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Consider the well-known language: $$L = \{\langle M,x,1^t \rangle \ \ |\ \text{$M$ accepts $x$ after $t$ steps}\}$$

I've encountered some technical problem while trying to show a logarithmic reduction from some $L'\in P$ (which will show completeness of $L$)

Basically, the reduction is: $$f(x) = \langle M_{L'},x,n^c\rangle $$

Where $n^c$ is the time it takes for $M_{L'}$ to decide some input, $x$.

At first everything seems legitimate, but... How do we construct $M_{L'}$ at the first place?

In other words, do we have an "access" (as the reduction function) to the machine description?

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You need to show that for every $L'\in P$ there exists a logspace reduction from $L'$ to $L$. Your $f$ has $\langle M_{L'}\rangle$ hardcoded into it. The encoding was part of its definition, $f$ does not ask what is the encoding of the machine for $L'$ when it tries to compute $f(x)$.

One way to think of this is as follows: consider the sequence of functions $\left\{f_{s ,c}\right\}_{c\in\mathbb{N}, s\in\{0,1\}^*}$ where $f_{s, c}(x)=\left(s, x, 1^{|x|^c}\right)$. Since $L'\in P$, there exists a function in this sequence which is a reduction from $L'$ to $L$ (the specific index depends on the machine deciding $L'$, but we don't really care about it).

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  • $\begingroup$ Oh I see what made me confused about it. Thanks Ariel! $\endgroup$ – Covvar Jul 19 '17 at 20:06

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