TMP

Reviewed-on: #8
Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>

.gitignore

@@ -3,6 +3,7 @@ venv
 __pycache__
 *.ignore
 .mypy_cache
+.DS_Store
 
 # Output files
 /output
@@ -10,7 +11,7 @@ __pycache__
 *.pdf
 
 # TeX build files
-*.synctex.gz*
+*.synctex*
 *.latexmk
 *.aux
 *.out

.vscode/settings.json

@@ -1,5 +1,6 @@
 {
 	"latex-workshop.latex.recipe.default": "latexmk (xelatex)",
 	"tinymist.formatterPrintWidth": 80,
-	"tinymist.typstExtraArgs": ["--package-path=./lib/typst"]
+	"tinymist.typstExtraArgs": ["--package-path=./lib/typst"],
+	"tinymist.formatterMode": "typstyle"
 }

README.md

@@ -0,0 +1,128 @@
+[tinymist]: https://marketplace.visualstudio.com/items?itemName=myriad-dreamin.tinymist
+[latex-workshop]: https://marketplace.visualstudio.com/items?itemName=James-Yu.latex-workshop
+[CC BY-NC-SA 4.0]: https://creativecommons.org/licenses/by-nc-sa/4.0
+[betalupi.com/handouts]: https://static.betalupi.com/ormc
+[ORMC]: https://circles.math.ucla.edu/circles/
+[Overleaf]: https://overleaf.com
+[Typst.app]: https://typst.app
+[vscode]: https://code.visualstudio.com
+[vscodium]: https://vscodium.com
+[homebrew]: https://brew.sh
+
+# Mark's Handout Library
+
+This is a collection of math circle handouts that I (and many others) have written. \
+They are used regularly at the [ORMC].
+
+For more information, visit [betalupi.com/handouts]. \
+The latest version of each handout is available at that page.
+
+## License
+
+Unless otherwise stated, all documents in this repository are licensed under [CC BY-NC-SA 4.0]. \
+Each document has its own authors. See `meta.toml` in each project directory for details.
+
+By submitting or editing a handout in this repository, you agree to release it under this license.
+
+## 🛠️ Contributing
+
+If you want to use one of these handouts for a class, see [`betalupi.com/handouts`](https://betalupi.com/handouts). \
+You only need to read this section if you want to edit these handouts.
+
+### Setup
+
+Use git to clone this repository, then open the root folder in [vscode] or [vscodium].
+
+We use the [latex-workshop] and [tinymist] extensions. Install them before continuing.
+[`./vscode/settings.json`](./vscode/settings.json) will automatically configure them to work with this repository. \
+You may need to install texlive and typst:
+
+-   If you use Linux, you'll figure it out.
+-   On macos, use [homebrew]: \
+     `brew install texlive typst typstyle`
+-   On Windows, I don't know. I may write instructions later.
+
+### Editing
+
+This repository is organized as follows:
+
+-   All handouts are in [`./src`](./src). \
+    Every handout is stored in its own directory, even if it only consists of one file. \
+    Handouts are organized by group (see [betalupi.com/handouts] for details).
+
+-   Packages are stored in [`./lib`](./lib) \
+    You shouldn't need to modify any library files, but you may want to read them to see how they work.
+
+-   [`./tools`](./tools) contains build scripts, [`./.github`](./.github) configures automation. \
+    You can ignore everything in these directories.
+
+All handouts in this repository are based on `ormc_handout.cls` or `handout@0.1.0`.
+
+-   If you're using Typst (preferred), read [`docs-typst.md`](./docs-typst.md)
+-   If you're still using LaTeX, read [`docs-latex.md`](./docs-latex.md).
+
+### Metadata
+
+Every handout directory should contain a file called `meta.toml` with the following contents:
+
+```toml
+# This is a sample `meta.toml`.
+# A copy of this file should exist in every handout directory.
+# All keys are required.
+
+[metadata]
+title = "title of this handout"
+
+
+[publish]
+# Should we publish this handout?
+# If `false`, no part of this handout is published.
+handout = true
+
+# Should we publish an "instructor's" version of this handout?
+# This key has no effect if `publish.handout` is false.
+#
+# If `true`, publish a second version of this handout with solutions.
+# Set this to `false` if solutions haven't been written.
+solutions = true
+```
+
+## 💾 Out-of-band compilation
+
+If you want to compile these handouts _without_ this repository (e.g, on [Overleaf] or [Typst.app]), do the following: \
+_(I do not recommend this. The default toolchain makes it easier to share improvements to these handouts.)_
+
+### For LaTeX:
+
+1. Get the handout's directory (i.e, download the whole repo as a zip and extract the folder you want.)
+2. Download [`./resources/ormc_handout.cls`](./resources/ormc_handout.cls)
+3. Put this `ormc_handout.cls` in the same directory as the handout.
+4. Fix the include path at the top of `main.tex`:
+
+You'll need to replace
+
+```latex
+\documentclass[
+	...
+]{../../../lib/tex/ormc_handout}
+```
+
+with
+
+```latex
+\documentclass[
+	...
+]{ormc_handout}
+```
+
+5. Make a new overleaf project with the resulting directory.
+6. **Do not use pdflatex**, it misbehaves with `ormc_handout`. Tell Overleaf to use XeLaTeX.
+
+### For Typst:
+
+Out-of-band typst compilation isn't supported. Clone the repository and use vscode. \
+This is because typst can't import packages from a relative path.
+
+If you _really_ want it, standalone typst compilation _is_ possible. \
+Follow the LaTeX instructions, but fix `handout@0.1.0` instead of `ormc_handout`. \
+You'll figure it out.

