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