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