411 lines
11 KiB
C
411 lines
11 KiB
C
/*
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* Copyright (c) 2003, 2007-14 Matteo Frigo
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* Copyright (c) 2003, 2007-14 Massachusetts Institute of Technology
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*
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* This program is free software; you can redistribute it and/or modify
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* it under the terms of the GNU General Public License as published by
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* the Free Software Foundation; either version 2 of the License, or
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* (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program; if not, write to the Free Software
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* Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
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*
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*/
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/* Do an R{E,O}DFT{01,10} problem via an R2HC problem, with some
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pre/post-processing ala FFTPACK. */
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#include "reodft/reodft.h"
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typedef struct {
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solver super;
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} S;
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typedef struct {
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plan_rdft super;
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plan *cld;
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twid *td;
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INT is, os;
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INT n;
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INT vl;
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INT ivs, ovs;
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rdft_kind kind;
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} P;
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/* A real-even-01 DFT operates logically on a size-4N array:
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I 0 -r(I*) -I 0 r(I*),
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where r denotes reversal and * denotes deletion of the 0th element.
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To compute the transform of this, we imagine performing a radix-4
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(real-input) DIF step, which turns the size-4N DFT into 4 size-N
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(contiguous) DFTs, two of which are zero and two of which are
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conjugates. The non-redundant size-N DFT has halfcomplex input, so
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we can do it with a size-N hc2r transform. (In order to share
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plans with the re10 (inverse) transform, however, we use the DHT
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trick to re-express the hc2r problem as r2hc. This has little cost
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since we are already pre- and post-processing the data in {i,n-i}
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order.) Finally, we have to write out the data in the correct
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order...the two size-N redundant (conjugate) hc2r DFTs correspond
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to the even and odd outputs in O (i.e. the usual interleaved output
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of DIF transforms); since this data has even symmetry, we only
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write the first half of it.
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The real-even-10 DFT is just the reverse of these steps, i.e. a
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radix-4 DIT transform. There, however, we just use the r2hc
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transform naturally without resorting to the DHT trick.
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A real-odd-01 DFT is very similar, except that the input is
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0 I (rI)* 0 -I -(rI)*. This format, however, can be transformed
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into precisely the real-even-01 format above by sending I -> rI
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and shifting the array by N. The former swap is just another
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transformation on the input during preprocessing; the latter
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multiplies the even/odd outputs by i/-i, which combines with
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the factor of -i (to take the imaginary part) to simply flip
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the sign of the odd outputs. Vice-versa for real-odd-10.
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The FFTPACK source code was very helpful in working this out.
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(They do unnecessary passes over the array, though.) The same
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algorithm is also described in:
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John Makhoul, "A fast cosine transform in one and two dimensions,"
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IEEE Trans. on Acoust. Speech and Sig. Proc., ASSP-28 (1), 27--34 (1980).
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Note that Numerical Recipes suggests a different algorithm that
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requires more operations and uses trig. functions for both the pre-
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and post-processing passes.
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*/
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static void apply_re01(const plan *ego_, R *I, R *O)
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{
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const P *ego = (const P *) ego_;
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INT is = ego->is, os = ego->os;
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INT i, n = ego->n;
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INT iv, vl = ego->vl;
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INT ivs = ego->ivs, ovs = ego->ovs;
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R *W = ego->td->W;
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R *buf;
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buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
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for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
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buf[0] = I[0];
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for (i = 1; i < n - i; ++i) {
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E a, b, apb, amb, wa, wb;
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a = I[is * i];
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b = I[is * (n - i)];
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apb = a + b;
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amb = a - b;
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wa = W[2*i];
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wb = W[2*i + 1];
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buf[i] = wa * amb + wb * apb;
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buf[n - i] = wa * apb - wb * amb;
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}
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if (i == n - i) {
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buf[i] = K(2.0) * I[is * i] * W[2*i];
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}
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{
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plan_rdft *cld = (plan_rdft *) ego->cld;
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cld->apply((plan *) cld, buf, buf);
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}
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O[0] = buf[0];
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for (i = 1; i < n - i; ++i) {
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E a, b;
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INT k;
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a = buf[i];
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b = buf[n - i];
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k = i + i;
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O[os * (k - 1)] = a - b;
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O[os * k] = a + b;
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}
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if (i == n - i) {
