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