diff options
Diffstat (limited to 'src/crypto/fipsmodule/ec/util.c')
| -rw-r--r-- | src/crypto/fipsmodule/ec/util.c | 175 |
1 files changed, 163 insertions, 12 deletions
diff --git a/src/crypto/fipsmodule/ec/util.c b/src/crypto/fipsmodule/ec/util.c index 7303a151..4f39f18b 100644 --- a/src/crypto/fipsmodule/ec/util.c +++ b/src/crypto/fipsmodule/ec/util.c @@ -18,6 +18,7 @@ #include "internal.h" + // This function looks at 5+1 scalar bits (5 current, 1 adjacent less // significant bit), and recodes them into a signed digit for use in fast point // multiplication: the use of signed rather than unsigned digits means that @@ -43,13 +44,13 @@ // of a nonnegative integer (b_k in {0, 1}), rewrite it in digits 0, 1, -1 // by using bit-wise subtraction as follows: // -// b_k b_(k-1) ... b_2 b_1 b_0 -// - b_k ... b_3 b_2 b_1 b_0 -// ------------------------------------- -// s_k b_(k-1) ... s_3 s_2 s_1 s_0 +// b_k b_(k-1) ... b_2 b_1 b_0 +// - b_k ... b_3 b_2 b_1 b_0 +// ----------------------------------------- +// s_(k+1) s_k ... s_3 s_2 s_1 s_0 // // A left-shift followed by subtraction of the original value yields a new -// representation of the same value, using signed bits s_i = b_(i+1) - b_i. +// representation of the same value, using signed bits s_i = b_(i-1) - b_i. // This representation from Booth's paper has since appeared in the // literature under a variety of different names including "reversed binary // form", "alternating greedy expansion", "mutual opposite form", and @@ -73,7 +74,7 @@ // (1961), pp. 67-91), in a radix-2^5 setting. That is, we always combine five // signed bits into a signed digit: // -// s_(4j + 4) s_(4j + 3) s_(4j + 2) s_(4j + 1) s_(4j) +// s_(5j + 4) s_(5j + 3) s_(5j + 2) s_(5j + 1) s_(5j) // // The sign-alternating property implies that the resulting digit values are // integers from -16 to 16. @@ -81,14 +82,164 @@ // Of course, we don't actually need to compute the signed digits s_i as an // intermediate step (that's just a nice way to see how this scheme relates // to the wNAF): a direct computation obtains the recoded digit from the -// six bits b_(4j + 4) ... b_(4j - 1). +// six bits b_(5j + 4) ... b_(5j - 1). // -// This function takes those five bits as an integer (0 .. 63), writing the +// This function takes those six bits as an integer (0 .. 63), writing the // recoded digit to *sign (0 for positive, 1 for negative) and *digit (absolute -// value, in the range 0 .. 8). Note that this integer essentially provides the -// input bits "shifted to the left" by one position: for example, the input to -// compute the least significant recoded digit, given that there's no bit b_-1, -// has to be b_4 b_3 b_2 b_1 b_0 0. +// value, in the range 0 .. 16). Note that this integer essentially provides +// the input bits "shifted to the left" by one position: for example, the input +// to compute the least significant recoded digit, given that there's no bit +// b_-1, has to be b_4 b_3 b_2 b_1 b_0 0. +// +// DOUBLING CASE: +// +// Point addition formulas for short Weierstrass curves are often incomplete. +// Edge cases such as P + P or P + ∞ must be handled separately. This +// complicates constant-time requirements. P + ∞ cannot be avoided (any window +// may be zero) and is handled with constant-time selects. P + P (where P is not +// ∞) usually is not. Instead, windowing strategies are chosen to avoid this +// case. Whether this happens depends on the group order. +// +// Let w be the window width (in this function, w = 5). The non-trivial doubling +// case in single-point scalar multiplication may occur if and only if the +// 2^(w-1) bit of the group order is zero. +// +// Note the above only holds if the scalar is fully reduced and the group order +// is a prime that is much larger than 2^w. It also only holds when windows +// are applied from most significant to least significant, doubling between each +// window. It does not apply to more complex table strategies such as +// |EC_GFp_nistz256_method|. +// +// PROOF: +// +// Let n be the group order. Let l be the number of bits needed to represent n. +// Assume there exists some 0 <= k < n such that signed w-bit windowed +// multiplication hits the doubling case. +// +// Windowed multiplication consists of iterating over groups of s_i (defined +// above based on k's binary representation) from most to least significant. At +// iteration i (for i = ..., 3w, 2w, w, 0, starting from the most significant +// window), we: +// +// 1. Double the accumulator A, w times. Let A_i be the value of A at this +// point. +// +// 2. Set A to T_i + A_i, where T_i is a precomputed multiple of P +// corresponding to the window