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