Hi Nathann, First, thank you for taking time to give a very detailed reply. I'm sorry but I'm not yet done bothering you :-] I think this is wrong: > When v is considered for removal, it is a leaf of the lex-BFS tree. > Its father (and first discoverer) is named x, and we suppose that > there is a vertex y which is not a neighbor of x, otherwise v is > removed without any problem. (I also mentioned this in message #3, but it seems it still holds) Consider: g = graph({1:[2,3,4,5],2:[3,5],4:[3,5]}). g is like [the 4-cycle 2--3--4--5] + [(1,x) for x in [2 .. 5]] g is not chordal. A LexBFS ordering (reverse elimination ordering) could be [1, 2, 3, 5, 4]. Notice that all v in [2 .. 5] have father equal to 1. But then x is adjacent to every other vertex, i.e. we can't find y.
Regards, Jan -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org