Промежуточные таблицы истинности:A≡B:
C⊕B:
A→(C⊕B):
A | C | B | C⊕B | A→(C⊕B) |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 |
B∧C:
(B∧C)∧A:
B | C | A | B∧C | (B∧C)∧A |
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:
(A≡B)∧V:
A | B | V | A≡B | (A≡B)∧V |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 1 |
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)∧V)∧(A→(C⊕B)):
A | B | V | C | A≡B | (A≡B)∧V | C⊕B | A→(C⊕B) | ((A≡B)∧V)∧(A→(C⊕B)) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
(((A≡B)∧V)∧(A→(C⊕B)))∧V:
A | B | V | C | A≡B | (A≡B)∧V | C⊕B | A→(C⊕B) | ((A≡B)∧V)∧(A→(C⊕B)) | (((A≡B)∧V)∧(A→(C⊕B)))∧V |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
((((A≡B)∧V)∧(A→(C⊕B)))∧V)∧((B∧C)∧A):
A | B | V | C | A≡B | (A≡B)∧V | C⊕B | A→(C⊕B) | ((A≡B)∧V)∧(A→(C⊕B)) | (((A≡B)∧V)∧(A→(C⊕B)))∧V | B∧C | (B∧C)∧A | ((((A≡B)∧V)∧(A→(C⊕B)))∧V)∧((B∧C)∧A) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
(((((A≡B)∧V)∧(A→(C⊕B)))∧V)∧((B∧C)∧A))→(A∧B):
A | B | V | C | A≡B | (A≡B)∧V | C⊕B | A→(C⊕B) | ((A≡B)∧V)∧(A→(C⊕B)) | (((A≡B)∧V)∧(A→(C⊕B)))∧V | B∧C | (B∧C)∧A | ((((A≡B)∧V)∧(A→(C⊕B)))∧V)∧((B∧C)∧A) | A∧B | (((((A≡B)∧V)∧(A→(C⊕B)))∧V)∧((B∧C)∧A))→(A∧B) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
Общая таблица истинности:
A | B | V | C | A≡B | C⊕B | A→(C⊕B) | B∧C | (B∧C)∧A | A∧B | (A≡B)∧V | ((A≡B)∧V)∧(A→(C⊕B)) | (((A≡B)∧V)∧(A→(C⊕B)))∧V | ((((A≡B)∧V)∧(A→(C⊕B)))∧V)∧((B∧C)∧A) | (A≡B)∧V∧(A→C⊕B)∧V∧(B∧C∧A)→(A∧B) |
0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
A | B | V | C | 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∧¬B∧¬V∧¬C ∨ ¬A∧¬B∧¬V∧C ∨ ¬A∧¬B∧V∧¬C ∨ ¬A∧¬B∧V∧C ∨ ¬A∧B∧¬V∧¬C ∨ ¬A∧B∧¬V∧C ∨ ¬A∧B∧V∧¬C ∨ ¬A∧B∧V∧C ∨ A∧¬B∧¬V∧¬C ∨ A∧¬B∧¬V∧C ∨ A∧¬B∧V∧¬C ∨ A∧¬B∧V∧C ∨ A∧B∧¬V∧¬C ∨ A∧B∧¬V∧C ∨ A∧B∧V∧¬C ∨ A∧B∧V∧C
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
A | B | V | C | 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 | B | V | C | 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∧B ⊕ C
0010∧V ⊕ C
0001∧C ⊕ C
1100∧A∧B ⊕ C
1010∧A∧V ⊕ C
1001∧A∧C ⊕ C
0110∧B∧V ⊕ C
0101∧B∧C ⊕ C
0011∧V∧C ⊕ C
1110∧A∧B∧V ⊕ C
1101∧A∧B∧C ⊕ C
1011∧A∧V∧C ⊕ C
0111∧B∧V∧C ⊕ C
1111∧A∧B∧V∧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 = 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