It's impossible for a problem to require exponential space without being exponential-time.

  1. Consider that if an $EXPSPACE~~complete$ problem can be solved in $2^n$ time. It will now fall into the class $EXPTIME$.
  2. Then $EXPSPACE~~complete$ problems are in $EXP$ if they can be solved in $2^n$ time. This means they can reduce into $EXP~~complete$ problems and vice versa.

To me, this should be easy to write a proof that $EXPTIME$ = $EXPSPACE$.

My intuition tells me that if $Exptime$ = $Expspace$; then $PSPACE$ != $EXPTIME$,

Because $PSPACE$ already is not equal to $EXPSPACE$.


As an amateur, what would make this reasoning be wrong or right?


1 Answer 1


I did not understand the first part of the question, but just so you know, its unknown whether $EXPTIME=EXPSPACE$ or not. What is known is that $EXPSPACE\subseteq \bigcup _cDTIME(2^{2^{n^c}})$ from the relations shown here.

The second part of the question: If $EXPTIME=EXPSPACE$ then $PSPACE\neq EXPTIME$, is absolutely correct: by the space hierarchy theorem, $PSPACE\subsetneq EXPSPACE$ and therefore $PSPACE\subsetneq EXPTIME$ as $EXPSPACE=EXPTIME$ (by our assumption).


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