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. 

CPP

#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;
}

Java

import java.io.*;

class GFG {

    static float InverseSquareRoot(float x)
    {
        return 1 / (float)Math.sqrt(x);
    }

    public static void main(String[] args)
    {
        System.out.println(InverseSquareRoot(0.5f));
        System.out.println(InverseSquareRoot(3.6f));
        System.out.println(InverseSquareRoot(1.0f));
    }
}

// This code is contributed by souravmahato348.

Python3

# Python code for the above approach
from math import ceil, sqrt

def InverseSquareRoot(x) :

    return 1/sqrt(x)

# Driver Code
print(InverseSquareRoot(0.5) )
print(InverseSquareRoot(3.6) )
print(InverseSquareRoot(1.0) )

# This code is contributed by code_hunt.

C#

using System;

class GFG {

    static float InverseSquareRoot(float x)
    {
        return 1 / (float)Math.Sqrt(x);
    }

    public static void Main()
    {
        Console.WriteLine(InverseSquareRoot(0.5f));
        Console.WriteLine(InverseSquareRoot(3.6f));
        Console.WriteLine(InverseSquareRoot(1.0f));
    }
}

// This code is contributed by subham348.

Javascript

<script>
        // JavaScript code for the above approach

     function InverseSquareRoot(x)
    {
        return 1 / Math.sqrt(x);
    }

        // Driver Code
        document.write(InverseSquareRoot(0.5) + "<br/>");
        document.write(InverseSquareRoot(3.6) + "<br/>");
        document.write(InverseSquareRoot(1.0) + "<br/>");

        // This code is contributed by sanjoy_62.
    </script>

Output: 

1.41421
0.527046
1

Following is a fast and interesting method based for the same. See this for detailed explanation.

C

#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