"Anton Vredegoor" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] | Raymond Hettinger wrote: | | > To make the solutions equi-probable, a simple approach is to | > recursively enumerate all possibilities and then choose one of them | > with random.choice(). | | Maybe it is possible to generate the possibilities by an indexing | function and then use randint to pick one of them. I suppose this is | like the bricks and bins problem this thread was about: | | http://groups.google.nl/group/comp.lang.python/browse_thread/thread/4782b54fa39b3bad | | Except that the bins now have at least 1 brick in them (if we have | positive numbers). | | I posted a rather simplistic solution (but working I think) after Steven | Taschuk made some insightful remarks. I believe it is possible to | generate the list of numbers directly instead of permuting a list of '0' | and '1' characters and then finding the positions of the '1' elements.
Partitioning positive count m into n positive counts that sum to m is a standard combinatorial problem at least 300 years old. The number of such partitions, P(m,n) has no known exact formula but can be computed inductively rather easily. The partitions for m and n can be ranked in lexicographic order from 0 to P(m,n)-1. Given a rank r in that range, one can calculate the particular partition that has that rank. So a equi-probable random count in the range can be turned into a equi-probable random partition. This topic is section 3.1 in Combinatorial Algorithms: Generation, Enumeration, and Search by Kreher and Stinson. The authors reference several other books as their sources. I plan to someday rewrite many of their pseudocode algorithms in Python. Terry Jan Reedy -- http://mail.python.org/mailman/listinfo/python-list
