> Probably that won't be needed. At some point I need to know if a vertex > is part of an oriented cycle. That's all. I need to know it only once.
Okay. Then try that: sage: any(v == x for x in d.breadth_first_search(d.neighbors_out(v))) Algorithmically it is optimal, but it is written at higher level than the strongly connected components routines (and may be slower as a result). Nathann -- You received this message because you are subscribed to the Google Groups "sage-combinat-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-combinat-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-combinat-devel@googlegroups.com. Visit this group at http://groups.google.com/group/sage-combinat-devel. For more options, visit https://groups.google.com/d/optout.
