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authorscuri <scuri>2008-10-17 06:10:15 +0000
committerscuri <scuri>2008-10-17 06:10:15 +0000
commit5a422aba704c375a307a902bafe658342e209906 (patch)
tree5005011e086bb863d8fb587ad3319bbec59b2447 /src/fftw3/reodft/rodft00e-r2hc.c
First commit - moving from LuaForge to SourceForge
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+/*
+ * 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());
+}