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

[momentum]

Here is the minimize the function using K-Map given in my book

I have marked in red . Is that correct in book ? It seems wrong to me.Here is my solution.

Who is correct ?

thanks

 

[tnich]

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.

 

[momentum]

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.

 

[tnich]

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.

 

[Jody]

I see it. It's circled even though it's enclosed in one of the groups of four probably to emphasize it.

 

[momentum]

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 ?

But $\bar B \bar C \bar D + \bar B \bar C \bar D\bar E = \bar B \bar C \bar D$

How ?

 

[momentum]

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.

 

[tnich]

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

Circled in blue

 

[tnich]

How ?

$\bar B \bar C \bar D+\bar B \bar C \bar D\bar E=\bar B \bar C \bar D(1+\bar E)=\bar B \bar C \bar D(1)=\bar B \bar C \bar D$

 

[momentum]

$\bar B \bar C \bar D+\bar B \bar C \bar D\bar E=\bar B \bar C \bar D(1+\bar E)=\bar B \bar C \bar D(1)=\bar B \bar C \bar D$

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 ?

 

[Tom.G]

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.

 

[tnich]

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.