GATE CSE 1998 | Question: 2.7

Question

The function represented by the Karnaugh map given below is

• A. $A.B$
• B. $AB+BC+CA$
• C. $\overline{B \oplus C}$
• D. $A.BC$

Comments

should the K-map have BC as 00, 01, 11, 10?

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yes. It must be a typo.

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it isn't typo it is tricky i would say . Just to confuse the people . if we rewrite it according to the standrard formula . asnwer is so simple as that . BUT min term m1 is 0 not 1 , this is a typo

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what trick is that?? isnt it a typo?? If we consider normal kmap with 00,01,11,10 we get answer as C. now what trick are you talking about?

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Hey I found out slight mistake in this question added, I checked original 1998 question , the minterm value at  '001' is '0' ..so by clearing that mistake I solved further question..

    Here The k-map is not given in cyclic form/gray code form. so after converting it into cyclic code format we got two essential prime implicants ie B'C' and BC.
So
  SOP(Sum of Product)=B'C'+BC
                                        =(B xOR C) '
Answer : C

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not clear....???

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sir can uh please explain the meaning of typo in this question ??

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I got

A' B' +B' C' +A' C + BC after solving the given c k-map is it correct or not can anyone clarify me ??

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A\BC	00	01	11	10
0	1	1	1	0
1	1	0	1	0

BC+B'C'+A'B' OR BC+B'C'+A'C

plz correct if anything is wrong? 

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meghna $m_1$ is $0$

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@ Mk Utkarsh I tried it as given in question considering m1=1, if m1=0 then B(XNOR)C.

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why m1 will 0? is there  mistake in question ?

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I can't understand the question is right or wrong

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Question is perfectly right but its not in the standard form which we are used to of it.

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Just a small doubt In case of 3 variables, if all the min terms are in corner, is it allowed to make a pair of 4.

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@AshishB Yes allowed.

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Complement of xor is xnor.

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@ritiksri8 Only when there are even number of boolean variables. When there are odd number of boolean variables, they are not complements of each other.

 

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Correct answer is option c✅

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Watch Detailed Solution   

Answer 1

The given K-map is not standard as after "01" we have "10" and two variables are changing for consecutive column. This means it is not safe to merge adjacent $1s.$ By converting the K-map to standard form we get

which gives

$BC + \bar B\bar C = B \text{ XNOR } C = B \odot C$

This can be represented as negation of $\text{XOR} = \overline{B \oplus C}$

Option C is correct.

Comments

what about minterm formed by postion m0 and m1 ie A'B'

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This is typo minterm m1 is 0 not 1

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why can't we make a pair of 4??? if we have all the ones then we must have taken 8 pairs. why here is a problem with 4 ones?

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@Gurjot_Singh

• Read the comments after the question
• Closely obserb the given image and compare with the standard diagram used to represent K-Map.

This would solve your doubt

Answer 2

On observation , wherever the value of B and C are equal ,
 

we find that the function value becomes 1,

this represents an XNOR gate which is the complement of XOR gate

and hence option c ,

(B XOR C)'