[curves] Implementation

[Trevor Perrin]

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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