Промежуточные таблицы истинности:X2∧X3:
¬X1:
X4⊕(¬X1):
X4 | X1 | ¬X1 | X4⊕(¬X1) |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
X2∨(X4⊕(¬X1)):
X2 | X4 | X1 | ¬X1 | X4⊕(¬X1) | X2∨(X4⊕(¬X1)) |
0 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 1 | 0 | 1 | 1 |
¬(X2∧X3):
X2 | X3 | X2∧X3 | ¬(X2∧X3) |
0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 |
¬(X2∨(X4⊕(¬X1))):
X2 | X4 | X1 | ¬X1 | X4⊕(¬X1) | X2∨(X4⊕(¬X1)) | ¬(X2∨(X4⊕(¬X1))) |
0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 0 |
(¬(X2∧X3))∧(¬(X2∨(X4⊕(¬X1)))):
X2 | X3 | X4 | X1 | X2∧X3 | ¬(X2∧X3) | ¬X1 | X4⊕(¬X1) | X2∨(X4⊕(¬X1)) | ¬(X2∨(X4⊕(¬X1))) | (¬(X2∧X3))∧(¬(X2∨(X4⊕(¬X1)))) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
X1∨((¬(X2∧X3))∧(¬(X2∨(X4⊕(¬X1))))):
X1 | X2 | X3 | X4 | X2∧X3 | ¬(X2∧X3) | ¬X1 | X4⊕(¬X1) | X2∨(X4⊕(¬X1)) | ¬(X2∨(X4⊕(¬X1))) | (¬(X2∧X3))∧(¬(X2∨(X4⊕(¬X1)))) | X1∨((¬(X2∧X3))∧(¬(X2∨(X4⊕(¬X1))))) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
Общая таблица истинности:
X1 | X2 | X3 | X4 | X2∧X3 | ¬X1 | X4⊕(¬X1) | X2∨(X4⊕(¬X1)) | ¬(X2∧X3) | ¬(X2∨(X4⊕(¬X1))) | (¬(X2∧X3))∧(¬(X2∨(X4⊕(¬X1)))) | X1∨¬(X2∧X3)∧¬(X2∨(X4⊕¬X1)) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
X1 | X2 | X3 | X4 | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
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 | 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 = ¬X1∧¬X2∧¬X3∧X4 ∨ ¬X1∧¬X2∧X3∧X4 ∨ X1∧¬X2∧¬X3∧¬X4 ∨ X1∧¬X2∧¬X3∧X4 ∨ X1∧¬X2∧X3∧¬X4 ∨ X1∧¬X2∧X3∧X4 ∨ X1∧X2∧¬X3∧¬X4 ∨ X1∧X2∧¬X3∧X4 ∨ X1∧X2∧X3∧¬X4 ∨ X1∧X2∧X3∧X4
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
X1 | X2 | X3 | X4 | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
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 | 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
скнф = (X1∨X2∨X3∨X4) ∧ (X1∨X2∨¬X3∨X4) ∧ (X1∨¬X2∨X3∨X4) ∧ (X1∨¬X2∨X3∨¬X4) ∧ (X1∨¬X2∨¬X3∨X4) ∧ (X1∨¬X2∨¬X3∨¬X4)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
X1 | X2 | X3 | X4 | Fж |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
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 | 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∧X1 ⊕ C
0100∧X2 ⊕ C
0010∧X3 ⊕ C
0001∧X4 ⊕ C
1100∧X1∧X2 ⊕ C
1010∧X1∧X3 ⊕ C
1001∧X1∧X4 ⊕ C
0110∧X2∧X3 ⊕ C
0101∧X2∧X4 ⊕ C
0011∧X3∧X4 ⊕ C
1110∧X1∧X2∧X3 ⊕ C
1101∧X1∧X2∧X4 ⊕ C
1011∧X1∧X3∧X4 ⊕ C
0111∧X2∧X3∧X4 ⊕ C
1111∧X1∧X2∧X3∧X4
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 = 0 => С
0010 = 0 ⊕ 0 = 0
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 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 1 => С
1010 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(1001) = С
0000 ⊕ С
1000 ⊕ С
0001 ⊕ С
1001 = 1 => С
1001 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
F
ж(0110) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0110 = 0 => С
0110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(0101) = С
0000 ⊕ С
0100 ⊕ С
0001 ⊕ С
0101 = 0 => С
0101 = 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
F
ж(0011) = С
0000 ⊕ С
0010 ⊕ С
0001 ⊕ С
0011 = 1 => С
0011 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
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 ⊕ 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 = 0 ⊕ 1 ⊕ 0 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 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 ⊕ 0 ⊕ 0 ⊕ 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 = 1 => С
1111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = X1 ⊕ X4 ⊕ X1∧X4 ⊕ X2∧X4 ⊕ X1∧X2∧X4
Логическая схема, соответствующая полиному Жегалкина: