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 are $m$ 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].

I found a 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?

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    $\begingroup$ Both problems are $NP$-complete so there must be a reduction. (for the subset problem we can find a polynomial verifier). But I haven't found an elegant reduction yet. $\endgroup$ – plshelp Sep 8 '20 at 21:09
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    $\begingroup$ I’m voting to close this question because it was cross-posted. $\endgroup$ – D.W. Oct 25 '20 at 18:12