<SYSTEM_TASK:>
Given the following text description, write Python code to implement the functionality described below step by step
<END_TASK>
<USER_TASK:>
Description:
 automaton.shortest(num = None, len = None)
Return a (finite) approximation of the behavior of the automaton controled by the length of the words.  In other words, compute the polynomial of the first accepted words (according to the shortlex order), and their associated weight.
Arguments
Step1:  Boolean Automata
Step2:  Calling a.shortest() is equivalent to a.shortest(1) which is equivalent to a.shortest(1, 0)
Step3:  To get the first four words, use a.shortest(4) (or a.shortest(len = 4))
Step4:  The words with at most four letters
Step5:  At most 10 words of at most 4 letters
Step6:  At most 10 words of at most 3 letters
Step7:  Weighted Automata
The following automaton decodes binary numbers.
Step8:  Transducers
The following automaton maps words to words.  The length of a multitape words is the largest length of its components
<ASSISTANT_TASK:>
Python Code:
import vcsn
Explanation: automaton.shortest(num = None, len = None)
Return a (finite) approximation of the behavior of the automaton controled by the length of the words.  In other words, compute the polynomial of the first accepted words (according to the shortlex order), and their associated weight.
Arguments:
- num the number of words to find (there might be fewer).  None stands for $\infty$ if len is specified, 1 otherwise.
- len the maximum length of the words to find. None stands for $\infty$.
Default values are handled as follows:
<table>
    <tr>
        <td></td>
        <td>len = None</td>
        <td>len = $\ell$</td>
    </tr>
    <tr>
        <td>num = None</td>
        <td>$1, \infty$</td>
        <td>$\infty, \ell$</td>
    </tr>
    <tr>
        <td>num = $n$</td>
        <td>$n, \infty$</td>
        <td>$n, \ell$</td>
    </tr>
</table>
Preconditions:
- the automaton must not have spontaneous cycles
See also:
- automaton.eval
- automaton.lightest
- expression.shortest
Examples
End of explanation
a = vcsn.context('lal_char(ab), b').de_bruijn(1)
a
Explanation: Boolean Automata
End of explanation
a.shortest()
Explanation: Calling a.shortest() is equivalent to a.shortest(1) which is equivalent to a.shortest(1, 0): find the shortest word, whatever its length:
End of explanation
a.shortest(4)
Explanation: To get the first four words, use a.shortest(4) (or a.shortest(len = 4)):
End of explanation
a.shortest(len = 4)
Explanation: The words with at most four letters:
End of explanation
a.shortest(num = 10, len = 4)
Explanation: At most 10 words of at most 4 letters:
End of explanation
a.shortest(num = 10, len = 3)
Explanation: At most 10 words of at most 3 letters:
End of explanation
%%automaton -s bin
context = "lal_char(01), z"
$ -> 0
0 -> 0 0, 1
0 -> 1 1
1 -> $
1 -> 1 <2>0, <2>1
bin.shortest(len = 3)
bin.shortest(num = 10)
Explanation: Weighted Automata
The following automaton decodes binary numbers.
End of explanation
%%automaton -s t
context = "lat<law<char>, law<char>>, q"
$ -> 0
0 -> 1 <1>one|un
0 -> 2 <2>two|deux
0 -> 3 <3>three|trois
1 -> 0 \e|\e
2 -> 0 \e|\e
3 -> 0 \e|\e
1 -> $
2 -> $
3 -> $
t.shortest(num = 10, len = 9)
Explanation: Transducers
The following automaton maps words to words.  The length of a multitape words is the largest length of its components: $\mathsf{len}(abc|x) = 3$.
End of explanation
<END_TASK>