NP-complete problems are the "hardest problems" in NP. This means that all other problems in NP reduce (in polytime) to these problems. A consequence of this is if we were to find some polynomial time algorithm to solve one of these problems, we could prove $P = NP$. And if we could prove that no polynomial time algorithm could exist, this implies $P \neq NP$.

Are there other problems in $NP$ that have this property? That is,

Is it possible for a problem $X$ in $NP$ to not be NP-complete, but we could still solve $P$ vs. $NP$ by determining if it has polynomial time lower bounds.

Of course, proving that there is an exponential lower bound on any NP problem proves $P \neq NP$, but are there problems in NP where finding a polynomial time lower bound (besides NP complete problems) shows $P = NP$?

  • $\begingroup$ Did you mean to write $\mathsf P \neq \mathsf{NP}$ in your last sentence? $\endgroup$ Apr 7, 2023 at 9:25
  • $\begingroup$ @Watercrystal I'm asking for something more along the lines of a class of problems that act like NP-complete problems, where if you find a polynomial time solution to them, you can prove $P = NP$. If you show exponential lower bounds on problems in this class, then $P \neq NP$, yes. But that is true for any problem in $NP$. I think my comment makes sense, please let me know if I am not being clear. Thanks. $\endgroup$
    – Loic Stoic
    Apr 7, 2023 at 9:44
  • $\begingroup$ I think this analogy might be clearer actually: If one found a solution to graph isomorphism, that would not prove $P$ = $NP$. But if one were to find a solution to SAT, that would prove $P$ = $NP$. Is there a problem that is not NP-complete that has this same property of proving $P = NP$? $\endgroup$
    – Loic Stoic
    Apr 7, 2023 at 9:58
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    $\begingroup$ If $GI \notin P$, then $P \neq NP$. $\endgroup$
    – Pål GD
    Apr 7, 2023 at 11:39
  • $\begingroup$ @PålGD yes, but I’m asking for problems that satisfy the inverse. A problem $X$, such that $X \in P$ then $P = NP$. $\endgroup$
    – Loic Stoic
    Apr 7, 2023 at 17:43

2 Answers 2



Ladner's theorem shows that for any problem $X \in \mathrm{NP} \setminus \mathrm{P}$ there is a problem $Y \in \mathrm{NP} \setminus \mathrm{P}$ such that $Y$ is reducible to $X$ but not vice versa.

If we apply the construction from Ladner's theorem to e.g.~$\mathrm{SAT}$ we obtain a problem $Z \in \mathrm{NP}$ such that either $Z$ is neither in $\mathrm{P}$ nor $\mathrm{NP}$-complete (in case that $\mathrm{P} \neq \mathrm{NP}$); or $Z$ is simulatenously in $\mathrm{P}$ and $\mathrm{NP}$-complete (if $\mathrm{P} = \mathrm{NP}$).

In other words $Z$ is not $\mathrm{NP}$-complete unless for trivial reasons, and proving that $Z$ belongs to $\mathrm{P}$ entails $\mathrm{P} = \mathrm{NP}$.


The answer is no.

Consider a problem $X$ in $\mathsf{NP}$ that is not $\mathsf{NP}$-complete. Suppose that finding out some bound on the complexity of $X$ would show $\mathsf{P}=\mathsf{NP}$.

The thing is that if $\mathsf{P}=\mathsf{NP}$, then all problems in $\mathsf{NP}$ are $\mathsf{NP}$-complete, except for the two trivial problems “always yes” and “always no”.

So $X$ would have to be either “always yes” or “always no”, whose complexity is well-known (they are constant time) and has no bearing on the $\mathsf{P}$ vs $\mathsf{NP}$ problem.


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