Chris@10: /*
Chris@10:  * Copyright (c) 2003, 2007-11 Matteo Frigo
Chris@10:  * Copyright (c) 2003, 2007-11 Massachusetts Institute of Technology
Chris@10:  *
Chris@10:  * This program is free software; you can redistribute it and/or modify
Chris@10:  * it under the terms of the GNU General Public License as published by
Chris@10:  * the Free Software Foundation; either version 2 of the License, or
Chris@10:  * (at your option) any later version.
Chris@10:  *
Chris@10:  * This program is distributed in the hope that it will be useful,
Chris@10:  * but WITHOUT ANY WARRANTY; without even the implied warranty of
Chris@10:  * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
Chris@10:  * GNU General Public License for more details.
Chris@10:  *
Chris@10:  * You should have received a copy of the GNU General Public License
Chris@10:  * along with this program; if not, write to the Free Software
Chris@10:  * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301  USA
Chris@10:  *
Chris@10:  */
Chris@10: 
Chris@10: 
Chris@10: /* Do a RODFT00 problem via an R2HC problem, with some pre/post-processing.
Chris@10: 
Chris@10:    This code uses the trick from FFTPACK, also documented in a similar
Chris@10:    form by Numerical Recipes.  Unfortunately, this algorithm seems to
Chris@10:    have intrinsic numerical problems (similar to those in
Chris@10:    reodft11e-r2hc.c), possibly due to the fact that it multiplies its
Chris@10:    input by a sine, causing a loss of precision near the zero.  For
Chris@10:    transforms of 16k points, it has already lost three or four decimal
Chris@10:    places of accuracy, which we deem unacceptable.
Chris@10: 
Chris@10:    So, we have abandoned this algorithm in favor of the one in
Chris@10:    rodft00-r2hc-pad.c, which unfortunately sacrifices 30-50% in speed.
Chris@10:    The only other alternative in the literature that does not have
Chris@10:    similar numerical difficulties seems to be the direct adaptation of
Chris@10:    the Cooley-Tukey decomposition for antisymmetric data, but this
Chris@10:    would require a whole new set of codelets and it's not clear that
Chris@10:    it's worth it at this point.  However, we did implement the latter
Chris@10:    algorithm for the specific case of odd n (logically adapting the
Chris@10:    split-radix algorithm); see reodft00e-splitradix.c. */
Chris@10: 
Chris@10: #include "reodft.h"
Chris@10: 
Chris@10: typedef struct {
Chris@10:      solver super;
Chris@10: } S;
Chris@10: 
Chris@10: typedef struct {
Chris@10:      plan_rdft super;
Chris@10:      plan *cld;
Chris@10:      twid *td;
Chris@10:      INT is, os;
Chris@10:      INT n;
Chris@10:      INT vl;
Chris@10:      INT ivs, ovs;
Chris@10: } P;
Chris@10: 
Chris@10: static void apply(const plan *ego_, R *I, R *O)
Chris@10: {
Chris@10:      const P *ego = (const P *) ego_;
Chris@10:      INT is = ego->is, os = ego->os;
Chris@10:      INT i, n = ego->n;
Chris@10:      INT iv, vl = ego->vl;
Chris@10:      INT ivs = ego->ivs, ovs = ego->ovs;
Chris@10:      R *W = ego->td->W;
Chris@10:      R *buf;
Chris@10: 
Chris@10:      buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
Chris@10: 
Chris@10:      for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
Chris@10: 	  buf[0] = 0;
Chris@10: 	  for (i = 1; i < n - i; ++i) {
Chris@10: 	       E a, b, apb, amb;
Chris@10: 	       a = I[is * (i - 1)];
Chris@10: 	       b = I[is * ((n - i) - 1)];
Chris@10: 	       apb =  K(2.0) * W[i] * (a + b);
Chris@10: 	       amb = (a - b);
Chris@10: 	       buf[i] = apb + amb;
Chris@10: 	       buf[n - i] = apb - amb;
Chris@10: 	  }
Chris@10: 	  if (i == n - i) {
Chris@10: 	       buf[i] = K(4.0) * I[is * (i - 1)];
Chris@10: 	  }
Chris@10: 	  
Chris@10: 	  {
Chris@10: 	       plan_rdft *cld = (plan_rdft *) ego->cld;
Chris@10: 	       cld->apply((plan *) cld, buf, buf);
Chris@10: 	  }
Chris@10: 	  
Chris@10: 	  /* FIXME: use recursive/cascade summation for better stability? */
Chris@10: 	  O[0] = buf[0] * 0.5;
Chris@10: 	  for (i = 1; i + i < n - 1; ++i) {
