Quake edits

src/Advanced/Fast Inverse Root/main.tex

@@ -37,28 +37,28 @@
 
 	\maketitle
 
-
+	%\begin{center}
-	\begin{center}
+	%\begin{minipage}{6cm}
-	\begin{minipage}{6cm}
+	%	Dad says that anyone who can't use
-		Dad says that anyone who can't use
+	%	a slide rule is a cultural illiterate
-		a slide rule is a cultural illiterate
+	%and should not be allowed to vote.
-		and should not be allowed to vote.
+	%
-
+	%	\vspace{1ex}
-		\vspace{1ex}
+	%
-
+	%	\textit{Have Space Suit --- Will Travel, 1958}
-		\textit{Have Space Suit --- Will Travel, 1958}
+	%\end{minipage}
-	\end{minipage}
+	%\end{center}
-	\end{center}
+	%\hfill
-	\hfill
 
 	%\input{parts/0 logarithms.tex}
 	%\input{parts/1 intro.tex}
 	%\input{parts/2 multiplication.tex}
 	%\input{parts/3 division.tex}
-	\pagebreak
+	%\pagebreak
-	\input{parts/4 float.tex}
+	\input{parts/4 int.tex}
-	\input{parts/5 approximate.tex}
+	\input{parts/5 float.tex}
-	\input{parts/6 quake.tex}
+	\input{parts/6 approximate.tex}
+	\input{parts/7 quake.tex}
 
 	% Make sure the slide rule is on an odd page,
 	% so that double-sided prints won't require

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

@@ -0,0 +1,101 @@
+\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/5 float.tex

@@ -1,99 +1,44 @@
-
+\section{Floats}
-\section{\texttt{int}s and \texttt{float}s}
 
 \definition{}
-A \textit{signed 32-bit integer} (equivalently, a \texttt{long int}) consists of thirty-two binary digits, \par
+\textit{Binary decimals}\footnotemark{} are very similar to base-10 decimals. \par
-and is used represent a subset of the integers.
+In base 10, we interpret place value as follows:
-
-\vspace{2mm}
-
-The first bit of a \texttt{long} tells us its sign:
 \begin{itemize}
-	\item if the first bit of a \texttt{long} is \texttt{1}, it represents a negative number;
+	\item $0.1 = 10^{-1}$
-	\item if the first bit is \texttt{0}, it represents a positive number.
+	\item $0.03 = 3 \ \times 10^{-2}$
-\end{itemize}
+	\item $0.0008 = 8 \times 10^{-4}$
-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}
-
-We'll denote binary strings with the prefix \texttt{0b}. \par
-Underscores are added between every eight digits for readability, and have no meaning.
-
-\vspace{2mm}
-
-
-The value of a positive signed \texttt{long} is simply the value of its binary digits. \par
-For example:
-\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}
 
-Remember---we only need positive integers today. Assume the \say{sign} bit is always \texttt{0}.
+\footnotetext{this 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{}
-What is the largest number that can be represented with a \texttt{long}?
+Rewrite the following binary decimals in base 10: \par
-
+\note{You may leave your answer as a fraction}
-\begin{solution}
-	$\texttt{0b01111111\_11111111\_11111111\_11111111} = 2^{31}$
-\end{solution}
-
-\vfill
-
-
-
-\problem{}
-What is the smallest possible number that can be represented with a \texttt{long}? \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.
-
-	\vspace{2mm}
-
-	We therefore have access to $2^{64} - 1 - 2^{31}$ negative numbers,
-	giving us a minimum representable value of $-2^{31} + 1$.
-\end{solution}
-
-\problem{}
-What is the value of the following longs?
 \begin{itemize}
-	\item \texttt{0b00000000\_00000000\_00000101\_00111001}
+	\item $\texttt{1011.101}$
-	\item \texttt{0b00000000\_00000000\_00000001\_00101100}
+	\item $\texttt{110.1101}$
-	\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
 \pagebreak
 
-
 \definition{}
-A \textit{signed 32-bit floating-point decimal} (equivalently, a \textit{float})
+Another way we can interpret a bit string is as a \textit{signed floating-point decimal}, or a \texttt{float} for short. \par
-consists of 32 binary digits, and is used to represent a subset of the real numbers.
+Floats represent a subset of the real numbers, and are interpreted as follows: \par
-These 32 bits are interpreted as follows:
+\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}}};
@@ -144,76 +89,98 @@ These 32 bits are interpreted as follows:
 	\end{tikzpicture}
 \end{center}
 
-In other words:
+\begin{itemize}[itemsep = 2mm]
-\begin{itemize}[itemsep = 1mm]
 	\item The first bit denotes the sign of the float's value. We'll label it $s$. \par
-	If $s = 1$, this float is negative; if $s = 0$, it is positive.
+	If $s = \texttt{1}$, this float is negative; if $s = \texttt{0}$, it is positive.
 
