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/*
* Copyright © 2019 Keith Packard <keithp@keithp.com>
*
* 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 3 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 St, Fifth Floor, Boston, MA 02110-1301, USA.
*/
#include "snek.h"
#include <math.h>
#ifdef AVR
#define float double
#endif
/*
* Compute expf using the traditional power series
*
* expf(a) = 1 + a/1 + a²/2! + a³/3! …
*/
float
expf(float a)
{
if (a < 0.0f)
return 1/expf(-a);
uint8_t e = 0;
/*
* Values larger than 0.5 need to be scaled so that the series
* converges in a reasonable amount of time We scale them to
* between 0.25 and 0.5, which makes the series converge in
* no more than 8 iterations
*/
if (a >= 0.5) {
int _e;
a = frexpf(a, &_e)/2.0;
e = _e + 1;
if (e > 7)
return INFINITY;
}
float term = 1.0f;
float sum = 0.0f;
uint8_t loop;
/*
* It would be better to perform the addition starting with
* the smallest terms, as that captures the contributions of
* those values better. But, that takes extra code and stack
* space.
*
* This loop converges after 8 iterations in the worst case,
* which is when a is the largest value less than 0.5. Instead
* of putting conditions in the loop, just run the loop for
* 9 iterations each time.
*/
for (loop = 1; loop < 9; loop++) {
/* Compute next term in the sum */
term *= a / (float) loop;
/* Add it in */
sum += term;
}
/* sum = sum ** (2 ** e) */
sum += 1.0f;
while (e--)
sum *= sum;
return sum;
}
/*
* newtons method computation of log
*
* a = exp(x)
* f(x) = exp(x) - a
* f'(x) = exp(x)
*
* x' = x - (exp(x) - a) / exp(x)
*/
static const float _log2 = 0.693147180559945f;
float
logf(float a)
{
int e;
a = frexpf(a, &e);
float l = -_log2;
if (a != 0.5f) {
uint8_t loop;
/* Worst case, we need 5 iterations to converge */
for (loop = 0; loop < 5; loop++) {
float ex = expf(l);
l = l - (ex - a) / ex;
}
}
return l + (float) e * _log2;
}
/* A more accurate version when the exponent is an integer. This
* makes integer ** integer give an integer where possible
*/
static inline float
ipow(float a, uint32_t x)
{
float r = 1;
while (x) {
if (x & 1)
r *= a;
x >>= 1;
a *= a;
}
return r;
}
float
powf(float a, float b)
{
uint32_t u = b;
if (u == b)
return ipow(a, u);
return expf(logf(a) * b);
}
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