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<<<-------- Simone Gabbriellini-------->>>
> *****  To join INSNA, visit http://www.insna.org  *****
>
> Dear Vlado,
>
> maybe a rough example can clarify what I would like to achieve... take
> a first network A, with nodes={a,b,c} and no links, then possible
> power-sets are of course:
> a
> b
> c
> ab
> ac
> bc
> abc
>
> take a second network B, with nodes={a,b,c} and one link={a<->b}, then
> possible solutions to my problem, let's call them "conflict-free"
> sets, are:
> a
> b
> c
> ac
> bc
>
> is this sufficient to answer your question?
>
  Yes, now I understand your question.
  Note that the sets can be very large.
  For an empty graph on n vertices you get 2**n sets
    2**20 = 1048576 = 1.049e+6
    2**30 = 1073741824 = 1.074e+9
    2**50 = 1125899906842624 = 1.126e+15
    2**100 = 1267650600228229401496703205376 = 1.268e+30
  So you are limited to small graphs.

  Vlado
-- 
Vladimir Batagelj, University of Ljubljana, FMF, Department of Mathematics
  Jadranska 19, 1000 Ljubljana, Slovenia
http://vlado.fmf.uni-lj.si

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