Quake edits

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{pdfpages}
+\usepackage{sliderule}
+\usepackage{changepage}
+\usepackage{listings}
+
+% Args:
+% x, top scale y, label
+\newcommand{\slideruleind}[3]{
+	\draw[
+		line width=1mm,
+		draw=black,
+		opacity=0.3,
+		text opacity=1
+	]
+		({#1}, {#2 + 1})
+		--
+		({#1}, {#2 - 1.1})
+		node [below] {#3};
+}
+
+
+\uptitlel{Advanced 2}
+\uptitler{\smallurl{}}
+\title{Fast Inverse Square Root}
+\subtitle{Prepared by Mark on \today}
+
+\begin{document}
+
+	\maketitle
+
+	%\begin{center}
+	%\begin{minipage}{6cm}
+	%	Dad says that anyone who can't use
+	%	a slide rule is a cultural illiterate
+	%and should not be allowed to vote.
+	%
+	%	\vspace{1ex}
+	%
+	%	\textit{Have Space Suit --- Will Travel, 1958}
+	%\end{minipage}
+	%\end{center}
+	%\hfill
+
+	%\input{parts/0 logarithms.tex}
+	%\input{parts/1 intro.tex}
+	%\input{parts/2 multiplication.tex}
+	%\input{parts/3 division.tex}
+	%\pagebreak
+	\input{parts/4 int.tex}
+	\input{parts/5 float.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
+	% students to tear off problems.
+	\checkoddpage
+	\ifoddpage\else
+		\vspace*{\fill}
+		\begin{center}
+		{
+			\Large
+			\textbf{This page unintentionally left blank.}
+		}
+		\end{center}
+		\vspace{\fill}
+		\pagebreak
+	\fi
+
+	%\includepdf[
+	%	pages=1,
+	%	fitpaper=true
+	%]{resources/rule.pdf}
+
+\end{document}

src/Advanced/Fast Inverse Root/parts/0 logarithms.tex

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+\section{Logarithms}
+
+\definition{}<logdef>
+The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \par
+In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?''
+
+\vspace{2mm}
+
+In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
+
+
+\problem{}
+Evaluate the following by hand:
+
+\begin{enumerate}
+	\item $\log_{10}{(1000)}$
+	\vfill
+	\item $\log_2{(64)}$
+	\vfill
+	\item $\log_2{(\frac{1}{4})}$
+	\vfill
+	\item $\log_x{(x)}$ for any $x$
+	\vfill
+	\item $log_x{(1)}$ for any $x$
+	\vfill
+\end{enumerate}
+
+\pagebreak
+
+
+
+\problem{}<logids>
+Prove the following identities:
+
+\begin{enumerate}[itemsep=2mm]
+	\item $\log_b{(b^x)} = x$
+	\item $b^{\log_b{x}} = x$
+	\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
+	\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
+	\item $\log_b{(x^y)} = y \log_b{(x)}$
+\end{enumerate}
+
+\vfill
+
+\begin{instructornote}
+	A good intro to the following sections is the linear slide rule:
+
+	\begin{center}
+		\begin{tikzpicture}[scale=1]
+			\linearscale{2}{1}{}
+			\linearscale{0}{0}{}
+
+			\slideruleind
+				{5}
+				{1}
+				{2 + 3 = 5}
+	\end{tikzpicture}
+	\end{center}
+
+	Take two linear rulers, offset one, and you add. \par
+	If you do the same with a log scale, you multiply!
+
+	\vspace{2mm}
+
+	Note that the slide rules above start at 0.
+
+	\linehack{}
+
+	After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
+\end{instructornote}
+
+\pagebreak

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

@@ -0,0 +1,44 @@
+\section{Slide Rules}
+
+Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
+
+\vspace{2mm}
+
+The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. \par
+Before you continue, tear off the last page of this handout and assemble your slide rule.
+
+\vspace{2mm}
+
+There are four scales on your slide rule, each labeled with a letter on the left side:
+
+\def\sliderulewidth{13}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\tscale{0}{9}{T}
+	\kscale{0}{8}{K}
+	\abscale{0}{7}{A}
+
+	\abscale{0}{5.5}{B}
+	\ciscale{0}{4.5}{CI}
+	\cdscale{0}{3.5}{C}
+
+	\cdscale{0}{2}{D}
+	\lscale{0}{1}{L}
+	\sscale{0}{0}{S}
+\end{tikzpicture}
+\end{center}
+
+Each scale's ``generating function'' is on the right:
+\begin{itemize}
+	\item T: $\tan$
+	\item K: $x^3$
+	\item A,B: $x^2$
+	\item CI: $\frac{1}{x}$
+	\item C, D: $x$
+	\item L: $\log_{10}(x)$
+	\item S: $\sin$
+\end{itemize}
+
+Once you understand the layout of your slide rule, move on to the next page.
+
+\pagebreak

