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Таблица истинности ONLINE
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Таблица истинности для функции ((C∧¬A)∨(C∧A))∧¬((D∧¬B)↓(D∧B)):
Промежуточные таблицы истинности:¬A: C∧(¬A): C | A | ¬A | C∧(¬A) | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |
C∧A: (C∧(¬A))∨(C∧A): C | A | ¬A | C∧(¬A) | C∧A | (C∧(¬A))∨(C∧A) | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
¬B: D∧(¬B): D | B | ¬B | D∧(¬B) | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |
D∧B: (D∧(¬B))↓(D∧B): D | B | ¬B | D∧(¬B) | D∧B | (D∧(¬B))↓(D∧B) | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
¬((D∧(¬B))↓(D∧B)): D | B | ¬B | D∧(¬B) | D∧B | (D∧(¬B))↓(D∧B) | ¬((D∧(¬B))↓(D∧B)) | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 |
((C∧(¬A))∨(C∧A))∧(¬((D∧(¬B))↓(D∧B))): C | A | D | B | ¬A | C∧(¬A) | C∧A | (C∧(¬A))∨(C∧A) | ¬B | D∧(¬B) | D∧B | (D∧(¬B))↓(D∧B) | ¬((D∧(¬B))↓(D∧B)) | ((C∧(¬A))∨(C∧A))∧(¬((D∧(¬B))↓(D∧B))) | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
Общая таблица истинности:C | A | D | B | ¬A | C∧(¬A) | C∧A | (C∧(¬A))∨(C∧A) | ¬B | D∧(¬B) | D∧B | (D∧(¬B))↓(D∧B) | ¬((D∧(¬B))↓(D∧B)) | ((C∧¬A)∨(C∧A))∧¬((D∧¬B)↓(D∧B)) | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: C | A | D | B | F | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
F сднф = C∧¬A∧D∧¬B ∨ C∧¬A∧D∧B ∨ C∧A∧D∧¬B ∨ C∧A∧D∧B Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: C | A | D | B | F | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
F скнф = (C∨A∨D∨B) ∧ (C∨A∨D∨¬B) ∧ (C∨A∨¬D∨B) ∧ (C∨A∨¬D∨¬B) ∧ (C∨¬A∨D∨B) ∧ (C∨¬A∨D∨¬B) ∧ (C∨¬A∨¬D∨B) ∧ (C∨¬A∨¬D∨¬B) ∧ (¬C∨A∨D∨B) ∧ (¬C∨A∨D∨¬B) ∧ (¬C∨¬A∨D∨B) ∧ (¬C∨¬A∨D∨¬B) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции C | A | D | B | Fж | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
Построим полином Жегалкина: F ж = C 0000 ⊕ C 1000∧C ⊕ C 0100∧A ⊕ C 0010∧D ⊕ C 0001∧B ⊕ C 1100∧C∧A ⊕ C 1010∧C∧D ⊕ C 1001∧C∧B ⊕ C 0110∧A∧D ⊕ C 0101∧A∧B ⊕ C 0011∧D∧B ⊕ C 1110∧C∧A∧D ⊕ C 1101∧C∧A∧B ⊕ C 1011∧C∧D∧B ⊕ C 0111∧A∧D∧B ⊕ C 1111∧C∧A∧D∧B Так как F ж(0000) = 0, то С 0000 = 0. Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы: F ж(1000) = С 0000 ⊕ С 1000 = 0 => С 1000 = 0 ⊕ 0 = 0 F ж(0100) = С 0000 ⊕ С 0100 = 0 => С 0100 = 0 ⊕ 0 = 0 F ж(0010) = С 0000 ⊕ С 0010 = 0 => С 0010 = 0 ⊕ 0 = 0 F ж(0001) = С 0000 ⊕ С 0001 = 0 => С 0001 = 0 ⊕ 0 = 0 F ж(1100) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 1100 = 0 => С 1100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(1010) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 1010 = 1 => С 1010 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1 F ж(1001) = С 0000 ⊕ С 1000 ⊕ С 0001 ⊕ С 1001 = 0 => С 1001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(0110) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0110 = 0 => С 0110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(0101) = С 0000 ⊕ С 0100 ⊕ С 0001 ⊕ С 0101 = 0 => С 0101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(0011) = С 0000 ⊕ С 0010 ⊕ С 0001 ⊕ С 0011 = 0 => С 0011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(1110) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0010 ⊕ С 1100 ⊕ С 1010 ⊕ С 0110 ⊕ С 1110 = 1 => С 1110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(1101) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0001 ⊕ С 1100 ⊕ С 1001 ⊕ С 0101 ⊕ С 1101 = 0 => С 1101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(1011) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 0001 ⊕ С 1010 ⊕ С 1001 ⊕ С 0011 ⊕ С 1011 = 1 => С 1011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(0111) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0001 ⊕ С 0110 ⊕ С 0101 ⊕ С 0011 ⊕ С 0111 = 0 => С 0111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(1111) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0001 ⊕ С 1100 ⊕ С 1010 ⊕ С 1001 ⊕ С 0110 ⊕ С 0101 ⊕ С 0011 ⊕ С 1110 ⊕ С 1101 ⊕ С 1011 ⊕ С 0111 ⊕ С 1111 = 1 => С 1111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0 Таким образом, полином Жегалкина будет равен: F ж = C∧D Логическая схема, соответствующая полиному Жегалкина:
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