/* mpfr_agm -- arithmetic-geometric mean of two floating-point numbers

Copyright 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013 Free Software Foundation, Inc.
Contributed by the AriC and Caramel projects, INRIA.

This file is part of the GNU MPFR Library.

The GNU MPFR Library is free software; you can redistribute it and/or modify
it under the terms of the GNU Lesser General Public License as published by
the Free Software Foundation; either version 3 of the License, or (at your
option) any later version.

The GNU MPFR Library is distributed in the hope that it will be useful, but
WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU Lesser General Public
License for more details.

You should have received a copy of the GNU Lesser General Public License
along with the GNU MPFR Library; see the file COPYING.LESSER.  If not, see
http://www.gnu.org/licenses/ or write to the Free Software Foundation, Inc.,
51 Franklin St, Fifth Floor, Boston, MA 02110-1301, USA. */

#define MPFR_NEED_LONGLONG_H
#include "mpfr-impl.h"

/* agm(x,y) is between x and y, so we don't need to save exponent range */
int
mpfr_agm (mpfr_ptr r, mpfr_srcptr op2, mpfr_srcptr op1, mpfr_rnd_t rnd_mode)
{
  int compare, inexact;
  mp_size_t s;
  mpfr_prec_t p, q;
  mp_limb_t *up, *vp, *ufp, *vfp;
  mpfr_t u, v, uf, vf, sc1, sc2;
  mpfr_exp_t scaleop = 0, scaleit;
  unsigned long n; /* number of iterations */
  MPFR_ZIV_DECL (loop);
  MPFR_TMP_DECL(marker);
  MPFR_SAVE_EXPO_DECL (expo);

  MPFR_LOG_FUNC
    (("op2[%Pu]=%.*Rg op1[%Pu]=%.*Rg rnd=%d",
      mpfr_get_prec (op2), mpfr_log_prec, op2,
      mpfr_get_prec (op1), mpfr_log_prec, op1, rnd_mode),
     ("r[%Pu]=%.*Rg inexact=%d",
      mpfr_get_prec (r), mpfr_log_prec, r, inexact));

  /* Deal with special values */
  if (MPFR_ARE_SINGULAR (op1, op2))
    {
      /* If a or b is NaN, the result is NaN */
      if (MPFR_IS_NAN(op1) || MPFR_IS_NAN(op2))
        {
          MPFR_SET_NAN(r);
          MPFR_RET_NAN;
        }
      /* now one of a or b is Inf or 0 */
      /* If a and b is +Inf, the result is +Inf.
         Otherwise if a or b is -Inf or 0, the result is NaN */
      else if (MPFR_IS_INF(op1) || MPFR_IS_INF(op2))
        {
          if (MPFR_IS_STRICTPOS(op1) && MPFR_IS_STRICTPOS(op2))
            {
              MPFR_SET_INF(r);
              MPFR_SET_SAME_SIGN(r, op1);
              MPFR_RET(0); /* exact */
            }
          else
            {
              MPFR_SET_NAN(r);
              MPFR_RET_NAN;
            }
        }
      else /* a and b are neither NaN nor Inf, and one is zero */
        {  /* If a or b is 0, the result is +0 since a sqrt is positive */
          MPFR_ASSERTD (MPFR_IS_ZERO (op1) || MPFR_IS_ZERO (op2));
          MPFR_SET_POS (r);
          MPFR_SET_ZERO (r);
          MPFR_RET (0); /* exact */
        }
    }

  /* If a or b is negative (excluding -Infinity), the result is NaN */
  if (MPFR_UNLIKELY(MPFR_IS_NEG(op1) || MPFR_IS_NEG(op2)))
    {
      MPFR_SET_NAN(r);
      MPFR_RET_NAN;
    }

  /* Precision of the following calculus */
  q = MPFR_PREC(r);
  p = q + MPFR_INT_CEIL_LOG2(q) + 15;
  MPFR_ASSERTD (p >= 7); /* see algorithms.tex */
  s = MPFR_PREC2LIMBS (p);

  /* b (op2) and a (op1) are the 2 operands but we want b >= a */
  compare = mpfr_cmp (op1, op2);
  if (MPFR_UNLIKELY( compare == 0 ))
    {
      mpfr_set (r, op1, rnd_mode);
      MPFR_RET (0); /* exact */
    }
  else if (compare > 0)
    {
      mpfr_srcptr t = op1;
      op1 = op2;
      op2 = t;
    }

  /* Now b (=op2) > a (=op1) */

  MPFR_SAVE_EXPO_MARK (expo);

  MPFR_TMP_MARK(marker);

  /* Main loop */
  MPFR_ZIV_INIT (loop, p);
  for (;;)
    {
      mpfr_prec_t eq;
      unsigned long err = 0;  /* must be set to 0 at each Ziv iteration */
      MPFR_BLOCK_DECL (flags);

