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