Condidering the proof, NLOGSPACE$\subset$PSPACE

I wrote following proof:

NLOGSPACE = NSPACE$(\log n)$ $\hspace{15pt} \because$ Definition of NLOGSPACE

NSPACE$(\log n)$ $\subseteq$ DSPACE$(\log^2 n)$ $\hspace{15pt} \because$ Theorem(1)

We let $m = \log n$, then,

DSPACE$(m^2)$ $\subset$PSPACE $\hspace{15pt} \because$ Definition(2)

Therefore, NLOGSPACE$\subset$PSPACE.

Theorem(1): If $S$ is a space constructible function and $S(n) \ge \log n$, then NSPACE$(S(n)) \subseteq$ DSPACE$(S^2(n))$.

definition(2): PSPACE = DSPACE$(n)$$\cup$DSPACE$(n^2)$$\cup$DSPACE$(n^3)$$\cup$...$\cup$DSPACE$(n^k)$$\cup$...

But I am still not that sure if that "We let $m = \log n$" is ok, is this proving NLOGSPACE$\subset$PSPACE?


1 Answer 1


What would you think about a proof that says the following?

  • $NP \subset DTIME(2^n)$
  • let $m = \log n$, then $DTIME(2^n) = DTIME(m) \subset P$
  • Therefore $P = NP$

Here, you have indeed $DPSACE(log^2n)\subset PSPACE$ but the argument is weird.

Also please note that Savitch's theorem already proves that $NPSPACE \subset PSPACE$, and you have the result you want, since $NLOGSPACE \subset NPSPACE$.

  • $\begingroup$ Thanks for answering! How Savitch's theorem proves NPSPACE⊂PSPACE? also, where is NLOGSPACE⊂NPSPACE? $\endgroup$
    – t24akeru
    Commented Apr 24, 2021 at 9:31
  • $\begingroup$ The first question is true because if $f$ is a polynomial function, then $f^2$ is a polynomial function too. And $NPSPACE = \underset{f\text{ polynomial}}{NSPACE}(f(n)) \subset \underset{f\text{ polynomial}}{DSPAC}E(f(n)^2) = PSPACE$. $\endgroup$
    – Nathaniel
    Commented Apr 24, 2021 at 9:40
  • $\begingroup$ The second question is true because $\log n = O(n)$, and so $NLOGSPACE = NSPACE(\log n) \subset DSPACE((\log n)^2) \subset DSPACE(n^2) \subset PSPACE$ $\endgroup$
    – Nathaniel
    Commented Apr 24, 2021 at 9:43
  • $\begingroup$ I see. At $DSPACE(f(n)^2) = PSPACE$, why can you write "=", not "$\subset$"? $\endgroup$
    – t24akeru
    Commented Apr 24, 2021 at 10:06
  • $\begingroup$ Because the union is done on all polynomial functions. $\endgroup$
    – Nathaniel
    Commented Apr 24, 2021 at 10:13

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