# Soddalashtirish

 Sana 03.12.2020 Hajmi 103.69 Kb.

1. Soddalashtirish:

⌐((x v ⌐y)→((x ⌐y) ↓y)) ↓z

⌐((x v ⌐y)→( ⌐xyY)) ↓z

⌐((x v ⌐y)→ Y) ↓z

⌐((⌐x v ⌐y) ↓z

(xyz)

X v Y v Z

 x y z ⌐y xv(⌐y) x↓(⌐y) (x↓(⌐y))↓y (xv(⌐y))→((x↓(⌐y))↓y) ⌐((xv(⌐y))→((x↓(⌐y))↓y)) (⌐((xv(⌐y))→((x↓(⌐y))↓y)))↓z 0 0 0 1 1 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 1 1 1 0 1 0 0 0 1 0

1. Mukammal Dizyunktiv Normal Shakl ni toppish uchun rostlik jadvalidagi qiymati “1” ga teng qatorlarni yozib olamiz.

 X

 Y

 Z

 Y

 X ∨ Y

 Y

 X ↓ Y

 ( X ↓ Y ) ↓ Y

 ( X ∨ Y ) → ( ( X ↓ Y ) ↓ Y )

 ( ( X ∨ Y ) → ( ( X ↓ Y ) ↓ Y ) ) ↓ Z

 ( ( X ∨ Y ) → ( ( X ↓ Y ) ↓ Y ) ) ↓ Z

0

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0

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Bundan tashqari, har bir satr uchun biz barcha algoritmlarning bog'lanishini quyidagi algoritmga muvofiq yozamiz: agar bu satrdagi o'zgaruvchining qiymati 1 ga teng bo'lsa, u holda o'zgaruvchining o'zi qo'shma qismga yozamiz, agar u 0 bo'lsa, bu o'zgaruvchining inkori.

MDNSH :

1. MKNSH ni topish uchun rostlik jadvalidan faqat qiymati 0 bo'lgan qatorlarni tanlash kerak, bu funktsiya uchun qatorlar to'plami quyidagicha bo'ladi:

 X

 Y

 Z

 Y

 X ∨ Y

 Y

 X ↓ Y

 ( X ↓ Y ) ↓ Y

 ( X ∨ Y ) → ( ( X ↓ Y ) ↓ Y )

 ( ( X ∨ Y ) → ( ( X ↓ Y ) ↓ Y ) ) ↓ Z

 ( ( X ∨ Y ) → ( ( X ↓ Y ) ↓ Y ) ) ↓ Z

1

1

0

0

1

0

0

0

0

1

0

X v Y v Z.

1. ⌐((x v ⌐y)→((x ↓ ⌐y) ↓y)) ↓z
ushbu mantiqiy ifodamizni ⌐X v ⌐Y v Z ko’rinishli soda shaklga keltirib olgan edik.

Bu ifoda uchun mantiqiy sxema chizamiz:

X Y

F(A,B,C)

C

1. Karno kartasini tuzish.

 X\YZ 00 01 11 10 0 1 1 1 1 1 1 1 1 0