Add slide rule warm-up

src/Advanced/Fast Inverse Root/main.typ

@@ -13,16 +13,19 @@
   by: "Mark",
 )
 
-#include "parts/00 int.typ"
+#include "parts/00 intro.typ"
 #pagebreak()
 
-#include "parts/01 float.typ"
+#include "parts/01 int.typ"
 #pagebreak()
 
-#include "parts/02 approx.typ"
+#include "parts/02 float.typ"
 #pagebreak()
 
-#include "parts/03 quake.typ"
+#include "parts/03 approx.typ"
 #pagebreak()
 
-#include "parts/04 bonus.typ"
+#include "parts/04 quake.typ"
+#pagebreak()
+
+#include "parts/05 bonus.typ"

src/Advanced/Fast Inverse Root/parts/00 intro.typ

@@ -0,0 +1,45 @@
+#import "@local/handout:0.1.0": *
+
+= Introduction
+
+In 2005, ID Software published the source code of _Quake III Arena_, a popular game released in 1999. \
+This caused quite a stir: ID Software was responsible for many games popular among old-school engineers (most notably _Doom_, which has a place in programmer humor even today).
+
+#v(2mm)
+
+Naturally, this community immediately began dissecting _Quake_'s source. \
+One particularly interesting function is reproduced below, with original comments: \
+
+#v(3mm)
+
+```c
+float Q_rsqrt( float number ) {
+	long i;
+	float x2, y;
+	const float threehalfs = 1.5F;
+
+	x2 = number * 0.5F;
+	y  = number;
+	i  = * ( long * ) &y;						// evil floating point bit level hacking
+	i  = 0x5f3759df - ( i >> 1 );               // [redacted]
+	y  = * ( float * ) &i;
+	y  = y * ( threehalfs - ( x2 * y * y ) );   // 1st iteration
+//	y  = y * ( threehalfs - ( x2 * y * y ) );   // 2nd iteration, this can be removed
+
+	return y;
+}
+```
+
+#v(3mm)
+
+This code defines a function `Q_sqrt`, which was used as a fast approximation of the inverse square root in graphics routines. (in other words, `Q_sqrt` efficiently approximates $1 div sqrt(x)$)
+
+#v(3mm)
+
+The key word here is "fast": _Quake_ ran on very limited hardware, and traditional approximation techniques (like Taylor series)#footnote[Taylor series aren't used today, and for the same reason. There are better ways.] were too computationally expensive to be viable.
+
+#v(3mm)
+
+Our goal today is to understand how `Q_sqrt` works. \
+To do that, we'll first need to understand how computers represent numbers. \
+We'll start with simple binary integers---turn the page.

src/Advanced/Fast Inverse Root/parts/01 int.typ

@@ -5,7 +5,8 @@
 #definition()
 A _bit string_ is a string of binary digits. \
 In this handout, we'll denote bit strings with the prefix `0b`. \
-That is, $1010 =$ "one thousand and one," while $#text([`0b1001`]) = 2^3 + 2^0 = 9$
+#note[This prefix is only notation---it is _not_ part of the string itself.] \
+For example, $1001$ is the number "one thousand and one," while $#text([`0b1001`])$ is the string of bits "1 0 0 1".
 
 #v(2mm)
 We will separate long bit strings with underscores for readability. \
@@ -40,7 +41,7 @@ The value of a `uint` is simply its value as a binary number:
 What is the largest number we can represent with a 32-bit `uint`?
 
 #solution([
-  $#text([`0b01111111_11111111_11111111_11111111`]) = 2^(31)$
+  $#text([`0b11111111_11111111_11111111_11111111`]) = 2^(32)-1$
 ])
 
 #v(1fr)
@@ -53,6 +54,10 @@ Find the value of each of the following 32-bit unsigned integers:
 - `0b00000000_00000000_00000100_10110000`
 #hint([The third conversion is easy---look carefully at the second.])
 
