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src/Advanced/Fast Inverse Root/main.tex

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+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	singlenumbering,
+	solutions
+]{../../../lib/tex/ormc_handout}
+\usepackage{../../../lib/tex/macros}
+
+\usepackage{listings}
+
+\uptitlel{Advanced 2}
+\uptitler{\smallurl{}}
+\title{Fast Inverse Square Root}
+\subtitle{Prepared by Mark on \today}
+
+\begin{document}
+
+	\maketitle
+
+	\input{parts/1 int.tex}
+	\input{parts/2 float.tex}
+	\input{parts/3 approximate.tex}
+	\input{parts/4 quake.tex}
+\end{document}

src/Advanced/Fast Inverse Root/parts/1 int.tex

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+\section{Integers}
+
+\definition{}
+A \textit{bit string} is a string of binary digits. \par
+In this handout, we'll denote bit strings with the prefix \texttt{0b}. \par
+That is, $1010 =$ \say{one thousand and one,} while $\texttt{0b1001} = 2^3 + 2^0 = 9$
+
+\vspace{2mm}
+We will seperate long bit strings with underscores for readability. \par
+Underscores have no meaning: $\texttt{0b1111\_0000} = \texttt{0b11110000}$.
+
+\problem{}
+What is the value of the following bit strings, if we interpret them as integers in base 2? \par
+\begin{itemize}
+	\item \texttt{0b0001\_1010}
+	\item \texttt{0b0110\_0001}
+\end{itemize}
+
+\begin{solution}
+	\begin{itemize}
+		\item $\texttt{0b0001\_1010} = 2 + 8 + 16 = 26$
+		\item $\texttt{0b0110\_0001} = 1 + 32 + 64 = 95$
+	\end{itemize}
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+\definition{}
+We can interpret a bit string in any number of ways. \par
+One such interpretation is the \textit{signed integer}, or \texttt{int} for short. \par
+\texttt{ints} allow us to represent negative and positive integers using 32-bit strings.
+
+\vspace{2mm}
+
+The first bit of an \texttt{int} tells us its sign:
+\begin{itemize}
+	\item if the first bit is \texttt{1}, the \textit{int} represents a negative number;
+	\item if the first bit is \texttt{0}, it represents a positive number.
+\end{itemize}
+We do not need negative numbers today, so we will assume that the first bit is always zero. \par
+\note{If you'd like to know how negative integers are written, look up \say{two's complement} after class.}
+
+\vspace{2mm}
+
+The value of a positive signed \texttt{long} is simply the value of its binary digits:
+\begin{itemize}
+	\item $\texttt{0b00000000\_00000000\_00000000\_00000000} = 0$
+	\item $\texttt{0b00000000\_00000000\_00000000\_00000011} = 3$
+	\item $\texttt{0b00000000\_00000000\_00000000\_00100000} = 32$
+	\item $\texttt{0b00000000\_00000000\_00000000\_10000010} = 130$
+\end{itemize}
+
+\problem{}
+What is the largest number we can represent with a 32-bit \texttt{int}?
+
+\begin{solution}
+	$\texttt{0b01111111\_11111111\_11111111\_11111111} = 2^{31}$
+\end{solution}
+
+\vfill
+
+\problem{}
+What is the smallest possible number we can represented with a 32-bit \texttt{int}? \par
+\hint{
+	You do not need to know \textit{how} negative numbers are represented. \par
+	Assume that we do not skip any integers, and don't forget about zero.
+}
+
+\begin{solution}
+	There are $2^{64}$ possible 32-bit patterns,
+	of which 1 represents zero and $2^{31}$ represent positive numbers.
+	We therefore have access to $2^{64} - 1 - 2^{31}$ negative numbers,
+	giving us a minimum representable value of $-2^{31} + 1$.
+\end{solution}
+
+\vfill
+
+\problem{}
+Find the value of each of the following 32-bit \texttt{int}s:
+\begin{itemize}
+	\item \texttt{0b00000000\_00000000\_00000101\_00111001}
+	\item \texttt{0b00000000\_00000000\_00000001\_00101100}
+	\item \texttt{0b00000000\_00000000\_00000100\_10110000}
+\end{itemize}
+\hint{The third conversion is easy---look carefully at the second.}
+
+\begin{solution}
+	\begin{itemize}
+		\item $\texttt{0b00000000\_00000000\_00000101\_00111001} = 1337$
+		\item $\texttt{0b00000000\_00000000\_00000001\_00101100} = 300$
+		\item $\texttt{0b00000000\_00000000\_00000010\_01011000} = 1200$
+	\end{itemize}
+
+	Notice that the third long is the second shifted left twice (i.e, multiplied by 4)
+\end{solution}
+
+\vfill
+\vfill
+\pagebreak

