There is a better algorithm, which needs at most 6 iterations to converge to maximum precision for double numbers: #include <math.h> double sqrt(double x) { if (x <= 0) return 0; // if negative number throw an exception? int exp = 0; x = frexp(x, &exp); // extract binary exponent from x if (exp & … Read more
yeah, I know fsqrt. But how the CPU does it? I can’t debug hardware Typical div/sqrt hardware in modern CPUs uses a power of 2 radix to calculate multiple result bits at once. e.g. http://www.imm.dtu.dk/~alna/pubs/ARITH20.pdf presents details of a design for a Radix-16 div/sqrt ALU, and compares it against the design in Penryn. (They claim … Read more
It depends on the implementation. CPython is using math functions from the standard C library. Jython is most likely using Java’s math methods. And so on. In fact, Python has nothing to do with the actual implementation of math functions. Those are more related to IEEE 754 which is used almost exclusively to represent floating … Read more
You need to link with the math library (use a ‘-lm’ on the command line). In the constant case, the compiler is probably being smart and precomputing sqrt(2.0) (so the code that is compiled is essentially ‘b = 1.414…;’)
[Edit1] repaired code Recently found the results where off even if tests determine all was OK so I dig deeper and found out that I had a silly bug in my equation and due to name conflicts with my pgm environment the tests got false positives so I overlooked it before. Now it work in … Read more
Newton’s method works perfectly well on integers: def isqrt(n): x = n y = (x + 1) // 2 while y < x: x = y y = (x + n // x) // 2 return x This returns the largest integer x for which x * x does not exceed n. If you want … Read more