Re: [jgrapht-developers] Adding multiple edges and loops

[Barak N. <ba...@3p...>]

> > for a similar reason, if the user tries to "forcibly" add multiple ed=
ges
> to
> > graph that forbids multiple edges (this can be done by abusing the
> > addEdge(Edge e) method) then an exception should be thrown -- the gra=
ph
> > needs to defend its consistency. if this is not currently the case, i=
t
> > should be fixed.
>=20
> This is currently not the case. addEdge(e) behaves exactly like=20
> addEdge(e.getSource(), e.getTarget()). It throws exception on forbidden
> loops and silently ignores forbidden multiple edges.=20

hmm, my goof then.. i think it should be corrected
(http://sourceforge.net/tracker/?func=3Ddetail&atid=3D579687&aid=3D103311=
1&group_id=3D86459)

> >
> > However, if the user attempts to add an edge which is already existin=
g
> > using addEdge(v1,v2), then it is not a mistake. he or she wants to in=
clude
> > the specified edge in the edge set. if the edge already in the edge s=
et,
> > there is nothing to do. if it is not, it is included.  in other words=
, the
> > call can satisfy the user wish, without violating any graph consisten=
cy
> > criteria and therefore there is no reason to throw an exception.
> >
> > is my reasoning making sense?
>=20
> Partly. I think it is more reasonible to let addEdge(e) behave exactly=20
> like addEdge(v1, v2), because on graphs with multiple edges allowed
> your argumentation would fail. What does the user want when calling
> addEdge(v1,v2) when such an edge is already present. Add another=20
> one or just ensure that an edge is present? I think a user calling=20
> addEdge really wants to add an Edge and following your reasoning
> both versions of addEdge should always throw an exception when the=20
> two vertices are already connected and multiedges are forbidden.

On graphs with multiple edges allowed the argumentation indeed becomes we=
aker,
or at least question raising. As a user i would want addEdge(v1,v2) to al=
ways
add a new edge if multiple edges are allowed. If multiple edges are forbi=
den, i
would want it to add a new edge only if no edge exists between v1 and v2.=
 Such
behavior is (almost:) consistent with the fact that the edge set is, well=
, a
Set. As such, it behaves differently when multiple edges are allowed --pa=
rallel
edges are equal in the non-multiple case, but aren't in the multiple case=
.

Not sure if it's the best solution, but i suppose it's better than a sile=
nt
ignoring. What do you think? Other ideas? John?

barak