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According to answers here Are there subexponential-time algorithms for NP-complete problems? $\mathsf{NP}$ complete problems can be in $DTIME[2^{n^{1/\alpha}}]$ for $\alpha>1$.

Now supposing $DTIME[2^{n^{1/\alpha}}]\subseteq\mathsf{P/Poly}$ holds for every $\alpha>1$ then does it mean $\mathsf{NP}\subseteq\mathsf{P/Poly}$ holds?

What are the consequences of $DTIME[2^{n^{1/\alpha}}]\subseteq\mathsf{P/Poly}$?

Would this impact $\mathsf{subexp}\subseteq\mathsf{P/poly}$ problem?

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Turbo
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If $DTIME\left[2^{n^{\frac{1}{\alpha}}}\right]\subseteq P/Poly$ for some $\alpha>1$ then $NP\subseteq P/Poly$, since this means we can put an NP-complete problem in $P/Poly$, because as you mentioned, for all $\alpha>1$ there exists some NP complete problem in $DTIME\left[2^{n^{\frac{1}{\alpha}}}\right]$. To see why the inclusion holds in that case, notice that for any other language in $NP$ you can first compute the reduction in polynomial time, and then work with the machine taking advice for the complete problem.

In order for the above to work, the reduction $f$ to our NP-Complete problem must have the property that $|f(x)|$ depends only on $|x|$ (the length of the output depends only on the length of the input, and not structure). However, we need not worry, since the reduction in Cook-Levin theorem has this property (verify), and the reduction from $SAT$ to padded SAT:

$SAT' = \{\langle \varphi,w\rangle \mid \varphi\in SAT \text{ and } |w|=|\varphi|^k \}$

also has this property.

A well known consequence of $NP\subseteq P/Poly$ is the Karp-Lipton theorem.

As for your second question, use the fact that $SUBEXP=\bigcap\limits_{\epsilon>0}DTIME\left[2^{n^\epsilon}\right]\subseteq DTIME\left[2^{n^{\frac{1}{\alpha}}}\right] $, for all $\alpha>1$.

Ariel
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  • I seriously doubt this. I will wait for other clearer answers. – Turbo Oct 23 '16 at 08:06
  • Well, there is nothing much i can do with this kind of comment. If you see a flaw in my proof, please point it out, I don't see any. – Ariel Oct 23 '16 at 08:58
  • I do not see any. I am not sure if it is possible though seems weird. – Turbo Oct 23 '16 at 08:59
  • I think $SUBEXP^{NP}\subsetneq P/poly$ so either $SUBEXP\subsetneq P/poly$ or ${NP}\subsetneq P/poly$ should hold. Here we get both. – Turbo Oct 23 '16 at 09:47
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    @AJ. That's the beauty of mathematical proof: it's equipped to convince you of the falsehood of your intuitions. – Raphael Oct 23 '16 at 10:20
  • Where does $SUBEXP^{NP}\subseteq P/Poly$ come from? We don't expect $NP$ to lie inside $P/Poly$, let alone $SUBEXP^{NP}$. – Ariel Oct 23 '16 at 13:41
  • @Ariel $\subsetneq$. – Turbo Oct 23 '16 at 18:34
  • Obviously they are not equal, since $P/Poly$ contains undecidable problems. The inclusion is what's not known. – Ariel Oct 23 '16 at 18:41
  • @Ariel no I think $subexp^{NP}$ is not in $P/poly$ is known so either one of what I said should be true. – Turbo Oct 23 '16 at 19:04
  • I'm not sure @Ariel, but is it possible to use padding argument and translational lemma to show this results. I suspect yes! – user777 Apr 10 '21 at 23:48