/*

 * 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());

}