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Re: comments on client auth
Bill Stewart wrote:
> >> Perhaps someone more proficient with numbers could calculate how many
> >> certificates would have to exist in the world, each generated with perfect
> >> 1024 bit entropy, before there was, say, a 1% chance of a collision.
> >I can't quite be bothered to solve this for n=2^1024 and P=.99 but for the
> >sake of illustration, lets take k=2^100 (=10^33, roughly, or 2*10^23
> >certificates per person in the world). Then P=1-2^200/2^1025=1-2^(-825).
> >In other words, the probability of a collision is roughly 1 in 10^275.
> >Vanishingly small, I would say.
> Keys aren't distributed evenly, because prime numbers aren't distributed
> evenly; density is roughly proportional to 1/logn, so if you have 1024 bits
> of entropy picking a random number, and searching upwards for primes from there,
> you'll probably have to pass 693 of them on average (logs base e) to get a
> So it's maybe 2^1015 keys instead of 2^1024, and collisions 26 times more
> likely :-)
Good point. Thanks for the correction ;-)
Perhaps we should use 1033 bit keys just to be safe <really big grin>.
BTW, I also got my base 2 to base 10 conversion wrong. 2^825 is nearer 10^250.
Allowing for the correction, that should be around 10^247.
Also, I don't quite see the factor of 26 ... for the low likelihood in the
example, the factor is 693, not 693^.5, surely?
> The effect is much more severe if you're using overly-short keys,
> like 384-bits, but it's still excessively unlikely that you'll get any
Even at 384 bits we are still looking at around 10^50:1 for 2^100 keys. This is
still an exceedingly large number. BTW, can anyone remember the number of atoms
in the known universe? (I seem to remember around 10^30) - I'm still trying to
find a vaguely graspable comparison...
> unless something's seriously wrong with your RNGs.
> # Thanks; Bill
> # Bill Stewart +1-415-442-2215 email@example.com
> # http://www.idiom.com/~wcs
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