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Таблица истинности для функции (X∨Z)∧(X∨¬Y)∧(X∨Y∨¬Z)∧(X∨Y∨Z)∧(X∨Y)∧(Y∨¬Z):
Промежуточные таблицы истинности:X∨Z: ¬Y: X∨(¬Y): X | Y | ¬Y | X∨(¬Y) | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
¬Z: X∨Y: (X∨Y)∨(¬Z): X | Y | Z | X∨Y | ¬Z | (X∨Y)∨(¬Z) | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
(X∨Y)∨Z: X | Y | Z | X∨Y | (X∨Y)∨Z | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Y∨(¬Z): Y | Z | ¬Z | Y∨(¬Z) | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
(X∨Z)∧(X∨(¬Y)): X | Z | Y | X∨Z | ¬Y | X∨(¬Y) | (X∨Z)∧(X∨(¬Y)) | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)): X | Z | Y | X∨Z | ¬Y | X∨(¬Y) | (X∨Z)∧(X∨(¬Y)) | X∨Y | ¬Z | (X∨Y)∨(¬Z) | ((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)) | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
(((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z): X | Z | Y | X∨Z | ¬Y | X∨(¬Y) | (X∨Z)∧(X∨(¬Y)) | X∨Y | ¬Z | (X∨Y)∨(¬Z) | ((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)) | X∨Y | (X∨Y)∨Z | (((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z) | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
((((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z))∧(X∨Y): X | Z | Y | X∨Z | ¬Y | X∨(¬Y) | (X∨Z)∧(X∨(¬Y)) | X∨Y | ¬Z | (X∨Y)∨(¬Z) | ((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)) | X∨Y | (X∨Y)∨Z | (((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z) | X∨Y | ((((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z))∧(X∨Y) | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(((((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z))∧(X∨Y))∧(Y∨(¬Z)): X | Z | Y | X∨Z | ¬Y | X∨(¬Y) | (X∨Z)∧(X∨(¬Y)) | X∨Y | ¬Z | (X∨Y)∨(¬Z) | ((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)) | X∨Y | (X∨Y)∨Z | (((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z) | X∨Y | ((((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z))∧(X∨Y) | ¬Z | Y∨(¬Z) | (((((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z))∧(X∨Y))∧(Y∨(¬Z)) | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
Общая таблица истинности:X | Z | Y | X∨Z | ¬Y | X∨(¬Y) | ¬Z | X∨Y | (X∨Y)∨(¬Z) | (X∨Y)∨Z | Y∨(¬Z) | (X∨Z)∧(X∨(¬Y)) | ((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)) | (((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z) | ((((X∨Z)∧(X∨(¬Y)))∧((X∨Y)∨(¬Z)))∧((X∨Y)∨Z))∧(X∨Y) | (X∨Z)∧(X∨¬Y)∧(X∨Y∨¬Z)∧(X∨Y∨Z)∧(X∨Y)∧(Y∨¬Z) | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: X | Z | Y | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
F сднф = X∧¬Z∧¬Y ∨ X∧¬Z∧Y ∨ X∧Z∧Y Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: X | Z | Y | F | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
F скнф = (X∨Z∨Y) ∧ (X∨Z∨¬Y) ∧ (X∨¬Z∨Y) ∧ (X∨¬Z∨¬Y) ∧ (¬X∨¬Z∨Y) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции X | Z | Y | Fж | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
Построим полином Жегалкина: F ж = C 000 ⊕ C 100∧X ⊕ C 010∧Z ⊕ C 001∧Y ⊕ C 110∧X∧Z ⊕ C 101∧X∧Y ⊕ C 011∧Z∧Y ⊕ C 111∧X∧Z∧Y 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 = 1 => С 100 = 0 ⊕ 1 = 1 F ж(010) = С 000 ⊕ С 010 = 0 => С 010 = 0 ⊕ 0 = 0 F ж(001) = С 000 ⊕ С 001 = 0 => С 001 = 0 ⊕ 0 = 0 F ж(110) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 110 = 0 => С 110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1 F ж(101) = С 000 ⊕ С 100 ⊕ С 001 ⊕ С 101 = 1 => С 101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(011) = С 000 ⊕ С 010 ⊕ С 001 ⊕ С 011 = 0 => С 011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(111) = С 000 ⊕ С 100 ⊕ С 010 ⊕ С 001 ⊕ С 110 ⊕ С 101 ⊕ С 011 ⊕ С 111 = 1 => С 111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1 Таким образом, полином Жегалкина будет равен: F ж = X ⊕ X∧Z ⊕ X∧Z∧Y Логическая схема, соответствующая полиному Жегалкина:
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