+#instructornote[
+  Consider making a list of the powers of two $>= 1024$ on the board.
+]
+
 #solution([
   - $#text([`0b00000000_00000000_00000101_00111001`]) = 1337$
   - $#text([`0b00000000_00000000_00000001_00101100`]) = 300$
@@ -64,20 +69,20 @@ Find the value of each of the following 32-bit unsigned integers:
 
 
 #definition()
-In general, division of `uints` is nontrivial#footnote([One may use repeated subtraction, but that isn't efficient.]). \
+In general, fast division of `uints` is difficult#footnote([One may use repeated subtraction, but this isn't efficient.]). \
 Division by powers of two, however, is incredibly easy: \
 To divide by two, all we need to do is shift the bits of our integer right.
 
 #v(2mm)
 
 For example, consider $#text[`0b0000_0110`] = 6$. \
-If we insert a zero at the left end of this bit string and delete the digit at the right \
+If we insert a zero at the left end of this string and delete the zero at the right \
 (thus "shifting" each bit right), we get `0b0000_0011`, which is 3. \
 
 #v(2mm)
 
-Of course, we loose the remainder when we left-shift an odd number: \
-$9 div 2 = 4$, since `0b0000_1001` shifted right is `0b0000_0100`.
+Of course, we lose the remainder when we right-shift an odd number: \
+$9$ shifted right is $4$, since `0b0000_1001` shifted right is `0b0000_0100`.
 
 #problem()
 Right shifts are denoted by the `>>` symbol: \
@@ -86,6 +91,7 @@ Find the value of the following:
 - $12 #text[`>>`] 1$
 - $27 #text[`>>`] 3$
 - $16 #text[`>>`] 8$
+#note[Naturally, you'll have to convert these integers to binary first.]
 
 #solution[
   - $12 #text[`>>`] 1 = 6$

src/Advanced/Fast Inverse Root/parts/02 float.typ

@@ -3,7 +3,7 @@
 
 = Floats
 #definition()
-_Binary decimals_#footnote["decimal" is a misnomer, but that's ok.] are very similar to base-10 decimals. \
+_Binary decimals_#footnote([Note that "binary decimal" is a misnomer---"deci" means "ten"!]) are very similar to base-10 decimals.\
 In base 10, we interpret place value as follows:
 - $0.1 = 10^(-1)$
 - $0.03 = 3 times 10^(-2)$
@@ -107,11 +107,13 @@ Floats represent a subset of the real numbers, and are interpreted as follows: \
 - The next eight bits represent the _exponent_ of this float.
   #note([(we'll see what that means soon)]) \
   We'll call the value of this eight-bit binary integer $E$. \
-  Naturally, $0 <= E <= 255$ #note([(since $E$ consist of eight bits.)])
+  Naturally, $0 <= E <= 255$ #note([(since $E$ consist of eight bits)])
 
-- The remaining 23 bits represent the _fraction_ of this float, which we'll call $F$. \
-  These 23 bits are interpreted as the fractional part of a binary decimal. \
-  For example, the bits `0b10100000_00000000_00000000` represents $0.5 + 0.125 = 0.625$.
+- The remaining 23 bits represent the _fraction_ of this float. \
+  They are interpreted as the fractional part of a binary decimal. \
+  For example, the bits `0b10100000_00000000_00000000` represent $0.5 + 0.125 = 0.625$. \
+  We'll call the value of these bits as a binary integer $F$. \
+  Their value as a binary decimal is then $F div 2^23$. #note([(convince yourself of this)])
 
 
 #problem(label: "floata")
@@ -135,12 +137,17 @@ $
   (-1)^s times 2^(E - 127) times (1 + F / (2^(23)))
 $
 
-Notice that this is very similar to decimal scientific notation, which is written as
+Notice that this is very similar to base-10 scientific notation, which is written as
 
 $
   (-1)^s times 10^(e) times (f)
 $
 
+#note[
+  We subtract 127 from $E$ so we can represent positive and negative numbers. \
+  $E$ is an eight bit binary integer, so $0 <= E <= 255$ and thus $-127 <= (E - 127) <= 127$.
+]
+
 #problem()
 Consider `0b01000001_10101000_00000000_00000000`. \
 This is the same bit string we used in @floata. \

src/Advanced/Fast Inverse Root/parts/03 approx.typ

@@ -5,7 +5,7 @@
 = Integers and Floats
 
 #generic("Observation:")
-For small values of $x$, $log_2(1 + x)$ is approximately equal to $x$. \
+If $x$ is smaller than 1, $log_2(1 + x)$ is approximately equal to $x$. \
 Note that this equality is exact for $x = 0$ and $x = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
 
