Mantiqiy element «VA» - konyuksiya, mantiqiy ko‘paytirish amali, AND

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Mantiqiy element «VA» - konyuksiya, mantiqiy ko‘paytirish amali, AND
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bet | 2/2 | Sana | 25.07.2024 | Hajmi | 134,31 Kb. | | #268595 |
Bog'liq 4-Topshiriq fsggegege (1) Mantiqiy element «VA» - konyuksiya, mantiqiy ko‘paytirish amali, AND
Ta’rif: A va B mulohozaning ikkalasi rost bo‘lganda rost qolgan hollarda yolg‘on (1-jadval). (A & B, A and B, A ∩ B)
A
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B
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A ∩ B
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0
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0
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0
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0
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1
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0
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1
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0
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0
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1
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1
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1
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Mantiqiy element «YOKI» - dizyuksiya, mantiqiy qo‘shish, OR
Ta’rif: A va B mulohozaning kamida bittasi rost bo‘lganda rost qolgan hollarda yolg‘on (2-jadval). (A | B, A yoki B, A or B, A 𝖴 B)
A
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B
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A 𝖴 B
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0
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0
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0
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0
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1
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1
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1
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0
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1
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1
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1
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1
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Mantiqiy element «INKOR» - inkor, invertor, NOT
Ta’rif: A mulohoza rost bo‘lganda yolg‘on, yolg‘on bo‘lganda rost (3- jadval). (not A, emas A, 𝑨̅)
Mantiqiy element «VA-INKOR» - inkor bilan bog‘lanish (mantiqiy
ko‘paytirish), NAND
ya’ni,
Ta’rif: A va B mulohozaning ikkalasi rost bo‘lganda rost qolgan hollarda yolg‘on bo‘lgan mantiqiy qo‘shish amalining inkori (2-jadval).
A
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B
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A & B
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̅𝑨̅̅&̅̅̅̅𝑩̅
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0
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0
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1
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0
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0
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1
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1
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0
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1
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0
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1
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0
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1
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1
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0
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1
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Mantiqiy element «YOKI-INKOR» - inkor bilan dizyunksiya (mantiqiy
qo‘shish), NOR
ya’ni
Ta’rif: A va B mulohozaning kamida bittasi rost bo‘lganda rost qolgan hollarda yolg‘on bo‘lgan mantiqiy amalining inkori (2-jadval).
A
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B
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A 𝖴 B
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̅𝑨̅̅̅𝖴̅̅̅̅𝐁̅
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0
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0
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0
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1
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0
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1
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1
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0
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1
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0
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1
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0
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1
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1
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1
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0
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Mantiqiy amallarni quyidagi ketma-ketlikda yechamiz:
Invertor (YO‘Q) mantiqiy inkor amali yechiladi.
Konyuksiya (VA) mantiqiy ko‘paytirish amali bajariladi.
Dizyuksiya (YOKI) mantiqiy qo‘shish amali bajariladi.
Implekatsiya (VA-YO‘Q) mantiqiy amali.
Ekvivalinsiya (YOKI-YO‘Q) mantiqiy amali bajariladi.
