# How to reduce the original partition problem to one of its variation?

Here's a statement of the set partition problem:

The set partition problem takes as input a set $$S = \{ a_1, a_2, ..., a_n \}$$(all positive integers). Can $$S$$ be partitioned into two sets $$A$$ and $$B$$ such that the sum of the numbers in $$A$$ is equal to the sum of the numbers in $$B$$?

Here's a statement of a variation of the set partition problem:

Given an input set $$S = \{ a_1, a_2, ..., a_n \}$$ (all positive integers), can you partition $$S$$ into $$A$$ and $$B$$ such that $$| \sum_{i \in A} a_i - \sum_{i \in B} a_i| < 1000$$?

My goal is to show that the variation is NP-complete by reducing the set partition problem to the variation, and here's what I've tried:

1. Given an instance of the set partition, $$S$$, let the total sum of its elements be $$s$$, then create a set $$S \cup \{ s + 1000, s \}$$ as an instance for the variation. But this doesn't seem to do the trick.
2. Given an instance of the set partition, $$S$$, do something to the elements in $$S$$ (for example, multiply by 2, subtract by 1000). This path feels promising but I'm again stuck.

Thanks to D.W.'s generous help, I managed to figure it out. Given an instance of the Partition problem, $$S$$, we can just scale every element in $$S$$ by 1000, and the new $$S'$$ becomes an instance of the variation. Then the reduction mapping holds, because scaling increases the difference between numbers. If $$S$$ is a YES instance, $$S'$$ can also be partitioned into two sets with equal sum, but if $$S$$ is a NO instance, all of the possible partitions of $$S'$$ must have difference $$\geq 1000$$. The reduction is clearly polynomial, and the variation is clearly in NP, so proof is complete.
If $$k$$ is an integer and $$|1000k| < 1000$$, then what can you conclude about $$k$$? How could that be useful?