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