/*  $OpenBSD: bn_bpsw.c,v 1.12 2025/02/13 11:10:01 tb Exp $ */

/*

 * Copyright (c) 2022 Martin Grenouilloux <martin.grenouilloux@lse.epita.fr>

 * Copyright (c) 2022 Theo Buehler <tb@openbsd.org>

 *

 * Permission to use, copy, modify, and 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 <openssl/bn.h>

#include "bn_local.h"

#include "bn_prime.h"

/*

 * For an odd n compute a / 2 (mod n). If a is even, we can do a plain

 * division, otherwise calculate (a + n) / 2. Then reduce (mod n).

 */

static int

bn_div_by_two_mod_odd_n(BIGNUM *a, const BIGNUM *n, BN_CTX *ctx)

{

  if (!BN_is_odd(n))

    return 0;

  if (BN_is_odd(a)) {

    if (!BN_add(a, a, n))

      return 0;

  }

  if (!BN_rshift1(a, a))

    return 0;

  if (!BN_mod_ct(a, a, n, ctx))

    return 0;

  return 1;

}

/*

 * Given the next binary digit of k and the current Lucas terms U and V, this

 * helper computes the next terms in the Lucas sequence defined as follows:

 *

 *   U' = U * V                  (mod n)

 *   V' = (V^2 + D * U^2) / 2    (mod n)

 *

 * If digit == 0, bn_lucas_step() returns U' and V'. If digit == 1, it returns

 *

 *   U'' = (U' + V') / 2         (mod n)

 *   V'' = (V' + D * U') / 2     (mod n)

 *

 * Compare with FIPS 186-4, Appendix C.3.3, step 6.

 */

static int

bn_lucas_step(BIGNUM *U, BIGNUM *V, int digit, const BIGNUM *D,

    const BIGNUM *n, BN_CTX *ctx)

{

  BIGNUM *tmp;

  int ret = 0;

  BN_CTX_start(ctx);

  if ((tmp = BN_CTX_get(ctx)) == NULL)

    goto err;

  /* Calculate D * U^2 before computing U'. */

  if (!BN_sqr(tmp, U, ctx))

    goto err;

  if (!BN_mul(tmp, D, tmp, ctx))

    goto err;

  /* U' = U * V (mod n). */

  if (!BN_mod_mul(U, U, V, n, ctx))

    goto err;

  /* V' = (V^2 + D * U^2) / 2 (mod n). */

  if (!BN_sqr(V, V, ctx))

    goto err;

  if (!BN_add(V, V, tmp))

    goto err;

  if (!bn_div_by_two_mod_odd_n(V, n, ctx))

    goto err;

  if (digit == 1) {

    /* Calculate D * U' before computing U''. */

    if (!BN_mul(tmp, D, U, ctx))

      goto err;

    /* U'' = (U' + V') / 2 (mod n). */

    if (!BN_add(U, U, V))

      goto err;

    if (!bn_div_by_two_mod_odd_n(U, n, ctx))

      goto err;

    /* V'' = (V' + D * U') / 2 (mod n). */

    if (!BN_add(V, V, tmp))

      goto err;

    if (!bn_div_by_two_mod_odd_n(V, n, ctx))

      goto err;

  }

  ret = 1;

 err:

  BN_CTX_end(ctx);

  return ret;

}

/*

 * Compute the Lucas terms U_k, V_k, see FIPS 186-4, Appendix C.3.3, steps 4-6.

 */

static int

bn_lucas(BIGNUM *U, BIGNUM *V, const BIGNUM *k, const BIGNUM *D,

    const BIGNUM *n, BN_CTX *ctx)

{

  int digit, i;

  int ret = 0;

  if (!BN_one(U))

    goto err;

  if (!BN_one(V))

    goto err;

  /*

   * Iterate over the digits of k from MSB to LSB. Start at digit 2

   * since the first digit is dealt with by setting U = 1 and V = 1.

   */

  for (i = BN_num_bits(k) - 2; i >= 0; i--) {

    digit = BN_is_bit_set(k, i);

    if (!bn_lucas_step(U, V, digit, D, n, ctx))

      goto err;

  }

  ret = 1;

 err:

  return ret;

}

/*

 * This is a stronger variant of the Lucas test in FIPS 186-4, Appendix C.3.3.

 * Every strong Lucas pseudoprime n is also a Lucas pseudoprime since

 * U_{n+1} == 0 follows from U_k == 0 or V_{k * 2^r} == 0 for 0 <= r < s.

 */

static int

bn_strong_lucas_test(int *is_pseudoprime, const BIGNUM *n, const BIGNUM *D,

    BN_CTX *ctx)

{

  BIGNUM *k, *U, *V;

  int r, s;

  int ret = 0;

  BN_CTX_start(ctx);

  if ((k = BN_CTX_get(ctx)) == NULL)

    goto err;

  if ((U = BN_CTX_get(ctx)) == NULL)

    goto err;

  if ((V = BN_CTX_get(ctx)) == NULL)

    goto err;

  /*

   * Factorize n + 1 = k * 2^s with odd k: shift away the s trailing ones

   * of n and set the lowest bit of the resulting number k.

