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Is there a functional problem for which there is an algorithm that can decide if a solution is a solution or not in polynomial time but we can't find a solution in polynomial time?

Let FP be the class of functional problems.

Update: Every problem in FP, say $a$, has a correspondent decisione problem $a'$ (the associated decision problem $a'$ is the problem where we ask if a possible solution is a solution), I ask if there is a functional problem $a$ for which $a'$ is in P but $a$ is NOT in FP.

In this link: Hamiltonian cycle, verifying and finding

we have instead a functional problem HC (hamiltonian cycle) for which HC' is P and HC is in FP.

Intuitively I think that the kind of problems I search, should be 'more' the problem of kind of HC. But maybe the hard part, given $a$ (with $a'$ in P), is to prove that $a$ is NOT in FP.

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  • $\begingroup$ I refer to this: en.wikipedia.org/wiki/Function_problem The class P is the class of decision problem which are solvable in polynomial time in a Turing Machine. I ask for a functional problem for which the associated decision problem is in P (the associated decision problem is the problem where we ask if a possible solution is a solution), but the functional problem is not solvable in polynomial time. $\endgroup$
    – asv
    Dec 2, 2019 at 12:18
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    $\begingroup$ Thanks for your explanation. Are you asking whether FP = FNP? $\endgroup$
    – D.W. ♦
    Dec 2, 2019 at 17:14
  • $\begingroup$ Every problem in FP, say $a$, has a correspondent decisione problem $a'$ (the associated decision problem $a'$ is the problem where we ask if a possible solution is a solution), I ask if there is a functional problem $a$ for which $a'$ is in P but $a$ is NOT in FP. $\endgroup$
    – asv
    Dec 3, 2019 at 9:32
  • $\begingroup$ In this link: cs.stackexchange.com/questions/117894/… we have instead a functional problem HC (hamiltonian cycle) for which HC' is P and HC is in FP. $\endgroup$
    – asv
    Dec 3, 2019 at 9:33
  • $\begingroup$ I don't know if asking FP=FNP is the same of my question. $\endgroup$
    – asv
    Dec 3, 2019 at 9:38

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It appears that you are asking whether FP = FNP. That is an open problem. It is known that FP = FNP if and only if P = NP. I suspect that most theoreticians expect that it's likely that P != NP, which would imply FP != FNP. If FP != FNP, there exists a function problem whose solutions can be verified in polynomial time, but where you can't find a solution in polynomial time. In particularly, any FNP-complete problem -- such as FSAT, say -- will have that property, if FP != FNP. So, the FNP-complete problems are particularly good candidates for having the property you articulate. See https://en.wikipedia.org/wiki/Function_problem for an overview.

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  • $\begingroup$ What if $2^n$ as a function problem is shown to require $2^n$ digits to display its solutions? What if I could verify the solutions for $2^n$ by converting the solution into binary and comparing the trailing $0-bits$ to the value of $n$ Where, $2^n$ = $solution$. pastebin.com/fGcSRGdU $\endgroup$
    – Travis Wells
    Mar 6, 2020 at 19:13
  • $\begingroup$ @TravisWells, I don't see how that's relevant to my answer. Computing the function $n \mapsto 2^n$ is not in FP (assuming the input is represented in binary). Please don't use comments to raise new questions or seek feedback on your ideas. Thank you. $\endgroup$
    – D.W. ♦
    Mar 6, 2020 at 19:42

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