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