Boolean Algebra: Minimum Sum-Of-Products Expression

AI Thread Summary
The discussion focuses on finding the minimum Sum-Of-Products expression for the Boolean function f = ab'c' + abd + ab'cd'. The user attempts to introduce missing variables to achieve an expression that includes all variables a, b, c, and d, resulting in f = ab'c'd + ab'c'd' + abcd + abc'd + ab'cd'. After combining terms, the user arrives at f = ab'c' + ab'd' + abd, which they feel is not significantly minimized. It is clarified that while the original expression has a summand complexity of 3 and a literal complexity of 10, the minimized form retains the same summand complexity but reduces the literal complexity to 9. The discussion concludes with the acknowledgment that achieving a simpler form is challenging, and the user appreciates the confirmation of their understanding.
Ithryndil
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Homework Statement



Find the minimum Sum-Of-Product Expression for:
f = ab'c' + abd + ab'cd'

The Attempt at a Solution



By introducing the missing variable in term 1 and term 2 I can get an expression that has all the variables: a, b, c, and d.

I do so by:

f = ab'c'd + ab'c'd' + abcd + abc'd + ab'cd'

I can combine terms like so: (1 & 2),( 2 & 5), (3 & 4) I get:

f = ab'c' + ab'd' + abd

This hardly seems minimized from the original expression. Thanks for any help.
 
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What you've stated is one of two equivalent minsum forms of that Boolean expression. There are several methods for arriving at these (consensus, Karaugh maps, Quine-McCluskey). I'd examine them for more info.

The original expression has a summand complexity (SC) of 3 and a literal complexity (LC) of 10. The minsum has an SC of 3 and an LC of 9 (as does the other). It isn't much simpler but it as simple as one can get.

--Elucidus
 
Thanks, when you're learning about these concepts it is nice to have confirmation that you are doing things right. Normally it class we get the function down a term or two...or even to one term. So, when I got this down to three terms, with three variables in each term, it didn't really seem minimized. Thanks again!
 
Karnaugh product of sums answer:

a(b+d)(b+c'+d')
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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