pigweed / third_party / boringssl / boringssl / a810d82575ecbde26406fa583371f807f8721ed7 / . / crypto / fipsmodule / ec / util.c

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#include <openssl/base.h> | |

#include <openssl/ec.h> | |

#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 | |

// 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_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; | |

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; | |

} |