Unfortunately, the government imagines that people are using their home 
computers to compute hashes and try and decrypt stuff.  Look at what is 
happening with GPUs these days.  People are hooking up 4 GPUs in their 
computers and getting huge performance gains.  5-6 char password space covered 
in a few days.  12 or so chars would take one machine a couple of years if I 
recall.  So, if we had 20 people with that class of machine, we'd be down to a 
few months.   I'm just suggesting that while the compute space is still huge, 
it's not actually undoable, it just requires some thought into how to approach 
the problem, and then some time to do the computations.

Huge space, but still finite…

Gregg Wonderly

On Jul 11, 2012, at 9:13 AM, Edward Ned Harvey wrote:

>> From: zfs-discuss-boun...@opensolaris.org [mailto:zfs-discuss-
>> boun...@opensolaris.org] On Behalf Of Gregg Wonderly
>> Since there is a finite number of bit patterns per block, have you tried
> to just
>> calculate the SHA-256 or SHA-512 for every possible bit pattern to see if
> there
>> is ever a collision?  If you found an algorithm that produced no
> collisions for
>> any possible block bit pattern, wouldn't that be the win?
> Maybe I misunderstand what you're saying, but if I got it right, what you're
> saying is physically impossible to do in the time of the universe...  And
> guaranteed to fail even if you had all the computational power of God.
> I think you're saying ... In a block of 128k, sequentially step through all
> the possible values ... starting with 0, 1, 2, ... 2^128k ... and compute
> the hashes of each value, and see if you ever find a hash collision.  If
> this is indeed what you're saying, recall, the above operation will require
> on order 2^128k operations to complete.  But present national security
> standards accept 2^256 operations as satisfactory to protect data from brute
> force attacks over the next 30 years.  Furthermore, in a 128k block, there
> exist 2^128k possible values, while in a 512bit hash, there exist only 2^512
> possible values (which is still a really huge number.)  This means there
> will exist at least 2^127.5k collisions.  However, these numbers are so
> astronomically universally magnanimously huge, it will still take more than
> a lifetime to find any one of those collisions.  So it's impossible to
> perform such a computation, and if you could, you would be guaranteed to
> find a LOT of collisions.
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