# Tag Info

### Reduction from $2$-Partitioning to (simple) pairwise $2$-Partitioning
I think you took the problem the wrong way, but your idea is good. Let $A = \{a_1, …, a_n\}$ be an instance of $2$-$\texttt{Partition}$. Let $B = \{(a_1 + 1, 1), …, (a_n+1, 1)\}$. Then $A$ is a ...