s_(i+w-1) ... s_i. +// +// Let j be the index such that A_j = T_j ≠ ∞. Looking at A_i and T_i as +// multiples of P, define a_i and t_i to be scalar coefficients of A_i and T_i. +// Thus a_j = t_j ≠ 0 (mod n). Note a_i and t_i may not be reduced mod n. t_i is +// the value of the w signed bits s_(i+w-1) ... s_i. a_i is computed as a_i = +// 2^w * (a_(i+w) + t_(i+w)). +// +// t_i is bounded by -2^(w-1) <= t_i <= 2^(w-1). Additionally, we may write it +// in terms of unsigned bits b_i. t_i consists of signed bits s_(i+w-1) ... s_i. +// This is computed as: +// +// b_(i+w-2) b_(i+w-3) ... b_i b_(i-1) +// - b_(i+w-1) b_(i+w-2) ... b_(i+1) b_i +// -------------------------------------------- +// t_i = s_(i+w-1) s_(i+w-2) ... s_(i+1) s_i +// +// Observe that b_(i+w-2) through b_i occur in both terms. Let x be the integer +// represented by that bit string, i.e. 2^(w-2)*b_(i+w-2) + ... + b_i. +// +// t_i = (2*x + b_(i-1)) - (2^(w-1)*b_(i+w-1) + x) +// = x - 2^(w-1)*b_(i+w-1) + b_(i-1) +// +// Or, using C notation for bit operations: +// +// t_i = (k>>i) & ((1<<(w-1)) - 1) - (k>>i) & (1<<(w-1)) + (k>>(i-1)) & 1 +// +// Note b_(i-1) is added in left-shifted by one (or doubled) from its place. +// This is compensated by t_(i-w)'s subtraction term. Thus, a_i may be computed +// by adding b_l b_(l-1) ... b_(i+1) b_i and an extra copy of b_(i-1). In C +// notation, this is: +// +// a_i = (k>>(i+w)) << w + ((k>>(i+w-1)) & 1) << w +// +// Observe that, while t_i may be positive or negative, a_i is bounded by +// 0 <= a_i < n + 2^w. Additionally, a_i can only be zero if b_(i+w-1) and up +// are all zero. (Note this implies a non-trivial P + (-P) is unreachable for +// all groups. That would imply the subsequent a_i is zero, which means all +// terms thus far were zero.) +// +// Returning to our doubling position, we have a_j = t_j (mod n). We now +// determine the value of a_j - t_j, which must be divisible by n. Our bounds on +// a_j and t_j imply a_j - t_j is 0 or n. If it is 0, a_j = t_j. However, 2^w +// divides a_j and -2^(w-1) <= t_j <= 2^(w-1), so this can only happen if +// a_j = t_j = 0, which is a trivial doubling. Therefore, a_j - t_j = n. +// +// Now we determine j. Suppose j > 0. w divides j, so j >= w. Then, +// +// n = a_j - t_j = (k>>(j+w)) << w + ((k>>(j+w-1)) & 1) << w - t_j +// <= k/2^j + 2^w - t_j +// < n/2^w + 2^w + 2^(w-1) +// +// n is much larger than 2^w, so this is impossible. Thus, j = 0: only the final +// addition may hit the doubling case. +// +// Finally, we consider bit patterns for n and k. Divide k into k_H + k_M + k_L +// such that k_H is the contribution from b_(l-1) .. b_w, k_M is the +// contribution from b_(w-1), and k_L is the contribution from b_(w-2) ... b_0. +// That is: +// +// - 2^w divides k_H +// - k_M is 0 or 2^(w-1) +// - 0 <= k_L < 2^(w-1) +// +// Divide n into n_H + n_M + n_L similarly. We thus have: +// +// t_0 = (k>>0) & ((1<<(w-1)) - 1) - (k>>0) & (1<<(w-1)) + (k>>(0-1)) & 1 +// = k & ((1<<(w-1)) - 1) - k & (1<<(w-1)) +// = k_L - k_M +// +// a_0 = (k>>(0+w)) << w + ((k>>(0+w-1)) & 1) << w +// = (k>>w) << w + ((k>>(w-1)) & 1) << w +// = k_H + 2*k_M +// +// n = a_0 - t_0 +// n_H + n_M + n_L = (k_H + 2*k_M) - (k_L - k_M) +// = k_H + 3*k_M - k_L +// +// k_H - k_L < k and k < n, so k_H - k_L ≠ n. Therefore k_M is not 0 and must be +// 2^(w-1). Now we consider k_H and n_H. We know k_H <= n_H. Suppose k_H = n_H. +// Then, +// +// n_M + n_L = 3*(2^(w-1)) - k_L +// > 3*(2^(w-1)) - 2^(w-1) +// = 2^w +// +// Contradiction (n_M + n_L is the bottom w bits of n). Thus k_H < n_H. Suppose +// k_H < n_H - 2*2^w. Then, +// +// n_H + n_M + n_L = k_H + 3*(2^(w-1)) - k_L +// < n_H - 2*2^w + 3*(2^(w-1)) - k_L +// n_M + n_L < -2^(w-1) - k_L +// +// Contradiction. Thus, k_H = n_H - 2^w. (Note 2^w divides n_H and k_H.) Thus, +// +// n_H + n_M + n_L = k_H + 3*(2^(w-1)) - k_L +// = n_H - 2^w + 3*(2^(w-1)) - k_L +// n_M + n_L = 2^(w-1) - k_L +// <= 2^(w-1) +// +// Equality would mean 2^(w-1) divides n, which is impossible if n is prime. +// Thus n_M + n_L < 2^(w-1), so n_M is zero, proving our condition. +// +// This proof constructs k, so, to show the converse, let k_H = n_H - 2^w, +// k_M = 2^(w-1), k_L = 2^(w-1) - n_L. This will result in a non-trivial point +// doubling in the final addition and is the only such scalar. +// +// COMMON CURVES: +// +// The group orders for common curves end in the following bit patterns: +// +// P-521: ...00001001; w = 4 is okay +// P-384: ...01110011; w = 2, 5, 6, 7 are okay +// P-256: ...01010001; w = 5, 7 are okay +// P-224: ...00111101; w = 3, 4, 5, 6 are okay void ec_GFp_nistp_recode_scalar_bits(uint8_t *sign, uint8_t *digit, uint8_t in) { uint8_t s, d; |