 #v(5mm)
@@ -18,7 +18,7 @@ This allows us to improve the average error of our linear approximation:
   align: center,
   columns: (1fr, 1fr),
   inset: 5mm,
-  [$log(1+x)$ and $x + 0$]
+  [$log_2(1+x)$ and $x + 0$]
     + cetz.canvas({
       import cetz.draw: *
 
@@ -64,7 +64,7 @@ This allows us to improve the average error of our linear approximation:
       Max error: 0.086 \
       Average error: 0.0573
     ],
-  [$log(1+x)$ and $x + 0.045$]
+  [$log_2(1+x)$ and $x + 0.045$]
     + cetz.canvas({
       import cetz.draw: *
 
@@ -125,7 +125,7 @@ We won't bother with this---we'll simply leave the correction term as an opaque
   [
     "Average error" above is simply the area of the region between the two graphs:
     $
-      integral_0^1 abs( #v(1mm) log(1+x) - (x+epsilon) #v(1mm))
+      integral_0^1 abs( #v(1mm) log(1+x)_2 - (x+epsilon) #v(1mm))
     $
     Feel free to ignore this note, it isn't a critical part of this handout.
   ],

src/Advanced/Fast Inverse Root/parts/04 quake.typ

@@ -2,10 +2,9 @@
 
 = The Fast Inverse Square Root
 
+A simplified version of the _Quake_ routine we are studying is reproduced below.
 
-The following code is present in _Quake III Arena_ (1999):
-
-#v(5mm)
+#v(2mm)
 
 ```c
 float Q_rsqrt( float number ) {
@@ -15,20 +14,20 @@ float Q_rsqrt( float number ) {
 }
 ```
 
-#v(5mm)
+#v(2mm)
 
 This code defines a function `Q_rsqrt` that consumes a float `number` and approximates its inverse square root.
 If we rewrite this using notation we're familiar with, we get the following:
 $
   #text[`Q_sqrt`] (n_f) =
-  #h(5mm)
   6240089 - (n_i div 2)
-  #h(5mm)
+  #h(10mm)
   approx 1 / sqrt(n_f)
 $
 
 #note[
   `0x5f3759df` is $6240089$ in hexadecimal. \
+  Ask an instructor to explain if you don't know what this means. \
   It is a magic number hard-coded into `Q_sqrt`.
 ]
 
@@ -56,7 +55,7 @@ For those that are interested, here are the details of the "code-to-math" transl
 
 
 - Notice the right-shift in the second line of the function. \
-  We translated `(i >> i)` into $(n_i div 2)$.
+  We translated `(i >> 1)` into $(n_i div 2)$.
 #v(2mm)
 
 - "`return * (float *) &i`" is again C magic. \
@@ -64,17 +63,17 @@ For those that are interested, here are the details of the "code-to-math" transl
 #pagebreak()
 
 #generic("Setup:")
-We are now ready to show that $#text[`Q_sqrt`] (x) approx 1/sqrt(x)$. \
+We are now ready to show that $#text[`Q_sqrt`] (x)$ effectively approximates $1/sqrt(x)$. \
 For convenience, let's call the bit string of the inverse square root $r$. \
 In other words,
 $
   r_f := 1 / (sqrt(n_f))
 $
-This is the value we want to approximate.
+This is the value we want to approximate. \
 
 #problem(label: "finala")
 Find an approximation for $log_2(r_f)$ in terms of $n_i$ and $epsilon$ \
-#note[Remember, $epsilon$ is the correction constant in our approximation of $log_2(1 + a)$.]
+#note[Remember, $epsilon$ is the correction constant in our approximation of $log_2(1 + x)$.]
 
 #solution[
   $
@@ -92,7 +91,11 @@ Let's call the "magic number" in the code above $kappa$, so that
 $
   #text[`Q_sqrt`] (n_f) = kappa - (n_i div 2)
 $
-Use @convert and @finala to show that $#text[`Q_sqrt`] (n_f) approx r_i$
+Use @convert and @finala to show that $#text[`Q_sqrt`] (n_f) approx r_i$ \
+#note(type: "Note")[
+  If we know $r_i$, we know $r_f$. \
+  We don't even need to convert between the two---the underlying bits are the same!
+]
 
 #solution[
   From @convert, we know that
@@ -164,8 +167,7 @@ though it is fairly close to the ideal $epsilon$.
 
