Таблица истинности для функции F≡A∨B∧C∨D:


Промежуточные таблицы истинности:
B∧C:
BCB∧C
000
010
100
111

A∨(B∧C):
ABCB∧CA∨(B∧C)
00000
00100
01000
01111
10001
10101
11001
11111

(A∨(B∧C))∨D:
ABCDB∧CA∨(B∧C)(A∨(B∧C))∨D
0000000
0001001
0010000
0011001
0100000
0101001
0110111
0111111
1000011
1001011
1010011
1011011
1100011
1101011
1110111
1111111

F≡((A∨(B∧C))∨D):
FABCDB∧CA∨(B∧C)(A∨(B∧C))∨DF≡((A∨(B∧C))∨D)
000000001
000010010
000100001
000110010
001000001
001010010
001101110
001111110
010000110
010010110
010100110
010110110
011000110
011010110
011101110
011111110
100000000
100010011
100100000
100110011
101000000
101010011
101101111
101111111
110000111
110010111
110100111
110110111
111000111
111010111
111101111
111111111

Общая таблица истинности:

FABCDB∧CA∨(B∧C)(A∨(B∧C))∨DF≡A∨B∧C∨D
000000001
000010010
000100001
000110010
001000001
001010010
001101110
001111110
010000110
010010110
010100110
010110110
011000110
011010110
011101110
011111110
100000000
100010011
100100000
100110011
101000000
101010011
101101111
101111111
110000111
110010111
110100111
110110111
111000111
111010111
111101111
111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FABCDF
000001
000010
000101
000110
001001
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100011
100100
100111
101000
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111
Fсднф = ¬F∧¬A∧¬B∧¬C∧¬D ∨ ¬F∧¬A∧¬B∧C∧¬D ∨ ¬F∧¬A∧B∧¬C∧¬D ∨ F∧¬A∧¬B∧¬C∧D ∨ F∧¬A∧¬B∧C∧D ∨ F∧¬A∧B∧¬C∧D ∨ F∧¬A∧B∧C∧¬D ∨ F∧¬A∧B∧C∧D ∨ F∧A∧¬B∧¬C∧¬D ∨ F∧A∧¬B∧¬C∧D ∨ F∧A∧¬B∧C∧¬D ∨ F∧A∧¬B∧C∧D ∨ F∧A∧B∧¬C∧¬D ∨ F∧A∧B∧¬C∧D ∨ F∧A∧B∧C∧¬D ∨ F∧A∧B∧C∧D
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FABCDF
000001
000010
000101
000110
001001
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100011
100100
100111
101000
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111
Fскнф = (F∨A∨B∨C∨¬D) ∧ (F∨A∨B∨¬C∨¬D) ∧ (F∨A∨¬B∨C∨¬D) ∧ (F∨A∨¬B∨¬C∨D) ∧ (F∨A∨¬B∨¬C∨¬D) ∧ (F∨¬A∨B∨C∨D) ∧ (F∨¬A∨B∨C∨¬D) ∧ (F∨¬A∨B∨¬C∨D) ∧ (F∨¬A∨B∨¬C∨¬D) ∧ (F∨¬A∨¬B∨C∨D) ∧ (F∨¬A∨¬B∨C∨¬D) ∧ (F∨¬A∨¬B∨¬C∨D) ∧ (F∨¬A∨¬B∨¬C∨¬D) ∧ (¬F∨A∨B∨C∨D) ∧ (¬F∨A∨B∨¬C∨D) ∧ (¬F∨A∨¬B∨C∨D)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FABCDFж
000001
000010
000101
000110
001001
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100011
100100
100111
101000
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111

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

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 = 0 => С10000 = 1 ⊕ 0 = 1
Fж(01000) = С00000 ⊕ С01000 = 0 => С01000 = 1 ⊕ 0 = 1
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 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
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 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 1 => С11010 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 1 => С10011 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 1 => С11110 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 1 => С11101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 1 => С11011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 1 => С10111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 0 => С01111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 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 = 1 => С11111 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0

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

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