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