Retrograde edits

Reviewed-on: #26

.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, if_solutions_else, instructornote,
-  if_no_solutions,
+  sample_solution, solution,
-  if_solutions_else,
-  solution,
-  instructornote,
 )
 
 
@@ -38,10 +35,7 @@
     margin: 20mm,
     width: 8.5in,
     height: 11in,
-    footer: align(
+    footer: align(center, context counter(page).display()),
-      center,
-      context counter(page).display(),
-    ),
     footer-descent: 5mm,
   )
 
@@ -102,8 +96,8 @@
 
   // Make handout title
   {
-    import "header.typ": make_header, solution_warning, short_solution_warning
+    import "header.typ": make_header, short_solution_warning, solution_warning
-    import "solution.typ": solutions_state, reset_solutions
+    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(
+  block(above: 8mm, below: 2mm, text(weight: "bold", obj_content))
-    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(
+  block(above: 8mm, below: 2mm, text(weight: "bold", obj_content))
-    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.
@@ -61,10 +61,7 @@
 }
 
 #let solution(content) = {
-  if_solutions(
+  if_solutions(align(center, stack(
-    align(
-      center,
-      stack(
     block(
       width: 100%,
       breakable: false,
@@ -83,16 +80,35 @@
       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 instructornote(content) = {
-  if_solutions(
+  if_solutions(align(center, stack(
-    align(
-      center,
-      stack(
     block(
       width: 100%,
       breakable: false,
@@ -111,7 +127,5 @@
       inset: 3mm,
       align(left, content),
     ),
-      ),
+  )))
-    ),
-  )
 }

src/Advanced/De Bruijn/parts/0 intro.tex

@@ -21,7 +21,7 @@ Unlock this lock with only 5 keypresses.
 \end{solution}
 \vfill
 
-Now, consider the same lock, now set with a three-digit binary code.
+Now consider the same lock, but configured with a three-digit binary code.
 \problem{}
 How many codes are possible?
 \vfill

src/Advanced/De Bruijn/parts/1 words.tex

@@ -20,7 +20,11 @@ We say $v$ is a \textit{subword} of $w$ if $v$ is contained in $w$. \par
 For example, \texttt{11} is a subword of \texttt{011}, but \texttt{00} is not.
 
 \definition{}
-Recall \ref{lockproblem}. Let's generalize this to the \textit{$n$-subword problem}: \par
+Recall the lock problem from the previous page.
+Let's generalize this to the \textit{$n$-subword problem}:
+
+\vspace{1mm}
+
 Given an alphabet $A$ and a positive integer $n$,
 we want a word over $A$ that contains all possible length-$n$ subwords.
 The shortest word that solves a given $n$-subword problem is called the \textit{optimal solution}.
@@ -67,7 +71,7 @@ Find the following:
 
 \problem{}<sbounds>
 Let $w$ be a word over an alphabet of size $k$. \par
-Prove the following:
+Show that all of the following are true:
 \begin{itemize}
 	\item $\mathcal{S}_n(w) \leq k^n$
 	\item $\mathcal{S}_n(w) \geq \mathcal{S}_{n-1}(w) - 1$
@@ -103,7 +107,7 @@ Prove the following:
 
 \definition{}
 Let $v$ and $w$ be words over the same alphabet. \par
-The word $vw$ is the word formed by writing $v$ after $w$. \par
+The word $vw$ is the word formed by writing $w$ after $v$. \par
 For example, if $v = \texttt{1001}$ and $w = \texttt{10}$, $vw$ is $\texttt{100110}$.
 
 \problem{}
@@ -116,7 +120,6 @@ We'll call this the \textit{Fibonacci word} of order $k$.
 	\item What are $F_3$, $F_4$, and $F_5$?
 	\item Compute $\mathcal{S}_0$ through $\mathcal{S}_5$ for $F_5$.
 	\item Show that the length of $F_k$ is the $(k + 2)^\text{th}$ Fibonacci number. \par
-	\hint{Induction.}
 \end{itemize}
 
 \begin{solution}

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()
@@ -31,13 +31,11 @@ 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.])
+#note(
+  [The following only applies to floats that consist of 32 bits. We won't encounter any others today.],
+)
 
-#align(
+#align(center, box(inset: 2mm, cetz.canvas({
-  center,
-  box(
-    inset: 2mm,
-    cetz.canvas({
   import cetz.draw: *
 
   let chars = (
@@ -96,9 +94,7 @@ Floats represent a subset of the real numbers, and are interpreted as follows: \
 
   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 $s$. \

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:0.4.2"
-#import "@preview/cetz-plot:0.1.0": plot, chart
+#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(
+          plot.add(f1, domain: domain, label: $log(1+x)$, style: (
-            f1,
+            stroke: ogrape,
-            domain: domain,
+          ))
-            label: $log(1+x)$,
-            style: (stroke: ogrape),
-          )
 
-          plot.add(
+          plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
-            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(
+          plot.add(f1, domain: domain, label: $log(1+x)$, style: (
-            f1,
+            stroke: ogrape,
-            domain: domain,
+          ))
-            label: $log(1+x)$,
-            style: (stroke: ogrape),
-          )
 
-          plot.add(
+          plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
-            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(
+#note(type: "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))
+    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(x_f) & = log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
-    &= log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
+               & = E - 127 + log_2(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)(2^23 E + F) - 127 + epsilon \
-    &= 1 / (2^23)(x_i) - 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 \
+  (3 times 2^22) (127 - epsilon) & = 6240089 \
-  (127 - epsilon) &= 126.955 \
+                 (127 - epsilon) & = 126.955 \
-  epsilon &= 0.0450466
+                         epsilon & = 0.0450466
 $
 
 So, $0.045$ is the $epsilon$ used by Quake. \

src/Advanced/Generating Functions/main.tex

@@ -1,7 +1,7 @@
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[
-	solutions,
+	%solutions,
 	singlenumbering
 ]{../../../lib/tex/handout}
 \usepackage{../../../lib/tex/macros}
@@ -19,4 +19,5 @@
 	\input{parts/01 fibonacci.tex}
 	\input{parts/02 dice.tex}
 	\input{parts/03 coins.tex}
+	\input{parts/04 bonus.tex}
 \end{document}

src/Advanced/Generating Functions/parts/01 fibonacci.tex

@@ -77,7 +77,7 @@ A \textit{rational function} $f$ is a function that can be written as a quotient
 That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials.
 
