Using canonical and boolean algebra

- Find a minimum cost implementation of the following function: f(x1,x2,x3)=M(0,4,5,6,7) - In this situation, cost is defined as the number of logic gates (ANDs and ORs), plus the number of gate inputs. - However, complementing thelinputs (! x1,!x2, and !x3) is a zero-cost operation, meaning do not count the NOT gate and its input for ! x1,!x2, and !x3. - For example, the cost of !x1+x2 is 3,1 OR gate plus 2 inputs, !x1 and x2. - Hint 1: You could start by finding the canonical SOP or POS (whichever is better for this situation), and then, if necessary, use Boolean algebra to simplify further