User talk:Jorgenev/MFD'd/Wikipedia:Recent Changes coefficient

something for meta? Martin

It is a little addendum to Wikipedia:Are You a Wikipediholic Test, so I have put in the same namespace. - Patrick 01:34 19 Jul 2003 (UTC)

I don't really get this. Shouldn't it be the other way around? If you have a low C value you're adept at following up on recent changes. If you have a score(rate of addiction) of almost 1 then you are very slow at getting up to speed! Correct me if am wrong... You could reverse the c value by making the formula something like this n/(c-1)=time needed. Eruantalon

The larger c (below 1) the longer it takes, even my wrong version, which I have now corrected, had that right.--Patrick 23:40, 15 Apr 2005 (UTC)

If c<1, after being away for n hours it takes n/(c-1) hours to catch up.

This is impossible - for c<1, this results in a negative number. Daniel

You are right, my formula was wrong, but the way it was changed was not an improvement. I have corrected it now. To process the changes which occurred while you were away takes nc hours, but in the meantime there are more changes, taking nc², etc. The sum of the geometric series is nc/(1-c).--Patrick 23:34, 15 Apr 2005 (UTC)

c=1 or more? So, essentially, c>1 would imply you're such a wikipedaholic that you can anticipate and react to changes before they're made. Yes, that is indeed scary. --Saforrest 23:30, 10 March 2006 (UTC)[reply]

Switch to watchlist

Perhaps, since RC is so overwhelming these days, we should be more concerned with one's watchlist? If one is keeping up with one's watchlist a good portion of the day, then that is addicted indeed. --maru (talk) contribs 06:55, 29 May 2006 (UTC)[reply]

Well that would not be too efficient as the number of watchlists users have varies considerably. --S iva1979^{Talk to me} 20:34, 9 June 2006 (UTC)[reply]

My reversal

I have changed the old c to $(c)^{-1}$ as if you take less than 1 hour to catch up on the RC of 1 hour, you're anticipating and reacting to changes before they're made. 4 T C 14:58, 29 December 2009 (UTC)[reply]

No, you're not. The scenario is that you go away for an hour, then return and start to catch up. --Tango (talk) 09:23, 21 March 2010 (UTC)[reply]

Equation is wrong?

The expression given is negative for c>1. Is this a mistake? NotAnonymous0 (talk) 06:17, 21 March 2010 (UTC)[reply]

Yes. I've reverted the most recent change. There has been a very slow edit war on this page about the maths... people keep making it not make any sense and I keep fixing it... --Tango (talk) 09:22, 21 March 2010 (UTC)[reply]