# Natural candidates for the hierarchy inside NPI

Let's assume that $\mathsf{P} \neq \mathsf{NP}$. $\mathsf{NPI}$ is the class of problems in $\mathsf{NP}$ which are neither in $\mathsf{P}$ nor in $\mathsf{NP}$-hard. You can find a list of problems conjectured to be $\mathsf{NPI}$ here.

Ladner's theorem tells us that if $\mathsf{NP}\neq\mathsf{P}$ then there is an infinite hierarchy of $\mathsf{NPI}$ problems, i.e. there are $\mathsf{NPI}$ problems which are harder than other $\mathsf{NPI}$ problems.

I am looking for candidates of such problems, i.e. I am interested in pairs of problems
- $A,B \in \mathsf{NP}$,
- $A$ and $B$ are conjectured to be $\mathsf{NPI}$,
- $A$ is known to reduce to $B$,
- but there are no known reductions from $B$ to $A$.

Even better if there are arguments for supporting these, e.g. there are results that $B$ does not reduce to $A$ assuming some conjectures in complexity theory or cryptography.

Are there any natural examples of such problems?

Example: Graph Isomorphism problem and Integer Factorization problem are conjectured to be in $\mathsf{NPI}$ and there are argument supporting these conjectures. Are there any decision problems harder than these two but not known to be $\mathsf{NP}$-hard?

• So you are looking for problems $P \in \mathrm{NP}$ such that $P_1 \leq_p P \leq_p P_2$ with $P_1 \in \mathrm{NPI}$ and $P_2 \in \mathrm{NPC}$? – Raphael Mar 7 '12 at 7:46
• Yes but there is no known reduction from P to P1 (similarly no known reduction from P2 to P). – Mohammad Al-Turkistany Mar 7 '12 at 7:58
• there are several problems with a status similar to factoring, see this paper by Papadimitriou theory.stanford.edu/~megiddo/pdf/papadimX.pdf – Marcos Villagra Mar 7 '12 at 8:42
• besides, we have a very nice list in cstheory cstheory.stackexchange.com/questions/79/… – Marcos Villagra Mar 7 '12 at 8:43
• why is the list that Marcos links to not the answer to your question ? – Suresh Mar 7 '12 at 16:21

I've found a nice problem called ModularFactorial. Take as input two $n$-digit integers $x$ and $y$, and output $x! \mod y$. This problem is at least as hard as Factoring and not know to be hard for FNP. The reference is the recent (and beautiful) book by Cristopher Moore and Stephan Mertens The Nature of Computation, page 79.
• Factoring can be reformulated into a decision problem as follows: given an integer $n$ and an integer $k$, does $n$ contain a factor $d$ with $1 < d \le k$? Is there an analogous decision version of the ModularFactorial problem? – Zach Langley Mar 9 '12 at 14:46
• @ZachLangley, yes of course I agree, but I was thinking in another decision version, namely, "does x has a factor?". The answer there is just, "yes" always. You can do the same with modularfactorial, give an integer k and decide if $x!\mod y$ is greater than $k$ or not. – Marcos Villagra Mar 9 '12 at 22:43