A \ B
ni toping?
1
1
1
1
а) A \ B (x b) A \ B (x с) A \ B (x e) A \ B (x
| 0.35),(x2
| 0.28),(x2
| 0.35),(x2
| 0.35),(x2
| 0.45),(x3
| 0.10),(x3
| 0.45),(x3
| 0.55),(x3
|1),(x4
|1),(x4
| 0),(x4
| 0),(x4
| 0);
| 0);
|1);
| 0).
32. Bizga ܷ = {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)}.
berilgan bo’lsin.
1
1
1
1
а) B \ A (x b) B \ A (x с) B \ A (x e) B \ A (x
| 0.35),(x2
| 0.10),(x2
| 0.60),(x2
| 0.35),(x2
| 0.27),(x3
| 0.27),(x3
| 0.14),(x3
| 0.27),(x3
| 0),(x4
|1),(x4
| 0),(x4
| 0),(x4
|1);
|1);
|1);
| 0).
Bizga
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
A=
|
1
|
0
|
1
|
1
|
1
|
0
|
1
|
B=
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
berilgan bo’lsin. A va B noravshan to’plamostilar orasidagi Xemming masofasini toping?
а) ݀ (ܣ, ܤ ) = 6; b) ݀ (ܣ, ܤ ) = 7; с) ݀ (ܣ, ܤ ) = 5; e) ݀ (ܣ, ܤ ) = 4.
Bizga
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ݑ଼
|
A=
|
1
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
B=
|
1
|
1
|
0
|
0
|
1
|
1
|
0
|
0
|
berilgan bo’lsin. n=8 quvvatga ega bo’lgan chekli U to’plamdagi ikkita A va B
noravshan to’plamostilar orasidagi Xemming masofasini toping?
а) ߜ(ܣ, ܤ) = ; b) ߜ(ܣ, ܤ)
଼
= ସ ;
଼
с) ߜ(ܣ, ܤ) = ଷ ;
଼
e) ߜ(ܣ, ܤ)
= ଼. ହ
Bizga A noravshan to’plamosti
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
A=
|
0.5
|
0.2
|
0
|
0.1
|
0.3
|
0.7
|
0.9
|
berilgam bo’lsin. A ga α-pog’onali to’plamostini qo’llab ܣ .ଶ va ܣ . ravshan to’plamostilarini toping?
а)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
b)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
с)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
e)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
Bizga A noravshan to’plamosti
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
A=
|
0.5
|
0.2
|
0
|
0.1
|
0.3
|
0.7
|
0.9
|
berilgam bo’lsin. A ga α-pog’onali to’plamostini qo’llab ܣ .ଶ va ܣ . ravshan to’plamostilarini toping?
а)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
b)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
с)
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
e)
X Y to’plamda da beriladigan noravshan munosabat tavsiflanadi.
|
ݑଵ
|
ݑଶ
|
ݑଷ
|
ݑସ
|
ݑହ
|
ݑ
|
ݑ
|
ܣ.ଷ =
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
ܣ.
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
а) x X , y Y : xRy;
с) x X , y Y : xRy;
b) x X , y Y : xRy;
e) x X , y Y : xRy.
X Y
to’plamda da beriladigan noravshan munosabat tavsiflanadi.
а) x X , y Y : xRy;
с) x X , y Y : xRy;
b) x X , y Y : xRy;
e) x X , y Y : xRy.
R munosabatning birinchi proyeksiyasini aniqlovchi ifodani to’g’ri ko’rsating?
а) (1) (x) (x, y); b) (1) (x)
(x, y);
R y R R y R
с) (1) (x) (x, y); e) (1) (x)
(x, y);
R y x R R y x R
R munosabatning ikinchi proyeksiyasini aniqlovchi ifodani to’g’ri ko’rsating?
а) (1) (x) (x, y); b) (1) (x)
(x, y);
R y R R y R
с) (1) (x) (x, y); e) (1) (x)
(x, y);
R y x R R y x R
R munosabatning global proyeksiyasini aniqlovchi ifodani to’g’ri ko’rsating?
