Efficient implementation of natural logarithm (ln) and exponentiation

Question

I\'m looking for implementation of log() and exp() functions provided in C library . I\'m working with 8 bit microcontro

Answer 1

The log function can be implemented with O(1)

First, let's make use of logn(x) = ln(x)/ln(n). ln(x) can be expressed as (x^h-1)/h, where h approaches 0. You can derive this with the differential quotient.

float log(x, n) {
    return ((x^h-1)/h)/((n^h-1)/h)
}

So now choose a small value for h, it gets more accurate as you make it smaller. Constant runtime and small code size, I always prefer this method.

Answer 2

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 3

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;
}

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

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 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.