A class of languages $C$ is recursively presentable if there is an effective enumeration of Turing machines $\mathcal{M}_1,\mathcal{M}_2,\ldots$ such that $C=\{L(\mathcal{M}_i)\mid i=1,2,\ldots\}$. Uwe Schöning considers this notion in his elegant generalisation of Ladner's theorem ("A Uniform Approach to Obtain Diagonal Sets in Complexity Classes", Theor. Comp. Sci. 18, 1982).

In a previous answer on this site, Ryan Williams argues that the set of all P-hard languages can be recursively presented "similar to how Schoening does it for the NP-hard sets". I would like to know how this works, since I cannot find this in the paper. Schöning works with Turing reductions and shows that the NP-complete sets are recursively presentable. We can adapt his argument to NP-complete sets under polynomial many-one reductions as follows:

  1. Take an enumeration $(\mathcal{M}_i,\mathcal{R}_i)_{i\geq 1}$ of pairs of polynomially time-bounded non-deterministic Turing machines $\mathcal{M}_i$ and polynomial reductions $\mathcal{R}_i$ (we can enumerate polytime machines and reductions by enumerating all machines and polynomials, and augmenting them with a polytime "stop clock" that enforces timely halting).
  2. For each pair of a machine $\mathcal{M}_i$ and reduction $\mathcal{R}_i$, the let $\mathcal{M}'_i$ denote a machine that does the following on input word $w$:
    • For all input words $v$ with $|v|<|w|$, test whether $\mathrm{SAT}(v)=\mathcal{M}_i(\mathcal{R}_i(v))$, i.e., whether $\mathcal{R}_i$ reduces $\mathrm{SAT}$ to $L(\mathcal{M}_i)$ on these instances.
    • If yes, accept iff $\mathcal{M}_i$ accepts $w$
    • If no, accept iff $w\in\mathrm{SAT}$

The required enumeration is $\mathcal{M}'_1,\mathcal{M}'_2,\ldots$, since each of those either decides $\mathrm{SAT}$ (up to finitely many cases) or some other language in NP to which $\mathrm{SAT}$ can truly be reduced.

I do not see how to apply this idea to recursively present all P-hard languages. One could readily recursively present, e.g., all P-hard languages in ExpTime (which would suffice as an answer in the other thread), but I don't see why the P-hard languages in general (including undecidable ones) should even be countable.

  • $\begingroup$ In Ryan's answer, it's enough to enumerate P-hard languages in EXP, which you already see how to do. If you want all (computable) P-hard languages, this method won't show that the set of P-hard computable languages is recursively presentable, since I'm pretty sure the set of (total) computable languages is not recursively presentable. $\endgroup$ Dec 20, 2017 at 23:11
  • $\begingroup$ Yes, the set of total recursive functions is not recursively enumerable. I also agree on the other part (as mentioned in my post). The fact that P-hard languages are not countable also seems to be quite easy t show. $\endgroup$
    – mak
    Dec 20, 2017 at 23:18
  • $\begingroup$ Ok, self-answered this one now and will fix the other thread. Apologies for the unnecessary commotion. $\endgroup$
    – mak
    Dec 20, 2017 at 23:30

1 Answer 1


After some further thought, partly triggered by writing a long question, it seems clear that the set of P-hard languages is not countable, and therefore not recursively presentable either.

To see why, consider languages $L$ that satisfy $w\in 2SAT$ iff $ww\in L$. 2SAT can be log-space reduced to any such language by definition. However, there are countably many words that do not have the form $ww$, and the languages might differ on their membership status. Hence we obtain uncountably many P-hard languages of this form.

Moreover, it does not seem possible to recursively present all decidable P-hard languages either, since there is no effective enumeration of all Turing machines that are guaranteed to halt (this seems to be the minimal requirement for applying Schöning's construction).


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