Для функции F≡B∧V∧(C∧¬A)∧V∧(A∧B):


Промежуточные таблицы истинности:
¬A:
A¬A
01
10

C∧(¬A):
CA¬AC∧(¬A)
0010
0100
1011
1100

A∧B:
ABA∧B
000
010
100
111

B∧V:
BVB∧V
000
010
100
111

(B∧V)∧(C∧(¬A)):
BVCAB∧V¬AC∧(¬A)(B∧V)∧(C∧(¬A))
00000100
00010000
00100110
00110000
01000100
01010000
01100110
01110000
10000100
10010000
10100110
10110000
11001100
11011000
11101111
11111000

((B∧V)∧(C∧(¬A)))∧V:
BVCAB∧V¬AC∧(¬A)(B∧V)∧(C∧(¬A))((B∧V)∧(C∧(¬A)))∧V
000001000
000100000
001001100
001100000
010001000
010100000
011001100
011100000
100001000
100100000
101001100
101100000
110011000
110110000
111011111
111110000

(((B∧V)∧(C∧(¬A)))∧V)∧(A∧B):
BVCAB∧V¬AC∧(¬A)(B∧V)∧(C∧(¬A))((B∧V)∧(C∧(¬A)))∧VA∧B(((B∧V)∧(C∧(¬A)))∧V)∧(A∧B)
00000100000
00010000000
00100110000
00110000000
01000100000
01010000000
01100110000
01110000000
10000100000
10010000010
10100110000
10110000010
11001100000
11011000010
11101111100
11111000010

F≡((((B∧V)∧(C∧(¬A)))∧V)∧(A∧B)):
FBVCAB∧V¬AC∧(¬A)(B∧V)∧(C∧(¬A))((B∧V)∧(C∧(¬A)))∧VA∧B(((B∧V)∧(C∧(¬A)))∧V)∧(A∧B)F≡((((B∧V)∧(C∧(¬A)))∧V)∧(A∧B))
0000001000001
0000100000001
0001001100001
0001100000001
0010001000001
0010100000001
0011001100001
0011100000001
0100001000001
0100100000101
0101001100001
0101100000101
0110011000001
0110110000101
0111011111001
0111110000101
1000001000000
1000100000000
1001001100000
1001100000000
1010001000000
1010100000000
1011001100000
1011100000000
1100001000000
1100100000100
1101001100000
1101100000100
1110011000000
1110110000100
1111011111000
1111110000100

Общая таблица истинности:

FBVCA¬AC∧(¬A)A∧BB∧V(B∧V)∧(C∧(¬A))((B∧V)∧(C∧(¬A)))∧V(((B∧V)∧(C∧(¬A)))∧V)∧(A∧B)F≡B∧V∧(C∧¬A)∧V∧(A∧B)
0000010000001
0000100000001
0001011000001
0001100000001
0010010000001
0010100000001
0011011000001
0011100000001
0100010000001
0100100100001
0101011000001
0101100100001
0110010010001
0110100110001
0111011011101
0111100110001
1000010000000
1000100000000
1001011000000
1001100000000
1010010000000
1010100000000
1011011000000
1011100000000
1100010000000
1100100100000
1101011000000
1101100100000
1110010010000
1110100110000
1111011011100
1111100110000

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FBVCAF
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∧¬B∧¬V∧¬C∧¬A ∨ ¬F∧¬B∧¬V∧¬C∧A ∨ ¬F∧¬B∧¬V∧C∧¬A ∨ ¬F∧¬B∧¬V∧C∧A ∨ ¬F∧¬B∧V∧¬C∧¬A ∨ ¬F∧¬B∧V∧¬C∧A ∨ ¬F∧¬B∧V∧C∧¬A ∨ ¬F∧¬B∧V∧C∧A ∨ ¬F∧B∧¬V∧¬C∧¬A ∨ ¬F∧B∧¬V∧¬C∧A ∨ ¬F∧B∧¬V∧C∧¬A ∨ ¬F∧B∧¬V∧C∧A ∨ ¬F∧B∧V∧¬C∧¬A ∨ ¬F∧B∧V∧¬C∧A ∨ ¬F∧B∧V∧C∧¬A ∨ ¬F∧B∧V∧C∧A
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FBVCAF
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∨B∨V∨C∨A) ∧ (¬F∨B∨V∨C∨¬A) ∧ (¬F∨B∨V∨¬C∨A) ∧ (¬F∨B∨V∨¬C∨¬A) ∧ (¬F∨B∨¬V∨C∨A) ∧ (¬F∨B∨¬V∨C∨¬A) ∧ (¬F∨B∨¬V∨¬C∨A) ∧ (¬F∨B∨¬V∨¬C∨¬A) ∧ (¬F∨¬B∨V∨C∨A) ∧ (¬F∨¬B∨V∨C∨¬A) ∧ (¬F∨¬B∨V∨¬C∨A) ∧ (¬F∨¬B∨V∨¬C∨¬A) ∧ (¬F∨¬B∨¬V∨C∨A) ∧ (¬F∨¬B∨¬V∨C∨¬A) ∧ (¬F∨¬B∨¬V∨¬C∨A) ∧ (¬F∨¬B∨¬V∨¬C∨¬A)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FBVCAFж
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∧B ⊕ C00100∧V ⊕ C00010∧C ⊕ C00001∧A ⊕ C11000∧F∧B ⊕ C10100∧F∧V ⊕ C10010∧F∧C ⊕ C10001∧F∧A ⊕ C01100∧B∧V ⊕ C01010∧B∧C ⊕ C01001∧B∧A ⊕ C00110∧V∧C ⊕ C00101∧V∧A ⊕ C00011∧C∧A ⊕ C11100∧F∧B∧V ⊕ C11010∧F∧B∧C ⊕ C11001∧F∧B∧A ⊕ C10110∧F∧V∧C ⊕ C10101∧F∧V∧A ⊕ C10011∧F∧C∧A ⊕ C01110∧B∧V∧C ⊕ C01101∧B∧V∧A ⊕ C01011∧B∧C∧A ⊕ C00111∧V∧C∧A ⊕ C11110∧F∧B∧V∧C ⊕ C11101∧F∧B∧V∧A ⊕ C11011∧F∧B∧C∧A ⊕ C10111∧F∧V∧C∧A ⊕ C01111∧B∧V∧C∧A ⊕ C11111∧F∧B∧V∧C∧A

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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