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The 3SUM problem is defined as follows:

Given three sets of integers $A,B,C \subseteq \{-W,\ldots,W\}$ for $W = n^3$, each containing $n$ integers, determine whether there exsists some $a \in A, b \in B, c \in C$ such that $a+b+c=0$.

The X+Y problem is defined as follows:

Given two sets of integers $X$ and $Y$ , each containing $n$ integers, determine whether the sumset $X + Y = \{a + b \ | \ a \in X, b \in Y \}$ (duplicates allowed) contains $n^2$ distinct integers.

I want to show that this problem is $3SUM$-hard, that is, if the $X+Y$ problem can be solved in $O(n^{2-\epsilon})$ for some $\epsilon > 0$ then $3SUM$ can also be solved in $O(n^{2-\delta})$ for some $\delta > 0$. In particular, I want to show this by reducing the $3SUM$ problem to the $X+Y$ problem. However, I am not able to come up with a sophisticated solution. Any advice would be appreciated.

dport
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1 Answers1

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I first describe an incorrect reduction, then explain how to fix it.

We are given an instance of the 3-SUM problem $(A, B, C)$. First, let $M := 4W + 1$ be a sufficiently large number. Define $A' := M + A$, $B' := 2 M + B$, $C' := 3M - C$. Then, define $X := A' \cup \{ 0 \}$ and $Y := B' \cup C'$ to be an instance of your X+Y problem.

The sumset $X + Y$ contains sums of the forms $a' + b'$, $a' + c'$, $0 + b'$, and $0 + c'$. If there is a solution $(a, b, c)$ to the 3-SUM problem (i.e. $a + b + c = 0$), then we have $(M+a) + (2M+b) = 3M + a + b = 3M - c$, meaning $a' + b' = 0 + c'$.

Conversely, sums of different forms other than $a' + b' = 0 + c'$ cannot be equal due to the different multiples of $M$. However, we could have multiple equal sums within a single form, like $a' + b'$ if the sums within $A + B$ (or $A + C$) are not unique, invalidating the reduction.

To fix this issue, we rely on a result from a recent paper. A set $S$ is called a Sidon set if all pairwise sums $x + y$ for $x, y \in S$ are all distinct. Jin and Xu [1] proved that the following problem is 3-SUM hard:

3SUM on Sidon Set: Given a Sidon set $S$, determine if there exist elements $a, b, c \in S$ such that $a + b + c = 0$.

This is exactly what we need for the previous reduction to work. If we let $S = A = B = C$ be a Sidon set, then the sums in $X + Y$ are distinct if and only if the 3-SUM problem has a solution. Therefore, the X+Y problem is 3-SUM hard.

pcpthm
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