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furnace/extern/fftw/reodft/reodft010e-r2hc.c
tildearrow 54e93db207 GUI: try using FFTW for per-chan osc wave center 
not reliable yet
2022-05-31 03:24:29 -05:00

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/*
* Copyright (c) 2003, 2007-14 Matteo Frigo
* Copyright (c) 2003, 2007-14 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., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
*
*/
/* Do an R{E,O}DFT{01,10} problem via an R2HC problem, with some
pre/post-processing ala FFTPACK. */
#include "reodft/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;
rdft_kind kind;
} P;
/* A real-even-01 DFT operates logically on a size-4N array:
I 0 -r(I*) -I 0 r(I*),
where r denotes reversal and * denotes deletion of the 0th element.
To compute the transform of this, we imagine performing a radix-4
(real-input) DIF step, which turns the size-4N DFT into 4 size-N
(contiguous) DFTs, two of which are zero and two of which are
conjugates. The non-redundant size-N DFT has halfcomplex input, so
we can do it with a size-N hc2r transform. (In order to share
plans with the re10 (inverse) transform, however, we use the DHT
trick to re-express the hc2r problem as r2hc. This has little cost
since we are already pre- and post-processing the data in {i,n-i}
order.) Finally, we have to write out the data in the correct
order...the two size-N redundant (conjugate) hc2r DFTs correspond
to the even and odd outputs in O (i.e. the usual interleaved output
of DIF transforms); since this data has even symmetry, we only
write the first half of it.
The real-even-10 DFT is just the reverse of these steps, i.e. a
radix-4 DIT transform. There, however, we just use the r2hc
transform naturally without resorting to the DHT trick.
A real-odd-01 DFT is very similar, except that the input is
0 I (rI)* 0 -I -(rI)*. This format, however, can be transformed
into precisely the real-even-01 format above by sending I -> rI
and shifting the array by N. The former swap is just another
transformation on the input during preprocessing; the latter
multiplies the even/odd outputs by i/-i, which combines with
the factor of -i (to take the imaginary part) to simply flip
the sign of the odd outputs. Vice-versa for real-odd-10.
The FFTPACK source code was very helpful in working this out.
(They do unnecessary passes over the array, though.) The same
algorithm is also described in:
John Makhoul, "A fast cosine transform in one and two dimensions,"
IEEE Trans. on Acoust. Speech and Sig. Proc., ASSP-28 (1), 27--34 (1980).
Note that Numerical Recipes suggests a different algorithm that
requires more operations and uses trig. functions for both the pre-
and post-processing passes.
*/
static void apply_re01(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] = I[0];
for (i = 1; i < n - i; ++i) {
E a, b, apb, amb, wa, wb;
a = I[is * i];
b = I[is * (n - i)];
apb = a + b;
amb = a - b;
wa = W[2*i];
wb = W[2*i + 1];
buf[i] = wa * amb + wb * apb;
buf[n - i] = wa * apb - wb * amb;
}
if (i == n - i) {
buf[i] = K(2.0) * I[is * i] * W[2*i];
}
{
plan_rdft *cld = (plan_rdft *) ego->cld;
cld->apply((plan *) cld, buf, buf);
}
O[0] = buf[0];
for (i = 1; i < n - i; ++i) {
E a, b;
INT k;
a = buf[i];
b = buf[n - i];
k = i + i;
O[os * (k - 1)] = a - b;
O[os * k] = a + b;
}
if (i == n - i) {
O[os * (n - 1)] = buf[i];
}
}
X(ifree)(buf);
}
/* ro01 is same as re01, but with i <-> n - 1 - i in the input and
the sign of the odd output elements flipped. */
static void apply_ro01(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] = I[is * (n - 1)];
for (i = 1; i < n - i; ++i) {
E a, b, apb, amb, wa, wb;
a = I[is * (n - 1 - i)];
b = I[is * (i - 1)];
apb = a + b;
amb = a - b;
wa = W[2*i];
wb = W[2*i+1];
buf[i] = wa * amb + wb * apb;
