Snakes!

Reviewed-on: #23

.gitea/workflows/ci.yml

@@ -26,7 +26,7 @@ jobs:
 
       - name: "Download Typstyle"
         run: |
-          wget -q "https://github.com/Enter-tainer/typstyle/releases/download/v0.12.14/typstyle-x86_64-unknown-linux-musl"
+          wget -q "https://github.com/Enter-tainer/typstyle/releases/download/v0.13.17/typstyle-x86_64-unknown-linux-musl"
           chmod +x typstyle-x86_64-unknown-linux-musl
 
       - name: Check typst formatting
@@ -62,7 +62,7 @@ jobs:
       # more control anyway.
       - name: "Download Typst"
         run: |
-          wget -q "https://github.com/typst/typst/releases/download/v0.12.0/typst-x86_64-unknown-linux-musl.tar.xz"
+          wget -q "https://github.com/typst/typst/releases/download/v0.13.1/typst-x86_64-unknown-linux-musl.tar.xz"
           tar -xf "typst-x86_64-unknown-linux-musl.tar.xz"
           mv "typst-x86_64-unknown-linux-musl/typst" .
           rm "typst-x86_64-unknown-linux-musl.tar.xz"

lib/typst/local/handout/0.1.0/lib.typ

@@ -3,13 +3,10 @@
 // Re-exports
 // All functions that maybe used by client code are listed here
 #import "misc.typ": *
-#import "object.typ": problem, definition, theorem, example, remark, generic
+#import "object.typ": definition, example, generic, problem, remark, theorem
 #import "solution.typ": (
-  if_solutions,
-  if_no_solutions,
-  if_solutions_else,
-  solution,
-  instructornote,
+  if_no_solutions, if_solutions, if_solutions_else, instructornote, review_box,
+  sample_solution, solution,
 )
 
 
@@ -33,15 +30,17 @@
   by: none,
   subtitle: none,
   short_warning: false,
+  page_numbers: true,
 ) = {
   set page(
     margin: 20mm,
     width: 8.5in,
     height: 11in,
-    footer: align(
-      center,
-      context counter(page).display(),
-    ),
+    footer: {
+      if page_numbers {
+        align(center, context counter(page).display())
+      }
+    },
     footer-descent: 5mm,
   )
 
@@ -57,6 +56,8 @@
     justify: true,
   )
 
+  set math.mat(delim: "[")
+
   //
   // List style
   set list(
@@ -102,8 +103,8 @@
 
   // Make handout title
   {
-    import "header.typ": make_header, solution_warning, short_solution_warning
-    import "solution.typ": solutions_state, reset_solutions
+    import "header.typ": make_header, short_solution_warning, solution_warning
+    import "solution.typ": reset_solutions, solutions_state
 
     reset_solutions()
 

lib/typst/local/handout/0.1.0/object.typ

@@ -29,11 +29,7 @@
   }
 
   // Render the object
-  block(
-    above: 8mm,
-    below: 2mm,
-    text(weight: "bold", obj_content),
-  )
+  block(above: 8mm, below: 2mm, text(weight: "bold", obj_content))
 
   // Generate labeled metadata for this object.
   //
@@ -57,7 +53,7 @@
   if not (
     it.element != none
       and it.element.has("value")
-      and type(it.element.value) == "dictionary"
+      and type(it.element.value) == dictionary
       and it.element.value.keys().contains(magic_key)
   ) {
     // This label is not attached to object metadata
@@ -100,9 +96,5 @@
 #let remark = _mkobj("Remark")
 
 #let generic(obj_content) = {
-  block(
-    above: 8mm,
-    below: 2mm,
-    text(weight: "bold", obj_content),
-  )
+  block(above: 8mm, below: 2mm, text(weight: "bold", obj_content))
 }

lib/typst/local/handout/0.1.0/solution.typ

@@ -1,4 +1,4 @@
-#import "misc.typ": ored, oblue
+#import "misc.typ": oblue, ored
 
 
 /// If false, hide instructor info.
@@ -61,57 +61,94 @@
 }
 
 #let solution(content) = {
-  if_solutions(
-    align(
-      center,
-      stack(
-        block(
-          width: 100%,
-          breakable: false,
-          fill: ored,
-          stroke: ored + 2pt,
-          inset: 1.5mm,
-          align(left, text(fill: white, weight: "bold", [Solution:])),
-        ),
-
-        block(
-          width: 100%,
-          height: auto,
-          breakable: false,
-          fill: ored.lighten(80%).desaturate(10%),
-          stroke: ored + 2pt,
-          inset: 3mm,
-          align(left, content),
-        ),
-      ),
+  if_solutions(align(center, stack(
+    block(
+      width: 100%,
+      breakable: false,
+      fill: ored,
+      stroke: ored + 2pt,
+      inset: 1.5mm,
+      align(left, text(fill: white, weight: "bold", [Solution:])),
     ),
-  )
+
+    block(
+      width: 100%,
+      height: auto,
+      breakable: false,
+      fill: ored.lighten(80%).desaturate(10%),
+      stroke: ored + 2pt,
+      inset: 3mm,
+      align(left, content),
+    ),
+  )))
 }
 
+#let sample_solution(content) = {
+  align(center, stack(
+    block(
+      width: 100%,
+      breakable: false,
+      fill: oblue,
+      stroke: oblue + 2pt,
+      inset: 1.5mm,
+      align(left, text(fill: white, weight: "bold", [Sample Solution:])),
+    ),
+
+    block(
+      width: 100%,
+      height: auto,
+      breakable: false,
+      fill: oblue.lighten(80%).desaturate(10%),
+      stroke: oblue + 2pt,
+      inset: 3mm,
+      align(left, content),
+    ),
+  ))
+}
+
+#let review_box(title, content) = {
+  align(center, stack(
+    block(
+      width: 100%,
+      breakable: false,
+      fill: oblue,
+      stroke: oblue + 2pt,
+      inset: 1.5mm,
+      align(left, text(fill: white, weight: "bold", title)),
+    ),
+
+    block(
+      width: 100%,
+      height: auto,
+      breakable: false,
+      fill: oblue.lighten(80%).desaturate(10%),
+      stroke: oblue + 2pt,
+      inset: 3mm,
+      align(left, content),
+    ),
+  ))
+}
+
+
 #let instructornote(content) = {
-  if_solutions(
-    align(
-      center,
-      stack(
-        block(
-          width: 100%,
-          breakable: false,
-          fill: oblue,
-          stroke: oblue + 2pt,
-          inset: 1.5mm,
-          align(left, text(fill: white, weight: "bold", [Instructor note:])),
-        ),
-
-        block(
-          width: 100%,
-          height: auto,
-          breakable: false,
-          fill: oblue.lighten(80%).desaturate(10%),
-          stroke: oblue + 2pt,
-          inset: 3mm,
-          align(left, content),
-        ),
-      ),
+  if_solutions(align(center, stack(
+    block(
+      width: 100%,
+      breakable: false,
+      fill: oblue,
+      stroke: oblue + 2pt,
+      inset: 1.5mm,
+      align(left, text(fill: white, weight: "bold", [Instructor note:])),
     ),
-  )
+
+    block(
+      width: 100%,
+      height: auto,
+      breakable: false,
+      fill: oblue.lighten(80%).desaturate(10%),
+      stroke: oblue + 2pt,
+      inset: 3mm,
+      align(left, content),
+    ),
+  )))
 }

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

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = Floats
 #definition()
@@ -31,75 +31,71 @@ Rewrite the following binary decimals in base 10: \
 #definition(label: "floatbits")
 Another way we can interpret a bit string is as a _signed floating-point decimal_, or a `float` for short. \
 Floats represent a subset of the real numbers, and are interpreted as follows: \
-#note([The following only applies to floats that consist of 32 bits. We won't encounter any others today.])
-
-#align(
-  center,
-  box(
-    inset: 2mm,
-    cetz.canvas({
-      import cetz.draw: *
-
-      let chars = (
-        `0`,
-        `b`,
-        `0`,
-        `_`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `_`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `_`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `_`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-        `0`,
-      )
-
-      let x = 0
-      for c in chars {
-        content((x, 0), c)
-        x += 0.25
-      }
-
-      let y = -0.4
-      line((0.3, y), (0.65, y))
-      content((0.45, y - 0.2), [s])
-
-      line((0.85, y), (2.9, y))
-      content((1.9, y - 0.2), [exponent])
-
-      line((3.10, y), (9.4, y))
-      content((6.3, y - 0.2), [fraction])
-    }),
-  ),
+#note(
+  [The following only applies to floats that consist of 32 bits. We won't encounter any others today.],
 )
 
+#align(center, box(inset: 2mm, cetz.canvas({
+  import cetz.draw: *
+
+  let chars = (
+    `0`,
+    `b`,
+    `0`,
+    `_`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `_`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `_`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `_`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+    `0`,
+  )
+
+  let x = 0
+  for c in chars {
+    content((x, 0), c)
+    x += 0.25
+  }
+
+  let y = -0.4
+  line((0.3, y), (0.65, y))
+  content((0.45, y - 0.2), [s])
+
+  line((0.85, y), (2.9, y))
+  content((1.9, y - 0.2), [exponent])
+
+  line((3.10, y), (9.4, y))
+  content((6.3, y - 0.2), [fraction])
+})))
+
 - The first bit denotes the sign of the float's value
   We'll label it $s$. \
   If $s = #text([`1`])$, this float is negative; if $s = #text([`0`])$, it is positive.

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

@@ -1,6 +1,6 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
-#import "@preview/cetz-plot:0.1.0": plot, chart
+#import "@preview/cetz:0.4.2"
+#import "@preview/cetz-plot:0.1.2": chart, plot
 
 = Integers and Floats
 
@@ -44,19 +44,11 @@ This allows us to improve the average error of our linear approximation:
         {
           let domain = (0, 1)
 
-          plot.add(
-            f1,
-            domain: domain,
-            label: $log(1+x)$,
-            style: (stroke: ogrape),
-          )
+          plot.add(f1, domain: domain, label: $log(1+x)$, style: (
+            stroke: ogrape,
+          ))
 
-          plot.add(
-            f2,
-            domain: domain,
-            label: $x$,
-            style: (stroke: oblue),
-          )
+          plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
         },
       )
     })
@@ -90,19 +82,11 @@ This allows us to improve the average error of our linear approximation:
         {
           let domain = (0, 1)
 
-          plot.add(
-            f1,
-            domain: domain,
-            label: $log(1+x)$,
-            style: (stroke: ogrape),
-          )
+          plot.add(f1, domain: domain, label: $log(1+x)$, style: (
+            stroke: ogrape,
+          ))
 
-          plot.add(
-            f2,
-            domain: domain,
-            label: $x$,
-            style: (stroke: oblue),
-          )
+          plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
         },
       )
     })
@@ -120,16 +104,13 @@ We won't bother with this---we'll simply leave the correction term as an opaque
 
 #v(1fr)
 
-#note(
-  type: "Note",
-  [
-    "Average error" above is simply the area of the region between the two graphs:
-    $
-      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.
-  ],
-)
+#note(type: "Note", [
+  "Average error" above is simply the area of the region between the two graphs:
+  $
+    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.
+])
 
 
 #pagebreak()
@@ -149,12 +130,11 @@ $
   Let $E$ and $F$ be the exponent and float bits of $x_f$. \
   We then have:
   $
-    log_2(x_f)
-    &= log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
-    &= E - 127 + log_2(1 + F / (2^23)) \
-    & approx E-127 + F / (2^23) + epsilon \
-    &= 1 / (2^23)(2^23 E + F) - 127 + epsilon \
-    &= 1 / (2^23)(x_i) - 127 + epsilon
+    log_2(x_f) & = log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
+               & = E - 127 + log_2(1 + F / (2^23)) \
+               & approx E-127 + F / (2^23) + epsilon \
+               & = 1 / (2^23)(2^23 E + F) - 127 + epsilon \
+               & = 1 / (2^23)(x_i) - 127 + epsilon
   $
 ])
 

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

@@ -156,9 +156,9 @@ float Q_rsqrt( float number ) {
 Using a calculator and some basic algebra, we can find the $epsilon$ this code uses: \
 #note[Remember, #text[`0x5f3759df`] is $6240089$ in hexadecimal.]
 $
-  (3 times 2^22) (127 - epsilon) &= 6240089 \
-  (127 - epsilon) &= 126.955 \
-  epsilon &= 0.0450466
+  (3 times 2^22) (127 - epsilon) & = 6240089 \
+                 (127 - epsilon) & = 126.955 \
+                         epsilon & = 0.0450466
 $
 
 So, $0.045$ is the $epsilon$ used by Quake. \

src/Advanced/Introduction to Quantum/src/parts/01 bits.tex

@@ -27,7 +27,7 @@ We'll represent this probabilistic bit's \textit{state} as a vector:
 $\left[\begin{smallmatrix}
 	p_0 \\ p_1
 \end{smallmatrix}\right]$ \par
-We do \textbf{not} assume this coin is fair, and thus $p_0$ might not equal $p_1$.
+We do \textbf{not} assume this coin is fair---$p_0$ might not equal $p_1$.
 
 \note{
 	This may seem a bit redundant: since $p_0 + p_1$, we can always calculate one probability given the other. \\
@@ -45,15 +45,16 @@ The simplest probabilistic bit states are of course $[0]$ and $[1]$, defined as
 	\item $[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
 	\item $[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$
 \end{itemize}
-That is, $[0]$ represents a bit that we known to be \texttt{0}, \par
+That is, $[0]$ represents a bit that we know to be \texttt{0}, \par
 and $[1]$ represents a bit we know to be \texttt{1}.
 
 \vfill
+\pagebreak
 
 
 \definition{}
-$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states: \par
-Every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
+$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states. This means that \par
+every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
 
 \begin{equation*}
 	\begin{bmatrix} p_0 \\ p_1 \end{bmatrix}
@@ -66,7 +67,6 @@ Every other probabilistic bit can be written as a \textit{linear combination} of
 
 
 \vfill
-\pagebreak
 
 \problem{}
 Every possible state of a probabilistic bit is a two-dimensional vector. \par
@@ -118,6 +118,10 @@ Draw all possible states on the axis below.
 
 
 
+%
+% MARK: measure
+%
+
 
 
 \section{Measuring Probabilistic Bits}
@@ -125,7 +129,7 @@ Draw all possible states on the axis below.
 
 \definition{}
 As we noted before, a probabilistic bit represents a coin we've tossed but haven't looked at. \par
-We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of both of these outcomes. \par
+We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of each outcome. \par
 
 \vspace{2mm}
 
@@ -135,7 +139,7 @@ knowledge of its state is updated to either $[0]$ or $[1]$, since we now certain
 \vspace{2mm}
 
 Since measurement changes what we know about a probabilistic bit, it changes the probabilistic bit's state.
-When we measure a bit, it's state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
+When we measure a bit, its state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
 bit vanishes. We \textit{cannot} recover the state $[x_0, x_1]$ from a measured probabilistic bit.
 
 
@@ -165,7 +169,10 @@ Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nic
 	\item If we measure $y$ first and observe \texttt{1}, \par
 	what is the probability of getting each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}?
 \end{itemize}
-\note[Note]{$[x]$ and $[y]$ are column vectors, but I've written them horizontally to save space.}
+\note[Note]{
+	$[x]$ and $[y]$ are column vectors in the previous pages, and are still column vectors here. \par
+	I've written them horizontally to save space.
+}
 
 
 \vfill
@@ -189,7 +196,9 @@ What is the probability that $x$ and $y$ produce different outcomes?
 
 
 
-
+%
+% MARK: tensor product
+%
 
 \section{Tensor Products}
 
@@ -315,7 +324,7 @@ $?
 
 \problem{}
 What is the \textit{span} of the vectors we found in \ref{basistp}? \par
-In other words, what is the set of vectors that can be written as linear combinations of the vectors above?
+In other words, what is the set of all vectors that can be written as linear combinations of the vectors above?
 
 \vfill
 
@@ -387,7 +396,7 @@ Compute the following. Is the result what we'd expect?
 
 
 \problem{}<fivequant>
-Of course, writing $[0] \otimes [1]$ is a bit excessive. We'll shorten this notation to $[01]$. \par
+Writing $[0] \otimes [1]$ is tedious. We'll shorten this notation to $[01]$. \par
 
 \vspace{2mm}
 
@@ -429,9 +438,9 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
 
 
 
-
-
-
+%
+% MARK: ops
+%
 
 
 
@@ -443,25 +452,6 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
 Now that we can write probabilistic bits as vectors, we can represent operations on these bits
 with linear transformations---in other words, as matrices.
 
-\definition{}
-Consider the NOT gate, which operates as follows: \par
-\begin{itemize}
-	\item $\text{NOT}[0] = [1]$
-	\item $\text{NOT}[1] = [0]$
-\end{itemize}
-What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
-If we return to our coin analogy, we can think of the NOT operation as
-flipping a coin we have already tossed, without looking at its state.
-Thus,
-\begin{equation*}
-	\text{NOT} \begin{bmatrix}
-		x_0 \\ x_1
-	\end{bmatrix} = \begin{bmatrix}
-		x_1 \\ x_0
-	\end{bmatrix}
-\end{equation*}
-
-
 \begin{hobox}{Review: Multiplying Vectors by Matrices}{black!10!white}{black!65!white}
 	\begin{equation*}
 		Av =
@@ -482,6 +472,10 @@ Thus,
 	Note that each element of $Av$ is the dot product of a row in $A$ and a column in $v$.
 \end{hobox}
 
+\generic{Remark:}
+Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
+We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
+
 \problem{}
 Compute the following product:
 \begin{equation*}
@@ -496,12 +490,23 @@ Compute the following product:
 
 \vfill
 
-\generic{Remark:}
-Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
-We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
-
-\pagebreak
-
+\definition{}
+Consider the NOT gate, which operates as follows: \par
+\begin{itemize}
+	\item $\text{NOT}[0] = [1]$
+	\item $\text{NOT}[1] = [0]$
+\end{itemize}
+What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
+If we return to our coin analogy, we can think of the NOT operation as
+flipping a coin we have already tossed, without looking at its state.
+Thus,
+\begin{equation*}
+	\text{NOT} \begin{bmatrix}
+		x_0 \\ x_1
+	\end{bmatrix} = \begin{bmatrix}
+		x_1 \\ x_0
+	\end{bmatrix}
+\end{equation*}
 
 \problem{}
 Find the matrix that represents the NOT operation on one probabilistic bit.
@@ -516,6 +521,8 @@ Find the matrix that represents the NOT operation on one probabilistic bit.
 
 \vfill
 
+\pagebreak
+
 
 \problem{Extension by linearity}
 Say we have an arbitrary operation $M$. \par

src/Advanced/Introduction to Quantum/src/parts/02 qubit.tex

@@ -4,7 +4,8 @@ Quantum bits (or \textit{qubits}) are very similar to probabilistic bits, but ha
 probabilities are replaced with \textit{amplitudes}.
 
 \vspace{2mm}
-Of course, a qubit can take the values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
+
+As before a qubit can take the extremal values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
 Like probabilistic bits, a quantum bit is written as a linear combination of $\ket{0}$ and $\ket{1}$:
 \begin{equation*}
 	\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}
@@ -19,8 +20,8 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one
 
 \vspace{2mm}
 This is very similar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
-As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
-and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
+When we work with qubits, we will write $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ as $\ket{0}$
+and $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ as $\ket{1}$.
 
 
 \vspace{8mm}
@@ -28,7 +29,7 @@ and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
 Recall that probabilistic bits are subject to the restriction that $p_0 + p_1 = 1$. \par
 Quantum bits have a similar condition: $\psi_0^2 + \psi_1^2 = 1$. \par
 Note that this implies that $\psi_0$ and $\psi_1$ are both in $[-1, 1]$. \par
-Quantum amplitudes may be negative, but probabilistic bit probabilities cannot.
+Quantum amplitudes may be negative!
 
 \vspace{2mm}
 
@@ -89,8 +90,8 @@ Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.
 
 
 \definition{Measurement I}
-Just like a probabilistic bit, we must observed $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
-If we were to measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we'd observe either $\ket{0}$ or $\ket{1}$, \par
+Just like a probabilistic bit, we must observe $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
+If we measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we will see either $\ket{0}$ or $\ket{1}$,
 with the following probabilities:
 \begin{itemize}[itemsep = 2mm, topsep = 2mm]
 	\item $\mathcal{P}(\ket{1}) = \psi_1^2$
@@ -117,7 +118,7 @@ leaving no trace of the previous superposition. \par
 
 \problem{}
 \begin{itemize}
-	\item What is the probability we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
+	\item What is the probability that we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
 	\item What can we observe if we measure $\ket{\psi}$ a second time? \par
 	\item What are these probabilities for $\ket{\varphi}$?
 \end{itemize}
@@ -235,6 +236,10 @@ Consider the following qubit states:
 \pagebreak
 
 
+%
+% MARK: ops
+%
+
 \section{Operations on One Qubit}
 
 We may apply transformations to qubits just as we apply transformations to probabilistic bits.
@@ -277,7 +282,7 @@ Find the matrix $X$.
 \vfill
 
 \problem{}
-What is $X\ket{+}$ and $X\ket{-}$? \par
+What are $X\ket{+}$ and $X\ket{-}$? \par
 \hint{Remember that all matrices are linear maps. What does this mean?}
 
 \begin{solution}
@@ -331,11 +336,21 @@ As we noted earlier, any rotation about the center is a valid quantum gate. \par
 Let's derive all transformations of this form.
 \begin{itemize}[itemsep = 1mm]
 	\item Let $U_\phi$ be the matrix that represents a counterclockwise rotation of $\phi$ degrees. \par
-	What is $U\ket{0}$ and $U\ket{1}$?
+	What are $U\ket{0}$ and $U\ket{1}$?
 
