I am calculating g with e and s, which are all doubles. After that I want to cut off all digits after the second and save the result in x, for example:
g = 2.123 => x = 2.12
g = 5.34995 => x = 5.34
and so on. I Use...
g = 0.5*e + 0.5*s;
x = floor(g*100)/100;
...and it works fine most of the time. But sometimes I get strange results. For example:
e = 3.0
s = 1.6
g = 2.30
but x = 2.29!!!
So I tried to track down the error:
g = 0.5*e + 0.5*s;
NSLog(#"%f",g);
gives me g = 2.30
g = g * 100;
NSLog(#"%f",g);
gives me g = 230.0
x = floor(g);
NSLog(#"%f",x);
results in x = 229.0 !!!
I don't get it! Help please! :-)
This will be due to floating point calculations.
Your calculation
g * 100
already brings back
229.99999999999997
From where your issue stems.
Have a look at INFO: Precision and Accuracy in Floating-Point Calculations
Also have a look at Floating point
Accuracy problems
The fact that floating-point numbers
cannot precisely represent all real
numbers, and that floating-point
operations cannot precisely represent
true arithmetic operations, leads to
many surprising situations. This is
related to the finite precision with
which computers generally represent
numbers.
As others have already mentioned, this is due to the limited precision of floating point numbers in computers. These imprecisions show up everywhere a hard yes/no decision about a floating point number is made. In order to resolve the problem, you can add/subtract a small number to find an answer that is correct up to a certain accuracy.
You may find functions like these useful:
#define ACC 1e-7
double floorAcc( double x ) { return floor(x + ACC);}
double ceilAcc( double x ) { return ceil(x - ACC); }
bool isLessThanAcc( double x, double y ) { return (x + ACC) < y; }
bool isEqualAcc( double x, double y ) { return (x + ACC) > y && (x - ACC) < y; }
Of course, these work only in a limited number range. When working with very small or very large numbers, you need to pick another value for ACC.
Note that the value of 'ACC' is in general dependent on the accuracy of the numbers in your application, not on the value of x. For example, comparing two numbers a and b for equality can be done in two ways: isEqualAcc(a, b) and isEqualAcc(a-b, 0). You would want the same result from both ways, even though in the second way the number x is likely much smaller.
Here is a possible approach using intermediate integer results:
double e = 3.0;
double s = 1.6;
NSInteger e1 = e * .5 * 100.0; // 150
NSInteger s1 = s * .5 * 100.0; // 80
double x = (e1 + s1)/100.0; // 2.3
Related
I want to know whether the program defined below can return 1 assuming:
IEEE754 floating point arithmetics
no overflow (neither in max/x nor in f*x)
no nan or inf (obviously)
0 < x and 0 < n < 32
no unsafe math optimization
int canfail(int n, double x) {
double max = 1ULL << n; // 2^n
double f = max / x;
return f * x > max;
}
In my opinion, it should sometime return 1, as roundToNearest(max / x) can in general be greater than max/x.
I'm able to find numbers for the opposite case, where f * x < max, but I have no examples of input that show f * x > max and I have no idea of how to find one. Can somebody help ?
EDIT:
I know the value of x if in a range between 10^(-6) and 10^6 (that still leaves a lot (too much possible double values), but I know I will not have to deal with overflow, underflow or sub-normal numbers !
In addition, I just realized that because max is a power of two and we don't deal with overflow, the solution will be the same by fixing max=1 as it is exactly the same computation, but shifted.
Therefore, the problem correspond to finding a positive, normal double value x such that `(1/x) * x > 1.0 !!
