Таблица истинности для функции A⊕(B∧¬C∨(¬D→E)):


Промежуточные таблицы истинности:
¬D:
D¬D
01
10

(¬D)→E:
DE¬D(¬D)→E
0010
0111
1001
1101

¬C:
C¬C
01
10

B∧(¬C):
BC¬CB∧(¬C)
0010
0100
1011
1100

(B∧(¬C))∨((¬D)→E):
BCDE¬CB∧(¬C)¬D(¬D)→E(B∧(¬C))∨((¬D)→E)
000010100
000110111
001010011
001110011
010000100
010100111
011000011
011100011
100011101
100111111
101011011
101111011
110000100
110100111
111000011
111100011

A⊕((B∧(¬C))∨((¬D)→E)):
ABCDE¬CB∧(¬C)¬D(¬D)→E(B∧(¬C))∨((¬D)→E)A⊕((B∧(¬C))∨((¬D)→E))
00000101000
00001101111
00010100111
00011100111
00100001000
00101001111
00110000111
00111000111
01000111011
01001111111
01010110111
01011110111
01100001000
01101001111
01110000111
01111000111
10000101001
10001101110
10010100110
10011100110
10100001001
10101001110
10110000110
10111000110
11000111010
11001111110
11010110110
11011110110
11100001001
11101001110
11110000110
11111000110

Общая таблица истинности:

ABCDE¬D(¬D)→E¬CB∧(¬C)(B∧(¬C))∨((¬D)→E)A⊕(B∧¬C∨(¬D→E))
00000101000
00001111011
00010011011
00011011011
00100100000
00101110011
00110010011
00111010011
01000101111
01001111111
01010011111
01011011111
01100100000
01101110011
01110010011
01111010011
10000101001
10001111010
10010011010
10011011010
10100100001
10101110010
10110010010
10111010010
11000101110
11001111110
11010011110
11011011110
11100100001
11101110010
11110010010
11111010010

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABCDEF
000000
000011
000101
000111
001000
001011
001101
001111
010001
010011
010101
010111
011000
011011
011101
011111
100001
100010
100100
100110
101001
101010
101100
101110
110000
110010
110100
110110
111001
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 ∨ ¬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:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABCDEF
000000
000011
000101
000111
001000
001011
001101
001111
010001
010011
010101
010111
011000
011011
011101
011111
100001
100010
100100
100110
101001
101010
101100
101110
110000
110010
110100
110110
111001
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)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABCDEFж
000000
000011
000101
000111
001000
001011
001101
001111
010001
010011
010101
010111
011000
011011
011101
011111
100001
100010
100100
100110
101001
101010
101100
101110
110000
110010
110100
110110
111001
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 = 1 => С01000 = 0 ⊕ 1 = 1
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 0 ⊕ 0 = 0
Fж(00010) = С00000 ⊕ С00010 = 1 => С00010 = 0 ⊕ 1 = 1
Fж(00001) = С00000 ⊕ С00001 = 1 => С00001 = 0 ⊕ 1 = 1
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 0 => С11000 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 0 => С10001 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 1 => С01010 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 1 => С01001 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 1 => С00101 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 1 => С00011 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 1 => С01110 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 1 => С01101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 1 => С01011 = 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 1 => С00111 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 0 => С10111 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 1 => С01111 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0

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

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