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Таблица истинности ONLINE
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Таблица истинности для функции (K∧L∧M)→(¬M∧¬N)≡1:
Промежуточные таблицы истинности:K∧L: (K∧L)∧M: K | L | M | K∧L | (K∧L)∧M | 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: ¬N: (¬M)∧(¬N): M | N | ¬M | ¬N | (¬M)∧(¬N) | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
((K∧L)∧M)→((¬M)∧(¬N)): K | L | M | N | K∧L | (K∧L)∧M | ¬M | ¬N | (¬M)∧(¬N) | ((K∧L)∧M)→((¬M)∧(¬N)) | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
(((K∧L)∧M)→((¬M)∧(¬N)))≡1: K | L | M | N | K∧L | (K∧L)∧M | ¬M | ¬N | (¬M)∧(¬N) | ((K∧L)∧M)→((¬M)∧(¬N)) | (((K∧L)∧M)→((¬M)∧(¬N)))≡1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
Общая таблица истинности:K | L | M | N | K∧L | (K∧L)∧M | ¬M | ¬N | (¬M)∧(¬N) | ((K∧L)∧M)→((¬M)∧(¬N)) | (K∧L∧M)→(¬M∧¬N)≡1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: K | L | M | N | F | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 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 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
F сднф = ¬K∧¬L∧¬M∧¬N ∨ ¬K∧¬L∧¬M∧N ∨ ¬K∧¬L∧M∧¬N ∨ ¬K∧¬L∧M∧N ∨ ¬K∧L∧¬M∧¬N ∨ ¬K∧L∧¬M∧N ∨ ¬K∧L∧M∧¬N ∨ ¬K∧L∧M∧N ∨ K∧¬L∧¬M∧¬N ∨ K∧¬L∧¬M∧N ∨ K∧¬L∧M∧¬N ∨ K∧¬L∧M∧N ∨ K∧L∧¬M∧¬N ∨ K∧L∧¬M∧N Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: K | L | M | N | F | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 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 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
F скнф = (¬K∨¬L∨¬M∨N) ∧ (¬K∨¬L∨¬M∨¬N) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции K | L | M | N | Fж | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 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 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина: F ж = C 0000 ⊕ C 1000∧K ⊕ C 0100∧L ⊕ C 0010∧M ⊕ C 0001∧N ⊕ C 1100∧K∧L ⊕ C 1010∧K∧M ⊕ C 1001∧K∧N ⊕ C 0110∧L∧M ⊕ C 0101∧L∧N ⊕ C 0011∧M∧N ⊕ C 1110∧K∧L∧M ⊕ C 1101∧K∧L∧N ⊕ C 1011∧K∧M∧N ⊕ C 0111∧L∧M∧N ⊕ C 1111∧K∧L∧M∧N Так как F ж(0000) = 1, то С 0000 = 1. 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 = 1 ⊕ 1 = 0 F ж(0100) = С 0000 ⊕ С 0100 = 1 => С 0100 = 1 ⊕ 1 = 0 F ж(0010) = С 0000 ⊕ С 0010 = 1 => С 0010 = 1 ⊕ 1 = 0 F ж(0001) = С 0000 ⊕ С 0001 = 1 => С 0001 = 1 ⊕ 1 = 0 F ж(1100) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 1100 = 1 => С 1100 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(1010) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 1010 = 1 => С 1010 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(1001) = С 0000 ⊕ С 1000 ⊕ С 0001 ⊕ С 1001 = 1 => С 1001 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(0110) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0110 = 1 => С 0110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(0101) = С 0000 ⊕ С 0100 ⊕ С 0001 ⊕ С 0101 = 1 => С 0101 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(0011) = С 0000 ⊕ С 0010 ⊕ С 0001 ⊕ С 0011 = 1 => С 0011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(1110) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0010 ⊕ С 1100 ⊕ С 1010 ⊕ С 0110 ⊕ С 1110 = 0 => С 1110 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1 F ж(1101) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0001 ⊕ С 1100 ⊕ С 1001 ⊕ С 0101 ⊕ С 1101 = 1 => С 1101 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(1011) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 0001 ⊕ С 1010 ⊕ С 1001 ⊕ С 0011 ⊕ С 1011 = 1 => С 1011 = 1 ⊕ 0 ⊕ 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 = 1 ⊕ 0 ⊕ 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 = 0 => С 1111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 Таким образом, полином Жегалкина будет равен: F ж = 1 ⊕ K∧L∧M Логическая схема, соответствующая полиному Жегалкина:
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