On Sat, Nov 01, 2003 at 06:25:45AM -0800, Martin Stone Davis wrote:
> Matthew Toseland wrote:
> 
> >Update of /cvsroot/freenet/freenet/src/freenet
> >In directory sc8-pr-cvs1:/tmp/cvs-serv25429/src/freenet
> >
> >Modified Files:
> >     Version.java 
> >Log Message:
> >6295:
> >If HTL == maxHTL, 50% chance of not decrementing. I believe this
> >converges to an extra 2 hops, somebody please check my math. This
> >provides some plausible deniability when you send a request at max
> >HTL (perturb runs only on client level...).
> Nope, it's one extra hop.  To make it two extra hops, you need to 
> increase the chance of not decrementing to 66.67%.

Cool. I didn't particularly want extra hops. I just wanted to fix the
current behaviour, which is bad.
> 
> EXPLANATION:
> 
> Given a request at HTL==25, there's a 50% chance that it will 
> immediately decrement.  There's a 25% chance that it will decrement 
> after 1 hop.  There's a 12.5% chance that it will decrement after 2 
> hops.  6.25% ... 3 hops, etc.
> 
> So, in terms of expected # of extra hops, E:
> 
> E=0.5*0 +
>   0.25*1 +
>   0.125*2 +
>   0.0625*3 +
>   ...
>  =0.5*Sum[k=0 to infintiy,k*(0.5^k)]
>  =0.5*0.5/(1-0.5)^2
>  =1
> 
> To derive that, I used the formula:
> 
> Sum[k=0 to infinity,k*q^k]
> = q/(1-q)^2,  (for |q|<1)
> 
> 
> So, if we want it to converge to 2 extra nodes, and q is the chance that 
> we don't decrement, we have to solve:
> 
> 2=(1-q)*0 +
>   q*(1-q)*1 +
>   q^2*(1-q)*2 +
>   ...
>  =(1-q)*Sum[k=0 to infinity,k*q^k]
>  =(1-q)*q/(1-q)^2
>  =q/(1-q)
> 
> Solving for q: q=2/3.  So, make the chance of not decrementing=2/3.
> 
> -Martin
> 
> 
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-- 
Matthew J Toseland - [EMAIL PROTECTED]
Freenet Project Official Codemonkey - http://freenetproject.org/
ICTHUS - Nothing is impossible. Our Boss says so.

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