 	\item Find the matrix $U_\phi$ for an arbitrary $\phi$.
 \end{itemize}
 
+
+\begin{solution}
+	\begin{equation*}
+		\begin{bmatrix}
+			cos(\theta) & -sin(\theta) \\
+			sin(\theta) & cos(\theta)
+		\end{bmatrix}
+	\end{equation*}
+\end{solution}
+
 \vfill
 
 
@@ -343,5 +358,10 @@ Let's derive all transformations of this form.
 Say we have a qubit that is either $\ket{+}$ or $\ket{-}$. We do not know which of the two states it is in. \par
 Using one operation and one measurement, how can we find out, for certain, which qubit we received? \par
 
+
+\begin{solution}
+	Use a 45 degree ccw rotation.
+\end{solution}
+
 \vfill
 \pagebreak

src/Advanced/Introduction to Quantum/src/parts/03 two qubits.tex

@@ -60,7 +60,7 @@ $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3
 Again, consider the two-qubit state
 $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par
 If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\psi}$? \par
-What would the state be if we'd measured $\ket{1}$ instead?
+What would that state be if we'd measured $\ket{1}$ instead?
 
 \vfill
 
@@ -85,7 +85,7 @@ Say we measure the first two qubits and get $\ket{00}$. What is the resulting st
 
 
 \definition{Entanglement}
-Some product states can be factored into a tensor product of individual qubit states. For example,
+Some compound states can be factored into a tensor product of individual qubit states. For example,
 \begin{equation*}
 	\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
 	= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes

src/Advanced/Introduction to Quantum/src/parts/04 logic gates.tex

@@ -1,40 +1,42 @@
 \section{Logic Gates}
 
-\definition{Matrices}
-Throughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
-and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
-map, we can write it as follows:
-\begin{equation*}
-		f\left(
-			\ket{x}
-		\right)
-	=
-		\begin{bmatrix}
-			m_1 & m_2 \\
-			m_3 & m_4
-		\end{bmatrix}
-		\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
-	=
-		\left[
-		\begin{matrix}
-			m_1x_1 + m_2x_2 \\
-			m_3x_1 + m_4x_2
-		\end{matrix}
-		\right]
-\end{equation*}
-
+%\definition{Matrices}
+%cThroughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
+%and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
+%map, we can write it as follows:
+%\begin{equation*}
+%		f\left(
+%			\ket{x}
+%		\right)
+%	=
+%		\begin{bmatrix}
+%			m_1 & m_2 \\
+%			m_3 & m_4
+%		\end{bmatrix}
+%		\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
+%	=
+%		\left[
+%		\begin{matrix}
+%			m_1x_1 + m_2x_2 \\
+%			m_3x_1 + m_4x_2
+%		\end{matrix}
+%		\right]
+%\end{equation*}
+%
+%A classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$
 
 \definition{}
-Before we discussing multi-qubit quantum gates, we need to review to classical logic. \par
-Of course, a classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$
+Before we discussing multi-qubit quantum gates, we need to review classical logic. \par
+In this section, let's return to probabilistic bits.
+
 
 
 \problem{}<notgatex>
 The \texttt{not} gate is a map defined by the following table: \par
 
 \begin{itemize}
-	\item $X\ket{0} = \ket{1}$
-	\item $X\ket{1} = \ket{0}$
+	\item $X[0] = [1]$
+	\item $X[1] = [0]$
 \end{itemize}
 
 Write the \texttt{not} gate as a matrix that operates on single-bit vector states. \par
@@ -93,11 +95,11 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
 
 		\vspace{2mm}
 		For example, if we look at the first column of $A$ (which is $[1, 0]$), we see: \par
-		$A\ket{00} = A[1,0,0,0] = [1,0] = \ket{0}$
+		$A[00] = A[1,0,0,0] = [1,0] = [0]$
 
 		\vspace{2mm}
 		Also with the last column (which is $[0,1]$): \par
-		$A\ket{00} = A[0,0,0,1] = [0,1] = \ket{1}$
+		$A[00] = A[0,0,0,1] = [0,1] = [1]$
 	\end{instructornote}
 \end{solution}
 
@@ -244,10 +246,10 @@ Say we want to invert the first bit of a two-bit state. That is, we want a trans
 
 In other words, we want a matrix $T$ satisfying the following equalities:
 \begin{itemize}
-	\item $T\ket{00} = \ket{10}$
-	\item $T\ket{01} = \ket{11}$
-	\item $T\ket{10} = \ket{00}$
-	\item $T\ket{11} = \ket{01}$
+	\item $T[00] = [10]$
+	\item $T[01] = [11]$
+	\item $T[10] = [00]$
+	\item $T[11] = [01]$
 \end{itemize}
 
 
@@ -256,8 +258,8 @@ In other words, we want a matrix $T$ satisfying the following equalities:
 Find the matrix that corresponds to the above transformation. \par
 \hint{
 	Remember that
-	$\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
-	$\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
+	$[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
+	$[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
 	Also, we found earlier that $X = \left[\begin{smallmatrix} 0 && 1 \\ 1 && 0 \end{smallmatrix}\right]$,
 	and of course $I = \left[\begin{smallmatrix} 1 && 0 \\ 0 && 1 \end{smallmatrix}\right]$.
 }
@@ -279,25 +281,26 @@ Find the matrix that corresponds to the above transformation. \par
 We could draw the above transformation as a combination $X$ and $I$ (identity) gate:
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
-	\node[qubit] (a) at (0, 0) {$\ket{0}$};
-	\node[qubit] (b) at (0, -1) {$\ket{0}$};
+	\node[qubit] (a) at (0, 0) {$[0]$};
+	\node[qubit] (b) at (0, -1) {$[0]$};
 
-	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
-	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
+	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$[1]$};
+	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$[0]$};
 
 	\qubox{a}{1}{a}{2}{$X$}
 	\qubox{b}{1}{b}{2}{$I$}
 \end{tikzpicture}
 \end{center}
 
-We can even omit the $I$ gate, since we now know that transformations affect the whole state: \par
+We can even omit the $I$ gate. We know that transformations affect the whole state,
+and can assume the empty space uncer $X$ implies $I$. \par
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
-	\node[qubit] (a) at (0, 0) {$\ket{0}$};
-	\node[qubit] (b) at (0, -1) {$\ket{0}$};
+	\node[qubit] (a) at (0, 0) {$[0]$};
+	\node[qubit] (b) at (0, -1) {$[0]$};
 
-	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
-	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
+	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$[1]$};
+	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$[0]$};
 
 	\qubox{a}{1}{a}{2}{$X$}
 \end{tikzpicture}

src/Advanced/Introduction to Quantum/src/parts/06 hxh.tex

@@ -1,7 +1,7 @@
 \section{HXH}
 
 Let's return to the quantum circuit diagrams we discussed a few pages ago. \par
-Keep in mind that we're working with quantum gates and proper half-qubits---not classical bits, as we were before.
+Keep in mind that we're working with quantum gates and proper qubits---not classical bits, as we were before.
 
 \definition{Controlled Inputs}
 A \textit{control input} or \textit{inverted control input} may be attached to any gate. \par
@@ -289,7 +289,7 @@ $\ket{-} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} - \ket{1}\Bigr)$
 \pagebreak
 
 \generic{Remark:}
-Now, consider the following circuit:
+Consider the following circuit:
 
 \begin{center}
 	\begin{tikzpicture}[scale=0.8]
@@ -386,21 +386,21 @@ but the state $\ket{cd}$ on the right is $\ket{11} = [0,0,0,1]$. \par
 \vspace{4mm}
 
 How does this make sense? \par
-Remember that a two-bit quantum state is \textit{not} equivalent to a pair of one-qubit quantum states.
-We must treat a multi-qubit state as a single unit.
+Remember that a two-bit quantum state is \textit{not} equivalent to a pair of disjoint one-qubit quantum states.
+We cannot treat a multi-qubit state as a combination of $n$ independent bits.
 
-Recall that a two-bit state $\ket{ab}$ comes with four probabilities:
-$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$.
+\vspace{2mm}
+
+A two-bit state $\ket{ab}$ comes with four probabilities:
+$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$. \par
 If we change the probabilities of only $\ket{a}$, \textit{all four of these change!}
 
 
 \vfill
 
-Because of this fact, \say{controlled gates} may not work as you expect. They may seem
-to \say{read} their controlling qubit without affecting its state, but remember---a
-controlled gate still affects the \textit{entire} state. As we noted before, it is
-not possible to apply a transformation to one bit of a quantum state.
-
+Because of this fact, \say{controlled gates} behave in a somewhat counterintuitive way.
+They do not simply \say{read} their controlling qubit without affecting its state---
+they mutate the entire state they are applied to, and may change all its bits!
 
 \begin{center}
 	\begin{tikzpicture}[scale=1]

src/Advanced/Introduction to Quantum/src/parts/07 superdense.tex

@@ -12,7 +12,7 @@ Consider the following entangled two-qubit states, called the \textit{bell state
 The probabilistic bits we get when measuring any of the above may be called \textit{anticorrelated bits}. \par
 If we measure the first bit of any of these states and observe $1$, what is the resulting compound state? \par
 What if we observe $0$ instead? \par
-Do you see why we can call these bits anticorrelated?
+Do you see why we call these bits \say{anticorrelated}?
 
 \vfill
 

src/Advanced/Lattices/parts/0 intro.tex

@@ -1,5 +1,6 @@
 \definition{}
-The \textit{integer lattice} $\mathbb{Z}^n \subset \mathbb{R}^n$ is the set of points with integer coordinates.
+The \textit{integer lattice} $\mathbb{Z}^n$ is the set of points with integer coordinates in $n$ dimensions. \par
+For example, $\mathbb{Z}^3$ is the set of points $(a, b, c)$ where $a$, $b$, and $c$ are integers.
 
 \problem{}
 Draw $\mathbb{Z}^2$.
@@ -8,12 +9,13 @@ Draw $\mathbb{Z}^2$.
 
 
 \definition{}
-We say a set of vectors $\{v_1, v_2, ..., v_k\}$ \textit{generates} $\mathbb{Z}^n$ if every lattice point can be written uniquely as
+We say a set of vectors $\{v_1, v_2, ..., v_k\}$ \textit{generates} $\mathbb{Z}^n$
+if every lattice point can be written as
 $$
 	a_1v_1 + a_2v_2 + ... + a_kv_k
 $$
 for integer coefficients $a_i$. \par
-It is fairly easy to show that $k$ must be at least $n$.
+\textbf{Bonus:} show that $k$ must be at least $n$.
 
 \problem{}
 Which of the following generate $\mathbb{Z}^2$?
@@ -30,8 +32,7 @@ Which of the following generate $\mathbb{Z}^2$?
 \vfill
 
 \problem{}
-Find a set of two vectors that generates $\mathbb{Z}^2$. \\
-Don't say $\{ (0, 1), (1, 0) \}$, that's too easy.
+Find a set of two vectors other than $\{ (0, 1), (1, 0) \}$ that generates $\mathbb{Z}^2$. \\
 
 \vfill
 
@@ -44,7 +45,8 @@ Find a set of vectors that generates $\mathbb{Z}^n$.
 \pagebreak
 
 \definition{}
-A \textit{fundamental region} of a lattice is the parallelepiped spanned by a generating set. The exact shape of this region depends on the generating set we use.
+Say we have a generating set of a lattice. \par
+The \textit{fundamental region} of this set is the $n$-dimensional parallelogram spanned by its members. \par
 
 \problem{}
 Draw two fundamental regions of $\mathbb{Z}^2$ using two different generating sets. Verify that their volumes are the same.

src/Advanced/Lattices/parts/1 minkowski.tex

@@ -1,6 +1,6 @@
 \section{Minkowski's Theorem}
 
-\theorem{Blichfeldt's Theorem}
+\theorem{Blichfeldt's Theorem}<blich>
 Let $X$ be a finite connected region. If the volume of $X$ is greater than $1$, $X$ must contain two distinct points that differ by an element of $\mathbb{Z}^n$. In other words, there exist distinct $x, y \in X$ so that $x - y \in \mathbb{Z}^n$.
 
 \vspace{2mm}
@@ -9,14 +9,22 @@ Intuitively, this means that you can translate $X$ to cover two lattice points a
 
 
 \problem{}
-Draw a region in $\mathbb{R}^2$ with volume greater than 1 that contains no lattice points. Find two points in that region which differ by an integer vector.
+Draw a connected region in $\mathbb{R}^2$ with volume greater than 1 that contains no lattice points. Find two points in that region which differ by an integer vector.
 \hint{Area is two-dimensional volume.}
 
 \vfill
 
 
 \problem{}
-The following picture gives an idea for the proof of Blichfeldt's theorem in $\mathbb{Z}^2$. Explain the picture and complete the proof.
+Draw a \textit{disconnected} region in $\mathbb{R}^2$ with volume greater than 1 that contains no lattice points, \par
+and show that no two points in that region differ by an integer vector.
+\note{In other words, show that \ref{blich} indeed requires a connected region.}
+
+\vfill
+
+\problem{}
+The following picture gives an idea for the proof of Blichfeldt's theorem in $\mathbb{Z}^2$. \par
+Explain the picture and complete the proof.
 
 \begin{center}
 	\includegraphics[angle=90,width=0.5\linewidth]{proof.png}
@@ -48,10 +56,8 @@ Let $X$ be a region $\in \mathbb{R}^2$ of volume $k$. How many integral points m
 A region $X$ is \textit{convex} if the line segment connecting any two points in $X$ lies entirely in $X$.
 
 \problem{}
-\begin{itemize}
-	\item Draw a convex region in the plane.
-	\item Draw a region that is not convex.
-\end{itemize}
+Draw a convex region in two dimensions. \par
+Then, draw a two-dimensional region that is \textit{not} convex.
 
 \vfill
 \pagebreak
@@ -59,23 +65,28 @@ A region $X$ is \textit{convex} if the line segment connecting any two points in
 
 
 \definition{}
-We say a region $X$ is \textit{symmetric} if for all points $x \in X$, $-x$ is also in $X$.
+We say a region $X$ is \textit{symmetric with respect to the origin} if for all points $x \in X$, $-x$ is also in $X$. \par
+In the following problems, \say{\textit{symmetric}} means \say{symmetric with respect to the origin.}
 
 \problem{}
-\begin{itemize}
-	\item Draw a symmetric region.
-	\item Draw an asymmetric region.
-\end{itemize}
+Draw a symmetric region. \par
+Then, draw an asymmetric region.
+
+\vfill
+
+\problem{}
+Show that a convex symmetric set always contains the origin.
 
 \vfill
 
 \theorem{Minkowski's Theorem}<mink>
-Every convex set in $\mathbb{R}^n$ that is symmetric with respect to the origin and which has a volume greater than $2^n$ contains an integral point that isn't zero.
+Every convex set in $\mathbb{R}^n$ that is symmetric and has a volume \par
+greater than $2^n$ contains an integral point that isn't zero.
 
 
 \problem{}
 Draw a few sets that satisfy \ref{mink} in $\mathbb{R}^2$. \par
-What is the simplest region that has the properties listed above?
+What is a simple class of regions that has the properties listed above?
 
 \vfill
 

src/Advanced/Symmetric Groups/macros.typ

@@ -0,0 +1,134 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+#let markline(s, from, to, c: oblue) = {
+  import cetz.draw: *
+  let del = 0.4 // small line
+
+  let arrow = (
+    symbol: ")>",
+    scale: s * 2.2,
+    fill: c,
+    stroke: c,
+  )
+
+  line(
+    from + ".south",
+    (v => cetz.vector.add(v, (0, -del * s)), from + ".south"),
+    (v => cetz.vector.add(v, (0, (del + 0.4) * s)), to + ".north"),
+    (v => cetz.vector.add(v, (0, 0.4 * s)), to + ".north"),
+    stroke: c + s * 1mm,
+  )
+  mark(
+    to + ".north",
+    270deg,
+    ..arrow,
+  )
+}
+
+
+#let marklinetop(s, from, to, del: 0.8, c: oblue) = {
+  import cetz.draw: *
+
+  let arrow = (
+    symbol: ")>",
+    scale: s * 2.2,
+    fill: c,
+    stroke: c,
+  )
+
+  line(
+    from + ".north",
+    (v => cetz.vector.add(v, (0, del * s)), from + ".north"),
+    (v => cetz.vector.add(v, (0, del * s)), to + ".north"),
+    (v => cetz.vector.add(v, (0, 0.2 * s)), to + ".north"),
+    stroke: c + s * 1mm,
+  )
+  mark(
+    to + ".north",
+    270deg,
+    ..arrow,
+  )
+}
+
+
+
+#let marklinebot(s, from, to, del: 0.8, c: oblue) = {
+  import cetz.draw: *
+
+  let arrow = (
+    symbol: ")>",
+    scale: s * 2.2,
+    fill: c,
+    stroke: c,
+  )
+
+  line(
+    from + ".south",
+    (v => cetz.vector.add(v, (0, -del * s)), from + ".south"),
+    (v => cetz.vector.add(v, (0, -del * s)), to + ".south"),
+    (v => cetz.vector.add(v, (0, -0.4 * s)), to + ".south"),
+    stroke: c + s * 1mm,
+  )
+  mark(
+    to + ".south",
+    90deg,
+    ..arrow,
+  )
+}
+
+
+
+#let marklinebotswap(s, from, to, del: 0.8, c: oblue) = {
+  import cetz.draw: *
+
+  let arrow = (
+    symbol: ")>",
+    scale: s * 2.2,
+    fill: c,
+    stroke: c,
+  )
+
+  line(
+    from + ".south",
+    (v => cetz.vector.add(v, (0, -del * s)), from + ".south"),
+    (v => cetz.vector.add(v, (0.5 * s, -del * s)), to + ".south"),
+    (v => cetz.vector.add(v, (0.5 * s, -del * s)), to + ".south"),
+    (v => cetz.vector.add(v, (0.5 * s, del * s)), to + ".north"),
+    (v => cetz.vector.add(v, (0, del * s)), to + ".north"),
+    (v => cetz.vector.add(v, (0, 0.2 * s)), to + ".north"),
+    stroke: c + s * 1mm,
+  )
+  mark(
+    to + ".north",
+    270deg,
+    ..arrow,
+  )
+}
+
+#let marklinetopswap(s, from, to, del: 0.8, c: oblue) = {
+  import cetz.draw: *
+
+  let arrow = (
+    symbol: ")>",
+    scale: s * 2.2,
+    fill: c,
+    stroke: c,
+  )
+
+  line(
+    from + ".north",
+    (v => cetz.vector.add(v, (0, del * s)), from + ".north"),
+    (v => cetz.vector.add(v, (0.5 * s, del * s)), to + ".north"),
+    (v => cetz.vector.add(v, (0.5 * s, del * s)), to + ".north"),
+    (v => cetz.vector.add(v, (0.5 * s, -del * s)), to + ".south"),
+    (v => cetz.vector.add(v, (0, -del * s)), to + ".south"),
+    (v => cetz.vector.add(v, (0, -0.2 * s)), to + ".south"),
+    stroke: c + s * 1mm,
+  )
+  mark(
+    to + ".south",
+    90deg,
+    ..arrow,
+  )
+}

