This is the Karnaugh map which I draw for F = A'B'C'+AB'C+A'BC+ABC'
But I am not sure how to simplify the Boolean expression for this as there are no groups here
Electrical – How to convert following Karnaugh map to a simplified Boolean expression
karnaugh map
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Best Answer
It cannot be further simplified with a Karnaugh map, but with a little Boolean algebra:
\$F = \bar A \bar B \bar C + A \bar B C + \bar A B C + A B \bar C\$
\$F = \bar A( \bar B \bar C + B C) + A (\bar B C + B \bar C)\$
\$F = \bar A . \overline {(B \oplus C)} + A . (B \oplus C)\$
let \$(B \oplus C) = X\$
\$F = \bar A \bar X + A X\$
\$F = \overline {(A \oplus X)}\$
now substituting X:
\$F = \overline { (A \oplus B \oplus C) }\$