diff src/fftw-3.3.3/reodft/reodft010e-r2hc.c @ 10:37bf6b4a2645

Add FFTW3

--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/fftw-3.3.3/reodft/reodft010e-r2hc.c	Wed Mar 20 15:35:50 2013 +0000
@@ -0,0 +1,410 @@
+/*
+ * Copyright (c) 2003, 2007-11 Matteo Frigo
+ * Copyright (c) 2003, 2007-11 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.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());
+}