Таблица истинности для функции (¬X1∨X2∨¬X3)∧(¬X4∧X5):


Промежуточные таблицы истинности:
¬X1:
X1¬X1
01
10

¬X3:
X3¬X3
01
10

(¬X1)∨X2:
X1X2¬X1(¬X1)∨X2
0011
0111
1000
1101

((¬X1)∨X2)∨(¬X3):
X1X2X3¬X1(¬X1)∨X2¬X3((¬X1)∨X2)∨(¬X3)
0001111
0011101
0101111
0111101
1000011
1010000
1100111
1110101

¬X4:
X4¬X4
01
10

(¬X4)∧X5:
X4X5¬X4(¬X4)∧X5
0010
0111
1000
1100

(((¬X1)∨X2)∨(¬X3))∧((¬X4)∧X5):
X1X2X3X4X5¬X1(¬X1)∨X2¬X3((¬X1)∨X2)∨(¬X3)¬X4(¬X4)∧X5(((¬X1)∨X2)∨(¬X3))∧((¬X4)∧X5)
000001111100
000011111111
000101111000
000111111000
001001101100
001011101111
001101101000
001111101000
010001111100
010011111111
010101111000
010111111000
011001101100
011011101111
011101101000
011111101000
100000011100
100010011111
100100011000
100110011000
101000000100
101010000110
101100000000
101110000000
110000111100
110010111111
110100111000
110110111000
111000101100
111010101111
111100101000
111110101000

Общая таблица истинности:

X1X2X3X4X5¬X1¬X3(¬X1)∨X2((¬X1)∨X2)∨(¬X3)¬X4(¬X4)∧X5(¬X1∨X2∨¬X3)∧(¬X4∧X5)
000001111100
000011111111
000101111000
000111111000
001001011100
001011011111
001101011000
001111011000
010001111100
010011111111
010101111000
010111111000
011001011100
011011011111
011101011000
011111011000
100000101100
100010101111
100100101000
100110101000
101000000100
101010000110
101100000000
101110000000
110000111100
110010111111
110100111000
110110111000
111000011100
111010011111
111100011000
111110011000

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
X1X2X3X4X5F
000000
000011
000100
000110
001000
001011
001100
001110
010000
010011
010100
010110
011000
011011
011100
011110
100000
100011
100100
100110
101000
101010
101100
101110
110000
110011
110100
110110
111000
111011
111100
111110
Fсднф = ¬X1∧¬X2∧¬X3∧¬X4∧X5 ∨ ¬X1∧¬X2∧X3∧¬X4∧X5 ∨ ¬X1∧X2∧¬X3∧¬X4∧X5 ∨ ¬X1∧X2∧X3∧¬X4∧X5 ∨ X1∧¬X2∧¬X3∧¬X4∧X5 ∨ X1∧X2∧¬X3∧¬X4∧X5 ∨ X1∧X2∧X3∧¬X4∧X5
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
X1X2X3X4X5F
000000
000011
000100
000110
001000
001011
001100
001110
010000
010011
010100
010110
011000
011011
011100
011110
100000
100011
100100
100110
101000
101010
101100
101110
110000
110011
110100
110110
111000
111011
111100
111110
Fскнф = (X1∨X2∨X3∨X4∨X5) ∧ (X1∨X2∨X3∨¬X4∨X5) ∧ (X1∨X2∨X3∨¬X4∨¬X5) ∧ (X1∨X2∨¬X3∨X4∨X5) ∧ (X1∨X2∨¬X3∨¬X4∨X5) ∧ (X1∨X2∨¬X3∨¬X4∨¬X5) ∧ (X1∨¬X2∨X3∨X4∨X5) ∧ (X1∨¬X2∨X3∨¬X4∨X5) ∧ (X1∨¬X2∨X3∨¬X4∨¬X5) ∧ (X1∨¬X2∨¬X3∨X4∨X5) ∧ (X1∨¬X2∨¬X3∨¬X4∨X5) ∧ (X1∨¬X2∨¬X3∨¬X4∨¬X5) ∧ (¬X1∨X2∨X3∨X4∨X5) ∧ (¬X1∨X2∨X3∨¬X4∨X5) ∧ (¬X1∨X2∨X3∨¬X4∨¬X5) ∧ (¬X1∨X2∨¬X3∨X4∨X5) ∧ (¬X1∨X2∨¬X3∨X4∨¬X5) ∧ (¬X1∨X2∨¬X3∨¬X4∨X5) ∧ (¬X1∨X2∨¬X3∨¬X4∨¬X5) ∧ (¬X1∨¬X2∨X3∨X4∨X5) ∧ (¬X1∨¬X2∨X3∨¬X4∨X5) ∧ (¬X1∨¬X2∨X3∨¬X4∨¬X5) ∧ (¬X1∨¬X2∨¬X3∨X4∨X5) ∧ (¬X1∨¬X2∨¬X3∨¬X4∨X5) ∧ (¬X1∨¬X2∨¬X3∨¬X4∨¬X5)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
X1X2X3X4X5Fж
000000
000011
000100
000110
001000
001011
001100
001110
010000
010011
010100
010110
011000
011011
011100
011110
100000
100011
100100
100110
101000
101010
101100
101110
110000
110011
110100
110110
111000
111011
111100
111110