а) H (R) (x, y); b) H (R)
(x, y);
y x R y x R
с) H (R) (x, y); e) H (R)
(x, y);
y x R y R
X = {x1 , x2 , x3 , x4 }; Y = {y1 , y2 , y3 , y4 } to’plamlar va XRY munosabat matritsasi jadval ko’rinishda berilgan.
|
y
1
|
y
2
|
y
3
|
y
4
|
1-proyeksiya
|
x
1
|
0.3
|
0.2
|
0
|
0.5
|
?
|
x
2
|
0.7
|
0.8
|
1
|
0.1
|
?
|
x
3
|
1
|
0.5
|
0.3
|
0.6
|
?
|
x
4
|
0.8
|
0.1
|
0.9
|
0
|
?
|
|
2-proyeksiya
|
Global
proyeksiya
|
|
?
|
?
|
?
|
?
|
?
|
1va 2-proyeksiyalar hamda global proyeksiyani toping?
а)
|
y
1
|
y
2
|
y
3
|
y
4
|
1-proyeksiya
|
x
1
|
0.3
|
0.2
|
0
|
0.5
|
0.5
|
x
2
|
0.7
|
0.8
|
1
|
0.1
|
1
|
x
3
|
1
|
0.5
|
0.3
|
0.6
|
1
|
x
4
|
0.8
|
0.1
|
0.9
|
0
|
0.9
|
|
2-proyeksiya
|
Global proyeksiya
|
|
1
|
0.8
|
1
|
0.6
|
1
|
b)
|
y
1
|
y
2
|
y
3
|
y
4
|
1-proyeksiya
|
x
1
|
0.3
|
0.2
|
0
|
0.5
|
0
|
x
2
|
0.7
|
0.8
|
1
|
0.1
|
0.1
|
x
3
|
1
|
0.5
|
0.3
|
0.6
|
0.3
|
x
4
|
0.8
|
0.1
|
0.9
|
0
|
0.9
|
|
2-proyeksiya
|
Global proyeksiya
|
|
1
|
0.8
|
1
|
0.6
|
1
|
с)
|
y
1
|
y
2
|
y
3
|
y
4
|
1-proyeksiya
|
x
1
|
0.3
|
0.2
|
0
|
0.5
|
0.5
|
x
2
|
0.7
|
0.8
|
1
|
0.1
|
1
|
x
3
|
1
|
0.5
|
0.3
|
0.6
|
1
|
x
4
|
0.8
|
0.1
|
0.9
|
0
|
0.9
|
|
2-proyeksiya
|
Global proyeksiya
|
|
0.1
|
0.1
|
0
|
0
|
1
|
e)
|
y
1
|
y
2
|
y
3
|
y
4
|
1-proyeksiya
|
x
1
|
0.3
|
0.2
|
0
|
0.5
|
0.5
|
x
2
|
0.7
|
0.8
|
1
|
0.1
|
1
|
x
3
|
1
|
0.5
|
0.3
|
0.6
|
1
|
x
4
|
0.8
|
0.1
|
0.9
|
0
|
0.9
|
|
2-proyeksiya
|
Global proyeksiya
|
|
1
|
0.8
|
1
|
0.6
|
0.5
|
R1 va R2 munosabatlarning “max-min” – kompozitsiyasini ifodalovchi formulani ko’rsating?
R oR R R
а) (x, z) [ (x, y) ( y, z)] max[min(
1 2 y 1 2
R oR R R
b) (x, z) [ (x, z) ( y, z)] max[min(
1 2 y 1 2
(x, y), ( y, z))].
R R
1 2
R R
(x, z), ( y, z))].
1 2
R oR R R R R
с) (x, z) [ (x, y) ( y, z)] min[max( (x, y), ( y, z))].
1 2 y 1 2 1 2
R oR R R
e) (x, z) [ (x, y) (x, z)] max[min(
1 2 y 1 2
(x, y), (x, z))].