I made a little program to try to find a solution:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdint.h>
#include <omp.h>
int main( void ) {
#pragma omp parallel
{
unsigned short int xsubi[3] = {
omp_get_thread_num(),
omp_get_thread_num(),
omp_get_thread_num()
};
#pragma omp for
for(int64_t i=0; i<INT64_MAX; i++) {
double x = fmod(nrand48(xsubi), 1048576.0);
if(x<0.000001)
continue;
double f = 1.0 / x;
if(f * x > 1.0) {
printf("found !!! x=%.30f\n", x);
fflush(stdout);
}
}
}
return 1;
}
If you change the sign of the comparison, you will find some value quickly. However, it seems to run forever with f * x > 1.0
In the absence of underflow or overflow, the exponents are irrelevant; if M/x*x > M, then (M/p) / (x/q) * (x/q) > (M/p) for any powers of two p and q. So let’s consider 252 ≤ x < 253 and M = 2105. We can eliminate x = 252 since this yields exact floating-point arithmetic, so 252 < x < 253.
Division of 2105 by x yields integer quotient q and integer remainder r, with 252 q < 253, 0 < r < x, and 2105 = q•x + r.
In order for M/x*x to exceed M, both the division and the multiplication must round up. Since the division rounds up, x/2 ≤ r.
With rounding up, the result of floating-point division of 2105 by x yields q+1. Then the exact (not rounded) multiplication yields (q+1)•x = q•x + x = q•x + x + r - r = q•x + r + x − r = 2105 + x − r. Since x/2 < r, x − r ≤ x/2, so rounding this exact result rounds down, yielding 2105. (The “<” case always rounds down, and the “=” case rounds down because 2105 has the low even bit.)
Therefore, for powers of two M and all arithmetic within exponent bounds, M/x*x > M never occurs with round-to-nearest-ties-to-even.
Multiplication by a power of two is just a scaling of exponent, it does not change the problem: so it's the same as finding x such that (1/x) * x > 1.
One solution is brute force search.
For same reasons, we can limit the search of such x in the interval (1.0,2.0(
A better approach is to analyze error bounds without brute force.
Let's note ix the nearest floating point to 1/x.
Considering xand ixas exact fractions, we can write the integer division: 1 = ix * x + r where ris the remainder
(these are all fractions with denominators being powers of 2, so we have to multiply the whole equation by appropriate power of 2 to really have integer division).
In other words, ix = 1/x - r/x, where -r/x is the rounding error of inversion.
When we multiply the inverse approximation by x, the exact value is ix*x = 1 - r.
We know that the floating point result will be rounded to the nearest float to that exact value.
So, assumming default rounding mode to nearest, tie to even, the question asked is whether -r can exceed 0.5 ulp.
The short answer is never!
Suppose |r| > 0.5 ulp, then the rounding error -r/x does exceed half ulp of exact result 1/x.
This is not a proper answer, because the exact result is not a floating point and does not have an ulp, but you get the idea...
I might come back with a correct proof if i have time, but my bet is that you can find it already done, possibly on SO
EDIT
Why can you find (1/x) * x < 1?
Simply because 1.0 is at a binade limit, so below 1, we have to prove that r<0.25 ulp, what we cannot...
canfail(1, pow(2, 1023) * (2 - pow(2, -51))) will return 1.
I'm working via a basic 'Programming in C' book.
I have written the following code based off of it in order to calculate the square root of a number:
#include <stdio.h>
float absoluteValue (float x)
{
if(x < 0)
x = -x;
return (x);
}
float squareRoot (float x, float epsilon)
{
float guess = 1.0;
while(absoluteValue(guess * guess - x) >= epsilon)
{
guess = (x/guess + guess) / 2.0;
}
return guess;
}
int main (void)
{
printf("SquareRoot(2.0) = %f\n", squareRoot(2.0, .00001));
printf("SquareRoot(144.0) = %f\n", squareRoot(144.0, .00001));
printf("SquareRoot(17.5) = %f\n", squareRoot(17.5, .00001));
return 0;
}
An exercise in the book has said that the current criteria used for termination of the loop in squareRoot() is not suitable for use when computing the square root of a very large or a very small number.
Instead of comparing the difference between the value of x and the value of guess^2, the program should compare the ratio of the two values to 1. The closer this ratio gets to 1, the more accurate the approximation of the square root.
