2
$\begingroup$

I'd like your help with the following question:

Assume we proved that $\mbox{BPP}\subseteq \Pi_2$ -What conclusions can you make?

BPP is the class of decision problems solvable by a probabilistic Turing machine in polynomial time, with an error probability of at most 1/3 for all instances,

$\Pi_2$ is the class of all languages $L$ such that there's a polynomial algorithm $M$ and a polynom $p$ so that $\forall x.x\in L\Leftrightarrow \forall u\in \{ 0,1 \}^*.\exists v \in \{ 0,1 \}^*.M(x,u,v)=1$.

We already know that $\mbox{BPP}\subseteq \Sigma_2$, so $\mbox{BPP}\subseteq \Pi_2\cap \Sigma_2$.

$\endgroup$
5
$\begingroup$

It is known. As your final statement says, $\mbox{BPP} \subseteq \Pi_2 \cap \Sigma_2$. It is called the Sipser–Gács–Lautemann theorem. (All though your "so" is somewhat misleading.)

It can be strengthened to $\mbox{BPP} \subseteq \mbox{MA} \subseteq \mbox{S}^P_2 \subseteq \Pi_2 \cap \Sigma_2 \subseteq \Pi_2$ (see also Arthur-Merlin protocol).

$\endgroup$
1
$\begingroup$

Just to add a minor point to Pal GD's answer, $BPP$ is closed under complementation. So, once you prove that $BPP \subseteq \Sigma_2$, you can easily deduce $BPP \subseteq \Sigma_2 \cap \Pi_2$. Also, (Just to improve the inclusions) even better $BPP \subseteq NP^{BPP} \subseteq MA \subseteq S_{2}^{P} \cdots$ is known.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.