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Relation between $P$ vs $\mathit{NP}$ and $P^{\mathit{NP}}$ vs $\mathit{NP}^{\mathit{NP}}$ questions?

Submitted by: @import:stackexchange-cs··
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Problem

Given that both $P \subset \mathit{NP}\ ?$ and $P^{\mathit{NP}} \subset \mathit{NP}^{\mathit{NP}}\ ?$ are open questions.

Is it possible that
$P \subset \mathit{NP}$ and $P^{\mathit{NP}} =\mathit{NP}^{\mathit{NP}}$?

Or is it the case that $P \subset \mathit{NP}$ iff $P^{\mathit{NP}} \subset \mathit{NP}^{\mathit{NP}}$?

Solution

We know that $\mathsf{P} \subseteq \mathsf{NP}$ and that $\mathsf{P}^{\mathsf{NP}} \subseteq \mathsf{NP}^{\mathsf{NP}}$; this follows directly from the definitions.

More interesting are the two reverse inclusions, $\mathsf{NP} \subseteq \mathsf{P}$ and $\mathsf{NP^{NP}} \subseteq \mathsf{P^{NP}}$. The former implies the latter: indeed, if $\mathsf{NP} \subseteq \mathsf{P}$ then $\mathsf{P} = \mathsf{NP} = \mathsf{P^{NP}} = \mathsf{NP^{NP}}$. The latter is not known to imply the former, and moreover, there are relativized worlds with respect to which the former is false but the latter is true. This means that there is an oracle $O$ (constructed by diagonalization) such that if all Turing machines have additional oracle access to $O$, then $\mathsf{P} \neq \mathsf{NP}$ but $\mathsf{P^{NP}} = \mathsf{NP^{NP}}$. Such an oracle was constructed by Heller, Relativized Polynomial Hierarchies Extending Two Levels.

Context

StackExchange Computer Science Q#148038, answer score: 4

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