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