      /* Init temporary vars */
      MPFR_TMP_INIT (up, u, p, s);
      MPFR_TMP_INIT (vp, v, p, s);
      MPFR_TMP_INIT (ufp, uf, p, s);
      MPFR_TMP_INIT (vfp, vf, p, s);

      /* Calculus of un and vn */
    retry:
      MPFR_BLOCK (flags,
                  mpfr_mul (u, op1, op2, MPFR_RNDN);
                  /* mpfr_mul(...): faster since PREC(op) < PREC(u) */
                  mpfr_add (v, op1, op2, MPFR_RNDN);
                  /* mpfr_add with !=prec is still good */);
      if (MPFR_UNLIKELY (MPFR_OVERFLOW (flags) || MPFR_UNDERFLOW (flags)))
        {
          mpfr_exp_t e1 , e2;

          MPFR_ASSERTN (scaleop == 0);
          e1 = MPFR_GET_EXP (op1);
          e2 = MPFR_GET_EXP (op2);

          /* Let's determine scaleop to avoid an overflow/underflow. */
          if (MPFR_OVERFLOW (flags))
            {
              /* Let's recall that emin <= e1 <= e2 <= emax.
                 There has been an overflow. Thus e2 >= emax/2.
                 If the mpfr_mul overflowed, then e1 + e2 > emax.
                 If the mpfr_add overflowed, then e2 = emax.
                 We want: (e1 + scale) + (e2 + scale) <= emax,
                 i.e. scale <= (emax - e1 - e2) / 2. Let's take
                 scale = min(floor((emax - e1 - e2) / 2), -1).
                 This is OK, as:
                 1. emin <= scale <= -1.
                 2. e1 + scale >= emin. Indeed:
                    * If e1 + e2 > emax, then
                      e1 + scale >= e1 + (emax - e1 - e2) / 2 - 1
                                 >= (emax + e1 - emax) / 2 - 1
                                 >= e1 / 2 - 1 >= emin.
                    * Otherwise, mpfr_mul didn't overflow, therefore
                      mpfr_add overflowed and e2 = emax, so that
                      e1 > emin (see restriction below).
                      e1 + scale > emin - 1, thus e1 + scale >= emin.
                 3. e2 + scale <= emax, since scale < 0. */
              if (e1 + e2 > MPFR_EXT_EMAX)
                {
                  scaleop = - (((e1 + e2) - MPFR_EXT_EMAX + 1) / 2);
                  MPFR_ASSERTN (scaleop < 0);
                }
              else
                {
                  /* The addition necessarily overflowed. */
                  MPFR_ASSERTN (e2 == MPFR_EXT_EMAX);
                  /* The case where e1 = emin and e2 = emax is not supported
                     here. This would mean that the precision of e2 would be
                     huge (and possibly not supported in practice anyway). */
                  MPFR_ASSERTN (e1 > MPFR_EXT_EMIN);
                  scaleop = -1;
                }

            }
          else  /* underflow only (in the multiplication) */
            {
              /* We have e1 + e2 <= emin (so, e1 <= e2 <= 0).
                 We want: (e1 + scale) + (e2 + scale) >= emin + 1,
                 i.e. scale >= (emin + 1 - e1 - e2) / 2. let's take
                 scale = ceil((emin + 1 - e1 - e2) / 2). This is OK, as:
                 1. 1 <= scale <= emax.
                 2. e1 + scale >= emin + 1 >= emin.
                 3. e2 + scale <= scale <= emax. */
              MPFR_ASSERTN (e1 <= e2 && e2 <= 0);
              scaleop = (MPFR_EXT_EMIN + 2 - e1 - e2) / 2;
              MPFR_ASSERTN (scaleop > 0);
            }

          MPFR_ALIAS (sc1, op1, MPFR_SIGN (op1), e1 + scaleop);
          MPFR_ALIAS (sc2, op2, MPFR_SIGN (op2), e2 + scaleop);
          op1 = sc1;
          op2 = sc2;
          MPFR_LOG_MSG (("Exception in pre-iteration, scale = %"
                         MPFR_EXP_FSPEC "d\n", scaleop));
          goto retry;
        }

      mpfr_clear_flags ();
      mpfr_sqrt (u, u, MPFR_RNDN);
      mpfr_div_2ui (v, v, 1, MPFR_RNDN);

      scaleit = 0;
      n = 1;
      while (mpfr_cmp2 (u, v, &eq) != 0 && eq <= p - 2)
        {
          MPFR_BLOCK_DECL (flags2);

          MPFR_LOG_MSG (("Iteration n = %lu\n", n));

        retry2:
          mpfr_add (vf, u, v, MPFR_RNDN);  /* No overflow? */
          mpfr_div_2ui (vf, vf, 1, MPFR_RNDN);
          /* See proof in algorithms.tex */
          if (4*eq > p)
            {
              mpfr_t w;
              MPFR_BLOCK_DECL (flags3);

