Given a 32 bit floating point number x stored in IEEE 754 floating point format, find inverse square root of x, i.e., x-1/2.
A simple solution is to do floating point arithmetic. Following is example function.
#include <iostream>#include <cmath>using namespace std; float InverseSquareRoot(float x){ return 1/sqrt(x);} int main(){ cout << InverseSquareRoot(0.5) << endl; cout << InverseSquareRoot(3.6) << endl; cout << InverseSquareRoot(1.0) << endl; return 0;} |
Output:
1.41421 0.527046 1
Following is a fast and interesting method based for the same. See this for detailed explanation.
#include <iostream>using namespace std; // This is fairly tricky and complex process. For details, see float InverseSquareRoot(float x){ float xhalf = 0.5f*x; int i = *(int*)&x; i = 0x5f3759d5 - (i >> 1); x = *(float*)&i; x = x*(1.5f - xhalf*x*x); return x;} int main(){ cout << InverseSquareRoot(0.5) << endl; cout << InverseSquareRoot(3.6) << endl; cout << InverseSquareRoot(1.0) << endl; return 0;} |
Output:
1.41386 0.526715 0.998307
Source:
http://en.wikipedia.org/wiki/Fast_inverse_square_root
This article is contributed by Shalki Agarwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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