Efficient implementation of natural logarithm (ln) and exponentiation - c

Question

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.

Answer 1

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;
}

Answer 2

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...

Answer 3

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.

Answer 4

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?

Answer 5

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)

Answer 6

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.

Answer 7

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

Answer 8

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;
}