O’zgaruvchilar

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, ܷ = ^{ݔ_ଵ, ݔ_ଶ, ݔ_ଷ, ݔ_ସ^}, ܯ = ^[0,1^].
ܣ = ^{(ݔ_ଵ^|0.25⁾, ⁽ݔ_ଶ^|0.73⁾, ⁽ݔ_ଷ^|1⁾, ⁽ݔ_ସ^|0^)}.
ܤ = ^{(ݔ_ଵ^|0.35⁾, ⁽ݔ_ଶ^|0.87⁾, ⁽ݔ_ଷ^|0⁾, ⁽ݔ_ସ^|1^)}.

1

B

B
_B (x)  1 _B (x)
asosan

B  
(x)  1 

10. 
     (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

AB  { 


AB
(x)  / x},
x  X ,

bu erda
 _AB (x)   _A\B (x)  _B\A (x)
ko`rinishdagi to`plam.

Misol. Aytaylik, ܷ = ^{ݔ_ଵ, ݔ_ଶ, ݔ_ଷ, ݔ_ସ^}, ܯ = ^[0,1^].
ܣ = ^{(ݔ_ଵ^|0.25⁾, ⁽ݔ_ଶ^|0.73⁾, ⁽ݔ_ଷ^|1⁾, ⁽ݔ_ସ^|0^)}.

ܤ = ^{(ݔ_ଵ^|0.35⁾, ⁽ݔ_ଶ^|0.87⁾, ⁽ݔ_ଷ^|0⁾, ⁽ݔ_ସ^|1^)}.

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).

2. 
   _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

10. 
     B \ A  (x
    

| 0.35),(x₂
| 0.27),(x₃
| 0),(x₄
|1).

AB 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

10. 
     (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, AB,
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))
_xX 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

• 
  Zade mantiqi bo`yicha
  

 _A (x)   _A
(x)  (1   _A (x))  ( _A (x)   _A
(x)) _;

1 2

• 
  Mamdani mantiqi bo`yicha
  

1 1 2

_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  {x}₁ ^{, x}₂ ^{, x}₃ ^{, x}₄ ^{, x}₅ ^} 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:

1. 
   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 _xX 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:

1. 
   Agar
   

v( A₁,
A₂ )  0.5 ^{, u holda}
A₁ noravshan sifatida
A₂ ^{kiritiladi, ya’ni}


^A₁ ^{ A}₂ , aks holda
A₁ noravshan sifatida
A₂ ^{kiritilmaydi (ya’ni}
A₁ _
A₂ ^{). Shuni}

ta’kidlash kerakki, ^ va  qiymatlar [0,1] oraliqda o`zgaradi.

2. 
   Agar
   

v( A₁,
A₂ )  0.5, ^{u holda}
A₁ va A₂
noravshan sifatida teng, ya’ni

1 2
^A₁ ^{ A}₂ ^{, 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 ,
A_{2 } 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