Image Processing Reference
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1
/
λ
⎛
⎞
n
n
∏
(
)
∏
w
⎜
⎜
j
w
⎟
⎟
λ
λ
GIFWA
w
(,,
aaa
,
…
=− −
, )
a
1
1
μ
,
(
11
−−
(
ν
) )
j
123
n
a
a
j
j
⎝
⎠
j
=
1
j
=
1
⎛
1 λ
/
⎞
1
/
λ
⎛
⎞
⎛
⎞
n
n
⎜
⎜
⎜
⎟
⎟
⎟
∏
∏
(
)
w
j
w
⎜
⎜
λ
⎟
⎟
⎜
⎜
λ
⎟
⎟
=− −
1
1
μ
,
11 11
−− −−
(
(
ν
)
)
j
a
a
j
j
⎝
⎠
⎝
⎠
j
=
1
j
=
1
⎝
⎠
The following cases hold from the GIFWA operator.
Case 1: If λ = 1, then GIFWA reduces to IFWA:
aaa
a
wa wa wa
wa
IFWA
w
(,,
,
…
, )(
= ⊕⊕⊕⊕
)
n
nn
123
1 1
2 2
3 3
⎛
⎞
n
n
∏∏
⎜
⎜
w
w
⎟
⎟
=− −μ
=
1
(
1
)
j
,
(( ))
11
−−
ν
j
a
a
j
j
⎝
⎠
j
1
j
=
1
⎛
⎞
n
n
∏
∏
w
⎜
⎜
w
⎟
⎟
=− −
1
(
1
μ
)
,
ν
j
j
a
a
j
j
⎝
⎠
j
=
1
j
=
1
T
111 1
,
=
⎛
⎞
⎟
Case 2: If
w
, then GIFWA reduces to the intuitionistic
,
,
…
,
⎜
nnn n
fuzzy averaging operator: IFA (
a
1
,
a
2
,
a
3
, …,
a
n
) = (1/
n
)(
a
1
⊕
a
2
⊕
a
3
⊕⋯⊕
a
n
)
Case 3: If λ → ∞, then GIFWA reduces to the intuitionistic fuzzy maximum
operator:
IFMAX
w
(,,
aaa
123
…
=
,
, )max(
a
a
)
n
j
j
An example is given to calculate GIFWA of the four intuitionistic fuzz
values.
Example 3.1
Let us consider four intuitionistic fuzzy values
a
1
= (0.1, 0.6),
a
2
= (0.4, 0.3),
a
3
= (0.6, 0.2),
a
4
= (0.2, 0.5) with weight vector
be
w
= (0.2, 0.3, 0.1, 0.4)
T
of
a
j
(
j
= 1, 2, 3, 4) and λ = 2
Solution
From the intuitionistic fuzzy values, we have
μ
=
01
.,
μ
=
04
.,
μ
=
06
.,
μ
=
02
.
a
a
a
a
1
2
3
4
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