automaton.is_cycle_ambiguous

Whether the automaton is cycle ambiguous. It means that there exist a state $s$ and a label $x$ such that there is more than one cycle in $s$ labeled with $x$.

Preconditions:

the labelset is free.

Examples



In [1]:

    
import vcsn









    



:0: FutureWarning: IPython widgets are experimental and may change in the future.



In [2]:

    
a = vcsn.automaton('''digraph
{
  vcsn_context = "lal_char(ab), b"
  I0 -> 0
  0 -> F0
  0 -> 1 [label = "a"]
  0 -> 2 [label = "a"]
  1 -> 0 [label = "b"]
  2 -> 0 [label = "b"]
}''')
a









    Out[2]:

At state $0$, the automaton has two cycles "ab" so it is cycle ambiguous.



In [3]:

    
a.is_cycle_ambiguous()









    Out[3]:





True



In [4]:

    
a = vcsn.automaton('''digraph
{
  vcsn_context = "lal_char(abc), b"
  I0 -> 0
  0 -> F0
  0 -> 1 [label = "a"]
  0 -> 2 [label = "a"]
  1 -> 0 [label = "b"]
  2 -> 0 [label = "c"]
}''')
a









    Out[4]:

At state $0$, the automaton has two cycles with different label "ab", "ac" so it is not cycle ambiguous (it is cycle-unambiguous).



In [5]:

    
a.is_cycle_ambiguous()









    Out[5]:





False



In [6]:

    
a = vcsn.context("lal_char(abc), b").ladybird(3)
a









    Out[6]:

Two cycles in $0$ with label "acac".



In [7]:

    
a.is_cycle_ambiguous()









    Out[7]:





True



In [8]:

    
a = vcsn.automaton('''digraph {
  vcsn_context = "lal_char(abc), b"
  I0 -> 0
  0 -> F0
  0 -> 1 [label = "a"]
  0 -> 2 [label = "a"]
  1 -> 0 [label = "b"]
  1 -> 3 [label = "b"]
  2 -> 1 [label = "c"]
  2 -> 2 [label = "b"]
  3 -> 1 [label = "c"]
}''')
a









    Out[8]:

Two cycles in $0$ with label "babc".



In [9]:

    
a.is_cycle_ambiguous()









    Out[9]:





True



In [10]:

    
a = vcsn.automaton('''digraph {
  vcsn_context = "lal_char(ab), b"
  I0 -> 0
  0 -> F0
  1 -> 2 [label = "a"]
  1 -> 3 [label = "a"]
  2 -> 1 [label = "b"]
  3 -> 1 [label = "b"]
}''')
a









    Out[10]:



In [11]:

    
a.is_cycle_ambiguous()









    Out[11]:





True