I'm currently following a course on Complexity Theory, and whilst studying, I came across a rather counterintuitive statement:

If $\textbf{P}=\textbf{NP}$, the following holds:

For every $A \in \textbf{NP}$, there is a $B \in \textbf{NP}$ such that $A \leq B$ (where $\leq$ means Karp reducible).

However, I do not understand how this applies to the empty problem $\emptyset$, and it's complement $\Sigma^*$, as these only have no-instances and yes-instances, respectively.

Are there other problems in NP such that these two are reducible to them?

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    $\begingroup$ You don't even need to assume $P=NP$ for this. Just take $A=B$. $\endgroup$ – Tom van der Zanden Apr 10 '19 at 12:14
  • $\begingroup$ Hey Tom, I think that the statement from the course meant that $\textbf{A} \neq \textbf{B}$. Otherwise, it is indeed an irrelevant requirement. $\endgroup$ – R. dV Apr 10 '19 at 12:26
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    $\begingroup$ @R.dV It's irrelevant even if you assume $A\neq B$. $\endgroup$ – David Richerby Apr 10 '19 at 14:32
  • $\begingroup$ @R.dV Adding to David Richerby's comment: $B \neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A \setminus \{ w \}$, where $w \in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A \cup \{ w \}$, where $w \not\in A$ and $A \neq \Sigma^\ast \setminus \{ w \}$). $\endgroup$ – dkaeae Apr 10 '19 at 15:59

Of course there is.

Just take any non-trivial language $L$ (i.e., $L \neq \varnothing$ and $L \neq \Sigma^\ast$). Then there are concrete words $x \in L$ and $y \not\in L$.

To reduce $\varnothing$ to $L$, simply map everything to $y$. Then the input is in $\varnothing$ (which is false) if and only if $y \in L$ (which is also false). Hence, the reduction is correct.

For $\Sigma^\ast$, do the same but use $x$ instead.

As a note: I assume you are puzzled about $A$ being reduced to $B$. Obviously, in the statement you cite $B$ should at the very least be a non-trivial set (and it seems $\textbf{P} = \textbf{NP}$ is redundant, as Tom van der Zanden notes in the comments; in fact, the statement is rather fishy, see David Richerby's answer); note you cannot reduce non-trivial sets to $\varnothing$ or $\Sigma^\ast$ (and you cannot reduce either to one another, as David Richerby points out in the comments).

  • $\begingroup$ Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $\in$ L) to anything in $\emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $\leq$ to be correct, you need two things: For I $\in$ as a yes-instance, f(i) $\in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none? $\endgroup$ – R. dV Apr 10 '19 at 12:03
  • $\begingroup$ You seem to have it the other way around. If you are reducing $\varnothing$ to $L$, then you should be mapping yes-instances of $\varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $\varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y \not\in L$. $\endgroup$ – dkaeae Apr 10 '19 at 13:12
  • $\begingroup$ Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :) $\endgroup$ – R. dV Apr 10 '19 at 13:52
  • $\begingroup$ Re your note at the end, you can't reduce between $\emptyset$ and $\Sigma^*$, either. $\endgroup$ – David Richerby Apr 10 '19 at 14:33

The statement is basically vacuous. Every language is reducible to itself (the reduction is the identity function), so you can just take $B=A$.

  • $\begingroup$ Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $\textbf{B}\neq\textbf{A}$ $\endgroup$ – R. dV Apr 10 '19 at 15:18
  • $\begingroup$ @R.dV OK but that's your assumption and it's not included in the question you say you came across. $\endgroup$ – David Richerby Apr 10 '19 at 15:24
  • $\begingroup$ Or $A$ union a finite set if $A$ is a singleton. $\endgroup$ – David Richerby Apr 10 '19 at 15:59
  • $\begingroup$ Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :) $\endgroup$ – R. dV Apr 10 '19 at 18:01

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