NTlarning ayirmasi. U to’plamdagi ikkita A va B NTostilarining ayirmasi - bu
A \ B { A\B (x) / x}, x X ,
bu erda
(x) (x) (x)
ko`rinishdagi to`plam.
A\ B A B
Misol. Aytaylik, ܷ = {ݔଵ, ݔଶ, ݔଷ, ݔସ}, ܯ = [0,1].
ܣ = {(ݔଵ|0.25), (ݔଶ|0.73), (ݔଷ|1), (ݔସ|0)}.
ܤ = {(ݔଵ|0.35), (ݔଶ|0.87), (ݔଷ|0), (ݔସ|1)}.
1
B
B
B ( x) 1 B ( x)
asosan
B
( x) 1
(x
| 0.65),( x2
| 0.13),( x3
| 1),( x4
| 0)
Endi
A \ B { A\ B (x) / x} A\ B (x) A (x) B (x) min(A (x) B (x))
asosan
1
A \ B (x | 0.25),(x | 0.13),(x |1),(x | 0).
2 3 4
NTlarning simmetrik ayirmasi. U to’plamdagi ikkita A va B NTostilarining simmetrik ayirmasi - bu
AB {
AB
(x) / x},
x X ,
bu erda
AB (x) A\B (x) B\A (x)
ko`rinishdagi to`plam.
Misol. Aytaylik, ܷ = {ݔଵ, ݔଶ, ݔଷ, ݔସ}, ܯ = [0,1].
ܣ = {(ݔଵ|0.25), (ݔଶ|0.73), (ݔଷ|1), (ݔସ|0)}.
ܤ = {(ݔଵ|0.35), (ݔଶ|0.87), (ݔଷ|0), (ݔସ|1)}.
B (x) 1 B (x) asosan
B (x) 1
(x) (x
| 0.65),(x2
| 0.13),(x3
|1),(x4
| 0).
Endi
B
1
B
A \ B { A\ B (x) / x} A\ B (x) A (x) B (x) min(A (x), B (x))
asosan
1
A\ B
(x) A \ B (x
| 0.25),(x2
| 0.13),(x3
|1),(x4
| 0).
A (x) 1 A (x) asosan
A (x) 1
(x) (x
| 0.75),(x2
| 0.27),(x3
| 0),(x4
|1).
Endi
A
1
A
B \ A { B\ A (x) / x} B\ A (x) B (x) A (x) min(B (x), A (x))
asosan
1
B\ A
B \ A (x
| 0.35),( x2
| 0.27),( x3
| 0),( x4
|1).
AB A\B B\ A
3) (x) (x) (x) va (x) (x) max( (x), (x))
A B A B
formulalarga asosan
1
A B
( x)
A\ B
( x)
B\ A
(x
| 0.35),( x2
| 0.27),( x3
|1),( x4
|1).
Algebraik (diz’yunktivli) yig’indi. U to’plamdagi ikkita A va B NTostilarining algebraik (diz’yunktivli) yig’indisi A⊕B birlashma va kesishma amallari yordamida quyidagicha aniqlanadi:
× )ݑ(ߤ − )ݑ(ߤ + )ݑ(ߤ = )ܤ ∩ ܣ( ∪ )ܤ ∩ ܣ( = )ܤ ∩ ܣ( = ܤ ⊕ ܣ ).ݑ(ߤ
1
Misol. U x , x , x , x , x ,
2 3 4 5
1
A (x | 0.4),(x | 0.8),(x
2 3
1
B (x | 0.6),(x | 0.6),(x
2 3
| 0),(x4
| 0),(x4
|1),(x5
| 0),(x5
| 0.3),
| 0.7),
1
A B (x | 0.76),(x | 0.92),(x
2 3
| 0),(x4 |1),(x5
| 0.79).
1
Topshiriqlar. Quyidagilar berilgan: U x , x , x
2 3
, x4
, x5 ,
1
A (x | 0.2), (x | 0.4), (x |1), (x
2 3 4
| 0), (x5
| 0.6),
1
B (x | 0.4), (x | 0.2), (x | 0), (x
2 3 4
|1), (x5
| 0.8).
A va B NTostilari ustida
A \ B, AB,
A B, A B, ∼A=B, ∼B=A, A ⊂ B, A = B,
A × B, A⊕B amallarini bajaring?