src/Advanced/Symmetric Groups/main.tex

@@ -1,58 +0,0 @@
-% use [nosolutions] flag to hide solutions.
-% use [solutions] flag to show solutions.
-\documentclass[
-	solutions,
-	singlenumbering
-]{../../../lib/tex/handout}
-\usepackage{../../../lib/tex/macros}
-\usetikzlibrary{calc}
-
-\uptitlel{Advanced 2}
-\uptitler{\smallurl{}}
-\title{Symmetric Groups}
-\subtitle{Prepared by Mark on \today{}}
-
-
-
-\def\line#1#2{
-	\draw[line width = 0.3mm, ->, ocyan]
-		(#1)
-		-- ($(#1) + (0, -1)$)
-		-- ($(#2) + (0,1)$)
-		-- (#2);
-}
-
-\begin{document}
-
-	\maketitle
-
-	\input{parts/0 intro}
-	\input{parts/1 cycle}
-	\input{parts/2 groups}
-	\input{parts/3 subgroup}
-
-
-	\section{Bonus problems}
-
-	\problem{}
-	Show that $x \in \mathbb{Z}^+$ has a multiplicative inverse mod $n$ iff $\text{gcd}(x, n) = 1$
-
-	\vfill
-
-	\problem{}
-	Let $\sigma = (\sigma_1 \sigma_2 ... \sigma_k)$ be a $k$-cycle in $S_n$, and let $\tau$ be an arbitrary element of $S_n$. \par
-	Show that $\tau \sigma \tau^{-1}$ = $\bigl(\tau(\sigma_1), \tau(\sigma_2), ..., \tau(\sigma_k)\bigr)$ \par
-	\hint{As usual, $\tau$ is a permutation. Thus, $\tau(x)$ is the value at position $x$ after applying $\tau$.}
-
-	\vfill
-
-	\problem{}
-	Show that the set $\Bigl\{ (1, 2),~ (1,2,...,n) \Bigr\}$ generates $S_n$.
-	\vfill
-
-	% TODO: (a second day?)
-	% alternating group
-	% type and sign and conjugation
-	% isomorphisms & automorphisms
-	% automorphism groups
-\end{document}

src/Advanced/Symmetric Groups/main.typ

@@ -0,0 +1,35 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Symmetric Groups],
+  by: "Mark",
+)
+
+#include "parts/00 intro.typ"
+#pagebreak()
+
+#include "parts/01 cycle.typ"
+#pagebreak()
+
+#include "parts/02 groups.typ"
+#pagebreak()
+
+#include "parts/03 subgroup.typ"
+#pagebreak()
+
+= Bonus problems
+
+#problem()
+Show that $x in ZZ^+$ has a multiplicative inverse mod $n$ if and only if $gcd(x, n) = 1$
+
+#v(1fr)
+
+#problem()
+Let $sigma = (sigma_1 sigma_2 ... sigma_k)$ be a $k$-cycle in $S_n$, and let $tau$ be an arbitrary element of $S_n$. \
+Show that $tau sigma tau^(-1)$ = $(tau(sigma_1), tau(sigma_2), ..., tau(sigma_k))$ \
+#hint[$tau$ is a permutation, so $tau(x)$ is the value at position $x$ after applying $tau$.]
+
+#v(1fr)
+#problem()
+Show that the set ${ (1, 2), (1,2,...,n)}$ generates $S_n$.
+#v(1fr)

src/Advanced/Symmetric Groups/parts/0 intro.tex

@@ -1,199 +0,0 @@
-\section{Introduction}
-
-
-\definition{}
-Informally, a \textit{permutation} of a collection of $n$ objects is an ordering of these $n$ objects. \par
-For example, a few permutations of $\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}$ are $\texttt{ABCD}$,
-$\texttt{BCDA}$, and $\texttt{DACB}$. \par
-
-\vspace{2mm}
-
-This, however, isn't the definition we'll use today. Instead of defining permutations as \say{ordered lists,}
-(as we do above), we'll define them as functions. Our first goal today is to make sense of this definition.
-
-
-
-\definition{Permutations}
-Let $\Omega$ be an arbitrary set of $n$ objects. \par
-A \textit{permutation} on $\Omega$ is a map from $\Omega$ to itself that produces a \textit{unique} output for each input. \par
-\note{In other words, if $a$ and $b$ are different, $f(a)$ and $f(b)$ must also be different.}
-
-
-\footnotetext{The words \say{function} and \say{map} are equivalent.}
-
-\vspace{2mm}
-
-For example, consider $\{1, 2, 3\}$. \par
-One permutation on this set can be defined as follows: \par
-\begin{itemize}
-	\item $f(1) = 3$
-	\item $f(2) = 1$
-	\item $f(3) = 2$
-\end{itemize}
-
-If we take the array $123$ and apply
-
-
-\problem{}
-List all permutations on three objects. \par
-How many permutations of $n$ objects are there?
-
-\vfill
-
-
-\problem{}
-What map corresponds to the permutation $[321]$?
-
-
-\vfill
-
-\problem{}
-What map corresponds to the \say{do-nothing} permutation? \par
-Write it as a function and in square-bracket notation. \par
-\note[Note]{We usually call this the \textit{trivial permutation}}
-
-
-\vfill
-\pagebreak
-
-
-We can visualize permutations with a \textit{string diagram}, shown below. \par
-The arrows in this diagram denote the image of $f$ for each possible input.
-Two examples are below:
-
-\vspace{2mm}
-\hfill
-\begin{tikzpicture}[scale=0.5]
-	\node (1a) at (0, 0.5) {1};
-	\node (2a) at (1, 0.5) {2};
-	\node (3a) at (2, 0.5) {3};
-	\node (4a) at (3, 0.5) {4};
-
-	\node (1b) at (0, -2) {1};
-	\node (3b) at (1, -2) {3};
-	\node (4b) at (2, -2) {4};
-	\node (2b) at (3, -2) {2};
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-	\line{4a}{4b}
-\end{tikzpicture}
-\hfill
-\begin{tikzpicture}[scale=0.5]
-	\node (1a) at (0, 0.5) {1};
-	\node (2a) at (1, 0.5) {2};
-	\node (3a) at (2, 0.5) {3};
-	\node (4a) at (3, 0.5) {4};
-
-	\node (2b) at (0, -2) {2};
-	\node (1b) at (1, -2) {1};
-	\node (3b) at (2, -2) {3};
-	\node (4b) at (3, -2) {4};
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-	\line{4a}{4b}
-\end{tikzpicture}
-\hfill\null
-\vspace{2mm}
-
-
-Note that in all our examples thus far, the objects in our set have an implicit order.
-This is only for convenience. The elements of $\Omega$ are not ordered (it is a \textit{set}, after all),
-and we may present them however we wish.
-
-
-
-
-
-
-
-\vspace{1cm}
-
-For example, consider the diagrams below. \par
-On the left, 1234 are ordered as usual. In the middle, they are ordered alphabetically. \par
-The rightmost diagram uses arbitrary, meaningless labels.
-
-\vspace{2mm}
-\hfill
-\begin{tikzpicture}[scale=0.5]
-	\node (1a) at (0, 0.5) {1};
-	\node (2a) at (1, 0.5) {2};
-	\node (3a) at (2, 0.5) {3};
-	\node (4a) at (3, 0.5) {4};
-
-	\node (2b) at (0, -2) {2};
-	\node (1b) at (1, -2) {1};
-	\node (3b) at (2, -2) {3};
-	\node (4b) at (3, -2) {4};
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-	\line{4a}{4b}
-\end{tikzpicture}
-\hfill
-\begin{tikzpicture}[scale=0.5]
-	\node (4a) at (0, 0.5) {4};
-	\node (1a) at (1, 0.5) {1};
-	\node (3a) at (2, 0.5) {3};
-	\node (2a) at (3, 0.5) {2};
-
-	\node (1b) at (0, -2) {1};
-	\node (4b) at (1, -2) {4};
-	\node (3b) at (2, -2) {3};
-	\node (2b) at (3, -2) {2};
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-	\line{4a}{4b}
-\end{tikzpicture}
-\hfill
-\begin{tikzpicture}[scale=0.5]
-	\node (1a) at (0, 0.5) {$\triangle$};
-	\node (2a) at (1, 0.5) {$\divideontimes$};
-	\node (3a) at (2, 0.5) {$\circledcirc$};
-	\node (4a) at (3, 0.5) {$\boxdot$};
-
-	\node (2b) at (0, -2) {$\divideontimes$};
-	\node (1b) at (1, -2) {$\triangle$};
-	\node (3b) at (2, -2) {$\circledcirc$};
-	\node (4b) at (3, -2) {$\boxdot$};
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-	\line{4a}{4b}
-\end{tikzpicture}
-\hfill\null
-\vspace{2mm}
-
-
-It shouldn't be hard to see that despite the different \say{output} order (2134 and 1432), \par
-the same permutation is depicted in all three diagrams. This example demonstrates two things:
-\begin{itemize}[itemsep=2mm]
-	\item First, the names of the items in our set do not have any meaning. \par
-	$\Omega$ is just a set of $n$ arbitrary things, which we may label however we like.
-
-	\item Second, permutations are verbs. We do not care about the \say{output} of a certain permutation,
-	we care about what it \textit{does}. We could, for example, describe the permutation above as
-	\say{swap the first two of four elements.}
-\end{itemize}
-
-\vspace{2mm}
-
-
-Why, then, do we order our elements when we talk about permutations? As noted before, this is for convenience.
-If we assign a natural order to the elements of $\Omega$ (say, 1234), we can identify permutations by simply listing
-their output:
-Clearly, $[1234]$ represents the trivial permutation, $[2134]$ represents \say{swap first two,}
-and $[4123]$ represents \say{cycle right.}
-
-\problem{}
-Draw string diagrams for $[4123]$ and $[2341]$.
-
-\vfill
-\pagebreak

src/Advanced/Symmetric Groups/parts/00 intro.typ

@@ -0,0 +1,378 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+#import "../macros.typ": *
+
+= Introduction
+
+#definition()
+Informally, a _permutation_ on a collection of $n$ objects is an ordering of these $n$ objects.
+
+For example, a few permutations of `A`, `B`, `C`, and `D` are `ABCD`, `BCDA`, and `DACB`.
+
+#v(2mm)
+
+This, however, isn't the definition we'll use today.
+Instead of defining permutations as "ordered lists" \
+(like we do above), we'll define them as _functions_ on finite sets. \
+Our first goal today is to make sense of this definition.
+
+
+#definition("Permutations")
+Let $Omega$ be a set of $n$ arbitrary objects.
+
+A _permutation_ $f$ on $Omega$ is a map#footnote[The words "function" and "map" are equivalent.]
+from $Omega$ to itself that produces a _unique_ output for each input.
+
+#note[This means that if $a$ and $b$ are different, $f(a)$ and $f(b)$ must also be different.]
+
+
+
+#v(2mm)
+
+For example, consider ${1, 2, 3}$. \
+One permutation on this set can be defined as follows:
+- $f(1) = 3$
+- $f(2) = 1$
+- $f(3) = 2$
+
+If we take the array $123$ and apply $f$, we get $312$.
+
+#problem()
+List all permutations on three objects. \
+How many permutations of $n$ objects are there?
+
+
+#v(1fr)
+
+#problem()
+What map corresponds to the permutation that produces the array `231` from the array `123`?
+
+#v(1fr)
+
+#problem()
+What map corresponds to the "do-nothing" permutation?
+
+Write it as a function and in square-bracket notation.
+
+#note([We will call this the _trivial permutation_])
+
+
+#v(1fr)
+#pagebreak()
+
+We can visualize a permutation using a _string diagram_.
+The arrows in this diagram denote \
+the output of $f$ for each possible input.
+Two examples are below:
+
+#table(
+  columns: (1fr, 1fr),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+    let s = 0.5 // scale
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    content((0 * s, 0 * s), $1$, name: "1b")
+    content((1 * s, 0 * s), $3$, name: "3b")
+    content((2 * s, 0 * s), $4$, name: "4b")
+    content((3 * s, 0 * s), $2$, name: "2b")
+
+    markline(s, "1a", "1b")
+    markline(s, "2a", "2b")
+    markline(s, "3a", "3b")
+    markline(s, "4a", "4b")
+  })),
+
+  align(center, cetz.canvas({
+    import cetz.draw: *
+    let s = 0.5 // scale
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    content((0 * s, 0 * s), $2$, name: "2b")
+    content((1 * s, 0 * s), $1$, name: "1b")
+    content((2 * s, 0 * s), $3$, name: "3b")
+    content((3 * s, 0 * s), $4$, name: "4b")
+
+    markline(s, "1a", "1b")
+    markline(s, "2a", "2b")
+    markline(s, "3a", "3b")
+    markline(s, "4a", "4b")
+  })),
+)
+
+Note that the elements of the set we are permuting are not ordered. (it is a _set_, after all!) \
+For example, consider the diagrams below.
+On the left, 1234 are ordered as usual. \
+In the middle, they are ordered alphabetically. \
+The rightmost diagram uses arbitrary, meaningless labels.
+
+#v(2mm)
+
+#table(
+  columns: (1fr, 1fr, 1fr),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.5 // scale
+    let del = 0.2 // small line
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    content((0 * s, 0 * s), $2$, name: "2b")
+    content((1 * s, 0 * s), $1$, name: "1b")
+    content((2 * s, 0 * s), $3$, name: "3b")
+    content((3 * s, 0 * s), $4$, name: "4b")
+
+    line(
+      "1a.south",
+      (v => cetz.vector.add(v, (0, -del)), "1a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "1b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "1b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "1b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "2a.south",
+      (v => cetz.vector.add(v, (0, -del)), "2a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "2b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "2b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "2b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "3a.south",
+      (v => cetz.vector.add(v, (0, -del)), "3a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "3b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "3b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "3b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "4a.south",
+      (v => cetz.vector.add(v, (0, -del)), "4a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "4b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "4b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "4b.north",
+      270deg,
+      ..arrow,
+    )
+  })),
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.5 // scale
+    let del = 0.2 // small line
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $4$, name: "4a")
+    content((1 * s, 3 * s), $1$, name: "1a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $2$, name: "2a")
+
+    content((0 * s, 0 * s), $1$, name: "1b")
+    content((1 * s, 0 * s), $4$, name: "4b")
+    content((2 * s, 0 * s), $3$, name: "3b")
+    content((3 * s, 0 * s), $2$, name: "2b")
+
+    line(
+      "1a.south",
+      (v => cetz.vector.add(v, (0, -del)), "1a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "1b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "1b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "1b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "2a.south",
+      (v => cetz.vector.add(v, (0, -del)), "2a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "2b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "2b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "2b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "3a.south",
+      (v => cetz.vector.add(v, (0, -del)), "3a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "3b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "3b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "3b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "4a.south",
+      (v => cetz.vector.add(v, (0, -del)), "4a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "4b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "4b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "4b.north",
+      270deg,
+      ..arrow,
+    )
+  })),
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.5 // scale
+    let del = 0.2 // small line
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $ast.circle$, name: "4a")
+    content((1 * s, 3 * s), $hexa.stroked$, name: "1a")
+    content((2 * s, 3 * s), $triangle.stroked.b$, name: "3a")
+    content((3 * s, 3 * s), $\#$, name: "2a")
+
+    content((0 * s, 0 * s), $hexa.stroked$, name: "1b")
+    content((1 * s, 0 * s), $ast.circle$, name: "4b")
+    content((2 * s, 0 * s), $triangle.stroked.b$, name: "3b")
+    content((3 * s, 0 * s), $\#$, name: "2b")
+
+    line(
+      "1a.south",
+      (v => cetz.vector.add(v, (0, -del)), "1a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "1b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "1b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "1b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "2a.south",
+      (v => cetz.vector.add(v, (0, -del)), "2a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "2b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "2b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "2b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "3a.south",
+      (v => cetz.vector.add(v, (0, -del)), "3a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "3b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "3b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "3b.north",
+      270deg,
+      ..arrow,
+    )
+
+    line(
+      "4a.south",
+      (v => cetz.vector.add(v, (0, -del)), "4a.south"),
+      (v => cetz.vector.add(v, (0, del + 0.2)), "4b.north"),
+      (v => cetz.vector.add(v, (0, 0.2)), "4b.north"),
+      stroke: oblue + s * 1mm,
+    )
+    mark(
+      "4b.north",
+      270deg,
+      ..arrow,
+    )
+  })),
+)
+
+#v(2mm)
+
+It shouldn't be hard to see that despite the different "output" each diagram displays \
+($2134$, $1432$, and $hexa.stroked ast.circle triangle.stroked.b \#$), the same permutation ("swap first two") is shown in each.
+
+Observe the following:
+- The "names" of the items in our set do not have any meaning. \
+  We are interested in sets of $n$ arbitrary things, which we may label however we like.
+- Permutations are _verbs_. \
+  We do not care about the "output" of a certain permutation. Rather, we care about what it _does_. \
+  We could, for example, describe the permutation in the above three diagrams as "swap the first two elements."
+
+
+#definition("Square Brackets")
+However, elements with an implicit order (1, 2, 3, ...) are convenient. \
+Such sets let us denote a permutation by writing the array it produces \
+after transforming the "reference order" $123...n$.
+
+We will call this _square-bracket notation_. \
+$[312]$ denotes the permutation that produces $312$ when applied to $123$.
+
+#problem()
+Draw string diagrams for $[4123]$ and $[2341]$.
+
+#v(1fr)