buf[n - i] = wa * apb - wb * amb;
}
if (i == n - i) {
buf[i] = K(2.0) * I[is * (i - 1)] * W[2*i];
}
{
plan_rdft *cld = (plan_rdft *) ego->cld;
cld->apply((plan *) cld, buf, buf);
}
O[0] = buf[0];
for (i = 1; i < n - i; ++i) {
E a, b;
INT k;
a = buf[i];
b = buf[n - i];
k = i + i;
O[os * (k - 1)] = b - a;
O[os * k] = a + b;
}
if (i == n - i) {
O[os * (n - 1)] = -buf[i];
}
}
X(ifree)(buf);
}
static void apply_re10(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] = I[0];
for (i = 1; i < n - i; ++i) {
E u, v;
INT k = i + i;
u = I[is * (k - 1)];
v = I[is * k];
buf[n - i] = u;
buf[i] = v;
}
if (i == n - i) {
buf[i] = I[is * (n - 1)];
}
{
plan_rdft *cld = (plan_rdft *) ego->cld;
cld->apply((plan *) cld, buf, buf);
}
O[0] = K(2.0) * buf[0];
for (i = 1; i < n - i; ++i) {
E a, b, wa, wb;
a = K(2.0) * buf[i];
b = K(2.0) * buf[n - i];
wa = W[2*i];
wb = W[2*i + 1];
O[os * i] = wa * a + wb * b;
O[os * (n - i)] = wb * a - wa * b;
}
if (i == n - i) {
O[os * i] = K(2.0) * buf[i] * W[2*i];
}
}
X(ifree)(buf);
}
/* ro10 is same as re10, but with i <-> n - 1 - i in the output and
the sign of the odd input elements flipped. */
static void apply_ro10(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] = I[0];
for (i = 1; i < n - i; ++i) {
E u, v;
INT k = i + i;
u = -I[is * (k - 1)];
v = I[is * k];
buf[n - i] = u;
buf[i] = v;
}
if (i == n - i) {
buf[i] = -I[is * (n - 1)];
}
{
plan_rdft *cld = (plan_rdft *) ego->cld;
cld->apply((plan *) cld, buf, buf);
}
O[os * (n - 1)] = K(2.0) * buf[0];
for (i = 1; i < n - i; ++i) {
E a, b, wa, wb;
a = K(2.0) * buf[i];
b = K(2.0) * buf[n - i];
wa = W[2*i];
wb = W[2*i + 1];
O[os * (n - 1 - i)] = wa * a + wb * b;
O[os * (i - 1)] = wb * a - wa * b;
}
if (i == n - i) {
O[os * (i - 1)] = K(2.0) * buf[i] * W[2*i];
}
}
X(ifree)(buf);
}
static void awake(plan *ego_, enum wakefulness wakefulness)
{
P *ego = (P *) ego_;
static const tw_instr reodft010e_tw[] = {
{ TW_COS, 0, 1 },
{ TW_SIN, 0, 1 },
{ TW_NEXT, 1, 0 }
};
X(plan_awake)(ego->cld, wakefulness);
X(twiddle_awake)(wakefulness, &ego->td, reodft010e_tw,
4*ego->n, 1, ego->n/2+1);
}
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, "(%se-r2hc-%D%v%(%p%))",
X(rdft_kind_str)(ego->kind), ego->n, ego->vl, ego->cld);
}
static int applicable0(const solver *ego_, const problem *p_)
{
const problem_rdft *p = (const problem_rdft *) p_;
UNUSED(ego_);
return (1
&& p->sz->rnk == 1
&& p->vecsz->rnk <= 1
&& (p->kind[0] == REDFT01 || p->kind[0] == REDFT10
|| p->kind[0] == RODFT01 || p->kind[0] == RODFT10)
);
}
static int applicable(const solver *ego, const problem *p, const planner *plnr)
{
return (!NO_SLOWP(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;
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;
switch (p->kind[0]) {
case REDFT01: pln = MKPLAN_RDFT(P, &padt, apply_re01); break;
case REDFT10: pln = MKPLAN_RDFT(P, &padt, apply_re10); break;
case RODFT01: pln = MKPLAN_RDFT(P, &padt, apply_ro01); break;
case RODFT10: pln = MKPLAN_RDFT(P, &padt, apply_ro10); break;
default: A(0); return (plan*)0;
}
pln->n = n;
pln->is = p->sz->dims[0].is;
pln->os = p->sz->dims[0].os;
pln->cld = cld;
pln->td = 0;
pln->kind = p->kind[0];
X(tensor_tornk1)(p->vecsz, &pln->vl, &pln->ivs, &pln->ovs);
X(ops_zero)(&ops);
ops.other = 4 + (n-1)/2 * 10 + (1 - n % 2) * 5;
if (p->kind[0] == REDFT01 || p->kind[0] == RODFT01) {
ops.add = (n-1)/2 * 6;
ops.mul = (n-1)/2 * 4 + (1 - n % 2) * 2;
}
else { /* 10 transforms */
ops.add = (n-1)/2 * 2;
ops.mul = 1 + (n-1)/2 * 6 + (1 - n % 2) * 2;
}
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 = { PROBLEM_RDFT, mkplan, 0 };
S *slv = MKSOLVER(S, &sadt);
return &(slv->super);
}
void X(reodft010e_r2hc_register)(planner *p)
{
REGISTER_SOLVER(p, mksolver());
}
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