Question

Reduce the following Boolean expression using k-map.

F (a, b, c) = a'b'c'+ a'bc'+ a 'bc' + a'bc + ab'c' + abc' to SOP form

Answer 1

Given: F (a, b, c) = a'b'c'+ a'bc'+ a 'bc' + a'bc + ab'c' + abc'

Ignoring the repeated term a′bc′,

F = Σm(0, 2, 3, 4, 6)

K-map groups

• Group m(0, 2, 4, 6) ⇒ c'
• Group m(2, 3) ⇒ a′b

Therefore, F(a, b, c) = c' + a'b