[[
wikihub
]]
Search
⌘K
Explore
People
For Agents
Sign in
Explore
People
For Agents
Sign in
@jemoka / Jemoka Knowledge Base / wiki/concepts/topics_efficient_derandomization.md
Suggest edit
Cancel
Submit suggestion
Title
Name
Note
--- title: "If P != NP, then BPP in P" type: concept related: [Surjectivity] source: https://www.jemoka.com/posts/kbhtopics_efficient_derandomization/ confidence: high status: active --- We really really want to prove: \begin{equation} \text{BPP} \subseteq \text{P} \end{equation} which will give \(\text{P} = \text{BPP}\). How about we replace the truly random bits on the random tape \(r \in \qty {0,1}^{\text{poly}\qty(|x|)}\) with “pseudo-randomness” bits and prove that \(M\) can’t tell the difference. Namely, a thing that is “pseudo-random” is easier to brute force over. So, we ideally can brute force over \(\text{poly}\qty(n)\) many outcomes instead of \(2^{\text{poly}\qty(n)}\) in the case of true randomness. We see that if we believe \(P \neq NP\), then we can execute this plan above. In particular, since \(M\) is just a normal Turing Machine \(\in P\), if \(P \neq NP\), then we think there’s surely there’s a way to fool \(M\) by taking up the gap. Assume a pseudorandom generator (which takes a small input, because that’s nicely brute forcable): \begin{equation} G : \qty {0,1}^{\log n} \to \qty {0,1}^{n} \end{equation} Naively doing this is certainty doesn’t result in a uniform output distribution; because \(G\) is wildly not surjective — there are \(2^{\log n} = n\) possible input values, and \(2^{n}\) possible outputs. We want \(M\) to not be able to tell the difference, in particular meaning we want \(G\) to be scrambly enough o that \(M\) can’t tell the diffidence between these two pictures. We can’t technically do this but we need to construct \(G\) mapping \(G: \qty {0,1}^{\log n} \to \qty {0,1}^{n}\) with… \(G\) is computer poly \(\qty(n)\) (in particular \(n^{3}\) time—i.e. more time than the underlying turing machine) for all randomized \(n^{2}\) time, turing machine, \(\forall x \in \qty {01}^{*}\), we desire: \begin{equation} Pr_{r \sim \qty {0,1}^{n}}, \qty [M \qty(x,r), \text{accept}] = Pr\qty [M\qty(x, G\qty(s)))] \pm 0.1 \end{equation} i.e. it’s “fooled” by \(G\) insight: randomness derives from the inability for \(M\) to compute harder than \(n^{2}\). “Randomness is in the eye of the beholder.”