-	\item The next 8 bits represent the \textit{exponent} of this float. \par
+	\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 these eight bits $E$. \par
+	We'll call the value of this eight-bit binary integer $E$. \par
-	Naturally, $0 \leq E \leq 255$
+	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. \par
+	\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.}
 
-\vspace{2mm}
+\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*}
-	(\pm 1) ~\times~ 10^{e} ~\times~ (f)
+	(-1)^s ~\times~ 10^{e} ~\times~ (f)
 \end{equation*}
 
-\vfill
-\pagebreak
-
 \problem{}
-What is the value of \texttt{0b01000001\_10101000\_00000000\_00000000} if it is interpreted as a float? \par
+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{align*}
+	\begin{equation*}
-		&=~ 2^{131} \times \biggl(1 + \frac{2^{21} + 2^{19}}{2^{23}}\biggr) \\
+		2^{131} \times \biggl(1 + \frac{2^{21} + 2^{19}}{2^{23}}\biggr)
-		&=~ 2^{4} \times (1 + 0.25 + 0.0625) \\
+		~=~ 2^{4} \times (1 + 0.25 + 0.0625)
-		&=~ 16 \times (1.3125) \\
+		~=~ 16 \times (1.3125)
-		&=~ 21
+		~=~ 21
-	\end{align*}
+	\end{equation*}
 \end{solution}
 
 \vfill
+\pagebreak
 
 \problem{}
 Encode $12.5$ as a float. \par
 \hint{$12.5 \div 8 = 1.5625$}
 
-\vspace{2mm}
-
-What is the value of the resulting 32 bits if they are interpreted as a long? \par
-\hint{A sum of powers of two is fine.}
-
 \begin{solution}
-	\begin{align*}
+	\begin{equation*}
 		12.5
-		&=~ 8 \times 1.5625 \\
+		~=~ 8 \times 1.5625
-		&=~ 2^{3} \times \biggl(1 + (0.5 + 0.0625)\biggr) \\
+		~=~ 2^{3} \times \biggl(1 + (0.5 + 0.0625)\biggr)
-		&=~ 2^{130} \times \biggl(1 + \frac{2^{22} + 2^{19}}{2^{23}}\biggr)
+		~=~ 2^{130} \times \biggl(1 + \frac{2^{22} + 2^{19}}{2^{23}}\biggr)
-	\end{align*}
+	\end{equation*}
 
-	\linehack{}
+	which is \texttt{0b01000001\_01001000\_00000000\_00000000}. \par
-
-	This is \texttt{0b01000001\_01001000\_00000000\_00000000}, \par
-	which is $2^{30} + 2^{24} + 2^{22} + 2^{19} = 11,095,237,632$
 \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/6 approximate.tex

@@ -1,21 +1,5 @@
 \section{Approximation}
 
-\definition{}
-Say we have a bit string $x$. \par
-Let $x_f$ denote the value of $x$ interpreted as a float, \par
-and let $x_i$ denote the value of $x$ interpreted as an integer.
-
-\problem{}
-Let $x = \texttt{0b01000001\_01001000\_00000000\_00000000}$. \
-What are $x_f$ and $x_i$?
-
-\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}
-
-
 \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 $\log_2(2) = 1$.
@@ -27,7 +11,15 @@ We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $
 TODO: why? Graphs.
 
 \problem{}
-Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$, \par
+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{1}{2^{23}}(x_i) - 127 + \varepsilon
+\end{equation*}
+for some error term $\varepsilon$
 
 
 \begin{solution}

src/Advanced/Fast Inverse Root/parts/7 quake.tex

@@ -1,13 +1,11 @@
 \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;
@@ -16,45 +14,55 @@ float Q_rsqrt( float number ) {
 }
 \end{lstlisting}
 
-It defines a method \texttt{Q\_rsqrt} that consumes a float named \texttt{number} and quickly approximates its inverse
+This code defines
-square root (in other words, \texttt{Q\_rsqrt} computes $1/\sqrt{\texttt{number}}$).
+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{8mm}
+\vspace{2mm}
 
 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)
+	\texttt{Q\_sqrt}(n_i) = 6240089 - (n_i \div 2) \approx \frac{1}{\sqrt{n_f}}
 \end{equation*}
-Where $\kappa$ is the magic constant $6240089$ (which is \texttt{0x5f3759df} in hexadecimal)
+$6240089$ is the decimal value of hex \texttt{0x5f3759df}. \par
+It is a magic number hard-coded into the function.
+
+\vspace{2mm}
+
+Our goal in this section is to understand why this works. \par
+How are we able to approximate $\frac{1}{\sqrt{x}}$ by only subtracting and dividing by two??
+
+\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
+
 
 \problem{}
 Find the exact value of $\kappa$ in terms of $\varepsilon$. \par
-\hint{Remember that $\varepsilon$ is the correction term in the approximation $\log_2(1 + a) = a + \varepsilon$.}
+\note{Remember, $\varepsilon$ 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
+	Say $g_f = \frac{1}{\sqrt{n_f}}$ (i.e, $g_f$ is the value we want to compute). We then have:
-	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) \\
@@ -65,7 +73,9 @@ Rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2{x}$.
 		&=2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
 	\end{align*}
 
-	thus, $\kappa = 2^{23}\frac{3}{2}(\varepsilon - 127)$.
+	and thus $\kappa = 2^{23}\frac{3}{2}(\varepsilon - 127)$.
 \end{solution}
 
+\vfill
+
 \pagebreak