What is an algorithm to split a group of items into 3 separate groups fairly?

Question

I have this problem in my textbook: Given a group of n items, each with a distinct value V(i), what is the best way to divide the items into 3 groups so the group with the highest value is minimIzed? Give the value of this largest group.

I know how to do the 2 pile variant of this problem: it just requires running the knapsack algorithm backwards on the problem. However, I am pretty puzzled as how to solve this problem. Could anyone give me any pointers?

Answer: Pretty much the same thing as the 0-1 knapsack, although 2D

Answer 1

Tough homework problem. This is essentially the optimization version of the 3-partition problem.

http://en.wikipedia.org/wiki/3-partition_problem

It is closely related to bin packing, partition, and subset-sum (and, as you noted, knapsack). However, it happens to be strongly NP-Complete, which makes it a harder than its cousins. Anyway, I suggest you start by looking at dynamic programming solutions to the related problems (I'd start with partition, but find a non-wikipedia explanation of the DP solution).

Update: I apologize. I have mislead you. The 3-partition problem splits the input into sets of 3, not 3 sets. The rest of what I said still applies, but with the renewed hope that your variant isn't strongly np-complete.

Answer 2

Let f[i][j][k] denotes whether it is possible to have value j in the first set and value k in the second set, with the first i items.

So we have f[i][j][k] = f[i-1][j-v[i]][k] or f[i-1][j][k-v[i]].

and initially we have f[0][0][0] = True.

for every f[i][j][k] = True, update your answer depends on how you defines fairly.

Answer 3

I don't know about "The Best" mathematically speaking, but one obvious approach would be to build a population of groups initially with one item in each group. Then, for as long as you have more groups than the desired number of final groups, extract the two groups with the lowest values and combine them into a new group that you add back into the collection. This is similar to how Huffman compression trees are built.

Example:

1 3 7 9 10
becomes
4(1+3) 7 9 10
becomes
9 10 11(1+3+7)