(agar qavslar kelsa shu holatda qavslarni ichi birinchi bajariladi)
Berilgan misol:
𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥̅1𝑥̅2𝑥3)(𝑥̅1𝑥2𝑥̅3)
bunda
Yechish:
𝑥1 = 1, 𝑥2 = 0, 𝑥3 = 1
𝑦1 = (1 ∩ 0 ∩ 0) 𝖴 (0 ∩ 1 ∩ 1) 𝖴 (0 ∩ 0 ∩ 0) = (1 ∩ 0) 𝖴 (0 ∩ 1) 𝖴 (0 ∩ 0) = 0 𝖴 0 𝖴 0 = 0
Javob:
𝒀𝟏 = 𝟎
1-misol yechishda sxema asosidagi mantiqiy elementlar
AMALIY ISHNI BAJARISH BO‘YICHA TOPSHIRIQ
№
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Variantlar
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Berilgan qiymat
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1
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𝑦̅̅̅̅=̅̅̅(̅̅𝑥̅̅̅̅̅𝑥̅̅̅̅̅̅𝑥̅̅̅ ̅)̅̅̅̅(̅𝑥̅̅̅̅̅𝑥̅̅̅ ̅̅̅𝑥̅̅̅)̅̅̅̅(̅𝑥̅̅̅ ̅̅̅𝑥̅̅̅̅̅̅𝑥̅̅ ̅̅)
1 1 2 3 1 2 3 1 2 3
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x1=0,x2=1, x3=1
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2
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𝑦̅̅̅̅=̅̅̅(̅̅𝑥̅̅̅̅̅̅𝑥̅̅̅̅̅𝑥̅̅̅ ̅)̅̅̅̅(̅𝑥̅̅̅̅̅̅𝑥̅̅̅ ̅̅̅𝑥̅̅̅)̅̅̅̅(̅𝑥̅̅̅ ̅̅̅𝑥̅̅̅̅̅̅𝑥̅̅ ̅̅)
1 1 2 3 1 2 3 1 2 3
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x1=1,x2=0, x3=1
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3
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𝑦1 = (𝑥1̅ 𝑥̅2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=1
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4
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥̅3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=1,x2=0, x3=0
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5
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥̅2𝑥̅3)
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x1=0,x2=1, x3=0
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6
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𝑦1 = (𝑥1 ∩ 𝑥2 𝖴 𝑥3)(𝑥̅1𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=1,x2=1, x3=0
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7
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𝑦1 = (𝑥1 𝖴 𝑥2 𝖴 𝑥̅3)(𝑥̅1𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=1
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8
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=1,x2=0, x3=1
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9
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥̅1 𝖴 𝑥̅2 𝖴 𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=1
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10
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥̅1𝑥̅2𝑥3)(𝑥1̅ 𝖴 𝑥2 𝖴 𝑥̅3)
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x1=1,x2=1, x3=0
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11
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𝑦1 = (𝑥1 𝖴 𝑥2 𝖴 𝑥̅3) 𝖴 (𝑥̅1 𝖴 𝑥̅2 𝖴 𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=1
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12
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𝑦1 = (𝑥1 𝖴 𝑥2 𝖴 𝑥̅3)(𝑥̅1 𝖴 𝑥̅2 𝖴 𝑥3)(𝑥1̅ 𝖴 𝑥2 𝖴 𝑥̅3)
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x1=1,x2=1, x3=0
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13
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𝑦1 = (𝑥1̅ 𝑥̅2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=1
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14
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=1,x2=1, x3=0
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15
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=0, x3=1
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16
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥̅1𝑥̅2𝑥3)(𝑥2𝑥3)
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x1=0,x2=0, x3=0
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17
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𝑦1 = (𝑥1𝑥̅2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥̅2𝑥̅3)
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x1=1,x2=1, x3=1
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18
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=0
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19
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=1
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20
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𝑦1 = (𝑥1𝑥̅2 𝑥3)(𝑥̅2𝑥3)(𝑥̅1𝑥2𝑥̅3)
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x1=0,x2=0, x3=0
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21
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𝑦1 = (𝑥1𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥3)
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x1=1,x2=1, x3=1
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22
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=0,x2=1, x3=0
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23
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=1,x2=0, x3=1
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24
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥3)(𝑥̅1𝑥2𝑥̅3)
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x1=0,x2=1, x3=0
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25
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥2𝑥3)(𝑥1̅ 𝑥2𝑥̅3)
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x1=1,x2=1, x3=1
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26
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥̅2𝑥3)(𝑥̅1𝑥2𝑥̅3)
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x1=0,x2=0, x3=0
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27
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𝑦1 = (𝑥1𝑥2𝑥3)⬚(𝑥1̅ 𝑥2𝑥3)(𝑥1̅ 𝑥̅3)
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x1=0,x2=1, x3=1
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28
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥3)(𝑥̅1𝑥2𝑥̅3)
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x1=1,x2=1, x3=0
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29
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𝑦1 = (𝑥2𝑥̅3)(𝑥̅1𝑥̅2𝑥3)(𝑥1̅ 𝑥̅2𝑥̅3)
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x1=0,x2=1, x3=1
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30
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𝑦1 = (𝑥1𝑥2𝑥̅3)(𝑥1̅ 𝑥̅2𝑥3)(𝑥2𝑥̅3)
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x1=0,x2=1, x3=0
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