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Say that a class $C_1$ is superior to a class $C_2$ if there is a machine $M_1$ in class $C_1$ such that for every machine $M_2$ in class $C_2$ and every large enough $n$, there is an input of size between $n$ and $n^2$ on which $M_1$ and $M_2$ answer differently.

  1. Is DTIME($n^{1.1}$) superior to DTIME($n$)?

  2. Is NTIME($n^{1.1}$) superior to NTIME($n$)?

This is an exercise from book Computational Complexity: A Modern Approach. But I have no idea to handle it. Does it have relationship with theorem that DTIME$(n)$ $\subsetneqq$ DTIME($n^{1.5}$)?

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  • $\begingroup$ Is the large enough $n$ independent of $M_2$? Or can it be different for every $M_2$? $\endgroup$ Commented Mar 23, 2016 at 4:59
  • $\begingroup$ I think they are independent. So the negation is $\exists M_2$ and $n$, s.t. $M_1(x)=M_2(x), \forall x \in \{x: n<|x|<n^2\}$ $\endgroup$
    – Atom Lee
    Commented Mar 23, 2016 at 5:46
  • $\begingroup$ Take a closer look at the proof of the time hierarchy theorem. $\endgroup$ Commented Mar 23, 2016 at 6:44
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    $\begingroup$ What have you tried? Where did you get stuck? We do not want to just do your (home-)work for you; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for a relevant discussion. If you are uncertain how to improve your question, why not ask around in Computer Science Chat? You may also want to check out our reference questions. $\endgroup$
    – Raphael
    Commented Mar 23, 2016 at 9:08

1 Answer 1

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The proof of the time hierarchy theorems can be paraphrased like this. We want to show that $\mathsf{X} \subsetneq \mathsf{Y}$ for appropriate $\mathsf{X},\mathsf{Y}$. The idea is to construct a diagonalizing machine $M \in \mathsf{Y}$ (slight abuse of notation here) whose language is different from all languages in $\mathsf{X}$. On input $x$, the machine $M$ runs $x$ on $x$ up to some specified time bound, and then answers the opposite. If the simulation is efficient enough, $M \in \mathsf{Y}$.

Your question is very similar. Go over the proof of the time hierarchy theorems, and check whether the diagonalizing machines there fit your bill.

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  • $\begingroup$ If $n$ is indeed independent of $M_2$ then I think it is tough to prove the propositions. At least $DTIME(n^{1.1})$ won't be superior to $DTIME( (1+\epsilon)n+2)$ by linear speedup theorem. $\endgroup$ Commented Mar 23, 2016 at 11:41
  • $\begingroup$ Say $n=N$ is the large enough $n$ in question. Then we can construct a machine that runs $M_1$ for strings of length between $N$ and $N^2$ and any other $DTIME((1+\epsilon)n+2)$ machine for other lengths. The resulting machine is $DTIME( \max\{(1+\epsilon)n+2, N^{2.2}\})$ which is a $DTIME((1+\epsilon)n+2)$ by linear speedup theorem. $\endgroup$ Commented Mar 23, 2016 at 11:47
  • $\begingroup$ The value of $n$ is allowed to depend on $M_2$. That's the order of quantifiers in the question: for every $M_2$ there is a large enough $n$. If they wanted "large enough" to be uniform, they would have switched these quantifiers. $\endgroup$ Commented Mar 23, 2016 at 13:39

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