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298 lines
9.0 KiB
298 lines
9.0 KiB
/* Karatsuba convolution
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*
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* Copyright (C) 1999 Ralph Loader <suckfish@ihug.co.nz>
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*
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* This program is free software; you can redistribute it and/or modify
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* it under the terms of the GNU General Public License as published by
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* the Free Software Foundation; either version 2 of the License, or
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* (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program; if not, write to the Free Software
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* Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */
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/* The algorithm is based on the following. For the convolution of a pair
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* of pairs, (a,b) * (c,d) = (0, a.c, a.d+b.c, b.d), we can reduce the four
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* multiplications to three, by the formulae a.d+b.c = (a+b).(c+d) - a.c -
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* b.d. A similar relation enables us to compute a 2n by 2n convolution
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* using 3 n by n convolutions, and thus a 2^n by 2^n convolution using 3^n
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* multiplications (as opposed to the 4^n that the quadratic algorithm
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* takes. */
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/* For large n, this is slower than the O(n log n) that the FFT method
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* takes, but we avoid using complex numbers, and we only have to compute
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* one convolution, as opposed to 3 FFTs. We have good locality-of-
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* reference as well, which will help on CPUs with tiny caches. */
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/* E.g., for a 512 x 512 convolution, the FFT method takes 55 * 512 = 28160
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* (real) multiplications, as opposed to 3^9 = 19683 for the Karatsuba
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* algorithm. We actually want 257 outputs of a 256 x 512 convolution;
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* that doesn't appear to give an easy advantage for the FFT algorithm, but
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* for the Karatsuba algorithm, it's easy to use two 256 x 256
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* convolutions, taking 2 x 3^8 = 12312 multiplications. [This difference
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* is that the FFT method "wraps" the arrays, doing a 2^n x 2^n -> 2^n,
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* while the Karatsuba algorithm pads with zeros, doing 2^n x 2^n -> 2.2^n
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* - 1]. */
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/* There's a big lie above, actually... for a 4x4 convolution, it's quicker
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* to do it using 16 multiplications than the more complex Karatsuba
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* algorithm... So the recursion bottoms out at 4x4s. This increases the
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* number of multiplications by a factor of 16/9, but reduces the overheads
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* dramatically. */
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/* The convolution algorithm is implemented as a stack machine. We have a
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* stack of commands, each in one of the forms "do a 2^n x 2^n
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* convolution", or "combine these three length 2^n outputs into one
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* 2^{n+1} output." */
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#include <stdlib.h>
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#include "convolve.h"
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/*
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* Initialisation routine - sets up tables and space to work in.
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* Returns a pointer to internal state, to be used when performing calls.
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* On error, returns NULL.
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* The pointer should be freed when it is finished with, by convolve_close().
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*/
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convolve_state *convolve_init(void)
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{
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return (convolve_state *) malloc (sizeof(convolve_state));
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}
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/*
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* Free the state allocated with convolve_init().
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*/
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void convolve_close(convolve_state *state)
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{
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if (state)
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free(state);
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}
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static void convolve_4 (double * out, const double * left, const double * right)
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/* This does a 4x4 -> 7 convolution. For what it's worth, the slightly odd
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* ordering gives about a 1% speed up on my Pentium II. */
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{
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double l0, l1, l2, l3, r0, r1, r2, r3;
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double a;
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l0 = left[0];
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r0 = right[0];
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a = l0 * r0;
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l1 = left[1];
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r1 = right[1];
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out[0] = a;
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a = (l0 * r1) + (l1 * r0);
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l2 = left[2];
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r2 = right[2];
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out[1] = a;
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a = (l0 * r2) + (l1 * r1) + (l2 * r0);
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l3 = left[3];
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r3 = right[3];
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out[2] = a;
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out[3] = (l0 * r3) + (l1 * r2) + (l2 * r1) + (l3 * r0);
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out[4] = (l1 * r3) + (l2 * r2) + (l3 * r1);
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out[5] = (l2 * r3) + (l3 * r2);
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out[6] = l3 * r3;
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}
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static void convolve_run (stack_entry * top, unsigned size, double * scratch)
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/* Interpret a stack of commands. The stack starts with two entries; the
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* convolution to do, and an illegal entry used to mark the stack top. The
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* size is the number of entries in each input, and must be a power of 2,
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* and at least 8. It is OK to have out equal to left and/or right.
