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