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