GATE CSE 2008 | Question: 5

Question

In the Karnaugh map shown below, $X$ denotes a don’t care term. What is the minimal form of the function represented by the Karnaugh map?

• A. $\bar{b}.\bar{d} + \bar{a}.\bar{d}$

• B. $\bar{a}.\bar{b} + \bar{b}.\bar{d} + \bar{a}.b.\bar{d}$

• C. $\bar{b}.\bar{d} + \bar{a}.b.\bar{d}$

• D. $\bar{a}.\bar{b} + \bar{b}.\bar{d} + \bar{a}.\bar{d}$

Comments

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Only two square with size 4

Option A is correct

Answer 1

$2$ quads are getting formed:                                                            

Value for first one is $a'd'$ and value for 2$^{\text{nd}}$ one is $b'd'.$

Answer is Option A.

Comments

Also, only 2 essential PI are formed and they cover all the minterms.The size of each EPI is of $2^2$ so final minimum expression must have exactly 2 terms and in each term you'll have only $4-2=2$ variables.Only (A) satisfies.

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we can pair top right 1 with top left 1 what is the need to consider dont care for making quad, the minimal expression would be a'd'+c'd'b'

Answer 2

m0 ,m1,m8 ,m9 form one quad a'd'

m0,m2,m8,m10 form one quad b'd'

So f=a'd'+b'd'

Ans is a

Comments

hi, why we are not taking m0,m4,m12,m8 means with dont care

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Yeah how to know when to take don't cares or not to take them?

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use don't care only when you can form a bigger group with the help of don't cares.

Answer 3

X denotes a don’t care term ,means we can take them whenever necessary.Here, if we take 1 don't care,it is sufficient (m10)

m0 ,m1,m8 ,m9 -> a'd'

m0,m2,m8,m10 -> b'd'

So, Option A is correct

Answer 4

Answer – A ( Easiest Question in 2008).

If you know How to apply K-Map

Answer 5

Answer is Option A 

 

 

m0,m3,m8,m10 gives B'D'
 

m0,m4,m3,m6 gives A'D'

therefore the minimum SOP would be B'D'+A'D'