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