If the ratio is just guess^2/x, shouldn't my code inside of the while loop:
guess = (x/guess + guess) / 2.0;
be replaced by:
guess = ((guess * guess) / x ) / 1 ; ?
This compiles but nothing is printed out into the terminal. Surely I'm doing exactly what the exercise is asking?
To calculate the ratio just do (guess * guess / x) that could be either higher or lower than 1 depending on your implementation. Similarly, your margin of error (in percent) would be absoluteValue((guess * guess / x) - 1) * 100
All they want you to check is how close the square root is. By squaring the number you get and dividing it by the number you took the square root of you are just checking how close you were to the original number.
Example:
sqrt(4) = 2
2 * 2 / 4 = 1 (this is exact so we get 1 (2 * 2 = 4 = 4))
margin of error = (1 - 1) * 100 = 0% margin of error
Another example:
sqrt(4) = 1.999 (lets just say you got this)
1.999 * 1.999 = 3.996
3.996/4 = .999 (so we are close but not exact)
To check margin of error:
.999 - 1 = -.001
absoluteValue(-.001) = .001
.001 * 100 = .1% margin of error
How about applying a little algebra? Your current criterion is:
|guess2 - x| >= epsilon
You are elsewhere assuming that guess is nonzero, so it is algebraically safe to convert that to
|1 - x / guess2| >= epsilon / guess2
epsilon is just a parameter governing how close the match needs to be, and the above reformulation shows that it must be expressed in terms of the floating-point spacing near guess2 to yield equivalent precision for all evaluations. But of course that's not possible because epsilon is a constant. This is, in fact, exactly why the original criterion gets less effective as x diverges from 1.
Let us instead write the alternative expression
|1 - x / guess2| >= delta
Here, delta expresses the desired precision in terms of the spacing of floating point values in the vicinity of 1, which is related to a fixed quantity sometimes called the "machine epsilon". You can directly select the required precision via your choice of delta, and you will get the same precision for all x, provided that no arithmetic operations overflow.
Now just convert that back into code.
Suggest a different point of view.
As this method guess_next = (x/guess + guess) / 2.0;, once the initial approximation is in the neighborhood, the number of bits of accuracy doubles. Example log2(FLT_EPSILON) is about -23, so 6 iterations are needed. (Think 23, 12, 6, 3, 2, 1)
The trouble with using guess * guess is that it may vanish, become 0.0 or infinity for a non-zero x.
To form a quality initial guess:
assert(x > 0.0f);
int expo;
float signif = frexpf(x, &expo);
float guess = ldexpf(signif, expo/2);
Now iterate N times (e.g. 6), (N based on FLT_EPSILON, FLT_DECIMAL_DIG or FLT_DIG.)
for (i=0; i<N; i++) {
guess = (x/guess + guess) / 2.0f;
}
The cost of perhaps an extra iteration is saved by avoiding an expensive termination condition calculation.
If code wants to compare a/b nearest to 1.0f
Simply use some epsilon factor like 1 or 2.
float a = guess;
float b = x/guess;
assert(b);
float q = a/b;
#define FACTOR (1.0f /* some value 1.0f to maybe 2,3 or 4 */)
if (q >= 1.0f - FLT_EPSILON*N && q <= 1.0f + FLT_EPSILON*N) {
close_enough();
}
First lesson in numerical analysis: for floating point numbers x+y has the potential for large relative errors, especially when the sum is near zero, but x*y has very limited relative errors.
Hey everybody I am working on a program in c that tells you the least number of coins needed for any given amount of money. I have a program written that works for for every amount I have tested except for $4.20.
Here is my code:
#include <cs50.h>
#include <stdio.h>
#include <math.h>
int main(void)
{
float f;
int n, x, y, z, q, s, d, t;
do {
printf("How much change do you need?\n");
f = GetFloat();
} while(f <= 0);
{
n = (f * 100);
}
q = (n / 25);
x = (n % 25);
y = (x / 10);
z = (x % 10);
s = (z / 5);
d = (z % 5);
t = (q + y + s + d);
{
printf("%d\n" ,t);
}
}
The strange thing is when I input 4.20 the output is 22 instead of 18 (16 quarters and 2 dimes). I did some sleuthing and found that the problem is with my variable x. When I input 4.2, x gives me 19 and not 20 like it should. I tried other cases that I thought should have produced the same problem like 5.2 and 1.2 but it worked correctly in those cases. It might be a rounding issue but I would think that same error would also happen with those similar values.
