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@serg06
Last active August 14, 2024 15:39
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A short pow(float x, float y) = x^y approximation that doesn't use any libraries (not even math.h.)
#include <stdio.h>
// Iterations for exp(x) approximation. Higher = more accurate, but higher likelihood of running into inf.
#define EXP_ITERATIONS 100
// Iterations for ln(x) approximation. Higher = more accurate, but slower.
#define LN_ITERATIONS 10000
// Returned if invalid input entered.
#define ERROR_RESULT -999999999
// Super fast x^p (integer power) using Exponentiation by Squaring.
double powfi(double x, unsigned p) {
double result = 1;
while (p) {
if (p & 0x1) {
result *= x;
}
x *= x;
p >>= 1;
}
return result;
}
// Factorial
double fact2(unsigned n) {
if (n < 0 || n >= 172) return ERROR_RESULT; // error out (fact(172) = inf)
double result = 1;
for (int i = 2; i <= n; i++) {
result *= i;
}
return result;
}
// Approximate e^x calculation using Taylor series.
// e^x = sum_{n=0}^{inf} (x^n / n!) for x >= 0(?)
double exp2(double x) {
if (x < 0) return ERROR_RESULT; // error out (note: it might work for negative numbers too so this line might be useless)
double result = 0;
for (int n = 0; n < EXP_ITERATIONS; n++) {
result += powfi(x, n) / fact2(n);
}
return result;
}
// Approximate ln(x) calculation using Taylor series
// ln(x) = 2 * sum_{n=0}^{inf} (((x-1)/(x+1))^(2n-1))/(2n-1) for x > 0
double ln2(double x) {
if (x <= 0) return ERROR_RESULT; // error out
double result = 0;
for (int n = 1; n < LN_ITERATIONS; n++) {
result += powfi((x-1)/(x+1), 2*n-1)/(2*n-1);
}
return 2*result;
}
// Finally, the custom pow(float x, float y) function.
// Uses the fact that pow(x,y) = exp(y*log(x))
double powff(double x, double p) {
if (x <= 0) return ERROR_RESULT; // error out
// split the calculation into two to discourage inf
// y = y - (int)y + (int)y
// so
// x^y = x^((int)y) * x^(y-((int)y))
// first calculate x^((int)y)
int int_power = (unsigned)p;
double int_pow_result = powfi(x, (unsigned)int_power);
// then calculate x^(y-((int)y))
double remaining_power = p - int_power;
double remaining_result = exp2(remaining_power*ln2(x));
// Then multiply them together!
return int_pow_result * remaining_result;
}
// Example
int main() {
float a = 1305.3968;
float b = 12.530006;
printf("%f ^ %f = %f\n", a, b, powff(a, b));
return 0;
}
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