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