Chris@10: 	       INT k = i + i;
Chris@10: 	       O[os * (k - 1)] = -buf[n - i];
Chris@10: 	       O[os * k] = O[os * (k - 2)] + buf[i];
Chris@10: 	  }
Chris@10: 	  if (i + i == n - 1) {
Chris@10: 	       O[os * (n - 2)] = -buf[n - i];
Chris@10: 	  }
Chris@10:      }
Chris@10: 
Chris@10:      X(ifree)(buf);
Chris@10: }
Chris@10: 
Chris@10: static void awake(plan *ego_, enum wakefulness wakefulness)
Chris@10: {
Chris@10:      P *ego = (P *) ego_;
Chris@10:      static const tw_instr rodft00e_tw[] = {
Chris@10:           { TW_SIN, 0, 1 },
Chris@10:           { TW_NEXT, 1, 0 }
Chris@10:      };
Chris@10: 
Chris@10:      X(plan_awake)(ego->cld, wakefulness);
Chris@10: 
Chris@10:      X(twiddle_awake)(wakefulness,
Chris@10: 		      &ego->td, rodft00e_tw, 2*ego->n, 1, (ego->n+1)/2);
Chris@10: }
Chris@10: 
Chris@10: static void destroy(plan *ego_)
Chris@10: {
Chris@10:      P *ego = (P *) ego_;
Chris@10:      X(plan_destroy_internal)(ego->cld);
Chris@10: }
Chris@10: 
Chris@10: static void print(const plan *ego_, printer *p)
Chris@10: {
Chris@10:      const P *ego = (const P *) ego_;
Chris@10:      p->print(p, "(rodft00e-r2hc-%D%v%(%p%))", ego->n - 1, ego->vl, ego->cld);
Chris@10: }
Chris@10: 
Chris@10: static int applicable0(const solver *ego_, const problem *p_)
Chris@10: {
Chris@10:      const problem_rdft *p = (const problem_rdft *) p_;
Chris@10:      UNUSED(ego_);
Chris@10: 
Chris@10:      return (1
Chris@10: 	     && p->sz->rnk == 1
Chris@10: 	     && p->vecsz->rnk <= 1
Chris@10: 	     && p->kind[0] == RODFT00
Chris@10: 	  );
Chris@10: }
Chris@10: 
Chris@10: static int applicable(const solver *ego, const problem *p, const planner *plnr)
Chris@10: {
Chris@10:      return (!NO_SLOWP(plnr) && applicable0(ego, p));
Chris@10: }
Chris@10: 
Chris@10: static plan *mkplan(const solver *ego_, const problem *p_, planner *plnr)
Chris@10: {
Chris@10:      P *pln;
Chris@10:      const problem_rdft *p;
Chris@10:      plan *cld;
Chris@10:      R *buf;
Chris@10:      INT n;
Chris@10:      opcnt ops;
Chris@10: 
Chris@10:      static const plan_adt padt = {
Chris@10: 	  X(rdft_solve), awake, print, destroy
Chris@10:      };
Chris@10: 
Chris@10:      if (!applicable(ego_, p_, plnr))
Chris@10:           return (plan *)0;
Chris@10: 
Chris@10:      p = (const problem_rdft *) p_;
Chris@10: 
Chris@10:      n = p->sz->dims[0].n + 1;
Chris@10:      buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
Chris@10: 
Chris@10:      cld = X(mkplan_d)(plnr, X(mkproblem_rdft_1_d)(X(mktensor_1d)(n, 1, 1),
Chris@10:                                                    X(mktensor_0d)(),
Chris@10:                                                    buf, buf, R2HC));
Chris@10:      X(ifree)(buf);
Chris@10:      if (!cld)
Chris@10:           return (plan *)0;
Chris@10: 
Chris@10:      pln = MKPLAN_RDFT(P, &padt, apply);
Chris@10: 
Chris@10:      pln->n = n;
Chris@10:      pln->is = p->sz->dims[0].is;
Chris@10:      pln->os = p->sz->dims[0].os;
Chris@10:      pln->cld = cld;
Chris@10:      pln->td = 0;
Chris@10:      
Chris@10:      X(tensor_tornk1)(p->vecsz, &pln->vl, &pln->ivs, &pln->ovs);
Chris@10:      
Chris@10:      X(ops_zero)(&ops);
Chris@10:      ops.other = 4 + (n-1)/2 * 5 + (n-2)/2 * 5;
Chris@10:      ops.add = (n-1)/2 * 4 + (n-2)/2 * 1;
Chris@10:      ops.mul = 1 + (n-1)/2 * 2;
Chris@10:      if (n % 2 == 0)
Chris@10: 	  ops.mul += 1;
Chris@10: 
Chris@10:      X(ops_zero)(&pln->super.super.ops);
Chris@10:      X(ops_madd2)(pln->vl, &ops, &pln->super.super.ops);
Chris@10:      X(ops_madd2)(pln->vl, &cld->ops, &pln->super.super.ops);
Chris@10: 
Chris@10:      return &(pln->super.super);
Chris@10: }
Chris@10: 
Chris@10: /* constructor */
Chris@10: static solver *mksolver(void)
Chris@10: {
Chris@10:      static const solver_adt sadt = { PROBLEM_RDFT, mkplan, 0 };
Chris@10:      S *slv = MKSOLVER(S, &sadt);
Chris@10:      return &(slv->super);
Chris@10: }
Chris@10: 
Chris@10: void X(rodft00e_r2hc_register)(planner *p)
Chris@10: {
Chris@10:      REGISTER_SOLVER(p, mksolver());
Chris@10: }