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intuitionistic fuzzy minimum spanning tree (MST) clustering algorithm to deal with
intuitionistic fuzzy information. Zhao et al. (2012b) gave a measure for calculating
the association coefficient between IFVs, and presented an algorithm for clustering
IFVs. Moreover, they extended the algorithm to cluster IVIFVs. Wang et al. (2011)
proposed a formula to derive the intuitionistic fuzzy similarity degree between two
IFSs and developed an approach to constructing an intuitionistic fuzzy similarity
matrix. Then, they presented a netting method to make cluster analysis of IFSs via
intuitionistic fuzzy similarity matrix. Wang et al. (2012) developed an intuitionistic
fuzzy implication operator and extended the Lukasiewicz implication operator to
intuitionistic fuzzy environments, and then defined an intuitionistic fuzzy triangle
product and an intuitionistic fuzzy square product. Furthermore, they used the intu-
itionistic fuzzy square product to construct an intuitionistic fuzzy similarity matrix,
based on which a direct method for intuitionistic fuzzy cluster analysis was given.
Considering their wide range of application prospects of the intuitionistic fuzzy
clustering techniques in the fields of medical diagnosis, pattern recognition, etc. (Xu
and Cai 2012), in this chapter, we shall give a detailed introduction of the above
intuitionistic fuzzy clustering algorithms.
2.1 Clustering Algorithms Based on Intuitionistic Fuzzy
Similarity Matrices
α = α ,
v α )
α 1 = α 1 ,
v α 1 )
α 2 = α 2 ,
v α 2 )
Let
,
, and
be three IFVs, Zhang et al.
(2007) defined some basic operational laws as below:
(1)
c
α
= (
v α α )
;
(2)
α 1 α 2 = (
min
{ μ α 1 α 2 } ,
max
{
v α 1 ,
v α 2 } )
;
(3)
;
Based on the operational laws above, Zhang et al. (2007) derived the following
conclusions:
α 1 α 2 = (
max
{ μ α 1 α 2 } ,
min
{
v α 1 ,
v α 2 } )
α i
= α 1 ,
v α 1 )(
=
,
,
)
Theorem 2.1 (Zhang et al. 2007) Let
i
1
2
3
be the IFVs,
then
(1)
1 α 2 ) α 3 = 1 α 3 ) 2 α 3 )
.
(2)
1 α 2 ) α 3 = 1 α 3 ) 2 α 3 )
.
(3)
1 α 2 ) α 3 = α 1 2 α 3 )
.
(4)
1 α 2 ) α 3 = α 1 2 α 3 )
.
Proof
(1)
1 α 2 ) α 3 = (
min
{
max
{ μ α 1 α 2 } α 3 } ,
max
{
min
{
v
,
v
} ,
v
} )
α
α
α
1
2
3
= (
max
{
min
{ μ α 1 α 3 } ,
min
{ μ α 2 , μ α 3 }} ,
min
{
max
{
v
,
v
} ,
max
{
v
,
v
}} )
α
α
α
α
1
3
2
3
= 1 α 3 ) 2 α 3 )
 
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