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Таблица истинности для функции Z∧(X≡Y)∨Z∧¬((X↓¬Y)↓(¬X↓Y))⊕X:
Промежуточные таблицы истинности:X≡Y: ¬Y: X↓(¬Y): X | Y | ¬Y | X↓(¬Y) | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
¬X: (¬X)↓Y: X | Y | ¬X | (¬X)↓Y | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |
(X↓(¬Y))↓((¬X)↓Y): X | Y | ¬Y | X↓(¬Y) | ¬X | (¬X)↓Y | (X↓(¬Y))↓((¬X)↓Y) | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
¬((X↓(¬Y))↓((¬X)↓Y)): X | Y | ¬Y | X↓(¬Y) | ¬X | (¬X)↓Y | (X↓(¬Y))↓((¬X)↓Y) | ¬((X↓(¬Y))↓((¬X)↓Y)) | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
Z∧(X≡Y): Z | X | Y | X≡Y | Z∧(X≡Y) | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
Z∧(¬((X↓(¬Y))↓((¬X)↓Y))): Z | X | Y | ¬Y | X↓(¬Y) | ¬X | (¬X)↓Y | (X↓(¬Y))↓((¬X)↓Y) | ¬((X↓(¬Y))↓((¬X)↓Y)) | Z∧(¬((X↓(¬Y))↓((¬X)↓Y))) | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
(Z∧(X≡Y))∨(Z∧(¬((X↓(¬Y))↓((¬X)↓Y)))): Z | X | Y | X≡Y | Z∧(X≡Y) | ¬Y | X↓(¬Y) | ¬X | (¬X)↓Y | (X↓(¬Y))↓((¬X)↓Y) | ¬((X↓(¬Y))↓((¬X)↓Y)) | Z∧(¬((X↓(¬Y))↓((¬X)↓Y))) | (Z∧(X≡Y))∨(Z∧(¬((X↓(¬Y))↓((¬X)↓Y)))) | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
((Z∧(X≡Y))∨(Z∧(¬((X↓(¬Y))↓((¬X)↓Y)))))⊕X: Z | X | Y | X≡Y | Z∧(X≡Y) | ¬Y | X↓(¬Y) | ¬X | (¬X)↓Y | (X↓(¬Y))↓((¬X)↓Y) | ¬((X↓(¬Y))↓((¬X)↓Y)) | Z∧(¬((X↓(¬Y))↓((¬X)↓Y))) | (Z∧(X≡Y))∨(Z∧(¬((X↓(¬Y))↓((¬X)↓Y)))) | ((Z∧(X≡Y))∨(Z∧(¬((X↓(¬Y))↓((¬X)↓Y)))))⊕X | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
Общая таблица истинности:Z | X | Y | X≡Y | ¬Y | X↓(¬Y) | ¬X | (¬X)↓Y | (X↓(¬Y))↓((¬X)↓Y) | ¬((X↓(¬Y))↓((¬X)↓Y)) | Z∧(X≡Y) | Z∧(¬((X↓(¬Y))↓((¬X)↓Y))) | (Z∧(X≡Y))∨(Z∧(¬((X↓(¬Y))↓((¬X)↓Y)))) | Z∧(X≡Y)∨Z∧¬((X↓¬Y)↓(¬X↓Y))⊕X | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: Z | X | Y | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
F сднф = ¬Z∧X∧¬Y ∨ ¬Z∧X∧Y ∨ Z∧¬X∧¬Y ∨ Z∧¬X∧Y Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: Z | X | Y | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
F скнф = (Z∨X∨Y) ∧ (Z∨X∨¬Y) ∧ (¬Z∨¬X∨Y) ∧ (¬Z∨¬X∨¬Y) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции Z | X | Y | Fж | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина: F ж = C 000 ⊕ C 100∧Z ⊕ C 010∧X ⊕ C 001∧Y ⊕ C 110∧Z∧X ⊕ C 101∧Z∧Y ⊕ C 011∧X∧Y ⊕ C 111∧Z∧X∧Y Так как F ж(000) = 0, то С 000 = 0. Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы: F ж(100) = С 000 ⊕ С 100 = 1 => С 100 = 0 ⊕ 1 = 1 F ж(010) = С 000 ⊕ С 010 = 1 => С 010 = 0 ⊕ 1 = 1 F ж(001) = С 000 ⊕ С 001 = 0 => С 001 = 0 ⊕ 0 = 0 F ж(110) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 110 = 0 => С 110 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0 F ж(101) = С 000 ⊕ С 100 ⊕ С 001 ⊕ С 101 = 1 => С 101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(011) = С 000 ⊕ С 010 ⊕ С 001 ⊕ С 011 = 1 => С 011 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(111) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 001 ⊕ С 110 ⊕ С 101 ⊕ С 011 ⊕ С 111 = 0 => С 111 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 Таким образом, полином Жегалкина будет равен: F ж = Z ⊕ X Логическая схема, соответствующая полиному Жегалкина:
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