Промежуточные таблицы истинности:X1∨X2:
F∧X:
X1∧X2:
X2∧X1:
(X1∧X2)∨(X1∨X2):
X1 | X2 | X1∧X2 | X1∨X2 | (X1∧X2)∨(X1∨X2) |
0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 |
((X1∧X2)∨(X1∨X2))∨(X1∧X2):
X1 | X2 | X1∧X2 | X1∨X2 | (X1∧X2)∨(X1∨X2) | X1∧X2 | ((X1∧X2)∨(X1∨X2))∨(X1∧X2) |
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
(((X1∧X2)∨(X1∨X2))∨(X1∧X2))∨(X2∧X1):
X1 | X2 | X1∧X2 | X1∨X2 | (X1∧X2)∨(X1∨X2) | X1∧X2 | ((X1∧X2)∨(X1∨X2))∨(X1∧X2) | X2∧X1 | (((X1∧X2)∨(X1∨X2))∨(X1∧X2))∨(X2∧X1) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(F∧X)≡((((X1∧X2)∨(X1∨X2))∨(X1∧X2))∨(X2∧X1)):
F | X | X1 | X2 | F∧X | X1∧X2 | X1∨X2 | (X1∧X2)∨(X1∨X2) | X1∧X2 | ((X1∧X2)∨(X1∨X2))∨(X1∧X2) | X2∧X1 | (((X1∧X2)∨(X1∨X2))∨(X1∧X2))∨(X2∧X1) | (F∧X)≡((((X1∧X2)∨(X1∨X2))∨(X1∧X2))∨(X2∧X1)) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Общая таблица истинности:
F | X | X1 | X2 | X1∨X2 | F∧X | X1∧X2 | X2∧X1 | (X1∧X2)∨(X1∨X2) | ((X1∧X2)∨(X1∨X2))∨(X1∧X2) | (((X1∧X2)∨(X1∨X2))∨(X1∧X2))∨(X2∧X1) | F∧(X)≡X1∧X2∨(X1∨X2)∨X1∧X2∨X2∧X1 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
F | X | X1 | X2 | F |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
F
сднф = ¬F∧¬X∧¬X1∧¬X2 ∨ ¬F∧X∧¬X1∧¬X2 ∨ F∧¬X∧¬X1∧¬X2 ∨ F∧X∧¬X1∧X2 ∨ F∧X∧X1∧¬X2 ∨ F∧X∧X1∧X2
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
F | X | X1 | X2 | F |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
F
скнф = (F∨X∨X1∨¬X2) ∧ (F∨X∨¬X1∨X2) ∧ (F∨X∨¬X1∨¬X2) ∧ (F∨¬X∨X1∨¬X2) ∧ (F∨¬X∨¬X1∨X2) ∧ (F∨¬X∨¬X1∨¬X2) ∧ (¬F∨X∨X1∨¬X2) ∧ (¬F∨X∨¬X1∨X2) ∧ (¬F∨X∨¬X1∨¬X2) ∧ (¬F∨¬X∨X1∨X2)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
F | X | X1 | X2 | Fж |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
Построим полином Жегалкина:
F
ж = C
0000 ⊕ C
1000∧F ⊕ C
0100∧X ⊕ C
0010∧X1 ⊕ C
0001∧X2 ⊕ C
1100∧F∧X ⊕ C
1010∧F∧X1 ⊕ C
1001∧F∧X2 ⊕ C
0110∧X∧X1 ⊕ C
0101∧X∧X2 ⊕ C
0011∧X1∧X2 ⊕ C
1110∧F∧X∧X1 ⊕ C
1101∧F∧X∧X2 ⊕ C
1011∧F∧X1∧X2 ⊕ C
0111∧X∧X1∧X2 ⊕ C
1111∧F∧X∧X1∧X2
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 = 0 => С
0010 = 1 ⊕ 0 = 1
F
ж(0001) = С
0000 ⊕ С
0001 = 0 => С
0001 = 1 ⊕ 0 = 1
F
ж(1100) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
1100 = 0 => С
1100 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 0 => С
1010 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(1001) = С
0000 ⊕ С
1000 ⊕ С
0001 ⊕ С
1001 = 0 => С
1001 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(0110) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0110 = 0 => С
0110 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(0101) = С
0000 ⊕ С
0100 ⊕ С
0001 ⊕ С
0101 = 0 => С
0101 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(0011) = С
0000 ⊕ С
0010 ⊕ С
0001 ⊕ С
0011 = 0 => С
0011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1
F
ж(1110) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
1100 ⊕ С
1010 ⊕ С
0110 ⊕ С
1110 = 1 => С
1110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
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 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(1011) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
0001 ⊕ С
1010 ⊕ С
1001 ⊕ С
0011 ⊕ С
1011 = 0 => С
1011 = 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(0111) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
0111 = 0 => С
0111 = 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = 1 ⊕ X1 ⊕ X2 ⊕ F∧X ⊕ X1∧X2
Логическая схема, соответствующая полиному Жегалкина: