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I want to know whether there is a decision problem, written EasyProblem, satisfying the follow property:

  • For every valid instance $x$, $x$ is a Yes-instance for EasyProblem (if we construct EasyProblem as a nature problem). Formally speaking, maybe we can define EasyProblem as a language $L \in \mathrm{DTIME}(n)$ or even $L \in \mathrm{DTIME}(1)$.

  • The counting version #EasyProblem is in #$P$-complete.

Really what I'm asking is: can we construct a very easy decision problem, but its counting version is too hard? Or can we construct a very hard counting poblem, but its decision version is too easy?

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An example of a such problem in $\mathrm{DTIME}(n)$ is DNF satisfiability [1]. Its counting version is #P-complete [2], but its decision version can be solved by simply checking if there is a conjunction that is satisfiable.

For $\mathrm{DTIME}(1)$ I think the answer is that there are no such problems because we can iterate over all computation paths in $O(1)$ so counting is easy.

Note that in general, disallowing instances with answer 0 does not change the #P-completeness of any problem whose decision version is in P because we can check this case in polynomial time.

[1] https://en.m.wikipedia.org/wiki/Disjunctive_normal_form

[2] https://en.m.wikipedia.org/wiki/%E2%99%AFP-complete

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  • $\begingroup$ If we disallow the instance with answer 0, then we can construct an algorithm that returns Yes in $O(1)$ time (since we can encode all the invalid input to as a string). How to understand the second paragraph in your answer? It make sense, but there is a contradiction. $\endgroup$ – TeamBright Dec 7 '20 at 8:23
  • $\begingroup$ Yeah, that is a good point. I think the contradiction arises from non-rigorous definitions. In particular, I think the definition that there are Yes-instances, No-instances, and invalid instances is incompatible with the standard definitions of #P-completeness. For the standard definitions there are only Yes-instances and No-instances (invalid instances are No-instances). $\endgroup$ – Laakeri Dec 7 '20 at 8:38
  • $\begingroup$ Let me give an example, suppose that $L = \{ y \vee \varphi \mid \varphi \text{ is a DNF with varibles } x_{1}, x_{2}, \ldots, x_{n} \}$. Since #DNF is #$\mathrm{P}$-complete, so is $L$. But $L \in \mathrm{DTIME}(1)$ right? Otherwise, there is not a $\mathrm{DTIME}(1)$ language anymore. $\endgroup$ – TeamBright Dec 7 '20 at 11:45
  • $\begingroup$ You need to use at least $\Theta(n)$ computation to check if the input is a valid representation of DNF (note that inputs are binary strings, not arbitrary data). Yeah I guess there is not many interesting languages in DTIME(1). $\endgroup$ – Laakeri Dec 7 '20 at 12:22
  • $\begingroup$ I think I have got the answer that I want. That is there do exist a decision problem in $\mathrm{DTIME}(1)$, but the counting version is complete in #$\mathrm{P}$. (Thanks to your answer,) $L = \{ y \vee \varphi \mid \varphi \text{ is a DNF with variables }x_{1}, x_{2}, \ldots, x_{n} \}$. $\endgroup$ – TeamBright Dec 10 '20 at 7:03

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