix}) \end{matrix}$

The clustering results obtained according to different confidence levels are shown in Table 11. The results of two cluster categories: {A₁,A₂,A₃,A₄,A₅} and {A₆,A₇,A₈,A₉,A₁₀,A₁₁,A₁₂,A₁₃,A₁₄,A₁₅} are confirmed by [7]. We can find that the results of the three cluster categories:{A₁,A₂,A₃,A₄,A₅}, {A₆,A₇,A₈,A₉,A₁₀} and {A₁₁,A₁₂,A₁₃,A₁₄,A₁₅} are the same as [7][9][11][14]. The results are supported by many previous works, which shows the effectiveness of the proposed method.

7 Conclusions

In this paper, we propose a new method for measuring the degree of distance and similarity between IFSs. Different from our previous work, this method is derived from the sum of areas of two right triangles that are constructed by the diagonal line in square area and the transformed isosceles triangles of two IFSs. The new similarity measure method is proposed based on a geometrical model in a rectangular coordinate system with unit one, which has a favorable interpretability. A lot of mathematical derivation are conducted to prove the validity of the structured similarity measure model.

In addition, we also compare many the existing methods with the proposed method through numerical experiments. In terms of the measure of degree of similarity and distance, the proposed method has a strong superiority on our previous work and other existing methods. Subsequently, we apply the proposed method to pattern recognition, medical diagnosis, and cluster analysis to further illustrate the practicality of the proposed method. Our next research is to explore how to extend the similarity/distance measure method between IFSs to more applications, and we will also try to introduce the ideas of geometrical modeling into other fuzzy measure model fields.

Footnotes

Acknowledgments

This study was supported in part by the National Natural Science Foundation of China (No. 62002313), Key Laboratory in Software Engineering of Yunnan Province (No. 2020SE408), Postdoctoral Science Foundation ofYunnan Province in China, Doctoral Candidate Academic Award of Yunnan Province in China, and Yunnan University’s Research Innovation Fund for Graduate Students (No. 2019164).

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