src/Advanced/Symmetric Groups/parts/01 cycle.typ

@@ -0,0 +1,635 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+#import "../macros.typ": *
+
+= Cycle Notation
+
+#definition("Order")
+The _order_ of a permutation $f$ is the smallest positive $n$ where $f^n (x) = x$ for all $x$. \
+In other words, if we repeatedly apply a permutation with order $n$, \
+we will get back to where we started after $n$ steps. \
+
+#v(2mm)
+
+For example, consider $[2134]$. This permutation has order $2$, as we can see below:
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.5 // scale
+    let del = 0.2 // small line
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    content((0 * s, 0 * s), $2$, name: "2b")
+    content((1 * s, 0 * s), $1$, name: "1b")
+    content((2 * s, 0 * s), $3$, name: "3b")
+    content((3 * s, 0 * s), $4$, name: "4b")
+
+    content((0 * s, -3 * s), $1$, name: "1c")
+    content((1 * s, -3 * s), $2$, name: "2c")
+    content((2 * s, -3 * s), $3$, name: "3c")
+    content((3 * s, -3 * s), $4$, name: "4c")
+
+    markline(s, "1a", "1b")
+    markline(s, "2a", "2b")
+    markline(s, "3a", "3b")
+    markline(s, "4a", "4b")
+
+    markline(s, "1b", "1c")
+    markline(s, "2b", "2c")
+    markline(s, "3b", "3c")
+    markline(s, "4b", "4c")
+  })),
+)
+
+Swapping the first two elements of a list twice changes nothing. \
+Thus, $[2134]$ has an order of two.
+
+
+#problem()
+What is the order of $[2314]$? \
+How about $[4321]$? \
+#note(type: "Note")[Try to solve this problem without drawing any strings!]
+
+#v(1fr)
+
+#problem()
+Find a permutation on five elements with order 4.
+
+#v(1fr)
+
+
+#problem(label: "finiteorder")
+Show that all permutations on a finite set have a well-defined order. \
+In other words, show that there must always be an integer $n$ where $f^n (x) = x$.
+
+#v(1fr)
+#pagebreak()
+
+
+
+#definition("Composition", label: "compdef")
+The _composition_ of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \
+We'll denote this by simply writing the permutations we're composing next to each other, like $f g$. \
+Note that $g$ is applied _before_ $f$ in $f g$.
+
+#problem()
+Show that function composition is associative. \
+That is, show that $f(g h) = (f g)h$.
+
+#v(1fr)
+
+#problem()
+What is $[1324][4321]$? \
+How about $[321][213][231]$? \
+Rewrite these compositions as one permutation in square brackets.
+
+#solution([
+  - $[1324][4321]$ is $[4231]$
+  - $[321][213][231]$ is $[123]$
+])
+
+
+#v(1fr)
+
+
+As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
+Permutations are verbs---but we've been referring to them using a noun (i.e, their output).
+
+Square-bracket notation fails to capture the structure of the permutation it identifies.
+
+#v(2mm)
+
+Is the permutation $[1234]$ different than the permutation $[12345]$? \
+These permutations operate on different sets---but they are both the identity! \
+Are $[5342761]$ and $[1342567][5234761]$ similar? What are their orders?
+
+
+#v(2mm)
+
+Good notation should help us understand the objects we are studying. \
+We need something better than square brackets.
+
+
+#pagebreak()
+
+
+#remark("Cycles")
+Any permutation is composed of a number of _cycles_. \
+Reread @finiteorder to convince yourself of this fact.
+
+
+#example()
+Consider the permutation $[2134]$. \
+It consists of one two-cycle: $1 arrow.r 2 arrow.r 1$, which we can see in the diagram below. \
+#note(
+  type: "Note",
+)[
+  $3 arrow.r 3$ and $4 arrow.r 4$ are also cycles, but we'll ignore them.
+  One-cycles aren't interesting.
+]
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.6 // scale
+    let del = 0.4 // small line
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    marklinetop(s, "1a", "2a")
+    marklinebot(s, "2a", "1a")
+  })),
+)
+
+#v(4mm)
+
+The permutation $[431265]$ is a bit more interesting---it contains two cycles: \
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.6 // scale
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+    content((4 * s, 3 * s), $5$, name: "5a")
+    content((5 * s, 3 * s), $6$, name: "6a")
+
+
+    marklinetop(s, "3a", "2a", del: 0.8)
+    marklinebot(s, "2a", "4a", del: 1.3)
+    marklinetop(s, "4a", "1a", del: 1.3)
+    marklinebot(s, "1a", "3a", del: 0.8)
+    marklinebot(s, "5a", "6a", del: 0.8, c: ogreen)
+    marklinetop(s, "6a", "5a", del: 0.8, c: ogreen)
+  })),
+)
+
+#remark()
+Two-cycles may also be called _transpositions_. \
+Any permutation that swaps two elements is a transposition.
+
+#problem()
+Find all cycles in $[5342761]$.
+
+#solution[
+  #table(
+    columns: (1fr,),
+    align: center,
+    stroke: none,
+    align(center, cetz.canvas({
+      import cetz.draw: *
+
+      let s = 0.5 // scale
+      let arrow = (
+        symbol: ")>",
+        scale: s * 2.2,
+        fill: oblue,
+        stroke: oblue,
+      )
+
+      set-style(content: (
+        frame: "rect",
+        stroke: none,
+        padding: .1,
+      ))
+      content((0 * s, 3 * s), $1$, name: "1a")
+      content((1 * s, 3 * s), $2$, name: "2a")
+      content((2 * s, 3 * s), $3$, name: "3a")
+      content((3 * s, 3 * s), $4$, name: "4a")
+      content((4 * s, 3 * s), $5$, name: "5a")
+      content((5 * s, 3 * s), $6$, name: "6a")
+      content((6 * s, 3 * s), $7$, name: "7a")
+
+
+      marklinetop(s, "1a", "7a", del: 1.6)
+      marklinebot(s, "7a", "5a", del: 1.2)
+      marklinetopswap(s, "5a", "1a", del: 1.2)
+
+
+      marklinebot(s, "2a", "4a", del: 1.2, c: ogreen)
+      marklinetop(s, "4a", "3a", del: 0.8, c: ogreen)
+      marklinebotswap(s, "3a", "2a", del: 0.8, c: ogreen)
+    })),
+  )
+
+  There are two non-trivial cycles:
+  - $4 arrow.r 3 arrow.r 2 arrow.r 4$
+  - $1 arrow.r 7 arrow.r 5 arrow.r 1$
+]
+
+
+
+#v(1fr)
+
+
+#problem()
+What permutation on five objects is formed by the cycles $3 arrow.r 5 arrow.r 3$ and $1 arrow.r 2 arrow.r 4 arrow.r 1$? \
+Write it in square-bracket notation.
+
+#solution[
+  #table(
+    columns: (1fr,),
+    align: center,
+    stroke: none,
+    align(center, cetz.canvas({
+      import cetz.draw: *
+
+      let s = 0.6 // scale
+      let arrow = (
+        symbol: ")>",
+        scale: s * 2.2,
+        fill: oblue,
+        stroke: oblue,
+      )
+
+      set-style(content: (
+        frame: "rect",
+        stroke: none,
+        padding: .1,
+      ))
+      content((0 * s, 3 * s), $1$, name: "1a")
+      content((1 * s, 3 * s), $2$, name: "2a")
+      content((2 * s, 3 * s), $3$, name: "3a")
+      content((3 * s, 3 * s), $4$, name: "4a")
+      content((4 * s, 3 * s), $5$, name: "5a")
+
+      marklinetop(s, "3a", "5a", del: 0.8, c: ogreen)
+      marklinebot(s, "5a", "3a", del: 0.8, c: ogreen)
+
+      marklinebot(s, "1a", "2a", del: 0.8)
+      marklinetop(s, "2a", "4a", del: 1.2)
+
+      marklinebotswap(s, "4a", "1a", del: 1.2)
+    })),
+  )
+
+
+  This is $[41523]$.
+]
+
+
+#v(1fr)
+#pagebreak()
+
+
+#definition("Cycle Notation")
+We can use cycles to develop better notation: \
+Instead of identifying permutations using their output, we'll identify them using their _cycles_.
+
+#v(2mm)
+
+For example, we'll write $[2134]$ is $(12)$ in cycle notation, \
+since it consists only of the cycle $1 arrow.r 2 arrow.r 1$:
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.6 // scale
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (
+      frame: "rect",
+      stroke: none,
+      padding: .1,
+    ))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    marklinebot(s, "1a", "2a", del: 0.8)
+    marklinetop(s, "2a", "1a", del: 0.8)
+  })),
+)
+
+#v(2mm)
+
+
+Permutations that consist of more than one cycle are written as a composition. \
+$[2143]$ is written as $(12)(34)$. Applying the permutation $[2143]$ has the same effect as applying $(34)$, then applying $(12)$.
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.6 // scale
+    let arrow = (
+      symbol: ")>",
+      scale: s * 2.2,
+      fill: oblue,
+      stroke: oblue,
+    )
+
+    set-style(content: (
+      frame: "rect",
+      stroke: none,
+      padding: .1,
+    ))
+    content((0 * s, 3 * s), $1$, name: "1a")
+    content((1 * s, 3 * s), $2$, name: "2a")
+    content((2 * s, 3 * s), $3$, name: "3a")
+    content((3 * s, 3 * s), $4$, name: "4a")
+
+    marklinetop(s, "1a", "2a", del: 0.8)
+    marklinebot(s, "2a", "1a", del: 0.8)
+    marklinetop(s, "3a", "4a", del: 0.8, c: ogreen)
+    marklinebot(s, "4a", "3a", del: 0.8, c: ogreen)
+  })),
+)
+
+
+
+#remark()
+According to @finiteorder, any permutation may be written as a composition of disjoint cycles. \
+Convince yourself of this fact.
+
+#problem()
+Rewrite $[431265]$ in cycle notation.
+
+#solution[
+  $[431265]$ is $(1324)(56)$:
+
+  #table(
+    columns: (1fr,),
+    align: center,
+    stroke: none,
+    align(center, cetz.canvas({
+      import cetz.draw: *
+
+      let s = 0.6 // scale
+      let arrow = (
+        symbol: ")>",
+        scale: s * 2.2,
+        fill: oblue,
+        stroke: oblue,
+      )
+
+      set-style(content: (
+        frame: "rect",
+        stroke: none,
+        padding: .1,
+      ))
+      content((0 * s, 3 * s), $1$, name: "1a")
+      content((1 * s, 3 * s), $2$, name: "2a")
+      content((2 * s, 3 * s), $3$, name: "3a")
+      content((3 * s, 3 * s), $4$, name: "4a")
+      content((4 * s, 3 * s), $5$, name: "5a")
+      content((5 * s, 3 * s), $6$, name: "6a")
+
+      marklinetop(s, "1a", "3a", del: 0.8, c: ogreen)
+      marklinebot(s, "4a", "1a", del: 1.3, c: ogreen)
+      marklinebot(s, "3a", "2a", del: 0.8, c: ogreen)
+      marklinetop(s, "2a", "4a", del: 1.3, c: ogreen)
+
+      marklinetop(s, "5a", "6a", del: 0.8)
+      marklinebot(s, "6a", "5a", del: 0.8)
+    })),
+  )
+]
+
+
+
+
+#remark()
+The identity permutation $f(x) = x$ is written as $()$ in cycle notation.
+
+
+
+#problem()
+Convince yourself that disjoint cycles commute. \
+That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \
+
+
+#v(1fr)
+#pagebreak()
+
+#problem(label: "insquare")
+Write the following in square-bracket notation.
+
+- $(12)$      on a set of 2 elements
+- $(12)(435)$   on a set of 5 elements
+#v(2mm)
+- $(321)$       on a set of 3 elements
+- $(321)$       on a set of 6 elements
+#v(2mm)
+- $(1234)$       on a set of 4 elements
+- $(3412)$       on a set of 4 elements
+
+#note[
+  Note that $(12)$ refers the "swap first two" permutation on a set of _any_ size. \
+  We can use consistent notation for the same action on two different sets! \
+]
+
+#v(1fr)
+
+
+
+#problem()
+Write the following in square-bracket notation.
+Pay attention!
+- $(13)(243)$  on a set of 4 elements
+- $(243)(13)$  on a set of 4 elements
+
+#v(1fr)
+
+
+
+#problem()
+Consider the last two permutations in @insquare, $(1234)$ and $(3412)$. \
+These are _identical_---they are the same cycle written in two different ways. \
+List all other ways to write this cycle. \
+#hint[There are two more.]
+
+#v(1fr)
+#pagebreak()
+
+
+#definition("Inverse")
+The _inverse_ of a permutation $f$ is a permutation $g$ that "un-does" $f$. \
+This means that $g(f(x)) = x$ for all $x$.
+
+#problem()
+What is the inverse of $(12)$? \
+How about $(123)$? And $(4231)$? \
+#note[
+  Note we do not need to know the size of the set we are operating on. \
+  The inverse of $(12)$ is the same in sets of all sizes!
+]
+
+#v(1fr)
+
+
+#problem()
+Let $sigma$ be a permutation composed of disjoint cycles $sigma_1sigma_2...sigma_k$. \
+Say we know the order of all $sigma_i$. What is the order of $sigma$?
+
+#solution[
+  $
+    #text[lcm]\(#text[ord]\(sigma_1),#h(0.5em) #text[ord]\(sigma_2),#h(0.5em) ...,#h(0.5em) #text[ord]\(sigma_k))
+  $
+]
+
+
+#v(1fr)
+
+#problem(label: "cycletrans")
+Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
+
+#solution[
+  *Intuition:*\
+  $(123...n)$ is a right-shift. Swapping all pairs from right to left achieves the same effect.
+
+  #v(2mm)
+
+  *Complete solution:* \
+  Consider $n-1$. After applying $(123...n)$, it takes the position of $n$.
+
+  After applying $(n-1, n)$, $n-1$ moves to the same position _and is never moved again!_ \
+  Repeat this argument for all other $n$.
+]
+
+#v(1fr)
+
+#problem()
+Write $(7126453)$ as a product of transpositions. \
+
+#solution[
+  Move elements one at a time, using the last position as temporary storage.
+
+  We get $(71)(72)(76)(74)(75)(73)$.
+  Other solutions are possible. \
+
+  #v(2mm)
+
+  *Bonus:* How can we do this in the fewest number of transpositions?
+]
+
+#v(1fr)
+#pagebreak()
+
+#problem(label: "simpletrans")
+Show that any permutation is a product of transpositions.
+
+#solution[
+  Re-use the argument in @cycletrans. \
+  Pick an arbitrary "working slot," and re-build all cycles. \
+  Use the "not touched again" argument for a proper proof.
+]
+
+#v(1fr)
+
+
+#problem(label: "onetrans")
+Show that any permutation is a product of transpositions of the form $(1, k)$. \
+
+#solution[
+  Use @simpletrans to rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$. \
+  Showing that $(a, b) = (1, a)(1, b)(1, a)$ is fairly easy.
+]
+
+#v(1fr)
+#pagebreak()
+
+
+
+#problem(label: "oneplustrans")
+Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$ whenever $a + 1 != b$.
+
+#solution[
+  This is the same as @onetrans,
+  but we use $a + 1$ as a "working slot" instead of $1$.
+]
+
+#v(1fr)
+
+
+
+#problem()
+Show that any permutation is a product of adjacent transpositions. \
+An _adjacent transposition_ swaps two adjacent elements, and thus looks like $(n, n+1)$.
+
+#solution[
+  As before, we will use @simpletrans and rewrite the transpositions it produces in a convenient fashion.
+  To do this, we must show that every transposition $(a, b)$ is a product of adjacent transpositions.
+
+  #v(2mm)
+
+  In the proof below, assume that $a < b$ and perform induction on $b - a$. \
+
+  #v(4mm)
+
+
+  *Base Case:*\
+  If $b - a = 1$, $(a, b)$ is a product of adjacent transpositions. \
+  In fact, it _is_ an adjacent transposition.
+
+  #v(4mm)
+
+  *Induction:*\
+  Now, say $b - a = n + 1$. \
+  Assume that all $(a, b)$ where $b - a <= n$ are products of adjacent transpositions.\
+  By @oneplustrans, $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.
+
+  #v(2mm)
+
+  $(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \
+  Thus, $(a, b)$ is a product of adjacent transpositions.
+]
+
+#v(1fr)

src/Advanced/Symmetric Groups/parts/02 groups.typ

@@ -0,0 +1,148 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+#import "../macros.typ": *
+
+= Groups
+
+#definition()
+Before we continue, we must introduce a bit of notation:
+- $S_n$ is the set of permutations on $n$ objects.
+- $ZZ_n$ is the set of integers mod $n$.
+
+#problem()
+What are the elements of $S_3$? #hint[Use cycle notation] \
+How about $ZZ_8$?
+
+#v(1fr)
+
+#definition()
+A _group_ $(G, *)$ consists of a set $G$ and an operator $*$. \
+Groups always have the following properties:
+
++ $G$ is closed under $*$. In other words, $a, b in G => a * b in G$.
++ $*$ is _associative_: $(a * b) * c = a * (b * c)$ for all $a,b,c in G$
++ There is an _identity_ $e in G$, so that $a * e = e * a = a$ for all $a in G$.
++ For any $a in G$, there exists a $b in G$ so that $a * b = b * a = e$. $b$ is called the _inverse_ of $a$. \
+  This element is written as $-a$ if our operator is addition and $a^(-1)$ otherwise.
+
+Any pair $(G, *)$ that satisfies these properties is a group.
+
+#problem()
+Is $(ZZ_5, +)$ a group? \
+Is $(ZZ_5, -)$ a group? \
+#note[$+$ and $-$ refer to the usual operations in modular arithmetic.]
+
+#v(1fr)
+
+#problem()
+What is the group with the fewest number of elements?
+
+#solution[
+  Let $(G, star)$ be our group, where $G = {x}$ and $star$ is defined by $x star x = x$
+
+  Verifying that the trivial group is a group is trivial.
+]
+
+
+#definition()
+$ZZ_n^times$ is the set of integers mod $n$ with multiplicative inverses. \
+We can prove that this is the set of integers smaller than $n$ and coprime to $n$. \
+For example, $ZZ_12^times = {1, 5, 7, 11}$.
+
+#problem()
+What are the elements of $ZZ^times_8$? \
+How about $ZZ^times_23$? #hint[23 is prime.]
+
+#v(1fr)
+#pagebreak()
+
+
+
+
+
+#problem()
+Show that function composition is associative
+
+#v(1fr)
+
+#problem()
+Show that $S_n$ is a group under composition.
+
+#v(1fr)
+
+#problem()
+Let $(G, *)$ be a group with finitely many elements, and let $a in G$. \
+Show that there is an $n$ in $in ZZ$ so that $a^n = e$ \
+#hint[
+  $a^n = a * a * ... * a$ repeated $n$ times. \
+  $a^(-n) = a^(-1) * a^(-1) * ... * a^(-1)$, where $a^(-1)$ is the inverse of $a$. \
+]
+
+#v(2mm)
+
+The smallest such $n$ defines the _order_ of $g$.
+
+#hint[
+  We've already done a special case of this problem! \
+  Find it in this handout, then rewrite your proof for an arbitrary (finite) group.
+]
+
+#v(1fr)
+
+#problem()
+What is the order of 5 in $(ZZ_25, +)$? \
+What is the order of 2 in $(ZZ_17^times, times)$? \
+
+#v(1fr)
+#pagebreak()
+
+
+
+
+#definition("Generator", label: "gendef")
+Let $G$ be a group, and let $g$ be an element of $G$. \
+We say $g$ is a _generator_ if every other element of $G$ may be written as a power of $g$. \
+
+#problem()
+Let $G$ be a group of $n$ elements. \
+If $g$ is a generator, what is its order? \
+Provide a proof.
+
+#v(1fr)
+
+#problem()
+Find the two generators in $(ZZ, +)$ \
+Then, find all generators of $(ZZ_5, +)$
+
+#v(1fr)
+
+#problem()
+How many groups have only one generator?
+
+#solution[
+  Two: the trivial group and $(ZZ_2, +)$.
+]
+
+#v(1fr)
+
+#definition()
+Let $S$ be a subset of the elements in $G$. \
+We say that $S$ _generates_ $G$ if every element of $G$ may be written as a product of elements in $S$. \
+#note(type: "Note")[This is an extension of @gendef.]
+
+#problem()
+We've already found a few generating sets of $S_n$. What are they?
+
+#solution[
+  The following sets generate $S_n$:
+  - All transpositions
+  - All transpositions of the form $(1, k)$
+  - All adjacent transpositions
+
+  #v(2mm)
+
+  The smallest generating set of $S_n$ consists of the transposition $(12)$ and the $n$-cycle $(1,2,...,n)$. \
+  The proof of this is a bonus problem later in the handout.
+]
+
+#v(1fr)

src/Advanced/Symmetric Groups/parts/03 subgroup.typ

@@ -0,0 +1,172 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+#import "../macros.typ": *
+
+= Subgroups
+
+#problem(label: "s2s3share")
+What elements do $S_2$ and $S_3$ share?
+
+#v(2cm)
+
+
+Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} subset \{1, 2, 3\}$. \
+Can we say something similar about $S_2$ and $S_3$?
+
+#v(2mm)
+
+Looking at @s2s3share, we may want to say that $S_2 subset S_3$ since every element of $S_2$ is in $S_3$. \
+This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are _groups_, not _sets_: \
+their elements come with structure, which the "subset" relation does not capture.
+
+#v(2mm)
+
+To account for this, we'll define a similar relation: subgroups.
+
+#definition("Subgroup")
+Let $G$ and $G'$ be groups. We say $G'$ is a _subgroup_ of $G$ (and write $G' subset G$) if the following are true:\
+(Note that $x, y$ are elements of $G$, and $x y$ is multiplication in $G$)
+- the set of elements in $G'$ is a subset of the set of elements in $G$.
+- the identity of $G$ is in $G'$
+- $x,y in G' => x y in G'$
+- $x in G' => x^(-1) in G'$
+
+The above definition may look fairly scary, but the idea behind a subgroup is simple. \
+Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \
+
+#v(2mm)
+
+Say we have a set of four elements and only look at the first three. \
+$S_3$ fully describes all the ways we can arrange those three elements:
+
+#table(
+  columns: (1fr,),
+  align: center,
+  stroke: none,
+  align(center, cetz.canvas({
+    import cetz.draw: *
+
+    let s = 0.7
+
+    set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
+    content((0 * s, 0.5 * s), $1$, name: "1a")
+    content((1 * s, 0.5 * s), $2$, name: "2a")
+    content((2 * s, 0.5 * s), $3$, name: "3a")
+    content((3 * s, 0.5 * s), $4$, name: "4a")
+
+    content((0 * s, -2 * s), $2$, name: "2b")
+    content((1 * s, -2 * s), $3$, name: "3b")
+    content((2 * s, -2 * s), $1$, name: "1b")
+    content((3 * s, -2 * s), $4$, name: "4b")
+
+
+    // These arrows are wrong,
+    // but create a symmetric picture
+    markline(s, "1a", "1b")
+    markline(s, "2a", "3b")
+    markline(s, "3a", "2b")
+    markline(s, "4a", "4b", c: ogreen)
+
+    content(
+      (1 * s, -0.55 * s),
+      $S_3$,
+      fill: white,
+      stroke: oblue + 0.6mm,
+      padding: 1.3mm,
+    )
+  })),
+)
+
+
+#problem()
+Show that $S_3$ is a subgroup of $S_4$.
+
+#v(1fr)
+#pagebreak()
+
+
+
+#definition("Isomorphism")
+Let $G$ and $H$ be groups. We say that $G$ and $H$ are _isomorphic_ (and write $G tilde.equiv H$) \
+if there is a bijection $f: G -> H$ with the following properties:
+- $f(e_G) = e_H$, where $e_G$ is the identity in $G$
+- $f(x^(-1)) = f(x)^(-1)$ for all $x$ in $G$
+- $f(x y) = f(x) f(y)$ for all $x, y$ in $G$
+
+Intuitively, you can think of isomorphism as a form of equivalence. \
+If two groups are isomorphic, they only differ by the names of their elements. \
+The function $f$ above tells us how to map one set of labels to the other.
+
+
+
+#problem()
+Show that $ZZ_7^times$ and $ZZ_9^times$ are isomorphic.
+#hint[
+  Build a bijection with the above properties. \
+  Remember that a group is fully defined by its multiplication table.
+]
+#v(1fr)
+
+
+#problem()
+Show that $ZZ_10^times$, $ZZ_5^times$, and $ZZ_4$ are isomorphic.
+#hint[
+  Build a bijection with the above properties. \
+  Remember that a group is fully defined by its multiplication table.
+]
+
+
+#v(1fr)
+
+#problem()
+Show that isomorphism is transitive. \
+That is, if $A tilde.equiv B$ and $B tilde.equiv C$, then $A tilde.equiv C$.
+
+#v(1fr)
+#pagebreak()
+
+
+#problem(label: "firstindex")
+How many subgroups of $S_4$ are isomorphic to $S_3$? \
+
+#v(1fr)
+
+
+
+
+#problem()
+What are the orders of $S_3$ and $S_4$? \
+How is this related to @firstindex?
+
+#solution[
+  $|S_4| = |S_3| times [S_4 : S_3]$
+
+  #v(2mm)
+
+  This solution is written using index notation, \
+  but the class doesn't need to know what it means yet.
+]
+
+#v(1fr)
+
+#problem()
+$S_4$ also has $S_2$ and the trivial group as subgroups. \
+How many instances of each does $S_4$ contain?
+
+#v(1fr)
+
+
+#problem()
+$(ZZ_4, +)$ is also a subgroup of $S_4$. Find it! \
+How many subgroups of $ZZ_4$ are isomorphic to $S_4$?
+
+#solution[
+  A good hint is "look at generators."
+
+  #v(4mm)
+
+  There are four instances of $ZZ_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \
+  (i.e, the group generated by $(1234)$ is isomorphic to $ZZ_4$)
+]
+
+#v(1fr)