 \problem{}
-Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational function.
+Solve the equation from \ref{fibo} for $F(x)$, expressing it as a rational function.
 
 \begin{solution}
 	\begin{align*}
@@ -99,8 +99,8 @@ Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational funct
 
 
 \definition{}
-\textit{Partial fraction decomposition} is an algebreic technique that works as follows: \par
+\textit{Partial fraction decomposition} is an algebraic technique that works as follows: \par
-If $p(x)$ is a polynomial and $a$ and $b$ are constants,
+If $p(x)$ is a polynomial of degree 1 and $a$ and $b$ are constants,
 we can rewrite the rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
 \begin{equation*}
 	\frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b}
@@ -131,7 +131,7 @@ find a closed-form expression for its coefficients using partial fraction decomp
 
 \problem{}
 Using problems from the introduction and \ref{pfd}, find an expression
-for the coefficients of $F(x)$ (and this, for the Fibonacci numbers).
+for the coefficients of $F(x)$ (and thus, for the Fibonacci numbers).
 
 
 \begin{solution}

src/Advanced/Generating Functions/parts/02 dice.tex

@@ -76,7 +76,7 @@ the probability that the sum of the two dice is $k$.
 
 \problem{}
 Using generating functions, find two six-sided dice whose sum has the same
-distribution as the sum of two standard six-sided dice? \par
+distribution as the sum of two standard six-sided dice. \par
 
 That is, for any integer $k$, the number if ways that the sum of the two
 nonstandard dice rolls as $k$ is equal to the number of ways the sum of

src/Advanced/Generating Functions/parts/03 coins.tex

@@ -9,7 +9,7 @@ using pennies, nickels, dimes, quarters and half-dollars?}
 \vspace{2mm}
 
 Most ways of solving this involve awkward brute-force
-approache that don't reveal anything interesting about the problem:
+approaches that don't reveal anything interesting about the problem:
 how can we change our answer if we want to make change for
 \$0.51, or \$1.05, or some other quantity?
 

src/Advanced/Generating Functions/parts/04 bonus.tex

@@ -0,0 +1,57 @@
+\section{Extra Problems}
+
+
+\problem{USAMO 1996 Problem 6}
+Determine (with proof) whether there is a subset $X$ of
+the nonnegative integers with the following property: for any nonnegative integer $n$ there is exactly
+one solution of $a + 2b = n$ with $a, b \in X$.
+(The original USAMO question asked about all integers, not just nonnegative - this is harder,
+but still approachable with generating functions.)
+
+
+\vfill
+
+\problem{IMO Shortlist 1998}
+Let $a_0, a_1, ...$ be an increasing sequence of nonnegative integers
+such that every nonnegative integer can be
+expressed uniquely in the form $a_i + 2a_j + 4a_k$,
+where $i, j, k$ are not necessarily distinct.
+
+Determine $a_1998$.
+
+
+\vfill
+
+\problem{USAMO 1986 Problem 5}
+By a partition $\pi$ of an integer $n \geq 1$, we mean here a
+representation of $n$ as a sum of one or more positive integers where the summands must be put in
+nondecreasing order. (e.g., if $n = 4$, then the partitions $\pi$ are
+$1 + 1 + 1 + 1$, $1 + 1 + 2$, $1 + 3, 2 + 2$, and $4$).
+
+
+For any partition $\pi$, define $A(\pi)$ to be the number of ones which appear in $\pi$, and define $B(\pi)$
+to be the number of distinct integers which appear in $\pi$ (e.g, if $n = 13$ and $\pi$ is the partition
+$1 + 1 + 2 + 2 + 2 + 5$, then $A(\pi) = 2$ and $B(\pi) = 3$).
+
+Show that for any fixed $n$, the sum of $A(\pi)$ over all partitions of $\pi$ of $n$ is equal to the sum of
+$B(\pi)$ over all partitions of $\pi$ of $n$.
+
+\vfill
+
+\problem{USAMO 2017 Problem 2}
+Let $m_1, m_2, ..., m_n$ be a collection of $n$ distinct positive
+integers. For any sequence of integers $A = (a_1, ..., a_n)$ and any permutation $w = w_1, ..., w_n$ of
+$m_1, ..., m_n$, define an $A$-inversion of $w$ to be a pair of entries $w_i, w_j$ with $i < j$ for which one of the
+following conditions holds:
+\begin{itemize}
+	\item $ai \geq wi > wj$
+	\item $wj > ai \geq wi$
+	\item $wi > wj > ai$
+\end{itemize}
+
+Show that for any two sequences of integers $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ and for any
+positive integer $k$, the number of permutations of $m_1, ..., m_n$ having exactly $k$ $A$-inversions is equal
+to the number of permutations of $m_1, ..., m_n$ having exactly $k$ $B$-inversions.
+(The original USAMO problem allowed the numbers $m_1, ..., m_n$ to not necessarily be distinct.)
+
+\vfill

