There are already good answers from several perspectives regarding the "hardness" of $PSPACE$-complete problems, such as this: What is practical difference between NP and PSPACE-complete?

But what are the practical implications when we actually try to solve (decide) a $PSPACE$-complete problem on a deterministic Turing machine? If $PSPACE \neq NP$ (or $PSPACE \neq P$, should it make a difference) then obviously it will take super-polynomial (exponential?) time. But what about space? Is there some $k>1$ that we know of such that it will definitely take $\Omega(n^k)$ space? And do we know for sure that it won't take more than $O(n^l)$ space for some $l$?

(Or is the question itself stated badly for some reason?)


1 Answer 1


By definition, PSPACE consists of all languages decided by some Turing machine using polynomial space. So every language in PSPACE can be decided by some Turing machine using space $O(n^\ell)$ for some $\ell$. The space hierarchy theorem ensures moreover that for any $\ell$ there is a language in PSPACE which requires space $\Omega(n^\ell)$.

Any PSPACE-hard language requires space $\Omega(n^k)$ for some $k$, since otherwise any language in PSPACE would be decided by a Turing machine using subpolynomial space, and this contradicts the space hierarchy theorem.

For any $k > 0$ there is a PSPACE-hard language which can be decided in space $O(n^k)$. Indeed, take any PSPACE-complete language $L$. It can be decided using space $O(n^\ell)$ for some $\ell$. Consider now the language $L' = \{(x,0^{|x|^{\ell/k}}) : x \in L\}$, which is PSPACE-hard by reduction from $L$. An algorithm for $L$ can be converted to one for $L'$ which uses $\log n$ space to verify the input structure and $O(|x|^\ell)$ for the rest. Since $n \geq |x|^{\ell/k}$, $O(|x|^\ell) = O(n^{(k/\ell)\ell}) = O(n^k)$.

  • $\begingroup$ Regarding the "$l$", I meant if there is a fixed $l$ which no PSPACE-complete problem is above. But is that true - are there really PSPACE-complete problems in $DSPACE(n)$? How can that be? (I know, this kind of question makes no sense, but it just seems so very unintuitive!) $\endgroup$
    – Duke
    Commented Feb 13, 2017 at 20:21
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    $\begingroup$ My answer proves that there is a PSPACE-complete problem in $\mathsf{DSPACE}(n)$, or even in $\mathsf{DSPACE}(n^{0.1})$, for that matter. Regarding your second question, the space hierarchy theorem shows that no fixed $l$ works. $\endgroup$ Commented Feb 13, 2017 at 20:24
  • $\begingroup$ Yeah, I got your proof - more or less... It's just so unintuitive, it contradicts my sense of hardness of PSPACE-complete problems. Anyway, this is such a great answer!! Thanks a lot - I'd very much like to vote you up, but my reputation is too low. Sorry! $\endgroup$
    – Duke
    Commented Feb 13, 2017 at 20:29
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    $\begingroup$ Note that your "requires space $\Omega(n^k)$" requires either the weak definition of $\Omega$ or considering the bit-padded versions, since there are PSPACE-complete languages which are trivial on all even input lengths. ​ ​ $\endgroup$
    – user12859
    Commented Feb 13, 2017 at 21:19

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