Question

Related questions:

• Algorithm to Divide a list of numbers into 2 equal sum lists
• divide list in two parts that their sum closest to each other

Let’s assume I have a list, which contains exactly 2k elements. Now, I’m willing to split it into two parts, where each part has a length of k while trying to make the sum of the parts as equal as possible.

Quick example:
[3, 4, 4, 1, 2, 1] might be splitted to [1, 4, 3] and [1, 2, 4] and the sum difference will be 1

Now – if the parts can have arbitrary lengths, this is a variation of the Partition problem and we know that’s it’s weakly NP-Complete.

But does the restriction about splitting the list into equal parts (let’s say it’s always k and 2k) make this problem solvable in polynomial time? Any proofs to that (or a proof scheme for the fact that it’s still NP)?

Answer 1

It is still NP complete. Proof by reduction of PP (your full variation of the Partition problem) to QPP (equal parts partition problem):

Take an arbitrary list of length k plus additional k elements all valued as zero.

We need to find the best performing partition in terms of PP. Let us find one using an algorithm for QPP and forget about all the additional k zero elements. Shifting zeroes around cannot affect this or any competing partition, so this is still one of the best performing unrestricted partitions of the arbitrary list of length k.