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧X1 ⊕ C01000∧X2 ⊕ C00100∧X3 ⊕ C00010∧X4 ⊕ C00001∧X5 ⊕ C11000∧X1∧X2 ⊕ C10100∧X1∧X3 ⊕ C10010∧X1∧X4 ⊕ C10001∧X1∧X5 ⊕ C01100∧X2∧X3 ⊕ C01010∧X2∧X4 ⊕ C01001∧X2∧X5 ⊕ C00110∧X3∧X4 ⊕ C00101∧X3∧X5 ⊕ C00011∧X4∧X5 ⊕ C11100∧X1∧X2∧X3 ⊕ C11010∧X1∧X2∧X4 ⊕ C11001∧X1∧X2∧X5 ⊕ C10110∧X1∧X3∧X4 ⊕ C10101∧X1∧X3∧X5 ⊕ C10011∧X1∧X4∧X5 ⊕ C01110∧X2∧X3∧X4 ⊕ C01101∧X2∧X3∧X5 ⊕ C01011∧X2∧X4∧X5 ⊕ C00111∧X3∧X4∧X5 ⊕ C11110∧X1∧X2∧X3∧X4 ⊕ C11101∧X1∧X2∧X3∧X5 ⊕ C11011∧X1∧X2∧X4∧X5 ⊕ C10111∧X1∧X3∧X4∧X5 ⊕ C01111∧X2∧X3∧X4∧X5 ⊕ C11111∧X1∧X2∧X3∧X4∧X5

Так как Fж(00000) = 0, то С00000 = 0.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
Fж(10000) = С00000 ⊕ С10000 = 0 => С10000 = 0 ⊕ 0 = 0
Fж(01000) = С00000 ⊕ С01000 = 0 => С01000 = 0 ⊕ 0 = 0
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 0 ⊕ 0 = 0
Fж(00010) = С00000 ⊕ С00010 = 0 => С00010 = 0 ⊕ 0 = 0
Fж(00001) = С00000 ⊕ С00001 = 1 => С00001 = 0 ⊕ 1 = 1
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 0 => С11000 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 1 => С01001 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 1 => С00101 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 0 => С11100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 1 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 0 ⊕ 0 ⊕ 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 = 1 => С11101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 1
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 0 => С01111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 0 => С11111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1

Таким образом, полином Жегалкина будет равен:
Fж = X5 ⊕ X4∧X5 ⊕ X1∧X3∧X5 ⊕ X1∧X2∧X3∧X5 ⊕ X1∧X3∧X4∧X5 ⊕ X1∧X2∧X3∧X4∧X5
Логическая схема, соответствующая полиному Жегалкина:

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