/* Karatsuba convolution
 *
 *  Copyright (C) 1999 Ralph Loader <suckfish@ihug.co.nz>
 *
 *  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.  */

/* The algorithm is based on the following.  For the convolution of a pair
 * of pairs, (a,b) * (c,d) = (0, a.c, a.d+b.c, b.d), we can reduce the four
 * multiplications to three, by the formulae a.d+b.c = (a+b).(c+d) - a.c -
 * b.d.  A similar relation enables us to compute a 2n by 2n convolution
 * using 3 n by n convolutions, and thus a 2^n by 2^n convolution using 3^n
 * multiplications (as opposed to the 4^n that the quadratic algorithm
 * takes. */

/* For large n, this is slower than the O(n log n) that the FFT method
 * takes, but we avoid using complex numbers, and we only have to compute
 * one convolution, as opposed to 3 FFTs.  We have good locality-of-
 * reference as well, which will help on CPUs with tiny caches.  */

/* E.g., for a 512 x 512 convolution, the FFT method takes 55 * 512 = 28160
 * (real) multiplications, as opposed to 3^9 = 19683 for the Karatsuba
 * algorithm.  We actually want 257 outputs of a 256 x 512 convolution;
 * that doesn't appear to give an easy advantage for the FFT algorithm, but
 * for the Karatsuba algorithm, it's easy to use two 256 x 256
 * convolutions, taking 2 x 3^8 = 12312 multiplications.  [This difference
 * is that the FFT method "wraps" the arrays, doing a 2^n x 2^n -> 2^n,
 * while the Karatsuba algorithm pads with zeros, doing 2^n x 2^n -> 2.2^n
 * - 1]. */

/* There's a big lie above, actually... for a 4x4 convolution, it's quicker
 * to do it using 16 multiplications than the more complex Karatsuba
 * algorithm...  So the recursion bottoms out at 4x4s.  This increases the
 * number of multiplications by a factor of 16/9, but reduces the overheads
 * dramatically. */

/* The convolution algorithm is implemented as a stack machine.  We have a
 * stack of commands, each in one of the forms "do a 2^n x 2^n
 * convolution", or "combine these three length 2^n outputs into one
 * 2^{n+1} output." */

#include <stdlib.h>
#include "convolve.h"

/*
 * Initialisation routine - sets up tables and space to work in.
 * Returns a pointer to internal state, to be used when performing calls.
 * On error, returns NULL.
 * The pointer should be freed when it is finished with, by convolve_close().
 */
convolve_state *convolve_init(void)
{
	return (convolve_state *) malloc (sizeof(convolve_state));
}

/*
 * Free the state allocated with convolve_init().
 */
void convolve_close(convolve_state *state)
{
	if (state)
		free(state);
}

static void convolve_4 (double * out, const double * left, const double * right)
/* This does a 4x4 -> 7 convolution.  For what it's worth, the slightly odd
 * ordering gives about a 1% speed up on my Pentium II. */
{
	double l0, l1, l2, l3, r0, r1, r2, r3;
	double a;
	l0 = left[0];
	r0 = right[0];
	a = l0 * r0;
	l1 = left[1];
	r1 = right[1];
	out[0] = a;
	a = (l0 * r1) + (l1 * r0);
	l2 = left[2];
	r2 = right[2];
	out[1] = a;
	a = (l0 * r2) + (l1 * r1) + (l2 * r0);
	l3 = left[3];
	r3 = right[3];
	out[2] = a;

	out[3] = (l0 * r3) + (l1 * r2) + (l2 * r1) + (l3 * r0);
	out[4] = (l1 * r3) + (l2 * r2) + (l3 * r1);
	out[5] = (l2 * r3) + (l3 * r2);
	out[6] = l3 * r3;
}

static void convolve_run (stack_entry * top, unsigned size, double * scratch)
/* Interpret a stack of commands.  The stack starts with two entries; the
 * convolution to do, and an illegal entry used to mark the stack top.  The
 * size is the number of entries in each input, and must be a power of 2,
 * and at least 8.  It is OK to have out equal to left and/or right.
 * scratch must have length 3*size.  The number of stack entries needed is
 * 3n-4 where size=2^n. */
{
	do {
		const double * left;
		const double * right;
		double * out;

		/* When we get here, the stack top is always a convolve,
		 * with size > 4.  So we will split it.  We repeatedly split
		 * the top entry until we get to size = 4. */
			
		left = top->v.left;
		right = top->v.right;
		out = top->v.out;
		top++;

		do {
			double * s_left, * s_right;
			int i;

			/* Halve the size. */
			size >>= 1;

			/* Allocate the scratch areas. */
			s_left = scratch + size * 3;
			/* s_right is a length 2*size buffer also used for
			 * intermediate output. */
			s_right = scratch + size * 4;