   */

  s = 0;

  while (BN_is_bit_set(n, s))

    s++;

  if (!BN_rshift(k, n, s))

    goto err;

  if (!BN_set_bit(k, 0))

    goto err;

  /*

   * Calculate the Lucas terms U_k and V_k. If either of them is zero,

   * then n is a strong Lucas pseudoprime.

   */

  if (!bn_lucas(U, V, k, D, n, ctx))

    goto err;

  if (BN_is_zero(U) || BN_is_zero(V)) {

    *is_pseudoprime = 1;

    goto done;

  }

  /*

   * Calculate the Lucas terms U_{k * 2^r}, V_{k * 2^r} for 1 <= r < s.

   * If any V_{k * 2^r} is zero then n is a strong Lucas pseudoprime.

   */

  for (r = 1; r < s; r++) {

    if (!bn_lucas_step(U, V, 0, D, n, ctx))

      goto err;

    if (BN_is_zero(V)) {

      *is_pseudoprime = 1;

      goto done;

    }

  }

  /*

   * If we got here, n is definitely composite.

   */

  *is_pseudoprime = 0;

 done:

  ret = 1;

 err:

  BN_CTX_end(ctx);

  return ret;

}

/*

 * Test n for primality using the strong Lucas test with Selfridge's Method A.

 * Returns 1 if n is prime or a strong Lucas-Selfridge pseudoprime.

 * If it returns 0 then n is definitely composite.

 */

static int

bn_strong_lucas_selfridge(int *is_pseudoprime, const BIGNUM *n, BN_CTX *ctx)

{

  BIGNUM *D, *two;

  int is_perfect_square, jacobi_symbol, sign;

  int ret = 0;

  BN_CTX_start(ctx);

  /* If n is a perfect square, it is composite. */

  if (!bn_is_perfect_square(&is_perfect_square, n, ctx))

    goto err;

  if (is_perfect_square) {

    *is_pseudoprime = 0;

    goto done;

  }

  /*

   * Find the first D in the Selfridge sequence 5, -7, 9, -11, 13, ...

   * such that the Jacobi symbol (D/n) is -1.

   */

  if ((D = BN_CTX_get(ctx)) == NULL)

    goto err;

  if ((two = BN_CTX_get(ctx)) == NULL)

    goto err;

  sign = 1;

  if (!BN_set_word(D, 5))

    goto err;

  if (!BN_set_word(two, 2))

    goto err;

  while (1) {

    /* For odd n the Kronecker symbol computes the Jacobi symbol. */

    if ((jacobi_symbol = BN_kronecker(D, n, ctx)) == -2)

      goto err;

    /* We found the value for D. */

    if (jacobi_symbol == -1)

      break;

    /* n and D have prime factors in common. */

    if (jacobi_symbol == 0) {

      *is_pseudoprime = 0;

      goto done;

    }

    sign = -sign;

    if (!BN_uadd(D, D, two))

      goto err;

    BN_set_negative(D, sign == -1);

  }

  if (!bn_strong_lucas_test(is_pseudoprime, n, D, ctx))

    goto err;

 done:

  ret = 1;

 err:

  BN_CTX_end(ctx);

  return ret;

}

/*

 * Fermat criterion in Miller-Rabin test.

 *

 * Check whether 1 < base < n - 1 witnesses that n is composite. For prime n:

 *

 *  * Fermat's little theorem: base^(n-1) = 1 (mod n).

 *  * The only square roots of 1 (mod n) are 1 and -1.

 *

 * Calculate base^((n-1)/2) by writing n - 1 = k * 2^s with odd k. Iteratively

 * compute power = (base^k)^(2^(s-1)) by successive squaring of base^k.

 *

 * If power ever reaches -1, base^(n-1) is equal to 1 and n is a pseudoprime

 * for base. If power reaches 1 before -1 during successive squaring, we have

 * an unexpected square root of 1 and n is composite. Otherwise base^(n-1) != 1,

 * and n is composite.

 */

static int

bn_fermat(int *is_pseudoprime, const BIGNUM *n, const BIGNUM *n_minus_one,

    const BIGNUM *k, int s, const BIGNUM *base, BN_CTX *ctx, BN_MONT_CTX *mctx)

{

  BIGNUM *power;

  int ret = 0;

  int i;

  BN_CTX_start(ctx);

  if ((power = BN_CTX_get(ctx)) == NULL)

    goto err;

  /* Sanity check: ensure that 1 < base < n - 1. */

  if (BN_cmp(base, BN_value_one()) <= 0 || BN_cmp(base, n_minus_one) >= 0)

    goto err;

  if (!BN_mod_exp_mont_ct(power, base, k, n, ctx, mctx))

    goto err;

  if (BN_is_one(power) || BN_cmp(power, n_minus_one) == 0) {

    *is_pseudoprime = 1;

    goto done;

  }

  /* Loop invariant: power is neither 1 nor -1 (mod n). */

  for (i = 1; i < s; i++) {

    if (!BN_mod_sqr(power, power, n, ctx))

      goto err;

    /* n is a pseudoprime for base. */

    if (BN_cmp(power, n_minus_one) == 0) {

      *is_pseudoprime = 1;

      goto done;

    }

    /* n is composite: there's a square root of unity != 1 or -1. */

    if (BN_is_one(power)) {

      *is_pseudoprime = 0;

      goto done;

    }

  }

  /*

   * If we get here, n is definitely composite: base^(n-1) != 1.

