Таблица истинности для функции A∧C∨(B∧¬D)∨¬F:


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

B∧(¬D):
BD¬DB∧(¬D)
0010
0100
1011
1100

¬F:
F¬F
01
10

A∧C:
ACA∧C
000
010
100
111

(A∧C)∨(B∧(¬D)):
ACBDA∧C¬DB∧(¬D)(A∧C)∨(B∧(¬D))
00000100
00010000
00100111
00110000
01000100
01010000
01100111
01110000
10000100
10010000
10100111
10110000
11001101
11011001
11101111
11111001

((A∧C)∨(B∧(¬D)))∨(¬F):
ACBDFA∧C¬DB∧(¬D)(A∧C)∨(B∧(¬D))¬F((A∧C)∨(B∧(¬D)))∨(¬F)
00000010011
00001010000
00010000011
00011000000
00100011111
00101011101
00110000011
00111000000
01000010011
01001010000
01010000011
01011000000
01100011111
01101011101
01110000011
01111000000
10000010011
10001010000
10010000011
10011000000
10100011111
10101011101
10110000011
10111000000
11000110111
11001110101
11010100111
11011100101
11100111111
11101111101
11110100111
11111100101

Общая таблица истинности:

ACBDF¬DB∧(¬D)¬FA∧C(A∧C)∨(B∧(¬D))A∧C∨(B∧¬D)∨¬F
00000101001
00001100000
00010001001
00011000000
00100111011
00101110011
00110001001
00111000000
01000101001
01001100000
01010001001
01011000000
01100111011
01101110011
01110001001
01111000000
10000101001
10001100000
10010001001
10011000000
10100111011
10101110011
10110001001
10111000000
11000101111
11001100111
11010001111
11011000111
11100111111
11101110111
11110001111
11111000111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ACBDFF
000001
000010
000101
000110
001001
001011
001101
001110
010001
010010
010101
010110
011001
011011
011101
011110
100001
100010
100101
100110
101001
101011
101101
101110
110001
110011
110101
110111
111001
111011
111101
111111
Fсднф = ¬A∧¬C∧¬B∧¬D∧¬F ∨ ¬A∧¬C∧¬B∧D∧¬F ∨ ¬A∧¬C∧B∧¬D∧¬F ∨ ¬A∧¬C∧B∧¬D∧F ∨ ¬A∧¬C∧B∧D∧¬F ∨ ¬A∧C∧¬B∧¬D∧¬F ∨ ¬A∧C∧¬B∧D∧¬F ∨ ¬A∧C∧B∧¬D∧¬F ∨ ¬A∧C∧B∧¬D∧F ∨ ¬A∧C∧B∧D∧¬F ∨ A∧¬C∧¬B∧¬D∧¬F ∨ A∧¬C∧¬B∧D∧¬F ∨ A∧¬C∧B∧¬D∧¬F ∨ A∧¬C∧B∧¬D∧F ∨ A∧¬C∧B∧D∧¬F ∨ A∧C∧¬B∧¬D∧¬F ∨ A∧C∧¬B∧¬D∧F ∨ A∧C∧¬B∧D∧¬F ∨ A∧C∧¬B∧D∧F ∨ A∧C∧B∧¬D∧¬F ∨ A∧C∧B∧¬D∧F ∨ A∧C∧B∧D∧¬F ∨ A∧C∧B∧D∧F
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ACBDFF
000001
000010
000101
000110
001001
001011
001101
001110
010001
010010
010101
010110
011001
011011
011101
011110
100001
100010
100101
100110
101001
101011
101101
101110
110001
110011
110101
110111
111001
111011
111101
111111
Fскнф = (A∨C∨B∨D∨¬F) ∧ (A∨C∨B∨¬D∨¬F) ∧ (A∨C∨¬B∨¬D∨¬F) ∧ (A∨¬C∨B∨D∨¬F) ∧ (A∨¬C∨B∨¬D∨¬F) ∧ (A∨¬C∨¬B∨¬D∨¬F) ∧ (¬A∨C∨B∨D∨¬F) ∧ (¬A∨C∨B∨¬D∨¬F) ∧ (¬A∨C∨¬B∨¬D∨¬F)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ACBDFFж
000001
000010
000101
000110
001001
001011
001101
001110
010001
010010
010101
010110
011001
011011
011101
011110
100001
100010
100101
100110
101001
101011
101101
101110
110001
110011
110101
110111
111001
111011
111101
111111

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

Так как Fж(00000) = 1, то С00000 = 1.

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

Таким образом, полином Жегалкина будет равен:
Fж = 1 ⊕ F ⊕ B∧F ⊕ A∧C∧F ⊕ B∧D∧F ⊕ A∧C∧B∧F ⊕ A∧C∧B∧D∧F
Логическая схема, соответствующая полиному Жегалкина:

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