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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} $
asked in Digital Logic by Veteran (367k points)
edited by | 1.4k views

5 Answers

+25 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.

answered by Junior (617 points)
edited by
+9 votes

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

answer = option B

answered by Boss (30.9k points)
+1 vote
2 quads are getting formed which is b'd'+c'b'
answered by Boss (14.4k points)
+1 vote

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

so ans is B

answered by (29 points)
+1 vote
b'd' + b'c' option B
answered by Loyal (7.3k points)
Answer:

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