authors.toml

@@ -1,3 +0,0 @@
-[authors."mark"]
-email = "mark@betalupi.com"
-webpage = "betalupi.com"

docs-latex.md

@@ -0,0 +1,85 @@
+# LaTeX documentation
+
+All LaTeX handouts are based on [`ormc_handout.cls`](./lib/tex/ormc_handout.cls). \
+This class is based on `article.cls`, and should work with most LaTeX packages.
+
+The best way to start a new document is to make a copy of an existing one.
+- [Advanced/Cryptography](./src/Advanced/Cryptography) is a good example of a simple handout.
+- [Advanced/DFAs](./src/Advanced/DFAs) is a good example of a handout with graphs.
+- [Advanced/Geometric Optimization](./src/Advanced/Geometric%20Optimization) is a good example of a handout with geometry.
+
+
+## Notes
+- Compile your handouts with XeLaTeX. \
+  `pdflatex` is known to misbehave with `ormc_handout.cls`. \
+  This will happen by default if you use vscode. \
+  If you use Overleaf, you'll have to configure it manually (see document settings).
+
+## Document Options
+
+Document options are passed to `\documentclass`, as follows:
+```latex
+\documentclass[
+	% Show solutions is `solutions` is provided,
+	% hide them if `nosolutions` is provided.
+	%
+	% You should set only ONE of these flags at a time.
+	% Solutions are shown by default.
+	% All handouts are stored with `solutions` enabled.
+	solutions,
+	nosolutions,
+
+	% Enable this option if you need more space on the handout's first page.
+	% We use a long warning by default.
+	shortwarning,
+
+	% If present, hide page numbers.
+	% This should only be used for single-page handouts
+	% (e.g, warm-ups)
+	nopagenumber
+]{ormc_handout}
+```
+
+Use `geometry` to change margins and page dimensions. US letter is the default.
+
+
+## Utilities
+- `\say{text}`: Puts text in quotes, handling details like period spacing. Courtesy of `dirtytalk`.
+- `\note[Type]{text}`: Makes a note.
+- `\hint{text}`: Shorthand for `\note[Hint]{text}`
+
+## Sections
+
+The usual LaTeX title-customization techniques *WILL NOT WORK* with this class. \
+Don't even try to load `titlesec`.
+
+`ormc_handout.cls` supports two levels of sections:
+- `\section`, for large parts of the handout
+- `\definition`, `\theorem`, `\proposition`, `\example`, `\remark`, `\problem`, and `\problempart`
+
+All these macros have the following syntax: `\problem{title}<label>`
+- `title` is the problem's title, and may be empty.
+- `label` is the problem's label. This is optional. \
+  If a label is provided, this section may be referenced with `\ref{label}`.
+
+Examples:
+- `\problem{}`
+- `\problem{Bonus}`
+- `\problem{}<gcd>`, which may be referenced with `\ref{gcd}`
+
+Do **not** use `\begin{problem} ... \end{problem}`. \
+Sections are macros, not environments.
+
+## Environments:
+
+- `\begin{solution}`: A fancy red for solutions to problems. \
+	This is hidden if the `nosolutions` is provided.
+- `\begin{instrutornote}`: A fancy blue box for instructor notes. \
+	This is hidden if the `nosolutions` is provided.
+- `\begin{examplesolution}`: A fancy gray for sample solutions. \
+	This is never hidden.
+
+All the above environments break across pages and may safely be nested.
+
+Each of these environments also provides the `\linehack` macro, which draws a line across the box. \
+This is useful for, say, solutions to multipart problems.

