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 <[email protected]> wrote: > Could we expect a github? I'd love to see this! > > Trevor > > > On Tue, Feb 11, 2014 at 12:31 AM, Mike Hamburg <[email protected]> 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 >> >> >> >> >> _______________________________________________ >> Curves mailing list >> [email protected] >> https://moderncrypto.org/mailman/listinfo/curves >> _______________________________________________ Curves mailing list [email protected] https://moderncrypto.org/mailman/listinfo/curves
