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Does $\mathrm{EXP}^\mathrm{EXP}=\mathrm{EXP}$?

Here is my thought: $\mathrm{EXP}$ machine can ask $2^{O(n)}$ queries to the oracle, and each oracle would itself solve an exponential time problem in a single step. So the total power would be $2^{O(n)}\times2^{O(n)}$ which would still be in EXP.

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  • $\begingroup$ Your statement is not correct, and EXP with oracle in EXP is not equal to EXP. Can you credit the original source? $\endgroup$ – lox Aug 11 '19 at 18:02
  • $\begingroup$ Do you have a proof or any idea why the the statement is not correct?This is an question from a book I am reading and the original question did not implement that the classes are equal.Sorry for the misunderstanding. @lox $\endgroup$ – fgdhdfg Aug 11 '19 at 18:14
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    $\begingroup$ Can you name the book title and give the context in which this question appeared? $\endgroup$ – lox Aug 11 '19 at 18:41
  • $\begingroup$ I am copying the question from the book:"Show that EXP with oracle in EXP is not equal to EXP".I got the thought and posted it up their (posting wrong answer in the internet gets more attention) . Pay attention that In my question I did not state that they are equal. @lox $\endgroup$ – fgdhdfg Aug 11 '19 at 18:59
  • $\begingroup$ What about book title? $\endgroup$ – Evil Aug 11 '19 at 22:36
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Your definition of EXP is a bit off (you're thinking of $\textbf{E} = \textbf{DTIME}(2^{O(n)})$ instead of $\textbf{EXP} = \textbf{DTIME}(2^{n^{O(1)}})$), but either way the assertion that "each oracle would itself solve an exponential time problem in a single step" is false.

This is because, given exponential time, the machine can write (say) $2^n$ bits to the oracle's input tape, at which point the oracle is allowed to run exponential-time algorithms as measured relative to the size of the $2^n$ bit input (and therefore doubly-exponential in $n$). Thus, just by padding the input to an exponential size and calling the oracle once, you can solve any problem in 2EXP, violating the time hierarchy.

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