docs-typst.md

@@ -0,0 +1,94 @@
+# Typst documentation
+
+See [typst.app/docs](https://typst.app/docs) for typst's documentation. \
+All typst handouts are based on [`handout@0.1.0`](./lib/typst/local/handout/0.1.0).
+
+The best way to start a new document is to make a copy of an existing one.
+- [Advanced/Tropical Polynomials](./src/Advanced/Tropical%20Polynomials) is a good place to start.
+- [Warm-Ups/Painting](./src/Warm-Ups/Painting) is a good example of tikz-like pictures.
+
+
+## Notes
+- Typst's equivalent of tikz is cetz ([homepage](https://cetz-package.github.io), [docs](https://cetz-package.github.io/docs/api))
+- Typst handouts are always compiled with solutions. \
+  Handouts without solutions are automatically compiled and published at [betalupi.com/handouts](https://static.betalupi.com/ormc). \
+  If you'd like to compile a student handout manually, run the following command in a handout directory:
+```bash
+typst compile main.typ --package-path ../../../lib/typst --input show_solutions=false
+```
+Where `package_path` is a relative path to [./lib/typst](./lib/typst).
+
+## Document Options
+
+All typst handouts start with the following:
+```typst
+#show: handout.with(
+  // Should match `meta.toml`
+  title: [handout title],
+
+  // Authors
+  by: "Mark",
+
+  // Subtitle (optional)
+  subtitle: "Based on a handout by Bryant Mathews",
+
+  // Group (optional)
+  group: "Advanced 2",
+)
+```
+
+## Notable commands
+- `#v(1fr)`: Like LaTeX's `\vfill`. Creates whitespace that grows automatically. \
+  `fr` means "fraction". `#v(2fr)` will fill twice as much space as `#v(1fr)` on the same page.
+
+## Utilities
+- `#note([content], type: "Note type")`: Makes a note. `type` is optional.
+- `#hint([content])`: Shorthand for `#note([content], type: "Hint")`
+- `#solution([content])`: A pretty box for solutions. Hidden in student handouts.
+- `#examplesolution([content])`: Like `#solution()`, but is never hidden.
+- `#if_solutions([content])`: Shows content only if we are showing solutions.
+- `#if_no_solutions([content])`: Shows content only if we **aren't** showing solutions.
+
+## Sections
+High-level sections are denoted with `=`. \
+Subsections start with `==`, subsubsections with `===`, and so on. \
+**`handout@0.1.0` is only designed to use `=`, subsections might be ugly.**
+
+
+`handout@0.1.0` also provides the following commands:
+- `problem`
+- `definition`
+- `theorem`
+- `example`
+- `remark`
+
+These all have the same syntax: `#problem("title", label: "label")`
+- `title` is the problem's title, and may be omitted.
+- `label` is the problem's label. This is optional. \
+  If a label is provided, this problem can be referenced with `@label`
+
+**Examples:**
+- `#problem()`
+- `#problem("Bonus")`
+- `#problem(label: "gcd")`, which may be referenced with `@gcd`
+
+### Complete example:
+
+```typst
+// Import definition(), problem(), etc.
+// Must be at the top of each file.
+#import "@local/handout:0.1.0": *
+
+// Make a section called "Tropical Cubic Polynomials"
+= Tropical Cubic Polynomials
+
+// Make a problem with a label
+#problem(label: "imaproblem")
+Consider the polynomial $f(x) = x^3 + 1x^2 + 3x + 6$.
+- sketch a graph of this polynomial
+
+// Make an untitled problem that references `problem`.
+#problem()
+Recall @imaproblem.
+- use this graph to find the roots of $f$
+```

lib/tex/ormc_handout.cls

@@ -1,18 +1,3 @@
-% Copyright (C) 2023 Mark (mark@betalupi.com)
-%
-% This program is free software: you can redistribute it and/or modify
-% it under the terms of the GNU General Public License as published by
-% the Free Software Foundation, either version 3 of the License, or
-% (at your option) any later version.
-%
-% This program is distributed in the hope that it will be useful,
-% but WITHOUT ANY WARRANTY; without even the implied warranty of
-% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
-% GNU General Public License for more details.
-%
-% You should have received a copy of the GNU General Public License
-% along with this program. If not, see <https://www.gnu.org/licenses/>.
-
 \NeedsTeXFormat{LaTeX2e}
 \ProvidesClass{../../../lib/tex/ormc_handout}[2023/05/29 2.0.2 ORMC Handout]
 
@@ -58,7 +43,7 @@
 \ExecuteOptions{
 	10pt,
 	solutions,
-	multinumbering,
+	singlenumbering,
 	pagenumber,
 	longwarning,
 	yeswarning