Does anyone have an idea about why this might be happening?
PS I am fairly new to coding and I haven't gotten much formal instruction so I also welcome tips on better indentation and formatting if you see anything obvious.
IEEE 754 floating point is often slightly imprecise, and casting will truncate, not round. What's likely happening is that 4.20 * 100 evaluates to 419.999999999999994 (exact number is immaterial, point is, it's not quite 420), and the conversion to int drops the decimal portion, producing 419.
The simple approach is to just do:
n = f * 100 + 0.5;
or you can use a proper function:
n = round(f * 100);
If the number is "almost" exact, either one will be fine, you'd only get discrepancies when someone passed non-integer cents ("4.195" or the like), and if you're using float for monetary values, you've already accepted precision issues in the margins; if you want exact numbers, you'd use the decimal formats that have fixed precision for decimal values, and are intended for financial calculations.
Try this: Provides up to 2 digit precision.
//float f
double f
f *= 1000;
f = floor(f); /* optional */
f /= 10;
f = floor(f); /* optional */
n = f;
I wrote a code for calculating sin using its maclaurin series and it works but when I try to calculate it for large x values and try to offset it by giving a large order N (the length of the sum) - eventually it overflows and doesn't give me correct results. This is the code and I would like to know is there an additional way to optimize it so it works for large x values too (it already works great for small x values and really big N values).
Here is the code:
long double calcMaclaurinPolynom(double x, int N){
long double result = 0;
long double atzeretCounter = 2;
int sign = 1;
long double fraction = x;
for (int i = 0; i <= N; i++)
{
result += sign*fraction;
sign = sign*(-1);
fraction = fraction*((x*x) / ((atzeretCounter)*(atzeretCounter + 1)));
atzeretCounter += 2;
}
return result;
}
The major issue is using the series outside its range where it well converges.
As OP said "converted x to radX = (x*PI)/180" indicates the OP is starting with degrees rather than radians, the OP is in luck. The first step in finding my_sin(x) is range reduction. When starting with degrees, the reduction is exact. So reduce the range before converting to radians.
long double calcMaclaurinPolynom(double x /* degrees */, int N){
// Reduce to range -360 to 360
// This reduction is exact, no round-off error
x = fmod(x, 360);
// Reduce to range -180 to 180
if (x >= 180) {
x -= 180;
x = -x;
} else if (x <= -180) {
x += 180;
x = -x;
}
// Reduce to range -90 to 90
if (x >= 90) {
x = 180 - x;
} else if (x <= -90) {
x = -180 - x;
}
//now convert to radians.
x = x*PI/180;
// continue with regular code
Alternative, if using C11, use remquo(). Search SO for sample code.
As #user3386109 commented above, no need to "convert back to degrees".
[Edit]
With typical summation series, summing the least significant terms first improves the precision of the answer. With OP's code this can be done with
for (int i = N; i >= 0; i--)
Alternatively, rather than iterating a fixed number of times, loop until the term has no significance to the sum. The following uses recursion to sum the least significant terms first. With range reduction in the -90 to 90 range, the number of iterations is not excessive.
static double sin_d_helper(double term, double xx, unsigned i) {
if (1.0 + term == 1.0)
return term;
return term - sin_d_helper(term * xx / ((i + 1) * (i + 2)), xx, i + 2);
}
#include <math.h>
double sin_d(double x_degrees) {
// range reduction and d --> r conversion from above
double x_radians = ...
return x_radians * sin_d_helper(1.0, x_radians * x_radians, 1);
}
You can avoid the sign variable by incorporating it into the fraction update as in (-x*x).
