In the standard 3-partition problem, there are $3 m$ integers, their sum is $m T$, and they have to be partitioned into $m$ subsets of sum $T$ and size $3$.
Consider the variant without the restriction that the size is $3$ (but with the restriction that there must be $m$ subsets with a sum of $T$). Are these problems computationally equivalent?
[NOTE: often there is an additional restriction that each input number is in $(T/4 , T/2)$; in this question there is no such restriction - the numbers in both variants can be any positive integers].
There is an easy reduction from the triplet-variant to the subset-variant: given an instance of the triplet-variant with target-sum $T$, construct an instance of the subset-variant by adding $2 T$ to each element and changing the target-sum to $7 T$. Every solution to the triplet-instance is also obviously a solution to the subset-instance. Conversely, in every solution to the subset-instance, each subset must have exactly 3 elements, since the sum of any 2-element subset is at most $6 T$ and the sum of every 4-element subset is at least $8 T$.
Is there a reduction in the opposite direction? Particularly, given an instance of the subset-variant, how can I construct an instance of the triplet-variant, that has a solution whenever the original instance has one?
Note that, since standard 3-partition is NP-hard and the unconstrained partition problem is in NP, there must be a polynomial-time reduction between these problems, maybe going through some other problems. But I am looking for a more direct reduction. Such a direct reduction can be useful, for example, for applying approximation algorithms for one problem (for optimization variants of the problem) to the other one.
Note: the question was posted in cs.SE about two months ago - no answer there
