Copy the KDE 3.5 branch to branches/trinity for new KDE 3.5 features.

BUG:215923


git-svn-id: svn://anonsvn.kde.org/home/kde/branches/trinity/kdeaddons@1054174 283d02a7-25f6-0310-bc7c-ecb5cbfe19da

1 files changed, 297 insertions, 0 deletions

diff --git a/noatun-plugins/nexscope/convolve.c b/noatun-plugins/nexscope/convolve.c
new file mode 100644
index 0000000..03509eb
--- /dev/null
+++ b/noatun-plugins/nexscope/convolve.c
@@ -0,0 +1,297 @@
+/* Karatsuba convolution
+ *
+ *  Copyright (C) 1999 Ralph Loader <[email protected]>
+ *
+ *  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;
+}