I'm stuck on boolean algebra, it has 4 variables

[mr_coffee]

Hello everyone, I'm designing a IC that will result in a 1 output if the number is > 9. So I wrote out the truth table and or'ed all the min terms and got the following:
http://img282.imageshack.us/img282/482/lastscan45vd.jpg I'm stuck now! any help would be great!

 

[faust9]

AND the MSB with each of the two preceeding bits and then OR that.

(AB)+(AC) will yield a 1 when the output is greater than 9.

 

[mr_coffee]

Thanks for the reply but i finally figured out how k-maps work, and when i threw that big mess in the k-map it reduced down to:
z = AB + AB'C
note; B' = B complimented
here is a picture:
http://img43.imageshack.us/img43/3353/w0t1bp.jpg

 

[faust9]

You don't need the B! though. You're adding complexity by having it.

 

[mr_coffee]

I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?

 

[Kenneth Mann]

I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?

You have it backward. It's because you didn't use all possible overlaps that you have redundancies. Go back this time and group your "1's" so that each group is as large as possible, taking every overlap you can get, but the smallest number of groupings possible. (In this case there are obviously only two groupings needed, so there will be two 'AND' terms.) When you do that you will see the answer.

KM

By the way: your drawing was hard to see.

 

[faust9]

I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?

You do get the right answer if you arrange things logically:

______________________________ 
|  \cd                        |
|ab \ ________________________|
|     | cd  | cd! | c!d | c!d!|
|______________________________
|  ab |  1  |  1  |  1  |  1  |
|______________________________
| ab! |  1  |  1  |  0  |  0  |
|______________________________
| a!b |  0  |  0  |  0  |  0  | 
|______________________________
|a!b! |  0  |  0  |  0  |  0  |
_______________________________

You see AB is one part of the solution because it works for all combinations of CD. If you look in the columns you will see AC is the other solution because the solution AC does not depend on B at all.

 

[mr_coffee]

Thanks so much guys, my professor so told us that we should not overlap, infact we should avoid overlapping, what the heck is he talking about! I ended up with AB+AC like u guys said!