With your algorithm you do not have problems with integer overflow in the factorials.
As soon as x*x < (2*k)*(2*k+1) the error - assuming exact evaluation - is bounded by abs(fraction), i.e., the size of the next term in the series.
For large x the biggest source for errors is truncation resp. floating point errors that are magnified via cancellation of the terms of the alternating series. For k about x/2 the terms around the k-th term have the biggest size and have to be offset by other big terms.
Halving-and-Squaring
One easy method to deal with large x without using the value of pi is to employ the trigonometric theorems where
sin(2*x)=2*sin(x)*cos(x)
cos(2*x)=2*cos(x)^2-1=cos(x)^2-sin(x)^2
and first reduce x by halving, simultaneously evaluating the Maclaurin series for sin(x/2^n) and cos(x/2^n) and then employ trigonometric squaring (literal squaring as complex numbers cos(x)+i*sin(x)) to recover the values for the original argument.
cos(x/2^(n-1)) = cos(x/2^n)^2-sin(x/2^n)^2
sin(x/2^(n-1)) = 2*sin(x/2^n)*cos(x/2^n)
then
cos(x/2^(n-2)) = cos(x/2^(n-1))^2-sin(x/2^(n-1))^2
sin(x/2^(n-2)) = 2*sin(x/2^(n-1))*cos(x/2^(n-1))
etc.
See https://stackoverflow.com/a/22791396/3088138 for the simultaneous computation of sin and cos values, then encapsulate it with
def CosSinForLargerX(x,n):
k=0
while abs(x)>1:
k+=1; x/=2
c,s = getCosSin(x,n)
r2=0
for i in range(k):
s2=s*s; c2=c*c; r2=s2+c2
s = 2*c*s
c = c2-s2
return c/r2,s/r2
I'm looking for implementation of log() and exp() functions provided in C library <math.h>. I'm working with 8 bit microcontrollers (OKI 411 and 431). I need to calculate Mean Kinetic Temperature. The requirement is that we should be able to calculate MKT as fast as possible and with as little code memory as possible. The compiler comes with log() and exp() functions in <math.h>. But calling either function and linking with the library causes the code size to increase by 5 Kilobytes, which will not fit in one of the micro we work with (OKI 411), because our code already consumed ~12K of available ~15K code memory.
The implementation I'm looking for should not use any other C library functions (like pow(), sqrt() etc). This is because all library functions are packed in one library and even if one function is called, the linker will bring whole 5K library to code memory.
EDIT
The algorithm should be correct up to 3 decimal places.
Using Taylor series is not the simplest neither the fastest way of doing this. Most professional implementations are using approximating polynomials. I'll show you how to generate one in Maple (it is a computer algebra program), using the Remez algorithm.
For 3 digits of accuracy execute the following commands in Maple:
with(numapprox):
Digits := 8
minimax(ln(x), x = 1 .. 2, 4, 1, 'maxerror')
maxerror
Its response is the following polynomial:
-1.7417939 + (2.8212026 + (-1.4699568 + (0.44717955 - 0.056570851 * x) * x) * x) * x
With the maximal error of: 0.000061011436
We generated a polynomial which approximates the ln(x), but only inside the [1..2] interval. Increasing the interval is not wise, because that would increase the maximal error even more. Instead of that, do the following decomposition:
So first find the highest power of 2, which is still smaller than the number (See: What is the fastest/most efficient way to find the highest set bit (msb) in an integer in C?). That number is actually the base-2 logarithm. Divide with that value, then the result gets into the 1..2 interval. At the end we will have to add n*ln(2) to get the final result.