src/Advanced/Symmetric Groups/parts/1 cycle.tex

@@ -1,536 +0,0 @@
-
-\section{Cycle Notation}
-
-\definition{Order}
-The \textit{order} of a permutation $f$ is the \textbf{smallest} positive $n$ so that $f^n(x) = x$ for all $x$. \par
-If we repeatedly apply a permutation with order $n$, we will get back to where we started after $n$ steps. \par
-
-\vspace{2mm}
-
-For example, consider $[2134]$. This permutation has order $2$, as we clearly see below:
-
-\begin{center}
-\begin{tikzpicture}[scale=0.5]
-	\node (1a) at (0, 0.5) {1};
-	\node (2a) at (1, 0.5) {2};
-	\node (3a) at (2, 0.5) {3};
-	\node (4a) at (3, 0.5) {4};
-
-	\node (2b) at (0, -2) {2};
-	\node (1b) at (1, -2) {1};
-	\node (3b) at (2, -2) {3};
-	\node (4b) at (3, -2) {4};
-
-	\node (1c) at (0, -4.5) {1};
-	\node (2c) at (1, -4.5) {2};
-	\node (3c) at (2, -4.5) {3};
-	\node (4c) at (3, -4.5) {4};
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-	\line{4a}{4b}
-	\line{1b}{1c}
-	\line{2b}{2c}
-	\line{3b}{3c}
-	\line{4b}{4c}
-\end{tikzpicture}
-\end{center}
-Swapping the first two elements of a list twice changes nothing. \par
-Thus, $[2134]$ has an order of two.
-
-
-\problem{}
-What is the order of $[2314]$? \par
-How about $[4321]$? \par
-\note[Note]{You shouldn't need to draw any strings to solve this problem.}
-
-
-\vfill
-
-\problem{}
-Show that all permutations (on a finite set) have a well-defined order. \par
-In other words, show that there is always an integer $n$ so that $f^n(x) = x$.
-
-\vfill
-
-\definition{Composition}<compdef>
-The \textit{composition} of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \par
-We'll denote this as $fg$---that is, by simply writing the permutations we're composing next to each other.
-
-\problem{}
-Show that function composition is associative. \par
-That is, show that $f(gh) = (fg)h$.
-
-\vfill
-
-\problem{}
-What is $[1324][4321]$? \par
-How about $[321][213][231]$? \par
-
-
-\vfill
-\pagebreak
-
-
-As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
-Permutations are verbs---but we've been referring to them using a noun (namely, their output when
-applied to an ordered sequence of numbers). Our notation fails to capture the meaning of the
-underlying object.
-
-\vspace{2mm}
-
-Think about it: is the permutation $[1234]$ different than the permutation $[12345]$? \par
-Indeed, these permutations operate on different sets---but they are both the identity! \par
-What should we do if we want to talk about the identity on $\{1, 2, ..., 10\}$?
-
-\vspace{2mm}
-
-We need something better.
-
-
-
-
-\definition{Cycles}
-Any permutation is composed of a number of \textit{cycles}. \par
-
-For example, consider the permutation $[2134]$, which consists of one two-cycle: $1 \to 2 \to 1$ \par
-\note[Note]{$3 \to 3$ and $4 \to 4$ are also cycles, but we'll ignore them. One-cycles aren't aren't interesting.}
-
-\begin{center}
-\begin{tikzpicture}[scale=0.5]
-	\node (1) at (0, 0) {1};
-	\node (2) at (1, 0) {2};
-	\node (3) at (2, 0) {3};
-	\node (4) at (3, 0) {4};
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(1)
-		-- ($(1) + (0,-1)$)
-		-- ($(2) + (0,-1)$)
-		-- (2);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(2)
-		-- ($(2) + (0, 1)$)
-		-- ($(1) + (0, 1)$)
-		-- (1);
-\end{tikzpicture}
-\end{center}
-
-
-The permutation $[431265]$ is a bit more interesting---it contains two cycles: \par
-($1 \to 3 \to 2 \to 4 \to 1$ and $5 \to 6 \to 5$)
-
-
-\begin{center}
-\begin{tikzpicture}[scale=0.5]
-	\node (1) at (0, 0) {1};
-	\node (2) at (1, 0) {2};
-	\node (3) at (2, 0) {3};
-	\node (4) at (3, 0) {4};
-	\node (5) at (4, 0) {5};
-	\node (6) at (5, 0) {6};
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(3)
-		-- ($(3) + (0,-1)$)
-		-- ($(2) + (0,-1)$)
-		-- (2);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(2)
-		-- ($(2) + (0,1.5)$)
-		-- ($(4) + (0,1.5)$)
-		-- (4);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(4)
-		-- ($(4) + (0,-1.5)$)
-		-- ($(1) + (0,-1.5)$)
-		-- (1);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(1)
-		-- ($(1) + (0,1)$)
-		-- ($(3) + (0,1)$)
-		-- (3);
-
-	\draw[line width = 0.3mm, ->, ogreen]
-		(5)
-		-- ($(5) + (0,-1)$)
-		-- ($(6) + (0,-1)$)
-		-- (6);
-
-	\draw[line width = 0.3mm, ->, ogreen]
-		(6)
-		-- ($(6) + (0,1)$)
-		-- ($(5) + (0,1)$)
-		-- (5);
-
-\end{tikzpicture}
-\end{center}
-
-
-Another name we'll often use for two-cycles is \textit{transposition}. \par
-Any permutation that swaps two adjacent elements is called a transposition. \par
-
-
-\problem{}
-Find all cycles in $[5342761]$.
-
-\begin{solution}
-	\begin{center}
-	\begin{tikzpicture}[scale=0.5]
-		\node (1) at (0, 0) {1};
-		\node (2) at (1, 0) {2};
-		\node (3) at (2, 0) {3};
-		\node (4) at (3, 0) {4};
-		\node (5) at (4, 0) {5};
-		\node (6) at (5, 0) {6};
-		\node (7) at (6, 0) {7};
-
-		\draw[line width = 0.3mm, ->, ocyan]
-			(1)
-			-- ($(1) + (0,2)$)
-			-- ($(7) + (0,2)$)
-			-- (7);
-
-		\draw[line width = 0.3mm, ->, ocyan]
-			(7)
-			-- ($(7) + (0,-1.5)$)
-			-- ($(5) + (0,-1.5)$)
-			-- (5);
-
-		\draw[line width = 0.3mm, ->, ocyan]
-			(5)
-			-- ($(5) + (0,1.5)$)
-			-- ($(1) + (0.5,1.5)$)
-			-- ($(1) + (0.5,-1)$)
-			-- ($(1) + (0,-1)$)
-			-- (1);
-
-		\draw[line width = 0.3mm, ->, ogreen]
-			(2)
-			-- ($(2) + (0,-1.5)$)
-			-- ($(4) + (0,-1.5)$)
-			-- (4);
-
-		\draw[line width = 0.3mm, ->, ogreen]
-			(4)
-			-- ($(4) + (0,1)$)
-			-- ($(3) + (0,1)$)
-			-- (3);
-
-		\draw[line width = 0.3mm, ->, ogreen]
-			(3)
-			-- ($(3) + (0,-1)$)
-			-- ($(2) + (0.5,-1)$)
-			-- ($(2) + (0.5,1)$)
-			-- ($(2) + (0,1)$)
-			-- (2);
-	\end{tikzpicture}
-	\end{center}
-\end{solution}
-
-\vfill
-
-
-\problem{}
-What permutation (on five objects) is formed by the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?
-
-
-\begin{solution}
-	\begin{center}
-	\begin{tikzpicture}[scale=0.5]
-		\node (1) at (0, 0) {1};
-		\node (2) at (1, 0) {2};
-		\node (3) at (2, 0) {3};
-		\node (4) at (3, 0) {4};
-		\node (5) at (4, 0) {5};
-
-		\draw[line width = 0.3mm, ->, ocyan]
-			(3)
-			-- ($(3) + (0,1)$)
-			-- ($(5) + (0,1)$)
-			-- (5);
-
-		\draw[line width = 0.3mm, ->, ocyan]
-			(5)
-			-- ($(5) + (0,-1)$)
-			-- ($(3) + (0,-1)$)
-			-- (3);
-
-		\draw[line width = 0.3mm, ->, ogreen]
-			(1)
-			-- ($(1) + (0,-1)$)
-			-- ($(2) + (0,-1)$)
-			-- (2);
-
-		\draw[line width = 0.3mm, ->, ogreen]
-			(2)
-			-- ($(2) + (0,1.5)$)
-			-- ($(4) + (0,1.5)$)
-			-- (4);
-
-		\draw[line width = 0.3mm, ->, ogreen]
-			(4)
-			-- ($(4) + (0,-1.5)$)
-			-- ($(1) + (0.5,-1.5)$)
-			-- ($(1) + (0.5,1)$)
-			-- ($(1) + (0,1)$)
-			-- (1);
-	\end{tikzpicture}
-
-	This is $[41523]$
-	\end{center}
-\end{solution}
-
-
-
-\vfill
-\pagebreak
-
-\definition{Cycle Notation}
-We now have a solution to our problem of notation.
-Instead of referring to permutations using their output, we will refer to them using their \textit{cycles}.
-
-\vspace{2mm}
-
-For example, we'll write $[2134]$ as $(12)$, which denotes the cycle $1 \to 2 \to 1$:
-
-
-\begin{center}
-\begin{tikzpicture}[scale=0.5]
-	\node (1) at (0, 0) {1};
-	\node (2) at (1, 0) {2};
-	\node (3) at (2, 0) {3};
-	\node (4) at (3, 0) {4};
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(1)
-		-- ($(1) + (0,-1)$)
-		-- ($(2) + (0,-1)$)
-		-- (2);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(2)
-		-- ($(2) + (0, 1)$)
-		-- ($(1) + (0, 1)$)
-		-- (1);
-\end{tikzpicture}
-\end{center}
-
-
-
-As another example, $[431265]$ is $(1324)(56)$ in cycle notation. \par
-Note that we write $[431265]$ as a \textit{composition} of two cycles: \par
-applying the permutation $[431265]$ is the same as applying $(1324)$, then applying $(56)$.
-
-\begin{center}
-\begin{tikzpicture}[scale=0.5]
-	\node (1) at (0, 0) {1};
-	\node (2) at (1, 0) {2};
-	\node (3) at (2, 0) {3};
-	\node (4) at (3, 0) {4};
-	\node (5) at (4, 0) {5};
-	\node (6) at (5, 0) {6};
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(3)
-		-- ($(3) + (0,-1)$)
-		-- ($(2) + (0,-1)$)
-		-- (2);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(2)
-		-- ($(2) + (0,1.5)$)
-		-- ($(4) + (0,1.5)$)
-		-- (4);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(4)
-		-- ($(4) + (0,-1.5)$)
-		-- ($(1) + (0,-1.5)$)
-		-- (1);
-
-	\draw[line width = 0.3mm, ->, ocyan]
-		(1)
-		-- ($(1) + (0,1)$)
-		-- ($(3) + (0,1)$)
-		-- (3);
-
-	\draw[line width = 0.3mm, ->, ogreen]
-		(5)
-		-- ($(5) + (0,-1)$)
-		-- ($(6) + (0,-1)$)
-		-- (6);
-
-	\draw[line width = 0.3mm, ->, ogreen]
-		(6)
-		-- ($(6) + (0,1)$)
-		-- ($(5) + (0,1)$)
-		-- (5);
-
-\end{tikzpicture}
-\end{center}
-
-Any permutation $\sigma$ may be written as a product (i.e, composition) of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
-Make sure you believe this fact. If you don't, ask an instructor. \par
-Also, the identity $f(x) = x$ is written as $()$ in cycle notation.
-
-
-
-\problem{}
-Convince yourself that disjoint cycles commute. \par
-That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \par
-
-
-
-
-\problem{}<insquare>
-Write the following in square-bracket notation.
-\begin{itemize}
-	\item $(12)$       \tab~\tab on a set of 2 elements
-	\item $(12)(435)$  \tab on a set of 5 elements
-	\vspace{2mm}
-	\item $(321)$      \tab~\tab on a set of 3 elements
-	\item $(321)$      \tab~\tab on a set of 6 elements
-	\vspace{2mm}
-	\item $(1234)$      \tab on a set of 4 elements
-	\item $(3412)$      \tab on a set of 4 elements
-\end{itemize}
-\note{
-	Note that $(12)$ refers the \say{swap first two} permutation on a set of \textit{any} size. \\
-	We can now use the same name for the same permutation on two different sets! \\
-}
-
-\vfill
-
-
-
-\problem{}
-Write the following in square-bracket notation.
-Be careful.
-\begin{itemize}
-	\item $(13)(243)$ \tab on a set of 4 elements
-	\item $(243)(13)$ \tab on a set of 4 elements
-\end{itemize}
-
-\vfill
-
-
-
-\problem{}
-Look at the last two permutations in \ref{insquare}, $(1234)$ and $(3412)$. \par
-These are \textit{identical}---they are the same cycle written in two different ways. \par
-List all other ways to write this cycle. \hint{There are two more.} \par
-\note{Also, note that the last two permutations in \ref{insquare} are the same.}
-
-\pagebreak
-
-
-\problem{}
-What is the inverse of $(12)$? \par
-How about $(123)$? And $(4231)$? \par
-\note{
-	Note that again, we don't need to know how big our set is. \\
-	The inverse of $(12)$ is the same in all sets.
-}
-
-\vfill
-
-
-\problem{}
-Say $\sigma$ is a permutation composed of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
-Say we know the order of all $\sigma_i$. What is the order of $\sigma$?
-
-\begin{solution}
-	$\text{lcm}\Bigl(\text{ord}(\sigma_1),~ \text{ord}(\sigma_2),~ ..., ~ \text{ord}(\sigma_k)\Bigr)$
-\end{solution}
-
-
-\vfill
-
-\problem{}<cycletrans>
-Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
-
-\begin{solution}
-	TODO
-\end{solution}
-
-\vfill
-
-\problem{}
-Write $(7126453)$ as a product of transpositions. \par
-
-\vfill
-\pagebreak
-
-\problem{}<simpletrans>
-Show that any permutation is a product of transpositions.
-
-\begin{solution}
-	Use \ref{cycletrans}.
-\end{solution}
-
-\vfill
-
-
-\problem{}
-Show that any permutation is a product of transpositions of the form $(1, k)$. \par
-
-\begin{solution}
-	Use \ref{simpletrans} and rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$. \par
-	Showing that $(a, b) = (1, a)(1, b)(1, a)$ is fairly easy.
-\end{solution}
-
-\vfill
-\pagebreak
-
-
-
-\problem{}
-Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.
-
-\begin{solution}
-	This is the same as the $(1, a)(1, b)(1, a)$ case above, but we use $a + 1$
-	as a \say{working slot} instead of $1$.
-\end{solution}
-
-\vfill
-
-
-
-\problem{}
-Show that any permutation is a product of adjacent transpositions. \par
-(An \textit{adjacent transposition} swaps two adjacent elements, and thus looks like $(n, n+1)$)
-
-\begin{solution}
-	As before, we will use \ref{simpletrans} and rewrite the transpositions it produces in a form that fits the problem.
-	We thus need to show that every transposition $(a, b)$ is a product of adjacent transpositions.
-
-	\vspace{8mm}
-
-	In the proof below, assume that $a < b$ and perform induction on $b - a$. \par
-
-	\textbf{Base Case:}\par
-	If $b - a = 1$, we clearly see that $(a, b)$ is a product of adjacent. \par
-	In fact, it \textit{is} an adjacent transposition.
-
-	\vspace{4mm}
-
-	\textbf{Induction:}\par
-	Now, say $b - a = n + 1$. \par
-	Assume that all $(a, b)$ where $b - a \leq n$ are products of adjacent transpositions.\par
-	Note that $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.
-
-	\vspace{2mm}
-
-	$(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \par
-	Thus, $(a, b)$ is a product of adjacent transpositions.
-\end{solution}
-
-
-\vfill
-\pagebreak

src/Advanced/Symmetric Groups/parts/2 groups.tex

@@ -1,165 +0,0 @@
-\section{Groups (review)}
-
-\definition{}
-Before we continue, we must introduce a bit of notation:
-\begin{itemize}
-	\item $S_n$ is the set of permutations on $n$ objects.
-	\item $\mathbb{Z}_n$ is the set of integers mod $n$.
-
-	\item $\mathbb{Z}_n^\times$ is the set of integers mod $n$ with multiplicative inverses. \par
-	In other words, it is the set of integers smaller than $n$ and coprime to $n$.\footnotemark{} \par
-	For example, $\mathbb{Z}_{12}^\times = \{1, 5, 7, 11\}$.
-
-	\footnotetext{We proved this in another handout, but you may take it as fact here.}
-\end{itemize}
-
-\problem{}
-What are the elements of $S_3$? \tab\hint{Use cycle notation}\par
-How about $\mathbb{Z}_{17}^\times$?
-
-\vfill
-
-
-\definition{}
-A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \par
-Groups always have the following properties:
-
-\begin{enumerate}
-	\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
-	\item $\ast$ is \textit{associative}: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
-	\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
-	\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = b \ast a = e$. $b$ is called the \textit{inverse} of $a$. \par
-	This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
-\end{enumerate}
-
-Any pair $(G, \ast)$ that satisfies these properties is a group.
-
-\problem{}
-Is $(\mathbb{Z}_5, +)$ a group? \par
-Is $(\mathbb{Z}_5, -)$ a group? \par
-\note[Note]{$+$ and $-$ refer to the usual operations in modular arithmetic.}
-\vfill
-
-
-\problem{}
-What is the group with the fewest elements?
-
-\begin{solution}
-	Let $(G, \star)$ be our group, where $G = \{x\}$ and $\star$ is defined by $x \star x = x$
-
-	Verifying that the trivial group is a group is trivial.
-\end{solution}
-
-\vfill
-\pagebreak
-
-
-
-
-
-
-
-
-
-
-
-\problem{}
-Show that function composition is associative
-
-\vfill
-
-
-\problem{}
-Show that $S_n$ is a group under composition.
-
-\vfill
-
-\problem{}
-Let $(G, \ast)$ be a group with finitely many elements, and let $a \in G$. \par
-Show that $\exists n \in \mathbb{Z}^+$ so that $a^n = e$ \par
-\hint{$a^n = a \ast a \ast ... \ast a$ repeated $n$ times.}
-
-\vspace{2mm}
-
-The smallest such $n$ defines the \textit{order} of $g$.
-
-\begin{examplesolution}
-	We've already done a special case of this problem! \par
-	Find it in this handout, then rewrite your proof for an arbitrary (finite) group.
-\end{examplesolution}
-
-
-\vfill
-
-\problem{}
-What is the order of 5 in $(\mathbb{Z}_{25}, +)$? \par
-What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \par
-
-\vfill
-\pagebreak
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-\definition{}<gendef>
-Let $G$ be a group, and let $g$ be an element of $G$. \par
-We say $g$ is a \textit{generator} if every other element of $G$ may be written as a power of $g$. \par
-
-\problem{}
-Say the size of a group $G$ is $n$. \par
-If $g$ is a generator, what is its order? \par
-Provide a proof.
-\vfill
-
-
-
-\problem{}
-Find the two generators in $(\mathbb{Z}, +)$ \par
-Then, find all generators of $(\mathbb{Z}_5, +)$
-\vfill
-
-
-\problem{}
-How many groups have only one generator?
-
-\begin{solution}
-	Only one: the trivial group. The inverse of a generator is also a generator!
-\end{solution}
-
-\vfill
-
-
-\definition{}
-Let $S$ be a subset of the elements in $G$. \par
-We say that $S$ \textit{generates} $G$ if every element of $G$ may be written as a product of elements in $S$. \par
-\note{Note that this is an extension of \ref{gendef}.}
-
-\problem{}
-We've already found a few generating sets of $S_n$. What are they?
-
-\begin{solution}
-	The following sets generate $S_n$:
-	\begin{itemize}
-		\item All transpositions
-		\item All transpositions of the form $(1, k)$
-		\item All adjacent transpositions
-	\end{itemize}
-
-	\vspace{2mm}
-
-	The smallest generating set of $S_n$ consists of the transposition $(12)$ and the $n$-cycle $(1,2,...,n)$. \par
-	The proof of this is a bonus problem later in the handout.
-\end{solution}
-
-\vfill
-\pagebreak