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

@@ -1,282 +0,0 @@
-/// Typst handout library, used for all documents in this repository.
-
-
-/// If false, hide instructor info.
-///
-/// Compile with the following command to hide solutions:
-/// `typst compile main.typ --input show_solutions=false`
-///
-/// Solutions are shown by default. This behavior
-/// is less surprising than hiding content by default.
-#let show_solutions = {
-  if "show_solutions" in sys.inputs {
-    // Show solutions unless they're explicitly disabled
-    not (
-      sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
-    )
-  } else {
-    // Show solutions by default
-    true
-  }
-}
-
-// Colors
-#let ored = rgb("D62121")
-#let ogrape = rgb("9C36B5")
-#let ocyan = rgb("2288BF")
-#let oteal = rgb("12B886")
-#let ogreen = rgb("37B26D")
-#let oblue = rgb("1C7ED6")
-
-
-
-//
-// MARK: header
-//
-
-#let make_title(
-  group,
-  quarter,
-  title,
-  subtitle,
-) = {
-  align(
-    center,
-    block(
-      width: 60%,
-      height: auto,
-      breakable: false,
-      align(
-        center,
-        stack(
-          spacing: 7pt,
-          (
-            text(size: 10pt, group) + h(1fr) + text(size: 10pt, quarter)
-          ),
-          line(length: 100%, stroke: 0.2mm),
-          (
-            text(size: 20pt, title) + linebreak() + text(size: 10pt, subtitle)
-          ),
-          line(length: 100%, stroke: 0.2mm),
-        ),
-      ),
-    ),
-  )
-}
-
-#let warn = {
-  set text(ored)
-  align(
-    center,
-    block(
-      width: 60%,
-      height: auto,
-      breakable: false,
-      fill: rgb(255, 255, 255),
-      stroke: ored + 2pt,
-      inset: 3mm,
-      (
-        align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
-          + parbreak()
-          + align(
-            left,
-            text(
-              size: 10pt,
-              [This handout contains solutions and notes.]
-                + linebreak()
-                + [Recompile without solutions before distributing.],
-            ),
-          )
-      ),
-    ),
-  )
-}
-
-#let preparedby(name) = (
-  text(
-    size: 10pt,
-    [Prepared by ]
-      + name
-      + [ on ]
-      + datetime
-        .today()
-        .display("[month repr:long] [day padding:none], [year]"),
-  )
-)
-
-//
-// MARK: Solutions
-//
-
-#let solution(content) = {
-  if show_solutions {
-    align(
-      center,
-      stack(
-        block(
-          width: 100%,
-          breakable: false,
-          fill: ored,
-          stroke: ored + 2pt,
-          inset: 1.5mm,
-          (
-            align(left, text(fill: white, weight: "bold", [Solution:]))
-          ),
-        ),
-        block(
-          width: 100%,
-          height: auto,
-          breakable: false,
-          fill: ored.lighten(80%).desaturate(10%),
-          stroke: ored + 2pt,
-          inset: 3mm,
-          align(left, content),
-        ),
-      ),
-    )
-  }
-}
-
-#let notsolution(content) = {
-  if not show_solutions { content }
-}
-
-
-//
-// MARK: Sections
-//
-
-#let generic(t) = block(
-  above: 8mm,
-  below: 2mm,
-  text(weight: "bold", t),
-)
-
-#let _generic_base(kind, ..args) = {
-  counter("obj").step()
-  if args.pos().len() == 0 {
-    generic([
-      #kind
-      #context counter("obj").display():
-    ])
-  } else {
-    generic(
-      [
-        #kind
-        #context counter("obj").display():
-      ]
-        + " "
-        + args.pos().at(0),
-    )
-  }
-}
-
-#let problem(..args) = _generic_base("Problem", ..args)
-#let definition(..args) = _generic_base("Definition", ..args)
-#let theorem(..args) = _generic_base("Theorem", ..args)
-
-
-//
-// MARK: Misc
-//
-
-
-#let hint(content) = {
-  text(fill: rgb(100, 100, 100), style: "oblique", "Hint: ")
-  text(fill: rgb(100, 100, 100), content)
-}
-
-#let note(content) = {
-  text(fill: rgb(100, 100, 100), content)
-}
-
-#let examplesolution(content) = {
-  let c = oblue
-
-  align(
-    center,
-    stack(
-      block(
-        width: 100%,
-        breakable: false,
-        fill: c,
-        stroke: c + 2pt,
-        inset: 1.5mm,
-        (
-          align(left, text(fill: white, weight: "bold", [Example solution:]))
-        ),
-      ),
-      block(
-        width: 100%,
-        height: auto,
-        breakable: false,
-        fill: c.lighten(80%).desaturate(10%),
-        stroke: c + 2pt,
-        inset: 3mm,
-        align(left, content),
-      ),
-    ),
-  )
-}
-
-
-//
-// MARK: wrapper
-//
-
-#let handout(
-  doc,
-  group: none,
-  quarter: none,
-  title: none,
-  by: none,
-  subtitle: none,
-) = {
-  set par(leading: 0.55em, first-line-indent: 0mm, justify: true)
-  set text(font: "New Computer Modern")
-  set par(spacing: 0.5em)
-  show list: set block(spacing: 0.5em, below: 1em)
-  set heading(numbering: (..nums) => nums.pos().at(0))
-
-  set page(
-    margin: 20mm,
-    width: 8in,
-    height: 11.5in,
-    footer: align(
-      center,
-      context counter(page).display(),
-    ),
-    footer-descent: 5mm,
-  )
-
-
-  set list(
-    tight: false,
-    indent: 5mm,
-    spacing: 3mm,
-  )
-
-  show heading.where(level: 1): it => {
-    set align(center)
-    set text(weight: "bold")
-    block[
-      Section #counter(heading).display(): #text(it.body)
-    ]
-  }
-
-  make_title(
-    group,
-    quarter,
-    title,
-    {
-      if by == none { none } else { [#preparedby(by)\ ] }
-      if subtitle == none { none } else { subtitle }
-    },
-  )
-
-  if show_solutions {
-    warn
-  }
-  doc
-}
-

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

@@ -0,0 +1,78 @@
+#import "misc.typ": ored
+
+#let solution_warning() = {
+  set text(ored)
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      fill: rgb(255, 255, 255),
+      stroke: ored + 2pt,
+      inset: 3mm,
+      (
+        align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
+          + parbreak()
+          + align(
+            left,
+            text(
+              size: 10pt,
+              [
+                This handout contains solutions and notes. \
+                Recompile without solutions before distributing.
+              ],
+            ),
+          )
+      ),
+    ),
+  )
+}
+
+#let make_header(
+  title,
+  subtitle: none,
+  by: none,
+  top_left: "",
+  top_right: "",
+) = {
+  let date = datetime
+    .today()
+    .display("[month repr:long] [day padding:none], [year]")
+
+  if (by != none) {
+    by = text(size: 10pt, [Prepared by #by on #date])
+  }
+
+  let sub = ()
+  if (by != none) {
+    sub.push(by)
+  }
+  if (subtitle != none) {
+    sub.push(subtitle)
+  }
+
+  // Main title
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      align(
+        center,
+        stack(
+          spacing: 7pt,
+          // Top
+          text(size: 10pt, top_left) + h(1fr) + text(size: 10pt, top_right),
+          line(length: 100%, stroke: 0.2mm),
+          // Title
+          text(size: 20pt, title),
+          // Subtitle
+          ..sub,
+          line(length: 100%, stroke: 0.2mm),
+        ),
+      ),
+    ),
+  )
+}