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* scratch must have length 3*size. The number of stack entries needed is
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* 3n-4 where size=2^n. */
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{
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do {
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const double * left;
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const double * right;
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double * out;
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/* When we get here, the stack top is always a convolve,
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* with size > 4. So we will split it. We repeatedly split
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* the top entry until we get to size = 4. */
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left = top->v.left;
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right = top->v.right;
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out = top->v.out;
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top++;
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do {
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double * s_left, * s_right;
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int i;
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/* Halve the size. */
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size >>= 1;
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/* Allocate the scratch areas. */
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s_left = scratch + size * 3;
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/* s_right is a length 2*size buffer also used for
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* intermediate output. */
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s_right = scratch + size * 4;
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/* Create the intermediate factors. */
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for (i = 0; i < size; i++) {
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double l = left[i] + left[i + size];
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double r = right[i] + right[i + size];
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s_left[i + size] = r;
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s_left[i] = l;
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}
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/* Push the combine entry onto the stack. */
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top -= 3;
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top[2].b.main = out;
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top[2].b.null = NULL;
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/* Push the low entry onto the stack. This must be
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* the last of the three sub-convolutions, because
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* it may overwrite the arguments. */
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top[1].v.left = left;
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top[1].v.right = right;
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top[1].v.out = out;
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/* Push the mid entry onto the stack. */
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top[0].v.left = s_left;
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top[0].v.right = s_right;
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top[0].v.out = s_right;
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/* Leave the high entry in variables. */
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left += size;
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right += size;
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out += size * 2;
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} while (size > 4);
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/* When we get here, the stack top is a group of 3
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* convolves, with size = 4, followed by some combines. */
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convolve_4 (out, left, right);
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convolve_4 (top[0].v.out, top[0].v.left, top[0].v.right);
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convolve_4 (top[1].v.out, top[1].v.left, top[1].v.right);
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top += 2;
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/* Now process combines. */
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do {
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/* b.main is the output buffer, mid is the middle
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* part which needs to be adjusted in place, and
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* then folded back into the output. We do this in
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* a slightly strange way, so as to avoid having
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* two loops. */
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double * out = top->b.main;
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double * mid = scratch + size * 4;
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unsigned int i;
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top++;
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out[size * 2 - 1] = 0;
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for (i = 0; i < size-1; i++) {
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double lo;
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double hi;
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lo = mid[0] - (out[0] + out[2 * size]) + out[size];
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hi = mid[size] - (out[size] + out[3 * size]) + out[2 * size];
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out[size] = lo;
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out[2 * size] = hi;
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out++;
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mid++;
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}
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size <<= 1;
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} while (top->b.null == NULL);
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} while (top->b.main != NULL);
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}
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int convolve_match (float * lastchoice,
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float * input,
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convolve_state * state)
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/* lastchoice is a 256 sized array. input is a 512 array. We find the
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* contiguous length 256 sub-array of input that best matches lastchoice.
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* A measure of how good a sub-array is compared with the lastchoice is
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* given by the sum of the products of each pair of entries. We maximise
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* that, by taking an appropriate convolution, and then finding the maximum
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* entry in the convolutions. state is a (non-NULL) pointer returned by
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* convolve_init. */
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{
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double avg;
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double best;
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int p;
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int i;
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double * left = state->left;
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double * right = state->right;
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double * scratch = state->scratch;
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stack_entry * top = state->stack + STACK_SIZE - 1;
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for (i=0; i<512; i++)
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left[i]=input[i];
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avg = 0;
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for (i = 0; i < 256; i++)
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{
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double a = lastchoice[255 - i];
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right[i] = a;
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avg += a;
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}
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/* We adjust the smaller of the two input arrays to have average
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* value 0. This makes the eventual result insensitive to both
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* constant offsets and positive multipliers of the inputs. */
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avg /= 256;
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for (i = 0; i < 256; i++)
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right[i] -= avg;
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/* End-of-stack marker. */
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top[1].b.null = scratch;
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top[1].b.main = NULL;
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/* The low 256x256, of which we want the high 256 outputs. */
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top->v.left = left;
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top->v.right = right;
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top->v.out = right + 256;
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convolve_run (top, 256, scratch);
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/* The high 256x256, of which we want the low 256 outputs. */
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top->v.left = left + 256;
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top->v.right = right;
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top->v.out = right;
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convolve_run (top, 256, scratch);
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/* Now find the best position amoungs this. Apart from the first
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* and last, the required convolution outputs are formed by adding
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* outputs from the two convolutions above. */
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best = right[511];
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right[767] = 0;
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p = -1;
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for (i = 0; i < 256; i++) {
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double a = right[i] + right[i + 512];
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if (a > best) {
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best = a;
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p = i;
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}
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}
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p++;
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#if 0
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{
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/* This is some debugging code... */
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int bad = 0;
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best = 0;
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for (i = 0; i < 256; i++)
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best += ((double) input[i+p]) * ((double) lastchoice[i] - avg);
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for (i = 0; i < 257; i++) {
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double tot = 0;
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unsigned int j;
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for (j = 0; j < 256; j++)
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tot += ((double) input[i+j]) * ((double) lastchoice[j] - avg);
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if (tot > best)
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printf ("(%i)", i);
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if (tot != left[i + 255])
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printf ("!");
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}
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printf ("%i\n", p);
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}
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#endif
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return p;
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}
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