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