leif wrote: > Jori Mäntysalo wrote: >> On Fri, 26 Aug 2016, leif wrote: >> >>> Hmmm, does Posets.BooleanLattice() care whether you pass an int or an >>> Integer? (I.e., does the return type of its methods change?) >> >> No: >> >> sage: P = Posets.BooleanLattice(3) >> sage: Q = Posets.BooleanLattice(3r) >> sage: P == Q >> True >> sage: P is Q >> True > > I rather meant e.g. type(P.cardinality()), but since 'P is Q' holds, > they won't differ. > > No idea...
For me, it /does/ make a difference (using Python vs. Sage literals): $ ./sage ┌────────────────────────────────────────────────────────────────────┐ │ SageMath version 7.3, Release Date: 2016-08-04 │ │ Type "notebook()" for the browser-based notebook interface. │ │ Type "help()" for help. │ └────────────────────────────────────────────────────────────────────┘ sage: from sage.combinat.posets.poset_examples import Posets sage: from sage.misc.functional import log sage: P = Posets.BooleanLattice(3); n = P.cardinality(); s = P._hasse_diagram.size() sage: if 2*s > n*log(n, 2): print "True" sage: from sage.combinat.posets.poset_examples import Posets sage: from sage.misc.functional import log sage: P = Posets.BooleanLattice(3r); n = P.cardinality(); s = P._hasse_diagram.size() sage: if 2r*s > n*log(n, 2r): print "True" True sage: And it's caused by this: sage: type(n*log(n, 2)) <type 'sage.rings.integer.Integer'> sage: type(n*log(n, 2r)) <type 'sage.symbolic.expression.Expression'> (FWIW, while .cardinality() returns Integer, ._hasse_diagram.size() returns int, but that doesn't matter here.) -leif -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-devel@googlegroups.com. Visit this group at https://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
