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Using map, simplify the following expression, using sum-of-products form :

(a) A’B’C’+ AB’C’+ ABC + A’B’C

(b) ABCD + A’B’C’D + A’BCD + A’B’CD + ABCD

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(a) A’B’C’+ AB’C’+ ABC + A’B’C

Mapping the given function in a K-map, we get 2 Pairs i.e., Pair-1 is m0 + m1 and Pair-2 is m0 + m4

= A’B’C’ + A’B’C + A’B’C’ + AB’C’ + ABC

= A’B’(C’ + C) + B’C’(A’ + A) + ABC

= A’B’ + B’C’ + ABC 

Simplified Boolean expression as sum of products for given K-map is F(A, B, C) = A’B’ + B’C’ + ABC.

(b)  ABCD + A’B’C’D + A’BCD + A’B’CD + ABCD

ABCD = 0011 = A’B’CD = m3 and ABCD = 0110 = 

A’BCD’ = m6 ABCD = 1001 = AB’C’D = m9 and ABCD = 1110 = ABCD’ = m14

Mapping the given outputs in a K-map, we get 1 Pairs i.e., m5 + m14

You notice that there are a single 1 in a m3 and m9 because it have no adjacent 1 so it is not possible to make a pair.

= A’B’CD + A’BCD’ + ABCD’ + AB’C’D

= A’B’CD + BCD’(A’ + A) + AB’C’D’

= A’B’CD + BCD’ + AB’C’D

Simplified Boolean expression as sum of products for given K-map is F(A, B, C, D) = A’B’CD + BCD’ + AB’C’D

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