# Subset sum solver. Worth continue working on this method? [closed]

I have been working in a subset sum problem solver for some time.

The implementation is an exact/exhaustive search solver.

The variable determining the asymptomatic growth rate is just $N$ (the number of elements in the input).

Current upper bound is $O^{*}(2^{log(N)^{2}})$

$M$, the bit length of each input variable from the set, does not impact/change the asymptomatic growth rate, that is, it has the same upper bound for an input set with 32 elements/32 bits per value than an input set with 32 elements using 35,50,64 (or more bits) to represent the values.

By current implementation details this is true for $N \leq 33$. Higher values of $N$ will require to continue developing the method a bit.

Giving the above information.

Do you think it worth continue developing this method or better look to improve the upper bound before going into additional work?

## closed as unclear what you're asking by Kristoffer Arnsfelt Hansen, R B, Sasho Nikolov, András Salamon, Lev Reyzin♦Dec 17 '14 at 16:09

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• What am I reading wrong?: $O(2^{logN*logN})=O(N^{logN})$ right? so this is a sub-exponential, exact algorithm for subset sum? – Harry Dec 15 '14 at 11:47
• @Harry no, you are not reading wrong, just I forget the * for the polynomial factors (sorry about this) I edited the question to reflect this. The accurate upper bound is $O(N^4+2^{log(N)^{2}})$ however as I said, right now, is limited to $N \leq 33$ – John Flanders Dec 15 '14 at 18:05
• A quasipolynomial time algorithm for subset sum implies that all of NP can be solved in quasipolynomial time, and I consider that extremely unlikely. So I guess what I am saying is that if you can prove such a thing it would be a major result but I am apriori very skeptical that you can. – Sasho Nikolov Dec 15 '14 at 18:46
• @SashoNikolov I completely understand your skepticism. The method is not just 'magic' and the algorithm is already implemented and working. Although there is always a chance that the time analysis is somehow flawed and that will be my fault. I didn't want to open a discussion about how unlikely is this. Just raised a question that really matters me. You can treat the question context as hypothetical if you want. – John Flanders Dec 15 '14 at 22:26
• First it's hard to answer this question when parts of the setup verge on the absurd (for example it's meaningless to say that some asymptotic bound is "true" for small $N$, let alone to claim that an implementation can show this). Second, what even is the question? What do you mean by developing this method? If you are asking whether a $2^{\log^2 N}$ algorithm for an NP-hard problem is interesting, then 1) this is not research level question, 2) yes, it definitely is interesting. – Sasho Nikolov Dec 16 '14 at 18:32