annotate src/fftw-3.3.8/reodft/redft00e-r2hc.c @ 82:d0c2a83c1364

Add FFTW 3.3.8 source, and a Linux build
author Chris Cannam
date Tue, 19 Nov 2019 14:52:55 +0000
parents
children
rev   line source
Chris@82 1 /*
Chris@82 2 * Copyright (c) 2003, 2007-14 Matteo Frigo
Chris@82 3 * Copyright (c) 2003, 2007-14 Massachusetts Institute of Technology
Chris@82 4 *
Chris@82 5 * This program is free software; you can redistribute it and/or modify
Chris@82 6 * it under the terms of the GNU General Public License as published by
Chris@82 7 * the Free Software Foundation; either version 2 of the License, or
Chris@82 8 * (at your option) any later version.
Chris@82 9 *
Chris@82 10 * This program is distributed in the hope that it will be useful,
Chris@82 11 * but WITHOUT ANY WARRANTY; without even the implied warranty of
Chris@82 12 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
Chris@82 13 * GNU General Public License for more details.
Chris@82 14 *
Chris@82 15 * You should have received a copy of the GNU General Public License
Chris@82 16 * along with this program; if not, write to the Free Software
Chris@82 17 * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
Chris@82 18 *
Chris@82 19 */
Chris@82 20
Chris@82 21
Chris@82 22 /* Do a REDFT00 problem via an R2HC problem, with some pre/post-processing.
Chris@82 23
Chris@82 24 This code uses the trick from FFTPACK, also documented in a similar
Chris@82 25 form by Numerical Recipes. Unfortunately, this algorithm seems to
Chris@82 26 have intrinsic numerical problems (similar to those in
Chris@82 27 reodft11e-r2hc.c), possibly due to the fact that it multiplies its
Chris@82 28 input by a cosine, causing a loss of precision near the zero. For
Chris@82 29 transforms of 16k points, it has already lost three or four decimal
Chris@82 30 places of accuracy, which we deem unacceptable.
Chris@82 31
Chris@82 32 So, we have abandoned this algorithm in favor of the one in
Chris@82 33 redft00-r2hc-pad.c, which unfortunately sacrifices 30-50% in speed.
Chris@82 34 The only other alternative in the literature that does not have
Chris@82 35 similar numerical difficulties seems to be the direct adaptation of
Chris@82 36 the Cooley-Tukey decomposition for symmetric data, but this would
Chris@82 37 require a whole new set of codelets and it's not clear that it's
Chris@82 38 worth it at this point. However, we did implement the latter
Chris@82 39 algorithm for the specific case of odd n (logically adapting the
Chris@82 40 split-radix algorithm); see reodft00e-splitradix.c. */
Chris@82 41
Chris@82 42 #include "reodft/reodft.h"
Chris@82 43
Chris@82 44 typedef struct {
Chris@82 45 solver super;
Chris@82 46 } S;
Chris@82 47
Chris@82 48 typedef struct {
Chris@82 49 plan_rdft super;
Chris@82 50 plan *cld;
Chris@82 51 twid *td;
Chris@82 52 INT is, os;
Chris@82 53 INT n;
Chris@82 54 INT vl;
Chris@82 55 INT ivs, ovs;
Chris@82 56 } P;
Chris@82 57
Chris@82 58 static void apply(const plan *ego_, R *I, R *O)
Chris@82 59 {
Chris@82 60 const P *ego = (const P *) ego_;
Chris@82 61 INT is = ego->is, os = ego->os;
Chris@82 62 INT i, n = ego->n;
Chris@82 63 INT iv, vl = ego->vl;
Chris@82 64 INT ivs = ego->ivs, ovs = ego->ovs;
Chris@82 65 R *W = ego->td->W;
Chris@82 66 R *buf;
Chris@82 67 E csum;
Chris@82 68
Chris@82 69 buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
Chris@82 70
Chris@82 71 for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
Chris@82 72 buf[0] = I[0] + I[is * n];
Chris@82 73 csum = I[0] - I[is * n];
Chris@82 74 for (i = 1; i < n - i; ++i) {
Chris@82 75 E a, b, apb, amb;
Chris@82 76 a = I[is * i];
Chris@82 77 b = I[is * (n - i)];
Chris@82 78 csum += W[2*i] * (amb = K(2.0)*(a - b));
Chris@82 79 amb = W[2*i+1] * amb;
Chris@82 80 apb = (a + b);
Chris@82 81 buf[i] = apb - amb;
Chris@82 82 buf[n - i] = apb + amb;
Chris@82 83 }
Chris@82 84 if (i == n - i) {
Chris@82 85 buf[i] = K(2.0) * I[is * i];
Chris@82 86 }
Chris@82 87
Chris@82 88 {
Chris@82 89 plan_rdft *cld = (plan_rdft *) ego->cld;
Chris@82 90 cld->apply((plan *) cld, buf, buf);
Chris@82 91 }
Chris@82 92
