patternMinor
Are runtime bounds decidable for anything nontrivial?
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nontrivialareanythingruntimedecidableforbounds
Problem
Problem Given a Turing machine $M$ which has known runtime ${O}(g(n))$ with respect to input length $n$, is the runtime of $M \in {O}(f(n))$?
Is the above problem decidable for some nontrivial pairs of $g$ and $f$?A solution is trivial if $g(n) \in O(f(n))$.
This is related to the problem Are runtime bounds in P decidable? (answer: no). One can derive from Viola's answer that if $f(n)\not \in o(n)$ and $f(n)\not \in O(g(n))$ then the problem is undecidable.
The requirement that $f(n)\not \in o(n)$ is because the $M'$ in Viola's proof need $O(n)$ time to find its input size. Thus Viola's proof could not work when $f(n)=1$.
It would be interesting if we can decide on the run time of sublinear time algorithms. A special case is when we have arbitrary $g(n)$ and $f(n)=1$.
Is the above problem decidable for some nontrivial pairs of $g$ and $f$?A solution is trivial if $g(n) \in O(f(n))$.
This is related to the problem Are runtime bounds in P decidable? (answer: no). One can derive from Viola's answer that if $f(n)\not \in o(n)$ and $f(n)\not \in O(g(n))$ then the problem is undecidable.
The requirement that $f(n)\not \in o(n)$ is because the $M'$ in Viola's proof need $O(n)$ time to find its input size. Thus Viola's proof could not work when $f(n)=1$.
It would be interesting if we can decide on the run time of sublinear time algorithms. A special case is when we have arbitrary $g(n)$ and $f(n)=1$.
Solution
Here are a few remarks which could be relevant:
- Kobayashi proved that a TM running in time $o(n\log n)$ accepts a regular language (and so runs in time $O(n)$); recently this has been extended to non-deterministic TMs (Tadaki, Yamakami and Lin).
- Machines running in time $o(n)$ actually run in constant time (consider any $n$ for which the running time is less than $n$; adding characters to the end doesn't affect the TM).
Context
StackExchange Computer Science Q#12518, answer score: 7
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