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