              MPFR_LOG_MSG (("4*eq > p\n", 0));

              /* vf = V(k) */
              mpfr_init2 (w, (p + 1) / 2);
              MPFR_BLOCK
                (flags3,
                 mpfr_sub (w, v, u, MPFR_RNDN);       /* e = V(k-1)-U(k-1) */
                 mpfr_sqr (w, w, MPFR_RNDN);          /* e = e^2 */
                 mpfr_div_2ui (w, w, 4, MPFR_RNDN);   /* e*= (1/2)^2*1/4  */
                 mpfr_div (w, w, vf, MPFR_RNDN);      /* 1/4*e^2/V(k) */
                 );
              if (MPFR_LIKELY (! MPFR_UNDERFLOW (flags3)))
                {
                  mpfr_sub (v, vf, w, MPFR_RNDN);
                  err = MPFR_GET_EXP (vf) - MPFR_GET_EXP (v); /* 0 or 1 */
                  mpfr_clear (w);
                  break;
                }
              /* There has been an underflow because of the cancellation
                 between V(k-1) and U(k-1). Let's use the conventional
                 method. */
              MPFR_LOG_MSG (("4*eq > p -> underflow\n", 0));
              mpfr_clear (w);
              mpfr_clear_underflow ();
            }
          /* U(k) increases, so that U.V can overflow (but not underflow). */
          MPFR_BLOCK (flags2, mpfr_mul (uf, u, v, MPFR_RNDN););
          if (MPFR_UNLIKELY (MPFR_OVERFLOW (flags2)))
            {
              mpfr_exp_t scale2;

              scale2 = - (((MPFR_GET_EXP (u) + MPFR_GET_EXP (v))
                           - MPFR_EXT_EMAX + 1) / 2);
              MPFR_EXP (u) += scale2;
              MPFR_EXP (v) += scale2;
              scaleit += scale2;
              MPFR_LOG_MSG (("Overflow in iteration n = %lu, scaleit = %"
                             MPFR_EXP_FSPEC "d (%" MPFR_EXP_FSPEC "d)\n",
                             n, scaleit, scale2));
              mpfr_clear_overflow ();
              goto retry2;
            }
          mpfr_sqrt (u, uf, MPFR_RNDN);
          mpfr_swap (v, vf);
          n ++;
        }

      MPFR_LOG_MSG (("End of iterations (n = %lu)\n", n));

      /* the error on v is bounded by (18n+51) ulps, or twice if there
         was an exponent loss in the final subtraction */
      err += MPFR_INT_CEIL_LOG2(18 * n + 51); /* 18n+51 should not overflow
                                                 since n is about log(p) */
      /* we should have n+2 <= 2^(p/4) [see algorithms.tex] */
      if (MPFR_LIKELY (MPFR_INT_CEIL_LOG2(n + 2) <= p / 4 &&
                       MPFR_CAN_ROUND (v, p - err, q, rnd_mode)))
        break; /* Stop the loop */

      /* Next iteration */
      MPFR_ZIV_NEXT (loop, p);
      s = MPFR_PREC2LIMBS (p);
    }
  MPFR_ZIV_FREE (loop);

  if (MPFR_UNLIKELY ((__gmpfr_flags & (MPFR_FLAGS_ALL ^ MPFR_FLAGS_INEXACT))
                     != 0))
    {
      MPFR_ASSERTN (! mpfr_overflow_p ());   /* since mpfr_clear_flags */
      MPFR_ASSERTN (! mpfr_underflow_p ());  /* since mpfr_clear_flags */
      MPFR_ASSERTN (! mpfr_divby0_p ());     /* since mpfr_clear_flags */
      MPFR_ASSERTN (! mpfr_nanflag_p ());    /* since mpfr_clear_flags */
    }

  /* Setting of the result */
  inexact = mpfr_set (r, v, rnd_mode);
  MPFR_EXP (r) -= scaleop + scaleit;

  /* Let's clean */
  MPFR_TMP_FREE(marker);

  MPFR_SAVE_EXPO_FREE (expo);
  /* From the definition of the AGM, underflow and overflow
     are not possible. */
  return mpfr_check_range (r, inexact, rnd_mode);
  /* agm(u,v) can be exact for u, v rational only for u=v.
     Proof (due to Nicolas Brisebarre): it suffices to consider
     u=1 and v<1. Then 1/AGM(1,v) = 2F1(1/2,1/2,1;1-v^2),
     and a theorem due to G.V. Chudnovsky states that for x a
     non-zero algebraic number with |x|<1, then
     2F1(1/2,1/2,1;x) and 2F1(-1/2,1/2,1;x) are algebraically
     independent over Q. */
}
