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


Промежуточные таблицы истинности:
B∨C:
BCB∨C
000
011
101
111

D∧E:
DED∧E
000
010
100
111

¬A:
A¬A
01
10

(¬A)∧(B∨C):
ABC¬AB∨C(¬A)∧(B∨C)
000100
001111
010111
011111
100000
101010
110010
111010

((¬A)∧(B∨C))∨(D∧E):
ABCDE¬AB∨C(¬A)∧(B∨C)D∧E((¬A)∧(B∨C))∨(D∧E)
0000010000
0000110000
0001010000
0001110011
0010011101
0010111101
0011011101
0011111111
0100011101
0100111101
0101011101
0101111111
0110011101
0110111101
0111011101
0111111111
1000000000
1000100000
1001000000
1001100011
1010001000
1010101000
1011001000
1011101011
1100001000
1100101000
1101001000
1101101011
1110001000
1110101000
1111001000
1111101011

¬(((¬A)∧(B∨C))∨(D∧E)):
ABCDE¬AB∨C(¬A)∧(B∨C)D∧E((¬A)∧(B∨C))∨(D∧E)¬(((¬A)∧(B∨C))∨(D∧E))
00000100001
00001100001
00010100001
00011100110
00100111010
00101111010
00110111010
00111111110
01000111010
01001111010
01010111010
01011111110
01100111010
01101111010
01110111010
01111111110
10000000001
10001000001
10010000001
10011000110
10100010001
10101010001
10110010001
10111010110
11000010001
11001010001
11010010001
11011010110
11100010001
11101010001
11110010001
11111010110

Общая таблица истинности:

ABCDEB∨CD∧E¬A(¬A)∧(B∨C)((¬A)∧(B∨C))∨(D∧E)¬(¬A∧(B∨C)∨(D∧E))
00000001001
00001001001
00010001001
00011011010
00100101110
00101101110
00110101110
00111111110
01000101110
01001101110
01010101110
01011111110
01100101110
01101101110
01110101110
01111111110
10000000001
10001000001
10010000001
10011010010
10100100001
10101100001
10110100001
10111110010
11000100001
11001100001
11010100001
11011110010
11100100001
11101100001
11110100001
11111110010

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABCDEF
000001
000011
000101
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100110
101001
101011
101101
101110
110001
110011
110101
110110
111001
111011
111101
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
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABCDEF
000001
000011
000101
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100110
101001
101011
101101
101110
110001
110011
110101
110110
111001
111011
111101
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) ∧ (¬A∨¬B∨¬C∨¬D∨¬E)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABCDEFж
000001
000011
000101
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100110
101001
101011
101101
101110
110001
110011
110101
110110
111001
111011
111101
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) = 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 = 0 => С01000 = 1 ⊕ 0 = 1
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 1 ⊕ 0 = 1
Fж(00010) = С00000 ⊕ С00010 = 1 => С00010 = 1 ⊕ 1 = 0
Fж(00001) = С00000 ⊕ С00001 = 1 => С00001 = 1 ⊕ 1 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 1 => С11000 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 1 => С10010 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 1 => С11010 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 1 => С11110 = 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 0 => С10111 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 0 => С01111 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 => С11111 = 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1

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

Околостуденческое

Рейтинг@Mail.ru

© 2009-2021, Список Литературы