WIP ODD DICE

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

@@ -1,132 +0,0 @@
-\documentclass[
-	nosolutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\usepackage{tikz}
-\usetikzlibrary{arrows.meta}
-\usetikzlibrary{shapes.geometric}
-
-% We put nodes in a separate layer, so we can
-% slightly overlap with paths for a perfect fit
-\pgfdeclarelayer{nodes}
-\pgfdeclarelayer{path}
-\pgfsetlayers{main,nodes}
-
-% Layer settings
-\tikzset{
-	% Layer hack, lets us write
-	% later = * in scopes.
-	layer/.style = {
-		execute at begin scope={\pgfonlayer{#1}},
-		execute at end scope={\endpgfonlayer}
-	},
-	%
-	% Arrowhead tweak
-	>={Latex[ width=2mm, length=2mm ]},
-	%
-	% Nodes
-	main/.style = {
-		draw,
-		circle,
-		fill = white,
-		line width = 0.35mm
-	}
-}
-
-\title{Warm Up: Odd Dice}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-
-	We say a set of dice $\{A, B, C\}$ is \textit{nontransitive}
-	if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
-	In other words, we get a counterintuitive \say{rock - paper - scissors} effect.
-
-	\vspace{2mm}
-
-	Create a set of nontransitive six-sided dice. \par
-	\hint{All sides should be numbered with positive integers less than 10.}
-
-	\begin{solution}
-		One possible set can be numbered as follows:
-		\begin{itemize}
-			\item Die $A$: $2, 2, 4, 4, 9, 9$
-			\item Die $B$: $1, 1, 6, 6, 8, 8$
-			\item Die $C$: $3, 3, 5, 5, 7, 7$
-		\end{itemize}
-
-		\vspace{4mm}
-
-		Another solution is below:
-		\begin{itemize}
-			\item Die $A$: $3, 3, 3, 3, 3, 6$
-			\item Die $B$: $2, 2, 2, 5, 5, 5$
-			\item Die $C$: $1, 4, 4, 4, 4, 4$
-		\end{itemize}
-	\end{solution}
-
-	\vfill
-
-	\problem{}
-	Now, consider the set of six-sided dice below:
-	\begin{itemize}
-		\item Die $A$: $4, 4, 4, 4, 4, 9$
-		\item Die $B$: $3, 3, 3, 3, 8, 8$
-		\item Die $C$: $2, 2, 2, 7, 7, 7$
-		\item Die $D$: $1, 1, 6, 6, 6, 6$
-		\item Die $E$: $0, 5, 5, 5, 5, 5$
-	\end{itemize}
-	On average, which die beats each of the others? Draw a graph. \par
-
-	\begin{solution}
-		\begin{center}
-		\begin{tikzpicture}[scale = 0.5]
-			\begin{scope}[layer = nodes]
-				\node[main] (a) at (-2, 0.2) {$a$};
-				\node[main] (b) at (0, 2) {$b$};
-				\node[main] (c) at (2, 0.2) {$c$};
-				\node[main] (d) at (1, -2) {$d$};
-				\node[main] (e) at (-1, -2) {$e$};
-			\end{scope}
-
-			\draw[->]
-				(a) edge (b)
-				(b) edge (c)
-				(c) edge (d)
-				(d) edge (e)
-				(e) edge (a)
-
-				(a) edge (c)
-				(b) edge (d)
-				(c) edge (e)
-				(d) edge (a)
-				(e) edge (b)
-			;
-		\end{tikzpicture}
-		\end{center}
-	\end{solution}
-
-	\vfill
-
-	Now, say we roll each die twice. What happens to the graph above?
-
-	\begin{solution}
-		The direction of each edge is reversed!
-	\end{solution}
-
-	\vfill
-	\pagebreak
-
-\end{document}

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

@@ -0,0 +1,87 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Odd Dice],
+  by: "Mark",
+)
+
+#problem()
+We say a set of dice ${A, B, C}$ is _nontransitive_
+if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
+In other words, we get a counterintuitive "rock - paper - scissors" effect.
+
+#v(2mm)
+
+Create a set of nontransitive six-sided dice. \
+#hint([All sides should be numbered with positive integers less than 10.])
+
+#solution([
+  One possible set can be numbered as follows:
+  - Die $A$: $2, 2, 4, 4, 9, 9$
+  - Die $B$: $1, 1, 6, 6, 8, 8$
+  - Die $C$: $3, 3, 5, 5, 7, 7$
+
+  #v(4mm)
+
+  Another solution is below:
+  - Die $A$: $3, 3, 3, 3, 3, 6$
+  - Die $B$: $2, 2, 2, 5, 5, 5$
+  - Die $C$: $1, 4, 4, 4, 4, 4$
+
+])
+
+#v(1fr)
+
+#problem()
+Now, consider the set of six-sided dice below:
+- Die $A$: $4, 4, 4, 4, 4, 9$
+- Die $B$: $3, 3, 3, 3, 8, 8$
+- Die $C$: $2, 2, 2, 7, 7, 7$
+- Die $D$: $1, 1, 6, 6, 6, 6$
+- Die $E$: $0, 5, 5, 5, 5, 5$
+On average, which die beats each of the others? Draw a diagram.
+
+#solution([
+  /*
+  \begin{tikzpicture}[scale = 0.5]
+    \begin{scope}[layer = nodes]
+      \node[main] (a) at (-2, 0.2) {$a$};
+      \node[main] (b) at (0, 2) {$b$};
+      \node[main] (c) at (2, 0.2) {$c$};
+      \node[main] (d) at (1, -2) {$d$};
+      \node[main] (e) at (-1, -2) {$e$};
+    \end{scope}
+
+    \draw[->]
+      (a) edge (b)
+      (b) edge (c)
+      (c) edge (d)
+      (d) edge (e)
+      (e) edge (a)
+
+      (a) edge (c)
+      (b) edge (d)
+      (c) edge (e)
+      (d) edge (a)
+      (e) edge (b)
+    ;
+  \end{tikzpicture}
+  */
+])
+
+#v(1fr)
+
+#problem()
+Now, say we roll each die twice. What happens to the graph fromE the previous problem?
+
+#solution([
+  The direction of each edge is reversed!
+])
+
+#v(1fr)