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