Промежуточные таблицы истинности:A∨B:
¬A:
¬C:
(¬A)∧(¬C):
A | C | ¬A | ¬C | (¬A)∧(¬C) |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
A→K:
C∧B:
¬K:
(C∧B)∧(¬K):
C | B | K | C∧B | ¬K | (C∧B)∧(¬K) |
0 | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 |
(A∨B)∧((¬A)∧(¬C)):
A | B | C | A∨B | ¬A | ¬C | (¬A)∧(¬C) | (A∨B)∧((¬A)∧(¬C)) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
((A∨B)∧((¬A)∧(¬C)))∧(A→K):
A | B | C | K | A∨B | ¬A | ¬C | (¬A)∧(¬C) | (A∨B)∧((¬A)∧(¬C)) | A→K | ((A∨B)∧((¬A)∧(¬C)))∧(A→K) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
(((A∨B)∧((¬A)∧(¬C)))∧(A→K))∧((C∧B)∧(¬K)):
A | B | C | K | A∨B | ¬A | ¬C | (¬A)∧(¬C) | (A∨B)∧((¬A)∧(¬C)) | A→K | ((A∨B)∧((¬A)∧(¬C)))∧(A→K) | C∧B | ¬K | (C∧B)∧(¬K) | (((A∨B)∧((¬A)∧(¬C)))∧(A→K))∧((C∧B)∧(¬K)) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
C∧K:
((((A∨B)∧((¬A)∧(¬C)))∧(A→K))∧((C∧B)∧(¬K)))∧(C∧K):
A | B | C | K | A∨B | ¬A | ¬C | (¬A)∧(¬C) | (A∨B)∧((¬A)∧(¬C)) | A→K | ((A∨B)∧((¬A)∧(¬C)))∧(A→K) | C∧B | ¬K | (C∧B)∧(¬K) | (((A∨B)∧((¬A)∧(¬C)))∧(A→K))∧((C∧B)∧(¬K)) | C∧K | ((((A∨B)∧((¬A)∧(¬C)))∧(A→K))∧((C∧B)∧(¬K)))∧(C∧K) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
Общая таблица истинности:
A | B | C | K | A∨B | ¬A | ¬C | (¬A)∧(¬C) | A→K | C∧B | ¬K | (C∧B)∧(¬K) | (A∨B)∧((¬A)∧(¬C)) | ((A∨B)∧((¬A)∧(¬C)))∧(A→K) | (((A∨B)∧((¬A)∧(¬C)))∧(A→K))∧((C∧B)∧(¬K)) | C∧K | ((A∨B)∧(¬A∧¬C)∧(A→K)∧((C∧B)∧¬K))∧(C∧K) |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
A | B | C | K | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 |
В таблице истинности нет набора значений переменных при которых функция истинна!
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
A | B | C | K | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 |
F
скнф = (A∨B∨C∨K) ∧ (A∨B∨C∨¬K) ∧ (A∨B∨¬C∨K) ∧ (A∨B∨¬C∨¬K) ∧ (A∨¬B∨C∨K) ∧ (A∨¬B∨C∨¬K) ∧ (A∨¬B∨¬C∨K) ∧ (A∨¬B∨¬C∨¬K) ∧ (¬A∨B∨C∨K) ∧ (¬A∨B∨C∨¬K) ∧ (¬A∨B∨¬C∨K) ∧ (¬A∨B∨¬C∨¬K) ∧ (¬A∨¬B∨C∨K) ∧ (¬A∨¬B∨C∨¬K) ∧ (¬A∨¬B∨¬C∨K) ∧ (¬A∨¬B∨¬C∨¬K)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
A | B | C | K | Fж |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина:
F
ж = C
0000 ⊕ C
1000∧A ⊕ C
0100∧B ⊕ C
0010∧C ⊕ C
0001∧K ⊕ C
1100∧A∧B ⊕ C
1010∧A∧C ⊕ C
1001∧A∧K ⊕ C
0110∧B∧C ⊕ C
0101∧B∧K ⊕ C
0011∧C∧K ⊕ C
1110∧A∧B∧C ⊕ C
1101∧A∧B∧K ⊕ C
1011∧A∧C∧K ⊕ C
0111∧B∧C∧K ⊕ C
1111∧A∧B∧C∧K
Так как F
ж(0000) = 0, то С
0000 = 0.
Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F
ж(1000) = С
0000 ⊕ С
1000 = 0 => С
1000 = 0 ⊕ 0 = 0
F
ж(0100) = С
0000 ⊕ С
0100 = 0 => С
0100 = 0 ⊕ 0 = 0
F
ж(0010) = С
0000 ⊕ С
0010 = 0 => С
0010 = 0 ⊕ 0 = 0
F
ж(0001) = С
0000 ⊕ С
0001 = 0 => С
0001 = 0 ⊕ 0 = 0
F
ж(1100) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
1100 = 0 => С
1100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 0 => С
1010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1001) = С
0000 ⊕ С
1000 ⊕ С
0001 ⊕ С
1001 = 0 => С
1001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(0110) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0110 = 0 => С
0110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(0101) = С
0000 ⊕ С
0100 ⊕ С
0001 ⊕ С
0101 = 0 => С
0101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(0011) = С
0000 ⊕ С
0010 ⊕ С
0001 ⊕ С
0011 = 0 => С
0011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1110) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
1100 ⊕ С
1010 ⊕ С
0110 ⊕ С
1110 = 0 => С
1110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1101) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0001 ⊕ С
1100 ⊕ С
1001 ⊕ С
0101 ⊕ С
1101 = 0 => С
1101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1011) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
0001 ⊕ С
1010 ⊕ С
1001 ⊕ С
0011 ⊕ С
1011 = 0 => С
1011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(0111) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
0111 = 0 => С
0111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1111) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
1100 ⊕ С
1010 ⊕ С
1001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
1110 ⊕ С
1101 ⊕ С
1011 ⊕ С
0111 ⊕ С
1111 = 0 => С
1111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = 0