|
Список литературы
Генератор кроссвордов
Генератор титульных листов
Таблица истинности ONLINE
Прочие ONLINE сервисы
|
|
Таблица истинности для функции X⊕Y⊕M∧(X∧Y∧Z):
Промежуточные таблицы истинности:X∧Y: (X∧Y)∧Z: | X | Y | Z | X∧Y | (X∧Y)∧Z | | 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 | 1 | 0 | | 1 | 1 | 1 | 1 | 1 |
M∧((X∧Y)∧Z): | M | X | Y | Z | X∧Y | (X∧Y)∧Z | M∧((X∧Y)∧Z) | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 0 | 1 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 1 | 0 | 0 | 0 | | 0 | 1 | 1 | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 1 | 1 | 1 | 0 | | 1 | 0 | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
X⊕Y: (X⊕Y)⊕(M∧((X∧Y)∧Z)): | X | Y | M | Z | X⊕Y | X∧Y | (X∧Y)∧Z | M∧((X∧Y)∧Z) | (X⊕Y)⊕(M∧((X∧Y)∧Z)) | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
Общая таблица истинности:| X | Y | M | Z | X∧Y | (X∧Y)∧Z | M∧((X∧Y)∧Z) | X⊕Y | X⊕Y⊕M∧(X∧Y∧Z) | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: | X | Y | M | Z | 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 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | | 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 = ¬X∧Y∧¬M∧¬Z ∨ ¬X∧Y∧¬M∧Z ∨ ¬X∧Y∧M∧¬Z ∨ ¬X∧Y∧M∧Z ∨ X∧¬Y∧¬M∧¬Z ∨ X∧¬Y∧¬M∧Z ∨ X∧¬Y∧M∧¬Z ∨ X∧¬Y∧M∧Z ∨ X∧Y∧M∧Z Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: | X | Y | M | Z | 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 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | | 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 = (X∨Y∨M∨Z) ∧ (X∨Y∨M∨¬Z) ∧ (X∨Y∨¬M∨Z) ∧ (X∨Y∨¬M∨¬Z) ∧ (¬X∨¬Y∨M∨Z) ∧ (¬X∨¬Y∨M∨¬Z) ∧ (¬X∨¬Y∨¬M∨Z) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции | X | Y | M | Z | 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 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | | 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∧X ⊕ C 0100∧Y ⊕ C 0010∧M ⊕ C 0001∧Z ⊕ C 1100∧X∧Y ⊕ C 1010∧X∧M ⊕ C 1001∧X∧Z ⊕ C 0110∧Y∧M ⊕ C 0101∧Y∧Z ⊕ C 0011∧M∧Z ⊕ C 1110∧X∧Y∧M ⊕ C 1101∧X∧Y∧Z ⊕ C 1011∧X∧M∧Z ⊕ C 0111∧Y∧M∧Z ⊕ C 1111∧X∧Y∧M∧Z 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 = 1 => С 0100 = 0 ⊕ 1 = 1 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 ⊕ 1 ⊕ 1 ⊕ 0 = 0 F ж(1010) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 1010 = 1 => С 1010 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(1001) = С 0000 ⊕ С 1000 ⊕ С 0001 ⊕ С 1001 = 1 => С 1001 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(0110) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0110 = 1 => С 0110 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0 F ж(0101) = С 0000 ⊕ С 0100 ⊕ С 0001 ⊕ С 0101 = 1 => С 0101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 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 ⊕ 1 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(0111) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0001 ⊕ С 0110 ⊕ С 0101 ⊕ С 0011 ⊕ С 0111 = 1 => С 0111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 1 => С 1111 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1 Таким образом, полином Жегалкина будет равен: F ж = X ⊕ Y ⊕ X∧Y∧M∧Z Логическая схема, соответствующая полиному Жегалкина:
|
 |
|
 |
|
Вход на сайт
Информация
В нашем каталоге
Околостуденческое
|