src/Advanced/Symmetric Groups/parts/3 subgroup.tex

@@ -1,163 +0,0 @@
-\section{Subgroups}
-
-\problem{}<s2s3share>
-What elements do $S_2$ and $S_3$ share?
-\vspace{2cm}
-
-
-
-Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} \subset \{1, 2, 3\}$. \par
-Can we say something similar about $S_2$ and $S_3$?
-
-\vspace{2mm}
-
-Looking at \ref{s2s3share}, we may want to say that $S_2 \subset S_3$ since every element of $S_2$ is in $S_3$. \par
-This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are \textit{groups}, not \textit{sets}: \par
-their elements come with structure, which the \say{subset} relation does not capture.
-
-\vspace{2mm}
-
-To account for this, we'll define a similar relation: subgroups.
-
-\definition{}
-Let $G$ and $G'$ be groups. We say $G'$ is a \textit{subgroup} of $G$ (and write $G' \subset G$) if the following are true:\par
-(Note that $x, y$ are elements of $G$, and $xy$ is multiplication in $G$)
-\begin{itemize}
-	\item the set of elements in $G'$ is a subset of the set of elements in $G$.
-	\item the identity of $G$ is in $G'$
-	\item $x,y \in G' \implies xy \in G'$
-	\item $x \in G' \implies x^{-1} \in G'$
-\end{itemize}
-
-The above definition may look faily scary, but the idea behind a subgroup is simple. \par
-Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \par
-
-\vspace{2mm}
-
-Say we have a set of four elements and only look at the first three. \par
-$S_3$ fully describes all the ways we can arrange those three elements:
-
-\begin{center}
-\begin{tikzpicture}[scale=0.5]
-	\node (1a) at (0, 0.5) {1};
-	\node (2a) at (1, 0.5) {2};
-	\node (3a) at (2, 0.5) {3};
-	\node (4a) at (3, 0.5) {4};
-
-	\node (2b) at (0, -2) {2};
-	\node (3b) at (1, -2) {3};
-	\node (1b) at (2, -2) {1};
-	\node (4b) at (3, -2) {4};
-
-	\draw[line width = 0.3mm, ->, ogreen]
-		(4a)
-		-- ($(4a) + (0, -1)$)
-		-- ($(4b) + (0,1)$)
-		-- (4b);
-
-	\line{1a}{1b}
-	\line{2a}{2b}
-	\line{3a}{3b}
-
-	\node[fill=white,draw=oblue,line width=0.3mm] at (1, -0.75) {$S_3$};
-
-\end{tikzpicture}
-\end{center}
-
-
-\problem{}
-Show that $S_3$ is a subgroup of $S_4$.
-
-\vfill
-\pagebreak
-
-
-
-\definition{}
-Let $G$ and $H$ be groups. We say that $G$ and $H$ are \textit{isomorphic} (and write $A \simeq B$) \par
-if there is a bijection $f: G \to H$ with the following properties:
-\begin{itemize}
-	\item $f(e_G) = e_H$, where $e_G$ is the identity in $G$
-	\item $f(x^{-1}) = f(x)^{-1}$ for all $x$ in $G$
-	\item $f(xy) = f(x)f(y)$ for all $x, y$ in $G$
-\end{itemize}
-
-Intuitively, you can think of isomorphism as a form of equivalence. \par
-If two groups are isomorphic, they only differ by the names of their elements. \par
-The function $f$ above tells us how to map one set of labels to the other.
-
-
-
-\problem{}
-Show that $\mathbb{Z}_7^\times$ and $\mathbb{Z}_9^\times$ are isomorphic.
-\hint{
-	Build a bijection with the above properties. \\
-	Remember that a group is fully defined by its multiplication table.
-}
-\vfill
-
-
-\problem{}
-Show that $\mathbb{Z}_{10}^\times$ and $\mathbb{Z}_5^\times$, and $\mathbb{Z}_4$ are isomorphic.
-\hint{
-	Build a bijection with the above properties. \\
-	Remember that a group is fully defined by its multiplication table.
-}
-
-
-\vfill
-
-\problem{}
-Show that isomorphism is transitive. \par
-That is, if $A \simeq B$ and $B \simeq C$, then $A \simeq C$.
-
-\vfill
-\pagebreak
-
-
-\problem{}<firstindex>
-How many subgroups of $S_4$ are isomorphic to $S_3$? \par
-
-\vfill
-
-
-
-
-\problem{}
-What are the orders of $S_3$ and $S_4$? \par
-How is this related to \ref{firstindex}?
-
-\begin{solution}
-	$|S_4| = |S_3| \times [S_4 : S_3]$
-
-	\vspace{2mm}
-
-	This solution is written using index notation, \par
-	but the class doesn't need to know what it means yet.
-\end{solution}
-
-\vfill
-
-\problem{}
-$S_4$ also has $S_2$ and the trivial group as subgroups. \par
-How many instances of each does $S_4$ contain?
-
-\vfill
-
-
-\problem{}
-$(\mathbb{Z}_4, +)$ is also a subgroup of $S_4$. Find it! \par
-How many subgroups of $\mathbb{Z}_4$ are isomorphic to $S_4$?.
-
-\begin{solution}
-	A good hint is \say{look at generators.}
-
-	\vspace{4mm}
-
-	There are four instances of $\mathbb{Z}_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \par
-	(i.e, the group generated by $(1234)$ is isomorphic to $\mathbb{Z}_4$)
-\end{solution}
-
-
-\vfill
-\pagebreak

src/Advanced/Tropical Polynomials/macros.typ

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 
 // Shorthand, we'll be using these a lot.
@@ -7,35 +7,31 @@
 #let tm = sym.times.circle
 
 #let graphgrid(inner_content) = {
-  align(
-    center,
-    box(
-      inset: 3mm,
-      cetz.canvas({
-        import cetz.draw: *
-        let x = 5.25
+  align(center, box(inset: 3mm, cetz.canvas({
+    import cetz.draw: *
+    let x = 5.25
 
-        grid(
-          (0, 0), (x, x), step: 0.75,
-          stroke: luma(100) + 0.3mm
-        )
+    grid(
+      (0, 0),
+      (x, x),
+      step: 0.75,
+      stroke: luma(100) + 0.3mm,
+    )
 
-        if (inner_content != none) {
-          inner_content
-        }
+    if (inner_content != none) {
+      inner_content
+    }
 
-        mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
-        mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
+    mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
+    mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
 
-        line(
-          (0, x + 0.25),
-          (0, 0),
-          (x + 0.25, 0),
-          stroke: 0.75mm + black,
-        )
-      }),
-    ),
-  )
+    line(
+      (0, x + 0.25),
+      (0, 0),
+      (x + 0.25, 0),
+      stroke: 0.75mm + black,
+    )
+  })))
 }
 
 /// Adds extra padding to an equation.
@@ -48,23 +44,16 @@
 /// Note that there are newlines between the $ and content,
 /// this gives us display math (which is what we want when using this macro)
 #let eqnbox(eqn) = {
-  align(
-    center,
-    box(
-      inset: 3mm,
-      eqn,
-    ),
-  )
+  align(center, box(
+    inset: 3mm,
+    eqn,
+  ))
 }
 
 #let dotline(a, b) = {
-  cetz.draw.line(
-    a,
-    b,
-    stroke: (
-      dash: "dashed",
-      thickness: 0.5mm,
-      paint: ored,
-    ),
-  )
+  cetz.draw.line(a, b, stroke: (
+    dash: "dashed",
+    thickness: 0.5mm,
+    paint: ored,
+  ))
 }

src/Advanced/Tropical Polynomials/parts/00 arithmetic.typ

@@ -26,9 +26,9 @@ $
 
 #solution([
   - Is tropical addition commutative?\
-    Yes, $min(min(x,y),z) = min(x,y,z) = min(x,min(y,z))$
+    Yes, $min(min(x, y), z) = min(x, y, z) = min(x, min(y, z))$
   - Is tropical addition associative? \
-    Yes, $min(x,y) = min(y,x)$
+    Yes, $min(x, y) = min(y, x)$
   - Is there a tropical additive identity? \
     No. There is no $n$ where $x <= n$ for all real $x$
 ])
@@ -117,7 +117,7 @@ Do tropical multiplicative inverses always exist? \
 Is tropical multiplication distributive over addition? \
 #note([Does $x #tm (y #tp z) = x #tm y #tp x #tm z$?])
 
-#solution([Yes, $x + min(y,z) = min(x+y, x+z)$])
+#solution([Yes, $x + min(y, z) = min(x+y, x+z)$])
 
 #v(1fr)
 
@@ -134,14 +134,7 @@ Fill the following tropical addition and multiplication tables
     table(
       columns: (col, col, col, col, col, col),
       align: center,
-      table.header(
-        [$#tp$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-        [$#sym.infinity$],
-      ),
+      table.header([$#tp$], [$1$], [$2$], [$3$], [$4$], [$#sym.infinity$]),
 
       box(inset: 3pt, $1$), [], [], [], [], [],
       box(inset: 3pt, $2$), [], [], [], [], [],
@@ -152,14 +145,7 @@ Fill the following tropical addition and multiplication tables
     table(
       columns: (col, col, col, col, col, col),
       align: center,
-      table.header(
-        [$#tm$],
-        [$0$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-      ),
+      table.header([$#tm$], [$0$], [$1$], [$2$], [$3$], [$4$]),
 
       box(inset: 3pt, $0$), [], [], [], [], [],
       box(inset: 3pt, $1$), [], [], [], [], [],
@@ -178,14 +164,7 @@ Fill the following tropical addition and multiplication tables
     table(
       columns: (col, col, col, col, col, col),
       align: center,
-      table.header(
-        [$#tp$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-        [$#sym.infinity$],
-      ),
+      table.header([$#tp$], [$1$], [$2$], [$3$], [$4$], [$#sym.infinity$]),
 
       box(inset: 3pt, $1$),
       box(inset: 3pt, $1$),
@@ -225,14 +204,7 @@ Fill the following tropical addition and multiplication tables
     table(
       columns: (col, col, col, col, col, col),
       align: center,
-      table.header(
-        [$#tm$],
-        [$0$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-      ),
+      table.header([$#tm$], [$0$], [$1$], [$2$], [$3$], [$4$]),
 
       box(inset: 3pt, $0$),
       box(inset: 3pt, $0$),
@@ -281,10 +253,9 @@ Adjacent parenthesis imply tropical multiplication
 
 #solution([
   $
-    (x #tp 2)(x #tp 3)
-    &= x^2 #tp 2x #tp 3x #tp (2 #tm 3) \
-    &= x^2 #tp (2 #tp 3)x #tp (2 #tm 3) \
-    &= x^2 #tp 2x #tp 5
+    (x #tp 2)(x #tp 3) & = x^2 #tp 2x #tp 3x #tp (2 #tm 3) \
+                       & = x^2 #tp (2 #tp 3)x #tp (2 #tm 3) \
+                       & = x^2 #tp 2x #tp 5
   $
 
   Also, $f(1) = 2$ and $f(4) = 5$.

src/Advanced/Tropical Polynomials/parts/01 polynomials.typ

@@ -1,21 +1,18 @@
 #import "@local/handout:0.1.0": *
 #import "../macros.typ": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = Tropical Polynomials
 
 #definition()
 A _polynomial_ is an expression formed by adding and multiplying numbers and a variable $x$. \
 Every polynomial can be written as
-#align(
-  center,
-  box(
-    inset: 3mm,
-    $
-      c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
-    $,
-  ),
-)
+#align(center, box(
+  inset: 3mm,
+  $
+    c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
+  $,
+))
 for some nonnegative integer $n$ and coefficients $c_0, c_1, ..., c_n$. \
 The _degree_ of a polynomial is the largest $n$ for which $c_n$ is nonzero.
 
@@ -43,15 +40,12 @@ In this section, we will analyze tropical polynomials:
 #definition()
 A _tropical_ polynomial is a polynomial that uses tropical addition and multiplication. \
 In other words, it is an expression of the form
-#align(
-  center,
-  box(
-    inset: 3mm,
-    $
-      c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
-    $,
-  ),
-)
+#align(center, box(
+  inset: 3mm,
+  $
+    c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
+  $,
+))
 where all exponents represent repeated tropical multiplication.
 
 #pagebreak() // MARK: page
@@ -66,7 +60,7 @@ Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
 #if_no_solutions(graphgrid(none))
 
 #solution([
-  $f(x) = min(2x , 1+x, 4)$, which looks like:
+  $f(x) = min(2x, 1+x, 4)$, which looks like:
 
   #graphgrid({
     import cetz.draw: *
@@ -90,15 +84,12 @@ Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
 #problem()
 Now, factor $f(x) = x^2 #tp 1x #tp 4$ into two polynomials with degree 1. \
 In other words, find $r$ and $s$ so that
-#align(
-  center,
-  box(
-    inset: 3mm,
-    $
-      x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
-    $,
-  ),
-)
+#align(center, box(
+  inset: 3mm,
+  $
+    x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
+  $,
+))
 
 we will call $r$ and $s$ the _roots_ of $f$.
 
@@ -159,15 +150,19 @@ Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
 #solution([
   We (tropically) factor out $-2$ to get
 
-  #eqnbox($
-    f(x) = -2(x^2 #tp 2x #tp 10)
-  $)
+  #eqnbox(
+    $
+      f(x) = -2(x^2 #tp 2x #tp 10)
+    $,
+  )
 
 
   by the same process as the previous problem, we get
-  #eqnbox($
-    f(x) = -2(x #tp 2)(x #tp 8)
-  $)
+  #eqnbox(
+    $
+      f(x) = -2(x #tp 2)(x #tp 8)
+    $,
+  )
 ])
 
 #v(1fr)
@@ -236,11 +231,11 @@ Graph $f(x) = 1x^2 #tp 3x #tp 5$.
 #problem()
 Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
 
-#solution(
-  eqnbox($
+#solution(eqnbox(
+  $
     f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2
-  $),
-)
+  $,
+))
 
 #v(1fr)
 
@@ -263,23 +258,21 @@ Graph $f(x) = 2x^2 #tp 4x #tp 4$.
 
 #if_no_solutions(graphgrid(none))
 
-#solution(
-  graphgrid({
-    import cetz.draw: *
-    let step = 0.75
+#solution(graphgrid({
+  import cetz.draw: *
+  let step = 0.75
 
-    dotline((0, 2 * step), (3 * step, 8 * step))
-    dotline((0, 4 * step), (5 * step, 8 * step))
-    dotline((0, 4 * step), (8 * step, 4 * step))
+  dotline((0, 2 * step), (3 * step, 8 * step))
+  dotline((0, 4 * step), (5 * step, 8 * step))
+  dotline((0, 4 * step), (8 * step, 4 * step))
 
-    line(
-      (0, 2 * step),
-      (1 * step, 4 * step),
-      (7.5 * step, 4 * step),
-      stroke: 1mm + oblue,
-    )
-  }),
-)
+  line(
+    (0, 2 * step),
+    (1 * step, 4 * step),
+    (7.5 * step, 4 * step),
+    stroke: 1mm + oblue,
+  )
+}))
 
 
 #problem()
@@ -325,7 +318,7 @@ Find a formula for $B$ in terms of $a$, $b$, and $c$. \
 
 #solution([
   If we want to factor $a(x^2 #tp (b-a)x #tp (c-a))$, we need to find $r$ and $s$ so that
-  - $min(r,s) = b-a$, and
+  - $min(r, s) = b-a$, and
   - $r + s = c - a$
 
   #v(2mm)
@@ -341,9 +334,8 @@ Find a formula for $B$ in terms of $a$, $b$, and $c$. \
 
   *Case 2:* If $b > (a + c #sym.div) 2$, then
   $
-    accent(f, macron)(x)
-    &= a x^2 #tp ((a+c)/2)x #tp c \
-    &= a(x #tp (c-a)/2)^2
+    accent(f, macron)(x) & = a x^2 #tp ((a+c)/2)x #tp c \
+                         & = a(x #tp (c-a)/2)^2
   $
   has the same graph as $f$, and thus $B = (a+c) #sym.div 2$
 

src/Advanced/Tropical Polynomials/parts/02 cubic.typ

@@ -1,6 +1,6 @@
 #import "@local/handout:0.1.0": *
 #import "../macros.typ": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = Tropical Cubic Polynomials
 
@@ -131,15 +131,12 @@ Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b
 #problem()
 What are the roots of the following polynomial?
 
-#align(
-  center,
-  box(
-    inset: 3mm,
-    $
-      3 x^6 #tp 4 x^5 #tp 2 x^4 #tp x^3 #tp x^2 #tp 4 x #tp 5
-    $,
-  ),
-)
+#align(center, box(
+  inset: 3mm,
+  $
+    3 x^6 #tp 4 x^5 #tp 2 x^4 #tp x^3 #tp x^2 #tp 4 x #tp 5
+  $,
+))
 
 #solution([
   We have
@@ -169,9 +166,8 @@ Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
 
 #solution([
   $
-    A_j
-    &= min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
-    &= min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
+    A_j & = min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
+        & = min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
   $
 
   #v(2mm)

src/Advanced/Wallpaper/main.typ

@@ -0,0 +1,26 @@
+#import "@local/handout:0.1.0": *
+
+// Resources:
+//
+// https://eschermath.org/wiki/Wallpaper_Patterns.html
+// https://mathworld.wolfram.com/WallpaperGroups.html
+// https://en.wikipedia.org/wiki/Wallpaper_group
+
+#show: handout.with(
+  title: [Wallpaper Symmetry],
+  by: "Mark",
+)
+
+#include "parts/00 intro.typ"
+#pagebreak()
+
+#include "parts/01 reflect.typ"
+#pagebreak()
+
+#include "parts/02 rotate.typ"
+#pagebreak()
+
+#include "parts/03 problems.typ"
+#pagebreak()
+
+#include "parts/04 theorem.typ"

src/Advanced/Wallpaper/meta.toml

@@ -0,0 +1,7 @@
+[metadata]
+title = "Wallpaper Symmetries"
+
+
+[publish]
+handout = true
+solutions = true

src/Advanced/Wallpaper/parts/00 intro.typ

@@ -0,0 +1,153 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+= Wallpaper Symmetries
+
+#definition()
+A _Euclidean isometry_ is a transformation of the plane that preserves distances. \
+Intuitively, an isometry moves objects on the plane without deforming them.
+
+There are four classes of Euclidean isometries:
+- translations
+- reflections
+- rotations
+- glide reflections
+#note(
+  [We can prove there are no others, but this is beyond the scope of this handout.],
+) \
+A simple example of each isometry is shown below:
+
+#let demo(c) = {
+  let s = 0.5
+  cetz.draw.line(
+    (0, 0),
+    (3 * s, 0),
+    (3 * s, 1 * s),
+    (1 * s, 1 * s),
+    (1 * s, 2 * s),
+    (0, 2 * s),
+    close: true,
+    fill: c,
+    stroke: black + 0mm * s,
+  )
+}
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr),
+  rows: (3.5cm, 3.5cm),
+  row-gutter: 2mm,
+  [
+    #cetz.canvas({
+      import cetz.draw: *
+
+      demo(ored)
+      translate(x: -1.0, y: -1.0)
+      demo(oblue)
+    })
+    #v(1fr)
+    Translation
+  ],
+  [
+    #cetz.canvas({
+      import cetz.draw: *
+
+      circle((-2, 0), radius: 0.1, stroke: none, fill: black)
+      arc(
+        (-2, 0),
+        radius: 1,
+        anchor: "origin",
+        start: 0deg,
+        stop: -30deg,
+        mode: "PIE",
+      )
+
+      demo(ored)
+      rotate(z: -30deg, origin: (-2, 0))
+      demo(oblue)
+    })
+    #v(1fr)
+    Rotation
+  ],
+
+  [
+    #cetz.canvas({
+      import cetz.draw: *
+
+      line((-2, 0), (4, 0))
+
+      translate(x: 0, y: 0.25)
+      demo(ored)
+      set-transform(none)
+
+      set-transform((
+        (1, 0, 0, 0),
+        (0, 1, 0, 0),
+        (0, 0, 1, 0),
+        (0, 0, 0, 1),
+      ))
+
+      translate(x: 0, y: 0.25)
+      demo(oblue)
+    })
+    #v(1fr)
+    Reflection
+  ],
+  [
+    #cetz.canvas({
+      import cetz.draw: *
+
+
+      demo(ored)
+
+      set-transform((
+        (1, 0, 0, 0),
+        (0, 1, 0, 0),
+        (0, 0, 0, 0),
+        (0, 0, 0, 0),
+      ))
+      translate(x: 2, y: 0)
+
+      demo(oblue)
+
+      set-transform(none)
+      line((-1, 0), (5, 0))
+    })
+    #v(1fr)
+    Glide reflection
+  ],
+)
+
+#definition()
+A _wallpaper_ is a two-dimensional pattern that...
+- has translational symmetry in at least two non-parallel directions (and therefore fills the plane) \
+  #note[
+    "Translational symmetry" means that we can slide the entire wallpaper in some direction, \
+    eventually mapping the pattern to itself.]
+- has a countable number of reflection, rotation, or glide symmetries. \
+
+#v(1fr)
+#pagebreak()
+
+#problem()
+Is a plain square grid a valid wallpaper?
+
+#solution([
+  Yes!
+  - It has translational symmetry in the horizontal and vertical directions
+  - It has a countable number of symmetries---namely, six distinct mirror lines (horizontal, vertical, and diagonal) duplicated once per square.
+  - A square grid is #sym.convolve`442`
+])
+
+#v(1fr)
+
+
+#problem()
+Is the empty plane a valid wallpaper?
+
+#solution([
+  No, since it has uncountably many symmetries.
+])
+
+#v(1fr)