src/Advanced/Fast Inverse Root/parts/2 float.tex

@@ -0,0 +1,186 @@
+\section{Floats}
+
+\definition{}
+\textit{Binary decimals}\footnotemark{} are very similar to base-10 decimals. \par
+In base 10, we interpret place value as follows:
+\begin{itemize}
+	\item $0.1 = 10^{-1}$
+	\item $0.03 = 3 \ \times 10^{-2}$
+	\item $0.0208 = 2 \times 10^{-2} + 8 \times 10^{-4}$
+\end{itemize}
+
+\footnotetext{\say{decimal} is a misnomer, but that's ok.}
+
+\vspace{5mm}
+
+We can do the same in base 2:
+\begin{itemize}
+	\item $\texttt{0.1} = 2^{-1} = 0.5$
+	\item $\texttt{0.011} = 2^{-2} + 2^{-3} = 0.375$
+	\item $\texttt{101.01} = 5.125$
+\end{itemize}
+
+\vspace{5mm}
+
+\problem{}
+Rewrite the following binary decimals in base 10: \par
+\note{You may leave your answer as a fraction.}
+\begin{itemize}
+	\item $\texttt{1011.101}$
+	\item $\texttt{110.1101}$
+\end{itemize}
+
+\vfill
+\pagebreak
+
+\definition{}
+Another way we can interpret a bit string is as a \textit{signed floating-point decimal}, or a \texttt{float} for short. \par
+Floats represent a subset of the real numbers, and are interpreted as follows: \par
+\note{The following only applies to floats that consist of 32 bits. We won't encounter any others today.}
+\begin{center}
+	\begin{tikzpicture}
+		\node[anchor=south west] at (0, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (0.25, 0) {\texttt{\texttt{b}}};
+		\node[anchor=south west] at (0.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (0.75, 0) {\texttt{\texttt{\_}}};
+
+		\node[anchor=south west] at (1.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (1.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (1.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (1.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.75, 0) {\texttt{\texttt{0}}};
+
+		\node[anchor=south west] at (3.00, 0) {\texttt{\texttt{\_}}};
+		\node[anchor=south west] at (3.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (3.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (3.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (5.00, 0) {\texttt{\texttt{\_}}};
+
+		\node[anchor=south west] at (5.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (5.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (5.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (7.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (7.25, 0) {\texttt{\texttt{\_}}};
+
+		\node[anchor=south west] at (7.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (7.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (9.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (9.25, 0) {\texttt{\texttt{0}}};
+
+
+		\draw (0.50, 0) -- (0.95, 0) node [midway, below=1mm] {sign};
+		\draw (1.05, 0) -- (3.15, 0) node [midway, below=1mm] {exponent};
+		\draw (3.30, 0) -- (9.70, 0) node [midway, below=1mm] {fraction};
+	\end{tikzpicture}
+\end{center}
+
+\begin{itemize}[itemsep = 2mm]
+	\item The first bit denotes the sign of the float's value. We'll label it $s$. \par
+	If $s = \texttt{1}$, this float is negative; if $s = \texttt{0}$, it is positive.
+
+	\item The next eight bits represent the \textit{exponent} of this float. \note{(we'll see what that means soon)}\par
+	We'll call the value of this eight-bit binary integer $E$. \par
+	Naturally, $0 \leq E \leq 255$ \note{(since $E$ consist of eight bits.)}
+
+	\item The remaining 23 bits represent the \textit{fraction} of this float, which we'll call $F$. \par
+	These 23 bits are interpreted as the fractional part of a binary decimal. \par
+	For example, the bits \texttt{0b1010000\_00000000\_00000000} represents $0.5 + 0.125 = 0.625$.
+\end{itemize}
+
+\problem{}<floata>
+Consider \texttt{0b01000001\_10101000\_00000000\_00000000}. \par
+Find the $s$, $E$, and $F$ we get if we interpret this bit string as a \texttt{float}. \par
+\note[Note]{Leave $F$ as a sum of powers of two.}
+
+\begin{solution}
+	$s = 0$ \par
+	$E = 258$ \par
+	$F = 2^{31}+2^{19} = 2,621,440$
+\end{solution}
+
+\vfill
+
+
+\definition{}
+The final value of a float with sign $s$, exponent $E$, and fraction $F$ is
+\begin{equation*}
+	(-1)^s ~\times~ 2^{E - 127} ~\times~ \left(1 + \frac{F}{2^{23}}\right)
+\end{equation*}
+
+Notice that this is very similar to decimal scientific notation, which is written as
+\begin{equation*}
+	(-1)^s ~\times~ 10^{e} ~\times~ (f)
+\end{equation*}
+
+\problem{}
+Consider \texttt{0b01000001\_10101000\_00000000\_00000000}. \par
+This is the same bit string we used in \ref{floata}. \par
+
+\vspace{2mm}
+
+What value do we get if we interpret this bit string as a float? \par
+\hint{$21 \div 16 = 1.3125$}
+
+\begin{solution}
+	This is 21:
+	\begin{equation*}
+		2^{131} \times \biggl(1 + \frac{2^{21} + 2^{19}}{2^{23}}\biggr)
+		~=~ 2^{4} \times (1 + 0.25 + 0.0625)
+		~=~ 16 \times (1.3125)
+		~=~ 21
+	\end{equation*}
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}
+Encode $12.5$ as a float. \par
+\hint{$12.5 \div 8 = 1.5625$}
+
+\begin{solution}
+	\begin{equation*}
+		12.5
+		~=~ 8 \times 1.5625
+		~=~ 2^{3} \times \biggl(1 + (0.5 + 0.0625)\biggr)
+		~=~ 2^{130} \times \biggl(1 + \frac{2^{22} + 2^{19}}{2^{23}}\biggr)
+	\end{equation*}
+
+	which is \texttt{0b01000001\_01001000\_00000000\_00000000}. \par
+\end{solution}
+
+
+\vfill
+
+\definition{}
+Say we have a bit string $x$. \par
+We'll let $x_f$ denote the value we get if we interpret $x$ as a float, \par
+and we'll let $x_i$ denote the value we get if we interpret $x$ an integer.
+
+\problem{}
+Let $x = \texttt{0b01000001\_01001000\_00000000\_00000000}$. \par
+What are $x_f$ and $x_i$? \note{As always, you may leave big numbers as powers of two.}
+\begin{solution}
+	$x_f = 12.5$ \par
+	\vspace{2mm}
+	$x_i = 2^{30} + 2^{24} + 2^{22} + 2^{19} = 11,095,237,632$
+\end{solution}
+
+\vfill
+
+\pagebreak

