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Fast method to calculate inverse square root of a floating point number in IEEE 754 format
• Difficulty Level : Expert
• Last Updated : 20 May, 2017

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 ``#include ``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 ``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```

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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