# Reduction from $2$-Partitioning to (simple) pairwise $2$-Partitioning

I'm currently stuck showing $$NP$$-hardness of a problem of mine.

An instance of my problem (I call it (simple) pairwise $$2$$-Partitioning) is given by the following:

Given a set of tupels $$B=\{(b_1,1),\ldots,(b_n,1)\}$$, with $$b_i\in\mathbb{N}_{>0}$$.
Goal: A partition of $$B$$ into two subsets $$S_1, S_2$$ such that each element $$b_i$$ is exclusively either in $$S_1$$ or in $$S_2$$ and $$(\sum_{i\in S_1}b_i)+|S_2|=(\sum_{i\in S_2}b_i)+|S_1|$$

This can be seen in the following way. When $$b_i$$ is choosen from the pair $$(b_i,1)$$ to be in $$S_1$$ the other Element (in this case always $$1$$) must be in $$S_2$$.

I assume that this problem is $$NP$$-hard and that this can be shown by using the "classical" $$2$$-Partitioning-Problem where the goal is to partition a set $$A=\{a_1,\ldots,a_n\}$$ (with $$a_i\in\mathbb{N}_{>0}$$) into subsets $$S_1,S_2$$ s.t. the sum of these subset equals.

This far I wasn't able to show that the "classical" $$2$$-Partiting-Problem can be reduced to (simple) pairwise $$2$$-Partitioning.

My first attempt was to model some arbitrary but fixed instance of the $$2$$-Partitioning Problem with $$A=\{a_1,\ldots,a_n\}$$ to an instance of (simple) pairwise 2-Partitioning by constructing $$B=\{(a_1-1,1),\ldots,(a_n-1,1)\}$$ but somehow wasn't able to show that each $$YES$$-Instance of $$A$$ also is some $$YES$$-Instance of $$B$$.

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 positive instance of $$2$$-$$\texttt{Partition}$$ if and only if $$B$$ is a positive instance of $$\texttt{Pairwise}$$ $$2$$-$$\texttt{Partition}$$:
• suppose there exists $$I_1\subset \{1, …, n\}$$ such that $$\sum\limits_{i\in I_1}a_i = \sum\limits_{i\notin I_1}a_i$$. Then: \begin{align} \sum\limits_{i\in I_1}b_i + |\overline{I_1}| & =\sum\limits_{i\in I_1}(a_i +1)+ |\overline{I_1}|\\ & = \sum\limits_{i\in I_1}a_i + |I_1| + |\overline{I_1}|\\ & =\sum\limits_{i\notin I_1}a_i + |I_1| + |\overline{I_1}|\\ & = \sum\limits_{i\notin I_1}b_i + |I_1| \end{align}