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

@@ -0,0 +1,278 @@
+\section{Multiplication}
+
+We'll use the C and D scales of your slide rule to multiply. \par
+Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index}
+of the C scale over the smaller number, $2$:
+
+\def\sliderulewidth{10}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+\end{tikzpicture}
+\end{center}
+
+Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{1}
+		{6}
+
+\end{tikzpicture}
+\end{center}
+
+Of course, our answer is 6.
+
+\problem{}
+What is $1.15 \times 2.1$? \par
+Use your slide rule.
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=1]
+			\cdscale{\cdscalefn(1.15)}{1}{C}
+			\cdscale{0}{0}{D}
+
+			\slideruleind
+				{\cdscalefn(1.15)}
+				{1}
+				{1.15}
+
+			\slideruleind
+				{\cdscalefn(1.15) + \cdscalefn(2.1)}
+				{1}
+				{2.415}
+
+		\end{tikzpicture}
+		\end{center}
+\end{solution}
+
+\vfill
+
+Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within
+two decimal places is close enough for most practical applications.
+
+\pagebreak
+
+
+
+
+Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
+What should we do if we want to calculate $32 \times 210$? \par
+
+\problem{}
+Using your slide rule, calculate $32 \times 210$.
+
+
+\begin{solution}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(2.1)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(2.1)}
+			{1}
+			{2.1}
+
+		\slideruleind
+			{\cdscalefn(2.1) + \cdscalefn(3.2)}
+			{1}
+			{6.72}
+
+	\end{tikzpicture}
+	\end{center}
+
+		Placing the decimal point correctly is your job. \par
+		$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
+\end{solution}
+
+\vfill
+
+\problem{}
+Compute the following:
+\begin{enumerate}
+	\item $1.44 \times 52$
+	\item $0.38 \times 1.24$
+	\item $\pi \times 2.35$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $1.44 \times 52 = 74.88$
+		\item $0.38 \times 1.24 = 0.4712$
+		\item $\pi \times 2.35 = 7.382$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+
+\problem{}<provemult>
+Note that the numbers on your C and D scales are logarithmically spaced.
+
+\def\sliderulewidth{13}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{0}{1}{C}
+	\cdscale{0}{0}{D}
+\end{tikzpicture}
+\end{center}
+
+Why does our multiplication procedure work?
+
+\vfill
+\pagebreak
+
+Now we want to compute $7.2 \times 5.5$:
+
+\def\sliderulewidth{10}
+\begin{center}
+\begin{tikzpicture}[scale=0.8]
+	\cdscale{\cdscalefn(5.5)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(5.5)}
+		{1}
+		{5.5}
+
+	\slideruleind
+		{\cdscalefn(5.5) + \cdscalefn(7.2)}
+		{1}
+		{???}
+
+\end{tikzpicture}
+\end{center}
+
+No matter what order we go in, the answer ends up off the scale. There must be another way.
+
+\vspace{2mm}
+
+Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}.
+Move it over the \textit{larger} number, $7.2$:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(7.2)}
+		{1}
+		{7.2}
+
+\end{tikzpicture}
+\end{center}
+
+Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+
+	\slideruleind
+		{\cdscalefn(7.2)}
+		{1}
+		{7.2}
+
+	\slideruleind
+		{\cdscalefn(3.96)}
+		{1}
+		{3.96}
+
+\end{tikzpicture}
+\end{center}
+
+Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$.
+We can do this by hand to verify our answer.
+
+\vspace{2mm}
+
+\problem{}
+Why does this work?
+
+\begin{solution}
+	Consider the following picture, where we have two D scales next to each other:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=0.7]
+		\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+		\cdscale{0}{0}{}
+		\cdscale{-10}{0}{}
+
+		\draw[
+			draw=black,
+		]
+			(0, 0)
+			--
+			(0, -0.3)
+			node [below] {D};
+
+		\draw[
+			draw=black,
+		]
+			(-10, 0)
+			--
+			(-10, -0.3)
+			node [below] {D};
+
+		\slideruleind
+			{-10 + \cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(3.96)}
+			{1}
+			{3.96}
+
+	\end{tikzpicture}
+	\end{center}
+
+	\vspace{2mm}
+
+	The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$
+	smaller than it should be.
+
+	\vspace{2mm}
+
+	\vspace{2mm}
+	In other words, the answer we get from reverse multiplication is $\log{a} + \log{b} - \log{10}$. \par
+	This reduces to $\log{(\frac{a \times b}{10})}$, and explains the misplaced decimal point in $7.2 \times 5.5$.
+\end{solution}
+
+
+\vfill
+\pagebreak
+
+\problem{}
+Compute the following using your slide rule:
+\begin{enumerate}
+	\item $9 \times 8$
+	\item $15 \times 35$
+	\item $42.1 \times 7.65$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $9 \times 8 = 72$
+		\item $15 \times 35 = 525$
+		\item $42.1 \times 7.65 = 322.065$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+\pagebreak

