In other words, what is the likelihood that a recognizer of a given regular language will accept a random string of length $n$?


If there is only a single non-terminal $A$, then there are only two kinds of rules:

  1. Intermediate rules of the form $ S \to \sigma S $.
  2. Terminating rules of the form $ S \to \sigma $.

Such a grammar can then be rewritten in shorthand with exactly two rules, thusly:

$$\left\{\begin{align} &S \enspace \to \enspace \{\sigma, \tau, \dots\} S = ΤS\\ &S \enspace \to \enspace \{\sigma, \tau, \dots\} = Τ'\\ \end{align}\right.\\ \space \\ (Τ, Τ' \subset \Sigma) $$

So, we simply choose one of the $Τ$ (this is Tau) symbols at every position, except for the last one, which we choose from $Τ'$.

$$ d = \frac {\lvert Τ\rvert^{n - 1} \lvert Τ' \rvert} {\lvert\Sigma\rvert^n} $$

I will call an instance of such language $L_1$.


If there are two non-terminals, the palette widens:

  1. Looping rules of the form $ S \to \sigma S $.
  2. Alternating rules of the form $ S \to \sigma A $.
  3. Terminating rules of the form $ S \to \sigma $.
  4. Looping rules of the form $ A \to \sigma A $.
  5. Alternating rules of the form $ A \to \sigma S $.
  6. Terminating rules of the form $ A \to \sigma $.

In shorthand: $$\left\{\begin{align} &S \enspace \to \enspace Τ_{SS} S \\ &S \enspace \to \enspace Τ_{SA} A \\ &S \enspace \to \enspace Τ_{S\epsilon} \\ &A \enspace \to \enspace Τ_{AA} A \\ &A \enspace \to \enspace Τ_{AS} S \\ &A \enspace \to \enspace Τ_{S\epsilon} \\ \end{align}\right.\\ \space \\ (Τ_{SS}, Τ_{SA}, Τ_{S\epsilon}, Τ_{AA}, Τ_{AS}, Τ_{S\epsilon} \subset \Sigma) $$

Happily, we may deconstruct this complicated language into words of the simpler languages $L_1$ by taking only a looping rule and either an alternating or a terminating shorthand rule. This gives us four languages that I will intuitively denote $L_{1S}, L_{1S\epsilon}, L_{1A}, L_{1A\epsilon}$. I will also say $L^n$ meaning all the sentences of $L$ that are $n$ symbols long.

So, the sentences of this present language (let us call it $L_2$) consist of $k$ alternating words of $L_{1S}$ and $L_{1A}$ of lengths $m_1 \dots m_k, \sum_{i = 1 \dots k}m_i = n$, starting with $L_{1S}^{m_1}$ and ending on either $L_{1S\epsilon}^{m_k}$ if $k$ is odd or otherwise on $L_{1A\epsilon}^{m_k}$.

To compute the number of such sentences, we may start with the set $\{P\}$ of integer partitions of $n$, then from each partition $P = \langle m_1\dots m_k \rangle$ compute the following numbers:

  1. The number $p$ of distinct permutations $\left(^k_Q\right)$ of the constituent words, where $Q = \langle q_1\dots\ \rangle$ is the number of times each integer is seen in $P$. For instance, for $n = 5$ and $P = \langle 2, 2, 1 \rangle$, $Q = \langle 1, 2 \rangle$ and $p = \frac{3!}{2! \times 1!} = 3$

  2. The product $r$ of the number of words of lengths $m_i \in P$, given that the first word comes from $L_{1S}$, the second from $L_{1A}$, and so on (and accounting for the last word being of a slightly different form):

    $$ r = \prod_{i = 1, 3\dots k - 1}\lvert L_{1S}^{m_i} \rvert \times \prod_{i = 2, 4\dots k - 1}\lvert L_{1A}^{m_i} \rvert \times \begin{cases} & \lvert L_{1S\epsilon}^{m_k} \rvert &\text{if $m$ is odd}\\ & \lvert L_{1A\epsilon}^{m_k} \rvert &\text{if $m$ is even}\\ \end{cases} $$

If my thinking is right, the sum of $p \times r$ over the partitions of $n$ is the number of sentences of $L_2$ of length $n$, but this is a bit difficult for me.


My questions:

  • Is this the right way of thinking?
  • Can it be carried onwards to regular grammars of any complexity?
  • Is there a simpler way?
  • Is there prior art on this topic?
  • $\begingroup$ Maybe I should transplant this question to Mathematics? $\endgroup$ Oct 2, 2019 at 17:50

1 Answer 1


If I understand the question, your problem is the following:

Given a regular language $L$ over alphabet $\Sigma$ and a positive integer $n$, compute the probability that a word chosen uniformly at random from $\Sigma^n$ will be in $L$.

That's equivalent to computing $|L \cap \Sigma^n|$, i.e., the cardinality of the language $L \cap \Sigma^n$. Note that $L \cap \Sigma^n$ is regular, so your problem is equivalent to the problem of counting the number of words in a finite regular language. This is a standard problem that is well-studied; see https://cstheory.stackexchange.com/q/8200/5038, https://cstheory.stackexchange.com/q/32473/5038, Why isn't it simple to count the number of words in a regular language?, Counting the number of words accepted by an acyclic NFA. Here are some results:

  • If the language $L$ is specified as a DFA or as an unambiguous regexp, then the problem can be solved in polynomial time.

  • If the language $L$ is specified as a NFA and $n$ is specified in unary, the problem is $\#P$-complete. Thus, there is an exponential-time algorithm but you should not expect a polynomial-time algorithm.

  • If the language $L$ is specified as a NFA and $n$ is specified in binary, the problem is $PSPACE$-complete. Thus, you should not expect a polynomial-time algorithm.

There are approximation algorithms that you might be able to use in practice to estimate the probability you're seeking (using a SAT solver as a subroutine).

  • $\begingroup$ Awesome! However, can you also comment on my attempt at solution? I put in 2 days of meditation. $\endgroup$ Oct 2, 2019 at 18:59
  • $\begingroup$ @IgnatInsarov, sorry, it was too complex for me to follow every step in the amount of time I have available to spend on this answer. If I got the gist of it, it smells like your procedure might cause exponential blowup and thus take exponential time. I don't know if that's accurate or not. If it is: I link to exponential-time algorithms that are known to work, so I'm not super motivated to try to understand another way to do it in exponential time if that other way looks complicated to understand. Sorry about that, but I hope this is still some partial help. $\endgroup$
    – D.W.
    Oct 2, 2019 at 19:07
  • $\begingroup$ I'm sorry. I was tactless. $\endgroup$ Oct 2, 2019 at 19:11
  • $\begingroup$ @IgnatInsarov, oh gosh, not at all! I didn't see it that way -- I took it as honest curiousity. I hope my response didn't come off as harsh or cold. I hope to see you continuing to contribute here! $\endgroup$
    – D.W.
    Oct 2, 2019 at 19:17
  • $\begingroup$ It did make me consider how I could contribute more to the answering and the moderation effort. What you are managing here is amazing. $\endgroup$ Oct 2, 2019 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.