src/Advanced/Fast Inverse Root/parts/3 approximate.tex

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+\section{Integers and Floats}
+
+\generic{Observation:}
+For small values of $a$, $\log_2(1 + a)$ is approximately equal to $a$. \par
+Note that this equality is exact for $a = 0$ and $a = 1$, since $\log_2(1) = 0$ and $\log_2(2) = 1$.
+
+\vspace{2mm}
+
+We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $\log_2(1 + a) \approx a + \varepsilon$.
+
+TODO: why? Graphs.
+
+\problem{}<convert>
+Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$. \par
+
+\vspace{5mm}
+
+Namely, show that
+\begin{equation*}
+	\log_2(x_f) ~=~ \frac{x_i}{2^{23}} - 127 + \varepsilon
+\end{equation*}
+for some correction term term $\varepsilon$ \par
+\note{
+	In other words, we're finding an expression for $x$ as a float
+	in terms of $x$ as an int.
+}
+
+\begin{solution}
+	Let $E$ and $F$ be the exponent and float bits of $x_f$. \par
+	We then have:
+	\begin{align*}
+		\log_2(x_f)
+		&=~ \log_2 \left( 2^{E-127} \times \left(1 + \frac{F}{2^{23}}\right) \right) \\
+		&=~ E-127 + \log_2\left(1 + \frac{F}{2^{23}}\right) \\
+		&\approx~ E-127 + \frac{F}{2^{23}} + \varepsilon \\
+		&=~ \frac{1}{2^{23}}(2^{23}E + F) - 127 + \varepsilon \\
+		&=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
+	\end{align*}
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Fast Inverse Root/parts/4 quake.tex

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