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I understand that the language:

$\operatorname{SPACE-TMSAT} = \{⟨M, w, 1^n⟩ : \text{DTM $M$ accepts $w$ in space $n$}\}$

is in PSPACE since it doesn't use more than $n$ space. But to prove that it is PSPACE-complete, we need to do the reduction from any other PSPACE language to it. If we assume the language to run on a DTM using $O(n^k)$ space, how could we reduce it to SPACE-TMSAT?

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Suppose $L$ is accepted by some machine $M$ using up to $p(n)$ space, for some polynomial $p(n)$. We map $w$ to $\langle M, w, 1^{p(|w|)} \rangle$.

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  • $\begingroup$ thank you! I guess we assume in the definition of SPACE TMSAT, n is the size of the input. $\endgroup$ Apr 4, 2019 at 22:32
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    $\begingroup$ No, $n$ is completely arbitrary. $\endgroup$ Apr 5, 2019 at 4:03
  • $\begingroup$ So the input is a binary string consisting Turing machine, machine's input, and a number of ones at the end? $\endgroup$ Apr 5, 2019 at 13:39
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    $\begingroup$ That's just right. $\endgroup$ Apr 5, 2019 at 14:00

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