Промежуточные таблицы истинности:X∨Y:
¬(X∨Y):
X | Y | X∨Y | ¬(X∨Y) |
0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 |
Z∧F:
(¬(X∨Y))→(Z∧F):
X | Y | Z | F | X∨Y | ¬(X∨Y) | Z∧F | (¬(X∨Y))→(Z∧F) |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
¬Z:
F→(¬Z):
F | Z | ¬Z | F→(¬Z) |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 0 |
((¬(X∨Y))→(Z∧F))∧(F→(¬Z)):
X | Y | Z | F | X∨Y | ¬(X∨Y) | Z∧F | (¬(X∨Y))→(Z∧F) | ¬Z | F→(¬Z) | ((¬(X∨Y))→(Z∧F))∧(F→(¬Z)) |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
(((¬(X∨Y))→(Z∧F))∧(F→(¬Z)))→(X∨Y):
X | Y | Z | F | X∨Y | ¬(X∨Y) | Z∧F | (¬(X∨Y))→(Z∧F) | ¬Z | F→(¬Z) | ((¬(X∨Y))→(Z∧F))∧(F→(¬Z)) | X∨Y | (((¬(X∨Y))→(Z∧F))∧(F→(¬Z)))→(X∨Y) |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
Общая таблица истинности:
X | Y | Z | F | X∨Y | ¬(X∨Y) | Z∧F | (¬(X∨Y))→(Z∧F) | ¬Z | F→(¬Z) | ((¬(X∨Y))→(Z∧F))∧(F→(¬Z)) | ((¬(X∨Y))→Z∧F)∧(F→¬Z)→(X∨Y) |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
X | Y | Z | F | 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 = ¬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∧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∧F
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
X | Y | Z | F | 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
X | Y | Z | F | 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∧X ⊕ C
0100∧Y ⊕ C
0010∧Z ⊕ C
0001∧F ⊕ C
1100∧X∧Y ⊕ C
1010∧X∧Z ⊕ C
1001∧X∧F ⊕ C
0110∧Y∧Z ⊕ C
0101∧Y∧F ⊕ C
0011∧Z∧F ⊕ C
1110∧X∧Y∧Z ⊕ C
1101∧X∧Y∧F ⊕ C
1011∧X∧Z∧F ⊕ C
0111∧Y∧Z∧F ⊕ C
1111∧X∧Y∧Z∧F
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