Chris@82 93 /* FIXME: use recursive/cascade summation for better stability? */
Chris@82 94 O[0] = buf[0];
Chris@82 95 O[os] = csum;
Chris@82 96 for (i = 1; i + i < n; ++i) {
Chris@82 97 INT k = i + i;
Chris@82 98 O[os * k] = buf[i];
Chris@82 99 O[os * (k + 1)] = O[os * (k - 1)] - buf[n - i];
Chris@82 100 }
Chris@82 101 if (i + i == n) {
Chris@82 102 O[os * n] = buf[i];
Chris@82 103 }
Chris@82 104 }
Chris@82 105
Chris@82 106 X(ifree)(buf);
Chris@82 107 }
Chris@82 108
Chris@82 109 static void awake(plan *ego_, enum wakefulness wakefulness)
Chris@82 110 {
Chris@82 111 P *ego = (P *) ego_;
Chris@82 112 static const tw_instr redft00e_tw[] = {
Chris@82 113 { TW_COS, 0, 1 },
Chris@82 114 { TW_SIN, 0, 1 },
Chris@82 115 { TW_NEXT, 1, 0 }
Chris@82 116 };
Chris@82 117
Chris@82 118 X(plan_awake)(ego->cld, wakefulness);
Chris@82 119 X(twiddle_awake)(wakefulness,
Chris@82 120 &ego->td, redft00e_tw, 2*ego->n, 1, (ego->n+1)/2);
Chris@82 121 }
Chris@82 122
Chris@82 123 static void destroy(plan *ego_)
Chris@82 124 {
Chris@82 125 P *ego = (P *) ego_;
Chris@82 126 X(plan_destroy_internal)(ego->cld);
Chris@82 127 }
Chris@82 128
Chris@82 129 static void print(const plan *ego_, printer *p)
Chris@82 130 {
Chris@82 131 const P *ego = (const P *) ego_;
Chris@82 132 p->print(p, "(redft00e-r2hc-%D%v%(%p%))", ego->n + 1, ego->vl, ego->cld);
Chris@82 133 }
Chris@82 134
Chris@82 135 static int applicable0(const solver *ego_, const problem *p_)
Chris@82 136 {
Chris@82 137 const problem_rdft *p = (const problem_rdft *) p_;
Chris@82 138 UNUSED(ego_);
Chris@82 139
Chris@82 140 return (1
Chris@82 141 && p->sz->rnk == 1
Chris@82 142 && p->vecsz->rnk <= 1
Chris@82 143 && p->kind[0] == REDFT00
Chris@82 144 && p->sz->dims[0].n > 1 /* n == 1 is not well-defined */
Chris@82 145 );
Chris@82 146 }
Chris@82 147
Chris@82 148 static int applicable(const solver *ego, const problem *p, const planner *plnr)
Chris@82 149 {
Chris@82 150 return (!NO_SLOWP(plnr) && applicable0(ego, p));
Chris@82 151 }
Chris@82 152
Chris@82 153 static plan *mkplan(const solver *ego_, const problem *p_, planner *plnr)
Chris@82 154 {
Chris@82 155 P *pln;
Chris@82 156 const problem_rdft *p;
Chris@82 157 plan *cld;
Chris@82 158 R *buf;
Chris@82 159 INT n;
Chris@82 160 opcnt ops;
Chris@82 161
Chris@82 162 static const plan_adt padt = {
Chris@82 163 X(rdft_solve), awake, print, destroy
Chris@82 164 };
Chris@82 165
Chris@82 166 if (!applicable(ego_, p_, plnr))
Chris@82 167 return (plan *)0;
Chris@82 168
Chris@82 169 p = (const problem_rdft *) p_;
Chris@82 170
Chris@82 171 n = p->sz->dims[0].n - 1;
Chris@82 172 A(n > 0);
Chris@82 173 buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
Chris@82 174
Chris@82 175 cld = X(mkplan_d)(plnr, X(mkproblem_rdft_1_d)(X(mktensor_1d)(n, 1, 1),
Chris@82 176 X(mktensor_0d)(),
Chris@82 177 buf, buf, R2HC));
Chris@82 178 X(ifree)(buf);
Chris@82 179 if (!cld)
Chris@82 180 return (plan *)0;
Chris@82 181
Chris@82 182 pln = MKPLAN_RDFT(P, &padt, apply);
Chris@82 183
Chris@82 184 pln->n = n;
Chris@82 185 pln->is = p->sz->dims[0].is;
Chris@82 186 pln->os = p->sz->dims[0].os;
Chris@82 187 pln->cld = cld;
Chris@82 188 pln->td = 0;
Chris@82 189
Chris@82 190 X(tensor_tornk1)(p->vecsz, &pln->vl, &pln->ivs, &pln->ovs);
Chris@82 191
Chris@82 192 X(ops_zero)(&ops);
Chris@82 193 ops.other = 8 + (n-1)/2 * 11 + (1 - n % 2) * 5;
Chris@82 194 ops.add = 2 + (n-1)/2 * 5;
Chris@82 195 ops.mul = (n-1)/2 * 3 + (1 - n % 2) * 1;
Chris@82 196
Chris@82 197 X(ops_zero)(&pln->super.super.ops);
Chris@82 198 X(ops_madd2)(pln->vl, &ops, &pln->super.super.ops);
Chris@82 199 X(ops_madd2)(pln->vl, &cld->ops, &pln->super.super.ops);
Chris@82 200
Chris@82 201 return &(pln->super.super);
Chris@82 202 }
Chris@82 203
Chris@82 204 /* constructor */
Chris@82 205 static solver *mksolver(void)
Chris@82 206 {
Chris@82 207 static const solver_adt sadt = { PROBLEM_RDFT, mkplan, 0 };
Chris@82 208 S *slv = MKSOLVER(S, &sadt);
Chris@82 209 return &(slv->super);
Chris@82 210 }
Chris@82 211
Chris@82 212 void X(redft00e_r2hc_register)(planner *p)
Chris@82 213 {
Chris@82 214 REGISTER_SOLVER(p, mksolver());
Chris@82 215 }