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

@@ -0,0 +1,91 @@
+\section{Division}
+
+Now that you can multiply, division should be easy. All you need to do is work backwards. \\
+Let's look at our first example again: $3 \times 2 = 6$.
+
+\medskip
+
+We can easily see that $6 \div 3 = 2$
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{1}
+		{Align here}
+
+	\slideruleind
+		{\cdscalefn(2)}
+		{1}
+		{2}
+\end{tikzpicture}
+\end{center}
+
+and that $6 \div 2 = 3$:
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(3)}{-3}{C}
+	\cdscale{0}{-4}{D}
+
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{-3}
+		{Align here}
+
+	\slideruleind
+		{\cdscalefn(3)}
+		{-3}
+		{3}
+
+\end{tikzpicture}
+\end{center}
+
+If your left-hand index is off the scale, read the right-hand one. \\
+Consider $42.25 \div 6.5 = 6.5$:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+
+	\slideruleind
+		{\cdscalefn(4.225)}
+		{1}
+		{Align here}
+
+	\slideruleind
+		{\cdscalefn(6.5)}
+		{1}
+		{6.5}
+
+\end{tikzpicture}
+\end{center}
+
+Place your decimal points carefully.
+
+\vfill
+
+\problem{}
+Compute the following using your slide rule. \par
+
+\begin{enumerate}
+	\item $135 \div 15$
+	\item $68.2 \div 0.575$
+	\item $(118 \times 0.51) \div 6.6$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $135 \div 15 = 9$
+		\item $68.2 \div 0.575 = 118.609$
+		\item $(118 \times 0.51) \div 6.6 = 9.118$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+\pagebreak

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

@@ -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/6 approximate.tex

@@ -0,0 +1,46 @@
+\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 $\varepsilon$
+
+\vspace{5mm}
+
+Before you start, make sure you understand what we're trying to do: \par
+We're finding an expression for $\log_2(x_f)$ in terms of $x_i$ and an correction term $\varepsilon$. \par
+That is, we're finding a closed-form expression that connects the two interpretations \par
+(\texttt{float} and \texttt{int}) of the bit string $x$.
+
+
+\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/7 quake.tex

@@ -0,0 +1,132 @@
+\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 soluton to \ref{finalb}.
+\end{solution}
+
+\makeatletter
+\if@solutions\else
+	\vspace{2cm}
+\fi\makeatother
+
+\pagebreak

