Для функции A∧B∧(C∨¬E∨D)∧¬B:


Промежуточные таблицы истинности:
¬E:
E¬E
01
10

C∨(¬E):
CE¬EC∨(¬E)
0011
0100
1011
1101

(C∨(¬E))∨D:
CED¬EC∨(¬E)(C∨(¬E))∨D
000111
001111
010000
011001
100111
101111
110011
111011

¬B:
B¬B
01
10

A∧B:
ABA∧B
000
010
100
111

(A∧B)∧((C∨(¬E))∨D):
ABCEDA∧B¬EC∨(¬E)(C∨(¬E))∨D(A∧B)∧((C∨(¬E))∨D)
0000001110
0000101110
0001000000
0001100010
0010001110
0010101110
0011000110
0011100110
0100001110
0100101110
0101000000
0101100010
0110001110
0110101110
0111000110
0111100110
1000001110
1000101110
1001000000
1001100010
1010001110
1010101110
1011000110
1011100110
1100011111
1100111111
1101010000
1101110011
1110011111
1110111111
1111010111
1111110111

((A∧B)∧((C∨(¬E))∨D))∧(¬B):
ABCEDA∧B¬EC∨(¬E)(C∨(¬E))∨D(A∧B)∧((C∨(¬E))∨D)¬B((A∧B)∧((C∨(¬E))∨D))∧(¬B)
000000111010
000010111010
000100000010
000110001010
001000111010
001010111010
001100011010
001110011010
010000111000
010010111000
010100000000
010110001000
011000111000
011010111000
011100011000
011110011000
100000111010
100010111010
100100000010
100110001010
101000111010
101010111010
101100011010
101110011010
110001111100
110011111100
110101000000
110111001100
111001111100
111011111100
111101011100
111111011100

Общая таблица истинности:

ABCED¬EC∨(¬E)(C∨(¬E))∨D¬BA∧B(A∧B)∧((C∨(¬E))∨D)A∧B∧(C∨¬E∨D)∧¬B
000001111000
000011111000
000100001000
000110011000
001001111000
001011111000
001100111000
001110111000
010001110000
010011110000
010100000000
010110010000
011001110000
011011110000
011100110000
011110110000
100001111000
100011111000
100100001000
100110011000
101001111000
101011111000
101100111000
101110111000
110001110110
110011110110
110100000100
110110010110
111001110110
111011110110
111100110110
111110110110

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABCEDF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
В таблице истинности нет набора значений переменных при которых функция истинна!

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABCEDF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fскнф = (A∨B∨C∨E∨D) ∧ (A∨B∨C∨E∨¬D) ∧ (A∨B∨C∨¬E∨D) ∧ (A∨B∨C∨¬E∨¬D) ∧ (A∨B∨¬C∨E∨D) ∧ (A∨B∨¬C∨E∨¬D) ∧ (A∨B∨¬C∨¬E∨D) ∧ (A∨B∨¬C∨¬E∨¬D) ∧ (A∨¬B∨C∨E∨D) ∧ (A∨¬B∨C∨E∨¬D) ∧ (A∨¬B∨C∨¬E∨D) ∧ (A∨¬B∨C∨¬E∨¬D) ∧ (A∨¬B∨¬C∨E∨D) ∧ (A∨¬B∨¬C∨E∨¬D) ∧ (A∨¬B∨¬C∨¬E∨D) ∧ (A∨¬B∨¬C∨¬E∨¬D) ∧ (¬A∨B∨C∨E∨D) ∧ (¬A∨B∨C∨E∨¬D) ∧ (¬A∨B∨C∨¬E∨D) ∧ (¬A∨B∨C∨¬E∨¬D) ∧ (¬A∨B∨¬C∨E∨D) ∧ (¬A∨B∨¬C∨E∨¬D) ∧ (¬A∨B∨¬C∨¬E∨D) ∧ (¬A∨B∨¬C∨¬E∨¬D) ∧ (¬A∨¬B∨C∨E∨D) ∧ (¬A∨¬B∨C∨E∨¬D) ∧ (¬A∨¬B∨C∨¬E∨D) ∧ (¬A∨¬B∨C∨¬E∨¬D) ∧ (¬A∨¬B∨¬C∨E∨D) ∧ (¬A∨¬B∨¬C∨E∨¬D) ∧ (¬A∨¬B∨¬C∨¬E∨D) ∧ (¬A∨¬B∨¬C∨¬E∨¬D)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABCEDFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
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∧E ⊕ C00001∧D ⊕ C11000∧A∧B ⊕ C10100∧A∧C ⊕ C10010∧A∧E ⊕ C10001∧A∧D ⊕ C01100∧B∧C ⊕ C01010∧B∧E ⊕ C01001∧B∧D ⊕ C00110∧C∧E ⊕ C00101∧C∧D ⊕ C00011∧E∧D ⊕ C11100∧A∧B∧C ⊕ C11010∧A∧B∧E ⊕ C11001∧A∧B∧D ⊕ C10110∧A∧C∧E ⊕ C10101∧A∧C∧D ⊕ C10011∧A∧E∧D ⊕ C01110∧B∧C∧E ⊕ C01101∧B∧C∧D ⊕ C01011∧B∧E∧D ⊕ C00111∧C∧E∧D ⊕ C11110∧A∧B∧C∧E ⊕ C11101∧A∧B∧C∧D ⊕ C11011∧A∧B∧E∧D ⊕ C10111∧A∧C∧E∧D ⊕ C01111∧B∧C∧E∧D ⊕ C11111∧A∧B∧C∧E∧D

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 = 0 => С10000 = 0 ⊕ 0 = 0
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 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 0 => С10001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 0 => С11101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 0 => С10111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 0 => С01111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0

Таким образом, полином Жегалкина будет равен:
Fж = 0

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