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Таблица истинности ONLINE
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Таблица истинности для функции (A∨¬B)⊕(A⊕B⊕C∨¬A⊕¬B⊕¬C)⊕(¬A∨C)⊕(¬C∨B):
Промежуточные таблицы истинности:¬B: A∨(¬B): A | B | ¬B | A∨(¬B) | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
¬A: ¬C: C∨(¬A): C | A | ¬A | C∨(¬A) | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
A⊕B: (A⊕B)⊕(C∨(¬A)): A | B | C | A⊕B | ¬A | C∨(¬A) | (A⊕B)⊕(C∨(¬A)) | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
((A⊕B)⊕(C∨(¬A)))⊕(¬B): A | B | C | A⊕B | ¬A | C∨(¬A) | (A⊕B)⊕(C∨(¬A)) | ¬B | ((A⊕B)⊕(C∨(¬A)))⊕(¬B) | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
(((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C): A | B | C | A⊕B | ¬A | C∨(¬A) | (A⊕B)⊕(C∨(¬A)) | ¬B | ((A⊕B)⊕(C∨(¬A)))⊕(¬B) | ¬C | (((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C) | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
(¬A)∨C: A | C | ¬A | (¬A)∨C | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 |
(¬C)∨B: C | B | ¬C | (¬C)∨B | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 |
(A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)): A | B | C | ¬B | A∨(¬B) | A⊕B | ¬A | C∨(¬A) | (A⊕B)⊕(C∨(¬A)) | ¬B | ((A⊕B)⊕(C∨(¬A)))⊕(¬B) | ¬C | (((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C) | (A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)) | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
((A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)))⊕((¬A)∨C): A | B | C | ¬B | A∨(¬B) | A⊕B | ¬A | C∨(¬A) | (A⊕B)⊕(C∨(¬A)) | ¬B | ((A⊕B)⊕(C∨(¬A)))⊕(¬B) | ¬C | (((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C) | (A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)) | ¬A | (¬A)∨C | ((A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)))⊕((¬A)∨C) | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |
(((A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)))⊕((¬A)∨C))⊕((¬C)∨B): A | B | C | ¬B | A∨(¬B) | A⊕B | ¬A | C∨(¬A) | (A⊕B)⊕(C∨(¬A)) | ¬B | ((A⊕B)⊕(C∨(¬A)))⊕(¬B) | ¬C | (((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C) | (A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)) | ¬A | (¬A)∨C | ((A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)))⊕((¬A)∨C) | ¬C | (¬C)∨B | (((A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)))⊕((¬A)∨C))⊕((¬C)∨B) | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
Общая таблица истинности:A | B | C | ¬B | A∨(¬B) | ¬A | ¬C | C∨(¬A) | A⊕B | (A⊕B)⊕(C∨(¬A)) | ((A⊕B)⊕(C∨(¬A)))⊕(¬B) | (((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C) | (¬A)∨C | (¬C)∨B | (A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)) | ((A∨(¬B))⊕((((A⊕B)⊕(C∨(¬A)))⊕(¬B))⊕(¬C)))⊕((¬A)∨C) | (A∨¬B)⊕(A⊕B⊕C∨¬A⊕¬B⊕¬C)⊕(¬A∨C)⊕(¬C∨B) | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: A | B | C | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
F сднф = ¬A∧B∧¬C ∨ A∧¬B∧¬C ∨ A∧¬B∧C ∨ A∧B∧¬C Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: A | B | C | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
F скнф = (A∨B∨C) ∧ (A∨B∨¬C) ∧ (A∨¬B∨¬C) ∧ (¬A∨¬B∨¬C) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции A | B | C | Fж | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина: F ж = C 000 ⊕ C 100∧A ⊕ C 010∧B ⊕ C 001∧C ⊕ C 110∧A∧B ⊕ C 101∧A∧C ⊕ C 011∧B∧C ⊕ C 111∧A∧B∧C Так как F ж(000) = 0, то С 000 = 0. Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы: F ж(100) = С 000 ⊕ С 100 = 1 => С 100 = 0 ⊕ 1 = 1 F ж(010) = С 000 ⊕ С 010 = 1 => С 010 = 0 ⊕ 1 = 1 F ж(001) = С 000 ⊕ С 001 = 0 => С 001 = 0 ⊕ 0 = 0 F ж(110) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 110 = 1 => С 110 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1 F ж(101) = С 000 ⊕ С 100 ⊕ С 001 ⊕ С 101 = 1 => С 101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(011) = С 000 ⊕ С 010 ⊕ С 001 ⊕ С 011 = 0 => С 011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1 F ж(111) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 001 ⊕ С 110 ⊕ С 101 ⊕ С 011 ⊕ С 111 = 0 => С 111 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0 Таким образом, полином Жегалкина будет равен: F ж = A ⊕ B ⊕ A∧B ⊕ B∧C Логическая схема, соответствующая полиному Жегалкина:
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