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O[os * (n - 1)] = buf[i];
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}
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}
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X(ifree)(buf);
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}
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/* ro01 is same as re01, but with i <-> n - 1 - i in the input and
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the sign of the odd output elements flipped. */
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static void apply_ro01(const plan *ego_, R *I, R *O)
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{
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const P *ego = (const P *) ego_;
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INT is = ego->is, os = ego->os;
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INT i, n = ego->n;
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INT iv, vl = ego->vl;
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INT ivs = ego->ivs, ovs = ego->ovs;
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R *W = ego->td->W;
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R *buf;
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buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
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for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
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buf[0] = I[is * (n - 1)];
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for (i = 1; i < n - i; ++i) {
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E a, b, apb, amb, wa, wb;
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a = I[is * (n - 1 - i)];
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b = I[is * (i - 1)];
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apb = a + b;
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amb = a - b;
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wa = W[2*i];
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wb = W[2*i+1];
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buf[i] = wa * amb + wb * apb;
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buf[n - i] = wa * apb - wb * amb;
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}
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if (i == n - i) {
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buf[i] = K(2.0) * I[is * (i - 1)] * W[2*i];
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}
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{
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plan_rdft *cld = (plan_rdft *) ego->cld;
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cld->apply((plan *) cld, buf, buf);
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}
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O[0] = buf[0];
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for (i = 1; i < n - i; ++i) {
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E a, b;
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INT k;
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a = buf[i];
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b = buf[n - i];
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k = i + i;
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O[os * (k - 1)] = b - a;
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O[os * k] = a + b;
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}
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if (i == n - i) {
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O[os * (n - 1)] = -buf[i];
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}
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}
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X(ifree)(buf);
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}
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static void apply_re10(const plan *ego_, R *I, R *O)
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{
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const P *ego = (const P *) ego_;
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INT is = ego->is, os = ego->os;
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INT i, n = ego->n;
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INT iv, vl = ego->vl;
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INT ivs = ego->ivs, ovs = ego->ovs;
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R *W = ego->td->W;
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R *buf;
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buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
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for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
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buf[0] = I[0];
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for (i = 1; i < n - i; ++i) {
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E u, v;
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INT k = i + i;
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u = I[is * (k - 1)];
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v = I[is * k];
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buf[n - i] = u;
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buf[i] = v;
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}
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if (i == n - i) {
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buf[i] = I[is * (n - 1)];
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}
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{
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plan_rdft *cld = (plan_rdft *) ego->cld;
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cld->apply((plan *) cld, buf, buf);
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}
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O[0] = K(2.0) * buf[0];
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for (i = 1; i < n - i; ++i) {
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E a, b, wa, wb;
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a = K(2.0) * buf[i];
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b = K(2.0) * buf[n - i];
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wa = W[2*i];
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wb = W[2*i + 1];
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O[os * i] = wa * a + wb * b;
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O[os * (n - i)] = wb * a - wa * b;
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}
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if (i == n - i) {
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O[os * i] = K(2.0) * buf[i] * W[2*i];
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}
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}
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X(ifree)(buf);
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}
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/* ro10 is same as re10, but with i <-> n - 1 - i in the output and
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the sign of the odd input elements flipped. */
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static void apply_ro10(const plan *ego_, R *I, R *O)
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{
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const P *ego = (const P *) ego_;
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INT is = ego->is, os = ego->os;
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INT i, n = ego->n;
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INT iv, vl = ego->vl;
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INT ivs = ego->ivs, ovs = ego->ovs;
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R *W = ego->td->W;
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R *buf;
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buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
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for (iv = 0; iv < vl; ++iv, I += ivs, O += ovs) {
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buf[0] = I[0];
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for (i = 1; i < n - i; ++i) {
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E u, v;
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INT k = i + i;
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u = -I[is * (k - 1)];
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v = I[is * k];
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buf[n - i] = u;
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buf[i] = v;
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}
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if (i == n - i) {
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buf[i] = -I[is * (n - 1)];
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}
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{
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plan_rdft *cld = (plan_rdft *) ego->cld;
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cld->apply((plan *) cld, buf, buf);
