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-rw-r--r--src/fftw3/reodft/rodft00e-r2hc.c212
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diff --git a/src/fftw3/reodft/rodft00e-r2hc.c b/src/fftw3/reodft/rodft00e-r2hc.c
deleted file mode 100644
index 46bb299..0000000
--- a/src/fftw3/reodft/rodft00e-r2hc.c
+++ /dev/null
@@ -1,212 +0,0 @@
-/*
- * Copyright (c) 2003 Matteo Frigo
- * Copyright (c) 2003 Massachusetts Institute of Technology
- *
- * This program is free software; you can redistribute it and/or modify
- * it under the terms of the GNU General Public License as published by
- * the Free Software Foundation; either version 2 of the License, or
- * (at your option) any later version.
- *
- * This program 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 General Public License for more details.
- *
- * You should have received a copy of the GNU General Public License
- * along with this program; if not, write to the Free Software
- * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
- *
- */
-
-/* $Id: rodft00e-r2hc.c,v 1.1 2008/10/17 06:13:18 scuri Exp $ */
-
-/* Do a RODFT00 problem via an R2HC problem, with some pre/post-processing.
-
- This code uses the trick from FFTPACK, also documented in a similar
- form by Numerical Recipes. Unfortunately, this algorithm seems to
- have intrinsic numerical problems (similar to those in
- reodft11e-r2hc.c), possibly due to the fact that it multiplies its
- input by a sine, causing a loss of precision near the zero. For
- transforms of 16k points, it has already lost three or four decimal
- places of accuracy, which we deem unacceptable.
-
- So, we have abandoned this algorithm in favor of the one in
- rodft00-r2hc-pad.c, which unfortunately sacrifices 30-50% in speed.
- The only other alternative in the literature that does not have
- similar numerical difficulties seems to be the direct adaptation of
- the Cooley-Tukey decomposition for antisymmetric data, but this
- would require a whole new set of codelets and it's not clear that
- it's worth it at this point. */
-
-#include "reodft.h"
-
-typedef struct {
- solver super;
-} S;
-
-typedef struct {
- plan_rdft super;
- plan *cld;
- twid *td;
- int is, os;
- int n;
- int vl;
- int ivs, ovs;
-} P;
-
-static void apply(const plan *ego_, R *I, R *O)
-{
- const P *ego = (const P *) ego_;
- int is = ego->is, os = ego->os;
- int i, n = ego->n;
- int iv, vl = ego->vl;
- int ivs = ego->ivs, ovs = ego->ovs;
- R *W = ego->td->W;
- R *buf;
-
- buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
-
- for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
- buf[0] = 0;
- for (i = 1; i < n - i; ++i) {
- E a, b, apb, amb;
- a = I[is * (i - 1)];
- b = I[is * ((n - i) - 1)];
- apb = K(2.0) * W[i] * (a + b);
- amb = (a - b);
- buf[i] = apb + amb;
- buf[n - i] = apb - amb;
- }
- if (i == n - i) {
- buf[i] = K(4.0) * I[is * (i - 1)];
- }
-
- {
- plan_rdft *cld = (plan_rdft *) ego->cld;
- cld->apply((plan *) cld, buf, buf);
- }
-
- /* FIXME: use recursive/cascade summation for better stability? */
- O[0] = buf[0] * 0.5;
- for (i = 1; i + i < n - 1; ++i) {
- int k = i + i;
- O[os * (k - 1)] = -buf[n - i];
- O[os * k] = O[os * (k - 2)] + buf[i];
- }
- if (i + i == n - 1) {
- O[os * (n - 2)] = -buf[n - i];
- }
- }
-
- X(ifree)(buf);
-}
-
-static void awake(plan *ego_, int flg)
-{
- P *ego = (P *) ego_;
- static const tw_instr rodft00e_tw[] = {
- { TW_SIN, 0, 1 },
- { TW_NEXT, 1, 0 }
- };
-
- AWAKE(ego->cld, flg);
-
- X(twiddle_awake)(flg, &ego->td, rodft00e_tw, 2*ego->n, 1, (ego->n+1)/2);
-}
-
-static void destroy(plan *ego_)
-{
- P *ego = (P *) ego_;
- X(plan_destroy_internal)(ego->cld);
-}
-
-static void print(const plan *ego_, printer *p)
-{
- const P *ego = (const P *) ego_;
- p->print(p, "(rodft00e-r2hc-%d%v%(%p%))", ego->n - 1, ego->vl, ego->cld);
-}
-
-static int applicable0(const solver *ego_, const problem *p_)
-{
- UNUSED(ego_);
- if (RDFTP(p_)) {
- const problem_rdft *p = (const problem_rdft *) p_;
- return (1
- && p->sz->rnk == 1
- && p->vecsz->rnk <= 1
- && p->kind[0] == RODFT00
- );
- }
-
- return 0;
-}
-
-static int applicable(const solver *ego, const problem *p, const planner *plnr)
-{
- return (!NO_UGLYP(plnr) && applicable0(ego, p));
-}
-
-static plan *mkplan(const solver *ego_, const problem *p_, planner *plnr)
-{
- P *pln;
- const problem_rdft *p;
- plan *cld;
- R *buf;
- int n;
- opcnt ops;
-
- static const plan_adt padt = {
- X(rdft_solve), awake, print, destroy
- };
-
- if (!applicable(ego_, p_, plnr))
- return (plan *)0;
-
- p = (const problem_rdft *) p_;
-
- n = p->sz->dims[0].n + 1;
- buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
-
- cld = X(mkplan_d)(plnr, X(mkproblem_rdft_1_d)(X(mktensor_1d)(n, 1, 1),
- X(mktensor_0d)(),
- buf, buf, R2HC));
- X(ifree)(buf);
- if (!cld)
- return (plan *)0;
-
- pln = MKPLAN_RDFT(P, &padt, apply);
-
- pln->n = n;
- pln->is = p->sz->dims[0].is;
- pln->os = p->sz->dims[0].os;
- pln->cld = cld;
- pln->td = 0;
-
- X(tensor_tornk1)(p->vecsz, &pln->vl, &pln->ivs, &pln->ovs);
-
- X(ops_zero)(&ops);
- ops.other = 4 + (n-1)/2 * 5 + (n-2)/2 * 5;
- ops.add = (n-1)/2 * 4 + (n-2)/2 * 1;
- ops.mul = 1 + (n-1)/2 * 2;
- if (n % 2 == 0)
- ops.mul += 1;
-
- X(ops_zero)(&pln->super.super.ops);
- X(ops_madd2)(pln->vl, &ops, &pln->super.super.ops);
- X(ops_madd2)(pln->vl, &cld->ops, &pln->super.super.ops);
-
- return &(pln->super.super);
-}
-
-/* constructor */
-static solver *mksolver(void)
-{
- static const solver_adt sadt = { mkplan };
- S *slv = MKSOLVER(S, &sadt);
- return &(slv->super);
-}
-
-void X(rodft00e_r2hc_register)(planner *p)
-{
- REGISTER_SOLVER(p, mksolver());
-}
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