Can a nondeterministic finite automata (NDFA) be efficiently converted to a deterministic finite automata (DFA) in subexponential space/time?

Question

Twenty years ago, I built an regular expression package that included conversions from regular expressions to a finite state machine (DFA) and supported a host of closed regular expression operations (Kleene star, concatenation, reverse, set operations, etc). I was unsure about the worst case performance of my package.

A DFA has the same expressive power as an NDFA, because an n-state NDFA can be trivially converted to a DFA having 2^n states. However, are there any lower upper bound guarantees for such a conversion that do not require an exponential explosion in state?

I was unable to come up with examples mal-behaving regular expressions or NDFAs, but I didn't spend much time thinking about it. I am guessing a regular expression like ((((e|A|B|C)*(e|D|E|F))*(e|G|H|I))*(e|J|K|L|M))* which mixes a lot of alternations and Kleene stars would have a linearly sized NDFA but an expansive DFA.

Answer 1

It is known that for every pair of naturals numbers n,a such that n <= a <= 2^n, there exists a minimal NDFA with n states whose corresponding equivalent minimal dfa has a states (over a four letter alphabet).

See the paper here: Deterministic blow-ups of minimal nondeterministic finite automata over a fixed alphabet.

Abstract of the paper:

We show that for all integers n and α such that n ≤ α ≤ 2^n, there exists a minimal nondeterministic finite automaton of n states with a four-letter input alphabet whose equivalent minimal deterministic finite automaton has exactly a states. It follows that in the case of a four-letter alphabet, there are no "magic numbers", i.e., the holes in the hierarchy. This improves a similar result obtained by Geffert for a growing alphabet of size n + 2 (Proc. 7th DCFS, Como, Italy, 23-37).

So, I suppose the answer to your question is, no.

Answer 2

The classic example for a language with an exponential separation between DFA size and NFA size is the following finite language: binary strings of length exactly 2n in which the first half is not equal to the second half. A NFA would guess an index i in which the first and second half disagree. A lower bound for a DFA follows from communication complexity, for example.

Answer 3

The minimum DFA corresponding to an NFA has 2^n states in the worst case, so you can't garantee anything. Without having a constructive example, the reasoning is that in an NFA you can be at any arbitrary subset of states after reading a certain input string, and each such subset might behave differently when observing one character. Suppose a language with two characters in the alphabet (a and b), and an NFA N with n states that starts with an accepting state at s_0. Now enumerate all subsets of states of N, and build the transition table such that observing "a" from subset S_i takes you to subset S_i+1 and observing b takes you to subset S_i-1 (this is doable for some enumerations, I think). Now this automata has n states and accepts sequences of m a's and n b's such that m-n = 0 mod 2^|N|, and cannot be expressed with a DFA that has less than 2^|N| states (since it might need to cycle through all subsets of states of the NFA N).