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Diffstat (limited to 'src/fftw3/reodft/rodft00e-r2hc.c')
| -rw-r--r-- | src/fftw3/reodft/rodft00e-r2hc.c | 212 |
1 files changed, 212 insertions, 0 deletions
diff --git a/src/fftw3/reodft/rodft00e-r2hc.c b/src/fftw3/reodft/rodft00e-r2hc.c new file mode 100644 index 0000000..46bb299 --- /dev/null +++ b/src/fftw3/reodft/rodft00e-r2hc.c @@ -0,0 +1,212 @@ +/* + * 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()); +} |