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

@@ -0,0 +1,119 @@
+/// Typst handout library, used for all documents in this repository.
+
+// 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 "solution.typ": (
+  if_solutions,
+  if_no_solutions,
+  if_solutions_else,
+  solution,
+  instructornote,
+)
+
+
+/// Main handout wrapper.
+/// Use this as follows:
+///
+/// ```
+/// #show: handout.with(
+///  group: "Advanced 2",
+///  title: [Handout Title],
+///  by: "author",
+///  subtitle: "optional",
+/// )
+///
+/// <rest of document>
+/// ```
+#let handout(
+  doc,
+  group: none,
+  title: none,
+  by: none,
+  subtitle: none,
+) = {
+  set page(
+    margin: 20mm,
+    width: 8.5in,
+    height: 11in,
+    footer: align(
+      center,
+      context counter(page).display(),
+    ),
+    footer-descent: 5mm,
+  )
+
+  //
+  // Text style
+  set text(font: "New Computer Modern")
+  set par(
+    leading: 0.55em,
+    first-line-indent: 0mm,
+    justify: true,
+    spacing: 0.5em,
+  )
+
+  //
+  // List style
+  show list: set block(spacing: 0.5em, below: 1em)
+  set list(
+    tight: false,
+    indent: 5mm,
+    spacing: 3mm,
+  )
+
+  //
+  // Heading style
+  set heading(numbering: (..nums) => nums.pos().at(0))
+  show heading.where(level: 1): it => {
+    set align(center)
+    set text(weight: "bold")
+    block[
+      Section #counter(heading).display(): #text(it.body)
+    ]
+  }
+
+  //
+  // Hack for custom references
+  show ref: it => {
+    import "object.typ": ref_obj
+
+    let x = ref_obj(it) // Custom impl for object references
+    if (x != none) { return x }
+
+    return it // Use default `ref` implementation otherwise
+  }
+
+
+  //
+  // Begin content
+  //
+
+  // Make handout title
+  {
+    import "header.typ": make_header, solution_warning
+    import "solution.typ": show_solutions
+
+    let url = link(
+      "https://betalupi.com/handouts",
+      `betalupi.com/handouts`,
+    )
+
+    make_header(
+      title,
+      subtitle: subtitle,
+      by: by,
+      top_left: group,
+      top_right: url,
+    )
+
+    if show_solutions {
+      solution_warning()
+    }
+  }
+
+  // Include rest of document
+  doc
+}
+

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

@@ -0,0 +1,49 @@
+/// Miscellaneous utilities
+
+#let ored = rgb("D62121")
+#let oorange = rgb("#ffaa3b")
+#let ogrape = rgb("9C36B5")
+#let ocyan = rgb("2288BF")
+#let oteal = rgb("12B886")
+#let ogreen = rgb("37B26D")
+#let oblue = rgb("1C7ED6")
+
+
+#let note(content, type: none) = {
+  set text(fill: rgb(100, 100, 100))
+  if type != none {
+    text(style: "oblique", [#type: ])
+  }
+  text(content)
+}
+
+#let hint = note.with(type: "Hint")
+
+#let examplesolution(content) = {
+  let c = oblue
+
+  align(
+    center,
+    stack(
+      block(
+        width: 100%,
+        breakable: false,
+        fill: c,
+        stroke: c + 2pt,
+        inset: 1.5mm,
+        (
+          align(left, text(fill: white, weight: "bold", [Example solution:]))
+        ),
+      ),
+      block(
+        width: 100%,
+        height: auto,
+        breakable: false,
+        fill: c.lighten(80%).desaturate(10%),
+        stroke: c + 2pt,
+        inset: 3mm,
+        align(left, content),
+      ),
+    ),
+  )
+}

