Intuitionistic Fuzzy Clustering Algorithms - Intuitionistic Fuzzy Aggregation and Clustering - page 196

the interval-valued intuitionistic fuzzy aggregation operator ( 2.110 )) of the IVIFSs
assigned to the cluster, and the distance between two clusters is defined as the distance
between the centers of each clusters.

Example 2.3 (Xu 2009) Given five building materials: sealant, floor varnish, wall
paint, carpet, and polyvinyl chloride flooring, which are represented by the IFSs
A j ( j = 1 , 2 , 3 , 4 , 5 ) in the feature space X ={ x 1 , x 2 ,..., x 8 } . w = ( 0 . 15 , 0 . 10 ,

T

0 . 12 , 0 . 15 , 0 . 10 , 0 . 13 , 0 . 14 , 0 . 11 ) is the weight vector of x i ( i = 1 , 2 ,..., 8 ) , and

the given data are listed as follows:

A 1 ={ x 1 , 0 . 20 , 0 . 50 , x 2 , 0 . 10 , 0 . 80 , x 3 , 0 . 50 , 0 . 30 , x 4 , 0 . 90 , 0 . 00 ,

x 5 , 0 . 40 , 0 . 35 , x 6 , 0 . 10 , 0 . 90 , x 7 , 0 . 30 , 0 . 50 , x 8 , 1 . 00 , 0 . 00 }

A 2 ={ x 1 , 0 . 50 , 0 . 40 , x 2 , 0 . 60 , 0 . 15 , x 3 , 1 . 00 , 0 . 00 , x 4 , 0 . 15 , 0 . 65 ,

x 5 , 0 . 00 , 0 . 80 , x 6 , 0 . 70 , 0 . 15 , x 7 , 0 . 50 , 0 . 30 , x 8 , 0 . 65 , 0 . 20 }

A 3 ={ x 1 , 0 . 45 , 0 . 35 , x 2 , 0 . 60 , 0 . 30 , x 3 , 0 . 90 , 0 . 00 , x 4 , 0 . 10 , 0 . 80 ,

x 5 , 0 . 20 , 0 . 70 , x 6 , 0 . 60 , 0 . 20 , x 7 , 0 . 15 , 0 . 80 , x 8 , 0 . 20 , 0 . 65 }

A 4 ={ x 1 , 1 . 00 , 0 . 00 , x 2 , 1 . 00 , 0 . 00 , x 3 , 0 . 85 , 0 . 10 , x 4 , 0 . 75 , 0 . 15 ,

x 5 , 0 . 20 , 0 . 80 , x 6 , 0 . 15 , 0 . 85 , x 7 , 0 . 10 , 0 . 70 , x 8 , 0 . 30 , 0 . 70 }

A 5 ={ x 1 , 0 . 90 , 0 . 00 , x 2 , 0 . 90 , 0 . 10 , x 3 , 0 . 80 , 0 . 10 , x 4 , 0 . 70 , 0 . 20 ,

x 5 , 0 . 50 , 0 . 15 , x 6 , 0 . 30 , 0 . 65 , x 7 , 0 . 15 , 0 . 75 , x 8 , 0 . 40 , 0 . 30 }

Now we utilize Algorithm 2.3 to classify the building materials A j

( j = 1 , 2 , 3 , 4 , 5 ) :

Step 1 In the first stage, each of the IFSs A j ( j = 1 , 2 , 3 , 4 , 5 ) is considered as a

unique cluster:

{ A 1 } , { A 2 } , { A 3 } , { A 4 } , { A 5 }

Step 2 Compare each IFS A i with all the other four IFSs by using the weighted
Hamming distance ( 2.110 ):

d wH ( A 1 , A 2 ) = d wH ( A 2 , A 1 ) = 0 . 4915 , d wH ( A 1 , A 3 ) = d wH ( A 3 , A 1 ) = 0 . 5115

d wH ( A 1 , A 4 ) = d wH ( A 4 , A 1 ) = 0 . 4310 , d wH ( A 1 , A 5 ) = d wH ( A 5 , A 1 ) = 0 . 4045

d wH ( A 2 , A 3 ) = d wH ( A 3 , A 2 ) = 0 . 2170 , d wH ( A 2 , A 4 ) = d wH ( A 4 , A 2 ) = 0 . 4515

d wH ( A 2 , A 5 ) = d wH ( A 5 , A 2 ) = 0 . 4545 , d wH ( A 3 , A 4 ) = d wH ( A 4 , A 3 ) = 0 . 4480

d wH ( A 3 , A 5 ) = d 7 ( A 5 , A 3 ) = 0 . 3735 , d wH ( A 4 , A 5 ) = d wH ( A 5 , A 4 ) = 0 . 1875

Since

d wH ( A 1 , A 5 ) = min { d wH ( A 1 , A 2 ), d wH ( A 1 , A 3 ), d wH ( A 1 , A 4 ), d wH ( A 1 , A 5 ) }= 0 . 4045
d wH ( A 2 , A 3 ) =

{ d wH ( A 2 , A 1 ), d wH ( A 2 , A 3 ), d wH ( A 2 , A 4 ), d wH ( A 2 , A 5 ) }= .

min 0 2170

d wH ( A 4 , A 5 ) = min { d wH ( A 4 , A 1 ), d wH ( A 4 , A 2 ), d wH ( A 4 , A 3 ), d wH ( A 4 , A 5 ) }= 0 . 1875