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

@@ -230,7 +230,7 @@ The \textit{tensor product} of two vectors is defined as follows:
 That is, we take our first vector, multiply the second
 vector by each of its components, and stack the result.
 You could think of this as a generalization of scalar
-mulitiplication, where scalar mulitiplication is a
+multiplication, where scalar multiplication is a
 tensor product with a vector in $\mathbb{R}^1$:
 \begin{equation*}
 	a

src/Advanced/Mock a Mockingbird/main.tex

@@ -81,5 +81,6 @@
 	\input{parts/00 intro}
 	\input{parts/01 tmam}
 	\input{parts/02 kestrel}
+	\input{parts/03 bonus}
 
 \end{document}

src/Advanced/Mock a Mockingbird/parts/01 tmam.tex

@@ -23,7 +23,7 @@ Complete his proof.
 		\lineno{} let A           \cmnt{Let A be any any bird.}
 		\lineno{} let Cx = A(Mx)  \cmnt{Define C as the composition of A and M}
 		\lineno{} CC = A(MC)
-		\lineno{}    = A(CC)  \qed{}
+		\lineno{}    = A(CC)
 	\end{alltt}
 \end{solution}
 
@@ -46,7 +46,7 @@ Show that the laws of the forest guarantee that at least one bird is egocentric.
 		\lineno{}
 		\lineno{} ME = E     \cmnt{By definition of fondness}
 		\lineno{} ME = EE    \cmnt{By definition of M}
-		\lineno{} \thus{} EE = E  \qed{}
+		\lineno{} \thus{} EE = E
 	\end{alltt}
 \end{solution}
 
@@ -99,7 +99,7 @@ Show that if $C$ is agreeable, $A$ is agreeable.
 		\lineno{} let y so that Cy = Ey  \cmnt{Such a y must exist because C is agreeable}
 		\lineno{}
 		\lineno{} A(By) = Ey
-		\lineno{}       = D(By)  \qed{}
+		\lineno{}       = D(By)
 	\end{alltt}
 \end{solution}
 
@@ -129,6 +129,20 @@ Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$
 We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\
 Note that $x$ and $y$ may be the same bird. \\
 
+
+\problem{}
+Show that any bird that is fond of at least one bird is compatible with itself.
+
+\begin{solution}
+	\begin{alltt}
+		\lineno{} let A
+		\lineno{} let x so that Ax = x  \cmnt{A is fond of at least one other bird}
+		\lineno{} Ax = x
+	\end{alltt}
+\end{solution}
+
+\vfill
+
 \problem{}
 Show that any two birds in this forest are compatible. \\
 \begin{alltt}
@@ -144,7 +158,6 @@ Show that any two birds in this forest are compatible. \\
 \begin{solution}
 	\begin{alltt}
 		\lineno{} let A, B
-		\lineno{}
 		\lineno{} let Cx = A(Bx)  \cmnt{Composition}
 		\lineno{} let y = Cy      \cmnt{Let C be fond of y}
 		\lineno{}
@@ -152,24 +165,9 @@ Show that any two birds in this forest are compatible. \\
 		\lineno{}    = A(By)
 		\lineno{}
 		\lineno{} let x = By  \cmnt{Rename By to x}
-		\lineno{} Ax = y  \qed{}
+		\lineno{} Ax = y
 	\end{alltt}
 \end{solution}
 
-\vfill
-
-\problem{}
-Show that any bird that is fond of at least one bird is compatible with itself.
-
-\begin{solution}
-	\begin{alltt}
-		\lineno{} let A
-		\lineno{} let x so that Ax = x  \cmnt{A is fond of at least one other bird}
-		\lineno{} Ax = x  \qed{}
-	\end{alltt}
-
-	That's it.
-\end{solution}
-
 \vfill
 \pagebreak

src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex

@@ -18,7 +18,7 @@ Say $A$ is fixated on $B$. Is $A$ fond of $B$?
 	\begin{alltt}
 		\lineno{} let A
 		\lineno{} let B so that Ax = B
-		\lineno{} \thus{} AB = B \qed{}
+		\lineno{} \thus{} AB = B
 	\end{alltt}
 \end{solution}
 \vfill
@@ -39,7 +39,7 @@ Show that an egocentric Kestrel is hopelessly egocentric.
 	\begin{alltt}
 		\lineno{} KK = K
 		\lineno{} \thus{} (KK)y = K  \cmnt{By definition of the Kestrel}
-		\lineno{} \thus{} Ky = K \qed{}  \cmnt{By 01}
+		\lineno{} \thus{} Ky = K     \cmnt{By 01}
 	\end{alltt}
 \end{solution}
 
@@ -64,7 +64,7 @@ Given the Law of Composition and the Law of the Mockingbird, show that at least
 	\begin{alltt}
 		\lineno{} let A so that KA = A   \cmnt{Any bird is fond of at least one bird}
 		\lineno{} (KA)y = y              \cmnt{By definition of the kestrel}
-		\lineno{} \thus{} Ay = A \qed{}  \cmnt{By 01}
+		\lineno{} \thus{} Ay = A              \cmnt{By 01}
 	\end{alltt}
 \end{solution}
 
@@ -90,7 +90,7 @@ Show that $Kx = Ky \implies x = y$.
 		\lineno{} (Kx)z = x
 		\lineno{} (Ky)z = y
 		\lineno{}
-		\lineno{} \thus{} x = (Kx)z = (Ky)z = y \qed{}
+		\lineno{} \thus{} x = (Kx)z = (Ky)z = y
 	\end{alltt}
 \end{solution}
 
