In the ABC-partition problem, there are three sets $A, B, C$ with $m$ positive integers in each. The sum of all integers is $m T$. The goal is to construct $m$ triplets with the same sum $T$, each of which contains exactly one integer from $A, B$ and $C$.
The problem "feels" NP-hard, but I could not find a proof for this. I could only find reductions in the wrong direction:
- Reduction from ABC-partition to 3-partition is easy: replace each $a \in A$ with $8a+1$, each $b \in B$ with $8b+2$, and each $c \in C$ with $8c+4$. This forces the 3-partition oracle to construct with exactly one item from each of $A$, $B$ and $C$.
- Reduction from ABC-partition to 3-dimensional matching is easy too: keep the items in $A, B, C$ as they are, and add a hyperedge $\{a,b,c\}$ iff $a+b+c = T$.
- I also found a reduction from 3-dimensional matching to ABCD-partition, which is similar to ABC-partition but with four sets; and a reduction from 4-partition to 3-partition.
But, none of these reductions shows that ABC-partition is NP hard.
Previously I asked specifically about a reduction from 3-partition from ABC-partition; now I am not even sure if it is NP-hard. Is it?
