A language $L$ is PSpace-complete, if it meets two conditions:

  1. It is in PSpace.
  2. Every other PSpace-complete language reduces to it in polynomial time.

Question: suppose we change the second condition to polynomial space (instead of time)? Why is it so that then SAT would be PSpace-complete?


1 Answer 1


Every language $X$ in PSPACE would be complete under your proposed definition, except for $\emptyset$ and $\Sigma^*$. You could reduce any PSPACE language $Y$ to $X$ by a reduction that first decides whether its input $w$ is in $Y$ and then maps to a fixed "yes" instance of $X$ if so, and a fixed "no" instance if not.

Note also that the actual definition of PSPACE-completeness requires that every language in the whole of PSPACE reduces to the complete language. The definition you've given is circular, since it defines PSPACE-completeness in terms of itself.


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