#StackBounty: #np-hardness #partition-problem A partition problem with order constraints

Bounty: 50

In the OrderedPartition problem, the input is two sequences of $n$ positive integers, $(a_i){iin [n]}$ and $(b_i){iin [n]}$.
The output is a partition of the indices $[n]$ into two disjoint subsets, $I$ and $J$, such that:

  1. $sum_{iin I} a_i = sum_{jin J} a_i$
  2. For all $iin I$ and for all $jin J$: $b_ileq b_j$.

In other words, we have to first order the indices on a line such that the $b_i$ are weakly increasing, and then cut the line such that the sum of the $a_i$ in both sides is the same.

If all $b_i$ are the same, then condition 2 is irrelevant and we have an instance of the NP-hard Partition problem.
On the other hand, if all $b_i$ are different, then condition 2 imposes a single ordering on the indices, so there are only $n-1$ options to check, and the problem becomes polynomial. What happens in between these cases?

To formalize the question, define by OrderedPartition[n,d], for $1leq dleq n$,
the problem restricted to instances of size $n$, in which the largest subset of identical $b_i$-s is of size $d$. So the easy case, when all $b_i$-s are different, is OrderedPartition[n,1], and the hard case, when all $b_i$-s are identical, is OrderedPartition[n,n].

More generally, For any $n$ and $d$, in any OrderedPartition[n,d] instance, the number of possible partitions respecting condition 2 is $O(n 2^d)$. Hence, if $din O(log{n})$, then OrderedPartition[n,d] is still polynomial in $n$.

On the other hand, for any $n$ and $d$, we can reduce from a Partition problem with $d$ integers to OrderedPartition[n,d]. Let $p_1,ldots,p_d$ be an instance of Partition. Define the instance OrderedPartition[n,d] by:

  • For each $iin {1,ldots,d}$, let $a_i := 2ncdot p_i$ and $b_i := 1$.
  • For each $iin {d+1,ldots,n}$, let $a_i := 1$ and $b_i := i$
    [if $n-d$ is odd, make $a_n:=2$ such that the sum will be even].

Hence, if $dinOmega(n^{1/k})$, for any integer $kgeq 1$, then OrderedPartition[n,d] is NP-hard.

QUESTION: What happens in the intermediate cases, in which $d$ is super-logarithmic but sub-polynomial in $n$?


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