src/Advanced/Wallpaper/parts/01 reflect.typ

@@ -0,0 +1,132 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+= Mirror Symmetry
+
+#definition()
+A _reflection_ is a transformation of the plane obtained by reflecting all points about a line. \
+If this reflection maps the wallpaper to itself, we have a _mirror symmetry_. \
+
+If $n$ such mirror lines intersect at a point, they form a _mirror node of order $n$_. \
+#note[Mirror nodes with order 1 do not exist (i.e, $n >= 2$). A line does not intersect itself!]
+
+#v(2mm)
+
+Two mirror nodes on a wallpaper are identical if we can map one to the other with a translation and a rotation while preserving the pattern on that wallpaper.
+
+#problem(label: "pat333")
+Find all three distinct mirror nodes in the following pattern. \
+What is the order of each node? \
+#hint([
+  You may notice rotational symmetry in this pattern. \
+  Don't worry about that yet, we'll discuss it later.
+])
+
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 45mm,
+  image("../res/wolfram/p3m1.svg", height: 100%),
+)
+
+#solution([
+  The mirror nodes are:
+  - the center of the amber cross
+  - the center of each right-handed group of three adjacent hexagons
+  - the center of each left-handed group of three adjacent hexagons
+])
+
+#v(1fr)
+
+
+#definition()
+_Orbifold notation_ gives us a way to describe the symmetries of a wallpaper. \
+It defines a _signature_ that fully describes all the symmetries of a given pattern. \
+We will introduce orbifold notation one symmetry at a time.
+
+#definition()
+In orbifold notation, mirror nodes are denoted by a #sym.convolve followed by a list of integer. \
+Every integer $n$ following a #sym.convolve denotes a mirror node of order $n$.
+
+#v(2mm)
+
+The order of these integers doesn't matter. #sym.convolve`234` and #sym.convolve`423` are the same signature. \
+However, we usually denote $n$-fold symmetries in descending order (that is, like #sym.convolve`432`). \
+If we have many nodes of the same order, integers may be repeated.
+
+
+#problem()
+What is the signature of the wallpaper in @pat333? \
+#hint[Again, ignore rotational symmetry for now.]
+#solution([It is #sym.convolve`333`])
+
+
+// MARK: page
+#v(1fr)
+#pagebreak()
+
+#problem()
+Find the signature of the following pattern.
+
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 60mm,
+  image("../res/*632-a.png", height: 100%),
+)
+
+
+#solution([
+  It is #sym.convolve`632`:
+  #table(
+    stroke: none,
+    align: center,
+    columns: 1fr,
+    rows: 40mm,
+    image("../res/*632-b.png", height: 100%),
+  )
+])
+
+#v(1fr)
+
+#problem()
+Draw a wallpaper pattern with signature #sym.convolve`2222`
+
+#solution([
+  Sample solutions are below.
+
+  #table(
+    stroke: none,
+    align: center,
+    columns: (1fr, 1fr),
+    rows: 50mm,
+    image("../res/wolfram/pmm.svg", height: 100%),
+    image("../res/escher/pmm.svg", height: 100%),
+  )
+])
+
+#v(1fr)
+
+#pagebreak()
+
+
+#remark()
+In an exceptional case, we have two parallel mirror lines. \
+Consider the following pattern:
+
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 60mm,
+  image("../res/**.png", height: 100%),
+)
+
+The signature of this pattern is #sym.convolve#sym.convolve
+
+#problem()
+Draw another wallpaper pattern with signature #sym.convolve#sym.convolve.
+
+#v(1fr)

src/Advanced/Wallpaper/parts/02 rotate.typ

@@ -0,0 +1,181 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+= Rotational Symmetry
+
+
+#definition()
+A wallpaper may also have $n$-fold rotational symmetry about a point.
+#v(2mm)
+This means there are no more than $n$ rotations around that point that map the wallpaper to itself.
+#v(2mm)
+As before, two points of rotational symmetry are identical if we can perform a translation and rotation that maps one to the other without changing the wallpaper.
+
+#definition()
+In orbifold notation, rotation is specified similarly to reflection, but uses the prefix #sym.diamond.stroked.small. \
+For example:
+- #sym.diamond.stroked.small`333` denotes a pattern with three distinct centers of rotation of order 3.
+- #sym.diamond.stroked.small`4`#sym.convolve`2` denotes a pattern with one rotation center of order 4 and one mirror node of order 2.
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr),
+  rows: 50mm,
+  image("../res/333.png", height: 100%), image("../res/3*3.png", height: 100%),
+)
+
+
+#problem()
+Find the three rotation centers in the left wallpaper. \
+What are their orders?
+
+#solution([This is #sym.diamond.stroked.small`333`])
+
+#v(1fr)
+
+#problem()
+Find the signature of the pattern on the right.
+
+#solution([This is #sym.diamond.stroked.small`3`#sym.convolve`3`])
+
+#v(1fr)
+
+
+#remark()
+You may have noticed that we could have an ambiguous classification, since two reflections are equivalent to a translation and a rotation.
+We thus make the following distinction: _rotational symmetry that can be explained by reflection is not rotational symmetry._
+
+#v(2mm)
+
+In other words, when classifying a pattern...
+- we first find all mirror symmetries,
+- then all rotational symmetries that are not accounted for by reflection.
+
+#pagebreak()
+
+
+// MARK: glide
+
+= Glide Reflections
+
+#definition()
+Another type of symmetry is the _glide reflection_, denoted #sym.times.
+
+A glide reflection is the result of a translation along a line followed by reflection about that line.
+
+For example, consider the following pattern:
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 60mm,
+  image("../res/*x-a.png", height: 100%),
+)
+
+#problem()
+Convince yourself that all mirror lines in this pattern are _not_ distinct. /
+In other words, this pattern has only one mirror symmetry.
+
+#solution([
+  There may seem to be two, but they are identical. \
+  We can translate one onto the other.
+])
+
+#v(1fr)
+
+
+
+#problem()
+Use the following picture to find the glide reflection in the above pattern.
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 70mm,
+  image("../res/*x-b.png", height: 100%),
+)
+
+#v(1fr)
+
+#remark()
+The signature of this wallpaper is #sym.convolve#sym.times.
+
+#pagebreak()
+
+
+#definition()
+If none of the above symmetries appear in a pattern, then we only have simple translational symmetry. We denote this with the signature #sym.circle.small.
+
+#remark()
+In summary, to find the signature of a pattern:
+- find the mirror lines (#sym.convolve) and the distinct intersections;
+- then find the rotation centers (#sym.diamond.stroked.small) not explained by reflection;
+- then find all glide reflections (#sym.times) that do not cross a mirror line.
+- If we have none of the above, our pattern must be #sym.circle.small.
+
+
+
+#problem()
+Find the signature of the following pattern:
+
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 50mm,
+  image("../res/wiki/Wallpaper_group-cm-4.jpg", height: 100%),
+)
+
+#solution([
+  This is #sym.convolve#sym.times.
+])
+
+#v(1fr)
+
+
+#problem()
+Find the signature of the following pattern:
+
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 60mm,
+  image("../res/wiki/Wallpaper_group-p4g-2.jpg", height: 100%),
+)
+
+#solution([
+  This is #sym.diamond.stroked.small`4`#sym.convolve`2`
+])
+
+#v(1fr)
+
+
+
+#pagebreak()
+
+#problem()
+Find two glide reflections in the following pattern.\
+#note[(and thus show that its signature is #sym.times#sym.times.)]
+
+
+#table(
+  stroke: none,
+  align: center,
+  columns: 1fr,
+  rows: 70mm,
+  image("../res/xx-b.png", height: 100%),
+)
+
+#solution([
+  #table(
+    stroke: none,
+    align: center,
+    columns: 1fr,
+    rows: 40mm,
+    image("../res/xx-a.png", height: 100%),
+  )
+])
+
+#v(1fr)

src/Advanced/Wallpaper/parts/03 problems.typ

@@ -0,0 +1,67 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+#let pat(img, sol) = {
+  problem()
+
+  table(
+    stroke: none,
+    align: center,
+    columns: (1fr, 1fr),
+    rows: 80mm,
+    image(img, height: 100%), image(img, height: 100%),
+  )
+
+  solution(sol)
+  v(1fr)
+}
+
+= A few problems
+
+Find the signatures of the following patterns. Mark all mirror nodes, rotation centers, and glide reflections. \
+Each pattern is provided twice for convenience.
+
+
+#pat("../res/wolfram/cm.svg", [#sym.times#sym.convolve])
+#pat("../res/wolfram/cmm.svg", [#sym.diamond.stroked`2`#sym.convolve`22`])
+#pagebreak()
+
+
+
+
+#pat("../res/wolfram/p3.svg", [#sym.diamond.stroked`333`])
+#pat("../res/wolfram/p3m1.svg", [#sym.convolve`333`])
+#pagebreak()
+
+#pat("../res/wolfram/p4.svg", [#sym.diamond.stroked`442`])
+#pat("../res/wolfram/p4m.svg", [#sym.convolve`442`])
+#pagebreak()
+
+#pat("../res/wolfram/p6.svg", [#sym.diamond.stroked`632`])
+#pat("../res/wolfram/p6m.svg", [#sym.convolve`632`])
+#pagebreak()
+
+#pat("../res/wolfram/p4g.svg", [#sym.diamond.stroked`4`#sym.convolve`2`])
+#pat("../res/wolfram/p31m.svg", [#sym.diamond.stroked`3`#sym.convolve`3`])
+#pagebreak()
+
+#problem()
+Draw a wallpaper with the signature #sym.convolve`442` \
+#note[Make sure there are no other symmetries!]
+#v(1fr)
+#pagebreak()
+
+
+#pat("../res/wolfram/pgg.svg", [#sym.diamond.stroked`22`#sym.times])
+#pat("../res/wolfram/pmg.svg", [#sym.diamond.stroked`22`#sym.convolve])
+#pagebreak()
+
+#pat("../res/wolfram/pg.svg", [#sym.times#sym.times])
+#pat("../res/wolfram/pm.svg", [#sym.convolve#sym.convolve])
+#pagebreak()
+
+#pat("../res/wolfram/p2.svg", [#sym.diamond.stroked`2222`])
+#pat("../res/wolfram/pmm.svg", [#sym.convolve`2222`])
+#pagebreak()
+
+#pat("../res/wolfram/p1.svg", [#sym.circle.small])

src/Advanced/Wallpaper/parts/04 theorem.typ

@@ -0,0 +1,97 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+= The Signature-Cost Theorem
+
+#definition()
+First, we'll associate a _cost_ to each type of symmetry in orbifold notation:
+
+#v(4mm)
+#align(center, table(
+  stroke: (1pt, 1pt),
+  align: center,
+  columns: (auto, auto, auto, auto),
+  [*Symbol*], [*Cost*], [*Symbol*], [*Cost*],
+  [#sym.circle.small], [2], [#sym.times or #sym.convolve], [1],
+  [#sym.diamond.stroked.small`2`], [1/2], [#sym.convolve`2`], [1/4],
+  [#sym.diamond.stroked.small`3`], [2/3], [#sym.convolve`3`], [1/3],
+  [#sym.dots], [#sym.dots], [#sym.dots], [#sym.dots],
+  [#sym.diamond.stroked.small`n`],
+  [$(n-1) / n$],
+  [#sym.convolve`n`],
+  [$(n-1) / (2n)$],
+))
+
+
+We then calculate the total "cost" of a signature by adding up the costs of each component.
+
+For example, a pattern with signature #sym.convolve`333` has cost 2:
+
+#v(2mm)
+
+$
+  2 / 3 + 2 / 3 + 2 / 3 = 2
+$
+
+#problem()
+Calculate the costs of the following signatures:
+- #sym.diamond.stroked.small`3`#sym.convolve`3`
+- #sym.convolve#sym.convolve
+- #sym.diamond.stroked.small`4`#sym.convolve`2`:
+
+#solution([
+  - #sym.diamond.stroked.small`3`#sym.convolve`3`: $2/3 + 1 + 1/3 = 2$
+  - #sym.convolve#sym.convolve: $1 + 1 = 2$
+  - #sym.diamond.stroked.small`4`#sym.convolve`2`: $3/4 + 1 + 1/4 = 2$
+])
+
+#v(1fr)
+
+#theorem(name: "Signature Cost Theorem")
+The signatures of planar wallpaper patterns are exactly those with total cost 2. \
+#note([We will not prove this theorem today, accept it without proof.])
+
+#problem()
+Consider the 4 symmetries (translation, reflection, rotation, and glide reflection). \
+Which preserve orientation? Which reverse orientation?
+
+#solution([
+  - Reflections and glide reflections reverse orientation (directions of spirals).
+  - Translation and rotation preserve orientation.
+])
+
+#v(1fr)
+#pagebreak()
+
+#problem()
+Use the signature-cost theorem to find all the signatures consisting of only #sym.circle.small or rotational symmetries.
+
+#solution([
+  #sym.diamond.stroked.small`632`, #sym.diamond.stroked.small`442`, #sym.diamond.stroked.small`333`, #sym.diamond.stroked.small`2222`, #sym.circle.small
+])
+
+#v(1fr)
+
+
+#problem()
+Find all the signatures consisting of only mirror symmetries.
+
+#solution([
+  #sym.convolve`632`, #sym.convolve`442`, #sym.convolve`333`, #sym.convolve`2222`, #sym.convolve#sym.convolve
+])
+
+#v(1fr)
+
+
+#problem()
+Find all the remaining signatures. \
+Each must be a mix of of mirror symmetries, rotational symmetries, or glide reflections. \
+#hint([They are all shown in the problems section.])
+
+#solution([
+  #sym.diamond.stroked.small`3`#sym.convolve`3`, #sym.diamond.stroked.small`4`#sym.convolve`2`,
+  #sym.diamond.stroked.small`22`#sym.times, #sym.diamond.stroked.small`22`#sym.convolve,
+  #sym.times#sym.times, #sym.times#sym.convolve
+])
+
+#v(1fr)

src/Advanced/Wallpaper/res/escher/pmm.svg

@@ -0,0 +1,86 @@
+<?xml version="1.0" standalone="no"?>
+<!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 20001102//EN" "http://www.w3.org/TR/2000/CR-SVG-20001102/DTD/svg-20001102.dtd">
+<svg xmlns:svg="http://www.w3.org/2000/svg" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="10.16cm" height="5.08cm" viewBox="0 0 384 192">
+<title>pmm</title>
+<desc>Exported by Tess 1.70.</desc>
+<clipPath id="mainclip"><rect x="0" y="0" width="384" height="192"/></clipPath>
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+</g>
+</g>
+</svg>

src/Warm-Ups/Adders/main.typ

@@ -46,10 +46,10 @@ Use two half adders to construct a full adder.
 
 #solution([
   $
-    s_1, c_1 &= "HA"(a, b) \
-    s_2, c_2 &= "HA"(s_1, c_"in") \
-    s_"out" &= s_2 \
-    c_"out" &= "OR"(c_1, c_2)
+    s_1, c_1 & = "HA"(a, b) \
+    s_2, c_2 & = "HA"(s_1, c_"in") \
+     s_"out" & = s_2 \
+     c_"out" & = "OR"(c_1, c_2)
   $
 
   #v(2mm)

src/Warm-Ups/Big-Tac-Toe/main.typ

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 #show: handout.with(
   title: [Warm-Up: Big-Tac-Toe],
@@ -75,16 +75,13 @@ How does your strategy change? \
 #if extra_boards {
   pagebreak()
 
-  align(
-    center,
-    grid(
-      stroke: none,
-      align: center,
-      columns: (1fr, 1fr),
-      rows: (1fr, 1fr, 1fr),
-      btt(0.35), btt(0.35),
-      btt(0.35), btt(0.35),
-      btt(0.35), btt(0.35),
-    ),
-  )
+  align(center, grid(
+    stroke: none,
+    align: center,
+    columns: (1fr, 1fr),
+    rows: (1fr, 1fr, 1fr),
+    btt(0.35), btt(0.35),
+    btt(0.35), btt(0.35),
+    btt(0.35), btt(0.35),
+  ))
 }

src/Warm-Ups/Odd Dice/main.typ

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 
 #show: handout.with(
@@ -43,61 +43,56 @@ Now, consider the set of six-sided dice below:
 - Die $E$: $0, 5, 5, 5, 5, 5$
 On average, which die beats each of the others? Draw a diagram.
 
-#solution(
-  align(
-    center,
-    cetz.canvas({
-      import cetz.draw: *
+#solution(align(center, cetz.canvas({
+  import cetz.draw: *
 
-      let s = 0.8 // Scale
-      let t = 13pt * s // text size
-      let radius = 0.3 * s
+  let s = 0.8 // Scale
+  let t = 13pt * s // text size
+  let radius = 0.3 * s
 
-      // Points
-      let a = (-2 * s, 0.2 * s)
-      let b = (0 * s, 2 * s)
-      let c = (2 * s, 0.2 * s)
-      let d = (1.2 * s, -2.1 * s)
-      let e = (-1.2 * s, -2.1 * s)
+  // Points
+  let a = (-2 * s, 0.2 * s)
+  let b = (0 * s, 2 * s)
+  let c = (2 * s, 0.2 * s)
+  let d = (1.2 * s, -2.1 * s)
+  let e = (-1.2 * s, -2.1 * s)
 
-      set-style(
-        stroke: (thickness: 0.6mm * s),
-        mark: (
-          end: (
-            symbol: ">",
-            fill: black,
-            offset: radius + (0.025 * s),
-            width: 1.2mm * s,
-            length: 1.2mm * s,
-          ),
-        ),
-      )
+  set-style(
+    stroke: (thickness: 0.6mm * s),
+    mark: (
+      end: (
+        symbol: ">",
+        fill: black,
+        offset: radius + (0.025 * s),
+        width: 1.2mm * s,
+        length: 1.2mm * s,
+      ),
+    ),
+  )
 
-      line(a, b)
-      line(b, c)
-      line(c, d)
-      line(d, e)
-      line(e, a)
-      line(a, c)
-      line(b, d)
-      line(c, e)
-      line(d, a)
-      line(e, b)
+  line(a, b)
+  line(b, c)
+  line(c, d)
+  line(d, e)
+  line(e, a)
+  line(a, c)
+  line(b, d)
+  line(c, e)
+  line(d, a)
+  line(e, b)
 
-      circle(a, radius: radius, fill: oblue, stroke: none)
-      circle(b, radius: radius, fill: oblue, stroke: none)
-      circle(c, radius: radius, fill: oblue, stroke: none)
-      circle(d, radius: radius, fill: oblue, stroke: none)
-      circle(e, radius: radius, fill: oblue, stroke: none)
+  circle(a, radius: radius, fill: oblue, stroke: none)
+  circle(b, radius: radius, fill: oblue, stroke: none)
+  circle(c, radius: radius, fill: oblue, stroke: none)
+  circle(d, radius: radius, fill: oblue, stroke: none)
+  circle(e, radius: radius, fill: oblue, stroke: none)
 
-      content(a, text(fill: white, size: t, [*A*]))
-      content(b, text(fill: white, size: t, [*B*]))
-      content(c, text(fill: white, size: t, [*C*]))
-      content(d, text(fill: white, size: t, [*D*]))
-      content(e, text(fill: white, size: t, [*E*]))
-    }),
-  ),
-)
+  content(a, text(fill: white, size: t, [*A*]))
+  content(b, text(fill: white, size: t, [*B*]))
+  content(c, text(fill: white, size: t, [*C*]))
+  content(d, text(fill: white, size: t, [*D*]))
+  content(e, text(fill: white, size: t, [*E*]))
+})))
 
 #v(1fr)
 

src/Warm-Ups/Painting/main.typ

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 #show: handout.with(
   title: [Warm-Up: What's an AST?],
@@ -18,59 +18,56 @@ You may detach the string as you hang the painting, but it must be re-attached o
 
 #v(2mm)
 
-#align(
-  center,
-  cetz.canvas({
-    import cetz.draw: *
-    let s = 2.5
+#align(center, cetz.canvas({
+  import cetz.draw: *
+  let s = 2.5
 
 
-    line(
-      (0 * s, 1 * s),
-      (2 * s, 1 * s),
-      (2 * s, 0 * s),
-      (0 * s, 0 * s),
-      close: true,
-      stroke: (thickness: 0.8mm),
-    )
+  line(
+    (0 * s, 1 * s),
+    (2 * s, 1 * s),
+    (2 * s, 0 * s),
+    (0 * s, 0 * s),
+    close: true,
+    stroke: (thickness: 0.8mm),
+  )
 
