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Merge #404: Replace 3M + 4S doubling formula with 2M + 5S one
8ec49d8
Add note about 2M + 5S doubling formula (Andrew Poelstra)
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e2a8e92125
@ -249,7 +249,13 @@ static int secp256k1_ge_is_valid_var(const secp256k1_ge *a) {
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}
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}
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static void secp256k1_gej_double_var(secp256k1_gej *r, const secp256k1_gej *a, secp256k1_fe *rzr) {
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static void secp256k1_gej_double_var(secp256k1_gej *r, const secp256k1_gej *a, secp256k1_fe *rzr) {
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/* Operations: 3 mul, 4 sqr, 0 normalize, 12 mul_int/add/negate */
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/* Operations: 3 mul, 4 sqr, 0 normalize, 12 mul_int/add/negate.
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*
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* Note that there is an implementation described at
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* https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#doubling-dbl-2009-l
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* which trades a multiply for a square, but in practice this is actually slower,
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* mainly because it requires more normalizations.
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*/
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secp256k1_fe t1,t2,t3,t4;
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secp256k1_fe t1,t2,t3,t4;
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/** For secp256k1, 2Q is infinity if and only if Q is infinity. This is because if 2Q = infinity,
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/** For secp256k1, 2Q is infinity if and only if Q is infinity. This is because if 2Q = infinity,
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* 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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* 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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