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What is the minimal form of the Karnaugh map shown below? Assume that $X$ denotes a don’t care term

 

  1. $\bar{b} \bar{d}$
  2. $ \bar { b } \bar { d } + \bar{b} \bar{c} $
  3. $ \bar{b} \bar{d} + {a} \bar{b} \bar{c} {d}$
  4. $ \bar{b} \bar{d} + \bar{b} \bar{c} + \bar{c} \bar{d} $
in Digital Logic by Veteran (431k points)
edited by | 1.9k views

5 Answers

+28 votes
Best answer

$2$ quads are getting formed.
Value for First one is $b'd'$ and value for $2^{nd}$ one is $b'c'$. So, answer is option B.

by Junior (611 points)
edited by
+1
and those fully cover all minterm and they are also the essential PI.
+13 votes

Care for $1$'s; not for don't cares

answer = option B

by Boss (30.8k points)
+1 vote
2 quads are getting formed which is b'd'+c'b'
by Boss (14.4k points)
+1 vote

In dont cares we prioritize as...octacts< quads< dia

so ans is B

by (29 points)
0
Can explain more then it would easy to understand..
+1 vote
b'd' + b'c' option B
by Boss (11.7k points)
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