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Views: 45903
Kernel: Python 3

automaton.is_ambigous

Whether the automaton is ambiguous.

Preconditions:

  • the labelset is free.

Postconditions:

See also:

Examples

import vcsn a = vcsn.context('lal_char(abcd), b').expression('abc').standard() a
Image in a Jupyter notebook
a.is_ambiguous()
False
a + a
Image in a Jupyter notebook
(a+a).is_ambiguous()
True