spot/tests/python/stutter.py
Alexandre Duret-Lutz 36d20696bf word: introduce use_all_aps()
This allows fixing issue #388 reported by Victor Khomenko.

* spot/twaalgos/word.cc, spot/twaalgos/word.hh (use_all_aps): New
method.
* tests/python/stutter-inv.ipynb: Use it.
* tests/python/stutter.py: New file, with Victor's test case.
* tests/Makefile.am: Add python/stutter.py.
2019-05-24 23:26:43 +02:00

47 lines
1.7 KiB
Python

# -*- mode: python; coding: utf-8 -*-
# Copyright (C) 2019 Laboratoire de Recherche et Développement de
# l'Epita (LRDE).
#
# This file is part of Spot, a model checking library.
#
# Spot is free software; you can redistribute it and/or modify it
# under the terms of the GNU General Public License as published by
# the Free Software Foundation; either version 3 of the License, or
# (at your option) any later version.
#
# Spot is distributed in the hope that it will be useful, but WITHOUT
# ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
# or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public
# License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program. If not, see <http://www.gnu.org/licenses/>.
# Test that the spot.gen package works, in particular, we want
# to make sure that the objects created from spot.gen methods
# are usable with methods from the spot package.
import spot
def explain_stut(f):
f = spot.formula(f)
pos = spot.translate(f)
neg = spot.translate(spot.formula.Not(f))
word = spot.product(spot.closure(pos), spot.closure(neg)).accepting_word()
if word is None:
print(f, "is stutter invariant")
return
word.simplify()
# This line used to be missing, causing issue #388.
word.use_all_aps(pos.ap_vars())
waut = word.as_automaton()
aut = neg if waut.intersects(pos) else pos
word2 = spot.sl2(waut).intersecting_word(aut)
word2.simplify()
return(word, word2)
# Test from issue #388
w1, w2 = explain_stut('{(a:b) | (a;b)}|->Gc')
assert str(w1) == 'a & !b & !c; cycle{!a & b & !c}'
assert str(w2) == 'a & !b & !c; a & !b & !c; cycle{!a & b & !c}'