Incompletely Specified Functions (Cont.)

F(A,B,C) = A’B’C’ + A’B’C + A’BC’ + A’BC + ABC’ + ABC

However, if we assume that F(0,0,1) = 1 and

F(1,1,0) = 1, we obtain instead the equation:

= A’B’ ·1 + A’B ·1 + AB ·1

= A’B’(C’ + C) + A’B(C’ + C) + AB(C’ + C)

= A’B’ + A’B + AB

= A’B’ + A’B + A’B + AB

= A’(B’ + B) + (A’ + A)B

= A’·1 + 1·B

= A’ + B

Compare this with the other solution: F(A,B,C) = A’C’ + BC.

Which one is cheaper to implement?