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I am taking an algorithm analysis class and am stuck on one of my homework problems and would appreciate it if I could receive some guidance.

The problem I'm stuck on is proving that the empty language and $\{0, 1\}^*$ are the only languages in P that are not complete for P with respect to polynomial-time reductions (problem 34.3-6 in CLRS 3rd edition). The first part of the problem seems fairly straightforward enough (proving the empty language criteria). However, I'm not sure where to even begin when I have to prove the criteria for $\{0, 1\}^*$. I'm NOT looking for the answer, however I would appreciate some guidance on how I can begin to think about this problem.

David Richerby
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user30103927502
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    What is your proof for the empty language? – megas Mar 16 '15 at 02:44
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    To expand on megas' comment, the proof that ${0,1}^$ is not P-complete is the same as the proof that $\emptyset$ isn't. Don't forget that you also have to show that any other language is* P-complete with respect to polytime reductions. – Yuval Filmus Mar 16 '15 at 03:12
  • I'm fairly certain that we have covered this a couple of times already. Does somebody have a duplicate at their fingertips? – Raphael Mar 16 '15 at 07:30
  • @Raphael It's included in my answer here (that's dealing with what happens if P$,=,$NP so it talks about NP but applies to P). – David Richerby Mar 16 '15 at 08:25
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    @DavidRicherby Thanks. This is also a close fit. I could not find a perfect duplicate (i.e. with answers that focus on this issue) and are correct (there are a couple of answers that state P=P-complete under polytime many-one reductions). So maybe this is worth answering again, explicitly and clearly? – Raphael Mar 16 '15 at 09:50
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    @Raphael Maybe we should leave it a week, since the asker explicitly asks for hints, rather than a full solution? – David Richerby Mar 16 '15 at 10:48
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    (Warning work with David's hint before following the link, if you're stuck it takes it a bit further) .There's also a similar question here, but from the other end. – Luke Mathieson Mar 16 '15 at 12:07

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Hint. The definition of reduction from $X$ to $Y$ requires that "yes" instances of $X$ be mapped to "yes" instances of $Y$ and "no" instances to "no" instances.

David Richerby
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