Robert J. Hansen
rjh at sixdemonbag.org
Fri Nov 14 16:22:14 CET 2014
> Thanks for that (and the previous).... It makes the brain hurt but
> raises a few questions in my mind.
The real purpose of a Fermi estimate isn't to give you solid answers:
it's to give you an appreciation of the problem. If it does that, it's
done its job.
(Also, a listmember named Ineiev points out that I *may* be misapplying
the Margolus-Levitin theorem. It's ... interesting. I need to think on
it for a while. The objection is basically, "that much energy has to be
present, but not necessarily released as heat: Landauer still applies.
You need something really energetic, but that energy might not be
world-ending." I don't know: I need to think about that. :) )
> Does anything prevent the key breaker getting lucky and cracking it
> first try?
Nope. The odds are considerably worse than the lottery, though.
> It seems to me that all discussions on key breaking with their very
> large numbers always assume that the last try is THE ONE.
The assumption is the key is broken after exhausting 50% of the keyspace.
> And how does the cracker know he has succeeded ?
We're assuming the cracker has a crib -- a known message header or
something similar. This is usually a safe assumption to make.
> Does he have to pause between each iteration to see if he has
> 'something good' ?
Checking takes time, yes.
> And in the 10**38 key attempts, what's the chance of having multiple
> apparently 'GOOD ONES' ?
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