annotate src/fftw-3.3.3/reodft/rodft00e-r2hc.c @ 23:619f715526df sv_v2.1

Update Vamp plugin SDK to 2.5
author Chris Cannam
date Thu, 09 May 2013 10:52:46 +0100
parents 37bf6b4a2645
children
rev   line source
Chris@10 1 /*
Chris@10 2 * Copyright (c) 2003, 2007-11 Matteo Frigo
Chris@10 3 * Copyright (c) 2003, 2007-11 Massachusetts Institute of Technology
Chris@10 4 *
Chris@10 5 * This program is free software; you can redistribute it and/or modify
Chris@10 6 * it under the terms of the GNU General Public License as published by
Chris@10 7 * the Free Software Foundation; either version 2 of the License, or
Chris@10 8 * (at your option) any later version.
Chris@10 9 *
Chris@10 10 * This program is distributed in the hope that it will be useful,
Chris@10 11 * but WITHOUT ANY WARRANTY; without even the implied warranty of
Chris@10 12 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
Chris@10 13 * GNU General Public License for more details.
Chris@10 14 *
Chris@10 15 * You should have received a copy of the GNU General Public License
Chris@10 16 * along with this program; if not, write to the Free Software
Chris@10 17 * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
Chris@10 18 *
Chris@10 19 */
Chris@10 20
Chris@10 21
Chris@10 22 /* Do a RODFT00 problem via an R2HC problem, with some pre/post-processing.
Chris@10 23
Chris@10 24 This code uses the trick from FFTPACK, also documented in a similar
Chris@10 25 form by Numerical Recipes. Unfortunately, this algorithm seems to
Chris@10 26 have intrinsic numerical problems (similar to those in
Chris@10 27 reodft11e-r2hc.c), possibly due to the fact that it multiplies its
Chris@10 28 input by a sine, causing a loss of precision near the zero. For
Chris@10 29 transforms of 16k points, it has already lost three or four decimal
Chris@10 30 places of accuracy, which we deem unacceptable.
Chris@10 31
Chris@10 32 So, we have abandoned this algorithm in favor of the one in
Chris@10 33 rodft00-r2hc-pad.c, which unfortunately sacrifices 30-50% in speed.
Chris@10 34 The only other alternative in the literature that does not have
Chris@10 35 similar numerical difficulties seems to be the direct adaptation of
Chris@10 36 the Cooley-Tukey decomposition for antisymmetric data, but this
Chris@10 37 would require a whole new set of codelets and it's not clear that
Chris@10 38 it's worth it at this point. However, we did implement the latter
Chris@10 39 algorithm for the specific case of odd n (logically adapting the
Chris@10 40 split-radix algorithm); see reodft00e-splitradix.c. */
Chris@10 41
Chris@10 42 #include "reodft.h"
Chris@10 43
Chris@10 44 typedef struct {
Chris@10 45 solver super;
Chris@10 46 } S;
Chris@10 47
Chris@10 48 typedef struct {
Chris@10 49 plan_rdft super;
Chris@10 50 plan *cld;
Chris@10 51 twid *td;
Chris@10 52 INT is, os;
Chris@10 53 INT n;
Chris@10 54 INT vl;
Chris@10 55 INT ivs, ovs;
Chris@10 56 } P;
Chris@10 57
Chris@10 58 static void apply(const plan *ego_, R *I, R *O)
Chris@10 59 {
Chris@10 60 const P *ego = (const P *) ego_;
Chris@10 61 INT is = ego->is, os = ego->os;
Chris@10 62 INT i, n = ego->n;
Chris@10 63 INT iv, vl = ego->vl;
Chris@10 64 INT ivs = ego->ivs, ovs = ego->ovs;
Chris@10 65 R *W = ego->td->W;
Chris@10 66 R *buf;
Chris@10 67
Chris@10 68 buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
Chris@10 69
Chris@10 70 for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
Chris@10 71 buf[0] = 0;
Chris@10 72 for (i = 1; i < n - i; ++i) {
Chris@10 73 E a, b, apb, amb;
Chris@10 74 a = I[is * (i - 1)];
Chris@10 75 b = I[is * ((n - i) - 1)];
Chris@10 76 apb = K(2.0) * W[i] * (a + b);
Chris@10 77 amb = (a - b);
Chris@10 78 buf[i] = apb + amb;
Chris@10 79 buf[n - i] = apb - amb;
Chris@10 80 }
Chris@10 81 if (i == n - i) {
Chris@10 82 buf[i] = K(4.0) * I[is * (i - 1)];
Chris@10 83 }
Chris@10 84
Chris@10 85 {
Chris@10 86 plan_rdft *cld = (plan_rdft *) ego->cld;
Chris@10 87 cld->apply((plan *) cld, buf, buf);
Chris@10 88 }
Chris@10 89
