I know that $\mathsf{P}^A = \mathsf{EXP}$ for any $\mathsf{EXPTIME}$-complete language $A$.

Is it true that $\mathsf{DTIME}^A(n^k) = \mathsf{EXP}$ for any fixed $k$ and any $\mathsf{EXPTIME}$-complete oracle $A$?
If not, what do these complexity classes equal and why?

I am just confused because then it seems to me that we would then have $\mathsf{DTIME}^A(n^k) = \mathsf{DTIME}^A(n^j)$ for $k < j$ and this would contradict the fact that the Time Hierarchy Theorem holds under any oracle.


1 Answer 1


It is not true that for $A$ being $\sf EXP$-complete ${\sf DTIME}^A(n^k) = {\sf EXP}$, but you are right with ${\sf P}^A={\sf EXP}$.

Here is the reason for this. In order to make use of the oracle you have to transform your problem via a reduction. This reduction is a polynomial reduction. The running time of this particular reduction might need $\omega(n^k)$ steps. In this case you cannot use the oracle in the desired way.

As you already observed ${\sf DTIME}^A(n^k) \neq {\sf EXP}$ because this would imply for all $k,j$ that ${\sf DTIME}^A(n^k)={\sf DTIME}^A(n^j)$, which cannot be true since the time hierarchy theorem relativizes.

  • 1
    $\begingroup$ Thanks. I just wish that textbook writers were more explicit on how oracle Turing Machines work (or don't work) so I could have avoided this confusion in the first place. $\endgroup$
    – Ari
    Dec 7, 2014 at 19:44

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