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