Таблица истинности для функции F≡(A∧V∧¬B)∧V∧¬C∧B:


Промежуточные таблицы истинности:
¬B:
B¬B
01
10

A∧V:
AVA∧V
000
010
100
111

(A∧V)∧(¬B):
AVBA∧V¬B(A∧V)∧(¬B)
000010
001000
010010
011000
100010
101000
110111
111100

¬C:
C¬C
01
10

((A∧V)∧(¬B))∧V:
AVBA∧V¬B(A∧V)∧(¬B)((A∧V)∧(¬B))∧V
0000100
0010000
0100100
0110000
1000100
1010000
1101111
1111000

(((A∧V)∧(¬B))∧V)∧(¬C):
AVBCA∧V¬B(A∧V)∧(¬B)((A∧V)∧(¬B))∧V¬C(((A∧V)∧(¬B))∧V)∧(¬C)
0000010010
0001010000
0010000010
0011000000
0100010010
0101010000
0110000010
0111000000
1000010010
1001010000
1010000010
1011000000
1100111111
1101111100
1110100010
1111100000

((((A∧V)∧(¬B))∧V)∧(¬C))∧B:
нажмите на таблицу для просмотра*

F≡(((((A∧V)∧(¬B))∧V)∧(¬C))∧B):
нажмите на таблицу для просмотра*

Общая таблица истинности:

нажмите на таблицу для просмотра*

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FAVBCF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011101
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fсднф = ¬F∧¬A∧¬V∧¬B∧¬C ∨ ¬F∧¬A∧¬V∧¬B∧C ∨ ¬F∧¬A∧¬V∧B∧¬C ∨ ¬F∧¬A∧¬V∧B∧C ∨ ¬F∧¬A∧V∧¬B∧¬C ∨ ¬F∧¬A∧V∧¬B∧C ∨ ¬F∧¬A∧V∧B∧¬C ∨ ¬F∧¬A∧V∧B∧C ∨ ¬F∧A∧¬V∧¬B∧¬C ∨ ¬F∧A∧¬V∧¬B∧C ∨ ¬F∧A∧¬V∧B∧¬C ∨ ¬F∧A∧¬V∧B∧C ∨ ¬F∧A∧V∧¬B∧¬C ∨ ¬F∧A∧V∧¬B∧C ∨ ¬F∧A∧V∧B∧¬C ∨ ¬F∧A∧V∧B∧C
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FAVBCF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011101
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fскнф = (¬F∨A∨V∨B∨C) ∧ (¬F∨A∨V∨B∨¬C) ∧ (¬F∨A∨V∨¬B∨C) ∧ (¬F∨A∨V∨¬B∨¬C) ∧ (¬F∨A∨¬V∨B∨C) ∧ (¬F∨A∨¬V∨B∨¬C) ∧ (¬F∨A∨¬V∨¬B∨C) ∧ (¬F∨A∨¬V∨¬B∨¬C) ∧ (¬F∨¬A∨V∨B∨C) ∧ (¬F∨¬A∨V∨B∨¬C) ∧ (¬F∨¬A∨V∨¬B∨C) ∧ (¬F∨¬A∨V∨¬B∨¬C) ∧ (¬F∨¬A∨¬V∨B∨C) ∧ (¬F∨¬A∨¬V∨B∨¬C) ∧ (¬F∨¬A∨¬V∨¬B∨C) ∧ (¬F∨¬A∨¬V∨¬B∨¬C)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FAVBCFж
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011101
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110

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

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 = 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 ⊕ 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 = 0 => С11110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 = 0 => С11111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0

Таким образом, полином Жегалкина будет равен:
Fж = 1 ⊕ F
Логическая схема, соответствующая полиному Жегалкина:

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