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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.

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    $\begingroup$ What is your proof for the empty language? $\endgroup$ – megas Mar 16 '15 at 2:44
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    $\begingroup$ 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. $\endgroup$ – Yuval Filmus Mar 16 '15 at 3:12
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    $\begingroup$ @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? $\endgroup$ – Raphael Mar 16 '15 at 9:50
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    $\begingroup$ @Raphael Maybe we should leave it a week, since the asker explicitly asks for hints, rather than a full solution? $\endgroup$ – David Richerby Mar 16 '15 at 10:48
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    $\begingroup$ (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. $\endgroup$ – 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.

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    $\begingroup$ Since the question asks for hints only, I've done that. If I remember, I'll come back in a week or so (say, around 23rd March) and flesh this out into a full answer for reference purposes. $\endgroup$ – David Richerby Mar 16 '15 at 10:50
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    $\begingroup$ You didn't remember. :) $\endgroup$ – aste123 Jul 21 '16 at 0:14

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