Please explain how are you being confident of the option (a)?

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+5 votes

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$a\bar d + \bar a \bar c+b \bar c d = \overset{m_8}{a\bar b \bar c \bar d} + \overset{m_{10}}{a\bar b c \bar d} +\overset{m_{12}}{a b \bar c \bar d} + \overset{m_{14}}{a b c \bar d} $

$\qquad + \overset{m_0}{\bar a \bar b \bar c \bar d} + \overset{m_4}{\bar a b \bar c \bar d}+ \overset{m_1}{\bar a \bar b \bar c d}+ \overset{m_5}{\bar a b \bar c d}$

$\qquad + \overset{m_5}{\bar ab \bar c d}+ \overset{m_{13}}{ab\bar c d}$

When we minimize a K-map, we can assume either $0$ or $1$ for don't cares. But here they have asked for the expression represented by the K-map. So we can consider $X$ as $1$ and not as a don't care. Also the given expression is equivalent to the above K-map but not the minimal one. Minimal expression will be $\bar a \bar c + b\bar c+ a\bar d.$

Hence, Option A.

+10 votes

0

But we can still derive this expression using (c) and (d) as well right? They didn't ask for a minimal don't care set of K-map or something.

Please explain how are you being confident of the option (a)?

Please explain how are you being confident of the option (a)?

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