Промежуточные таблицы истинности:X1∨X2:
¬X1:
(¬X1)≡X3:
X1 | X3 | ¬X1 | (¬X1)≡X3 |
0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 |
¬((¬X1)≡X3):
X1 | X3 | ¬X1 | (¬X1)≡X3 | ¬((¬X1)≡X3) |
0 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
(X1∨X2)∧X4:
X1 | X2 | X4 | X1∨X2 | (X1∨X2)∧X4 |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 1 | 1 |
((X1∨X2)∧X4)∧(¬((¬X1)≡X3)):
X1 | X2 | X4 | X3 | X1∨X2 | (X1∨X2)∧X4 | ¬X1 | (¬X1)≡X3 | ¬((¬X1)≡X3) | ((X1∨X2)∧X4)∧(¬((¬X1)≡X3)) |
0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 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 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
(((X1∨X2)∧X4)∧(¬((¬X1)≡X3)))∧X4:
X1 | X2 | X4 | X3 | X1∨X2 | (X1∨X2)∧X4 | ¬X1 | (¬X1)≡X3 | ¬((¬X1)≡X3) | ((X1∨X2)∧X4)∧(¬((¬X1)≡X3)) | (((X1∨X2)∧X4)∧(¬((¬X1)≡X3)))∧X4 |
0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 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 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
Общая таблица истинности:
X1 | X2 | X4 | X3 | X1∨X2 | ¬X1 | (¬X1)≡X3 | ¬((¬X1)≡X3) | (X1∨X2)∧X4 | ((X1∨X2)∧X4)∧(¬((¬X1)≡X3)) | (X1∨X2)∧X4∧¬(¬X1≡X3)∧X4 |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
X1 | X2 | X4 | X3 | 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 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 |
F
сднф = ¬X1∧X2∧X4∧¬X3 ∨ X1∧¬X2∧X4∧X3 ∨ X1∧X2∧X4∧X3
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
X1 | X2 | X4 | X3 | 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 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 |
F
скнф = (X1∨X2∨X4∨X3) ∧ (X1∨X2∨X4∨¬X3) ∧ (X1∨X2∨¬X4∨X3) ∧ (X1∨X2∨¬X4∨¬X3) ∧ (X1∨¬X2∨X4∨X3) ∧ (X1∨¬X2∨X4∨¬X3) ∧ (X1∨¬X2∨¬X4∨¬X3) ∧ (¬X1∨X2∨X4∨X3) ∧ (¬X1∨X2∨X4∨¬X3) ∧ (¬X1∨X2∨¬X4∨X3) ∧ (¬X1∨¬X2∨X4∨X3) ∧ (¬X1∨¬X2∨X4∨¬X3) ∧ (¬X1∨¬X2∨¬X4∨X3)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
X1 | X2 | X4 | X3 | 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 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 |
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∧X1 ⊕ C
0100∧X2 ⊕ C
0010∧X4 ⊕ C
0001∧X3 ⊕ C
1100∧X1∧X2 ⊕ C
1010∧X1∧X4 ⊕ C
1001∧X1∧X3 ⊕ C
0110∧X2∧X4 ⊕ C
0101∧X2∧X3 ⊕ C
0011∧X4∧X3 ⊕ C
1110∧X1∧X2∧X4 ⊕ C
1101∧X1∧X2∧X3 ⊕ C
1011∧X1∧X4∧X3 ⊕ C
0111∧X2∧X4∧X3 ⊕ C
1111∧X1∧X2∧X4∧X3
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 = 1 => С
0110 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 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 = 1 => С
1011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Таким образом, полином Жегалкина будет равен:
F
ж = X2∧X4 ⊕ X1∧X2∧X4 ⊕ X1∧X4∧X3 ⊕ X2∧X4∧X3 ⊕ X1∧X2∧X4∧X3
Логическая схема, соответствующая полиному Жегалкина: