On Fri, Oct 17, 2008 at 12:56 AM, Ed Porter <[EMAIL PROTECTED]> wrote: > Vlad, > > If, as you below post indicates, it is easy to prove an example of how large > the ratio of the number of cell assemblies with say less than 5% overlap > with the population of any other assembly is compared to the number of nodes > out of which such cell assemblies are made --- could you provide me with a > few impressive examples. >
There are total of C(S,N) (binomial coefficient, N!/S!(N-S)!) sets of S nodes. If you include a certain node, it prohibits some of other nodes from being included. If you only allow intersection of up to O, all the sets of S nodes that have intersection greater than O get prohibited. Let's say there are T(N,S,O) such sets. It is a fixed number of sets prohibited by each new set accepted as cell assembly. Some sets can be prohibited by multiple included cell assemblies, so I'm getting a lower bound. At least C(S,N)/T(N,S,O) sets won't prohibit each other, it's a lower bound on number of assemblies. Now, T(N,S,O)=C(S,S)+C(S-1,S)*C(1,N-S)+C(S-2,S)*C(2,N-S)+...+C(O,S)*C(S-O,N-S). It's bound by polynomial of order S-O with respect to N, whereas C(S,N) is bound by polynomial of order S. Thus, even if you only allow the overlap of O=1 (so that no two cell assemblies are allowed to have even two nodes in common), you can get arbitrarily large number of cell assemblies (including in proportion to N) by choosing big enough N, for any given S. -- 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/?member_id=8660244&id_secret=117534816-b15a34 Powered by Listbox: http://www.listbox.com
