The partition problem is a well-known NP-complete problem. In the definitions I have seen, the input is assumed to be a multiset of integers, and we want to decide the existence of a partition into two sets that have the same sum. My question is:

Is the partition problem still NP-complete if all input integers are distinct (i.e., no integer is repeated)?


Here is an outline of a reduction from PARTITION to UNIQUE PARTITION. Suppose the original numbers are $x_1,\ldots,x_n$ and the target is $T$. I assume that all $x_i$ are positive integers. The new numbers are going to be $2^n x_i + i$, as well as $1,2,4,\ldots,2^{n/2}$, and the new target is $2^n T + 2^{n/2}$. (The numbers $2^n,2^{n/2}$ are quite arbitrary and could be made much smaller.)

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  • $\begingroup$ BTW, Is unique partition NP-hard in the strong sense?. Here, I want to to decide whether the input to partition problem has a unique solution. $\endgroup$ – Mohammad Al-Turkistany May 1 at 14:26
  • $\begingroup$ This sounds like a completely different question. $\endgroup$ – Yuval Filmus May 1 at 16:26
  • $\begingroup$ Ok I will post it as separate question. $\endgroup$ – Mohammad Al-Turkistany May 1 at 19:04

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