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