R R
1 2
R1 va R2 munosabatlarning “max-min” – kompozitsiyasini ifodalovchi formulani ko’rsating?
y
а) R1oR2 ( x, z) [ R1 ( x, y) R2 ( y, z)] max[min( R1 ( x, y), R2 ( y, z))].
y
b) R1oR2 ( x, z) [ R1 ( x, z) R2 ( y, z)] max[min( R1 ( x, z), R2 ( y, z))].
y
с) R1oR2 ( x, z) [ R1 ( x, y) R2 ( y, z)] min[max( R1 ( x, y), R2 ( y, z))].
y
e) R1oR2 ( x, z) [ R1 ( x, y) R2 ( x, z)] max[min( R1 ( x, y), R2 ( x, z))].
Mamdani bo`yicha noravshan mantiqiy xulosa chiqarish ifodasi qaysi formula bilan aniqlanadi?
C ( z) C ( z) C ( z) [ 1 C ( z)] [ 2 C ( z)]
а) 1 2 1 2
max[min( 1, c ( z)),min( 2 , c ( z))];
1 2
1
C (z) C (z) C (z) [1 C (z)] [2 C
(z)]
1
b) max[min(
, c1
2
(z)),max(2
, c2
1 2
(z))];
1
C (z) C (z) C (z) [1 C (z)] [2 C
(z)]
1
с) max[max(
, c1
2
(z)),min(2
, c2
1 2
(z))];
1
C (z) C (z) C (z) [1 C (z)] [2 C
(z)]
1
e) min[max(
, c1
2
(z)),max(2
, c2
1 2
(z))].
Larsen bo`yicha noravshan mantiqiy xulosa chiqarish qiymati qaysi formula bilan aniqlanadi?
а)
z
z
z;
C 1 C1 2 C2
b)
z
z
z;
C 1 C1 2 C2
с) C z 1 C z 2 C z;
1
2
1
2
e) C z 1 C z 2 C z .
Zade bo`yicha noravshan mantiqiy xulosa chiqarish qiymati qaysi formula bilan aniqlanadi?
1
1
2
а) C ( z) 1 Ñ z 1 Ñ z;
b) C
с)
(z) 1
1
1
(z) 1
Ñ
z 1
1
2
z 1
Ñ
z;
z;
C
e)
(z) 1
Ñ1
z 1
2 Ñ
z.
C 1 Ñ1 2 Ñ
Lukasevich bo`yicha noravshan mantiqiy xulosa chiqarish qiymati qaysi formula bilan aniqlanadi?
а) z 1 1 z 1 1 z;
Ñ 1 Ñ2 2 Ñ2
b) z 1 1 z 1 1
z;
Ñ 1 Ñ2 2 Ñ2
с) z 1 1 z 1 1
z;
Ñ 1 Ñ2 2 Ñ2
2
2
e) Ñ z 1 11 Ñ z 1 1 2 Ñ
z.
Lingvistik o’zgaruvchining strukturasini tavsivlash uchun quyidagi qanday qoidalardan foydalaniladi?
а) Sintaktikli va semantikli qoidalardan;
b) Grafikli va semantikli qoidalardan;
с) Chiziqli va sintaktikli qoidalardan;
e) Statistik li va ehtimolli qoidalardan.
Lingvistik o’zgaruvchi tushunchasining strukturasito’g’ri keltirilgan javobni ko’rsating?
а) ⟨A, T (A), U, V, M⟩; b) ⟨A, T (M), U, V, M⟩;
с) ⟨A, T (V), U, V, M⟩; e) ⟨A, T (U), U, V, M⟩.
Lingvistik o’zgaruvchi tushunchasining strukturasi to’g’ri keltirilgan javobni ko’rsating?
а) ⟨A, T (A), U, V, M⟩; b) ⟨A, T (M), U, V, M⟩;
с) ⟨A, T (V), U, V, M⟩; e) ⟨A, T (U), U, V, M⟩.
Do'stlaringiz bilan baham: |