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The top rated answer to Why, really, is the Halting Problem so important? lists a few examples for a noncomputable problem. However, these mostly involve an infinite search space. Are there noncomputable problems with a finite search space? If not, why not?

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  • $\begingroup$ It's been a long time since I had computational theory, arithmetic hierarchies and language complexity in college. So, I need a little fresh up. $\endgroup$
    – Karsten
    Commented Nov 5, 2017 at 19:55
  • $\begingroup$ What do you mean by a problem having an infinite search space? For example, the problem "Here is a Turing machine $M$ and a string $x$. Does $M$ halt when given input $x$?" is undecidable. Unlike a question such as "Is there any input for which $x$ halts?", this doesn't seem to have a "search space", as such. $\endgroup$ Commented Nov 5, 2017 at 20:39
  • $\begingroup$ As the answers in the linked questions says: The difficulty of a problem can be related to the number of quantifiers it has. Trying to show that there exists (∃) an object with an arbitrary property, you have to search until you find one. If none exists, there's no way (in general) to know this. So I was wondering if there are non-computable functions with a finite search space. If that is the right wording. $\endgroup$
    – Karsten
    Commented Nov 6, 2017 at 21:30
  • $\begingroup$ The standard halting problem, as described in my previous answer, isn't stated with any quantifiers at all. You might argue that there's an implicit existential quantifier (does there exist a computation path that halts?) but it's not really clear. $\endgroup$ Commented Nov 6, 2017 at 21:32

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Depends on what exactly you have in mind.

If the problem is to find an element with decidable (!) properties among a recursively enumerable set $A$ (of finitely represented elements), then yes: if $A$ is finite, the problem is trivially computable. Just check every element.

If $A$ is finite but unknown, it's more interesting but still: the problem is computable, since there is an algorithm that checks the every element in $A$, even though we can't point it out. See this for an example.

If the identifying property is undecidable, then the problem may be¹ uncomputable. For instance, the search space for the question "Does TM M halt on input x?" is finite -- just two possible answers -- but the problem is (in)famously undecidable.


  1. It's possible that the property as written down is not needed to identify the correct element.
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  • $\begingroup$ I should revise the question a little bit. Would you have a "real world" example / application for a case where $A$ is finite but the defining property is undecidable? I know the halting problem but it sometimes seem too abstract to explain it to other less technical versed persons. $\endgroup$
    – Karsten
    Commented Nov 6, 2017 at 20:58
  • $\begingroup$ @Karsten None of these? Or here? $\endgroup$
    – Raphael
    Commented Nov 6, 2017 at 21:52
  • $\begingroup$ Thanks for the links. I read up more on computable functions. I'm probably too hung up on this term "search space", i.e. the function should decide wether an element is in an infinite set. Practically all of the examples given in the linkes answers have this property. > Are there some inputs that make my program crash? E.g. here the function would have to check any possible input. Am I missing something? $\endgroup$
    – Karsten
    Commented Nov 6, 2017 at 22:26
  • $\begingroup$ @Karsten No, it wouldn't necessarily. The search space, as I understand it, is $\{0,1\}$ -- either the property is true or not. Just think of any correctness proof: we don't have to investigate all inputs individually to establish an algorithm is correct! The Pi question is my favorite example -- no, we don't have to check all decimal digits! $\endgroup$
    – Raphael
    Commented Nov 7, 2017 at 17:49
  • $\begingroup$ @Karsten But as for your question, yes, you'll get better answers if you can clarify what exactly you mean by "search space" respectively "search problem". $\endgroup$
    – Raphael
    Commented Nov 7, 2017 at 17:50

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