Let $M$ be an acyclic NFA.

Since $M$ is acyclic, $L(M)$ is finite.

Can we compute $|L(M)|$ in polynomial time?

If not, can we approximate it?

Note that the number of words is not the same as the number of accepting paths in $M$, which is easily computable.

Let me mention one obvious approach that doesn't work: convert the NFA to a DFA (which will also be acyclic), then count the number of accepting paths in the DFA. This doesn't result in a polynomial-time algorithm, since the conversion can cause an exponential blowup in the size of the DFA.

  • $\begingroup$ The techniques for arbitrary automata carry over, see e.g. over on cstheory.SE. $\endgroup$
    – Raphael
    Jul 17, 2014 at 12:28
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    $\begingroup$ @Raphael - I'm afraid I don't understand your answer there. Specifically, it doesn't seem to work for ambiguous NFA. Counting the number of words in UFA is the same as counting the number of accepting paths, which, as mentioned in the question, is simple. $\endgroup$
    – R B
    Jul 17, 2014 at 12:37

2 Answers 2


Here's one approach that I expect should give you a multiplicative-factor approximation, with polynomial time running time.

Let $L$ be a regular language that is a subset of $\{0,1\}^n$, e.g., $L=L(M) \cap \{0,1\}^n$. We will try to compute the approximate size of $L$.

At a high level, our approach to approximate $|L|$ will look something like this:

  1. Pick a fraction $p$, where $0<p<1$.

  2. Choose a regular language $R$ such that, roughly speaking, $R$ is a random subset of $\{0,1\}^n$ of size approximately $p2^n$ (i.e., $|R|\approx p2^n$).

  3. Check whether $L \cap R$ is non-empty. Note that this check can be done in polynomial time.

Repeatedly perform steps 1-3 for various values of $p$. This gives you some information that will let you approximate $|L|$.

In particular, if $|L|=m$, then we'd expect

$$\Pr[L \cap R = \emptyset] = (1-p)^m \approx e^{-pm}.$$

So, if you happen to choose $p=1/m$ and repeat steps 1-3 a bunch of times, you should expect to see an empty intersection about 37% of the time. If you see an empty intersection significantly more often then that, then increase $p$ and try again. If you see an empty intersection significantly less often then that, you might decrease $p$ and try again.

In this way, using something like binary search, you should be able to approximate $|L|$ to within a multiplicative approximation factor.

You'll still need to pick some way to choose $R$ so that it is regular but also behaves like a random subset. There are many possibilities, but one might good way might be to choose a random 2-universal hash $h:\{0,1\}^m \to \{0,1,2,\dots,k-1\}$, pick $y \in \{0,1,\dots,k-1\}$ randomly, and let $R=\{x \in \{0,1\}^n : h(x)=y\}$. Choosing $k=\lceil 1/p \rceil$ gives you a random set $R$ of approximately the right size, and because $h$ is 2-universal, all of the mathematics above should work out properly.

This should solve your problem for the case where all strings in the NFA have the same length, say $n$. If they have varying lengths, then you can handle each possible length separately. Since $M$ is acyclic, the maximum lengths of any string in $L(M){}$ is at most the number of states in $M$, so this does not increase the runtime too much.

(This construction might remind you of the Valiant-Vazirani theorem about unambiguous SAT.)


Assume that you can count in polynomial time the number of words of a language given by an acyclic NFA. In this case, considering two acyclic NFAs $A_1$ and $A_2$, you can compute in polynomial time the cardinal $n_1$ (resp. $n_2$) of the language of $A_1$ (resp. $A_2$). By a direct product (preserving acyclicity) you can also compute in polynomial time the cardinal $n_3$ of the intersection of these two languages. The two automata accepts the same language iff $n_1=n_2=n_3$. Therefore you can test the equality of two finite languages given by acyclic automata in polynomial, which is known to be an NP-complete problem. So, unless $P=NP$, you can't solve your problem in polynomial time.

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    $\begingroup$ citation needed for the hardness result $\endgroup$
    – A.Schulz
    Dec 15, 2014 at 11:14

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