On Tue, 21 Oct 2014 11:11:24 +0000
Vincent Delecroix <[email protected]> wrote:
> > G=Graph()
> > for i in range(0,10): G.add_vertex(randint(1,1000))
> > for x in G.vertex_iterator(): print x
> > G.vertices()
> >
> > gives them in about random order.
>
> Nope. The order is not random. It is the one you get by doing list(set(X)).
That, however, can depend on the order of X. For instance given
class C:
def __init__(self, v):
self.v = v
def __repr__(self):
return "C(%r)" % (self.v,)
def __hash__(self):
return int(0)
zero = C(0)
one = C(1)
when I do
print list(set([zero,one]))
print list(set([one,zero]))
I obtain
[C(0), C(1)]
[C(1), C(0)]
..
> sage: list(set(G.vertices())) == list(G.vertex_iterator())
> True
>
> You can have a look at how is coded vertex_iterator in the backend
> (i.e. G._backend.iterator_verts).
That can vary by backend, can it not?
>
> > I started thinking this when doing http://trac.sagemath.org/ticket/17173 .
> > Nathann suggested using directly (sub)graphs instead of (sub)poset. It is
> > OK with .vertices() but not with .vertex_iterator(); try
> >
> > G=Graph()
> > for i in range(2,50): G.add_vertex(i)
> > for i in range(2,50):
> > if not is_prime(i):
> > G.delete_vertex(i)
> > for x in G.vertex_iterator(): print x
> >
> > This seems to be quite open door for nasty bugs. .vertex_iterator might
> > give vertices in order when graph is small or has not been modified many
> > times.
>
> The code is not very clean, but nevertheless the order of
> .vertex_iterator does not depend on the way you built your graph. Do
> you have an example where it is not the case ?
Yes:
G = Graph()
G.add_vertex(zero)
G.add_vertex(one)
H = Graph()
H.add_vertex(one)
H.add_vertex(zero)
print G == H
print list(G.vertex_iterator()) == list(H.vertex_iterator())
print G.vertices() == H.vertices()
gives me
True
False
True
..
Cheers,
Erik Massop
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