[curves] Implementation

[Michael Hamburg]

Hi Trevor,

Not a github initially, because of Rambus legal and export control and all that.  I’ll see if I can set up something more private and get back to you.

Cheers,
— Mike

On Feb 12, 2014, at 11:22 AM, Trevor Perrin <trevp at trevp.net> wrote:

> Could we expect a github?  I'd love to see this!
> 
> Trevor
> 
> 
> On Tue, Feb 11, 2014 at 12:31 AM, Mike Hamburg <mike at shiftleft.org> wrote:
>> Hello curves,
>> 
>> I've been working on implementation for the new curves, and I'd like to
>> report status and some formulas and issues I found.
>> 
>> I'm aiming for a fairly generic C/intrinsics implementation which should
>> support any curves with minimal extra effort, but I'm starting with
>> Ed448-Goldilocks because it's mine.  I have Haswell and Sandy Bridge test
>> machines.  I also have a vectorless Cortex A9, but it doesn't work yet
>> because I'm using 64x64->128-bit multiply intrinsics.  Here's what I've
>> found so far.
>> 
>> If you have any suggestions on the formulas or algorithms, I'd definitely
>> appreciate it.
>> 
>> Field arithmetic:
>> * Karatsuba is beneficial for Ed448.
>> * Radix 2^56 in a 64-bit limb, 8 limbs.
>> * M ~ 153cy on Sandy Bridge, 125cy on Haswell
>> * square ~ 0.75M
>> * small fixed mul ~ 0.25M
>> * add/sub (unreduced) ~ 0.04M, a little cheaper on Haswell because of AVX
>> 
>> I'm using the 1/sqrt(x) point encoding for now, basically because I already
>> have code for that from an earlier project.  I'm not yet counting the time
>> to serialize and deserialize field elements, which is maybe 100 cyles at
>> most (counting the full reduce / checking that input is fully reduced).  I'm
>> not yet counting hashing or RNG times.
>> 
>> My earlier email about 1/sqrt(x) was slightly off: it encodes even points on
>> the curve, but odd points on the twist.
>> 
>> I haven't tried blind+EGCD for inverses or Legendre symbol checks.  It might
>> well be a win.  One inverse square root is 56k Sandy cycles (I don't
>> remember the Haswell number).
>> 
>> Full Montgomery ladder:
>> * Decompress.
>> * Constant-time ladder by 448-bit scalar.  The scalar should be even for
>> security.  It actually could be 447 bits.
>> * Recompress.  Reject points on the twist.  This is basically free, but
>> important because they can't be encoded with the 1/sqrt(x) encoding.
>> 
>> This takes about 571kcy on Haswell, and 688kcy on Sandy, corrected for
>> TurboBoost.
>> 
>> I'm using the formula from the thread on efficient laddering with the
>> isomorphic curve, but twisted.  Let (xd,zd) be the point to de doubled, and
>> (xa,za) be the point to be added.
>>    A = (xd+zd)
>>    B = (xd-zd)
>>    DA = (xa-za)*A
>>    BC = (xa+za)*B
>> 
>>    oxa = (DA+BC)^2
>>    oza = (DA-BC)^2 * xbase
>> 
>>    AA = A^2
>>    BB = B^2
>>    AAod = AA*(1-d)
>>    E = AA-BB
>> 
>>    oxd = AAod*BB
>>    ozd = E*(AAod-E)
>> 
>>    return (oxd,ozd,oxa,oza)
>> 
>> Except I'm actually using zbase instead of xbase, because of the 1/sqrt(x)
>> format.
>> 
>> Twisted Edwards (a=-1) windowed algorithm:
>> * Assumes that cofactor is canceled somehow.
>> * Recode scalar in signed form, because it's easy and I'm lazy.
>> * Compute 8 odd multiples of P.
>> * Constant-time add/sub chain with a 4-bit window, 448 bits.  Could be 446
>> bits, except that 446 isn't divisible by 4.
>> * No compress or decompress.
>> 
>> This takes slightly less time than the Montgomery ladder, some 530kcy on
>> Haswell and 636 kcy on Sandy.  A 5-bit window makes things maybe 1-2%
>> faster, but uses extra complexity and memory so I didn't think it was
>> worthwhile.
>> 
>> I'm using readdition coordinates:
>>  "Projective half-niels" for the tables, ((y-x)/2 : (y+x)/2 : dxy : 1) * z.
>>  "Lazy extended coordinates" for the accumulator, (x : y : z : t : u) where
