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