<html><head><meta http-equiv="Content-Type" content="text/html charset=us-ascii"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;">Hello curves,<div><br></div><div>I've been working on implementation for the new curves, and I'd like to report status and some formulas and issues I found.</div><div><br></div><div>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.</div><div><br></div><div>If you have any suggestions on the formulas or algorithms, I'd definitely appreciate it.</div><div><br></div><div><b>Field arithmetic:</b></div><div>* Karatsuba is beneficial for Ed448.</div><div>* Radix 2^56 in a 64-bit limb, 8 limbs.</div><div>* M ~ 153cy on Sandy Bridge, 125cy on Haswell</div><div>* square ~ 0.75M</div><div>* small fixed mul ~ 0.25M</div><div>* add/sub (unreduced) ~ 0.04M, a little cheaper on Haswell because of AVX</div><div><br></div><div>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.</div><div><br></div><div>My earlier email about 1/sqrt(x) was slightly off: it encodes even points on the curve, but odd points on the twist.</div><div><br></div><div>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).</div><div><br></div><div><b>Full Montgomery ladder:</b></div><div>* Decompress.</div><div>* Constant-time ladder by 448-bit scalar. The scalar should be even for security. It actually could be 447 bits.</div><div>* Recompress. Reject points on the twist. This is basically free, but important because they can't be encoded with the 1/sqrt(x) encoding.</div><div><br></div><div>This takes about 571kcy on Haswell, and 688kcy on Sandy, corrected for TurboBoost.</div><div><br></div><div>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.</div><div><div> A = (xd+zd)</div><div> B = (xd-zd)</div><div> DA = (xa-za)*A</div><div> BC = (xa+za)*B</div><div> </div><div> oxa = (DA+BC)^2</div><div> oza = (DA-BC)^2 * xbase</div><div> </div><div> AA = A^2</div><div> BB = B^2</div><div> AAod = AA*(1-d)</div><div> E = AA-BB</div><div><br></div><div> oxd = AAod*BB</div><div> ozd = E*(AAod-E)</div></div><div><br></div><div> return (oxd,ozd,oxa,oza)</div><div><br></div><div>Except I'm actually using zbase instead of xbase, because of the 1/sqrt(x) format.</div><div><br></div><div><b>Twisted Edwards (a=-1) windowed algorithm:</b></div><div>* Assumes that cofactor is canceled somehow.</div><div>* Recode scalar in signed form, because it's easy and I'm lazy.</div><div>* Compute 8 odd multiples of P.</div><div>* Constant-time add/sub chain with a 4-bit window, 448 bits. Could be 446 bits, except that 446 isn't divisible by 4.</div><div>* No compress or decompress.</div><div><br></div><div>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.</div><div><br></div><div>I'm using readdition coordinates:</div><div> "Projective half-niels" for the tables, ((y-x)/2 : (y+x)/2 : dxy : 1) * z.</div><div> "Lazy extended coordinates" for the accumulator, (x : y : z : t : u) where xy = tuz.</div><div><br></div><div>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.</div><div><br></div><div><b>Full constant-time scalarmul using twisted Edwards:</b></div><div>* Decompress points, rejecting those on the twist.</div><div>* Isogenize to the twisted curve, canceling the cofactor.</div><div>* Above windowed algorithm.</div><div>* Isogenize back to the main curve, effectively multiplying by 4.</div><div>* Recompress.</div><div><br></div><div>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.</div><div><br></div><div>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.</div><div><br></div><div><b>Formulas:</b></div><div>I'm making use of the "inverse square root trick":</div><div><br></div><div>def trick(a,b,i):</div><div> # assumes p==3 mod 4; similar trick exists for 1 mod 4</div><div> # returns sqrt(+-a/b), 1/i, is_square(a/b)</div><div> # assumes a,b,i are nonzero</div><div> ai = a*i</div><div> abi = b*ai</div><div> s = 1/sqrt(+- abi*i) # using a powering ladder</div><div> output sqrt(+-a/b) = s*ai</div><div> s2abi = s^2*abi</div><div> issquare = s2abi * i # = Legendre symbol</div><div> if you care about the result of 1/i when a/b is nonsquare:</div><div> output 1/i = s2abi*issquare</div><div> else:</div><div> output 1/i = s2abi</div><div><br></div><div>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.</div><div><b><br></b></div><div><div>The formula I'm using for point compression with Montgomery form is:</div><div><br></div><div>Let P1 + P2 = P3 and (u1,v1) = P1 etc. Then</div><div> 4*v1*v2*u3 = (u1*u2-1)^2 - u3^2*(u1-u2)^2</div></div><div><br></div><div>To compute the numerator of the RHS, do:</div><div><div> sa = (z2*z1 - x2*x1) * z3</div><div> sb = (x2*z1 - z2*x1) * x3</div></div><div> numerator = (sa + sb) * (sa - sb)</div><div>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.</div><div><br></div><div>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.</div><div><br></div><div>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:</div><div> 4y^2/(d-1) = x^3 + 2(d+1)/(d-1) * x^2 + x</div><div>where you can then scale y by sqrt(4/(d-1)) to get the standard curve.</div><div><br></div><div>To deserialize an Edwards point, compute</div><div> denominator = (u+1)^2 * (d-1) + 4u</div><div> x = 2 sqrt(u/denominator)</div><div> y = (1+u)/(1-u)</div><div>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.</div><div><br></div><div>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.</div><div><br></div><div><b>To do:</b></div><div>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.</div><div><br></div><div>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.</div><div><br></div><div>And of course there's API packaging, testing on ARM, etc.</div><div><br></div><div>Cheers,</div><div>-- Mike</div><div><br></div><div><br></div><div><br></div></body></html>