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Is there any language $L$ know to be complete for $NP \cap co-NP$, i.e. any language $L^{\prime} \in NP\cap co-NP$ reduces in polynomial-time to $L$ and it is known that $L\in NP\cap co-NP$?


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As far as I can find, there aren't any semantic classes that are known to have complete problems and are not known to be equal to a syntactic class. – Ricky Demer Aug 8 '10 at 5:29
up vote 16 down vote accepted

$NP \cap coNP$ is not known to have complete languages, but I don't know of any consequences as strong as what Marcos claims. Juris Hartmanis and his students worked extensively on this problem in the early 80's. Two references I know are:

Michael Sipser: On Relativization and the Existence of Complete Sets. ICALP 1982: 523-531

This paper shows that there is an oracle relative to which $NP \cap coNP$ does not have complete sets. So proving that it does have complete sets would at least require non-relativizing techniques. Also, as mentioned by Peter in Marcos' comment, there's also Hartmanis and Immerman's work:

Juris Hartmanis, Neil Immerman: On Complete Problems for NP$\cap$CoNP. ICALP 1985: 250-259

They give several interesting structural results about the problem. For one, they complement Sipser's result, giving an oracle relative to which $NP \cap coNP$ has complete languages yet $P \neq NP \cap coNP \neq NP$. They also show that $NP \cap coNP$ has complete languages under many-one reductions iff it has complete languages under Turing reductions. This paper also cites the following neat characterization by Kowalczyk (1985): $NP \cap coNP$ has a complete language iff there is a recursively enumerable list of pairs of $NP$ machines {$(N_{i,1}, N_{i,2})$} such that $\overline{L(N_{i,1})} = L(N_{i,2})$ and $\bigcup_i L(N_i) = NP \cap coNP$.

But since then, there hasn't been much progress on the question, to my knowledge. I'd be very happy if I were corrected...

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There are no complete problems for $NP\cap coNP$, unless the polynomial hierarchy collapses. You'll find that phrase in several textbooks on complexity theory.

Update: take a look at link talking about this kind of problems. Also, the book by Arora and Barak is a good reference.

Update: The claim above "unless the polynomial hierarchy collapses" is too strong. There is no evidence of such consequence.

A better way to put it would be, there are no problems known to be complete for $NP\cap coNP$. It seems that non-relativizing techniques are required to proof the existence or non-existence of complete sets.

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Got a reference?... – Ryan Williams Aug 8 '10 at 5:54
Please tell me the page on which this is proved in Arora and Barak. – Ryan Williams Aug 8 '10 at 6:01
in the draft version, chapter 5, the polynomial hierarchy and alternations, page 5.2(92), where it says "Note that $\sum_2^p contains both the classes NP and coNP". There is no proof, but this implies that a complete problem collapses the hierarchy to the second level. – Marcos Villagra Aug 8 '10 at 6:05
One would think that any such reference would have to cite Hartmanis and Immerman: but I can't find one that does. – Peter Shor Aug 8 '10 at 6:13
Marcos: Of course $Sigma_2 P$ contains $NP$ and $coNP$. That does not mean "$NP \cap coNP$ has a complete language implies the polynomial hierarchy collapses". – Ryan Williams Aug 8 '10 at 6:15

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