Given a list of positive integers with sum $s$, decide if there is a subset with sum $0.5s$. This is the well-known PARTITION problem, which is NP-hard.

What about the following: Given a list of positive integers with sum $s$, decide if there is a subset with sum in the range $[0.49s, 0.51s]$. Can this problem be solved in polynomial time?

  • $\begingroup$ There is a PTAS for a variant of the problem that minimizes the spread between the parts. I suspect that it could be used to solve the problem in polynomial time (might be wrong though). See en.wikipedia.org/wiki/…. $\endgroup$ Jun 13, 2022 at 13:54

1 Answer 1


Yes, this problem is polynomial-time solvable.

Let $A$ be the input numbers and let $S = \mathrm{sum}(A)$ be its sum. Let $T_1 = 0.49S$, $T_2 = 0.51 S$ be the target sum range. Let $\epsilon = 0.02$ so that $T_2 - T_1 = \epsilon S$.

Define $A_1 = \{ a \in A \mid a \geq \epsilon S \}$ to be large items and $A_2 = A \setminus A_1$ to be small items.

Proposition: For a subset of large items $X \subseteq A_1$, there exist a solution containing that subset if and only if $T_1 - \mathrm{sum}(A_2) \leq \mathrm{sum}(X) \leq T_2$.

The "only if" direction is from the fact that the range of possible sum containing the subset $X$ is $[\mathrm{sum}(X), \mathrm{sum}(X) + \mathrm{sum}(A_2)]$ (note the assumption of all items are nonnegative) and it must have a non-empty intersection with $[T_1, T_2]$.

To prove the "if" direction, consider adding small items arbitrary as long as the sum is less than the lower bound $T_1$. Because added items are small, it cannot overshoot the upper bound $T_2$.

Now, we have an algorithm solving the problem by brute-forcing all subsets $X \subseteq A_1$ of large items and checking its sum. The time complexity of the algorithm is $O(n + 2^{|A_1|} |A_1|)$. Because we have $|A_1| \leq \epsilon^{-1}$, the time complexity is polynomial in $n$ for a constant $\epsilon$. The exponent can be reduced to half using the meet-in-the-middle approach.


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