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Another tweak to the multiplication algorithm
Reduce the formula to calculate mantissa so that we reduce the net number of multiplications performed.
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@ -1,5 +1,8 @@
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2013-02-26 Siddhesh Poyarekar <siddhesh@redhat.com>
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* sysdeps/ieee754/dbl-64/mpa.c: Include alloca.h.
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(__mul): Reduce iterations for calculating mantissa.
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* sysdeps/ieee754/dbl-64/sincos32.c (__c32): Use MPONE and
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MPTWO.
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(__mpranred): Likewise.
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@ -43,6 +43,7 @@
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#include "endian.h"
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#include "mpa.h"
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#include <sys/param.h>
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#include <alloca.h>
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#ifndef SECTION
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# define SECTION
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@ -621,6 +622,7 @@ __mul (const mp_no *x, const mp_no *y, mp_no *z, int p)
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long p2 = p;
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double u, zk;
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const mp_no *a;
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double *diag;
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/* Is z=0? */
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if (__glibc_unlikely (X[0] * Y[0] == ZERO))
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@ -673,12 +675,33 @@ __mul (const mp_no *x, const mp_no *y, mp_no *z, int p)
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while (k > ip + ip2 + 1)
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Z[k--] = ZERO;
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zk = Z[k] = ZERO;
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zk = ZERO;
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/* Precompute sums of diagonal elements so that we can directly use them
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later. See the next comment to know we why need them. */
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diag = alloca (k * sizeof (double));
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double d = ZERO;
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for (i = 1; i <= ip; i++)
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{
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d += X[i] * Y[i];
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diag[i] = d;
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}
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while (i < k)
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diag[i++] = d;
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while (k > p2)
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{
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for (i = k - p2, j = p2; i < p2 + 1; i++, j--)
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zk += X[i] * Y[j];
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long lim = k / 2;
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if (k % 2 == 0)
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/* We want to add this only once, but since we subtract it in the sum
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of products above, we add twice. */
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zk += 2 * X[lim] * Y[lim];
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for (i = k - p2, j = p2; i < j; i++, j--)
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zk += (X[i] + X[j]) * (Y[i] + Y[j]);
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zk -= diag[k - 1];
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u = (zk + CUTTER) - CUTTER;
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if (u > zk)
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@ -687,11 +710,32 @@ __mul (const mp_no *x, const mp_no *y, mp_no *z, int p)
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zk = u * RADIXI;
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}
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/* The real deal. */
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/* The real deal. Mantissa digit Z[k] is the sum of all X[i] * Y[j] where i
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goes from 1 -> k - 1 and j goes the same range in reverse. To reduce the
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number of multiplications, we halve the range and if k is an even number,
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add the diagonal element X[k/2]Y[k/2]. Through the half range, we compute
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X[i] * Y[j] as (X[i] + X[j]) * (Y[i] + Y[j]) - X[i] * Y[i] - X[j] * Y[j].
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This reduction tells us that we're summing two things, the first term
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through the half range and the negative of the sum of the product of all
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terms of X and Y in the full range. i.e.
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SUM(X[i] * Y[i]) for k terms. This is precalculated above for each k in
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a single loop so that it completes in O(n) time and can hence be directly
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used in the loop below. */
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while (k > 1)
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{
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for (i = 1, j = k - 1; i < k; i++, j--)
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zk += X[i] * Y[j];
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long lim = k / 2;
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if (k % 2 == 0)
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/* We want to add this only once, but since we subtract it in the sum
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of products above, we add twice. */
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zk += 2 * X[lim] * Y[lim];
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for (i = 1, j = k - 1; i < j; i++, j--)
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zk += (X[i] + X[j]) * (Y[i] + Y[j]);
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zk -= diag[k - 1];
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u = (zk + CUTTER) - CUTTER;
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if (u > zk)
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