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I have this function:

$$f(A,B,C,D) = A'B' + CD' + ABC + A'B'CD' + ABCD'$$

I used a Karnaugh map to minimize the function to:

$$Minimum SOP = A'B' + C D' + A B C$$

How can I turn this into a product-of-sums?

My attempt was to first find the inverse:

$$(A'B' + C D' + A B C)'$$ $$= (A'B')'(C D')'(A B C)'$$ $$= (A + B)(C' + D)(A' + B' + C')$$ $$= (AC' + AD + BC' + CD)(A' + B' + C')$$ $$= A'(AC' + AD + BC') + B'(AC' + AD + BC') + C'(AC' + AD + BC')$$

I am stuck here because I see that I end up with this:

$$= A'AC' + A'AD + A'BC' + ...$$

How can this be correct?

Doesn't $A'A$ mean $A \wedge \neg A$ (which doesn't even make sense to me)?

If I made a mistake, where did I make it?

If I'm on the right track, what is my next step?

Søren
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1 Answers1

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$$\begin{array}{r|c|c|c|c} CD|AB&00&01&11&10 \\\hline 00&1&0&0&0\\\hline 01&1&0&0&0\\\hline 11&1&0&1&0\\\hline 10&1&1&1&1 \end{array}$$

From this K-map, those $0$'s (the negative of $f$) can be written as a sum of product:

$$[f(A, B, C, D)]' = BC' + AC' + A'BD + AB'D$$

Then, using de Morgan's laws, the negative of the whole formula is

$$\begin{align*} f(A,B,C,D) =& (BC' + AC' + A'BD + AB'D)'\\ =&(BC')'(AC')'(A'BD)'(AB'D)'\\ =&(B'+C)(A'+C)(A+B'+D')(A'+B+D') \end{align*}$$

In short, you should focus on the $0$'s when finding a product of sum form.

peterwhy
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  • How did you group the zeros? Two groups of 4? – Søren Dec 01 '13 at 09:59
  • Also, is there a way to turn my minimum SOP into POS, similar to the steps I wrote in my posted question? – Søren Dec 01 '13 at 09:59
  • First question: two groups of 4, two groups of 2. Second question: to continue with your steps above, eliminate those terms with $AA'$ (e.g. $A'AC$), since $AA' = 0$ alone and a further $0C = 0$. But still, it is difficult to be sure your terms after grouping are minimum. – peterwhy Dec 01 '13 at 10:04
  • By the way, the expansion in your question is missing the $CD$ term on the fifth/last line. – peterwhy Dec 01 '13 at 10:16