Charles-François Natali <neolo...@free.fr> added the comment: > Since we are are trying to fix a problem where hash(X) == hash(Y), you > can't make them orderable by using the hash-values and build a binary > out of the (equal) hash-values.
That's not what I suggested. Keys would be considered equal if they are indeed equal (__eq__). The hash value is just used to know if the key belongs to the left or the right child tree. With a self-balanced binary search tree, you'd still get O(log(N)) complexity. Anyway, I still think that the hash randomization is the right way to go, simply because it does solve the problem, whereas the collision counting doesn't: Martin made a very good point on python-dev with his database example. ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue13703> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
