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There is a question:

What is the number of canonical expressions that can be developed over a 3-valued boolean algebra?

I was trying to solve this. Canonical expression is the combination of minterms. In a 3 valued boolean algebra, the number of minterms is 2 to the power of 3 i.e 8.. so there are 8 minterms. But how do i proceed further? is it using permutation ?

draks ...
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  • How do you define two canonical expressions as different? So, I've got minterms like $a=1 \wedge b=u$-assumably this doesn't count as different from $(a=1 \wedge b=u) \vee (a=1 \wedge b=u)$. But is it different from $b=u \wedge a=1$, or do you want only the canonical expressions that define different functions on their variables? – Kevin Carlson Sep 22 '12 at 08:16

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