[curves] Finalizing XEdDSA

Ron Garret ron at flownet.com
Tue Nov 1 09:04:15 PDT 2016

On Nov 1, 2016, at 7:14 AM, Peter Schwabe <peter at cryptojedi.org> wrote:

> Trevor Perrin <trevp at trevp.net> wrote:
> Dear Trevor,
>> One last tweak to consider is clearing the cofactor in verification.
>> Currently XEdDSA does "cofactorless verification", i.e. it takes a
>> signature (R, s) and checks R == sB - hA.  We could change it to cR ==
>> c(sB - hA).  VXEdDSA would be unchanged.
>> This has no effect on valid signatures, but adding the cofactor
>> multiplication means signers could create signatures with a few
>> different values of R for the same s (which has no security relevance,
>> I think, and does not cause "malleability" because the signer's choice
>> of R is included in the hash).
>> Advantages to current "cofactorless" approach:
>> - matches existing code like (ref10, libsodium)
>> - less code, doesn't need a "point comparison" function (can encode,
>> then compare)
>> - less computation (by tiny amount, 1% or something)
>> Advantages to changing to "cofactor" approach:
>> - Allows batch verification of signatures (I'm told), that can give ~2x speedup
>> - Preferred approach in Ed25519 paper, "EdDSA for more curves" paper,
>> and CFRG draft
> The Ed25519 paper says 
>  "The verifier is /permitted/ to check this stronger equation and
>  to reject alleged signatures where the stronger equation does not
>  hold. However, this is not /required/; checking that
>  8SB=8R+8H(\encode{R},\encode{A},M)A is enough for security."
> You could decide to do the same; allowing both for verification in the
> specification and leaving the choice to the implementation. If I
> understand correctly, this gives you the advantages of both approaches,
> right?

Possibly naive question: What is “this stronger equation” that the paper refers to?  Because the immediately preceding equation is:

SB = rB + H(\encode{R},\encode{A},M)aB = R + H(\encode{R},\encode{A},M)A

which is actually two equations.  The first:

SB = rB + H(\encode{R},\encode{A},M)aB

relies on the secret key a, so that does not seem particularly useful.  But the second equation:

SB = R + H(\encode{R},\encode{A},M)A

Is (AFAICT) identical to the earlier “weaker” equation:

8SB = 8R+8H(\encode{R},\encode{A},M)A

except for factoring out the 8’s.  (Where did those 8’s come from anyway?  They seem completely unmotivated.)

What am I missing?


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