Fermi estimates

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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