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