# context.de_bruijn(n)¶

Create the "de Bruijn" automaton with $n+1$ states; it accepts word whose $n$-th letter before the end is an 'a'. This family of automata is close to be being a worst case for determinization: its determinized automaton has $2^n$ states (not $2^{n+1}$).

Preconditions:

• the labelset has at least two generators

Postconditions:

• the Result is isomorphic to the derived-term automaton of $(a+b)^*a(a+b)^n$.

## Examples¶

In [1]:
import vcsn
b = vcsn.context('lal_char(ab), b')

In [2]:
a = b.de_bruijn(3)
a

Out[2]:

The support of the determinized automaton is a de Bruijn graph:

In [3]:
a = b.de_bruijn(2)
a

Out[3]:
In [4]:
a.determinize()

Out[4]: