# 3-partition problem without the restriction to triplets [closed]

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?

• 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. – plshelp Sep 8 '20 at 21:09
• I’m voting to close this question because it was cross-posted. – D.W. Oct 25 '20 at 18:12