[RC5] RC5 question
Daniel Quintiliani
coredump0 at mercurylink.net
Fri Oct 3 22:32:56 EDT 2003
On Sat, 4 Oct 2003 09:05:04 +0800, Elektron <elektron_rc5 at yahoo.ca> wrote:
>
> 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.
>
Using yesterday's RC5-72 rate for a more accurate measure of time:
114327804047 Keys/s * 60 * 60 * 24 * 365.25
(goes upstairs and grabs TI-83+ from backpack)
k = 3.607911127 * 10^8
Plugging it into your formula:
t = ln(2^128/((3.607911127*10^8) ln(2^(1/1.5)))) / ln(2^(1/1.5))
t = 101.201789 years
--
-Dan
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