@@ -128,7 +128,7 @@ An egocentric Kestrel must be extremely lonely. Why is this?
 		\lineno{} Ky = K
 		\lineno{} Kx = Ky
 		\lineno{} x = y for all x, y  \cmnt{By \ref{leftcancel}}
-		\lineno{} x = y = K \qed{}  \cmnt{By 10, and since K exists}
+		\lineno{} x = y = K  \cmnt{By 10, and since K exists}
 	\end{alltt}
 \end{solution}
 

src/Advanced/Mock a Mockingbird/parts/03 bonus.tex

@@ -0,0 +1,102 @@
+\section{Bonus Problems}
+
+\definition{}
+The identity bird has sometimes been maligned, owing to
+the fact that whatever bird x you call to $I$, all $I$ does is to echo
+$x$ back to you.
+
+\vspace{2mm}
+
+Superficially, the bird $I$ appears to have no intelligence or imagination; all it can do is repeat what it hears.
+For this reason, in the past, thoughtless students of ornithology
+referred to it as the idiot bird. However, a more profound or-
+nithologist once studied the situation in great depth and dis-
+covered that the identity bird is in fact highly intelligent! The
+real reason for its apparently unimaginative behavior is that it
+has an unusually large heart and hence is fond of every bird!
+When you call $x$ to $I$, the reason it responds by calling back $x$
+is not that it can't think of anything else; it's just that it wants
+you to know that it is fond of $x$!
+
+\vspace{2mm}
+
+Since an identity bird is fond of every bird, then it is also
+fond of itself, so every identity bird is egocentric. However,
+its egocentricity doesn't mean that it is any more fond of itself
+than of any other bird!.
+
+
+\problem{}
+The laws of the forest no longer apply.
+
+Suppose we are told that the forest contains an identity bird
+$I$ and that $I$ is agreeable. \
+Does it follow that every bird must be fond of at least one bird?
+
+\vfill
+
+
+\problem{}
+Suppose we are told that there is an identity bird $I$ and that
+every bird is fond of at least one bird. \
+Does it necessarily follow that $I$ is agreeable?
+
+\vfill
+\pagebreak
+
+
+\problem{}
+Suppose we are told that there is an identity bird $I$, but we are
+not told whether $I$ is agreeable or not.
+
+However, we are told that every pair of birds is compatible. \
+Which of the following conclusiens can be validly drawn?
+
+\begin{itemize}
+	\item Every bird is fond of at least one bird
+	\item $I$ is agreeable.
+\end{itemize}
+
+\vfill
+
+\problem{}
+The identity bird $I$, though egocentric, is in general not hope-
+lessly egocentric. Indeed, if there were a hopelessly egocentric
+identity bird, the situation would be quite sad. Why?
+
+\vfill
+
+\definition{}
+A bird $L$ is called a lark if the following
+holds for any birds $x$ and $y$:
+
+\[
+	(Lx)y = x(yy)
+\]
+
+\problem{}
+Prove that if the forest contains a lark $L$ and an identity bird
+$I$, then it must also contain a mockingbird $M$.
+
+\vfill
+\pagebreak
+
+
+\problem{}
+Why is a hopelessly egocentric lark unusually attractive?
+
+\vfill
+
+
+\problem{}
+Assuming that no bird can be both a lark and a kestrel---as
+any ornithologist knows!---prove that it is impossible for a
+lark to be fond of a kestrel.
+
+\vfill
+
+\problem{}
+It might happen, however, that a kestrel is fond of a lark. \par
+Show that in this case, \textit{every} bird is fond of the lark.
+
+\vfill

src/Advanced/Retrograde Analysis/parts/02 easy.tex

@@ -127,10 +127,6 @@ Mate the king in one move. \par
 \pagebreak
 
 
-
-
-
-
 % Sherlock, a question of survival
 \problem{An empty board}
 \difficulty{2}{5}
@@ -161,42 +157,6 @@ There is one more piece on the board, which isn't shown. What color square does
 \pagebreak
 
 
-% Sherlock, another monochromatic
-\problem{The knight's grave}
-\difficulty{3}{5}
-
-In the game below, no pieces have moved from a black square to a white square, or from a white square to a black square.
-The white king has made less than fourteen moves. \par
-Use this information to show that a pawn was promoted. \par
-
-% spell:off
-\manyboards{
-	ke8,
-	Pb2,Pd2,
-	Ke1
-}
-% spell:on
-
-\begin{solution}
-	Knights always move to a different colored square, so all four missing knights must have been captured on their home square.
-	What pieces captured them?
-
-	\vspace{2mm}
-
-	We can easily account for the white knights and the black knight on G8, but who could've captured the knight from B8?
-	The only white pieces that can move to black squares are pawns, the Bishop (which is trapped on C1), the rook (which is stuck on column A and row 1), or the king (which would need at least 14 moves to do so).
-
-	\vspace{2mm}
-
-	If this knight was captured by a pawn, that pawn would be immediately promoted. If it was captured by a piece that wasn't a pawn, that piece must be a promoted pawn.
-\end{solution}
-
-\vfill
-\pagebreak
-
-
-
-
 
 
 % Arabian Knights, intro (given with solution)
@@ -373,3 +333,121 @@ Which bishop was it, and what did it capture? \par
 
