/*

* (C) 1999-2011,2016,2018,2019,2020 Jack Lloyd

*

* Botan is released under the Simplified BSD License (see license.txt)

*/

#include <botan/numthry.h>

#include <botan/internal/divide.h>

#include <botan/internal/ct_utils.h>

#include <botan/internal/mp_core.h>

#include <botan/internal/rounding.h>

namespace Botan {

namespace {

BigInt inverse_mod_odd_modulus(const BigInt& n, const BigInt& mod)

   {

   // Caller should assure these preconditions:

   BOTAN_DEBUG_ASSERT(n.is_positive());

   BOTAN_DEBUG_ASSERT(mod.is_positive());

   BOTAN_DEBUG_ASSERT(n < mod);

   BOTAN_DEBUG_ASSERT(mod >= 3 && mod.is_odd());

   /*

   This uses a modular inversion algorithm designed by Niels Möller

   and implemented in Nettle. The same algorithm was later also

   adapted to GMP in mpn_sec_invert.

   It can be easily implemented in a way that does not depend on

   secret branches or memory lookups, providing resistance against

   some forms of side channel attack.

   There is also a description of the algorithm in Appendix 5 of "Fast

   Software Polynomial Multiplication on ARM Processors using the NEON Engine"

   by Danilo Câmara, Conrado P. L. Gouvêa, Julio López, and Ricardo

   Dahab in LNCS 8182

      https://conradoplg.cryptoland.net/files/2010/12/mocrysen13.pdf

   Thanks to Niels for creating the algorithm, explaining some things

   about it, and the reference to the paper.

   */

   const size_t mod_words = mod.sig_words();

   BOTAN_ASSERT(mod_words > 0, "Not empty");

   secure_vector<word> tmp_mem(5*mod_words);

   word* v_w = &tmp_mem[0];

   word* u_w = &tmp_mem[1*mod_words];

   word* b_w = &tmp_mem[2*mod_words];

   word* a_w = &tmp_mem[3*mod_words];

   word* mp1o2 = &tmp_mem[4*mod_words];

   CT::poison(tmp_mem.data(), tmp_mem.size());

   copy_mem(a_w, n.data(), std::min(n.size(), mod_words));

   copy_mem(b_w, mod.data(), std::min(mod.size(), mod_words));

   u_w[0] = 1;

   // v_w = 0

   // compute (mod + 1) / 2 which [because mod is odd] is equal to

   // (mod / 2) + 1

   copy_mem(mp1o2, mod.data(), std::min(mod.size(), mod_words));

   bigint_shr1(mp1o2, mod_words, 0, 1);

   word carry = bigint_add2_nc(mp1o2, mod_words, u_w, 1);

   BOTAN_ASSERT_NOMSG(carry == 0);

   // Only n.bits() + mod.bits() iterations are required, but avoid leaking the size of n

   const size_t execs = 2 * mod.bits();

   for(size_t i = 0; i != execs; ++i)

      {

      const word odd_a = a_w[0] & 1;

      //if(odd_a) a -= b

      word underflow = bigint_cnd_sub(odd_a, a_w, b_w, mod_words);

      //if(underflow) { b -= a; a = abs(a); swap(u, v); }

      bigint_cnd_add(underflow, b_w, a_w, mod_words);

      bigint_cnd_abs(underflow, a_w, mod_words);

      bigint_cnd_swap(underflow, u_w, v_w, mod_words);

      // a >>= 1

      bigint_shr1(a_w, mod_words, 0, 1);

      //if(odd_a) u -= v;

      word borrow = bigint_cnd_sub(odd_a, u_w, v_w, mod_words);

      // if(borrow) u += p

      bigint_cnd_add(borrow, u_w, mod.data(), mod_words);

      const word odd_u = u_w[0] & 1;

      // u >>= 1

      bigint_shr1(u_w, mod_words, 0, 1);

      //if(odd_u) u += mp1o2;

      bigint_cnd_add(odd_u, u_w, mp1o2, mod_words);

      }

   auto a_is_0 = CT::Mask<word>::set();

   for(size_t i = 0; i != mod_words; ++i)

      a_is_0 &= CT::Mask<word>::is_zero(a_w[i]);

   auto b_is_1 = CT::Mask<word>::is_equal(b_w[0], 1);

   for(size_t i = 1; i != mod_words; ++i)

      b_is_1 &= CT::Mask<word>::is_zero(b_w[i]);

   BOTAN_ASSERT(a_is_0.is_set(), "A is zero");

   // if b != 1 then gcd(n,mod) > 1 and inverse does not exist

   // in which case zero out the result to indicate this

   (~b_is_1).if_set_zero_out(v_w, mod_words);

   /*

   * We've placed the result in the lowest words of the temp buffer.

