The Cost of an Equivalence Query for DFA Inspired by this question, I am curious about the following:


  What is the worst-case complexity of checking whether a given DFA accepts the same
  language as a given regular expression?


Is this known?  The hope would be that this problem is in P -- that there is an algorithm polynomial in the size of both.
According to Garey and Johnson (p. 174), REGULAR EXPRESSION NON-UNIVERSALITY is PSPACE-complete. This is the problem of deciding whether a regular expression over $\{0,1\}$ does not generate all strings. So your problem is also PSPACE-complete.

Here is one way to see that the OP's problem is in PSPACE. Given a DFA $A$ and a regular expression $r$, construct an NFA $B$ for $r$, and use the power set construction to virtually construct a DFA $C$ equivalent to $B$; we will not keep $C$ in memory, but we have oracle access to $C$ using only polynomial space. Now virtually construct a DFA $D$ for the symmetric difference of $A$ and $C$ using the product construction. This DFA accepts no strings (and so $L(A) = L(r)$) if there is no path from the starting state to an accepting state. Since reachability is in NL and $D$ has size $2^{\mathrm{poly}(n)}$, we can check whether $L(D) = \emptyset$ in $\mathrm{NSPACE}(\mathrm{poly}(n)) = \mathrm{NPSPACE} = \mathrm{PSPACE}$, the latter equality due to Savitch's theorem.