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