You're right. In A = floor((N-O)/(S-O)) * C(N,O) / (O+1), O is the maximum overlap.
--- On Thu, 10/16/08, Vladimir Nesov <[EMAIL PROTECTED]> wrote: > From: Vladimir Nesov <[EMAIL PROTECTED]> > Subject: Re: [agi] Who is smart enough to answer this question? > To: [email protected] > Date: Thursday, October 16, 2008, 6:24 PM > On Fri, Oct 17, 2008 at 5:04 AM, charles griffiths > <[EMAIL PROTECTED]> wrote: > > I think A = floor((N-O)/(S-O)) * C(N,O) / (O+1). > > > > Doesn't work for O=2 and S=2 where A=C(N,2). > > P.S. Is it a normal order to write arguments of C(,) this > way? I used > the opposite. > > P.P.S. In original problem, O-1 is the maximum allowed > overlap, in my > last reply I used O incorrectly in the first paragraph, but > checked > and used in correctly in the second with the lower bound. > > -- > Vladimir Nesov > [EMAIL PROTECTED] > http://causalityrelay.wordpress.com/ > > > ------------------------------------------- > agi > Archives: https://www.listbox.com/member/archive/303/=now > RSS Feed: https://www.listbox.com/member/archive/rss/303/ > Modify Your Subscription: > https://www.listbox.com/member/?&; > Powered by Listbox: http://www.listbox.com ------------------------------------------- agi Archives: https://www.listbox.com/member/archive/303/=now RSS Feed: https://www.listbox.com/member/archive/rss/303/ Modify Your Subscription: https://www.listbox.com/member/?member_id=8660244&id_secret=117534816-b15a34 Powered by Listbox: http://www.listbox.com
