Is it possible to determinise unambiguous finite automata without exponential blowup in the number of states? I think it should not be possible but I am unable to come up with counterexamples.


1 Answer 1


No, the exponential lower bound for determinization holds already for unambiguous NFAs. This is obtained as follows: Consider the alphabet $\{a,b\}$, and the language: $$L_k=\{w\in \{a,b\}^*:\text{the $k$-th before last letter in }w\text{ is }b\}$$ It's easy to construct an unambiguous NFA for $L_k$: the NFA guesses when the $k$ before last letter is, and then proceeds with verifying that the current letter is $b$, and that there are exactly $k$ letters remaining. This NFA has size $O(k)$.

However, to construct a DFA for $L_k$, you must keep in memory the last $k$ letters, yielding at least $2^{k}$ states.

  • $\begingroup$ For the sake of precision, what is $w$, a string from $\{a,b\}*$? $\endgroup$
    – chepner
    Sep 19, 2021 at 22:21
  • $\begingroup$ Also this is the exact example given in the Wikipedia page for Unambiguous Finite Automata, with some more helpful explanations there. $\endgroup$
    – justhalf
    Sep 20, 2021 at 3:04
  • $\begingroup$ @chepner - Yes. I'll add a clarification $\endgroup$
    – Shaull
    Sep 20, 2021 at 9:14
  • $\begingroup$ $k+1$ states suffice, so this is nearly optimal. $\endgroup$ Sep 20, 2021 at 17:57

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