 \vfill
 \pagebreak
+
+
+% Sherlock, appendix
+\problem{Moriarty's first}
+\difficulty{3}{5}
+
+
+No captures have been made in the last four moves. \par
+It is White's move. What was the previous move?
+
+% spell:off
+\manyboards{
+	Bc8,
+	pg6,
+	Pg5,kh5,
+	Pd4,Qg4,Bh4,
+	pd3,
+	Pd2,Be2,Bg2,
+	Nc1,rd1,Ne1,Kf1,Qg1,Rh1
+}
+% spell:on
+
+\begin{solution}
+	To see what the position was four moves ago,
+	move the Black queen to E4, the knight on E1 to F3,
+	the Black bishop to E1, and the White bishop on C8 to H3.
+
+	The following sequence of moves brought the game to the present position:
+
+	\begin{itemize}
+		\item bishop to c8, check
+		\item bishop to h4, check
+		\item knight to e1, check
+		\item queen to g4.
+	\end{itemize}
+
+	This is the only way the present position could have arisen,
+	so Black's last move was with the queen from E4 to G4.
+
+
+	Try any other last move, and you will find it impossible to play back three more moves.
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+% Sherlock, appendix
+\problem{Moriarty's second}
+\difficulty{3}{5}
+
+
+Neither the White king nor queen has moved
+during the last five moves, nor has any piece
+been captured during that time.
+What was the last move?
+
+% spell:off
+\manyboards{
+	kh8,
+	Kg6,Bh6,
+	pa4,
+	Qa2
+}
+% spell:on
+
+\begin{solution}
+	Put the Black pawn on A7, the Black king on G8, remove the
+	White bishop, and put a White pawn on d5; this was the position
+	five moves ago. The following sequence of moves brought the
+	game to its present position:
+
+	\begin{itemize}
+		\item White: P-d6
+		\item Black: K-h8
+		\item White: P-d7
+		\item Black: P-a6
+		\item White: P-d8 = B
+		\item Black: P-a5
+		\item White: B-g5
+		\item Black: P-a4
+		\item White: B-h6
+	\end{itemize}
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+% Sherlock, appendix
+\problem{Moriarty's third}
+\difficulty{3}{5}
+
+
+No pawn has moved, nor has any piece been
+captured in the last five moves. \par
+The Black king has been accidentally
+knocked off the board. \par
+On what square should he stand?
+
+% spell:off
+\manyboards{
+	rh8,
+	pa7,pb7,pc7,pd7,pe7,Kf7,pg7,Ph7,
+	Pg6,
+	na2
+}
+% spell:on
+
+\begin{solution}
+	The only way to avoid a retrograde stalemate for White is by
+	placing the Black king on C8. Black's last move was with
+	the rook from D8, White's move before that was with his
+	king from G8, and Black's move before that was to castle.
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Retrograde Analysis/parts/03 medium.tex

@@ -169,16 +169,156 @@ White to move. Which side of the board did each color start on? \par
 
 
 
+% Sherlock, another monochromatic
+\problem{Monochromatic}
+\difficulty{4}{5}
+
+In the game below, no pieces have moved from a black square to a white square or from a white square to a black square.
+The white king has made fewer than fourteen moves. \par
+Use this information to show that a pawn was promoted. \par
+
+% spell:off
+\manyboards{
+	ke8,
+	Pb2,Pd2,
+	Ke1
+}
+% spell:on
+
+\begin{solution}
+	Knights always move to a different colored square, so all four missing knights must have been captured on their home square.
+	What pieces captured them?
+
+	\vspace{2mm}
+
+	We can easily account for the white knights and the black knight on G8, but who could've captured the knight from B8?
+	The only white pieces that can move to black squares are pawns, the Bishop (which is trapped on C1), the rook (which is stuck on column A and row 1), or the king (which would need at least 14 moves to do so).
+
+	\vspace{2mm}
+
+	If this knight was captured by a pawn, that pawn would be immediately promoted. If it was captured by a piece that wasn't a pawn, that piece must be a promoted pawn.
+\end{solution}
+
+\vfill
+\pagebreak
 
 
 
 
+% Sherlock, another question of location
+\problem{Superposition}
+\difficulty{4}{5}
+
+A white pawn is missing; it is either on F2 or G2. \par
+Where is it?
+
+% spell:off
+\manyboards{
+	ke8,rh8,
+	pa7,pf7,pg7,
+	pa6,pb6,
+	pb5,
+	Pa4,Pb4,Pc4,
+	pa3,
+	Pa2,Pb2,
+	Ke1
+}
+% spell:on
+
+\vfill
+\pagebreak
+
+% Sherlock, another question of location
+\problem{Possibility}
+\difficulty{4}{5}
+
+Show that black can castle to either side. \par
+We know the following:
+
+\begin{itemize}
+	\item White started the game missing one rook.
+	\item White has not moved either knight
+	\item No promotions have been made
+	\item White's last move was from E2 to E4.
+\end{itemize}
+
+% spell:off
+\manyboards{
+	ra8,ke8,rh8,
+	pa7,bb7,pc7,pd7,pf7,pg7,ph7,
+	nc6,nh6,
+	pe5,qg5,
+	bb4,Pe4,
+	Pb2,Pc2,Pd2,Pf2,Pg2,Ph2,
+	Nb1,Bc1,Qd1,Ke1,Bf1,Ng1,Rh1
+}
+% spell:on
+
+
+\vfill
+\pagebreak
+
+
+
+
+
+% Sherlock, little exercise 2
+\problem{Kidnapping}
+\difficulty{4}{5}
+
+On which square was the White queen captured?. \par
+
+% spell:off
+\manyboards{
+	ra8,qd8,ke8,ng8,rh8,
+	pa7,pb7,pc7,pe7,pf7,ph7,
+	nc6,pe6,ph6,
+	Pb3,
+	Na2,Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2,
+	Ra1,Ke1,Rh1
+}
+% spell:on
+
+\begin{solution}
+	White is missing a queen, both bishops, and one knight. \par
+	The black pawns on E6 and H6 account for two captures.
+
+	\vspace{2mm}
+
+	Neither white bishop could've been captured by these pawns,
+	since both are trapped by their pawns. Thus, these black pawns must have captured a queen and a knight.
+
+	\vspace{4mm}
+
+	The white pawn on B3 must have captured a black bishop. \par
+	The white queen got onto the board through A2. \par
+	Therefore, the pawn on B3 made its capture before the queen escaped,
+	and the black bishop was captured before the white queen.
+
+	\vspace{4mm}
+
+	Similarly, the bishop from C8 must have been
+	captured on B3 after the capture on E6, since it
+	got on the board through D7.
+
+
+	\vspace{4mm}
+
+	The capture on E6 was made before the capture on B3 (black bishop),
+	which was made before the white queen was captured.
+	Therefore, the white queen was not captured on E6, and must
+	have been lost on H6.
+\end{solution}
+
+\vfill
+\pagebreak
+
 
