# Minimizing(optimizing) digital logic circuit with multiplexer(s)

(this by the way was an exam question)

For $F = \sum m(2, 3, 5, 7, 11, 13)$, I am trying to implement a digital logic circuit using only the following:

• 2:1 MUX (cost: 12, and must use at least one 2:1 MUX)
• 2-input AND gate (cost: 6)
• 2-input OR gate (cost: 6)
• Inverter (cost: 2)

It obviously is pretty easy to implement it without any restrictions -- that is, if I don't care about the cost at all. But if I were to minimize the cost, how should I go about it?

I first simplified $F$ using the K-map of it, and I got two representations of $F$, $A$ being the MSB:

• $F = BC'D + A'B'C + B'CD + A'CD$
• $F = BC'D + A'B'C + B'CD + A'BD$.

The second representation just looked nice to me because I can group all the product terms by $B$ like this: $F = BD(A'+C') + B'C(A'+D)$. And apparently my design is indeed the implementation with the least cost -- I got:

But the way I solved it cannot be used for other expressions in general, and I honestly think I just got lucky. Is there a better way to solve this problem, that is, a general way that guarantees (or at least checks) that the implementation is that of minimal cost?

• Interesting question – MujjinGun Jan 2 '18 at 17:02
• Only way I know is to brute force it. Find any solution so you have an upper bound. Then go through all possible circuits with less cost and see if they have the right truth table. Output the least expensive. – Goswin von Brederlow Jan 30 '18 at 13:13
• How you will get two expressions ? The first expression is the correct one. Not the second – Mitu Raj May 11 '18 at 10:50