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