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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(crypto_word_t *sign, crypto_word_t *digit,

                                     crypto_word_t in) {

  crypto_word_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;

}