 #remark()
 And now, we're done! \
-We've shown that `Q_sqrt(x)` approximates $1/sqrt(x)$ fairly well, \
-thanks to the approximation $log(1+a) = a + epsilon$.
+We've shown that `Q_sqrt(x)` approximates $1/sqrt(x)$ fairly well. \
 
 #v(2mm)
 

src/Warm-Ups/Slide Rules/main.tex

@@ -0,0 +1,78 @@
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	shortwarning
+]{../../../lib/tex/ormc_handout}
+\usepackage{../../../lib/tex/macros}
+
+
+\usepackage{pdfpages}
+\usepackage{sliderule}
+\usepackage{changepage}
+
+% 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}
+\uptitler{\smallurl{}}
+\title{Warm-Up: Slide Rules}
+\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}
+
+	% Make sure the slide rule is on an odd page,
+	% so that double-sided printing 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/Warm-Ups/Slide Rules/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Slide Rules"
+
+[publish]
+handout = false
+solutions = true

src/Warm-Ups/Slide Rules/parts/0 logarithms.tex

@@ -0,0 +1,59 @@
+\section{Logarithms}
+
+\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}
+
+\problem{}<logids>
+Prove the following:
+
+\begin{enumerate}[itemsep=2mm]
+	\item $\log_b{(b^x)} = x$
+	\vfill
+	\item $b^{\log_b{x}} = x$
+	\vfill
+	\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
+	\vfill
+	\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
+	\vfill
+	\item $\log_b{(x^y)} = y \log_b{(x)}$
+	\vfill
+\end{enumerate}
+
+
+\begin{instructornote}
+	A good intro to the following sections is the linear slide rule:
+	\note{Note that these rules start at 0.}
+	\begin{center}
+		\begin{tikzpicture}[scale=0.6]
+			\linearscale{2}{1}{}
+			\linearscale{0}{0}{}
+
+			\slideruleind
+				{5}
+				{1}
+				{2 + 3 = 5}
+	\end{tikzpicture}
+	\end{center}
+
+	Take two linear rules, offset one, and you add.
+	Do the same with a log scale, and you multiply! \\
+
+	\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/Warm-Ups/Slide Rules/parts/1 intro.tex

@@ -0,0 +1,43 @@
+\section{Introduction}
+
+Mathematicians, physicists, and engineers needed to quickly compute products long before computers conquered the world.
+
+\medskip
+
+The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
+
+\medskip
+
+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/Warm-Ups/Slide Rules/parts/2 multiplication.tex

@@ -0,0 +1,299 @@
+\section{Multiplication}
+
+We'll use the C and D scales of your slide rule to multiply. \\
+
+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 $1.15 \times 2.1$? \\
+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 $32 \times 210$? \\
+
+\problem{}
+Using your slide rule, calculate $32 \times 210$. \\
+%\hint{$32 = 3.2 \times 10^1$}
+
+\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. \\
+		$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
+\end{solution}
+
+\vfill
+
+%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
+
+%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
+
+
+\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
+\pagebreak
+
+\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? \\
+%\hint{See \ref{logids}}
+
+\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. \\
+
+\medskip
+
+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. \\
+
+\medskip
+
+\problem{}
+Why does this work? \par
+\hint{Add a second $D$ scale.}
+
+\begin{solution}
+	Consider the following picture, where I've put 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}
+
+	\medskip
+
+	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.
+
+	\medskip
+
+	\medskip
+	In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
+	This reduces to $\log{(\frac{a \times b}{10})}$, which 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 $42.1 \times 7.65$
+	\item $6.5^2$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $9 \times 8 = 72$
+		\item $15 \times 35 = 525$
+		\item $42.1 \times 7.65 = 322.065$
+		\item $6.5^2 = 42.25$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+
+\problem{}
+Compute the following using your slide rule. \\
+
+\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/Warm-Ups/Slide Rules/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);
+		}
+	}
+}