An example implementation for numbers >= 1:
float ln(float y) {
int log2;
float divisor, x, result;
log2 = msb((int)y); // See: https://stackoverflow.com/a/4970859/6630230
divisor = (float)(1 << log2);
x = y / divisor; // normalized value between [1.0, 2.0]
result = -1.7417939 + (2.8212026 + (-1.4699568 + (0.44717955 - 0.056570851 * x) * x) * x) * x;
result += ((float)log2) * 0.69314718; // ln(2) = 0.69314718
return result;
}
Although if you plan to use it only in the [1.0, 2.0] interval, then the function is like:
float ln(float x) {
return -1.7417939 + (2.8212026 + (-1.4699568 + (0.44717955 - 0.056570851 * x) * x) * x) * x;
}
The Taylor series for e^x converges extremely quickly, and you can tune your implementation to the precision that you need. (http://en.wikipedia.org/wiki/Taylor_series)
The Taylor series for log is not as nice...
If you don't need floating-point math for anything else, you may compute an approximate fractional base-2 log pretty easily. Start by shifting your value left until it's 32768 or higher and store the number of times you did that in count. Then, repeat some number of times (depending upon your desired scale factor):
n = (mult(n,n) + 32768u) >> 16; // If a function is available for 16x16->32 multiply
count<<=1;
if (n < 32768) n*=2; else count+=1;
If the above loop is repeated 8 times, then the log base 2 of the number will be count/256. If ten times, count/1024. If eleven, count/2048. Effectively, this function works by computing the integer power-of-two logarithm of n**(2^reps), but with intermediate values scaled to avoid overflow.
Would basic table with interpolation between values approach work? If ranges of values are limited (which is likely for your case - I doubt temperature readings have huge range) and high precisions is not required it may work. Should be easy to test on normal machine.
Here is one of many topics on table representation of functions: Calculating vs. lookup tables for sine value performance?
Necromancing.
I had to implement logarithms on rational numbers.
This is how I did it:
Occording to Wikipedia, there is the Halley-Newton approximation method
which can be used for very-high precision.
Using Newton's method, the iteration simplifies to (implementation), which has cubic convergence to ln(x), which is way better than what the Taylor-Series offers.
// Using Newton's method, the iteration simplifies to (implementation)
// which has cubic convergence to ln(x).
public static double ln(double x, double epsilon)
{
double yn = x - 1.0d; // using the first term of the taylor series as initial-value
double yn1 = yn;
do
{
yn = yn1;
yn1 = yn + 2 * (x - System.Math.Exp(yn)) / (x + System.Math.Exp(yn));
} while (System.Math.Abs(yn - yn1) > epsilon);
return yn1;
}
This is not C, but C#, but I'm sure anybody capable to program in C will be able to deduce the C-Code from that.
Furthermore, since
logn(x) = ln(x)/ln(n).
You have therefore just implemented logN as well.
public static double log(double x, double n, double epsilon)
{
return ln(x, epsilon) / ln(n, epsilon);
}
where epsilon (error) is the minimum precision.
Now as to speed, you're probably better of using the ln-cast-in-hardware, but as I said, I used this as a base to implement logarithms on a rational numbers class working with arbitrary precision.
Arbitrary precision might be more important than speed, under certain circumstances.
Then, use the logarithmic identities for rational numbers:
logB(x/y) = logB(x) - logB(y)
In addition to Crouching Kitten's answer which gave me inspiration, you can build a pseudo-recursive (at most 1 self-call) logarithm to avoid using polynomials. In pseudo code
ln(x) :=
If (x <= 0)
return NaN
Else if (!(1 <= x < 2))
return LN2 * b + ln(a)
Else
return taylor_expansion(x - 1)
This is pretty efficient and precise since on [1; 2) the taylor series converges A LOT faster, and we get such a number 1 <= a < 2 with the first call to ln if our input is positive but not in this range.
You can find 'b' as your unbiased exponent from the data held in the float x, and 'a' from the mantissa of the float x (a is exactly the same float as x, but now with exponent biased_0 rather than exponent biased_b). LN2 should be kept as a macro in hexadecimal floating point notation IMO. You can also use http://man7.org/linux/man-pages/man3/frexp.3.html for this.
Also, the trick
unsigned long tmp = *(ulong*)(&d);
for "memory-casting" double to unsigned long, rather than "value-casting", is very useful to know when dealing with floats memory-wise, as bitwise operators will cause warnings or errors depending on the compiler.
