Add note about 2M + 5S doubling formula

pull/11871/head
Andrew Poelstra 9 years ago
parent 65285a63ee
commit 8ec49d8a5a

@ -251,7 +251,13 @@ static int secp256k1_ge_is_valid_var(const secp256k1_ge *a) {
} }
static void secp256k1_gej_double_var(secp256k1_gej *r, const secp256k1_gej *a, secp256k1_fe *rzr) { static void secp256k1_gej_double_var(secp256k1_gej *r, const secp256k1_gej *a, secp256k1_fe *rzr) {
/* Operations: 3 mul, 4 sqr, 0 normalize, 12 mul_int/add/negate */ /* Operations: 3 mul, 4 sqr, 0 normalize, 12 mul_int/add/negate.
*
* Note that there is an implementation described at
* https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#doubling-dbl-2009-l
* which trades a multiply for a square, but in practice this is actually slower,
* mainly because it requires more normalizations.
*/
secp256k1_fe t1,t2,t3,t4; secp256k1_fe t1,t2,t3,t4;
/** For secp256k1, 2Q is infinity if and only if Q is infinity. This is because if 2Q = infinity, /** For secp256k1, 2Q is infinity if and only if Q is infinity. This is because if 2Q = infinity,
* Q must equal -Q, or that Q.y == -(Q.y), or Q.y is 0. For a point on y^2 = x^3 + 7 to have * Q must equal -Q, or that Q.y == -(Q.y), or Q.y is 0. For a point on y^2 = x^3 + 7 to have

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