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


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

A∧B:
ABA∧B
000
010
100
111

(A∧B)∧C:
ABCA∧B(A∧B)∧C
00000
00100
01000
01100
10000
10100
11010
11111

((A∧B)∧C)∧(¬D):
ABCDA∧B(A∧B)∧C¬D((A∧B)∧C)∧(¬D)
00000010
00010000
00100010
00110000
01000010
01010000
01100010
01110000
10000010
10010000
10100010
10110000
11001010
11011000
11101111
11111100

F≡(((A∧B)∧C)∧(¬D)):
FABCDA∧B(A∧B)∧C¬D((A∧B)∧C)∧(¬D)F≡(((A∧B)∧C)∧(¬D))
0000000101
0000100001
0001000101
0001100001
0010000101
0010100001
0011000101
0011100001
0100000101
0100100001
0101000101
0101100001
0110010101
0110110001
0111011110
0111111001
1000000100
1000100000
1001000100
1001100000
1010000100
1010100000
1011000100
1011100000
1100000100
1100100000
1101000100
1101100000
1110010100
1110110000
1111011111
1111111000

Общая таблица истинности:

FABCD¬DA∧B(A∧B)∧C((A∧B)∧C)∧(¬D)F≡A∧B∧C∧¬D
0000010001
0000100001
0001010001
0001100001
0010010001
0010100001
0011010001
0011100001
0100010001
0100100001
0101010001
0101100001
0110011001
0110101001
0111011110
0111101101
1000010000
1000100000
1001010000
1001100000
1010010000
1010100000
1011010000
1011100000
1100010000
1100100000
1101010000
1101100000
1110011000
1110101000
1111011111
1111101100

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FABCDF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011100
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111101
111110
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
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011100
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111101
111110
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
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011100
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111101
111110

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 = 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 = 1 => С00001 = 1 ⊕ 1 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 0 => С11000 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 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 = 0 => С10001 = 1 ⊕ 1 ⊕ 0 ⊕ 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 = 1 => С01001 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 1 => С00101 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 1 => С00011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 0 => С11100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 1 => С01101 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 1 => С01011 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 1 => С00111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 1 => С11110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 0 => С11101 = 1 ⊕ 1 ⊕ 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 = 1 ⊕ 1 ⊕ 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 = 1 ⊕ 1 ⊕ 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 = 1 => С01111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 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 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0

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

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