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I've been having problems simplifying Boolean expressions with Boolean algebra and Karnaugh maps, I tried to simplify the same Boolean Expression using both ways but the two answers I got were different. Can that happen? After any Boolean expression simplification whether we use Karnaugh maps or Boolean algebra, the answers should be same, right? Or can the answers be different?

I tried simplifying the expression $A'B'C'+A'B'C+A'BC'+ABC'+ABC$. With boolean algebra I got $A'B'C+AB$, and with Karnaugh maps I got $A' + B$.

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  • $\begingroup$ How many variables were present? $\endgroup$ – user1952500 Apr 26 at 3:25
  • $\begingroup$ 3 variables. ABC only $\endgroup$ – Bachi Nirosh Apr 26 at 3:26
  • $\begingroup$ It might help if you gave a specific example here. What was the expression you attempted to simplify, and what were the results? $\endgroup$ – Dan Uznanski Apr 26 at 3:26
  • $\begingroup$ A'B'C'+A'B'C+A'BC'+ABC'+ABC $\endgroup$ – Bachi Nirosh Apr 26 at 3:30
  • $\begingroup$ Kmap result was A'+B $\endgroup$ – Bachi Nirosh Apr 26 at 3:31
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Boolean algebra and Karnaugh map simplification should give the same or very similar answers, though without some creativity the boolean algebra one may end up not getting quite as simple an answer because things that can be used to simplify some parts may have been absorbed by other parts.

Some expressions have multiple simplified forms that are equivalent! In the specific case of $A'B'C' + A'B'C + A'BC'+ABC' + ABC$, there are two equivalent fully simplified forms:

  • $AB + A'B' + A'C'$
  • $AB + A'B' + BC'$

and a small variety of slightly less simplified forms (note the additional symbol on the last term):

  • $AB + A'B' + A'BC$
  • $AB + A'C' + A'B'C$
  • $A'B' + BC' + ABC$

Unfortunately, neither of your two proposed solutions shown in the comments are correct, as can be seen by simply counting 'on' bits: $A'+B$ has $6$ on bits, and $A'B'C+AB$ has $3$, where the original has $5$. I'm not sure how you got either of those, unfortunately: I can't find a place in my own workings where an error would create these.

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  • $\begingroup$ thank you so much $\endgroup$ – Bachi Nirosh Apr 26 at 4:10

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