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