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

@@ -0,0 +1,108 @@
+
+/// This module defines all handout "objects"
+/// (problems, theorems, definitions, etc)
+
+/// Core render code for all objects (problems, theorems, etc)
+/// This should never be used directly by client code.
+///
+/// Args:
+/// - kind: the kind of object to make ("Problem", "Definition", etc)
+/// - label_name: a string. If provided, generate metadata for this object
+///   under the given label. Labels must be unique.
+///   This label can then be used to reference this object.
+///
+/// For example:
+/// ```
+/// #problem(label: "problem1")
+/// This is @problem1
+/// ```
+#let _obj_base(kind, ..args, label_name: none) = {
+  counter("obj").step()
+  let n = context counter("obj").get().first()
+
+  // The complete title text of this object,
+  // like "Problem 5:" or "Theorem: "
+  let obj_content = if args.pos().len() == 0 {
+    [#kind #n:]
+  } else {
+    [#kind #n: #args.pos().at(0)]
+  }
+
+  // Render the object
+  block(
+    above: 8mm,
+    below: 2mm,
+    text(weight: "bold", obj_content),
+  )
+
+  // Generate labeled metadata for this object.
+  //
+  // This can be viewed directly with `#context query(<label>).first().value`,
+  // Or referenced with `@label` (we define a custom renderer for this metadata later)
+  if label_name != none {
+    let meta = (
+      "obj_meta_ref_kind": kind,
+      // "obj_content": obj_content,
+      "label": label(label_name),
+      "counter": counter("obj"),
+    )
+    [ #metadata(meta) #label(label_name) ]
+  }
+}
+
+// `ref` implementation for object meta-references.
+// Returns `none` if `it` is not object metadata.
+#let ref_obj(it) = {
+  let magic_key = "obj_meta_ref_kind"
+  if not (
+    it.element != none
+      and it.element.has("value")
+      and type(it.element.value) == "dictionary"
+      and it.element.value.keys().contains(magic_key)
+  ) {
+    // This label is not attached to object metadata
+    return none
+  }
+
+  let v = it.element.value
+  let obj_type = v.at(magic_key)
+
+  // The value of this object's counter at its label
+  let obj_count = v.counter.at(v.label).first()
+
+  // Produces text like "Problem 2",
+  // which takes you to the referenced object when clicked.
+  return link(v.label, [#obj_type #obj_count])
+}
+
+/// Factory function for objects.
+/// Provided for convenience, lets us define objects in one line.
+#let _mkobj(kind) = {
+  let out(..args, label: none) = _obj_base(
+    kind,
+    ..args,
+    label_name: label,
+  )
+
+  return out
+}
+
+
+//
+// MARK: export
+//
+// Functions for client code are defined below
+
+#let problem = _mkobj("Problem")
+#let definition = _mkobj("Definition")
+#let theorem = _mkobj("Theorem")
+#let example = _mkobj("Example")
+#let remark = _mkobj("Remark")
+
+#let generic(obj_content) = {
+  block(
+    above: 8mm,
+    below: 2mm,
+    text(weight: "bold", obj_content),
+  )
+}

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

@@ -0,0 +1,88 @@
+#import "misc.typ": ored, oblue
+
+/// If false, hide instructor info.
+///
+/// Compile with the following command to hide solutions:
+/// `typst compile main.typ --input show_solutions=false`
+///
+/// Solutions are shown by default. This behavior
+/// is less surprising than hiding content by default.
+#let show_solutions = {
+  if "show_solutions" in sys.inputs {
+    // Show solutions unless they're explicitly disabled
+    not (
+      sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
+    )
+  } else {
+    // Show solutions by default
+    true
+  }
+}
+
+#let if_solutions(content) = {
+  if show_solutions { content }
+}
+
+#let if_no_solutions(content) = {
+  if not show_solutions { content }
+}
+
+#let if_solutions_else(if_yes, if_no) = {
+  if show_solutions { if_yes } else { if_no }
+}
+
+#let solution(content) = {
+  if_solutions(
+    align(
+      center,
+      stack(
+        block(
+          width: 100%,
+          breakable: false,
+          fill: ored,
+          stroke: ored + 2pt,
+          inset: 1.5mm,
+          align(left, text(fill: white, weight: "bold", [Solution:])),
+        ),
+
+        block(
+          width: 100%,
+          height: auto,
+          breakable: false,
+          fill: ored.lighten(80%).desaturate(10%),
+          stroke: ored + 2pt,
+          inset: 3mm,
+          align(left, content),
+        ),
+      ),
+    ),
+  )
+}
+
+#let instructornote(content) = {
+  if_solutions(
+    align(
+      center,
+      stack(
+        block(
+          width: 100%,
+          breakable: false,
+          fill: oblue,
+          stroke: oblue + 2pt,
+          inset: 1.5mm,
+          align(left, text(fill: white, weight: "bold", [Instructor note:])),
+        ),
+
+        block(
+          width: 100%,
+          height: auto,
+          breakable: false,
+          fill: oblue.lighten(80%).desaturate(10%),
+          stroke: oblue + 2pt,
+          inset: 3mm,
+          align(left, content),
+        ),
+      ),
+    ),
+  )
+}

lib/typst/local/handout/0.1.0/typst.toml

@@ -1,6 +1,12 @@
 [package]
 name = "handout"
+description = "A library for math circle handouts"
 version = "0.1.0"
-entrypoint = "handout.typ"
-authors = []
-license = "GPL"
+entrypoint = "lib.typ"
+
+homepage = "https://betalupi.com/handouts"
+repository = "https://git.betalupi.com/Mark/handouts"
+authors = ["Mark <mark@betalupi.com>"]
+license = "GPL-3.0-only "
+disciplines = ["education", "mathematics"]
+categories = ["layout", "components"]

src/Advanced/Error-Correcting Codes/parts/00 detection.tex

@@ -19,7 +19,7 @@ Only one of the following ISBNs is valid. Which one is it?
 
 \begin{itemize}
 	\item \texttt{0-134-54896-2}
-	\item \texttt{0-895-77258-2}
+	\item \texttt{0-895-77258-2} % oliver twist
 \end{itemize}
 
 \begin{solution}
@@ -67,18 +67,19 @@ This is called a \textit{transposition error}.
 \vfill
 \pagebreak
 