			/* Create the intermediate factors. */
			for (i = 0; i < size; i++) {
				double l = left[i] + left[i + size];
				double r = right[i] + right[i + size];
				s_left[i + size] = r;
				s_left[i] = l;
			}
			
			/* Push the combine entry onto the stack. */
			top -= 3;
			top[2].b.main = out;
			top[2].b.null = NULL;

			/* Push the low entry onto the stack.  This must be
			 * the last of the three sub-convolutions, because
			 * it may overwrite the arguments. */
			top[1].v.left = left;
			top[1].v.right = right;
			top[1].v.out = out;

			/* Push the mid entry onto the stack. */
			top[0].v.left = s_left;
			top[0].v.right = s_right;
			top[0].v.out = s_right;

			/* Leave the high entry in variables. */
			left += size;
			right += size;
			out += size * 2;

		} while (size > 4);

		/* When we get here, the stack top is a group of 3
		 * convolves, with size = 4, followed by some combines.  */
		convolve_4 (out, left, right);
		convolve_4 (top[0].v.out, top[0].v.left, top[0].v.right);
		convolve_4 (top[1].v.out, top[1].v.left, top[1].v.right);
		top += 2;

		/* Now process combines. */
		do {
			/* b.main is the output buffer, mid is the middle
			 * part which needs to be adjusted in place, and
			 * then folded back into the output.  We do this in
			 * a slightly strange way, so as to avoid having
			 * two loops. */
			double * out = top->b.main;
			double * mid = scratch + size * 4;
			unsigned int i;
			top++;
			out[size * 2 - 1] = 0;
			for (i = 0; i < size-1; i++) {
				double lo;
				double hi;
				lo = mid[0] - (out[0] + out[2 * size]) + out[size];
				hi = mid[size] - (out[size] + out[3 * size]) + out[2 * size];
				out[size] = lo;
				out[2 * size] = hi;
				out++;
				mid++;
			}
			size <<= 1;
		} while (top->b.null == NULL);
	} while (top->b.main != NULL);
}

int convolve_match (float * lastchoice,
		    float * input,
		    convolve_state * state)
/* lastchoice is a 256 sized array.  input is a 512 array.  We find the
 * contiguous length 256 sub-array of input that best matches lastchoice.
 * A measure of how good a sub-array is compared with the lastchoice is
 * given by the sum of the products of each pair of entries.  We maximise
 * that, by taking an appropriate convolution, and then finding the maximum
 * entry in the convolutions.  state is a (non-NULL) pointer returned by
 * convolve_init.  */
{
	double avg;
	double best;
	int p;
	int i;
	double * left = state->left;
	double * right = state->right;
	double * scratch = state->scratch;
	stack_entry * top = state->stack + STACK_SIZE - 1;

	for (i=0; i<512; i++)
		left[i]=input[i];

	avg = 0;
    for (i = 0; i < 256; i++)
	{
		double a = lastchoice[255 - i];
		right[i] = a;
		avg += a;
	}

	/* We adjust the smaller of the two input arrays to have average
	 * value 0.  This makes the eventual result insensitive to both
	 * constant offsets and positive multipliers of the inputs. */
	avg /= 256;
	for (i = 0; i < 256; i++)
		right[i] -= avg;

	/* End-of-stack marker. */
	top[1].b.null = scratch;
	top[1].b.main = NULL;

	/* The low 256x256, of which we want the high 256 outputs. */
	top->v.left = left;
	top->v.right = right;
	top->v.out = right + 256;
	convolve_run (top, 256, scratch);
	
	/* The high 256x256, of which we want the low 256 outputs. */
	top->v.left = left + 256;
	top->v.right = right;
	top->v.out = right;
	convolve_run (top, 256, scratch);

	/* Now find the best position amoungs this.  Apart from the first
	 * and last, the required convolution outputs are formed by adding
	 * outputs from the two convolutions above. */
	best = right[511];
	right[767] = 0;
	p = -1;
	for (i = 0; i < 256; i++) {
		double a = right[i] + right[i + 512];
		if (a > best) {
			best = a;
			p = i;
		}
	}
	p++;

#if 0
	{
		/* This is some debugging code... */
		int bad = 0;
		best = 0;
		for (i = 0; i < 256; i++)
			best += ((double) input[i+p]) * ((double) lastchoice[i] - avg);
		
		for (i = 0; i < 257; i++) {
			double tot = 0;
			unsigned int j;
			for (j = 0; j < 256; j++)
				tot += ((double) input[i+j]) * ((double) lastchoice[j] - avg);
			if (tot > best)
				printf ("(%i)", i);
			if (tot != left[i + 255])
				printf ("!");
		}

		printf ("%i\n", p);
	}
#endif		

	return p;
}
