Example Code
The following Python program computes the binary logarithm using the recursive method outlined above, showing the steps along the way: (This code fails when finding the binary logarithm of any number 2^n where n is any integer.)
def lg(x): ip = 0 rem = x # Integer part while rem<1: ip -= 1 rem *= 2 while rem>=2: ip += 1 rem /= 2 print("lg(%g) = %d + lg(%g)" % (x, ip, rem)) # Brackets required for Python 3 # Fractional part res = ip + frac_lg(rem) return res def frac_lg(x, fp=1.0, tol=1e-10): assert 1<=x<2 # terminate the recursion if we're close enough if fpSince Python does not optimize tail recursion, this can be implemented more efficiently with iteration. Here is an iterative version of the same algorithm in Perl:
sub lg { my $x = shift; my $tol = shift || 1e-13; my $res = 0.0; while ($x < 1) { $res -= 1; $x *= 2; } while ($x >= 2) { $res += 1; $x /= 2; } my $fp = 1.0; while ($fp >= $tol) { $fp /= 2; $x *= $x; if ($x >= 2) { $x /= 2; $res += $fp; } } $res } printf "x = %g\nlg(x) = %g\n", 4.5, lg(4.5);Read more about this topic: Binary Logarithm
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