-\problem{}
+\definition{}
 ISBN-13 error checking is slightly different. Given a partial ISBN-13 $n_1 n_2 n_3 ... n_{12}$, the final digit is given by
 
 $$
  n_{13} = \Biggr[ \sum_{i=1}^{12} n_i \times (2 + (-1)^i) \Biggl] \text{ mod } 10
 $$
 
+\problem{}
 What is the last digit of the following ISBN-13? \par
-\texttt{978-0-380-97726-?}
+\texttt{978-030-7292-06*} % foundation
 
 \begin{solution}
-	The final digit is 0.
+	The final digit is 3.
 \end{solution}
 
 \vfill
@@ -127,7 +128,7 @@ Take a valid ISBN-13 and swap two adjacent digits. When will the result be a val
 \vfill
 
 \problem{}<isbn-nocorrect>
-\texttt{978-0-08-2066-46-6} was a valid ISBN until I changed a single digit. \par
+\texttt{978-008-2066-466} was a valid ISBN until I changed a single digit. \par
 Can you find the digit I changed? Can you recover the original ISBN?
 
 \begin{solution}

src/Advanced/Error-Correcting Codes/parts/02 hamming.tex

@@ -570,8 +570,45 @@ If we know which parity bits are inconsistent, how can we find where the error i
 \vfill
 
 \problem{}<generalize-hamming>
-Can you generalize this system for messages of 4, 64, or 256 bits?
+Generalize this system for messages of 4, 64, or 256 bits. \par
+\begin{itemize}
+	\item How does the resilience of this scheme change if we use a larger message size?
+	\item How does the efficiency of this scheme change if we send larger messages?
+\end{itemize}
+
+\vfill
+\pagebreak
+
+
+\definition{}
+A \textit{deletion} error occurs when one bit of the message is deleted. \par
+Likewise, an \textit{insertion} error consists of a random inserted bit. \par
+
+\definition{}
+A \textit{message stream} is an infinite string of binary digits.
+
+\problem{}
+Show that Hamming codes do not reliably detect bit deletions: \par
+\hint{
+	Create a 17-bit message whose first 16 bits are a valid Hamming block, \par
+	and which is still valid when a bit (chosen by you; not the $17^\text{th}$) is deleted.
+}
 
 \vfill
 
+\problem{}
+Convince yourself that Hamming codes cannot correct insertions. \par
+Then, create a 16-bit message that...
+\begin{itemize}
+	\item is a valid Hamming block, and
+	\item incorrectly "corrects" a single bit error when it encounters an insertion error.
+\end{itemize}
+
+\vfill
+
+
+As we have seen, Hamming codes effectively handle substitutions, but cannot reliably
+detect (or correct) insertions and deletions. Correcting those errors is a bit more difficult:
+if the number of bits we receive is variable, how can we split a stream into a series of messages? \par
+\note{This is a rhetorical question, which we'll discuss another day.}
 \pagebreak

src/Advanced/Fast Inverse Root/main.typ

@@ -0,0 +1,31 @@
+#import "@local/handout:0.1.0": *
+
+// Bonus:
+// - Floats vs fixed point
+// - Float density
+// - Find non-floatable rational numbers
+// - What if we use `n`-bit floats?
+
+#show: doc => handout(
+  doc,
+  group: "Advanced 2",
+  title: [Fast Inverse Square Root],
+  by: "Mark",
+)
+
+#include "parts/00 intro.typ"
+#pagebreak()
+
+#include "parts/01 int.typ"
+#pagebreak()
+
+#include "parts/02 float.typ"
+#pagebreak()
+
+#include "parts/03 approx.typ"
+#pagebreak()
+
+#include "parts/04 quake.typ"
+#pagebreak()
+
+#include "parts/05 bonus.typ"

src/Advanced/Fast Inverse Root/meta.toml

@@ -0,0 +1,7 @@
+[metadata]
+title = "Fast Inverse Square Root"
+
+
+[publish]
+handout = true
+solutions = true

