lec07 производная булевой функции
.pdfПусть g = ∂f /∂xi производная функции f (x1,x2,... xi,...,xn) по переменной xi, тогда существует множество БФ {f1,f2,...,fm}, называемое булевым интегралом функции g, таких что:
g = xi g h
где h – БФ, зависящая от n-1 переменных (x1,x2,... xi−1,xi+1,...,xn).
!!! Если не оговорено особо, то для операции .
Утверждение 7.1. Любая БФ f (x1,x2,...,xn) может быть получена в результате n-кратного логического интегрирования константы {0} :
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Пример 7.10.
n-кратное интегрирование C для получения функций f(x1,x2).
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Пример 7.10.
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Интегрируя каждую из полученных 4 функций ( С 1) по x2,
получили все 16 БФ f(x1,x2). Продолжая интегрирование по следующей переменной x3 получим 256 функций f(x1,x2,x3) и т. д.
24
Интеграл от булевой функции f (x1,x2,...,xn) по переменной xn+1 для операции р(a,b)
f (x1,x2,...,xn) +1
p
есть множество булевых функций {f1,f2,...,fm}, зависящих от n+1 переменных, таких, что
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Пример 7.12. Найти интеграл от функции f(x1) = x1 для операции . Найдем для (x1,x2) все ∂f /∂x2 для операции .
1) f0 ≡ 0.
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Пример 7.12. |
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Пример 7.12. |
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Пример 7.12. |
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9) f8 = 1 |
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f8(1, 2) |
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Пример 7.12. |
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12) f11 = 1 2. |
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= |
f |
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, 0 f |
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, 1 = |
( 1) ( 0) = |
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11 |
11 |
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2 |
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1 |
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1 |
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1 |
? |
1 |
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1 |
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f12(1, 2) |
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= f |
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1, 0 f |
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1, 1 = |
= |
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13) f12 = 1. |
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2 |
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12 |
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12 |
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? |
1 |
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1 |
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1 |
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14) f |
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= |
. |
f13(1, 2) |
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= f13 1, 0 |
f13 1, 1 |
= ( 1 |
0) ( 1 1) = |
13 |
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1 |
2 |
2 |
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? |
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= 1 |
30