Password length paranoia

[Peter Pentchev]

On Tue, Feb 07, 2006 at 08:14:29PM +0100, Ludwig H?gelsch?fer wrote:
> Hi,
> 
> On 07.02.2006 20:05 Uhr, Oskar L. wrote:
> 
> > This is of course only true if the attacker knows it is exactly 15
> > characters long. If not, then it should be calculated like this: 95^1 +
> > 95^2 + 95^3 + ... + 95^15.
> 
> Right, this gives exactly 95^16 - 1.

Errr... only for very small values of 95, I'd say ;)  Namely, only
if 95 equals 2... which it doesn't quite seem to :)

Oh, ooookay, so it's more like (95^16 - 1^16) / (95 - 1), which is
only 94 times less than your bid.  But still... :)

> This is not a dramatic improvement compared to 2^256.

Errrrrr... what?!

Straight from bc's mouth:

(95^16-1)/(95-1)
468219860267835848675991626496

2^256
115792089237316195423570985008687907853269984665640564039457584007913129639936

Okay, so it might not be quite in the feasible range - I'll leave
that for others to judge - but it does seem pretty dramatic to me.

G'luck,
Peter

-- 
Peter Pentchev	roam at ringlet.net    roam at cnsys.bg    roam at FreeBSD.org
PGP key:	http://people.FreeBSD.org/~roam/roam.key.asc
Key fingerprint	FDBA FD79 C26F 3C51 C95E  DF9E ED18 B68D 1619 4553
If I had finished this sentence,
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