Is there any known $NP$-complete problem which hardness proof is non-constructive.

A constructive $NP$-completeness proof is a proof that $L_1\leq_p L_2$ by a reduction $r$ and from the argument therein, one can extract two polynomial-time algorithms $f$ and $g$ such that:

For every instance $x\in L_1$, given a certificate $w_1$ for $x$, we must have that $w_2=f(w_1)$ is a certificate for $r(x)\in L_2$

For every instance $x\in L_1$, given a certificate $w_2$ for $r(x)\in L_2$, we must have that $w_1=g(w_2)$ is a certificate for $x\in L_1$.

In $NP$-completeness result, that does not describe $f$ and $g$, we can say that it is a non-constructive proof. Given a certificate for an instance of $L_1$, one does not know how to construct a certificate for the corresponding instance of $L_2$ and vice versa.

In a sense, the proof only establishes the existence of corresponding certificates.

More specifically, do we have an existence-only proof for some $NP$-complete problem?

  • $\begingroup$ You might want to have a look at this question on CSTheory. $\endgroup$
    – Juho
    Oct 11, 2018 at 19:18
  • $\begingroup$ That is the non-constructivity of the existence of a Karp reduction. Here, a Karp reduction must be presented already. $\endgroup$ Oct 12, 2018 at 3:39

1 Answer 1


If worst-case one-way permutation exists (which exists iff. $P\neq UP\cap coUP$, proven here), then there exists inefficient reduction proof for some $NP$-complete problem.

We denote $\pi$ the one-way permutation above.

Specifically, for each proven $NP$-complete $L$, we can define:

$PERMUTED-WITNESS(L)=\{\pi(x)\mid x\in L\}$

with the certificate for $y\in PERMUTED-WITNESS(L)$ as: $(x,c)$ where $\pi(x)=y$ and $\pi(c)$ is a certificate for $x\in L$ (the "original" certificate). In other words, $c$ is the pre-image of the "original" certificate.

So, given a certificate $c'$ for $x\in L$, we cannot efficiently find $c$ (the pre-image of $c'$) for $\pi(x)\in PERMUTED-WITNESS(L)$


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