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Таблица истинности для функции A1∧¬A2∧A3∨A1∧A2∧¬A3∨¬A1∧¬A2∧A3∨¬A1∧A2∧A3:
Промежуточные таблицы истинности:¬A2: ¬A3: ¬A1: A1∧(¬A2): A1 | A2 | ¬A2 | A1∧(¬A2) | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |
(A1∧(¬A2))∧A3: A1 | A2 | A3 | ¬A2 | A1∧(¬A2) | (A1∧(¬A2))∧A3 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
A1∧A2: (A1∧A2)∧(¬A3): A1 | A2 | A3 | A1∧A2 | ¬A3 | (A1∧A2)∧(¬A3) | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |
(¬A1)∧(¬A2): A1 | A2 | ¬A1 | ¬A2 | (¬A1)∧(¬A2) | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
((¬A1)∧(¬A2))∧A3: A1 | A2 | A3 | ¬A1 | ¬A2 | (¬A1)∧(¬A2) | ((¬A1)∧(¬A2))∧A3 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
(¬A1)∧A2: A1 | A2 | ¬A1 | (¬A1)∧A2 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
((¬A1)∧A2)∧A3: A1 | A2 | A3 | ¬A1 | (¬A1)∧A2 | ((¬A1)∧A2)∧A3 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)): A1 | A2 | A3 | ¬A2 | A1∧(¬A2) | (A1∧(¬A2))∧A3 | A1∧A2 | ¬A3 | (A1∧A2)∧(¬A3) | ((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)) | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
(((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)))∨(((¬A1)∧(¬A2))∧A3): A1 | A2 | A3 | ¬A2 | A1∧(¬A2) | (A1∧(¬A2))∧A3 | A1∧A2 | ¬A3 | (A1∧A2)∧(¬A3) | ((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)) | ¬A1 | ¬A2 | (¬A1)∧(¬A2) | ((¬A1)∧(¬A2))∧A3 | (((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)))∨(((¬A1)∧(¬A2))∧A3) | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
((((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)))∨(((¬A1)∧(¬A2))∧A3))∨(((¬A1)∧A2)∧A3): A1 | A2 | A3 | ¬A2 | A1∧(¬A2) | (A1∧(¬A2))∧A3 | A1∧A2 | ¬A3 | (A1∧A2)∧(¬A3) | ((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)) | ¬A1 | ¬A2 | (¬A1)∧(¬A2) | ((¬A1)∧(¬A2))∧A3 | (((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)))∨(((¬A1)∧(¬A2))∧A3) | ¬A1 | (¬A1)∧A2 | ((¬A1)∧A2)∧A3 | ((((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)))∨(((¬A1)∧(¬A2))∧A3))∨(((¬A1)∧A2)∧A3) | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Общая таблица истинности:A1 | A2 | A3 | ¬A2 | ¬A3 | ¬A1 | A1∧(¬A2) | (A1∧(¬A2))∧A3 | A1∧A2 | (A1∧A2)∧(¬A3) | (¬A1)∧(¬A2) | ((¬A1)∧(¬A2))∧A3 | (¬A1)∧A2 | ((¬A1)∧A2)∧A3 | ((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)) | (((A1∧(¬A2))∧A3)∨((A1∧A2)∧(¬A3)))∨(((¬A1)∧(¬A2))∧A3) | A1∧¬A2∧A3∨A1∧A2∧¬A3∨¬A1∧¬A2∧A3∨¬A1∧A2∧A3 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: A1 | A2 | A3 | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
F сднф = ¬A1∧¬A2∧A3 ∨ ¬A1∧A2∧A3 ∨ A1∧¬A2∧A3 ∨ A1∧A2∧¬A3 Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: A1 | A2 | A3 | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
F скнф = (A1∨A2∨A3) ∧ (A1∨¬A2∨A3) ∧ (¬A1∨A2∨A3) ∧ (¬A1∨¬A2∨¬A3) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции A1 | A2 | A3 | Fж | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина: F ж = C 000 ⊕ C 100∧A1 ⊕ C 010∧A2 ⊕ C 001∧A3 ⊕ C 110∧A1∧A2 ⊕ C 101∧A1∧A3 ⊕ C 011∧A2∧A3 ⊕ C 111∧A1∧A2∧A3 Так как F ж(000) = 0, то С 000 = 0. Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы: F ж(100) = С 000 ⊕ С 100 = 0 => С 100 = 0 ⊕ 0 = 0 F ж(010) = С 000 ⊕ С 010 = 0 => С 010 = 0 ⊕ 0 = 0 F ж(001) = С 000 ⊕ С 001 = 1 => С 001 = 0 ⊕ 1 = 1 F ж(110) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 110 = 1 => С 110 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1 F ж(101) = С 000 ⊕ С 100 ⊕ С 001 ⊕ С 101 = 1 => С 101 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0 F ж(011) = С 000 ⊕ С 010 ⊕ С 001 ⊕ С 011 = 1 => С 011 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0 F ж(111) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 001 ⊕ С 110 ⊕ С 101 ⊕ С 011 ⊕ С 111 = 0 => С 111 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0 Таким образом, полином Жегалкина будет равен: F ж = A3 ⊕ A1∧A2 Логическая схема, соответствующая полиному Жегалкина:
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