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

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.

 

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

 

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.

 

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

Code (Text):

______________________________
|  \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.