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Таблица истинности ONLINE
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Таблица истинности для функции F≡¬A∧(C∨B):
Промежуточные таблицы истинности:C∨B: ¬A: (¬A)∧(C∨B): | A | C | B | ¬A | C∨B | (¬A)∧(C∨B) | | 0 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 1 | 1 | 1 | | 0 | 1 | 0 | 1 | 1 | 1 | | 0 | 1 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 0 | 1 | 0 | | 1 | 1 | 0 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 |
F≡((¬A)∧(C∨B)): | F | A | C | B | ¬A | C∨B | (¬A)∧(C∨B) | F≡((¬A)∧(C∨B)) | | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
Общая таблица истинности:| F | A | C | B | C∨B | ¬A | (¬A)∧(C∨B) | F≡¬A∧(C∨B) | | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: | F | A | C | B | F | | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 |
F сднф = ¬F∧¬A∧¬C∧¬B ∨ ¬F∧A∧¬C∧¬B ∨ ¬F∧A∧¬C∧B ∨ ¬F∧A∧C∧¬B ∨ ¬F∧A∧C∧B ∨ F∧¬A∧¬C∧B ∨ F∧¬A∧C∧¬B ∨ F∧¬A∧C∧B Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: | F | A | C | B | F | | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 |
F скнф = (F∨A∨C∨¬B) ∧ (F∨A∨¬C∨B) ∧ (F∨A∨¬C∨¬B) ∧ (¬F∨A∨C∨B) ∧ (¬F∨¬A∨C∨B) ∧ (¬F∨¬A∨C∨¬B) ∧ (¬F∨¬A∨¬C∨B) ∧ (¬F∨¬A∨¬C∨¬B) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции | F | A | C | B | Fж | | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина: F ж = C 0000 ⊕ C 1000∧F ⊕ C 0100∧A ⊕ C 0010∧C ⊕ C 0001∧B ⊕ C 1100∧F∧A ⊕ C 1010∧F∧C ⊕ C 1001∧F∧B ⊕ C 0110∧A∧C ⊕ C 0101∧A∧B ⊕ C 0011∧C∧B ⊕ C 1110∧F∧A∧C ⊕ C 1101∧F∧A∧B ⊕ C 1011∧F∧C∧B ⊕ C 0111∧A∧C∧B ⊕ C 1111∧F∧A∧C∧B Так как F ж(0000) = 1, то С 0000 = 1. 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 = 1 ⊕ 0 = 1 F ж(0100) = С 0000 ⊕ С 0100 = 1 => С 0100 = 1 ⊕ 1 = 0 F ж(0010) = С 0000 ⊕ С 0010 = 0 => С 0010 = 1 ⊕ 0 = 1 F ж(0001) = С 0000 ⊕ С 0001 = 0 => С 0001 = 1 ⊕ 0 = 1 F ж(1100) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 1100 = 0 => С 1100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0 F ж(1010) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 1010 = 1 => С 1010 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0 F ж(1001) = С 0000 ⊕ С 1000 ⊕ С 0001 ⊕ С 1001 = 1 => С 1001 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0 F ж(0110) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0110 = 1 => С 0110 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1 F ж(0101) = С 0000 ⊕ С 0100 ⊕ С 0001 ⊕ С 0101 = 1 => С 0101 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1 F ж(0011) = С 0000 ⊕ С 0010 ⊕ С 0001 ⊕ С 0011 = 0 => С 0011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1 F ж(1110) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0010 ⊕ С 1100 ⊕ С 1010 ⊕ С 0110 ⊕ С 1110 = 0 => С 1110 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 F ж(1101) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0001 ⊕ С 1100 ⊕ С 1001 ⊕ С 0101 ⊕ С 1101 = 0 => С 1101 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 F ж(1011) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 0001 ⊕ С 1010 ⊕ С 1001 ⊕ С 0011 ⊕ С 1011 = 1 => С 1011 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0 F ж(0111) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0001 ⊕ С 0110 ⊕ С 0101 ⊕ С 0011 ⊕ С 0111 = 1 => С 0111 = 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 = 1 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 = 0 => С 1111 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 Таким образом, полином Жегалкина будет равен: F ж = 1 ⊕ F ⊕ C ⊕ B ⊕ A∧C ⊕ A∧B ⊕ C∧B ⊕ A∧C∧B Логическая схема, соответствующая полиному Жегалкина:
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