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