Fast method to calculate inverse square root of a floating point number in IEEE 754 format

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  
// http://en.wikipedia.org/wiki/Fast_inverse_square_root 
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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