/* Copyright (c) 2015, Google Inc. * * Permission to use, copy, modify, and/or distribute this software for any * purpose with or without fee is hereby granted, provided that the above * copyright notice and this permission notice appear in all copies. * * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY * SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN ACTION * OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF OR IN * CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE. */ #include #include "../../internal.h" #if defined(__GNUC__) #pragma GCC diagnostic ignored "-Wconversion" #pragma GCC diagnostic ignored "-Wsign-conversion" #endif // 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 // fewer points need to be precomputed, given that point inversion is easy (a // precomputed point dP makes -dP available as well). // // BACKGROUND: // // Signed digits for multiplication were introduced by Booth ("A signed binary // multiplication technique", Quart. Journ. Mech. and Applied Math., vol. IV, // pt. 2 (1951), pp. 236-240), in that case for multiplication of integers. // Booth's original encoding did not generally improve the density of nonzero // digits over the binary representation, and was merely meant to simplify the // handling of signed factors given in two's complement; but it has since been // shown to be the basis of various signed-digit representations that do have // further advantages, including the wNAF, using the following general // approach: // // (1) Given a binary representation // // b_k ... b_2 b_1 b_0, // // 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+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. // 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 // "sign-alternating {+-1}-representation". // // An interesting property is that among the nonzero bits, values 1 and -1 // strictly alternate. // // (2) Various window schemes can be applied to the Booth representation of // integers: for example, right-to-left sliding windows yield the wNAF // (a signed-digit encoding independently discovered by various researchers // in the 1990s), and left-to-right sliding windows yield a left-to-right // equivalent of the wNAF (independently discovered by various researchers // around 2004). // // To prevent leaking information through side channels in point multiplication, // we need to recode the given integer into a regular pattern: sliding windows // as in wNAFs won't do, we need their fixed-window equivalent -- which is a few // decades older: we'll be using the so-called "modified Booth encoding" due to // MacSorley ("High-speed arithmetic in binary computers", Proc. IRE, vol. 49 // (1961), pp. 67-91), in a radix-2^5 setting. That is, we always combine five // signed bits into a signed digit: // // 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. // // 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_(5j + 4) ... b_(5j - 1). // // 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 .. 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_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 static inline void recode_scalar_bits(crypto_word *sign, crypto_word *digit, crypto_word in) { crypto_word s, d; s = ~((in >> 5) - 1); /* sets all bits to MSB(in), 'in' seen as * 6-bit value */ d = (1 << 6) - in - 1; d = (d & s) | (in & ~s); d = (d >> 1) + (d & 1); *sign = s & 1; *digit = d; }