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I have regular expression R, and I want to find F(n) is the number of strings of length n that strings that contain substring matching Regular Expression. Suppose the alphabet-size is M.

We can apply the generating function to compute the number of strings of length n that matches R, but i can't find any answer for the above question.

Do you have any idea on that?

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  • $\begingroup$ Do you have a particular regular expression in mind? $\endgroup$ – Evan Aug 26 '13 at 21:20
  • $\begingroup$ So for example, suppose we are working with the bitstrings (M=2), R = (10|1)*1*. Can you compute number of bitstrings of length n that contains substring(s) that match R @Evan ? $\endgroup$ – Loi.Luu Aug 27 '13 at 7:58
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For the particular example $M=2$ and $R=(10 | 1)^\ast 1^\ast$, I suppose if we first convert it to a CFG then there appears to be a general formula that involves convolutions. For example:

  • $A \rightarrow \epsilon | 0A | 1A$
  • $C \rightarrow \epsilon | 10C | 0C$
  • $D \rightarrow \epsilon | 1D$
  • $E \rightarrow ACDA$.

Then to get a string of length $n$, concatenate a string from $A$ of length $k_1$, a string from $C$ of length $k_2$, a string from $D$ of length $k_3$ and a string of length $A$ of length $k_4$, then you get

$N_E(n) = \sum_{k_1+k_2+k_3+k_4 = n} N_A(k_1)N_C(k_2)N_D(k_3) N_A(k_4)$

where $0 \leq k_i \leq n$. Ok, so there was no need to use a CFG, we could arrive here directly by identifying the parts!

Note $N_A(k_1) = 2^{k_1}$, $N_D(k_3) = 1$, and $N_E(k_4) = 2^{k_4}$. $N_C(k_2) = N_C(k_2-1) + N_C(k_2-2)$, Fibonacci recurrence with $N_C(1) = 1$ and $N_C(2) = 2$.

$N_{E,1}(n) = \sum_{k_1+k_2+k_3+k_4=n} 2^{k_1+k_4} F(k_3) = \sum_{a+b \leq n} 2^{a} F(b) = \sum_{b\leq n} (2^{n-b+1}-1) F(b)$. Looks like this can be attacked with generating functions.

This approach seems to generalize to other regular expressions. Oh and you said you know how to do it for exact matching. So for substring, just add an $A$ before and an $A$ after :) ?

Oh drats... but then you might be double-counting. You'd have to use inclusion-exclusion to subtract the number of strings where the pattern appears at least twice, add back where it appears at least three times, etc. But maybe it is okay.

For $ACDACDA$ (string appears at least twice): $N_{E,2} = \sum_{k_1+k_2+k_3+k_4+k_5+k_6+k_7=n} = N_A(k_1)N_C(k_2)N_A(k_4)N_C(k_5)N_A(k_7) = \sum_{a+b+c<n}(2^a-1) F(b)F(c)$.

Look for a general pattern?

$N_E(n) = N_{E,1}(n) - N_{E,2}(n) + N_{E,3}(n) + \ldots + N_{E,n}(n)$.


I didn't realize you had a generating function for strings matching $R$. Let us assume this now. Given the generating function $G_R(x)$ that contains number of strings of length $n$ matching $R$, let us figure out some way to obtain the generating function encoding the number of strings of length $n$ containing a substring that matches $R$.

So first, we do the same trick above and pad the sides with arbitrary strings. So the number of strings of length $n$ containing at least 1 substring matching $R$ is the sum $\sum_{a+b \leq n} M^a N_R(b) = \sum_{b \leq n} \frac{M^{n-b+1}-1}{M-1} N_R(b)$.

In general, the number of strings of length $n$ containing at least k substrings matching $R$ is the sum $\sum_{b_1+\ldots+b_k} \frac{M^{n-(b_1+\ldots+b_k)+1}-1}{M-1} N_R(b_1)\cdots N_R(b_k)$.

As a generating function, let $H_k(x)$ correspond to containing at least $k$ substrings.

$H_k(x) = \sum_{j} N_{R,k}(j) x^j = \sum_{j} \sum_{a+b_1+\ldots+b_k = j} C(j-a) x^{j-a} N_R(b_1) x^{b_1} \cdots N_R(b_k) x^{b_k}$

This is a $k$-fold convolution between the generating function of all $M$-strings, and $k-1$ copies of $G_R$.

Now we encode inclusion exclusion:

$H(x) = \sum_{j} \sum_{i=1}^j (-1)^{i-1} N_{R,\ i}(j) x^j = \sum_i (-1)^{i-1} \sum_{j \geq i} N_{R,\ i}(j) x^j $, and this is just the alternating sum of the corresponding partial generating functions. So... I'm going to stop here for the moment.

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  • $\begingroup$ Thank you @Evan, you did give an excellent solution for the example case, but I need further than that, which can solve the general case with arbitrary R and M >= 2? Do you have any suggestion on that? $\endgroup$ – Loi.Luu Aug 28 '13 at 2:33
  • $\begingroup$ Well, I tried... $\endgroup$ – Evan Aug 28 '13 at 3:52
  • $\begingroup$ Yeah, thank you @Evan for your efforts, I know that its hard.. $\endgroup$ – Loi.Luu Aug 28 '13 at 6:52
  • $\begingroup$ I think you have a typo in this part Given the generating function GR(x) that contains number of strings of length n matching R $\endgroup$ – Loi.Luu Aug 28 '13 at 9:56
  • $\begingroup$ And because the padding strings are arbitrary strings, hence how can you prevent the total result from counting the duplication? $\endgroup$ – Loi.Luu Aug 28 '13 at 9:58

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