[rc5] We may only need 5 years...

Eric Gindrup gindrup at okway.okstate.edu
Thu Oct 23 15:51:20 EDT 1997


     As I posted on 30 September 1997,
     
     
     By eyeball at http://home.pacbell.net/cwax/bovineg.html, the Bovine 
     effort seems to double its number of completed keys count every 2.5 
     weeks.  Current keyrate is 5Gkps.
     Let's say:
     1.  We start RC5-64 right now.
     2.  The doubling rate is constant for the duration I'm about to 
     compute.
     3.  We have to exhaust all of the RC5-64 keyspace.
     
     Let f(w) be the number of completed keys with w in weeks.  (The 
     following keeps about one significant digit.)
     f'(0) = 5Gkps.  f(w+2.5) = 2*f(w).  These together mean f(w) = 
     1.4*2^(w/2.5).  To make this =2^64, w=160.
     
     So, in 3 years, we could exhaust the keyspace.  We'd "probably" find 
     the key in about 18 months.  This is so terrible?
             -- Eric Gindrup ! gindrup at okway.okstate.edu


______________________________ Reply Separator 
_________________________________
Subject: [rc5] We may only need 5 years... 
Author:  <rc5 at llamas.net > at SMTP
Date:    10/23/97 3:20 PM


I've done some math regarding rc5-64.

OK, let's first assume we need to check (on average) 50% of the key space.  
It's amazing how close we were to 50% when we found the 56-bit key.

For the first (not incredibly realistic) example, let's assume we have a 
keyrate of 7*10^9 keys/sec, and that rate stays constant.  We would work for 
about 41.8 years to finish the project.

Next, let's assume our rate starts the same (7*10^9 keys/sec), and we double it 
every year.  If that's the case, then our key rate (dk/dt) is expressed by 
r*2^t, where r is our initial keyrate in keys/year, or 2.2*10^17.  Taking the 
integral of t from 0 to y, we get:
     
k=(r*2^y - r)/ln(2).  That describes the number of keys we check per year.  We 
set k=2^63, and solve for y, yielding y=(ln(2^63*ln(2) + r) - ln(r))/ln(2). 
     
Whew. Typing that into my calculator tells me that we'll "only" need about 4.9 
years to finish 50% of the key space.  On the upside, we only need 5.9 years to 
check 100% of the key space.
     
If we double it every six months, we'll still need about 2.5 years to check 50%.
     
Anybody concur with this analysis?
     
  -- Paul Leskinen
     
     
     
One further note on RC5-64:  Assuming we start with our 7e9 keyrate, and we 
double the rate each year (accounting for some advances in technology and people
joining the project), we'll take about 4.9 years to check half the keys, and 5.9
years to check all the keys.
     
     
\=======================================/ 
|  Paul Leskinen    (paul at csfi.com)     | 
|   Sr. Systems Analyst                 | 
|    Consolidated Systems of Florida    | 
|     "The Open Systems Solution"       | 
/=======================================\ 
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