9 Mar
2017
9 Mar
'17
6:58 p.m.
Hi Ayrat,
On Thu, Mar 9, 2017 at 4:17 PM, Ayrat Khalimov <ayrat.khalimov(a)gmail.com> wrote:
Is there a way to get more than one accepting run of
the product automaton?
Ideally would be to have all simple accepting runs,
but some cheap heuristics would also do.
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.
how to map edges of the accepting run to the edges of
the original product?
Do you mean this:
prod = spot.product(left, right)
run = prod.accepting_run(prod)
left_run = run.project(left)
or equivalently:
left_run = left.intersecting_run(right)
?
Reading you again, I'm not sure if the question is about a product or
any automaton. 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 of equality
if the run has been projected as above).
PS: also I think currently none of the emptiness checks look at the label
of edges they assume that all labels are different from bddfalse. This might
be a flaw. Or we should at least document it. If you are altering
left or right
in a product, this is OK, because product() will not create bddfalse
transitions,
but if you are running the emptiness check directly on an automaton with
bddfalse-labeled transition, that will fail (I think).
--
Alexandre Duret-Lutz