[RC5] RC5 question

Elektron elektron_rc5 at yahoo.ca
Sat Oct 4 10:05:04 EDT 2003

> The problem is not export restrictions or anything (they've been 
> lifted, as
> far as I know); it's just the sheer amount of work that has to be 
> done. A
> book of mine states that since the inception of the computer through 
> the year
> 2000, there have been approximately a mole (6e23) of computer 
> operations
> performed on all computers on Earth. In comparison, there are 3.4e38 
> RC5-128
> keys to check. Even taking into account that only 50% of keyspace is 
> expected
> to be checked on average, that's 1.7e38 keys, or 1.7e40 cycles. 
> Dividing out
> 1.7e38 by 6e23, you get a number which is on the order of magnitude of 
> a
> million times the age of the universe.

Taking the latest RC5-64 numbers,
10,428,970,063,364,096 Keys were completed yesterday (0.056536% of the 
keyspace) at a sustained rate of 120,705,672,030 Keys/sec.

That's 3809103098378452560 Keys/year (let's call it k). If computing 
power doubles every 18 months (1.5 years), then a^1.5=2, or a=2^(1/1.5) 
or approx. 1.587401051968. So our keyrate with respect to time t 
(years) is ka^t, so our keys tested (big K) (by integration) is ka^t 
ln(a) + C, and since we have no keys tested when t=0, k ln(a) + C = 0, 
so K = ka^t ln(a) - k ln(a). K=2^128.

2^128 = k ln(a) (a^t-1)
2^128/(k ln(a)) = a^t-1
2^128/(k ln(a)) + 1 = a^t
t = ln(2^128/(k ln(a)))/ln(a) = 101.08435774324143022752

Of course, I'm assuming RC5-128 is as fast as RC5-64 (which it isn't), 
and that moore's law holds (which it might). Interestingly, the last 
time I did this, I got 600 years. Somebody hit me if I did my math 
wrong. But either way, 101 years is a lot more than my lifetime.

- Purr.

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