If $R$ is an arbitrary decision problem that is reducible to $S$, which is an NP-complete problem, what can be said about $R$?

I think we should be able to say that $R$ is in NP since an instance of $S$ can have a certificate verified in polynomial time and from the reduction $R \le S$, we could say the same about $R$. Is this reasoning valid? Would it also work if $S$ was only supposed to be in NP rather than NP-complete?

Thanks in advance.


1 Answer 1


Yep - your reasoning regarding your NP is correct. But not regarding NP-complete. The crux of the matter is that NP is inclusive, P is in NP and your problem R could be in P (and thus in NP).

NP-complete is not inclusive. It refers to the hardest problems in NP and a problem is in NP is there is no "simpler solution" (assuming P!=NP). In your case since R < S it may be (temporarily) true that R is in NP-complete, but then you will have to show that there is no simpler solution to R. What you have shown is that there is a solution to R going through S - this is a solution, but for NP-complete, you must show also there there is no simpler solution. (Caveat - all this language of simpler solution, hardest problem, and so on assumes that P!=NP.


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