Hey Christian, On Monday, December 16, 2013 2:24:05 AM UTC-8, Christian Stump wrote: > > Hi, > > do you consider the following a bug? > > sage: A = Core([2],4) > sage: B = Core([2],5) > sage: hash(A) == hash(B) > True > > I wouldn't necessarily call that a bug, as it is necessary since A == B.
The hash of a core only > depends on the list and not on the core. Since the same list considered as > an x- and as a y-core are fundamentally different objects, I would rather > suggest to have the hash depending not only on the list. > > Let me ask this, is being a k-core a property of a partition or is it a *distinct* combinatorial object? I think of it as a property since we look at partitions and classify them as being k-cores, and because of that, I think they should evaluate as equal. Moreover, the 4-core [2] is also a 5-core, so I think this is correct: sage: Core([2],4) in Cores(5,2) True Since the cores are elements, the "A is B" should be False. However if we do agree that these are not equal, we should print that these are k-cores; something like "[2] as a 4-core". Here's also a similar situation, suppose you have a 5-bounded partition [2], should it evaluate as equal to the 4-bounded partition [2], or the usual partition [2] in Partitions() (or Partitions(2))? They all have different parents, so should they not evaluate as equal and/or have the same hash? For the record, the core() method is actually about a different type of core (see the method's doc). Best, Travis -- 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/groups/opt_out.