src/Advanced/Fast Inverse Root/sliderule.sty

@@ -0,0 +1,534 @@
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesPackage{sliderule}[2022/08/22 Slide rule tools]
+
+\RequirePackage{tikz}
+\RequirePackage{ifthen}
+
+
+% Scale functions:
+% See https://sliderulemuseum.com/SR_Scales.htm
+%
+% l: length of the rule
+% n: the number on the rule
+%
+% A/B: (l/2) * log(n)
+% C/D: l / log(n)
+% CI: abs(l * log(10 / n) - l)
+% K: (l/3) * log(n)
+%
+% L: n * l
+% T: l * log(10 * tan(n))
+% S: l * log(10 * sin(n))
+
+\def\sliderulewidth{10}
+
+\def\abscalefn(#1){(\sliderulewidth/2) * log10(#1)}
+\def\cdscalefn(#1){(\sliderulewidth * log10(#1))}
+\def\ciscalefn(#1){(\sliderulewidth - \cdscalefn(#1))}
+\def\kscalefn(#1){(\sliderulewidth/3) * log10(#1)}
+\def\lscalefn(#1){(\sliderulewidth * #1)}
+\def\tscalefn(#1){(\sliderulewidth * log10(10 * tan(#1)))}
+\def\sscalefn(#1){(\sliderulewidth * log10(10 * sin(#1)))}
+
+
+% Arguments:
+% Label
+% x of start
+% y of start
+\newcommand{\linearscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {0,..., 10}{
+		\draw[black]
+			({#1 + (\sliderulewidth / 10) * \i}, #2) --
+			({#1 + (\sliderulewidth / 10) * \i}, #2 + 0.3)
+			node[above] {\i};
+	}
+
+	% Submarks
+	\foreach \n in {0, ..., 9} {
+		\foreach \i in {1,..., 9} {
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+
+% Arguments:
+% Label
+% x of start
+% y of start
+\newcommand{\abscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 1 - 9
+	\foreach \i in {1,..., 9}{
+		\draw[black]
+			({#1 + \abscalefn(\i)}, #2) --
+			({#1 + \abscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+	% Numbers and marks 10 - 100
+	\foreach \i in {1,..., 10}{
+		\draw[black]
+			({#1 + \abscalefn(10 * \i)}, #2) --
+			({#1 + \abscalefn(10 * \i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<5}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \abscalefn(\n + \i / 10)}, #2) --
+					({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \abscalefn(\n + \i / 10)}, #2) --
+					({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 10 - 100
+	\foreach \n in {10,20,...,90} {
+		\ifthenelse{\n<50}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \abscalefn(\n + \i)}, #2) --
+					({#1 + \abscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \abscalefn(\n + \i)}, #2) --
+					({#1 + \abscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+\newcommand{\cdscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 1 - 10
+	\foreach \i in {1,..., 10}{
+		\draw[black]
+			({#1 + \cdscalefn(\i)}, #2) --
+			({#1 + \cdscalefn(\i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<3}{
+			\foreach \i in {5,10,...,95}
+		} {
+			\foreach \i in {10,20,...,90}
+		}
+		{
+			\ifthenelse{\i=50}{
+				\draw[black]
+					({#1 + \cdscalefn(\n + \i / 100)}, #2) --
+					({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2);
+				\ifthenelse{\n=1}{
+					\draw
+						({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2)
+						node [above] {1.5};
+				}{}
+			} {
+				\ifthenelse{
+					\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
+					\i=60 \OR \i=70 \OR \i=80 \OR \i=90
+				}{
+					\draw[black]
+						({#1 + \cdscalefn(\n + \i / 100)}, #2) --
+						({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.15);
+				} {
+					\draw[black]
+						({#1 + \cdscalefn(\n + \i / 100)}, #2) --
+						({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.1);
+				}
+			}
+		}
+	}
+}
+
+\newcommand{\ciscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {1,...,10}{
+		\draw[black]
+			({#1 + \ciscalefn(\i)}, #2) --
+			({#1 + \ciscalefn(\i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<3}{
+			\foreach \i in {5,10,...,95}
+		} {
+			\foreach \i in {10,20,...,90}
+		}
+		{
+			\ifthenelse{\i=50}{
+				\draw[black]
+					({#1 + \ciscalefn(\n + \i / 100)}, #2) --
+					({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2);
+				\ifthenelse{\n=1}{
+					\draw
+						({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2)
+						node [above] {1.5};
+				}{}
+			} {
+				\ifthenelse{
+					\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
+					\i=60 \OR \i=70 \OR \i=80 \OR \i=90
+				}{
+					\draw[black]
+						({#1 + \ciscalefn(\n + \i / 100)}, #2) --
+						({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.15);
+				} {
+					\draw[black]
+						({#1 + \ciscalefn(\n + \i / 100)}, #2) --
+						({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.1);
+				}
+			}
+		}
+	}
+}
+
+\newcommand{\kscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 1 - 9
+	\foreach \i in {1,...,9}{
+		\draw[black]
+			({#1 + \kscalefn(\i)}, #2) --
+			({#1 + \kscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+	% Numbers and marks 10 - 90
