\section{The Fast Inverse Square Root}

The following code is present in \textit{Quake III Arena} (1999):

\lstset{
	breaklines=false,
	numbersep=5pt,
	xrightmargin=0in
}

\begin{lstlisting}[language=C]
float Q_rsqrt( float number ) {
    long i = * ( long * ) &number;
    i = 0x5f3759df - ( i >> 1 );
    return * ( float * ) &i;
}
\end{lstlisting}

It defines a method \texttt{Q\_rsqrt} that consumes a float named \texttt{number} and quickly approximates its inverse
square root (in other words, \texttt{Q\_rsqrt} computes $1/\sqrt{\texttt{number}}$).

\vspace{8mm}

If we rewrite this using the notation we're familiar with, we get the following:

\begin{equation*}
	\frac{1}{\sqrt{n_f}} \approx \kappa - (n_i \div 2)
\end{equation*}
Where $\kappa$ is the magic constant $6240089$ (which is \texttt{0x5f3759df} in hexadecimal)

\problem{}
Find the exact value of $\kappa$ in terms of $\varepsilon$. \par
\hint{Remember that $\varepsilon$ is the correction term in the approximation $\log_2(1 + a) = a + \varepsilon$.}

\problem{}
Rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2{x}$.

\begin{solution}
	Say $g_f = \frac{1}{\sqrt{n_f}}$---that is, $g_f$ is the value we want to compute. \par
	Then:

	\begin{align*}
		\log_2(g_f)
		&=\log_2(\frac{1}{\sqrt{n_f}}) \\
		&=\frac{-1}{2}\log_2(n_f) \\
		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right)
	\end{align*}

	But we also know that

	\begin{align*}
		\log_2(g_f)
		&=\frac{g_i}{2^{23}} + \varepsilon - 127
	\end{align*}

	So,

	\begin{align*}
		\frac{g_i}{2^{23}} + \varepsilon - 127
		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right) \\
		\frac{g_i}{2^{23}}
		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} \right) + \frac{3}{2}(\varepsilon - 127) \\
		g_i
		&=\frac{-1}{2}\left(n_i \right) + 2^{23}\frac{3}{2}(\varepsilon - 127) \\
		&=2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
	\end{align*}

	thus, $\kappa = 2^{23}\frac{3}{2}(\varepsilon - 127)$.
\end{solution}

\pagebreak