If we have $L \in p$ and $L' \neq \emptyset$ and , $L' \neq \Sigma^*$ . is $L \leq^p L' $ ?
I read this question Reduction between $\Sigma^*$ and $\emptyset$
Maybe this question is irrelevant to my question, but I'm little seconfused.
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ilen
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What's the definition of $p$? – Sebastian Oberhoff Oct 23 '18 at 12:49
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If $L \in p$, then we can define the reduction function by cases as follows
$$ f(x) = \begin{cases} g(x) & \mbox{if $x\in L$} \\ h(x) & \mbox{if $x\notin L$} \end{cases} $$
where $g,h$ are functions in the same class $p$. Indeed, under such premises, $f$ also belongs to class $p$: it can be computed by first testing whether the input $x$ belongs to $L$, and then computing $g(x)$ or $h(x)$ accordingly. Both of these computation steps belong to class $p$.
If $L'$ is not trivial, we can fix a point $y_0 \notin L'$ and a point $y_1 \in L'$, and then let $g(x)=y_1$ and $h(x)=y_0$. Since these are constant functions, they are trivial to compute, so they belong to any reasonable class $p$.
You can now verify that $x\in L \iff f(x)\in L'$, proving that $f$ is indeed a reduction.
chi
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Hi @chi , thanks a lot! For $f(x) \in L'$ then $x \in L$ , we use $g(x)=y_1 \in L'$ , am I right? – ilen Oct 23 '18 at 17:47
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@ilen Yes. The point is, by definition of $f$, $f(x)$ is either $y_0$ or $y_1$. The former happens only when $x\in L$, the latter only when $x\notin L$. – chi Oct 23 '18 at 18:03