3-PARTITION problem

I think by reduction it goes like this:

Reducing 2-partition to 3-partition:

Let S be the original set, and A be its total sum, then let S'=union({A/2},S). Hence, perform a 3-partition on the set S' yields three sets X, Y, Z. Among X, Y, Z, one of them must be {A/2}, say it's set Z, then X and Y is a 2-partition. The witnesses of 3-partition on S' is the witnesses of 2-partition on S, thus 2-partition reduces to 3-partition.

It's easy to generalize 2-sets solution for 3-sets case.

In original version, you create array of boolean sums where sums[i] tells whether sum i can be reached with numbers from the set, or not. Then, once array is created, you just see if sums[TOTAL/2] is true or not.

Since you said you know old version already, I'll describe only difference between them.

In 3-partition case, you keep array of boolean sums, where sums[i][j] tells whether first set can have sum i and second - sum j. Then, once array is created, you just see if sums[TOTAL/3][TOTAL/3] is true or not.

If original complexity is O(TOTAL*n), here it's O(TOTAL^2*n).
It may not be polynomial in the strictest sense of the word, but then original version isn't strictly polynomial too :)