Is there an efficient test for if an NFA accepts a subset of another NFA?

Problem Detail: 

So, I know that testing if a regular language $R$ is a subset of regular language $S$ is decidable, since we can convert them both to DFAs, compute $R \cap \bar{S}$, and then test if this language is empty.

However, since this requires converting to DFAs, it's possible that the DFAs, and thus the testing algorithm, will be exponential in terms of the number of states in the input NFAs.

Is there a known way to do this in polynomial time? Has this problem in general been proved Co-NP complete?

Note that the problem is in Co-NP since a word accepted by $R$ but not by $S$ would be a polynomial certifier that $R \not \subseteq S $.

EDIT: this is incorrect, as there is no guarantee that such a word would be polynomial in the number of states.

Asked By : jmite

Answered By : Shaull

The problem of deciding language containment in NFAs is $PSPACE$-complete. To prove this, it is easy to reduce from the universaility problem for NFAs (testing whether $L(A)=\Sigma^*$) So, in a way, you have to determinize, but you may do so on-the-fly.

Your observation about co-NP is wrong (but nice). Such a witness can indeed be checked in polynomial time in the witness, but the shortest witness itself may be exponential in the length of the input. Since $PSPACE=co-PSPACE$, then deciding non-containment is also $PSPACE$-complete.

To state things more carefully, deciding whether $L(A)\subseteq L(B)$ is $PSPACE$ in the size of $B$ (since only $B$ needs to be complemented), and $NLOGSPACE$ in the size of $A$.

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Question Source : http://cs.stackexchange.com/questions/9954