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It's possible that this question is trivial and I've overlooked something.

Let us impose the following constraint on the well-known PARTITION problem: in all inputs of a new problem the number of elements is fixed and equals $k$?

Does such constraint make PARTITION polynomially solvable?

P.S. There is pseudopolynomial algorithm for it with running time $\mathcal{O}(nA)$, where $A$ is the sum of the numbers in the given input. However, fixing $n$ doesn't seem to make it polynomial in $\log A$. Or am I missing something?

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up vote 2 down vote accepted

Yes. The problem becomes in $P$. The reason is that there would be a fixed number of partitions bounded by $O(2^k)$ which is a constant. You do not need the $O(nA)$ dynamic programming algorithm.

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Of course, you're right. Thanks. –  Oleksandr Bondarenko Feb 5 '11 at 1:06

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