Таблица истинности для функции F≡A⊕(N∧V∧B)∧V∧B⊕A:


Промежуточные таблицы истинности:
N∧V:
NVN∧V
000
010
100
111

(N∧V)∧B:
NVBN∧V(N∧V)∧B
00000
00100
01000
01100
10000
10100
11010
11111

((N∧V)∧B)∧V:
NVBN∧V(N∧V)∧B((N∧V)∧B)∧V
000000
001000
010000
011000
100000
101000
110100
111111

(((N∧V)∧B)∧V)∧B:
NVBN∧V(N∧V)∧B((N∧V)∧B)∧V(((N∧V)∧B)∧V)∧B
0000000
0010000
0100000
0110000
1000000
1010000
1101000
1111111

A⊕((((N∧V)∧B)∧V)∧B):
ANVBN∧V(N∧V)∧B((N∧V)∧B)∧V(((N∧V)∧B)∧V)∧BA⊕((((N∧V)∧B)∧V)∧B)
000000000
000100000
001000000
001100000
010000000
010100000
011010000
011111111
100000001
100100001
101000001
101100001
110000001
110100001
111010001
111111110

(A⊕((((N∧V)∧B)∧V)∧B))⊕A:
ANVBN∧V(N∧V)∧B((N∧V)∧B)∧V(((N∧V)∧B)∧V)∧BA⊕((((N∧V)∧B)∧V)∧B)(A⊕((((N∧V)∧B)∧V)∧B))⊕A
0000000000
0001000000
0010000000
0011000000
0100000000
0101000000
0110100000
0111111111
1000000010
1001000010
1010000010
1011000010
1100000010
1101000010
1110100010
1111111101

F≡((A⊕((((N∧V)∧B)∧V)∧B))⊕A):
FANVBN∧V(N∧V)∧B((N∧V)∧B)∧V(((N∧V)∧B)∧V)∧BA⊕((((N∧V)∧B)∧V)∧B)(A⊕((((N∧V)∧B)∧V)∧B))⊕AF≡((A⊕((((N∧V)∧B)∧V)∧B))⊕A)
000000000001
000010000001
000100000001
000110000001
001000000001
001010000001
001101000001
001111111110
010000000101
010010000101
010100000101
010110000101
011000000101
011010000101
011101000101
011111111010
100000000000
100010000000
100100000000
100110000000
101000000000
101010000000
101101000000
101111111111
110000000100
110010000100
110100000100
110110000100
111000000100
111010000100
111101000100
111111111011

Общая таблица истинности:

FANVBN∧V(N∧V)∧B((N∧V)∧B)∧V(((N∧V)∧B)∧V)∧BA⊕((((N∧V)∧B)∧V)∧B)(A⊕((((N∧V)∧B)∧V)∧B))⊕AF≡A⊕(N∧V∧B)∧V∧B⊕A
000000000001
000010000001
000100000001
000110000001
001000000001
001010000001
001101000001
001111111110
010000000101
010010000101
010100000101
010110000101
011000000101
011010000101
011101000101
011111111010
100000000000
100010000000
100100000000
100110000000
101000000000
101010000000
101101000000
101111111111
110000000100
110010000100
110100000100
110110000100
111000000100
111010000100
111101000100
111111111011

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FANVBF
000001
000011
000101
000111
001001
001011
001101
001110
010001
010011
010101
010111
011001
011011
011101
011110
100000
100010
100100
100110
101000
101010
101100
101111
110000
110010
110100
110110
111000
111010
111100
111111
Fсднф = ¬F∧¬A∧¬N∧¬V∧¬B ∨ ¬F∧¬A∧¬N∧¬V∧B ∨ ¬F∧¬A∧¬N∧V∧¬B ∨ ¬F∧¬A∧¬N∧V∧B ∨ ¬F∧¬A∧N∧¬V∧¬B ∨ ¬F∧¬A∧N∧¬V∧B ∨ ¬F∧¬A∧N∧V∧¬B ∨ ¬F∧A∧¬N∧¬V∧¬B ∨ ¬F∧A∧¬N∧¬V∧B ∨ ¬F∧A∧¬N∧V∧¬B ∨ ¬F∧A∧¬N∧V∧B ∨ ¬F∧A∧N∧¬V∧¬B ∨ ¬F∧A∧N∧¬V∧B ∨ ¬F∧A∧N∧V∧¬B ∨ F∧¬A∧N∧V∧B ∨ F∧A∧N∧V∧B
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FANVBF
000001
000011
000101
000111
001001
001011
001101
001110
010001
010011
010101
010111
011001
011011
011101
011110
100000
100010
100100
100110
101000
101010
101100
101111
110000
110010
110100
110110
111000
111010
111100
111111
Fскнф = (F∨A∨¬N∨¬V∨¬B) ∧ (F∨¬A∨¬N∨¬V∨¬B) ∧ (¬F∨A∨N∨V∨B) ∧ (¬F∨A∨N∨V∨¬B) ∧ (¬F∨A∨N∨¬V∨B) ∧ (¬F∨A∨N∨¬V∨¬B) ∧ (¬F∨A∨¬N∨V∨B) ∧ (¬F∨A∨¬N∨V∨¬B) ∧ (¬F∨A∨¬N∨¬V∨B) ∧ (¬F∨¬A∨N∨V∨B) ∧ (¬F∨¬A∨N∨V∨¬B) ∧ (¬F∨¬A∨N∨¬V∨B) ∧ (¬F∨¬A∨N∨¬V∨¬B) ∧ (¬F∨¬A∨¬N∨V∨B) ∧ (¬F∨¬A∨¬N∨V∨¬B) ∧ (¬F∨¬A∨¬N∨¬V∨B)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FANVBFж
000001
000011
000101
000111
001001
001011
001101
001110
010001
010011
010101
010111
011001
011011
011101
011110
100000
100010
100100
100110
101000
101010
101100
101111
110000
110010
110100
110110
111000
111010
111100
111111

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

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 = 0 => С00111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
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 = 1 => С10111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 ⊕ 1 = 0

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

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