Possible computation of ln(x) and expo(x) in C without <math.h> :
static double expo(double n) {
int a = 0, b = n > 0;
double c = 1, d = 1, e = 1;
for (b || (n = -n); e + .00001 < (e += (d *= n) / (c *= ++a)););
// approximately 15 iterations
return b ? e : 1 / e;
}
static double native_log_computation(const double n) {
// Basic logarithm computation.
static const double euler = 2.7182818284590452354 ;
unsigned a = 0, d;
double b, c, e, f;
if (n > 0) {
for (c = n < 1 ? 1 / n : n; (c /= euler) > 1; ++a);
c = 1 / (c * euler - 1), c = c + c + 1, f = c * c, b = 0;
for (d = 1, c /= 2; e = b, b += 1 / (d * c), b - e/* > 0.0000001 */;)
d += 2, c *= f;
} else b = (n == 0) / 0.;
return n < 1 ? -(a + b) : a + b;
}
static inline double native_ln(const double n) {
// Returns the natural logarithm (base e) of N.
return native_log_computation(n) ;
}
static inline double native_log_base(const double n, const double base) {
// Returns the logarithm (base b) of N.
return native_log_computation(n) / native_log_computation(base) ;
}
Try it Online
Building off #Crouching Kitten's great natural log answer above, if you need it to be accurate for inputs <1 you can add a simple scaling factor. Below is an example in C++ that i've used in microcontrollers. It has a scaling factor of 256 and it's accurate to inputs down to 1/256 = ~0.04, and up to 2^32/256 = 16777215 (due to overflow of a uint32 variable).
It's interesting to note that even on an STMF103 Arm M3 with no FPU, the float implementation below is significantly faster (eg 3x or better) than the 16 bit fixed-point implementation in libfixmath (that being said, this float implementation still takes a few thousand cycles so it's still not ~fast~)
#include <float.h>
float TempSensor::Ln(float y)
{
// Algo from: https://stackoverflow.com/a/18454010
// Accurate between (1 / scaling factor) < y < (2^32 / scaling factor). Read comments below for more info on how to extend this range
float divisor, x, result;
const float LN_2 = 0.69314718; //pre calculated constant used in calculations
uint32_t log2 = 0;
//handle if input is less than zero
if (y <= 0)
{
return -FLT_MAX;
}
//scaling factor. The polynomial below is accurate when the input y>1, therefore using a scaling factor of 256 (aka 2^8) extends this to 1/256 or ~0.04. Given use of uint32_t, the input y must stay below 2^24 or 16777216 (aka 2^(32-8)), otherwise uint_y used below will overflow. Increasing the scaing factor will reduce the lower accuracy bound and also reduce the upper overflow bound. If you need the range to be wider, consider changing uint_y to a uint64_t
const uint32_t SCALING_FACTOR = 256;
const float LN_SCALING_FACTOR = 5.545177444; //this is the natural log of the scaling factor and needs to be precalculated
y = y * SCALING_FACTOR;
uint32_t uint_y = (uint32_t)y;
while (uint_y >>= 1) // Convert the number to an integer and then find the location of the MSB. This is the integer portion of Log2(y). See: https://stackoverflow.com/a/4970859/6630230
{
log2++;
}
divisor = (float)(1 << log2);
x = y / divisor; // FInd the remainder value between [1.0, 2.0] then calculate the natural log of this remainder using a polynomial approximation
result = -1.7417939 + (2.8212026 + (-1.4699568 + (0.44717955 - 0.056570851 * x) * x) * x) * x; //This polynomial approximates ln(x) between [1,2]
result = result + ((float)log2) * LN_2 - LN_SCALING_FACTOR; // Using the log product rule Log(A) + Log(B) = Log(AB) and the log base change rule log_x(A) = log_y(A)/Log_y(x), calculate all the components in base e and then sum them: = Ln(x_remainder) + (log_2(x_integer) * ln(2)) - ln(SCALING_FACTOR)
return result;
}