view Lib/fftw-3.2.1/reodft/reodft010e-r2hc.c @ 1:e86e9c111b29

Updates stuff that potentially fixes the memory leak and also makes it work on Windows and Linux (Need to test). Still have to fix fftw include for linux in Jucer.

/*
 * Copyright (c) 2003, 2007-8 Matteo Frigo
 * Copyright (c) 2003, 2007-8 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
 *
 */


/* Do an R{E,O}DFT{01,10} problem via an R2HC problem, with some
   pre/post-processing ala FFTPACK. */

#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;
     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());
}