>> xy = tuz.
>> 
>> I might replace the lazy extended coordinates with Hisil et al's lookahead
>> extended-or-not coordinates, which use less memory but require more care.
>> 
>> Full constant-time scalarmul using twisted Edwards:
>> * Decompress points, rejecting those on the twist.
>> * Isogenize to the twisted curve, canceling the cofactor.
>> * Above windowed algorithm.
>> * Isogenize back to the main curve, effectively multiplying by 4.
>> * Recompress.
>> 
>> This takes slightly longer than the Montgomery ladder: something like 633kcy
>> on Haswell and 750kcy on Sandy.  So Edwards or twisted Edwards is best for
>> points you've already got in projective form, and Montgomery is best for
>> ECDH.  Unsurprising.
>> 
>> The total executable code size to test and bench the arithmetic and curve
>> routines is currently around 41k under clang -O4 -fPIC.  That'll get bigger
>> once there are precomputed tables.
>> 
>> Formulas:
>> I'm making use of the "inverse square root trick":
>> 
>> def trick(a,b,i):
>>    # assumes p==3 mod 4; similar trick exists for 1 mod 4
>>    # returns sqrt(+-a/b), 1/i, is_square(a/b)
>>    # assumes a,b,i are nonzero
>>    ai = a*i
>>    abi = b*ai
>>    s = 1/sqrt(+- abi*i) # using a powering ladder
>>    output sqrt(+-a/b) = s*ai
>>    s2abi = s^2*abi
>>    issquare = s2abi * i # = Legendre symbol
>>    if you care about the result of 1/i when a/b is nonsquare:
>>        output 1/i = s2abi*issquare
>>    else:
>>        output 1/i = s2abi
>> 
>> You can tweak the trick to change the Legendre symbol of the output
>> according to some other variable as well; this depends on the residue of p
>> mod 8.
>> 
>> The formula I'm using for point compression with Montgomery form is:
>> 
>> Let P1 + P2 = P3 and (u1,v1) = P1 etc.  Then
>>    4*v1*v2*u3 = (u1*u2-1)^2 - u3^2*(u1-u2)^2
>> 
>> To compute the numerator of the RHS, do:
>>    sa = (z2*z1 - x2*x1) * z3
>>    sb = (x2*z1 - z2*x1) * x3
>>    numerator = (sa + sb) * (sa - sb)
>> This is good enough to get the Legendre symbol.  It shouldn't be too hard to
>> convert this into a formula with some other sign bit using the inverse
>> square root trick.
>> 
>> This is on an untwisted (B=1) curve, but the same "ought" to be true of
>> 4*B*v1*v2*u3 on a twisted one.
>> 
>> To serialize an Edwards point, we have to deal with the fact that the
>> isomorphic curve you'd get from Wikipedia is twisted, because it sets B =
>> 4/(1-d) which isn't square, at least when p==3 mod 4.  So I'm negating x to
>> get to the curve:
>>    4y^2/(d-1) = x^3 + 2(d+1)/(d-1) * x^2 + x
>> where you can then scale y by sqrt(4/(d-1)) to get the standard curve.
>> 
>> To deserialize an Edwards point, compute
>>    denominator = (u+1)^2 * (d-1) + 4u
>>    x = 2 sqrt(u/denominator)
>>    y = (1+u)/(1-u)
>> using the inverse square root trick.  This lands you on E_(1,d), because it
>> scales the x-coordinate to get rid of the twisting that the obvious
>> decompression would give you.
>> 
>> You have to check if u=0 or u=1.  The latter isn't on the curve, but you
>> have to make sure it doesn't slip past the check due to the zero divide.
>> The former works in the 1/sqrt(x) encoding without any checks.
>> 
>> To do:
>> I'm planning to use WNAF for variable time scalar mul, WNAF for signature
>> verification, and a precomputed signed comb for key generation and Schnorr
>> signing.
>> 
>> I'm experimenting with the best way to implement Elligator.  I currently
>> only have the map to the curve done, and I might change the signs.  My
>> implementation maps directly to affine using the inverse square root trick.
>> I'll report the formula once I'm done messing around with it.
>> 
>> And of course there's API packaging, testing on ARM, etc.
>> 
>> Cheers,
>> -- Mike
>> 
>> 
>> 
>> 
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