Let some deterministic finite automata be given. There is a problem of determining whether the intersection of these DFA is empty, and I want to show its PSPACE-completeness.

It seems to me that I understand, why this problem lies in PSPACE: if $ n_1,\dots, n_k $ are the numbers of states, then one can take the product of these DFA and get that if there is a recognisable string, then there is a recognisable string of length $ \le n_1\dots n_k$. Then we can non-deterministically guess this string and use Savitch theorem.

So, the question is, how to prove PSPACE-hardness? Should I reduce any given TM to this problem, or maybe some known problem like QBF?


1 Answer 1


Your proof that the problem is in PSPACE is not quite correct. The problem is that the product $n_1 \cdots n_k$ is not bounded by a polynomial in the input length $n_1 + \dots + n_k$. The correct way to do it is to directly apply Savitch's theorem to the NPSPACE machine that nondeterministically guesses a path through the product graph. The difference is that, while an accepting path can indeed be as long as $n_1 \dots n_k$, the machine only guesses one symbol at a time and never stores the entire input string in memory, so we stay within the polynomial space limit.

It is a well-known fact that this problem is PSPACE-hard, but the only proof that I am aware of is not particularly trivial. I believe the first proof was given in:

Dexter Kozen. Lower Bounds for Natural Proof Systems. FOCS 1977: 254-266.

The proof involves constructing a mess of DFAs which collectively enforce all the necessary constraints to require that any string in the intersection language is a trace of the execution of a polynomial space Turing machine that leads to an accepting configuration.


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