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

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Given a 32-bit floating point number x stored in IEEE 754 floating point format, find the inverse square root of x, i.e., x^-1/2.
A simple solution is to do floating point arithmetic.

Following is an example function.

C++14

#include <bits/stdc++.h>
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

Time Complexity: O(1)
Auxiliary Space: O(1)

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

C++14

#include <bits/stdc++.h>
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;
}

Java

public class Main {
 
    // This is fairly tricky and complex process. For
    // details, see
    // http://en.wikipedia.org/wiki/Fast_inverse_square_root
    static float inverseSquareRoot(float x)
    {
        float xhalf = 0.5f * x;
        int i = Float.floatToIntBits(x);
        i = 0x5f3759d5 - (i >> 1);
        x = Float.intBitsToFloat(i);
        x = x * (1.5f - xhalf * x * x);
        return 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));
    }
}

Python3

# Python program for the above approach
import struct
 
def inverse_square_root(x):
    # This is fairly tricky and complex process. For
    # details, see
    # http://en.wikipedia.org/wiki/Fast_inverse_square_root
    xhalf = 0.5 * x
    i = struct.unpack('i', struct.pack('f', x))[0]
    i = 0x5f3759d5 - (i >> 1)
    x = struct.unpack('f', struct.pack('i', i))[0]
    x = x * (1.5 - xhalf * x * x)
    return x
 
print(inverse_square_root(0.5))
print(inverse_square_root(3.6))
print(inverse_square_root(1.0))
 
# Contributed by adityashrmadev01

C#

using System;
 
public class MainClass
{
    // This is fairly tricky and complex process. For
    // details, see
    // https://en.wikipedia.org/wiki/Fast_inverse_square_root
    static float inverseSquareRoot(float x)
    {
        float xhalf = 0.5f * x;
        int i = BitConverter.ToInt32(BitConverter.GetBytes(x), 0);
        i = 0x5f3759d5 - (i >> 1);
        x = BitConverter.ToSingle(BitConverter.GetBytes(i), 0);
        x = x * (1.5f - xhalf * x * x);
        return x;
    }
 
    public static void Main(string[] args)
    {
        Console.WriteLine(inverseSquareRoot(0.5f));
        Console.WriteLine(inverseSquareRoot(3.6f));
        Console.WriteLine(inverseSquareRoot(1.0f));
    }
}

Javascript

// This is a fairly tricky and complex process. For details,
// see http://en.wikipedia.org/wiki/Fast_inverse_square_root
function inverseSquareRoot(x) {
    let xhalf = 0.5 * x;
    let i = new Float32Array(1);
    let y = new Int32Array(i.buffer);
    i[0] = x;
    y[0] = 0x5f3759d5 - (y[0] >> 1);
    x = i[0];
    x = x * (1.5 - xhalf * x * x);
    return x;
}
 
console.log(inverseSquareRoot(0.5));
console.log(inverseSquareRoot(3.6));
console.log(inverseSquareRoot(1.0));

Output

1.41386
0.526715
0.998307

Time Complexity: O(1)
Auxiliary Space: O(1)

Source:
http://en.wikipedia.org/wiki/Fast_inverse_square_root
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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