# Introduction

This article introduces the fast inv-sqrt algorithm, which is firstly used in Quake III Arena.

# The Source Code

So here is the source code:

And this function takes a 32-bit floating-point number $x$, and output the reciprocal of the square root $\frac{1}{\sqrt{x}}$of this number.

# 32-bit Floating-Point Number

Since this algorithm was designed with the IEEE 754-1985 32-bit floating-point specification in mind, it is necessary for us to get to know the structure of a floating-point number.

You can refer to Single-precision floating-point format for more information and I’ll skip the explanation since this is not the point of this article.

In short, for a positive floating-point number $x$, we have

$$x = 2^{e_x}(1 + m_{x})$$

which $e_x$ is the 8-bit exponent and $m_x$ is the fraction part.

Moreover, to convert a floating-point number to a integer, we have

$$I_{x} = E_{X}L + M_{x} = L(e_{x} + B + m_{x})$$

which L is $2^{23}$ and B is $127$.

# Algorithm

## Newton’s Method

We can find successively better approximations to the roots of a real-valued function using Newton's method.

Suppose a single-variable function $f$ defined for a real variable $x$, if the function satisfied sufficient assumptions and the initial guess is close, then

$$x_{1} = x_{0} - \frac{f(x_{0})}{f^{‘}(x_{n})}$$

is a better approximation of the root than $x_{0}$.

The process is repeated as

$$x_{n + 1} = x_{n} - \frac{f(x_{n})}{f^{‘}(x_{n})}$$

until a sufficiently precise value is reached.

The process iteration is like:

## Black Magic

The magic is i = 0x5f3759df - ( i >> 1 ), but how does it work?

For function

$$y = \frac{1}{\sqrt{x}}$$

, firstly we have:

$$f(y) = \frac{1}{y^{2}} - x = 0$$

then we convert it to Newton's method style:

$$y_{n+1} = \frac{y_{n}(3 - xy^{2}_{n})}{2}$$

So for $x = 2^{e_{x}}(1 + m_{x})$, we have:
$$log_{2}{x} = e_{x} + \log{2}{1 + m_{x}}$$

and since $m_{x} \in [0, 1)$, we have:

$$log_{x}{1 + m_{x}} \approx m_{x} + \sigma$$

where $\sigma \approx 0.0430357$ yields the optimal approximation.

Thus there is the approximation
$$log_{2}{x} \approx e_{x} + m_{x} + \sigma$$

Finally we have:
$$I_{x} = E_{x}L + M_{x} = L(e_{x} + m_{x} + \sigma + B - \sigma) \ \approx Llog_{2}{(x)} + L(B - \sigma)$$.

So:
$$I_{y} \approx \frac{3}{2}L(B -sigma) - \frac{1}{2}I_{x}$$,

which is, in the form of code:

$$\frac{3}{2}L(B - \sigma) = 0x5f3759df$$

I am so cute, please give me money...