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In "Computational Complexity, A modern approach", Arora & Barak proof the following claim (Claim 5.11.2):

Suppose that $NTIME(n) \subseteq DTIME(n^{1.2})$. Then $\Sigma_2TIME(n^{8}) \subseteq NTIME(n^{9.6})$

Which they prove in the following way:

  • If $L \in \Sigma_2TIME(n^8)$, then $\exists$ TM M which runs in time $O(|x|^8)$ s.t. $x \in L \leftrightarrow \exists u \in \{0,1\}^{c|x|^8} \forall v \in \{0,1\}^{d|x|^8} M(x,u,v) = 1$
  • But if $NTIME(n) \subseteq DTIME(n^{1.2})$, then

by a simple padding argument, we have a deterministic algorithm $D$ that on inputs $x, u$ with $|x| = n$ and $|u| = cn^8$ runs in time $O((n^8)^{1.2}) = O(n^{9.6})$ and returns $1$ iff there exists some $v \in \{0,1\}^{dn^8}$ such that $M(x,u,v) = 0$

I struggle to see what this "simple padding argument" might be? That is, how do we get respective deterministic algorithm $D$ if we have the TM $M$?

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  • $\begingroup$ Isn't it clear enough that standard padding technique is able to demonstrate that $DTIME(n^{1.2})\supseteq NTIME(n)\implies DTIME(n^{9.6})\supseteq NTIME(n^8)$? Hint: $D$ mimics $M$ by first padding the input length to $n^8$ (with trailing dummy symbols) then enters the start state of $M$ and runs $M$ on the padded input. $\endgroup$ – Thinh D. Nguyen Aug 21 at 3:49
  • $\begingroup$ Then $D$ will play the role of a de-quantifier machine that peels off the $\forall v$ quantifier for you. $\endgroup$ – Thinh D. Nguyen Aug 21 at 3:55
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The language $$L_1 = \{(x,u) ~|~ \exists v\in \{0,1\}^{d|x|^8}M(x,u,v) = 0\}$$ belongs to $NTIME(n)$ by design, since $|v| = O(|x| + |u|) = O(|x|^{8})$. By the assumption of the theorem, $L_1$ is decidable by some deterministic $D$ in $O((|x|+|u|)^{1.2}) = O(|x|^{9.6})$ steps. The recipee for $D$ isn't important, but the existence of such algorithm $D$ leads to the last step of the proof.

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