# whether a integer set can be divided into equal sum subsets [closed]

Is there a way to check whether a set containing 1 to n integers can be partitioned in 2 equal sum subsets in minimum time complexity.

## closed as off-topic by Aqua, GNUSupporter 8964民主女神 地下教會, ahulpke, Morgan Rodgers, hardmathJan 8 '18 at 0:32

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• If $n$ is divisible by $4$, you can do it for sure. – AlexanderJ93 Jan 7 '18 at 20:15
• Are you asking about the set $\{1,2,3,\ldots,n\}$, or about arbitrary sets of $n$ integers? – Barry Cipra Jan 7 '18 at 21:07
• It would improve your Question (and clarify it) if you worked out a couple of small examples to include in the body of your Question. E.g. $\{1,2,3\}$ can be partitioned into two equal sum subsets, but no proper subset of it (except the empty set) can be so partitioned. – hardmath Jan 8 '18 at 0:32

Yes, it is possible if and only if $n\equiv 0,3\pmod 4$.

Because $1+2+\cdots+n=\frac{n(n+1)}{2}$, for this to be further divisible by $2$ we must have either $n$ or $n+1$ divisible by $4$, which means $n\equiv 0,3\pmod 4$.

Conversely, assume $n\equiv 0,3\pmod 4$. It is obvious that four consecutive numbers can be split into pairs giving the same sum. ($\{k,k+1,k+2,k+3\}$ can be split into $\{k+1,k+2\}$ and $\{k,k+3\}$, both giving sum $2k+3$.) Now, start taking groups of $4$ numbers off the end of the sequence $1,2,\ldots,n$ (last $4$ numbers, then the $4$ numbers before that, then the $4$ numbers before that etc.). Split those as shown above. What you end up at the end is:

• $n\equiv 0\pmod 4$: nothing. We've thus split the whole set into two sets with equal sums.
• $n\equiv 3\pmod 4$: remaining are $\{1,2,3\}$, which can be split into $\{1,2\}$ and $\{3\}$.

Examples for those two cases:

• $n=8$: $8+5+4+1=7+6+3+2$
• $n=11$: $11+8+7+4+3=10+9+6+5+2+1$