OK, I am having a bit of trouble with an equasion I have to simplify. I have to simplify an equasion small enough so I can build a circuit using the least ammount of Integrated Circuits as possible. Then I have to build it. Building is easy, I can look at the equasion, then write down a diagram, and build the circuit using the diagram. And I can use a truth table to make sure I built it correctly.

EDIT: I can't write down a boolean equasion. I tried using the line above for my not symbols, but it messes it up when I make the post. So I made another post showing the truth table, which one can use to come up with the equasion


I believe that there is not one answer. I am starting to think the answer I get will have to be the one I can use to build the cheapest circuit. I am pretending I have to make 10,000 of these so that means that finding something to use 1 less circuit means saving a whole lot of money.

 

If it helps, Here is my truth table that I was given. I build a circuit that returns these values and I hook up 4 data switches for the varables A, B, C, and D. and I connect the output to a light and when the light turns on, it's a 1. when the light is off, it is a 0. When I attempted this circuit, I messed up on my boolean calculation, and there was 2 instances when the light turned on when it was supposed to be off. Every one of my attempts comes up using 4 integrated circuits, and I need it to be down to 2 or 3. But I don't know how I can possibly get it down to 2, because I am pretty sure I am going to need one hex inverter.

A B C D X
0 0 0 0 1
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1

 

Hi, after looking at your truth table i came up with an expression that can be implemented with 2 four input AND gates (using only 3 inputs, fourth tied high) and 3 two input OR gates. You can obtain a dual 4 input AND in once IC and also a quad 2 input OR in one IC. There you go, only 2 ic's. If you wanna be picky to get the low and high assertion levels for each inputs you would add one ic with 4 inverters on it.

By the way, the boolean expression i got was:
X = AN.BN.(CN + DN) + A.B.(C + D)

where AN denotes the complement of A, A(NOT) - you know, A with the bar on top.