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