 
 
 % Arabian Knights 4
 \problem{A missing piece}
-\difficulty{4}{5}
+\difficulty{6}{8}
 
 
 There is a piece at G4, marked with a $\odot$. \par
@@ -251,7 +391,7 @@ What is it, and what is its color? \par
 	\textbf{Part 4:}
 
 	The promoted black bishop on H2 must have been promoted on G1. The pawn which was promoted must have come from G7,
-	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accouted for).
+	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accounted for).
 	The Pawn from E7 has promoted to the bishop on A2.
 
 	What happened was this: the white pawn from G2 made its capture on H3 while the pawn on G3 was still on H2. This allowed the black pawn

src/Advanced/Retrograde Analysis/parts/04 hard.tex

@@ -2,7 +2,7 @@
 
 % Arabian Knights 5
 \problem{The hidden castle}
-\difficulty{7}{7}
+\difficulty{8}{8}
 
 There is a white castle hidden on this board. Where is it? \par
 None of the royalty has moved or been under attack. \par
@@ -30,7 +30,7 @@ None of the royalty has moved or been under attack. \par
 
 % Arabian Knights 6
 \problem{Who moved last?}
-\difficulty{7}{7}
+\difficulty{8}{8}
 
 After many moves of chess, the board looks as follows. \par
 Who moved last? \par
@@ -58,7 +58,7 @@ Who moved last? \par
 
 % Arabian Knights 3
 \problem{The king in disguise}<kingdisguise>
-\difficulty{7}{7}
+\difficulty{8}{8}
 
 The white king is exploring his kingdom under a disguise. He could look like any piece of any color.\par
 Show that he must be on C7.
@@ -119,7 +119,7 @@ Show that he must be on C7.
 
 % Arabian Knights 3
 \problem{The king in disguise once more}
-\difficulty{5}{7}
+\difficulty{5}{8}
 
 The white king is again exploring his kingdom, now under a different disguise. Where is he? \par
 \hint{\say{different disguise} implies that the white king looks like a different piece!}

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 $n$ iff $gcd(x, n) = 1$
+
+#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 `312` 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 $[4321]$
+  - $[321][213][231]$ 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 permitation $f$ 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, and 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 $(a, a+1)(a+1, b)(a, a+1)$.
+
+#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 +1,139 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+#import "../macros.typ": *
+
+= Groups (review)
+
+#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 $n$.
+
+- $ZZ_n^times$ is the set of integers mod $n$ with multiplicative inverses. \
+  In other words, it is the set of integers smaller than $n$ and coprime to $n$.#footnote[We proved this in another handout, but you may take it as fact here.] \
+  For example, $ZZ_12^times = {1, 5, 7, 11}$.
+
+#problem()
+What are the elements of $S_3$? #hint[Use cycle notation] \
+How about $ZZ_17^times$?
+
+#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.
+]
+
+#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.]
+
+#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[
+  Only one: the trivial group. The inverse of a generator is also a generator!
+]
+
+#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.
@@ -7,17 +7,15 @@
 #let tm = sym.times.circle
 
 #let graphgrid(inner_content) = {
-  align(
+  align(center, box(inset: 3mm, cetz.canvas({
-    center,
-    box(
-      inset: 3mm,
-      cetz.canvas({
     import cetz.draw: *
     let x = 5.25
 
     grid(
-          (0, 0), (x, x), step: 0.75,
+      (0, 0),
-          stroke: luma(100) + 0.3mm
+      (x, x),
+      step: 0.75,
+      stroke: luma(100) + 0.3mm,
     )
 
     if (inner_content != none) {
@@ -33,9 +31,7 @@
       (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(
+  align(center, box(
-    center,
-    box(
     inset: 3mm,
     eqn,
-    ),
+  ))
-  )
 }
 
 #let dotline(a, b) = {
-  cetz.draw.line(
+  cetz.draw.line(a, b, stroke: (
-    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(
+      table.header([$#tp$], [$1$], [$2$], [$3$], [$4$], [$#sym.infinity$]),
-        [$#tp$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-        [$#sym.infinity$],
-      ),
 
       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(
+      table.header([$#tm$], [$0$], [$1$], [$2$], [$3$], [$4$]),
-        [$#tm$],
-        [$0$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-      ),
 
       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(
+      table.header([$#tp$], [$1$], [$2$], [$3$], [$4$], [$#sym.infinity$]),
-        [$#tp$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-        [$#sym.infinity$],
-      ),
 
       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(
+      table.header([$#tm$], [$0$], [$1$], [$2$], [$3$], [$4$]),
-        [$#tm$],
-        [$0$],
-        [$1$],
-        [$2$],
-        [$3$],
-        [$4$],
-      ),
 
       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 #tp 2)(x #tp 3) & = x^2 #tp 2x #tp 3x #tp (2 #tm 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 (2 #tp 3)x #tp (2 #tm 3) \
+                       & = x^2 #tp 2x #tp 5
-    &= 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(
+#align(center, box(
-  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(
+#align(center, box(
-  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(
+#align(center, box(
-  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($
+  #eqnbox(
+    $
       f(x) = -2(x^2 #tp 2x #tp 10)
-  $)
+    $,
+  )
 
 
   by the same process as the previous problem, we get
-  #eqnbox($
+  #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(
+#solution(eqnbox(
-  eqnbox($
+  $
     f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2
-  $),
+  $,
-)
+))
 
 #v(1fr)
 
@@ -263,8 +258,7 @@ Graph $f(x) = 2x^2 #tp 4x #tp 4$.
 