   */

  *is_pseudoprime = 0;

 done:

  ret = 1;

 err:

  BN_CTX_end(ctx);

  return ret;

}

/*

 * Miller-Rabin primality test for base 2 and for |rounds| of random bases.

 * On success: is_pseudoprime == 0 implies that n is composite.

 */

static int

bn_miller_rabin(int *is_pseudoprime, const BIGNUM *n, BN_CTX *ctx,

    size_t rounds)

{

  BN_MONT_CTX *mctx = NULL;

  BIGNUM *base, *k, *n_minus_one;

  size_t i;

  int s;

  int ret = 0;

  BN_CTX_start(ctx);

  if ((base = BN_CTX_get(ctx)) == NULL)

    goto err;

  if ((k = BN_CTX_get(ctx)) == NULL)

    goto err;

  if ((n_minus_one = BN_CTX_get(ctx)) == NULL)

    goto err;

  if (BN_is_word(n, 2) || BN_is_word(n, 3)) {

    *is_pseudoprime = 1;

    goto done;

  }

  if (BN_cmp(n, BN_value_one()) <= 0 || !BN_is_odd(n)) {

    *is_pseudoprime = 0;

    goto done;

  }

  if (!BN_sub(n_minus_one, n, BN_value_one()))

    goto err;

  /*

   * Factorize n - 1 = k * 2^s.

   */

  s = 0;

  while (!BN_is_bit_set(n_minus_one, s))

    s++;

  if (!BN_rshift(k, n_minus_one, s))

    goto err;

  /*

   * Montgomery setup for n.

   */

  if ((mctx = BN_MONT_CTX_create(n, ctx)) == NULL)

    goto err;

  /*

   * Perform a Miller-Rabin test for base 2 as required by BPSW.

   */

  if (!BN_set_word(base, 2))

    goto err;

  if (!bn_fermat(is_pseudoprime, n, n_minus_one, k, s, base, ctx, mctx))

    goto err;

  if (!*is_pseudoprime)

    goto done;

  /*

   * Perform Miller-Rabin tests with random 3 <= base < n - 1 to reduce

   * risk of false positives in BPSW.

   */

  for (i = 0; i < rounds; i++) {

    if (!bn_rand_interval(base, 3, n_minus_one))

      goto err;

    if (!bn_fermat(is_pseudoprime, n, n_minus_one, k, s, base, ctx,

        mctx))

      goto err;

    if (!*is_pseudoprime)

      goto done;

  }

  /*

   * If we got here, we have a Miller-Rabin pseudoprime.

   */

  *is_pseudoprime = 1;

 done:

  ret = 1;

 err:

  BN_MONT_CTX_free(mctx);

  BN_CTX_end(ctx);

  return ret;

}

/*

 * The Baillie-Pomerance-Selfridge-Wagstaff algorithm combines a Miller-Rabin

 * test for base 2 with a Strong Lucas pseudoprime test.

 */

int

bn_is_prime_bpsw(int *is_pseudoprime, const BIGNUM *n, BN_CTX *in_ctx,

    size_t rounds)

{

  BN_CTX *ctx = NULL;

  BN_ULONG mod;

  int i;

  int ret = 0;

  if (BN_is_word(n, 2)) {

    *is_pseudoprime = 1;

    goto done;

  }

  if (BN_cmp(n, BN_value_one()) <= 0 || !BN_is_odd(n)) {

    *is_pseudoprime = 0;

    goto done;

  }

  /* Trial divisions with the first 2048 primes. */

  for (i = 0; i < NUMPRIMES; i++) {

    if ((mod = BN_mod_word(n, primes[i])) == (BN_ULONG)-1)

      goto err;

    if (mod == 0) {

      *is_pseudoprime = BN_is_word(n, primes[i]);

      goto done;

    }

  }

  if ((ctx = in_ctx) == NULL)

    ctx = BN_CTX_new();

  if (ctx == NULL)

    goto err;

  if (!bn_miller_rabin(is_pseudoprime, n, ctx, rounds))

    goto err;

  if (!*is_pseudoprime)

    goto done;

  if (!bn_strong_lucas_selfridge(is_pseudoprime, n, ctx))

    goto err;

 done:

  ret = 1;

 err:

  if (ctx != in_ctx)

    BN_CTX_free(ctx);

  return ret;

}