Stackoverflow.ipynb

Kernel: Python 3

Questions taken from Stackoverflow

This page lists questions about automata and other rational/regular expressions that were asked on Stackoverflow, and where Vcsn seems to be an appropriate tool to compute the answer.

import os
# This trick ensures that we always use the same random seed,
# hence running this documentation always gives the same result.
os.environ['VCSN_SEED'] = '1'

Build a Regular Expression and Finite Automata

The set of all strings beginning with 101 and ending with 01010.

First, let's define our "context": we work with "labels are letters" (lal), on the alphabet $\{0, 1\}$. We don't use weights, or rather, we use the traditional Boolean weights: $\mathbb{B}$.

import vcsn
c = vcsn.context('lal_char(01), b')
c

$\{0, 1\}\to\mathbb{B}$

Then, we build our expression using an unusual operator: $\mathsf{E} \& \mathsf{F}$ denotes the conjunction of expressions $\mathsf{E}$ and $\mathsf{F}$. In this case (unweighted/Boolean automata), it denotes exactly the intersection of languages.

e = c.expression('(101[01]*)&([01]*01010)')
e

$1 \, 0 \, 1 \, \left(0 + 1\right)^{*} \& \left(0 + 1\right)^{*} \, 0 \, 1 \, 0 \, 1 \, 0$

We want to normalize this extended expression (it has conjunction and complement operators) into a basic expression. To this end, we first convert it to an automaton.

a = e.automaton()
a

and then we convert this automaton into a basic expression:

a.expression()

$1 \, \left(0 \, 1 + 0 \, 1 \, \left(0 + 1\right)^{*} \, 0 \, 1\right) \, 0 \, 1 \, 0$

Or, in ASCII:

print(a.expression())

1(01+01(0+1)*01)010

Regular expression to match text that doesn't contain a word?

I'd like to know if it's possible to match lines that don't contain a specific word (e.g. hede) using a regular expression?

First, let's define that alphabet we work on: from $a$ to $z$ for instance.

import vcsn
c = vcsn.context('lal_char(a-z), b')
c

$\{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z\}\to\mathbb{B}$

Then we define our expression, which is extended (it uses the complement operator), so to normalize it, we first convert it into automaton (with _expression_.automaton), from which we extract a basic expression (with _automaton_.expresion).

e = c.expression('(hede){c}')
e

$\left(h \, e \, d \, e\right)^{c}$

a = e.automaton()
a

a.expression()

$\varepsilon + h \, \left(\varepsilon + e \, \left(\varepsilon + d\right)\right) + \left([\hat{}h] + h \, \left([\hat{}e] + e \, \left([\hat{}d] + d \, \left([\hat{}e] + e \, [\hat{}]\right)\right)\right)\right) \, {[\hat{}]}^{*}$

Or, in ASCII (+ is usually denoted |; \e denotes the empty word; and [^] denotes any character, usually written .):

print(a.expression())

\e+h(\e+e(\e+d))+([^h]+h([^e]+e([^d]+d([^e]+e[^]))))[^]*

Convert finite state machine to regular expression

Is there a tool (or an algorithm) to convert a finite state machine into a regular expression?

Vcsn is tool for rational expressions and automata. See http://vcsn.lrde.epita.fr.

import vcsn

Build a random automaton $a$ of 4 states, labeled by letters to choose in ${a, b, c}$. Then "lift" it into an automaton $b$ labeled by expressions.

a = vcsn.context('lal(abc), b').random_automaton(4, num_final=2)
b = a.lift()
b

Eliminate state 2, then 3, etc. The order does not matter for correction (any order gives a correct result), but the quality of the result does depend on this order.

b = b.eliminate_state(2)
b

b = b.eliminate_state(3); b

b = b.eliminate_state(1); b

Eliminate the last state, and read the answer.

b = b.eliminate_state(0); b

Alternatively, you can use the method automaton.expression to ask for the result.

a.expression()

$\left(b + c + c \, \left(b + c + b \, b \, \left(a + b\right)\right)^{*} \, b\right)^{*} \, \left(\varepsilon + c \, \left(b + c + b \, b \, \left(a + b\right)\right)^{*} \, b \, b\right)$

You may see this algorithm run "interactively" using %demo.

a = vcsn.context('lal(abc), b').random_automaton(10)
b = a.lift()
%demo b eliminate_state

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Constructing a Regular Expression from a Finite Automata

I'm trying to construct a regular expression from a Finite Automaton but found my self completely stuck with this one.

import vcsn

%%automaton a
$  -> q3
q1 -> q2 a
q1 -> q3 b
q2 -> q2 a, b
q3 -> q4 a
q3 -> q3 b
q4 -> q4 a
q4 -> q1 b
q3 -> $
q4 -> $

a.expression()

$\left(b + a \, {a}^{*} \, b \, b\right)^{*} \, \left(\varepsilon + a \, {a}^{*}\right)$

How to find the right quotient of a language given two languages?

(This is the question with typos in $L_1$ and $L_2$ fixed.)

If $L_1= \{a^n b^m \mid n \geqslant 1, m \geqslant 0 \} \cup \{ba\}$ and $L_2= \{b^m \mid m \geqslant 1 \}$. I am not getting how the DFA for $L_1/L_2$ is constructed in the second figure using the DFA for $L_1$, please tell me the approach.

First, let's define our "context": we work with "labels are letters" (lal), on the alphabet $\{a, b\}$. We don't use weights, or rather, we use the traditional Boolean weights: $\mathbb{B}$.

import vcsn
c = vcsn.context('lal(ab), b')
c

$\{a, b\}\to\mathbb{B}$

Then define the first automaton, Figure 4.1.

%%automaton a1
$  -> q0
q1 -> $
q0 -> q1 a
q0 -> q3 b
q1 -> q1 a
q1 -> q2 b
q2 -> $
q2 -> q2 b
q2 -> q5 a
q3 -> q4 a
q3 -> q5 b
q4 -> $
q4 -> q5 a, b
q5 -> q5 a, b

a1.is_deterministic() and a1.is_complete()

True

Automaton $\mathcal{A}_2$ represents the language $L_2$:

a2 = c.expression('b{+}').automaton()
a2

We may now compute the quotient, which is indeed the automaton of Figure 4.2. It is better looking once trimmed.

a1 / a2

(a1 / a2).trim()