 #if_no_solutions(graphgrid(none))
 
-#solution(
+#solution(graphgrid({
-  graphgrid({
   import cetz.draw: *
   let step = 0.75
 
@@ -278,8 +272,7 @@ Graph $f(x) = 2x^2 #tp 4x #tp 4$.
     (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)
+    accent(f, macron)(x) & = a x^2 #tp ((a+c)/2)x #tp c \
-    &= a x^2 #tp ((a+c)/2)x #tp c \
+                         & = a(x #tp (c-a)/2)^2
-    &= 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(
+#align(center, box(
-  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
+    A_j & = min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
-    &= 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) )
-    &= min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
   $
 
   #v(2mm)

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

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = Wallpaper Symmetries
 
@@ -12,7 +12,9 @@ There are four classes of Euclidean isometries:
 - reflections
 - rotations
 - glide reflections
-#note([We can prove there are no others, but this is beyond the scope of this handout.]) \
+#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) = {

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

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = Mirror Symmetry
 

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

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = Rotational Symmetry
 

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

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 #let pat(img, sol) = {
   problem()

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

@@ -1,5 +1,5 @@
 #import "@local/handout:0.1.0": *
-#import "@preview/cetz:0.3.1"
+#import "@preview/cetz:0.4.2"
 
 = The Signature-Cost Theorem
 
@@ -7,9 +7,7 @@
 First, we'll associate a _cost_ to each type of symmetry in orbifold notation:
 
 #v(4mm)
-#align(
+#align(center, table(
-  center,
-  table(
   stroke: (1pt, 1pt),
   align: center,
   columns: (auto, auto, auto, auto),
@@ -22,8 +20,7 @@ First, we'll associate a _cost_ to each type of symmetry in orbifold notation:
   [$(n-1) / n$],
   [#sym.convolve`n`],
   [$(n-1) / (2n)$],
-  ),
+))
-)
 
 
 We then calculate the total "cost" of a signature by adding up the costs of each component.

src/Intermediate/Instant Insanity/main.tex

@@ -331,7 +331,7 @@
 	representing all four cubes. \\
 
 	\begin{center} \begin{small}
-	\begin{tikzpicture} \label{pic:II_comfiguration}
+	\begin{tikzpicture} \label{pic:II_configuration}
 		\filldraw [blue] (0,5) -- (1,5) -- (1,6) --
 		(0,6) -- (0,5);
 		\draw [line width = 1.5pt] (0,5) --

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_1, c_1 & = "HA"(a, b) \
-    s_2, c_2 &= "HA"(s_1, c_"in") \
+    s_2, c_2 & = "HA"(s_1, c_"in") \
-    s_"out" &= s_2 \
+     s_"out" & = s_2 \
-    c_"out" &= "OR"(c_1, c_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],
@@ -75,9 +75,7 @@ How does your strategy change? \
 #if extra_boards {
   pagebreak()
 
-  align(
+  align(center, grid(
-    center,
-    grid(
     stroke: none,
     align: center,
     columns: (1fr, 1fr),
@@ -85,6 +83,5 @@ How does your strategy change? \
     btt(0.35), btt(0.35),
     btt(0.35), btt(0.35),
     btt(0.35), btt(0.35),
-    ),
+  ))
-  )
 }

src/Warm-Ups/Bugs/main.typ

@@ -0,0 +1,11 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Bugs on a Log],
+  by: "Mark",
+)
+
+#problem()
+2013 bugs are on a meter-long line. Each walks to the left or right at a constant speed. \
+If two bugs meet, both turn around and continue walking in opposite directions. \
+What is the longest time it could take for all the bugs to walk off the end of the log?

src/Warm-Ups/Bugs/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Bugs on a Log"
+
+[publish]
+handout = true
+solutions = true

