Assume $NTime(2^n) \subseteq NSpace(n^k)$, for some fixed $k$. Is it possible to imply that $EXP = PSPACE$? and what about $NEXP = PSPACE$? It seems the answer might be YES, because this question seems to be equivalent to the open question $EXP=PSPACE$?
Can it be shown using some padding argument?

  • $\begingroup$ Are you aware of Savitch's theorem? $\endgroup$
    – dkaeae
    Commented Jun 14, 2019 at 14:28
  • $\begingroup$ I can see why it leads to NEXP=PSPACE. But how Savitch theorem is related to EXP=PSPACE? $\endgroup$
    – LioH
    Commented Jun 16, 2019 at 15:01

1 Answer 1


$\let\c\mathrm$The assumption $$\c{NTIME}(2^n)\subseteq\c{NSPACE}(n^k)\tag1$$ indeed implies $$\c{PSPACE=EXP=NEXP}$$ by a simple padding argument. The inclusions $$\c{PSPACE\subseteq EXP\subseteq NEXP}$$ hold unconditionally. In order to show $\c{NEXP\subseteq PSPACE}$, take an arbitrary language $L\in\c{NEXP}$, and let $c$ be such that $L\in\c{NTIME}(2^{n^c})$. Then the language $$L'=\{(w,1^n):w\in L,n\ge|w|^c\}$$ is in $\c{NTIME}(2^n)$, hence $$L'\in\c{NSPACE}(n^k)\subseteq\c{DSPACE}(n^{2k})\subseteq\c{PSPACE}$$ by (1) and Savitch’s theorem. Since $L$ is polynomial-time reducible to $L'$ by means of the function $w\mapsto(w,1^{|w|^c})$, it follows that $L\in\c{PSPACE}$, too.


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