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. |
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. |
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 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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