12 Mar
2017
12 Mar
'17
2:59 p.m.
thanks!
I used the enumeration of accepting runs that you suggested. (my tool
didn't perform as better as i hoped, but that is my task:)
> Do you mean this:
> prod = spot.product(left, right)
> run = prod.accepting_run(prod)
> left_run = run.project(left)
> Anyway to map a edge from a run to its automaton,
> you'd have to enumerate all the edges that leave the source of the said
> edge, and locate one that have the same label, acceptance mark, and
> destination (you might have to check label implication instead
you nailed it! I enumerated the edges of the original automaton and
compared the edges
as you suggested. That worked!
thanks for your help,
Ayrat
On Fri, Mar 10, 2017 at 10:45 AM, Alexandre Duret-Lutz <adl(a)lrde.epita.fr>
wrote:
On Thu, Mar 9, 2017 at 6:58 PM, Alexandre Duret-Lutz
<adl(a)lrde.epita.fr>
wrote:
Some of the NDFS-based emptiness checks (like
SE05) support having their
check() method called several times and they will restart from where
they stopped, possibly returning more accepting runs. But I'm not sure
we
even have a test for this.
I'll try to see if I can demonstrate this in Python and get back to you.
The following seems to report only non-overlapping cycles:
import spot
aut = spot.translate("G(p0 | (p0 R Xp0) | XF(!p0 & p1))", 'BA')
# Note: SE05 should only be used with Büchi acceptance.
ec = spot.make_emptiness_check_instantiator('SE05')[0].instantiate(aut)
while True:
res = ec.check()
if not res:
break
print(res.accepting_run())
--
Alexandre Duret-Lutz