Промежуточные таблицы истинности:¬A1:
(¬A1)∧A2:
A1 | A2 | ¬A1 | (¬A1)∧A2 |
0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 |
¬A2:
A3↓A4:
(¬A2)|A1:
A2 | A1 | ¬A2 | (¬A2)|A1 |
0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 |
((¬A1)∧A2)∧A3:
A1 | A2 | A3 | ¬A1 | (¬A1)∧A2 | ((¬A1)∧A2)∧A3 |
0 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 |
(A3↓A4)∨((¬A2)|A1):
A3 | A4 | A2 | A1 | A3↓A4 | ¬A2 | (¬A2)|A1 | (A3↓A4)∨((¬A2)|A1) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
A4⊕(((¬A1)∧A2)∧A3):
A4 | A1 | A2 | A3 | ¬A1 | (¬A1)∧A2 | ((¬A1)∧A2)∧A3 | A4⊕(((¬A1)∧A2)∧A3) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
(¬A1)←((A3↓A4)∨((¬A2)|A1)):
A1 | A3 | A4 | A2 | ¬A1 | A3↓A4 | ¬A2 | (¬A2)|A1 | (A3↓A4)∨((¬A2)|A1) | (¬A1)←((A3↓A4)∨((¬A2)|A1)) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
((¬A1)←((A3↓A4)∨((¬A2)|A1)))≡(A4⊕(((¬A1)∧A2)∧A3)):
A1 | A3 | A4 | A2 | ¬A1 | A3↓A4 | ¬A2 | (¬A2)|A1 | (A3↓A4)∨((¬A2)|A1) | (¬A1)←((A3↓A4)∨((¬A2)|A1)) | ¬A1 | (¬A1)∧A2 | ((¬A1)∧A2)∧A3 | A4⊕(((¬A1)∧A2)∧A3) | ((¬A1)←((A3↓A4)∨((¬A2)|A1)))≡(A4⊕(((¬A1)∧A2)∧A3)) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
Общая таблица истинности:
A1 | A3 | A4 | A2 | ¬A1 | (¬A1)∧A2 | ¬A2 | A3↓A4 | (¬A2)|A1 | ((¬A1)∧A2)∧A3 | (A3↓A4)∨((¬A2)|A1) | A4⊕(((¬A1)∧A2)∧A3) | (¬A1)←((A3↓A4)∨((¬A2)|A1)) | ¬A1←A3↓A4∨¬A2|A1≡A4⊕(¬A1∧A2)∧A3 |
0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 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 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
A1 | A3 | A4 | A2 | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
F
сднф = ¬A1∧¬A3∧A4∧¬A2 ∨ ¬A1∧¬A3∧A4∧A2 ∨ ¬A1∧A3∧¬A4∧A2 ∨ ¬A1∧A3∧A4∧¬A2 ∨ A1∧¬A3∧¬A4∧¬A2 ∨ A1∧¬A3∧¬A4∧A2 ∨ A1∧¬A3∧A4∧¬A2 ∨ A1∧A3∧¬A4∧A2 ∨ A1∧A3∧A4∧¬A2
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
A1 | A3 | A4 | A2 | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
F
скнф = (A1∨A3∨A4∨A2) ∧ (A1∨A3∨A4∨¬A2) ∧ (A1∨¬A3∨A4∨A2) ∧ (A1∨¬A3∨¬A4∨¬A2) ∧ (¬A1∨A3∨¬A4∨¬A2) ∧ (¬A1∨¬A3∨A4∨A2) ∧ (¬A1∨¬A3∨¬A4∨¬A2)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
A1 | A3 | A4 | A2 | Fж |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина:
F
ж = C
0000 ⊕ C
1000∧A1 ⊕ C
0100∧A3 ⊕ C
0010∧A4 ⊕ C
0001∧A2 ⊕ C
1100∧A1∧A3 ⊕ C
1010∧A1∧A4 ⊕ C
1001∧A1∧A2 ⊕ C
0110∧A3∧A4 ⊕ C
0101∧A3∧A2 ⊕ C
0011∧A4∧A2 ⊕ C
1110∧A1∧A3∧A4 ⊕ C
1101∧A1∧A3∧A2 ⊕ C
1011∧A1∧A4∧A2 ⊕ C
0111∧A3∧A4∧A2 ⊕ C
1111∧A1∧A3∧A4∧A2
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 = 1 => С
0010 = 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 = 1
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 1 => С
1010 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 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 = 1 => С
0110 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(0101) = С
0000 ⊕ С
0100 ⊕ С
0001 ⊕ С
0101 = 1 => С
0101 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
F
ж(0011) = С
0000 ⊕ С
0010 ⊕ С
0001 ⊕ С
0011 = 1 => С
0011 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 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 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F
ж(1101) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0001 ⊕ С
1100 ⊕ С
1001 ⊕ С
0101 ⊕ С
1101 = 1 => С
1101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(1011) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
0001 ⊕ С
1010 ⊕ С
1001 ⊕ С
0011 ⊕ С
1011 = 0 => С
1011 = 0 ⊕ 1 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Таким образом, полином Жегалкина будет равен:
F
ж = A1 ⊕ A4 ⊕ A1∧A3 ⊕ A1∧A4 ⊕ A3∧A2 ⊕ A1∧A3∧A4 ⊕ A1∧A4∧A2 ⊕ A1∧A3∧A4∧A2
Логическая схема, соответствующая полиному Жегалкина: