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