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