From: Max Williams Date: 2008-07-01T18:20:11+09:00 Subject: Re: how to - quickly make permutations? Frederick Cheung wrote: > > This is recursive with a shortcut: since we are anyway accumulating > the previous results, there is no point calculating them over and over > again (if not p(n,1) is calculated max_k times, p(n,2) is calculated > max_k - 1 times etc... > > > Fred Thanks everyone - Fred, this is perfect, having the different sized groups seperated into their own arrays is actually even better for me than what i specified. Thanks. Axel - you're right, i did mean subsets. oops. thanks again, max -- Posted via http://www.ruby-forum.com/.