Simplifying Boolean.. or Electronic alternative?

Question

Hey,

On an Electronics project I'm working on, I need to light and LED if at least two of the four inputs are at a logic 1. The boolean for this would be:

(A.B)+(A.C)+(A.D)+(B.C)+(B.D)+(C.D).

Obviously I can simplify this to:

(A.(B+C+D))+(B.(C+D))+(C.D)

But still, I'm using too many gates.. Is there any way to simplify this even more?

Or, as a bonus alternative, is there a chip that takes 4 inputs and gives a different output depending on how many inputs are at a logic 1?

I also need to do this for 3 LEDs, the boolean for which would be (A.B.C)+(A.B.D)+(A.C.D)+(B.C.D)

Which simplifies to (A.((B.C)+(B.D)+(C.D)))+(B.C.D)

-> (A.((B.(C+D))+(C.D)))+(B.C.D)

Which, again, takes too many gates.

So, simplification of the two boolean statements would be wonderful, or information of a [common] chip that takes 4 inputs and gives an output dependant on the number of inputs that are at a logic 1, please!

Answer 1

Have you ever heard of karnaugh maps? I learned them in my first digital logic class in electrical engineering. They are an easy way to simply complex boolean expressions using a special table. I just used one to analyze your problem. The simplest form of your problem turns out to be:

Y = AB+AC+AD+BC+BD+CD

That's the simplest form possible. So you are looking at 6 AND gates(no need to OR the gate outputs, just hook the 6 wires together) this is the maxterm form.

If you want to use the minterm form, it gets more complicated, but you might be able to do it with just 4 3-input NOR gates, and 1 4-input AND gate, but each NOR gate must be able to invert the inputs. otherwise, you will also need 4 NOT gates.

Check out the wikipedia site for karnaugh maps. Once you learn them, they make alot of those boolean problems really easy to figure out.