+	\foreach \i in {1,..., 9}{
+		\draw[black]
+			({#1 + \kscalefn(10 * \i)}, #2) --
+			({#1 + \kscalefn(10 * \i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+	% Numbers and marks 100 - 1000
+	\foreach \i in {1,..., 10}{
+		\draw[black]
+			({#1 + \kscalefn(100 * \i)}, #2) --
+			({#1 + \kscalefn(100 * \i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<4}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \kscalefn(\n + \i / 10)}, #2) --
+					({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \kscalefn(\n + \i / 10)}, #2) --
+					({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 10 - 90
+	\foreach \n in {10,20,...,90} {
+		\ifthenelse{\n<40}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 100 - 1000
+	\foreach \n in {100,200,...,900} {
+		\ifthenelse{\n<400}{
+			\foreach \i in {10,20,...,90}
+		} {
+			\foreach \i in {20,40,60,80}
+		}
+		{
+			\ifthenelse{\i=50}{
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+\newcommand{\lscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {0,..., 10}{
+		\draw[black]
+			({#1 + \lscalefn(\i / 10)}, #2) --
+			({#1 + \lscalefn(\i / 10)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
+	}
+
+	% Submarks
+	\foreach \n in {0, ..., 9} {
+		\foreach \i in {1,...,19} {
+			\ifthenelse{\i=10}{
+				\draw[black]
+					({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
+					({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.2);
+			} {
+				\ifthenelse{
+					\i=1 \OR \i=3 \OR \i=5 \OR \i=7 \OR
+					\i=9 \OR \i=11 \OR \i=13 \OR \i=15 \OR
+					\i=17 \OR \i=19
+				}{
+					\draw[black]
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.1);
+				} {
+					\draw[black]
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.15);
+				}
+			}
+		}
+	}
+}
+
+\newcommand{\tscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+	% First line
+	\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 6 - 10
+	\foreach \i in {6,...,9,10,15,...,45}{
+		\draw[black]
+			({#1 + \tscalefn(\i)}, #2) --
+			({#1 + \tscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+
+	% Submarks 6 - 10
+	\foreach \n in {6, ..., 9} {
+		\foreach \i in {1,...,9}{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 10)}, #2) --
+					({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 10)}, #2) --
+					({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 15 - 45
+	\foreach \n in {10, 15, ..., 40} {
+		\foreach \i in {1,...,24}{
+			\ifthenelse{
+				\i=5 \OR \i=10 \OR \i=15 \OR \i=20
+			} {
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 5)}, #2) --
+					({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 5)}, #2) --
+					({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+\newcommand{\sscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+	% First line
+	\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {6,...,9,10,15,...,30,40,50,...,60,90}{
+		\draw[black]
+			({#1 + \sscalefn(\i)}, #2) --
+			({#1 + \sscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+
+	% Submarks 6 - 10
+	\foreach \n in {6, ..., 9} {
+		\foreach \i in {1,...,9}{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 10)}, #2) --
+					({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 10)}, #2) --
+					({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 15 - 30
+	\foreach \n in {10, 15, ..., 25} {
+		\foreach \i in {1,...,24}{
+			\ifthenelse{
+				\i=5 \OR \i=10 \OR \i=15 \OR \i=20
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 5)}, #2) --
+					({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 5)}, #2) --
+					({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 30
+	\foreach \n in {30} {
+		\foreach \i in {1,...,19}{
+			\ifthenelse{
+				\i=2 \OR \i=4 \OR \i=6 \OR \i=8 \OR
+				\i=10 \OR \i=12 \OR \i=14 \OR \i=16 \OR
+				\i=18
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 2)}, #2) --
+					({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 2)}, #2) --
+					({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 40 - 50
+	\foreach \n in {40, 50} {
+		\foreach \i in {1,...,9}{
+			\ifthenelse{
+				\i=5 \OR \i=10 \OR \i=15 \OR \i=20
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i)}, #2) --
+					({#1 + \sscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i)}, #2) --
+					({#1 + \sscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 60
+	\foreach \i in {1,...,10}{
+		\ifthenelse{
+			\i=5 \OR \i=10
+		} {
+			\draw[black]
+				({#1 + \sscalefn(60 + \i * 2)}, #2) --
+				({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.2);
+		} {
+			\draw[black]
+				({#1 + \sscalefn(60 + \i * 2)}, #2) --
+				({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.1);
+		}
+	}
+}