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Date: Fri, 17 Apr 2015 16:41:18 +0200
From: Thomas Pornin <pornin@...et.org>
To: discussions@...sword-hashing.net
Subject: Re: [PHC] Information theoretic security for delegated hardening
 was: winner selection

On Fri, Apr 17, 2015 at 06:39:12AM -0700, Bill Cox wrote:
> This is simpler than Makwa, but I haven't put any thought into attacking
> it.  Thomas Pornin, what do you think of this again?

I am preparing an implementation and some documentation, to be
published some time next week.

Roughly put: for Makwa, and for Adam's RSA-based proposal, the core
hashing operation is computing x^e mod N for input x, a huge exponent e,
and a composite modulus N (in Makwa, e = 2^w, while Adam uses e =
2^w-1). Blinding is done by assembling a "mask pair" (a, a^-e), because:

   x^e = (a*x)^e * a^-e

(All computations are done modulo N.)
So "a*x" is what is sent to the delegation server, and that server
raises that value to exponent e.

We need a new mask pair for each delegation, so the problem is about
producing such pairs efficiently, and without leaking information about
the input x.

In the Makwa specification, I described (and implemented) a scheme by
which I precompute 300 mask pairs (a_i, a_i^-e), and generate a new mask
pair by multiplying some of them together (on average half of them):

   a = (prod_{b_i=1} a_i, prod_{b_i=1} a_i^-e)   for random bits b_i

What Adam suggests is starting with a single mask pair (g, g^-e) and
computing a new mask pair by using (g^b, (g^-e)^b) with a random
exponent b.

Adam's method can be said to be "information theoretic secure" if b is
large enough and g is a generator of the subgroup containing inputs,
because it implies that after blinding, every single element of the
subgroup may be reached with roughly equal probability for any input, so
the delegation server (the untrusted system to which you delegate the
core of the hashing) provably learns nothing from the exchange, even if
it has unlimited computing power. With my scheme, an evil delegation
server with unlimited computing power may solve the knapsack problem and
thus, for each password guess x, test whether that x would lead to a
value that can be generated from the a_i, and thus filter out bad
guesses. Note that the attacker would solve a O(2^150) problem for each
password guess, so when we say "unlimited power" we mean it.

Adam's scheme has an issue here, because for N = pq (and p != q), there
is no such thing as a generator for all invertible integers modulo N
(the problem here being that p-1 and q-1 are both even: the order of any
g modulo N will be at most lcm(p-1,q-1), and that value is lower than
N/2). However, this is fixed in Makwa, in which we work in the subgroup
of invertible quadratic residues (the input is always a square), of size
(p-1)(q-1)/4, and there we can have a generator g (*)(**).

Makwa's delegation, as currently specified, requires about 300
multiplications per instance. With a 2048-bit modulus, you need the
exponent b to be 2100-bit or so, so you end up with two modular
exponentiations that will cost, in total, roughly 16 times as much as my
implementation (depending on the kind of optimization implemented for
the modular exponentiations). If you want to speed up Adam's scheme then
you can precompute (g^2, g^-2e), (g^4, g^-4e), (g^8, g^-8e)... at which
point you are indeed back to the basic scheme, albeit with 2100 source
mask pairs instead of 300, thus implying a 7x overhead.

So Adam's scheme and mine are really the same, which has the nice
property that my implementation already supports Adam's scheme, provided
that the precomputed (g^(2^i), (g^-e)^(2^i)) values are input as
"delegation parameters". What needs to be updated is the support code
that generates these base mask pairs.



(*) Reliably computing a generator can be hard in all generality. It is
easy at modulus generation time, if factorizations of p-1 and q-1 are
known. If the p and q factors are "safe primes" ((p-1)/2 and (q-1)/2 are
also prime), then g = 4 will work; but generating random safe primes is
rather expensive. I think I will generate p as p = p1*p2+1 where p1 and
p2 are two distinct primes of roughly the same size; g = 4 will be a
generator with overwhelming probability and I can check that at
generation time.

(**) Note that we ignore non-invertible elements modulo N here.
Probability that a given input (padded password) is non-invertible is
about 2/sqrt(N), i.e. 2^-1023, and I am reasonably sure we can neglect
_that_.


	--Thomas Pornin

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