-    line(
-      (0.1 * s, 1 * s),
-      (0.5 * s, 1.5 * s),
-      (1.5 * s, 1.5 * s),
-      (1.9 * s, 1 * s),
-      stroke: (thickness: 0.5mm, dash: "dotted"),
-    )
+  line(
+    (0.1 * s, 1 * s),
+    (0.5 * s, 1.5 * s),
+    (1.5 * s, 1.5 * s),
+    (1.9 * s, 1 * s),
+    stroke: (thickness: 0.5mm, dash: "dotted"),
+  )
 
-    circle((0.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
-    circle((1.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
+  circle((0.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
+  circle((1.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
 
-    line(
-      (0.66 * s, 0.66 * s),
-      (0.66 * s, 0.35 * s),
-      (0.60 * s, 0.1 * s),
-    )
+  line(
+    (0.66 * s, 0.66 * s),
+    (0.66 * s, 0.35 * s),
+    (0.60 * s, 0.1 * s),
+  )
 
-    line(
-      (0.72 * s, 0.1 * s),
-      (0.66 * s, 0.35 * s),
-    )
+  line(
+    (0.72 * s, 0.1 * s),
+    (0.66 * s, 0.35 * s),
+  )
 
-    line(
-      (0.66 * s, 0.575 * s),
-      (0.6 * s, 0.475 * s),
-      (0.525 * s, 0.575 * s),
-    )
+  line(
+    (0.66 * s, 0.575 * s),
+    (0.6 * s, 0.475 * s),
+    (0.525 * s, 0.575 * s),
+  )
 
-    line(
-      (0.66 * s, 0.575 * s),
-      (0.72 * s, 0.475 * s),
-      (0.795 * s, 0.575 * s),
-    )
+  line(
+    (0.66 * s, 0.575 * s),
+    (0.72 * s, 0.475 * s),
+    (0.795 * s, 0.575 * s),
+  )
 
-    circle((0.66 * s, 0.66 * s), radius: 0.07 * s, fill: white)
-  }),
-)
+  circle((0.66 * s, 0.66 * s), radius: 0.07 * s, fill: white)
+}))
 
 #solution([
   Say we have a left nail and a right nail. The path of the string is as follows:

src/Warm-Ups/Partition Products/main.typ

@@ -7,7 +7,7 @@
 
 #problem()
 Take any positive integer $n$. \
-Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ...  a_k$ \
+Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... a_k$ \
 Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
 
 
@@ -17,7 +17,9 @@ Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
   Of course, all $a_i$ should be greater than $1$. \
   Also, all $a_i$ should be smaller than four, since $x <= x(x-2)$ if $x >= 4$. \
   Thus, we're left with sequences that only contain 2 and 3. \
-  #note([Note that two twos are the same as one four, but we exclude fours for simplicity.])
+  #note(
+    [Note that two twos are the same as one four, but we exclude fours for simplicity.],
+  )
 
   #v(2mm)
 

src/Warm-Ups/Passing Balls/main.typ

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 #show: handout.with(
   title: [Warm-Up: Passing Balls],
@@ -78,32 +78,25 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
 
         let i = 1
         for p in pts {
-          circle(
-            p,
-            radius: radius * s,
-            fill: if i == 1 {
-              ored
-            } else if i == 2 {
-              ogreen
-            } else if i == 3 {
-              oorange
-            } else if i == 4 {
-              oblue
-            } else { white },
-          )
+          circle(p, radius: radius * s, fill: if i == 1 {
+            ored
+          } else if i == 2 {
+            ogreen
+          } else if i == 3 {
+            oorange
+          } else if i == 4 {
+            oblue
+          } else { white })
 
-          content(
-            p,
-            text(
-              fill: if i <= 4 {
-                white
-              } else {
-                black
-              },
-              size: t,
-              [*#i*],
-            ),
-          )
+          content(p, text(
+            fill: if i <= 4 {
+              white
+            } else {
+              black
+            },
+            size: t,
+            [*#i*],
+          ))
           i = i + 1
         }
       }),
@@ -118,32 +111,25 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
           let l = calc.rem(((i - 1) * 5), 12) + 1
 
 
-          circle(
-            p,
-            radius: radius * s,
-            fill: if l == 1 {
-              ored
-            } else if l == 2 {
-              ogreen
-            } else if l == 3 {
-              oorange
-            } else if l == 4 {
-              oblue
-            } else { white },
-          )
+          circle(p, radius: radius * s, fill: if l == 1 {
+            ored
+          } else if l == 2 {
+            ogreen
+          } else if l == 3 {
+            oorange
+          } else if l == 4 {
+            oblue
+          } else { white })
 
-          content(
-            p,
-            text(
-              fill: if l <= 4 {
-                white
-              } else {
-                black
-              },
-              size: t,
-              [*#l*],
-            ),
-          )
+          content(p, text(
+            fill: if l <= 4 {
+              white
+            } else {
+              black
+            },
+            size: t,
+            [*#l*],
+          ))
           i = i + 1
         }
       }),
@@ -158,32 +144,25 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
           let l = calc.rem(((i - 1) * 5), 12) + 1
 
 
-          circle(
-            p,
-            radius: radius * s,
-            fill: if l == 1 {
-              oblue
-            } else if l == 2 {
-              oorange
-            } else if l == 3 {
-              ored
-            } else if l == 4 {
-              ogreen
-            } else { white },
-          )
+          circle(p, radius: radius * s, fill: if l == 1 {
+            oblue
+          } else if l == 2 {
+            oorange
+          } else if l == 3 {
+            ored
+          } else if l == 4 {
+            ogreen
+          } else { white })
 
-          content(
-            p,
-            text(
-              fill: if l <= 4 {
-                white
-              } else {
-                black
-              },
-              size: t,
-              [*#l*],
-            ),
-          )
+          content(p, text(
+            fill: if l <= 4 {
+              white
+            } else {
+              black
+            },
+            size: t,
+            [*#l*],
+          ))
           i = i + 1
         }
       }),

src/Warm-Ups/Snakes/0.svg

@@ -0,0 +1,81 @@
+<svg class="typst-doc" viewBox="0 0 8.236633858267716 14.150551181102362" width="8.236633858267716pt" height="14.150551181102362pt" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:h5="http://www.w3.org/1999/xhtml">
+    <g>
+        <g transform="translate(0 0)">
+            <g class="typst-group">
+                <g>
+                    <g transform="translate(0 2.2237677165354333)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 4.015 0 L 4.015 9.669 L 0.0000000000000007867722 9.669 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(-0.00000003149606268881308 2.2237676220472435)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 7.238)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gB0C69879DCB2DFE01E1A8092C6399B95" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(4.282303149606299 8.46992125984252)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6806297 L 0 5.6806297 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(4.282303007874017 8.46992125984252)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
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+                                                    <use xlink:href="#gDCE911778D5C1849302153FEF2645F85" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(29.79411417322835 34.01574803149607)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6125984 L 0 5.6125984 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(29.794114031496065 34.01574803149607)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#g642BCE8B7105B32AF084522D60B4E652" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(12.786240157480314 34.01574803149607)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6125984 L 0 5.6125984 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(12.786240015748032 34.01574803149607)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#g642BCE8B7105B32AF084522D60B4E652" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(8.503937007874017 19.231641732283467)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 4.015 0 L 4.015 9.669 L 0 9.669 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(8.50393697637795 19.231641637795278)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 7.238)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gB0C69879DCB2DFE01E1A8092C6399B95" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(12.786240157480314 25.477795275590555)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6806297 L 0 5.6806297 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(12.786240015748032 25.477795275590555)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gDCE911778D5C1849302153FEF2645F85" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(17.007874015748033 10.727704724409453)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 4.015 0 L 4.015 9.669 L 0.0000000000000031470889 9.669 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(17.007873984251972 10.72770462992126)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 7.238)">
+                                                <g class="typst-text" transform="scale(1, -1)">
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+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(25.511811023622048 19.231641732283467)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 4.015 0 L 4.015 9.669 L 0.0000000000000031470889 9.669 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(25.511810992125987 19.231641637795278)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 7.238)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gB0C69879DCB2DFE01E1A8092C6399B95" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(29.79411417322835 25.477795275590555)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6806297 L 0 5.6806297 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(29.794114031496065 25.477795275590555)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gDCE911778D5C1849302153FEF2645F85" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(29.79411417322835 17.007874015748033)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6125984 L 0 5.6125984 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(29.794114031496065 17.007874015748033)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#g642BCE8B7105B32AF084522D60B4E652" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(25.511811023622048 2.2237677165354355)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 4.015 0 L 4.015 9.669 L 0.0000000000000031470889 9.669 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(25.511810992125987 2.2237676220472435)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 7.238)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gB0C69879DCB2DFE01E1A8092C6399B95" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(29.79411417322835 8.469921259842522)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6806297 L 0 5.6806297 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(29.794114031496065 8.46992125984252)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#gDCE911778D5C1849302153FEF2645F85" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                    <g transform="translate(29.79411417322835 0)">
+                        <path class="typst-shape" fill="none" d="M 0 0 L 3.9543307 0 L 3.9543307 5.6125984 L 0 5.6125984 L 0 0 Z "/>
+                    </g>
+                    <g transform="translate(29.794114031496065 -0.0000000000000008951719505639057)">
+                        <g class="typst-group">
+                            <g>
+                                <g transform="translate(0 0)">
+                                    <g class="typst-group">
+                                        <g>
+                                            <g transform="translate(0 5.595590551181102)">
+                                                <g class="typst-text" transform="scale(1, -1)">
+                                                    <use xlink:href="#g642BCE8B7105B32AF084522D60B4E652" x="0" y="0" fill="#000000" fill-rule="nonzero"/>
+                                                </g>
+                                            </g>
+                                        </g>
+                                    </g>
+                                </g>
+                            </g>
+                        </g>
+                    </g>
+                </g>
+            </g>
+        </g>
+    </g>
+    <defs id="glyph">
+        <symbol id="gB0C69879DCB2DFE01E1A8092C6399B95" overflow="visible">
+            <path d="M 1.54 0.253 C 1.54 -0.594 1.375 -1.892 1.067 -1.892 C 0.88 -1.892 0.726 -1.551 0.429 -1.551 C 0.22 -1.551 0 -1.738 0 -1.958 C 0 -2.233 0.242 -2.431 0.594 -2.431 C 1.6719999 -2.431 2.376 -1.43 2.376 1.925 L 2.376 3.234 C 2.376 3.927 2.332 4.719 2.332 5.269 C 2.332 6.732 2.552 7.139 2.783 7.139 C 3.047 7.139 3.069 6.776 3.421 6.776 C 3.685 6.776 3.905 6.963 3.905 7.205 C 3.905 7.48 3.641 7.678 3.322 7.678 C 2.156 7.678 1.518 6.446 1.518 3.74 L 1.518 2.299 C 1.518 1.419 1.54 1.133 1.54 0.253 Z "/>
+        </symbol>
+        <symbol id="gDCE911778D5C1849302153FEF2645F85" overflow="visible">
+            <path d="M 1.9388977 -0.08503937 C 2.593701 -0.08503937 3.5886614 0.57826775 3.5886614 2.576693 C 3.5886614 3.4185827 3.384567 4.1414175 3.0103939 4.634646 C 2.7892914 4.9322834 2.432126 5.187402 1.9729134 5.187402 C 1.1310236 5.187402 0.33165354 4.183937 0.33165354 2.5001576 C 0.33165354 1.5902363 0.61228347 0.73984253 1.08 0.272126 C 1.3181102 0.03401575 1.6072441 -0.08503937 1.9388977 -0.08503937 Z M 1.9729134 4.889764 C 2.1174803 4.889764 2.2535434 4.8387403 2.3555906 4.745197 C 2.6192126 4.5240946 2.848819 3.8777955 2.848819 2.7382677 C 2.848819 1.9559056 2.823307 1.4201576 2.704252 0.9864567 C 2.5171654 0.28913388 2.1004725 0.21259843 1.9474016 0.21259843 C 1.1565355 0.21259843 1.0714961 1.6667717 1.0714961 2.4066143 C 1.0714961 4.498583 1.5902363 4.889764 1.9729134 4.889764 Z "/>
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+        </symbol>
+    </defs>
+</svg>

src/Warm-Ups/Snakes/main.typ

@@ -0,0 +1,89 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+#show: handout.with(
+  title: [Warm-Up: Snakes!],
+  by: "Mark",
+  page_numbers: false,
+)
+
+#box(place(box(width: 100%, height: 8in, align(horizon, image(
+  "7.svg",
+  width: 360mm,
+)))))
+
+#problem()
+Evaluate this integral. \
+All integrals are of the form $integral_a^b 1 #h(1mm) d x$.
+
+#if_solutions(pagebreak())
+
+#solution[
+  This integral is drawn recursively with the following code: \
+
+  #v(2mm)
+
+  #let snake(depth, x, y) = {
+    if depth == 0 {
+      math.attach(math.integral, br: x, tr: y)
+    } else {
+      let bot = snake(depth - 1, x, "1")
+      let top = snake(depth - 1, "0", y)
+      snake(depth - 1, bot, top)
+    }
+  }
+
+  ```typst
+  #let snake(depth, x, y) = {
+    if depth == 0 {
+      $integral_#x ^#y$
+    } else {
+      let bot = snake(depth - 1, x, "1")
+      let top = snake(depth - 1, "0", y)
+      snake(depth - 1, bot, top)
+    }
+  }
+  ```
+
+  #v(5mm)
+
+  For example:
+
+  $ snake\(0, x, y) = #snake(0, "x", "y") $
+
+  #v(2mm)
+
+  $ snake\(1, x, y) = #snake(1, "x", "y") $
+
+  #v(2mm)
+
+  $ snake\(2, x, y) = #snake(2, "x", "y") $
+
+  In other words, we want $f_7 (0, 1)$, where...
+  - $f_0(x, y) := integral_x^y$
+  - $f_(n+1)(x, y) := f_n [f_n (x, 1), f_n (0, y)]$
+
+  #v(5mm)
+  Expanding a few iterations, we find:
+  - $f_0(x, y) = integral_x^y = y-x$
+  - $f_1(x, y) = f_0 [f_0 (x, 1), f_0 (0, y)] = y+x-1$
+  - $f_2(x, y) = f_1 [f_1 (x, 1), f_1 (0, y)] = y+x-2$
+
+  #v(5mm)
+
+  We can use induction to show that this pattern continues. \
+  If $g_n = y+x+c$, then $g_(n+1) = y+x+(3c+1)$.
+
+
+  #v(5mm)
+  Finally, use this recusion to find that
+  $f_0, f_1, ..., f_7 = 1, 0, -1, -4, -13, -40, -121, -364$
+
+  One can also find an explicit formula for $g_n$:
+
+  $
+    f_(n+1) = g_n & = x + y + 3^n c + 3^0 + 3^1 + ... + 3^n \
+                  & = x + y + 3^n c + sum_(i=0)^n 3^i       \
+                  & = x + y + 3^n c + (3^(n+1) + 1)/2
+  $
+]

src/Warm-Ups/Snakes/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Snakes!"
+
+[publish]
+handout = true
+solutions = true

src/Warm-Ups/Snakes/svg.typ

@@ -0,0 +1,57 @@
+#import "@preview/cetz:0.4.2"
+
+// Compile with:
+// typst compile --package-path "../../../lib/typst/" svg.typ --format svg
+// You'll need to recompile typst with a higher recursion limit.
+
+#set page(
+  fill: none,
+  width: auto,
+  height: auto,
+  margin: 0mm,
+);
+
+#let cetz_snake(depth, b, t) = {
+  import cetz.draw: *
+
+  if depth == 0 {
+    content((0, 0), math.inline(math.integral))
+
+    group({
+      translate(x: 0.15, y: -0.15)
+      b
+    })
+
+    group({
+      translate(x: 0.15, y: 0.15)
+      t
+    })
+  } else {
+    let bot = group({
+      translate(x: 0.15, y: -0.15)
+      cetz_snake(depth - 1, b, content((0, 0), text("1", size: 3mm)))
+    })
+
+    let top = group({
+      translate(x: 0.15, y: 0.15)
+      cetz_snake(depth - 1, content((0, 0), text("0", size: 3mm)), t)
+    })
+
+    cetz_snake(depth - 1, bot, top)
+  }
+}
+
+
+#{
+  import cetz.draw: *
+
+  // any bigger than 7 overflows the stack.
+  // any bigger than 5 exceeds typst's recursion limit.
+  // You'll have to increase those limits and recompile to
+  // build this document.
+  let depth = 7
+
+  let b = content((0, 0), text("0", size: 3mm))
+  let t = content((0, 0), text("1", size: 3mm))
+  cetz.canvas(cetz_snake(depth, b, t))
+};

src/Warm-Ups/What's an AST/main.typ

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 #show: handout.with(
   title: [Warm-Up: What's an AST?],
@@ -24,51 +24,48 @@ respecting the order of operations $[and, times, div, +, -]$.
 
 #v(2mm)
 
-#align(
-  center,
-  cetz.canvas({
-    import cetz.draw: *
+#align(center, cetz.canvas({
+  import cetz.draw: *
 
-    // spell:off
-    content((0, 0), $+$, name: "r")
-    content((-0.5, -1), $3$, name: "a")
-    content((0.5, -1), $div$, name: "b")
-    content((-0.3, -2), $times$, name: "ba")
-    content((1.3, -2), $and$, name: "bb")
-    content((-0.8, -3), $9$, name: "baa")
-    content((0.2, -3), $8$, name: "bab")
-    content((0.8, -3), $5$, name: "bba")
-    content((1.8, -3), $6$, name: "bbb")
-    // spell:on
+  // spell:off
+  content((0, 0), $+$, name: "r")
+  content((-0.5, -1), $3$, name: "a")
+  content((0.5, -1), $div$, name: "b")
+  content((-0.3, -2), $times$, name: "ba")
+  content((1.3, -2), $and$, name: "bb")
+  content((-0.8, -3), $9$, name: "baa")
+  content((0.2, -3), $8$, name: "bab")
+  content((0.8, -3), $5$, name: "bba")
+  content((1.8, -3), $6$, name: "bbb")
+  // spell:on
 
-    // Zero-sized arrows are a hack for offset.
-    set-style(
-      stroke: (thickness: 0.3mm),
-      mark: (
-        start: (
-          symbol: "|",
-          offset: 0.25,
-          width: 0mm,
-          length: 0mm,
-        ),
-        end: (
-          symbol: "|",
-          offset: 0.25,
-          width: 0mm,
-          length: 0mm,
-        ),
+  // Zero-sized arrows are a hack for offset.
+  set-style(
+    stroke: (thickness: 0.3mm),
+    mark: (
+      start: (
+        symbol: "|",
+        offset: 0.25,
+        width: 0mm,
+        length: 0mm,
       ),
-    )
+      end: (
+        symbol: "|",
+        offset: 0.25,
+        width: 0mm,
+        length: 0mm,
+      ),
+    ),
+  )
 
-    // spell:off
-    line("r", "a")
-    line("r", "b")
-    line("b", "ba")
-    line("b", "bb")
-    line("ba", "baa")
-    line("ba", "bab")
-    line("bb", "bba")
-    line("bb", "bbb")
-    // spell:on
-  }),
-)
+  // spell:off
+  line("r", "a")
+  line("r", "b")
+  line("b", "ba")
+  line("b", "bb")
+  line("ba", "baa")
+  line("ba", "bab")
+  line("bb", "bba")
+  line("bb", "bbb")
+  // spell:on
+}))

src/Warm-Ups/Wild Tic-Tac-Toe/main.typ

@@ -1,24 +1,21 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 #show: handout.with(
   title: [Warm-Up: Wild Tic-Tac-Toe],
   by: "Mark",
 )
 
-#let ttt = align(
-  center,
-  cetz.canvas({
-    import cetz.draw: *
-    let s = 0.7 // scale
+#let ttt = align(center, cetz.canvas({
+  import cetz.draw: *
+  let s = 0.7 // scale
 
-    set-style(stroke: (thickness: 0.5mm * s))
-    line((-1 * s, 3 * s), (-1 * s, -3 * s))
-    line((1 * s, 3 * s), (1 * s, -3 * s))
-    line((3 * s, -1 * s), (-3 * s, -1 * s))
-    line((3 * s, 1 * s), (-3 * s, 1 * s))
-  }),
-)
+  set-style(stroke: (thickness: 0.5mm * s))
+  line((-1 * s, 3 * s), (-1 * s, -3 * s))
+  line((1 * s, 3 * s), (1 * s, -3 * s))
+  line((3 * s, -1 * s), (-3 * s, -1 * s))
+  line((3 * s, 1 * s), (-3 * s, 1 * s))
+}))
 
 
 #problem()

tools/typos.toml

@@ -1,6 +1,7 @@
 [default]
 extend-words."LSAT" = "LSAT"
 extend-words."ket" = "ket"
+extend-words."typ" = "typ"
 
 extend-ignore-re = [
 	# spell:disable-line