$\newcommand{\cc}[1]{\mathsf{#1}}$Does a theorem along the following lines hold: If $g(n)$ is a little bigger than $f(n)$, then $\cc{NTIME}(g) \cap \cc{coNTIME}(g) \neq \cc{NTIME}(f) \cap \cc{coNTIME}(f)$?

It's easy to show that $\cc{NP} \cap \cc{coNP} \neq \cc{NEXP} \cap \cc{coNEXP}$, at least. Proof: Assume not. Then $$\cc{NEXP} \cap \cc{coNEXP} \subseteq \cc{NP} \cap \cc{coNP} \subseteq \cc{NP} \cup \cc{coNP} \subseteq \cc{NEXP} \cap \cc{coNEXP},$$ so $\cc{NP} = \cc{coNP}$, and hence (by padding) $\cc{NEXP} = \cc{coNEXP}$. But then our assumption implies that $\cc{NP} = \cc{NEXP}$, contradicting the nondeterministic time hierarchy theorem. QED.

But I don't even see how to separate $\cc{NP} \cap \cc{coNP}$ from $\cc{NSUBEXP} \cap \cc{coNSUBEXP}$, as diagonalization seems tricky in this setting.


(This would've been a comment, but it doesn't render properly when I try that.)

"I don't even see how to separate" $\;\;\;$$\mathbf{U}$$\mathbf{q}$$\mathbf{uasiLIN}$ $\cap \; \mathbf{coUquasiLIN} \;\;\;$ from
$[$the linear-exponent version of QIP[2]$] \;\; \cap \;\; [$the linear-exponent version of coQIP[2]$] \;\;\;\;\;$.


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.