The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+11 votes

The function represented by the Karnaugh map is given below is

  1. $A.B$
  2. $AB+BC+CA$
  3. $\overline{B \oplus C}$
  4. $A.BC$
asked in Digital Logic by Veteran (59.8k points)
edited by | 2k views
should the K-map have BC as 00, 01, 11, 10?
yes. It must be a typo.
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
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?
Hey I found out slight mistake in this question added, I checked original 1998 question , the minterm value at  '001' is '0' 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.
  SOP(Sum of Product)=B'C'+BC
                                        =(B xOR C) '
Answer : C
not clear....???
sir can uh please explain the meaning of typo in this question ??
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 ??
A\BC 00 01 11 10
0 1 1 1 0
1 1 0 1 0


plz correct if anything is wrong? 


meghna $m_1$ is $0$


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

why m1 will 0? is there  mistake in question ?
I can't understand the question is right or wrong

3 Answers

+20 votes
BC + B'C' + A'B' or BC + B'C' + A'C
answered by Veteran (59.7k points)
+9 votes
Here, we cant make pair of $4$ so, we have to go with pair of $2$ each

$BC + B'C'$

$B (XNOR) C = B \odot C$

It can be represented as negation of $XOR$

$\overline{B \oplus C}$

Option C is correct .
answered by Boss (45.5k points)
edited by
what about minterm formed by postion m0 and m1 ie A'B'
This is typo minterm m1 is 0 not 1
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?


  • 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

+1 vote
Either answer is 'None of the above' or if we want (B⊕C)' then our minterm M1 should be 0.

and there is no problem with 00,01,10,11 kind boxing. All K-map tell us is , to pair only neighbours (those differ by 1 bit only)
answered by Boss (13k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,114 questions
53,223 answers
70,469 users