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+2 votes

Kindly have a try .... IN canonical POS form following equation is written as

F(A,B,C)=AB+BC+AC

(A)πM(0,1,2,4)

(B)πM(3,5,6,7)

(C)πM(0,1,2,3)

(D)πM(4,5,6,7)

+5 votes

Best answer

$F(A,B,C) = AB+BC+AC$

$= ABC' + ABC + ABC + A'BC + AB'C + ABC$

$= ABC' + ABC + A'BC + AB'C$

$=\Sigma m(6,7,3,5)$ (m - minterm and M-maxterm))

$=\pi M(0,1,2,4)$ (proof shown below)

$F(A,B,C) = \Sigma m(3,5,6,7)$

$\implies F'(A,B,C) = \Sigma m(0,1,2,4)$

$\implies F'(A,B,C)=m_0+m_1+m_2+m_4$

$\implies (F')'(A,B,C) = (m_0+m_1+m_2+m_4)'$

$\implies F(A,B,C) = (m_0' m_1' m_2' m_4)$ (DeMorgan's law)

$\implies F(A,B,C) = (M_0 M_1 M_2 M_4)$

$\implies F(A,B,C) = \pi M(0,1 2,4)$ (Complement of a minterm is its corresponding maxterm, for example, $m_0 = A'B'C'$ and $m_0' = (A'B'C')' = A+B+C = M_0$)

Ref: http://www.cs.uiuc.edu/class/sp08/cs231/lectures/04-Kmap.pdf

+2 votes

First take the function, this is in SOP so find the min terms of the equations

so min terms are 3,5,6,7

For Finding the MAX TERMS subtract these min terms from whole possible terms with 3 variable which are {0,1,2,3,4,5,6,7}- {3,5,6,7} =(0,1,2,4)

ans is option A

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