As the title says, I would like to know what is the time complexity of math.log2(x)
. I know that it is possible to write such a function in C in O(1) complexity, but I could not find any information about the implementation of this function in Python.
In the CPython implementation of Python log2 is implemented as the following C function, plus a layer above this in C that handles error reporting and handles integers specially but ultimately even in the integer case it is the code below that performs the logarithm.
The logic is basically to use a standard C log2 function if one is available otherwise compute log2 in terms of log. In any case it is O(1) but with a relatively high constant factor due to all the layers of checks and sanitization.
/*
log2: log to base 2.
Uses an algorithm that should:
(a) produce exact results for powers of 2, and
(b) give a monotonic log2 (for positive finite floats),
assuming that the system log is monotonic.
*/
static double
m_log2(double x)
{
if (!Py_IS_FINITE(x)) {
if (Py_IS_NAN(x))
return x; /* log2(nan) = nan */
else if (x > 0.0)
return x; /* log2(+inf) = +inf */
else {
errno = EDOM;
return Py_NAN; /* log2(-inf) = nan, invalid-operation */
}
}
if (x > 0.0) {
#ifdef HAVE_LOG2
return log2(x);
#else
double m;
int e;
m = frexp(x, &e);
/* We want log2(m * 2**e) == log(m) / log(2) + e. Care is needed when
* x is just greater than 1.0: in that case e is 1, log(m) is negative,
* and we get significant cancellation error from the addition of
* log(m) / log(2) to e. The slight rewrite of the expression below
* avoids this problem.
*/
if (x >= 1.0) {
return log(2.0 * m) / log(2.0) + (e - 1);
}
else {
return log(m) / log(2.0) + e;
}
#endif
}
else if (x == 0.0) {
errno = EDOM;
return -Py_HUGE_VAL; /* log2(0) = -inf, divide-by-zero */
}
else {
errno = EDOM;
return Py_NAN; /* log2(-inf) = nan, invalid-operation */
}
}