Languages decidable by weak models of computation often have certain necessary characteristics, e.g. the pumping lemma for regular languages or the pumping lemma for context-free languages. Such characteristics are useful for determining when a language cannot be decided by that weak model of computation.

Do Turing-decidable languages have any such necessary characteristics? If not, why not?

Such a characteristic should avoid mention of Turing machines.

Motivation: The only way I know to show that a language is not Turing-decidable is diagonalization, or reducing to a language where diagonalization has already been applied. It would be nice to have another method.

I suspect that there aren't any nice characteristics of this type known, or they'd probably be very well-known; but an argument for why they can't exist could be illuminating.

I'd also be interested in the same question with the phrase "Turing-decidable" replaced with any complexity class, e.g. NP. This question seems more tractable (as there are fewer languages under examination).

Edit: Given the comments, I'll be a bit more precise about what I mean by "characteristics." I would be thrilled with either of the following two possibilities:

1) A property that allowed one to show a language was not in the relevant class (Turing-decidable, NP, whatever) without using diagonalization.


2) A property that can be stated in a language too weak to define Turing machines.

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    $\begingroup$ @Daniel Litt: we have been using 'computability theory', which is the hip lingo for the field of recursion theory. I changed the tag to match. $\endgroup$ Jul 21, 2010 at 19:25
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    $\begingroup$ @Mikola: Rice's theorem says that for any nontrivial property of a computable function, the set of functions that have the property is undecidable given an index for the function. This does not say there are no nontrivial properties of computable functions, just no decidable ones. $\endgroup$ Jul 21, 2010 at 19:26
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    $\begingroup$ To the original question: my opinion is that the answer is going to be "no". The main difficulty is choosing a reasonable, rigorous definition of a "characteristic" in order to prove this negative result. I don't know of one, but maybe someone else here has one. $\endgroup$ Jul 21, 2010 at 19:31
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    $\begingroup$ @Carl Mummert: Thanks! @Everybody else: Let it be known that any hipness imbued on this question comes from Carl Mummert. $\endgroup$ Jul 21, 2010 at 19:33
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    $\begingroup$ Another way you can think about pumping-type lemmas is that they are universal objects in the categories of regular/context-free/whatever languages. Modulo local string substitutions, they can all be converted into some giant master language. So, if you take this perspective then the equivalent pumping lemma for a complete language is just reduction to some other complete language. $\endgroup$
    – Mikola
    Jul 21, 2010 at 19:33

2 Answers 2


As far as I know, the answer is no. We don't have an argument not based (directly or indirectly) on digonalization for even proving there is a language outside Dec (The set of Turing decidable languages).

Characterizing a class means having criteria for showing when a language is in the class and when a language is not. The second part would give us tools to prove lowerbounds on such classes.

For very small classes like $AC^0$ and $AC^0[p]$ there are combinatorial arguments (also for monotone-$AC^0$ circuits), but the current state of knowledge is consistent with $AC^0[6] = NP$. The Razbarov-Rudich (RR) Natural Proofs theorem is related to why the previous combinatorial arguments haven't been successful in separating larger classes. In simple terms, there is no "natural proof" for separating the larger classes unless there is no pseudo-random number generators (many believe they do exists), and all previous proofs were natural. Tim Gowers wrote a number of interesting blog posts on the topic: http://gowers.wordpress.com/2009/09/22/a-conversation-about-complexity-lower-bounds/

The reason that we have pumping lemmas for $Reg$ (regular languages) and $CFL$ (context-free languages) is that the machine models for them are very specific which gives the strings accepted by them special combinatorial structure. For classes defined by Turing machines this is not the case. If a complexity class contains some basic functions (say $AC^0$) and is closed under composition, proving lowerbounds for it becomes a rather difficult task. Note that $Reg$ does not contain $AC^0$ ($\{0^n1^n: n \in \omega \}$ is in $AC^0$ but not in $Reg$.)


part 1. I don't have a theoretical justification right now, but the history of complexity theory shows that the structure of much smaller complexity classes are at least very difficult to understand. Many experts "believe" that there are pseudo-random number generators (PRNG), so RR's Natural Proof theorem applies and there is no "natural" property of sets/functions that can be used to prove a lowerbound for the complexity classes satisfying the condition of the RR theorem. I think a person accepting the existing of PRNGs can at least claim that there are no simple combinatorial structures for these complexity classes.

