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On the one hand, every algorithm (by which I will mean a Turing machine which halts on all inputs) has computable runtime. On the other hand, by diagonalization for every computable function $f$ there is a computable set $X_f$ such that no Turing machine computing $X_f$ runs in time $O(f)$. This is a good exercise. The obvious approach doesn't work: the ...


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Yes, of course. $2^{f(n)}$ is asymptotically larger than $f(n)$, so you can come up with an unending sequence of larger and larger running times. The answer to your other questions are also yes, by the time hierarchy theorem.


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