src/Warm-Ups/Cosa Nostra/main.typ

@@ -0,0 +1,50 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Cosa Nostra],
+  by: "Mark",
+)
+
+
+#problem()
+There are 36 gangsters in a certain district of Chicago.
+Some pairs of gangsters have feuds.
+
+- Each gangster is part of at least one outfit, and no two outfits share the same members.
+- If two gangsters are both in one outfit, there is no feud between them.
+- A gangster that is not in a certain outfit must have a feud with at least one if its members.
+
+What is the maximum number of outfits that can exist in this district?
+
+
+
+#solution[
+  *Definition:* Let the _authority_ of a gangster be the number of oufits they are a part of.
+
+  #v(5mm)
+
+  *Lemma:* Say two gangsters have a feud. Label them $x$ and $y$ so that $#text(`authority`) (x) > #text(`authority`) (y)$. \
+  Then, replacing $y$ with a clone of $x$ will strictly increase the number of outfits. \
+  If $#text(`authority`) (x) = #text(`authority`) (y)$, replacing $y$ with $x$ will not change the number of outfits.
+
+  #v(5mm)
+
+  *Proof:*
+  Let $A$ be the set of outfits that $y$ is a part of, and $B$ its complement (that is, all outfits that $a$ is _not_ a part of). If we delete $y$...
+  - all outfits in $B$ remain outfits.
+  - some outfits in $A$ cease to be outfits (as they are no longer maximal)
+
+  Also, no new outfits are formed. If a new outfit $o$ contains any enemies of $y$, it existed previously and is a member of $B$. If $o$ contains no enemies of $y$, it must have contained $y$ prior to deletion, and is thus a member of $A$. Therefore, the number of outfits is reduced by at most `authority(y)` when $y$ is deleted.
+
+  If we add a clone of $x$ after deleting $y$ (this clone has a feud with $x$), All previous outfits remain outfits, and `authority(x)` new outfits are created.
+  Therefore, replacing $y$ with a clone of $x$ strictly increases the number of outfits that exist.
+  We thus conclude that in the maximal case, all pairs of feuding gangsters have equal authority.
+
+  #v(5mm)
+
+  *Solution:* Consider an arbitrary gangster $g$. By the previous lemma, we can replace all gangsters $g$ has a feud with clones of itself. Repeat this for all gangsters, and we are left with groups of feuding gangsters who are friends with everyone outside their group. The total number of outfits is the product of the sizes of these groups.
+
+  #v(5mm)
+
+  This problem is now equivalent to the "Partition Products" warm-up. We want the list of numbers whose sum is 36 and whose product is maximal. The solution is to form 12 groups of three gangsters for a total of $3^12$ outfits.
+]

src/Warm-Ups/Cosa Nostra/meta.toml

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

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(
@@ -43,10 +43,7 @@ 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(
+#solution(align(center, cetz.canvas({
-  align(
-    center,
-    cetz.canvas({
   import cetz.draw: *
 
   let s = 0.8 // Scale
@@ -95,9 +92,7 @@ On average, which die beats each of the others? Draw a diagram.
   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

@@ -1,10 +1,9 @@
 #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?],
+  title: [Warm-Up: The Painting],
   by: "Mark",
-  subtitle: "Based on a true story.",
 )
 
 #problem()
@@ -13,14 +12,13 @@ Hang the painting on two nails so that if either is removed, the painting falls.
 
 #v(2mm)
 You may detach the string as you hang the painting, but it must be re-attached once you're done. \
-#hint[The solution to this problem isn't a "think outside the box" trick, it's a clever wrapping of the string.]
+
+The solution to this problem isn't a trick, it's a clever wrapping of the string.
 
 
 #v(2mm)
 
-#align(
+#align(center, cetz.canvas({
-  center,
-  cetz.canvas({
   import cetz.draw: *
   let s = 2.5
 
@@ -69,8 +67,7 @@ You may detach the string as you hang the painting, but it must be re-attached o
   )
 
   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

@@ -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],
@@ -78,10 +78,7 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
 
         let i = 1
         for p in pts {
-          circle(
+          circle(p, radius: radius * s, fill: if i == 1 {
-            p,
-            radius: radius * s,
-            fill: if i == 1 {
             ored
           } else if i == 2 {
             ogreen
@@ -89,12 +86,9 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
             oorange
           } else if i == 4 {
             oblue
-            } else { white },
+          } else { white })
-          )
 
-          content(
+          content(p, text(
-            p,
-            text(
             fill: if i <= 4 {
               white
             } else {
@@ -102,8 +96,7 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
             },
             size: t,
             [*#i*],
-            ),
+          ))
-          )
           i = i + 1
         }
       }),
@@ -118,10 +111,7 @@ 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(
+          circle(p, radius: radius * s, fill: if l == 1 {
-            p,
-            radius: radius * s,
-            fill: if l == 1 {
             ored
           } else if l == 2 {
             ogreen
@@ -129,12 +119,9 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
             oorange
           } else if l == 4 {
             oblue
-            } else { white },
+          } else { white })
-          )
 
-          content(
+          content(p, text(
-            p,
-            text(
             fill: if l <= 4 {
               white
             } else {
@@ -142,8 +129,7 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
             },
             size: t,
             [*#l*],
-            ),
+          ))
-          )
           i = i + 1
         }
       }),
@@ -158,10 +144,7 @@ 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(
+          circle(p, radius: radius * s, fill: if l == 1 {
-            p,
-            radius: radius * s,
-            fill: if l == 1 {
             oblue
           } else if l == 2 {
             oorange
@@ -169,12 +152,9 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
             ored
           } else if l == 4 {
             ogreen
-            } else { white },
+          } else { white })
-          )
 
-          content(
+          content(p, text(
-            p,
-            text(
             fill: if l <= 4 {
               white
             } else {
@@ -182,8 +162,7 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
             },
             size: t,
             [*#l*],
-            ),
+          ))
-          )
           i = i + 1
         }
       }),

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?],
@@ -24,9 +24,7 @@ respecting the order of operations $[and, times, div, +, -]$.
 
 #v(2mm)
 
-#align(
+#align(center, cetz.canvas({
-  center,
-  cetz.canvas({
   import cetz.draw: *
 
   // spell:off
@@ -70,5 +68,4 @@ respecting the order of operations $[and, times, div, +, -]$.
   line("bb", "bba")
   line("bb", "bbb")
   // spell:on
-  }),
+}))
-)

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

@@ -1,14 +1,12 @@
 #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(
+#let ttt = align(center, cetz.canvas({
-  center,
-  cetz.canvas({
   import cetz.draw: *
   let s = 0.7 // scale
 
@@ -17,8 +15,7 @@
   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()