Промежуточные таблицы истинности:¬X3:
¬X4:
¬X2:
(¬X3)∧(¬X4):
X3 | X4 | ¬X3 | ¬X4 | (¬X3)∧(¬X4) |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
(¬X2)∧X3:
X2 | X3 | ¬X2 | (¬X2)∧X3 |
0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 |
((¬X3)∧(¬X4))∨((¬X2)∧X3):
X3 | X4 | X2 | ¬X3 | ¬X4 | (¬X3)∧(¬X4) | ¬X2 | (¬X2)∧X3 | ((¬X3)∧(¬X4))∨((¬X2)∧X3) |
0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
X∧(((¬X3)∧(¬X4))∨((¬X2)∧X3)):
X | X3 | X4 | X2 | ¬X3 | ¬X4 | (¬X3)∧(¬X4) | ¬X2 | (¬X2)∧X3 | ((¬X3)∧(¬X4))∨((¬X2)∧X3) | X∧(((¬X3)∧(¬X4))∨((¬X2)∧X3)) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Общая таблица истинности:
X | X3 | X4 | X2 | ¬X3 | ¬X4 | ¬X2 | (¬X3)∧(¬X4) | (¬X2)∧X3 | ((¬X3)∧(¬X4))∨((¬X2)∧X3) | X∧(¬X3∧¬X4∨¬X2∧X3) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
X | X3 | X4 | X2 | 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 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
F
сднф = X∧¬X3∧¬X4∧¬X2 ∨ X∧¬X3∧¬X4∧X2 ∨ X∧X3∧¬X4∧¬X2 ∨ X∧X3∧X4∧¬X2
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
X | X3 | X4 | X2 | 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 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
F
скнф = (X∨X3∨X4∨X2) ∧ (X∨X3∨X4∨¬X2) ∧ (X∨X3∨¬X4∨X2) ∧ (X∨X3∨¬X4∨¬X2) ∧ (X∨¬X3∨X4∨X2) ∧ (X∨¬X3∨X4∨¬X2) ∧ (X∨¬X3∨¬X4∨X2) ∧ (X∨¬X3∨¬X4∨¬X2) ∧ (¬X∨X3∨¬X4∨X2) ∧ (¬X∨X3∨¬X4∨¬X2) ∧ (¬X∨¬X3∨X4∨¬X2) ∧ (¬X∨¬X3∨¬X4∨¬X2)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
X | X3 | X4 | X2 | 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 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина:
F
ж = C
0000 ⊕ C
1000∧X ⊕ C
0100∧X3 ⊕ C
0010∧X4 ⊕ C
0001∧X2 ⊕ C
1100∧X∧X3 ⊕ C
1010∧X∧X4 ⊕ C
1001∧X∧X2 ⊕ C
0110∧X3∧X4 ⊕ C
0101∧X3∧X2 ⊕ C
0011∧X4∧X2 ⊕ C
1110∧X∧X3∧X4 ⊕ C
1101∧X∧X3∧X2 ⊕ C
1011∧X∧X4∧X2 ⊕ C
0111∧X3∧X4∧X2 ⊕ C
1111∧X∧X3∧X4∧X2
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 = 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 = 1 => С
1100 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 0 => С
1010 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
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 = 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 = 1 => С
1110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F
ж(1101) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0001 ⊕ С
1100 ⊕ С
1001 ⊕ С
0101 ⊕ С
1101 = 0 => С
1101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
F
ж(1011) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
0001 ⊕ С
1010 ⊕ С
1001 ⊕ С
0011 ⊕ С
1011 = 0 => С
1011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 => С
1111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = X ⊕ X∧X4 ⊕ X∧X3∧X4 ⊕ X∧X3∧X2
Логическая схема, соответствующая полиному Жегалкина: