/* Karatsuba convolution
*
* Copyright (C) 1999 Ralph Loader
*
* This library is free software; you can redistribute it and/or
* modify it under the terms of the GNU Library General Public
* License as published by the Free Software Foundation; either
* version 2 of the License, or (at your option) any later version.
*
* This library 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
* Library General Public License for more details.
*
* You should have received a copy of the GNU Library General Public
* License along with this library; if not, write to the
* Free Software Foundation, Inc., 51 Franklin St, Fifth Floor,
* Boston, MA 02110-1301, USA.
*
*
* Note: 7th December 2004: This file used to be licensed under the GPL,
* but we got permission from Ralp Loader to relicense it to LGPL.
*
* $Id$
*
*/
/* 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." */
#ifdef HAVE_CONFIG_H
#include "config.h"
#endif
#include
#include "convolve.h"
typedef union stack_entry_s
{
struct
{
const double *left, *right;
double *out;
}
v;
struct
{
double *main, *null;
}
b;
}
stack_entry;
#define STACK_SIZE (CONVOLVE_DEPTH * 3)
struct _struct_convolve_state
{
double left[CONVOLVE_BIG];
double right[CONVOLVE_SMALL * 3];
double scratch[CONVOLVE_SMALL * 3];
stack_entry stack[STACK_SIZE + 1];
};
/*
* 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 *) calloc (1, sizeof (convolve_state));
}
/*
* Free the state allocated with convolve_init().
*/
void
convolve_close (convolve_state * 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 (const int *lastchoice,
const short *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 = 0;
int i;
double *left = state->left;
double *right = state->right;
double *scratch = state->scratch;
stack_entry *top = state->stack + (STACK_SIZE - 1);
#if 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;
}
#endif
/* 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;
}