phc-discussions - Re: [PHC] Representing cost in a formated hash

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Message-ID: <2110842680.363068.1389040200377.open-xchange@email.1and1.com>
Date: Mon, 6 Jan 2014 14:30:00 -0600 (CST)
From: Steve Thomas <steve@...tu.com>
To: discussions@...sword-hashing.net
Subject: Re: [PHC] Representing cost in a formated hash

> On January 6, 2014 at 7:16 AM Krisztián Pintér <pinterkr@...il.com> wrote:
>
>
>
> > The question is how to represent the cost parameters in the formated
> > hash.
>
> here is what i think. representing the hash is the task of the
> platform, and not the hash function. regular hash functions don't have
> default representation either, they are just bits. it is not even sure
> that we want to include the cost or other parameters at all. it can be
> the case that some platforms (especially embedded or hw
> implementations) will have fixed parametrization, thus there is no
> need to store them at all.

Once this competition is over, the winner will more than likely have an
encoding like $1$, $2y$, $5$, $6$, $P$... If it is also a KDF then for that
use it will just output binary.


> but if we are adamant on giving finer grained exponentially growing
> tuning ability, using constants like sqrt(2) seems to be unnecessarily
> taxing for hw implementations (not very taxing, but unnecessarily
> taxing).

I must not of been clear these two are exactly the same:
cost = floor(pow(sqrt(2), cost));
and
if ((cost & 1) == 0)
  cost = 1 << (cost >> 1);
else
  cost = 0x5a827999 >> (30 - (cost >> 1));
Note: 0x5a827999 is floor(pow(sqrt(2), 30));


> so my minimalist recommendation is the following:
>
> (0b1000 + m) << e
>
> where m is a 3 bit number, and e is a 5 bit number. the cost parameter
> thus becomes 8 bits, in eeeeemmm order. the range is from 8 to
> 15*2^31, and the increment is below 12.5%. naturally expandable to 13
> bit e, which covers ranges up to unbelievable. also, the number itself
> can be treated as an integer, can be increased, behaves as one would
> expect, etc.

I like this it is just like a custom floating point number. The only thing is,
it
is no longer 6 bits. So it can't fit in as one base64 character. Which really
doesn't matter. If we changed it to eeeeem then it's very similar to the
second way I purposed. The other way would be eeeeeemmmmmm.
This makes it easier to calculate because the exponent and mantissa
would be their own base64 character. Also starting at 64 shouldn't matter
too much.
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