# [curves] Torsion-safe representatives (was: Ed25519 "clamping" and its effect on hierarchical key derivation)

Oleg Andreev oleganza at gmail.com
Mon Mar 27 11:04:40 PDT 2017

Henry, the technique you showed is fantastic, I love it!

I have a lame question, though. You mention that a*B = a'*B holds for the base point. But is it also true for any point in the B's subgroup? The reason I ask is that I need to have not just regular EdDSA signatures, but also DLEQs (discrete log equality proofs) with random generator points.

Thank you!

>
> Hi all,
>
> this subject came up today at the Tor meeting in a discussion with Ian
> Goldberg, George Kadianakis, Isis Lovecruft, and myself.
>
> As Tony noted, using bit-twiddling to mask away the low bits of a scalar is
> problematic because the bit-twiddling is not well-defined (mod l).  (Here l
> is the order of the basepoint, so the full group has order 8*l).  This means
> that the "clamping" is not compatible with any arithmetic operations on
> scalars.
>
> We (Ian, George, Isis, and myself) have the following suggestion.
>
> Define a "torsion-safe representative" of a \in Z to be an integer a' \in
> Z such that a ≡ a' (mod l) and a' ≡ 0 (mod 8).
>
> This means that a B = a' B for the basepoint B, but a' T = id for T a
> torsion point, so accidentally multiplying by a torsion point can't leak
> information.  However, unlike "clamping", this operation is actually
> well-defined, and leaves the scalar unchanged, in the sense that scalar
> multiplication by a and by a' are the same on the prime-order subgroup.
>
> How do we compute such a representative?  Since (using the CRT)
>
>    k := 3l + 1 ≡ 1 (mod l)
>                ≡ 0 (mod 8),
>
> k*a mod 8l is a torsion-safe representative of a.  Computing k*a mod 8l
> directly is a problem because an implementation may only have arithmetic modulo
> l, not 8l.  However,
>
>    k*a ≡ (3l+1)*a ≡ 3al + a (mod 8l),
>
> so it's sufficient to compute 3al (mod 8l) and add it to a.  Since
>
>     3a mod 8 = 3a + 8n,
>    (3a mod 8)*l = 3al + 8ln ≡ 3al (mod 8l),
>
> so it's sufficient to compute 3a mod 8 and use a lookup table of
>
>    [0, 1l, 2l, 3l, 4l, 5l, 6l, 7l]
>
> to get 3al (mod 8l).  Arranging the lookup table as
>
>    [0, 3l, 6l, 1l, 4l, 7l, 2l, 5l]
>
> means that a mod 8 indexes 3al (mod 8l).  Therefore, computing a
> torsion-safe representative for a scalar a just amounts to computing
>
>    a + permuted_lookup_table[a & 7]
>
> in constant time.  If the input a is reduced, so that a < l, then the
> torsion-safe representative is bounded by 8l < 2^256 and therefore fits in 32
> bytes.
>
> This ends up being slightly more work than just bit-twiddling, but not by much,
> and it's certainly insignificant compared to the cost of a scalar
> multiplication.  There's a prototype implementation here [0] in case anyone is
> curious to see what it looks like.
>
> Cheers,
> Henry de Valence
>
> [0]: https://github.com/hdevalence/curve25519-dalek/commit/2ae0bdb6df26a74ef46d4332b635c9f6290126c7
> (subject to rebasing...)
>
> The permuted lookup table is, explicitly:
>
> sage: l = 2^252 + 27742317777372353535851937790883648493
> sage: lookup = [((3 * i) % 8)*l for i in range(8)]
> sage: lookup
> [0,
> 21711016731996786641919559689128982722571349078139722818005852814856362752967,
> 43422033463993573283839119378257965445142698156279445636011705629712725505934,
> 7237005577332262213973186563042994240857116359379907606001950938285454250989,
> 28948022309329048855892746252171976963428465437519630424007803753141817003956,
> 50659039041325835497812305941300959685999814515659353242013656567998179756923,
> 14474011154664524427946373126085988481714232718759815212003901876570908501978,
> 36185027886661311069865932815214971204285581796899538030009754691427271254945]
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