132 lines
3.7 KiB
TeX
132 lines
3.7 KiB
TeX
\section{The Fast Inverse Square Root}
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The following code is present in \textit{Quake III Arena} (1999):
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\lstset{
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breaklines=false,
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numbersep=5pt,
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xrightmargin=0in
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}
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\begin{lstlisting}[language=C]
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float Q_rsqrt( float number ) {
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long i = * ( long * ) &number;
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i = 0x5f3759df - ( i >> 1 );
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return * ( float * ) &i;
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}
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\end{lstlisting}
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This code defines a function \texttt{Q\_rsqrt} that consumes a float named
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\texttt{number} and approximates its inverse square root (in other words, \texttt{Q\_rsqrt} computes $1/\sqrt{\texttt{number}}$).
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\vspace{2mm}
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If we rewrite this using notation we're familiar with, we get the following:
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\begin{equation*}
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\texttt{Q\_sqrt}(n_f) = 6240089 - (n_i \div 2) \approx \frac{1}{\sqrt{n_f}}
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\end{equation*}
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\note{
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\texttt{0x5f3759df} is $6240089$ in hexadecimal. \par
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It is a magic number hard-coded into \texttt{Q\_sqrt}.
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}
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\vspace{2mm}
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Our goal in this section is to understand why this works: \par
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\begin{itemize}
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\item How does Quake approximate $\frac{1}{\sqrt{x}}$ by simply subtracting and dividing by two?
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\item What's special about $6240089$?
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\end{itemize}
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\problem{}
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Using basic log rules, rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2(x)$.
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\begin{solution}
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\begin{equation*}
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\log_2(1 / \sqrt{x}) = \frac{-1}{2}\log_2(x)
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\end{equation*}
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\end{solution}
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\vfill
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\pagebreak
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\generic{Setup:}
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We are now ready to show that $\texttt{Q\_sqrt} \approx \frac{1}{\sqrt{n_f}}$. \par
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For convenience, let's call the bit string of the inverse square root $r$. \par
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In other words,
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\begin{equation*}
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r_f := \frac{1}{\sqrt{n_f}}
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\end{equation*}
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This is the value we want to approximate.
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\problem{}<finala>
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Find an approximation for $\log_2(r_f)$ in terms of $n_i$ and $\varepsilon$ \par
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\note{Remember, $\varepsilon$ is the correction constant in our approximation of $\log_2(1 + a)$.}
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\begin{solution}
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\begin{equation*}
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\log_2(r_f)
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~=~ \log_2(\frac{1}{\sqrt{n_f}})
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~=~ \frac{-1}{2}\log_2(n_f)
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~\approx~ \frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right)
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}<finalb>
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Let's call the \say{magic number} in the code above $\kappa$, so that
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\begin{equation*}
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\texttt{Q\_sqrt}(n_f) = \kappa - (n_i \div 2)
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\end{equation*}
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Use problems \ref{num:convert} and \ref{num:finala} to show that $\texttt{Q\_sqrt}(n_f) \approx r_i$. \par
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\begin{solution}
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From \ref{convert}, we know that
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\begin{equation*}
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\log_2(r_f) \approx \frac{r_i}{2^{23}} + \varepsilon - 127
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\end{equation*}
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\note{
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Our approximation of $\log_2(1+a)$ uses a fixed correction constant, \par
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so the $\varepsilon$ here is equivalent to the $\varepsilon$ in \ref{finala}.
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}
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Combining this with the result from \ref{finala}, we get:
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\begin{align*}
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\frac{r_i}{2^{23}} + \varepsilon - 127
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&~\approx~\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right) \\
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\frac{r_i}{2^{23}}
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&~\approx~\frac{-1}{2}\left( \frac{n_i}{2^{23}} \right) + \frac{3}{2}(\varepsilon - 127) \\
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r_i
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&~\approx~\frac{-1}{2}\left(n_i \right) + 2^{23}\frac{3}{2}(\varepsilon - 127)
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~=~ 2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
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\end{align*}
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\vspace{2mm}
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This is exactly what we need! If we set $\kappa$ to $\frac{2^{24}}{3}(\varepsilon - 127)$, then
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\begin{equation*}
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r_i ~\approx~ \kappa - (n_i \div 2) ~=~ \texttt{Q\_sqrt}(n_f)
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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What is the exact value of $\kappa$ in terms of $\varepsilon$? \par
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\hint{Look at \ref{finalb}. We already found it!}
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\begin{solution}
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This problem makes sure our students see that
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$\kappa = \frac{2^{24}}{3}(\varepsilon - 127)$. \par
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See the solution to \ref{finalb}.
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\end{solution}
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\makeatletter
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\if@solutions\else
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\vspace{2cm}
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\fi\makeatother
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\pagebreak |