# Equivalence of 3SUM problems

#### Wed, Apr 27, 2016 02:00 CEST

######
**Tags:**
Algorithms

Are different versions of the 3SUM problem equivalent?

## Reducing 3SUMx1 on S with multiple pick to 3SUMx3 on A,B,C

Trivially with \(A=B=C=S\).

## Reducing 3SUMx1 on S with unique pick to 3SUMx3 on A,B,C

Construct \(t=\Theta(\log n)\) instances of 3SUMx3 where S is partitioned into \(A,B,C\) of size \(\frac{n}{3}\) each uniformly at random.

By construction, no instance has a triple \(a + b + c = 0\) with \(a = b = s_i\) or \(b = c = s_i\) or \(a = c = s_i\).

If \(s_1 + s_2 + s_3 = 0\) then, with high probability, at least one of the \(\Theta(\log n)\) instances has \(s_1 \in A\), \(s_2 \in B\), and \(s_3 \in C\).

The probability that this is not the case is \((\frac{27-3!}{27})^m\).

Hence, with \(m = \Theta(\log n)\), the probability that at least one instance has the triple is \(1 - n^{-\Omega(1)}\).

## Reducing 3SUMx3 to 3SUMx1 (with multiple or unique pick)

\( 10 a + 1, 10 b + 2, 10 c - 3 \)

\(a + b + c = 0 \implies 10 a + 1 + 10 b + 2 + 10 c - 3 = 0\)

\(1 + 2 - 3 = 0\)

\(1 + 1 + 1 = 3\)

\(2 + 2 + 2 = 6\)

\(-3 - 3 - 3 = -9\)

\(1 + 1 + 2 = 4\)

\(1 + 2 + 2 = 5\)

\(1 + 1 - 3 = -1\)

\(1 - 3 - 3 = -5\)

\(2 + 2 - 3 = 1\)

\(2 - 3 - 3 = -4\)