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The function represented by the Karnaugh map is given below is

1. $A.B$
2. $AB+BC+CA$
3. $\overline{B \oplus C}$
4. $A.BC$
edited | 1.9k views
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should the K-map have BC as 00, 01, 11, 10?
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yes. It must be a typo.
+5
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) '
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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

BC + B'C' + A'B' or BC + B'C' + A'C
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 .
edited by
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what about minterm formed by postion m0 and m1 ie A'B'
+2
This is typo minterm m1 is 0 not 1
+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?
+1

@Gurjot_Singh

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

+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)

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