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}
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O[os * (n - 1)] = K(2.0) * buf[0];
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for (i = 1; i < n - i; ++i) {
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E a, b, wa, wb;
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a = K(2.0) * buf[i];
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b = K(2.0) * buf[n - i];
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wa = W[2*i];
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wb = W[2*i + 1];
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O[os * (n - 1 - i)] = wa * a + wb * b;
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O[os * (i - 1)] = wb * a - wa * b;
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}
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if (i == n - i) {
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O[os * (i - 1)] = K(2.0) * buf[i] * W[2*i];
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}
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}
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X(ifree)(buf);
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}
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static void awake(plan *ego_, enum wakefulness wakefulness)
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{
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P *ego = (P *) ego_;
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static const tw_instr reodft010e_tw[] = {
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{ TW_COS, 0, 1 },
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{ TW_SIN, 0, 1 },
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{ TW_NEXT, 1, 0 }
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};
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X(plan_awake)(ego->cld, wakefulness);
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X(twiddle_awake)(wakefulness, &ego->td, reodft010e_tw,
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4*ego->n, 1, ego->n/2+1);
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}
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static void destroy(plan *ego_)
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{
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P *ego = (P *) ego_;
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X(plan_destroy_internal)(ego->cld);
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}
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static void print(const plan *ego_, printer *p)
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{
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const P *ego = (const P *) ego_;
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p->print(p, "(%se-r2hc-%D%v%(%p%))",
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X(rdft_kind_str)(ego->kind), ego->n, ego->vl, ego->cld);
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}
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static int applicable0(const solver *ego_, const problem *p_)
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{
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const problem_rdft *p = (const problem_rdft *) p_;
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UNUSED(ego_);
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return (1
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&& p->sz->rnk == 1
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&& p->vecsz->rnk <= 1
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&& (p->kind[0] == REDFT01 || p->kind[0] == REDFT10
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|| p->kind[0] == RODFT01 || p->kind[0] == RODFT10)
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);
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}
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static int applicable(const solver *ego, const problem *p, const planner *plnr)
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{
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return (!NO_SLOWP(plnr) && applicable0(ego, p));
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}
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static plan *mkplan(const solver *ego_, const problem *p_, planner *plnr)
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{
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P *pln;
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const problem_rdft *p;
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plan *cld;
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R *buf;
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INT n;
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opcnt ops;
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static const plan_adt padt = {
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X(rdft_solve), awake, print, destroy
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};
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if (!applicable(ego_, p_, plnr))
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return (plan *)0;
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p = (const problem_rdft *) p_;
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n = p->sz->dims[0].n;
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buf = (R *) MALLOC(sizeof(R) * n, BUFFERS);
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cld = X(mkplan_d)(plnr, X(mkproblem_rdft_1_d)(X(mktensor_1d)(n, 1, 1),
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X(mktensor_0d)(),
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buf, buf, R2HC));
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X(ifree)(buf);
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if (!cld)
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return (plan *)0;
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switch (p->kind[0]) {
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case REDFT01: pln = MKPLAN_RDFT(P, &padt, apply_re01); break;
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case REDFT10: pln = MKPLAN_RDFT(P, &padt, apply_re10); break;
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case RODFT01: pln = MKPLAN_RDFT(P, &padt, apply_ro01); break;
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case RODFT10: pln = MKPLAN_RDFT(P, &padt, apply_ro10); break;
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default: A(0); return (plan*)0;
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}
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pln->n = n;
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pln->is = p->sz->dims[0].is;
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pln->os = p->sz->dims[0].os;
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pln->cld = cld;
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pln->td = 0;
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pln->kind = p->kind[0];
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X(tensor_tornk1)(p->vecsz, &pln->vl, &pln->ivs, &pln->ovs);
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X(ops_zero)(&ops);
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ops.other = 4 + (n-1)/2 * 10 + (1 - n % 2) * 5;
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if (p->kind[0] == REDFT01 || p->kind[0] == RODFT01) {
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ops.add = (n-1)/2 * 6;
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ops.mul = (n-1)/2 * 4 + (1 - n % 2) * 2;
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}
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else { /* 10 transforms */
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ops.add = (n-1)/2 * 2;
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ops.mul = 1 + (n-1)/2 * 6 + (1 - n % 2) * 2;
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}
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X(ops_zero)(&pln->super.super.ops);
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X(ops_madd2)(pln->vl, &ops, &pln->super.super.ops);
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X(ops_madd2)(pln->vl, &cld->ops, &pln->super.super.ops);
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return &(pln->super.super);
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}
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/* constructor */
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static solver *mksolver(void)
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{
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static const solver_adt sadt = { PROBLEM_RDFT, mkplan, 0 };
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S *slv = MKSOLVER(S, &sadt);
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return &(slv->super);
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}
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void X(reodft010e_r2hc_register)(planner *p)
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{
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REGISTER_SOLVER(p, mksolver());
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}
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