Vlad,
Thanks. In respone to your email I tried plugging different values into the Excel spread sheet I sent by a prior email under this subject line, and, and low and behold, got some interesting answers for the number A of assemblies (or sets) of nodes of uniform size S you can create from N nodes where no two assemblies have more than O overlapping nodes: N S O A ============================================== 100K 20 1 2.5x10^2 100K 20 2 1.3x10^5-10% overlap 100K 20 3 1.2x10^8 ============================================== 100K 50 2 3.3x10^3 100K 50 3 4.4x10^5 100K 50 4 7.9x10^7 100K 50 5 1.8x10^10-10% overlap ============================================== 100K 88 2 3.5x10^2 100K 88 3 1.4x10^4 100K 88 4 8.1x10^5 100K 88 5 5.7x10^7 100K 88 6 5.6x10^9 100K 88 7 5.2x10^11 100K 88 8 6.3x10^13-9.09% overlap ============================================== 50K 88 6 8.2x10^7 50K 88 8 1.6x10^11-9.09% overlap ============================================== 25K 88 6 1.4x10^6 25K 88 8 1.1x10^9-9.09% overlap >From this data, it is clear that for a given N, increasing S (as least when S is small relative to N), increases the number of nodes that have a given percent of maximum overlap, such as 5% or 10% max overlap. Doubling N from 25k, to 50K, to 100K, provides a little less than two orders of magnitude in increases in the number of cell assemblies of size 88 having overlaps of 6 or 8 at each doubling. 88 was the largest number S that Excel could produce a C(100K,S) for, enabling the calculations to be made. BUT EVEN AT THIS LIMIT WITH 100K NODES, YOU COULD CREATE 57 MILLION NODE ASSEMBLIES WITH AN OVERLAP OF LESS 5.7%. This is a 570x increase in the number of states that could be represented, relative to representing states with individual nodes. Since it is clear that the number of possible assemblies relative to percent overlap, grows with N and with S, it is likely that one could produce even larger multiplicative increases in number of assemblies relative to the number of nodes having even lower percentage overlap, with larger Ns and/or Ss. The one thing I don't understand is how you derived the formula I used in this spreadsheet, the one you described in your Thu 10/16/2008 7:50 PM email (with the position of the variables in the combinations function switched). Switching the variables in the combinatorial formula to the convention more commonly used in America this formula that is as follows: A =C(N,S)/T(N,S,O) Where T(N,S,O)= C(S,S) +C(S, S-1)*C(N-S, 1) +C(S, S-2)*C(N-S, 2) +... +C(S, O)*C(N-S, S-O) (note the first term in T(N,S,O), i.e., C(S,S), is the equivalent of C(S,S-0)*C(N-S,0) since C(X,0)=1, oddly enough, which makes all the terms in T(N,S,O) have the same form, differing only by the value of the iterator which goes from 0 to O) THE ONE PROBLEM I HAVE, IS THAT I DON'T UNDERSTAND THE DERIVATION OF THIS FORMULA, SO I CAN'T KNOW HOW MUST FAITH OR ACCURACY I SHOULD ATTRIBUTE TO ITS ESTIMATION OF A LOWER BOUND. If it is possible to give an explanation of why this formula is a proper lower bounds, in a little more detail than in the email in which you first presented it, I would appreciate it very much, it would let me know how much faith I should put into the above numerical results. Ed Porter -----Original Message----- From: Vladimir Nesov [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 21, 2008 1:14 AM To: [email protected] Subject: Re: [agi] Who is smart enough to answer this question? On Tue, Oct 21, 2008 at 12:07 AM, Ed Porter <[EMAIL PROTECTED]> wrote: > > I built an excel spread sheet to calculate this for various values of N,S, > and O. But when O = zero, the value of C(N,S)/T(N,S,O) doesn't make sense > for most values of N and S. For example if N = 100 and S = 10, and O = > zero, then A should equal 10, not one as it does on the spread sheet. > It's a lower bound. > I have attached the excel spreadsheet I made to play around with your > formulas, and a PDF of one page of it, in case you don't have access to > Excel. > Your spreadsheet doesn't catch it for S=100 and O=1, it explodes when you try to increase N. But at S=10, O=2, you can see how lower bound increases as you increase N. At N=5000, lower bound is 6000, at N=10^6, it's 2.5*10^8, and at N=10^9 it's 2.5*10^14. -- 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
