Dear Jean-Yves, > OK, but there is some inconsistent behavior, anyway: > > > sage: P=Partitions(4) > sage: len(P) > 5 > sage: C=Compositions(4) > sage: len(C) > --------------------------------------------------------------------------- > AttributeError Traceback (most recent call last) > > /home/jyt/monge/TEX/QSCHUR/<ipython console> in <module>() > > /home/jyt/src/sage-5.4.1/local/lib/python2.7/site-packages/sage/combinat/combinat.py > > in __len__(self) > 972 AttributeError: __len__ has been removed; use > .cardinality() instead > 973 """ > --> 974 raise AttributeError, "__len__ has been removed; use > .cardinality() instead" > 975 > 976 def count(self): > > AttributeError: __len__ has been removed; use .cardinality() instead > sage: R=reversed(P) > sage: list(R) > [[1, 1, 1, 1], [2, 1, 1], [2, 2], [3, 1], [4]]
Yes ! As you probably already guessed, everything is not yet in a clean state... The first call should raise the same error than the second one. We can't answer consistently to len because Python specified it as being a machine integer. This is much too small for cardinality of combinatorial set and prevent for returning infinity. So as the error message tell you, please write sage: P=Partitions(4) sage: P.cardinality() 5 sage: P=Partitions(4000) sage: P.cardinality() 1024150064776551375119256307915896842122498030313150910234889093895 sage: P=Partitions() sage: P.cardinality() +Infinity And indeed, I think that sage: len(Compositions(4000)) [...] AttributeError: __len__ has been removed; use .cardinality() instead Is a much clean behavior than: sage: len(Partitions(4000)) [...] RuntimeError: maximum recursion depth exceeded in cmp A++ Florent -- You received this message because you are subscribed to the Google Groups "sage-combinat-devel" group. To post to this group, send email to sage-combinat-devel@googlegroups.com. To unsubscribe from this group, send email to sage-combinat-devel+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sage-combinat-devel?hl=en.
