How does a circuit for a 2-level majority with 16 inputs (four groups of 4)?

Question

this is for my compute class and ive have no idea how the thing look like. i have been searching on yahoo for a long time. i found a picture that looks similar to what is asked but i dont think it is right. can some one help me or explain to me how it looks like?

Can you design a circuit for a 2-level majority with 16 inputs (four groups of 4)? That is, the output is asserted only if a majority of the groups (at least 3) have a majority among their members. This is somewhat like the Electoral College, except that the groups (states) have different weights. It also illustrates one of the most important principles of engineering: if you have something already designed that fits the spec, use it. Of course, it regularly runs in competition with the NIH principle (what is the NIH principle? It's not about the National Institutes of Health)

Answer 1

Q = AB(C + D) + CD(A + B)
Use four to accommodate the 16 inputs and a fifth to accommodate the outputs of the first four. Each segment requires (4) 2-input ANDs and (3) 2-input ORs.

You can work it out to use entirely NANDs or NORs, but that's no longer necessary with today's ICs.

Answer 2

Well, The "NIH principle" is a typical occurrence amongst scientists and engineers: It means "Not Invented Here" or in other words, if *I* didn't have this idea, then it MUST be wrong....!
Whether I understand the actual question is another thing.
I visualize a logic circuit with 4 NAND gates, each having 4 inputs (that represents the 16 inputs in 4 groups).
Now you need a circuit that takes the 4 outputs of those NAND gates and only switches if 3 of them are "high". That could be done by a current summing amplifier, like this one
http://www.elektropage.com/Opamps/?id=11
You dimension the resistor in such a way that the Op-Amp only switches through if any 3 (or all 4) inputs are active.