   * So just clear out the other values and then give that buffer to a

   * BigInt.

   */

   clear_mem(&tmp_mem[mod_words], 4*mod_words);

   CT::unpoison(tmp_mem.data(), tmp_mem.size());

   BigInt r;

   r.swap_reg(tmp_mem);

   return r;

   }

BigInt inverse_mod_pow2(const BigInt& a1, size_t k)

   {

   /*

   * From "A New Algorithm for Inversion mod p^k" by Çetin Kaya Koç

   * https://eprint.iacr.org/2017/411.pdf sections 5 and 7.

   */

   if(a1.is_even())

      return 0;

   BigInt a = a1;

   a.mask_bits(k);

   BigInt b = 1;

   BigInt X = 0;

   BigInt newb;

   const size_t a_words = a.sig_words();

   X.grow_to(round_up(k, BOTAN_MP_WORD_BITS) / BOTAN_MP_WORD_BITS);

   b.grow_to(a_words);

   /*

   Hide the exact value of k. k is anyway known to word length

   granularity because of the length of a, so no point in doing more

   than this.

   */

   const size_t iter = round_up(k, BOTAN_MP_WORD_BITS);

   for(size_t i = 0; i != iter; ++i)

      {

      const bool b0 = b.get_bit(0);

      X.conditionally_set_bit(i, b0);

      newb = b - a;

      b.ct_cond_assign(b0, newb);

      b >>= 1;

      }

   X.mask_bits(k);

   X.const_time_unpoison();

   return X;

   }

}

BigInt inverse_mod(const BigInt& n, const BigInt& mod)

   {

   if(mod.is_zero())

      throw Invalid_Argument("inverse_mod modulus cannot be zero");

   if(mod.is_negative() || n.is_negative())

      throw Invalid_Argument("inverse_mod: arguments must be non-negative");

   if(n.is_zero() || (n.is_even() && mod.is_even()))

      return 0;

   if(mod.is_odd())

      {

      /*

      Fastpath for common case. This leaks if n is greater than mod or

      not, but we don't guarantee const time behavior in that case.

      */

      if(n < mod)

         return inverse_mod_odd_modulus(n, mod);

      else

         return inverse_mod_odd_modulus(ct_modulo(n, mod), mod);

      }

   const size_t mod_lz = low_zero_bits(mod);

   BOTAN_ASSERT_NOMSG(mod_lz > 0);

   const size_t mod_bits = mod.bits();

   BOTAN_ASSERT_NOMSG(mod_bits > mod_lz);

   if(mod_lz == mod_bits - 1)

      {

      // In this case we are performing an inversion modulo 2^k

      return inverse_mod_pow2(n, mod_lz);

      }

   /*

   * In this case we are performing an inversion modulo 2^k*o for

   * some k > 1 and some odd (not necessarily prime) integer.

   * Compute the inversions modulo 2^k and modulo o, then combine them

   * using CRT, which is possible because 2^k and o are relatively prime.

   */

   const BigInt o = mod >> mod_lz;

   const BigInt n_redc = ct_modulo(n, o);

   const BigInt inv_o = inverse_mod_odd_modulus(n_redc, o);

   const BigInt inv_2k = inverse_mod_pow2(n, mod_lz);

   // No modular inverse in this case:

   if(inv_o == 0 || inv_2k == 0)

      return 0;

   const BigInt m2k = BigInt::power_of_2(mod_lz);

   // Compute the CRT parameter

   const BigInt c = inverse_mod_pow2(o, mod_lz);

   // Compute h = c*(inv_2k-inv_o) mod 2^k

   BigInt h = c * (inv_2k - inv_o);

   const bool h_neg = h.is_negative();

   h.set_sign(BigInt::Positive);

   h.mask_bits(mod_lz);

   const bool h_nonzero = h.is_nonzero();

   h.ct_cond_assign(h_nonzero && h_neg, m2k - h);

   // Return result inv_o + h * o

   h *= o;

   h += inv_o;

   return h;

   }

}