On Sun, 07 Jan 2007 16:08:23 +0100, endrazine said:
> > yes that's correct but don't forget that hashes can collide
> >
> > it could be the case that:
> >
> can ? could ? might ? Do you have any mathematical prouve or are you
> just guessing ?
It's a pretty easy proof actually. If your password input routine allows
more different passwords than there are possible hashes, you *will* have
collisions. For instance, if you use a 64-bit hash, and reasonable-length
passwords, you can create more than 2**64 of them, and 2 *have* to collide.
> > xhash("$Up3$tr0n9 # P@$sWoRD!!") == xhash("1234") and you don't even
> > need the original strong one ;)
> what hashing algorithm is being use ? Is a collision realistic ? How
> much time would it take to actually break a given hash ?
If you're using anything resembling a sane hash (such as MD5 or similar),
what happens is that you basically ignore the hash collisions - because
rather than "1234", your colliding password/phrase is probably a 32-byte or so
string, which is likely not even enterable at the keyboard (it ends up being
A # ctl-b 9 e alt-control-meta-$ etcetc - of the 32, likely only 10 or so
of the characters are from the 96-char printable ASCII set, and there's a good
chance that at least several of the bytes are ones you can't enter from the
keyboard at all....)
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