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If a language L logspace reduces to SAT, does L also reduce to SAT in polynomial time?

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Shortest answer: Yes.

Short answer: Yes, for the same reason that LOGSPACE $\subseteq$ P.

Less short: If the number of tape squares used is logarithmic, then the number of possible tape configurations is polynomial, and since the tape configurations mustn't repeat, the time taken is polynomial too.

In horrible detail: Let $M$ be a TM that performs a logspace reduction of $L$ to SAT. Say that $M$ has $N$ states and a tape alphabet with $A$ symbols. Since $M$ performs a logspace reduction of $L$, the space taken by $M$ in transforming any element $s$ of $L$ to a SAT instance must be bounded above by $c\cdot\log|s|$ for some constant $c$.

$\def\s#1{\langle T_{#1}, S_{#1}\rangle}$Consider a transformation by machine $M$ of a string $s$ to an instance of SAT. At time $n$, $M$ is in state $S_n$ and its tape is in state $T_n$. There can't be two different times at which both state and tape are the same, or $M$ would be in an infinite loop, which we know doesn't happen. So the running time of $M$ is bounded above by the number of possible (state, tape configuration) pairs, which is $N$ times the total number of possible tape configurations. But at most $c\cdot\log|s|$ tape squares are used, so the total number of tape configurations is at most $A^{c\cdot\log|s|}$, and thus the total time is bounded by $$N\cdot A^{c\cdot\log|s|}.$$ This is a polynomial in $|s|$, so the whole computation is time-bounded by a polynomial in $|s|$, for any $s\in L$, and therefore $L$ is polynomial-time reducible to SAT.

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