Hi,
Sorry for sending this email if you are busy. But I need to find answers to the questions below to make further progress in my work. So please let me know any answers you have for these.
Thanks and Warm Regards,
Stephan
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How do I (easily/efficiently) compute the CNF/POS [Conjunctive Normal
Form / Product-of-Sum] Form of a given boolean expression using the
ESPRESSO package [which is a public domain software by UC-Berkeley that
does logic minimization]? [The default minimization Espresso uses, is
the DNF/SOP representation]
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Also, lets say I had the following multi-valued minimization problem:
Say we have variables B1, B2, C1, C2, C3, D1, D2.
Some of the assumptions are:
All like variables (those having the same letters) ... say B1,
B2 belong to one group of universe, called B; similarly do C1, C2 and
C3 belong to C; and the variables D1, D2 belong to D.
This entails that:
B1 + B2 = 1
C1 + C2 + C3 = 1
D1 + D2 =1
(Note: The symbol "+" stands for logical OR and the symbol "." stands
for logical AND and "1" stands for "TRUE")
Lets say we have a function f where
f = B2.C1 + B2.C2 + B2.C3.D2 + C3.D1
= B2.C1 + B2.C2 + B2.C3.D2 + 1.C3.D1
= B2.C1 + B2.C2 + B2.C3.D2 + (B1+B2).C3.D1
= B2.C1 + B2.C2 + B2.C3.D2 + B1.C3.D1 + B2.C3.D1
= B2.C1 + B2.C2 + B2.C3.D2 + B2.C3.D1 + B1.C3.D1
= B2.C1 + B2.C2 + B2.C3.(D2 + D1) + B1.C3.D1
= B2.C1 + B2.C2 + B2.C3.1 + B1.C3.D1
= B2.C1 + B2.C2 + B2.C3 + B1.C3.D1
= B2.(C1 + C2 + C3) + B1.C3.D1
= B2 + B1.C3.D1
which is more minimal than the original expression.
How do I perform such multi-valued minimizations using the Espresso package?
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Any suggestions will be greatly appreciated,
Stephan
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