part 2. You are looking for a combinatorial property of sets of strings in a complexity class. The computational complexity of checking a property is a good measure. Any property stated in the first order logic (on a finite structure) is $AC^0$. The RR theorem applies to properties which are in $P$, i.e. can be checked in polynomial time given the truth table of the functions (characteristic functions of sets restricted to inputs of size n). Encoding Turing Machines (TM) is not difficult, and I think it can be done if one goes a little above $AC^0$, e.g. $TC^0$ is probably capable of checking if a string encodes a TM. You are looking for a property that holds for all functions in a class, but not for those outside it. The negation of it would be a property of the kind considered in RR's theorem. By these considerations, I think there are no properties of the kind you are looking for unless there are no PRNGs.


$AC^0$ is class of sets of strings accepted by (uniform) circuits of polynomial size and bounded depth, i.e. there is a constant $d$ and a polynomial $p(n)$ and a sequence of circuits $\{C_n\}_{n \in \omega}$ using unbounded fan-in $\land$,$\lor$,$\lnot$ gates and of depth $d$ and size $p(n)$ that accepts them.

$AC^0[n]$ is the same as $AC^0$ but allows $mod_n$ gates. $p$ is a prime number.

$TC^0$ is the same as $AC^0$ but allows majority/threshold gates.

A property $Q$ is natural if it is:

  1. useful: i.e. does not hold for simple functions (functions of low complexity) but holds for some function in the larger class,

  2. constructive: i.e. can be checked in polytime from the truth table of the functions,

  3. large: i.e. for some k and all large enough n, the property holds for at least $\frac{1}{n^k}$ of the functions.

  • $\begingroup$ Is a counting argument diagonalization? There are uncountably many languages, of which countably many are Turing decidable. $\endgroup$ Jul 21, 2010 at 22:44
  • $\begingroup$ @Qiaochu Yuan: The argument that the number of languages (which is $P(N)$) is not countable is based on diagonalization. $\endgroup$
    – Kaveh
    Jul 21, 2010 at 23:14
  • $\begingroup$ @Qiaochu Yuan: The criterion I'm looking for ought to be effective -- I'd like to have an example of a language not contained in some class, not simply an existence proof. @Kaveh: 1) It's not obvious to me that Turing machines contain no relevant combinatorial structure. Can you justify this claim? 2) Your definition of a "characterization" is a bit stronger than what I'm looking for. See the second half of the question. $\endgroup$ Jul 22, 2010 at 0:42
  • $\begingroup$ @Daniel: I will add my comment to the end of my answer since it does not fit here. $\endgroup$
    – Kaveh
    Jul 23, 2010 at 1:14

It should be noted that the idea of finding a "property" that can be used to show non-membership in a certain class of languages was a driving force in computability theory in the 1950's or so. Specifically, Emil Post posed the question: are there computably enumerable (aka "c.e.", also known as Turing-recognizable (but not necessarily decidable)) sets that are undecidable and strictly Turing-below the Halting problem (note that any c.e. set is Turing-below the halting problem). That is, we want a c.e. set A such that $\emptyset <_T A <_T H$, where $H$ is the halting problem. This became known as Post's Problem. Further, he proposed Post's Program, which was to find a property of c.e. sets that implied the required conditions, and then to find an A with this property. Examples of such candidate properties are being "simple" (its a technical term).

Post's problem was solved in the positive by Friedberg and Muchnik, but by using a clever sort of delayed diagonalization, a sort of "injury argument". However, this did not show that Post's program could be solved in the positive, but indeed Harrington and Soare showed in 1991 that there is such a property satisfying Post's conditions (and a corresponding set A satisfying this property).

See the wikipedia article: http://en.wikipedia.org/wiki/Turing_degree#Post.27s_problem as well as the Harrington-Soare article explaining some of the background (its open-access): http://www.ncbi.nlm.nih.gov/pmc/articles/PMC52904/

  • $\begingroup$ This looks very good. $\endgroup$ Jul 22, 2010 at 17:17
  • $\begingroup$ These only apply to sets that are already r.e., though. Also, the proofs of these results use the usual computability-theoretic techniques, including priority arguments, which are souped-up diagonalization arguments. So this still doesn't give a diagonalization-free method. $\endgroup$ Jul 22, 2010 at 19:46

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