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

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+10 votes

The function represented by the Karnaugh map is given below is

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

+4

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

0

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?

+1

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

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

0

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

0

+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 .

$BC + B'C'$

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

It can be represented as negation of $XOR$

$\overline{B \oplus C}$

Option C is correct .

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