/* * Copyright (c) 2003 Matteo Frigo * Copyright (c) 2003 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 * */ /* $Id: reodft010e-r2hc.c,v 1.1 2008/10/17 06:13:18 scuri Exp $ */ /* 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.) 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_, int flg) { P *ego = (P *) ego_; static const tw_instr reodft010e_tw[] = { { TW_COS, 0, 1 }, { TW_SIN, 0, 1 }, { TW_NEXT, 1, 0 } }; AWAKE(ego->cld, flg); X(twiddle_awake)(flg, &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_) { UNUSED(ego_); if (RDFTP(p_)) { const problem_rdft *p = (const problem_rdft *) p_; 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) ); } return 0; } static int applicable(const solver *ego, const problem *p, const planner *plnr) { return (!NO_UGLYP(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 = { mkplan }; S *slv = MKSOLVER(S, &sadt); return &(slv->super); } void X(reodft010e_r2hc_register)(planner *p) { REGISTER_SOLVER(p, mksolver()); }