Trouble with K-Map Minimizing Function: Who is Correct?

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SUMMARY

The forum discussion centers on minimizing Boolean functions using Karnaugh Maps (K-Maps). Participants debate the correctness of a solution presented in a textbook versus a user's alternative approach. The consensus is that the textbook solution is more efficient due to fewer AND operations, specifically highlighting that the expression ##\bar B \bar C D\bar E + \bar B \bar C \bar D\bar E## simplifies to ##\bar B \bar C \bar E##. Users emphasize the importance of maximizing 1's in K-Maps for optimal solutions.

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  • Understanding of Boolean algebra and simplification techniques
  • Familiarity with Karnaugh Maps (K-Maps) for function minimization
  • Knowledge of logical operators and their representations (e.g., AND, OR, NOT)
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Students and professionals in electrical engineering, computer science, and digital logic design who are looking to enhance their understanding of Boolean function minimization and K-Map techniques.

momentum
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Here is the minimize the function using K-Map given in my book
nzDkd5R.png
I have marked in red . Is that correct in book ? It seems wrong to me.Here is my solution.

R7m7vYh.png


Who is correct ?

thanks
 
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Your result is equivalent to the solution in the book since ##\bar B \bar C D\bar E + \bar B \bar C \bar D\bar E = \bar B \bar C \bar E##. I would choose the book's solution over yours since it has one less AND operation.
 
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tnich said:
Your result is equivalent to the solution in the book since ##\bar B \bar C D\bar E + \bar B \bar C \bar D\bar E = \bar B \bar C \bar E##. I would choose the book's solution over yours since it has one less AND operation.
You can not add ## \bar B \bar C \bar D\bar E## because you don't have it . Check again.
 
momentum said:
You can not add ## \bar B \bar C \bar D\bar E## because you don't have it . Check again.
You must be looking at a different diagram than the one you posted. The one I see has ##1##s in the squares for ##B=C=D=E=0##. You don't seem to have any argument with the ## \bar B \bar C \bar D## term since you have it in your solution. But ## \bar B \bar C \bar D + \bar B \bar C \bar D\bar E = \bar B \bar C \bar D##, so you can add ##\bar B \bar C \bar D\bar E## to your solution to get an equivalent form.
 
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I see it. It's circled even though it's enclosed in one of the groups of four probably to emphasize it.
 
tnich said:
The one I see has ##1##s in the squares for ##B=C=D=E=0##.

I don't see it. Can you please circle it in different color in my diagram ?

tnich said:
But ## \bar B \bar C \bar D + \bar B \bar C \bar D\bar E = \bar B \bar C \bar D##

How ?
 
Joshy said:
I see it. It's circled even though it's enclosed in one of the groups of four probably to emphasize it.
could you please mark in a different color in diagram . I don't see it.
 
momentum said:
I don't see it. Can you please circle it in different color in my diagram ?
Circled in blue
Kdiagram.png
 
Last edited:
momentum said:
How ?
##\bar B \bar C \bar D+\bar B \bar C \bar D\bar E=\bar B \bar C \bar D(1+\bar E)=\bar B \bar C \bar D(1)=\bar B \bar C \bar D##
 
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tnich said:
##\bar B \bar C \bar D+\bar B \bar C \bar D\bar E=\bar B \bar C \bar D(1+\bar E)=\bar B \bar C \bar D(1)=\bar B \bar C \bar D##

Okay. got it ... that works.

But in diagram in general , we should highlight max 1's in K-Map. ...right ? we should not be using Single 1s ( blue circle which you marked) if the max 1s is available ...am I right ?

my Solution is also correct ...right ?
 
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  • #11
momentum said:
my Solution is also correct ...right ?
Functionally, yes. Maximally efficient, no. Leans a little bit in the direction of using a shotgun to kill a fly. :wink:
 
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  • #12
momentum said:
But in diagram in general , we should highlight max 1's in K-Map. ...right ? we should not be using Single 1s ( blue circle which you marked) if the max 1s is available ...am I right ?
Yes, you are right, and that has been the point all along.
$$\begin{align}&\bar D \bar E+ \bar B \bar C \bar D + && \bar ACDE + BCD+\bar B\bar CD\bar E\nonumber\\
=&\bar D \bar E+ \bar B \bar C \bar D+\bar B \bar C \bar D \bar E + && \bar ACDE + BCD+\bar B\bar CD\bar E\nonumber\\
=&\bar D \bar E+ \bar B \bar C \bar D + && \bar ACDE + BCD+\bar B\bar CD\bar E+\bar B \bar C \bar D \bar E\nonumber\\
=&\bar D \bar E+ \bar B \bar C \bar D + && \bar ACDE + BCD+\bar B\bar C\bar E\nonumber\end{align}$$
replaces ##\bar B\bar CD\bar E## with ##\bar B\bar C\bar E## giving a more efficient form.
 

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