[rc5] We may only need 5 years...
Paul Leskinen
paul at csfi.com
Thu Oct 23 16:20:44 EDT 1997
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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