To`plamlarni noravshan kiritish (qo`shish) amali.
darajasi
A1 NTni A2
NTga kiritish
( A1 , A2 ) ( A ( x) A ( x))
xX 1 2
qiymat bilan belgilanadi. Bu yerda → (implikatsiya) amali quyidagi:
Lukasevich mantiqi bo`yicha - A (x) A (x) 1 (1 A (x) A (x));
1 2 1 2
A (x) A
(x) (1 A (x)) ( A (x) A
(x)) ;
A (x) A (x) A (x) A (x) min(A (x), A
(x)) ;
1 2 1 2 1 2
qoidalar yordamida belgilanadi.
Misol.Asosiy X {x1 , x2 , x3 , x4 , x5 } to’plamda NTostilari
A1 = {< 0.3/ x2 , < 0.6/x3 >, < 0.4/x5 >} va
A2 {< 0.8/x1 >, < 0.5/x2 >, < 0.7/x3 >, < 0.6/x5 >}
berilgan bo’lsin. Shu NTostilarining kiritish va tenglik darajasi qiymatlarini topish talab etiladi.
Lukasevich mantiqidan foydalangan holda quyidagi natijalarni olamiz:
kiritish darajasi qiymatlari
v( A1 , A2 ) = (0 0.8) (0.3 0.5)
(0.6 0.7)
(0 0)
(0.4 0.6) =
= (1
(1- 0 + 0.8))
(1 (1- 0.3 + 0.5))
(1 (1- 0.6 + 0.7)) (1 (1- 0 + 0))
(1 (1- 0.4 + 0.6) = 1 1 1 1 1 = 1;
v( A2 , A1 ) (0.8 0) (0.5 0.3)
(0.7 0.6)
(0 0) (0.6 0.4) =
(1
(1- 0.8 + 0)) (1 (1- 0.5 + 0.3)) (1 (1- 0.7 + 0.6))
(1 (1- 0 + 0))
(1 (1- 0.6 + 0.4)) 0.2 0.8 0.9 1 0.8 = 0,2.
NTlarning tengligi amali.
A1 NTni
A2 NTga tenglik darajasi
( A , A ) ( (x)
(x))
1 2 xX A1 A2
qiymat bilan belgilanadi. Bu yerda ↔ tenglik (ekvivalentlik) amali
A (x) A (x) (A (x) A (x)) (A (x) A (x))
1 2 1 2 2 1
yordamida belgilanadi.
1 2
( A , A )
noravshan kiritish ifodani inobatga olgan holda
(A , A ) (A , A ) (A , A )
hosil qilamiz.
1 2 1 2 2 1
Misol.Yuqoridagi misolda Lukasevich mantiqidan foydalangan holda quyidagi
kiritish darajasi va qiymatlari
v( A1 , A2 ) = 1 va
v(A2 , A1 ) = 0.2
aniqlangan edi.
Bulardan foydalanib endi tenglik darajasi qiymatini aniqlaymiz
(A , A ) v(A , A ) v( A , A ) 1 0,2 0,2.
1 2 1 2 2 1
Odatda quyidagilar faraz qilinadi:
Agar
v( A1,
A2 ) 0.5 , u holda
A1 noravshan sifatida
A2 kiritiladi, ya’ni
A1 A2 , aks holda
A1 noravshan sifatida
A2 kiritilmaydi (ya’ni
A1
A2 ). Shuni
ta’kidlash kerakki, va qiymatlar [0,1] oraliqda o`zgaradi.
Agar
v( A1,
A2 ) 0.5, u holda
A1 va A2
noravshan sifatida teng, ya’ni
1 2
A1 A2 , aks holda noravshan sifatida (agar v(A1 , A2 ) 0.5 ) teng emas, ya’ni
A1
A2 qabul qilinadi.
( A , A ) 0.5 - holatda
A1 va
A2 o`zaro indifferentli,
ya’ni
A1
A2 - ham teng, ham teng emas bo’ladi.
Keltirilgan misolda:
A1 A2 ,
A2 A1 ,
1 2
A1
A2 .
1 2 2 1
Topshiriq. ( A , A ) , ( A , A )
formulalar bo`yicha hisoblang?
va (A , A )
qiymatlarni Zade va Mamdani
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