Re: [Cfrg] Curve manipulation, revisited

[Michael Hamburg <[e-mail]>]

> On Jan 1, 2015, at 2:16 AM, Peter Dettman <[e-mail]> wrote:
> 
> On 29/12/2014 8:31 pm, Adam Langley wrote:
>> (An implementation that wants to use a windowed method with a Montgomery-X *input* will need to perform a square-root first, but that's the same cost as a design where compressed Edwards points are sent. Note: smarter people than I might be able to eliminate the square-root but I don't see it)
> 
> I believe I can give an example of how to eliminate the square-root for a random-base scalar-mult. I'll use short Weierstrass to describe it though, as I'm less familiar with Edwards and Montgomery. I assume we can convert easily enough between forms, and that twist security carries across.
> 
> - Given input X0, use curve equation to calculate {Y0}^2, let K = {Y0}^2.
> - Initial point P0 is (X0,{Y0}) where {Y0} is unevaluated.
> - Change of variables to an isomorphic curve: x' = u^2.x, y' = u^3.y, a' = u^4.a, with u = {Y0} (b' = u^6.b, but not typically needed).
> - Can now write P0' as (K.X0, K^2) without unknowns.
> - Proceed with scalar multiplication of P0' to result point Pn'.
> - Recover output Xn from Pn' as Xn'/(K.Zn^2).
> 
> Note that the doubling formula is affected by the changed a' curve parameter (unless a==0); assume modified-Jacobian coordinates are used.
> 
> In the case where the original curve has a==-3, the extra cost of the windowing using the isomorphism in modified-Jacobian coordinates is +2S(quares) per addition (doubling cost is unchanged). The total extra cost using width 5 windows is therefore < 40% of the best-case cost of a sqrt.
> 
> However, there are actually good performance reasons to already be using an isomorphism: to allow mixed addition with precomputed points without needing an inversion in the precomputation. In that case, there is very little additional overhead for the scheme above.
> 
> Regards,
> Pete Dettman

This is neat.  Have you tested it?  Does it work, and do you know if it’s better than a Montgomery ladder?  From your high-level description it looks like it should be competitive with the short Weierstrass XZ Montgomery ladder (8M + 7S + 5m IIUC).

Cheers,
— Mike