@@ -492,11 +492,11 @@ static void secp256k1_gej_add_ge(secp256k1_gej *r, const secp256k1_gej *a, const
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/* Operations: 7 mul, 5 sqr, 24 add/cmov/half/mul_int/negate/normalize_weak/normalizes_to_zero */
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secp256k1_fe zz , u1 , u2 , s1 , s2 , t , tt , m , n , q , rr ;
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secp256k1_fe m_alt , rr_alt ;
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- int infinity , degenerate ;
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+ int degenerate ;
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VERIFY_CHECK (!b -> infinity );
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VERIFY_CHECK (a -> infinity == 0 || a -> infinity == 1 );
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- /** In:
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+ /* In:
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* Eric Brier and Marc Joye, Weierstrass Elliptic Curves and Side-Channel Attacks.
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* In D. Naccache and P. Paillier, Eds., Public Key Cryptography, vol. 2274 of Lecture Notes in Computer Science, pages 335-345. Springer-Verlag, 2002.
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* we find as solution for a unified addition/doubling formula:
@@ -558,8 +558,8 @@ static void secp256k1_gej_add_ge(secp256k1_gej *r, const secp256k1_gej *a, const
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secp256k1_fe_negate (& m_alt , & u2 , 1 ); /* Malt = -X2*Z1^2 */
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secp256k1_fe_mul (& tt , & u1 , & m_alt ); /* tt = -U1*U2 (2) */
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secp256k1_fe_add (& rr , & tt ); /* rr = R = T^2-U1*U2 (3) */
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- /** If lambda = R/M = R/0 we have a problem (except in the "trivial"
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- * case that Z = z1z2 = 0, and this is special-cased later on). */
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+ /* If lambda = R/M = R/0 we have a problem (except in the "trivial"
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+ * case that Z = z1z2 = 0, and this is special-cased later on). */
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degenerate = secp256k1_fe_normalizes_to_zero (& m );
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/* This only occurs when y1 == -y2 and x1^3 == x2^3, but x1 != x2.
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* This means either x1 == beta*x2 or beta*x1 == x2, where beta is
@@ -587,7 +587,6 @@ static void secp256k1_gej_add_ge(secp256k1_gej *r, const secp256k1_gej *a, const
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secp256k1_fe_cmov (& n , & m , degenerate ); /* n = M^3 * Malt (2) */
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secp256k1_fe_sqr (& t , & rr_alt ); /* t = Ralt^2 (1) */
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secp256k1_fe_mul (& r -> z , & a -> z , & m_alt ); /* r->z = Z3 = Malt*Z (1) */
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- infinity = secp256k1_fe_normalizes_to_zero (& r -> z ) & ~a -> infinity ;
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secp256k1_fe_add (& t , & q ); /* t = Ralt^2 + Q (2) */
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r -> x = t ; /* r->x = X3 = Ralt^2 + Q (2) */
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secp256k1_fe_mul_int (& t , 2 ); /* t = 2*X3 (4) */
@@ -597,11 +596,28 @@ static void secp256k1_gej_add_ge(secp256k1_gej *r, const secp256k1_gej *a, const
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secp256k1_fe_negate (& r -> y , & t , 3 ); /* r->y = -(Ralt*(2*X3 + Q) + M^3*Malt) (4) */
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secp256k1_fe_half (& r -> y ); /* r->y = Y3 = -(Ralt*(2*X3 + Q) + M^3*Malt)/2 (3) */
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- /** In case a->infinity == 1, replace r with (b->x, b->y, 1). */
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+ /* In case a->infinity == 1, replace r with (b->x, b->y, 1). */
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secp256k1_fe_cmov (& r -> x , & b -> x , a -> infinity );
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secp256k1_fe_cmov (& r -> y , & b -> y , a -> infinity );
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secp256k1_fe_cmov (& r -> z , & secp256k1_fe_one , a -> infinity );
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- r -> infinity = infinity ;
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+
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+ /* Set r->infinity if r->z is 0.
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+ *
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+ * If a->infinity is set, then r->infinity = (r->z == 0) = (1 == 0) = false,
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+ * which is correct because the function assumes that b is not infinity.
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+ *
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+ * Now assume !a->infinity. This implies Z = Z1 != 0.
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+ *
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+ * Case y1 = -y2:
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+ * In this case we could have a = -b, namely if x1 = x2.
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+ * We have degenerate = true, r->z = (x1 - x2) * Z.
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+ * Then r->infinity = ((x1 - x2)Z == 0) = (x1 == x2) = (a == -b).
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+ *
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+ * Case y1 != -y2:
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+ * In this case, we can't have a = -b.
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+ * We have degenerate = false, r->z = (y1 + y2) * Z.
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+ * Then r->infinity = ((y1 + y2)Z == 0) = (y1 == -y2) = false. */
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+ r -> infinity = secp256k1_fe_normalizes_to_zero (& r -> z );
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}
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static void secp256k1_gej_rescale (secp256k1_gej * r , const secp256k1_fe * s ) {
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