Таблица истинности для вектора значений {00000011011010011}:


Т.к. вектор значений неполон - остальные значения функции будут доопределены нулями!
Новый вектор значений {00000011011010011000000000000000}


Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
ABCDEF
000000
000010
000100
000110
001000
001010
001101
001111
010000
010011
010101
010110
011001
011010
011100
011111
100001
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fсднф = ¬A∧¬B∧C∧D∧¬E ∨ ¬A∧¬B∧C∧D∧E ∨ ¬A∧B∧¬C∧¬D∧E ∨ ¬A∧B∧¬C∧D∧¬E ∨ ¬A∧B∧C∧¬D∧¬E ∨ ¬A∧B∧C∧D∧E ∨ A∧¬B∧¬C∧¬D∧¬E
Логическая cхема:


Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
ABCDEF
000000
000010
000100
000110
001000
001010
001101
001111
010000
010011
010101
010110
011001
011010
011100
011111
100001
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fскнф = (A∨B∨C∨D∨E) ∧ (A∨B∨C∨D∨¬E) ∧ (A∨B∨C∨¬D∨E) ∧ (A∨B∨C∨¬D∨¬E) ∧ (A∨B∨¬C∨D∨E) ∧ (A∨B∨¬C∨D∨¬E) ∧ (A∨¬B∨C∨D∨E) ∧ (A∨¬B∨C∨¬D∨¬E) ∧ (A∨¬B∨¬C∨D∨¬E) ∧ (A∨¬B∨¬C∨¬D∨E) ∧ (¬A∨B∨C∨D∨¬E) ∧ (¬A∨B∨C∨¬D∨E) ∧ (¬A∨B∨C∨¬D∨¬E) ∧ (¬A∨B∨¬C∨D∨E) ∧ (¬A∨B∨¬C∨D∨¬E) ∧ (¬A∨B∨¬C∨¬D∨E) ∧ (¬A∨B∨¬C∨¬D∨¬E) ∧ (¬A∨¬B∨C∨D∨E) ∧ (¬A∨¬B∨C∨D∨¬E) ∧ (¬A∨¬B∨C∨¬D∨E) ∧ (¬A∨¬B∨C∨¬D∨¬E) ∧ (¬A∨¬B∨¬C∨D∨E) ∧ (¬A∨¬B∨¬C∨D∨¬E) ∧ (¬A∨¬B∨¬C∨¬D∨E) ∧ (¬A∨¬B∨¬C∨¬D∨¬E)
Логическая cхема:



Построение полинома Жегалкина:
По таблице истинности функции
ABCDEFж
000000
000010
000100
000110
001000
001010
001101
001111
010000
010011
010101
010110
011001
011010
011100
011111
100001
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧A ⊕ C01000∧B ⊕ C00100∧C ⊕ C00010∧D ⊕ C00001∧E ⊕ C11000∧A∧B ⊕ C10100∧A∧C ⊕ C10010∧A∧D ⊕ C10001∧A∧E ⊕ C01100∧B∧C ⊕ C01010∧B∧D ⊕ C01001∧B∧E ⊕ C00110∧C∧D ⊕ C00101∧C∧E ⊕ C00011∧D∧E ⊕ C11100∧A∧B∧C ⊕ C11010∧A∧B∧D ⊕ C11001∧A∧B∧E ⊕ C10110∧A∧C∧D ⊕ C10101∧A∧C∧E ⊕ C10011∧A∧D∧E ⊕ C01110∧B∧C∧D ⊕ C01101∧B∧C∧E ⊕ C01011∧B∧D∧E ⊕ C00111∧C∧D∧E ⊕ C11110∧A∧B∧C∧D ⊕ C11101∧A∧B∧C∧E ⊕ C11011∧A∧B∧D∧E ⊕ C10111∧A∧C∧D∧E ⊕ C01111∧B∧C∧D∧E ⊕ C11111∧A∧B∧C∧D∧E

Так как Fж(00000) = 0, то С00000 = 0.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
Fж(10000) = С00000 ⊕ С10000 = 1 => С10000 = 0 ⊕ 1 = 1
Fж(01000) = С00000 ⊕ С01000 = 0 => С01000 = 0 ⊕ 0 = 0
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 0 ⊕ 0 = 0
Fж(00010) = С00000 ⊕ С00010 = 0 => С00010 = 0 ⊕ 0 = 0
Fж(00001) = С00000 ⊕ С00001 = 0 => С00001 = 0 ⊕ 0 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 0 => С11000 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 0 => С10001 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 1 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 1 => С01010 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 1 => С01001 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 0 => С11100 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 1 => С00111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 0 => С11101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 0 => С10111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 1 => С01111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11111) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С11100 ⊕ С11010 ⊕ С11001 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С11110 ⊕ С11101 ⊕ С11011 ⊕ С10111 ⊕ С01111 ⊕ С11111 = 0 => С11111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1

Таким образом, полином Жегалкина будет равен:
Fж = A ⊕ A∧B ⊕ A∧C ⊕ A∧D ⊕ A∧E ⊕ B∧C ⊕ B∧D ⊕ B∧E ⊕ C∧D ⊕ A∧C∧E ⊕ A∧D∧E ⊕ B∧C∧D ⊕ A∧B∧C∧E ⊕ A∧B∧D∧E ⊕ A∧C∧D∧E ⊕ A∧B∧C∧D∧E
Логическая схема, соответствующая полиному Жегалкина:

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