Chris@10 90 /* FIXME: use recursive/cascade summation for better stability? */
Chris@10 91 O[0] = buf[0] * 0.5;
Chris@10 92 for (i = 1; i + i < n - 1; ++i) {
Chris@10 93 INT k = i + i;
Chris@10 94 O[os * (k - 1)] = -buf[n - i];
Chris@10 95 O[os * k] = O[os * (k - 2)] + buf[i];
Chris@10 96 }
Chris@10 97 if (i + i == n - 1) {
Chris@10 98 O[os * (n - 2)] = -buf[n - i];
Chris@10 99 }
Chris@10 100 }
Chris@10 101
Chris@10 102 X(ifree)(buf);
Chris@10 103 }
Chris@10 104
Chris@10 105 static void awake(plan *ego_, enum wakefulness wakefulness)
Chris@10 106 {
Chris@10 107 P *ego = (P *) ego_;
Chris@10 108 static const tw_instr rodft00e_tw[] = {
Chris@10 109 { TW_SIN, 0, 1 },
Chris@10 110 { TW_NEXT, 1, 0 }
Chris@10 111 };
Chris@10 112
Chris@10 113 X(plan_awake)(ego->cld, wakefulness);
Chris@10 114
Chris@10 115 X(twiddle_awake)(wakefulness,
Chris@10 116 &ego->td, rodft00e_tw, 2*ego->n, 1, (ego->n+1)/2);
Chris@10 117 }
Chris@10 118
Chris@10 119 static void destroy(plan *ego_)
Chris@10 120 {
Chris@10 121 P *ego = (P *) ego_;
Chris@10 122 X(plan_destroy_internal)(ego->cld);
Chris@10 123 }
Chris@10 124
Chris@10 125 static void print(const plan *ego_, printer *p)
Chris@10 126 {
Chris@10 127 const P *ego = (const P *) ego_;
Chris@10 128 p->print(p, "(rodft00e-r2hc-%D%v%(%p%))", ego->n - 1, ego->vl, ego->cld);
Chris@10 129 }
Chris@10 130
Chris@10 131 static int applicable0(const solver *ego_, const problem *p_)
Chris@10 132 {
Chris@10 133 const problem_rdft *p = (const problem_rdft *) p_;
Chris@10 134 UNUSED(ego_);
Chris@10 135
Chris@10 136 return (1
Chris@10 137 && p->sz->rnk == 1
Chris@10 138 && p->vecsz->rnk <= 1
Chris@10 139 && p->kind[0] == RODFT00
Chris@10 140 );
Chris@10 141 }
Chris@10 142
Chris@10 143 static int applicable(const solver *ego, const problem *p, const planner *plnr)
Chris@10 144 {
Chris@10 145 return (!NO_SLOWP(plnr) && applicable0(ego, p));
Chris@10 146 }
Chris@10 147
Chris@10 148 static plan *mkplan(const solver *ego_, const problem *p_, planner *plnr)
Chris@10 149 {
Chris@10 150 P *pln;
Chris@10 151 const problem_rdft *p;
Chris@10 152 plan *cld;
Chris@10 153 R *buf;
Chris@10 154 INT n;
Chris@10 155 opcnt ops;
Chris@10 156
Chris@10 157 static const plan_adt padt = {
Chris@10 158 X(rdft_solve), awake, print, destroy
Chris@10 159 };
Chris@10 160
Chris@10 161 if (!applicable(ego_, p_, plnr))
Chris@10 162 return (plan *)0;
Chris@10 163
Chris@10 164 p = (const problem_rdft *) p_;
Chris@10 165
Chris@10 166 n = p->sz->dims[0].n + 1;
Chris@10 167 buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
Chris@10 168
Chris@10 169 cld = X(mkplan_d)(plnr, X(mkproblem_rdft_1_d)(X(mktensor_1d)(n, 1, 1),
Chris@10 170 X(mktensor_0d)(),
Chris@10 171 buf, buf, R2HC));
Chris@10 172 X(ifree)(buf);
Chris@10 173 if (!cld)
Chris@10 174 return (plan *)0;
Chris@10 175
Chris@10 176 pln = MKPLAN_RDFT(P, &padt, apply);
Chris@10 177
Chris@10 178 pln->n = n;
Chris@10 179 pln->is = p->sz->dims[0].is;
Chris@10 180 pln->os = p->sz->dims[0].os;
Chris@10 181 pln->cld = cld;
Chris@10 182 pln->td = 0;
Chris@10 183
Chris@10 184 X(tensor_tornk1)(p->vecsz, &pln->vl, &pln->ivs, &pln->ovs);
Chris@10 185
Chris@10 186 X(ops_zero)(&ops);
Chris@10 187 ops.other = 4 + (n-1)/2 * 5 + (n-2)/2 * 5;
Chris@10 188 ops.add = (n-1)/2 * 4 + (n-2)/2 * 1;
Chris@10 189 ops.mul = 1 + (n-1)/2 * 2;
Chris@10 190 if (n % 2 == 0)
Chris@10 191 ops.mul += 1;
Chris@10 192
Chris@10 193 X(ops_zero)(&pln->super.super.ops);
Chris@10 194 X(ops_madd2)(pln->vl, &ops, &pln->super.super.ops);
Chris@10 195 X(ops_madd2)(pln->vl, &cld->ops, &pln->super.super.ops);
Chris@10 196
Chris@10 197 return &(pln->super.super);
Chris@10 198 }
Chris@10 199
Chris@10 200 /* constructor */
Chris@10 201 static solver *mksolver(void)
Chris@10 202 {
Chris@10 203 static const solver_adt sadt = { PROBLEM_RDFT, mkplan, 0 };
Chris@10 204 S *slv = MKSOLVER(S, &sadt);
Chris@10 205 return &(slv->super);
Chris@10 206 }
Chris@10 207
Chris@10 208 void X(rodft00e_r2hc_register)(planner *p)
Chris@10 209 {
Chris@10 210 REGISTER_SOLVER(p, mksolver());
Chris@10 211 }