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

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

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

@@ -0,0 +1,102 @@
+#import "@local/handout:0.1.0": *
+
+= Integers
+
+#definition()
+A _bit string_ is a string of binary digits. \
+In this handout, we'll denote bit strings with the prefix `0b`. \
+#note[This prefix is only notation---it is _not_ part of the string itself.] \
+For example, $1001$ is the number "one thousand and one," while $#text([`0b1001`])$ is the string of bits "1 0 0 1".
+
+#v(2mm)
+We will separate long bit strings with underscores for readability. \
+Underscores have no meaning: $#text([`0b1111_0000`]) = #text([`0b11110000`])$.
+
+#problem()
+What is the value of the following bit strings, if we interpret them as integers in base 2?
+- `0b0001_1010`
+- `0b0110_0001`
+
+#solution([
+  - $#text([`0b0001_1010`]) = 2 + 8 + 16 = 26$
+  - $#text([`0b0110_0001`]) = 1 + 32 + 64 = 95$
+])
+
+#v(1fr)
+
+#definition()
+We can interpret a bit string in any number of ways. \
+One such interpretation is the _unsigned integer_, or `uint` for short. \
+`uint`s allow us to represent positive (hence "unsigned") integers using 32-bit strings.
+
+#v(2mm)
+
+The value of a `uint` is simply its value as a binary number:
+- $#text([`0b00000000_00000000_00000000_00000000`]) = 0$
+- $#text([`0b00000000_00000000_00000000_00000011`]) = 3$
+- $#text([`0b00000000_00000000_00000000_00100000`]) = 32$
+- $#text([`0b00000000_00000000_00000000_10000010`]) = 130$
+
+#problem()
+What is the largest number we can represent with a 32-bit `uint`?
+
+#solution([
+  $#text([`0b11111111_11111111_11111111_11111111`]) = 2^(32)-1$
+])
+
+#v(1fr)
+#pagebreak()
+
+#problem()
+Find the value of each of the following 32-bit unsigned integers:
+- `0b00000000_00000000_00000101_00111001`
+- `0b00000000_00000000_00000001_00101100`
+- `0b00000000_00000000_00000100_10110000`
+#hint([The third conversion is easy---look carefully at the second.])
+
+#instructornote[
+  Consider making a list of the powers of two $>= 1024$ on the board.
+]
+
+#solution([
+  - $#text([`0b00000000_00000000_00000101_00111001`]) = 1337$
+  - $#text([`0b00000000_00000000_00000001_00101100`]) = 300$
+  - $#text([`0b00000000_00000000_00000010_01011000`]) = 1200$
+  Notice that the third int is the second shifted left twice (i.e, multiplied by 4)
+])
+
+#v(1fr)
+
+
+#definition()
+In general, fast division of `uints` is difficult#footnote([One may use repeated subtraction, but this isn't efficient.]). \
+Division by powers of two, however, is incredibly easy: \
+To divide by two, all we need to do is shift the bits of our integer right.
+
+#v(2mm)
+
+For example, consider $#text[`0b0000_0110`] = 6$. \
+If we insert a zero at the left end of this string and delete the zero at the right \
+(thus "shifting" each bit right), we get `0b0000_0011`, which is 3. \
+
+#v(2mm)
+
+Of course, we lose the remainder when we right-shift an odd number: \
+$9$ shifted right is $4$, since `0b0000_1001` shifted right is `0b0000_0100`.
+
+#problem()
+Right shifts are denoted by the `>>` symbol: \
+$#text[`00110`] #text[`>>`] n$ means "shift `0b0110` right $n$ times." \
+Find the value of the following:
+- $12 #text[`>>`] 1$
+- $27 #text[`>>`] 3$
+- $16 #text[`>>`] 8$
+#note[Naturally, you'll have to convert these integers to binary first.]
+
+#solution[
+  - $12 #text[`>>`] 1 = 6$
+  - $27 #text[`>>`] 3 = 3$
+  - $16 #text[`>>`] 8 = 0$
+]
+
+#v(1fr)

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

@@ -0,0 +1,211 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+= Floats
+#definition()
+_Binary decimals_#footnote([Note that "binary decimal" is a misnomer---"deci" means "ten"!]) are very similar to base-10 decimals.\
+In base 10, we interpret place value as follows:
+- $0.1 = 10^(-1)$
+- $0.03 = 3 times 10^(-2)$
+- $0.0208 = 2 times 10^(-2) + 8 times 10^(-4)$
+
+#v(5mm)
+
+We can do the same in base 2:
+- $#text([`0.1`]) = 2^(-1) = 0.5$
+- $#text([`0.011`]) = 2^(-2) + 2^(-3) = 0.375$
+- $#text([`101.01`]) = 5.125$
+
+#v(5mm)
+
+#problem()
+Rewrite the following binary decimals in base 10: \
+#note([You may leave your answer as a fraction.])
+- `1011.101`
+- `110.1101`
+
+
+#v(1fr)
+#pagebreak()
+
+#definition(label: "floatbits")
+Another way we can interpret a bit string is as a _signed floating-point decimal_, or a `float` for short. \
+Floats represent a subset of the real numbers, and are interpreted as follows: \
+#note([The following only applies to floats that consist of 32 bits. We won't encounter any others today.])
+
+#align(
+  center,
+  box(
+    inset: 2mm,
+    cetz.canvas({
+      import cetz.draw: *
+
+      let chars = (
+        `0`,
+        `b`,
+        `0`,
+        `_`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `_`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `_`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `_`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+        `0`,
+      )
+
+      let x = 0
+      for c in chars {
+        content((x, 0), c)
+        x += 0.25
+      }
+
+      let y = -0.4
+      line((0.3, y), (0.65, y))
+      content((0.45, y - 0.2), [s])
+
+      line((0.85, y), (2.9, y))
+      content((1.9, y - 0.2), [exponent])
+
+      line((3.10, y), (9.4, y))
+      content((6.3, y - 0.2), [fraction])
+    }),
+  ),
+)
+
+- The first bit denotes the sign of the float's value
+  We'll label it $s$. \
+  If $s = #text([`1`])$, this float is negative; if $s = #text([`0`])$, it is positive.
+
+- The next eight bits represent the _exponent_ of this float.
+  #note([(we'll see what that means soon)]) \
+  We'll call the value of this eight-bit binary integer $E$. \
+  Naturally, $0 <= E <= 255$ #note([(since $E$ consist of eight bits)])
+
+- The remaining 23 bits represent the _fraction_ of this float. \
+  They are interpreted as the fractional part of a binary decimal. \
+  For example, the bits `0b10100000_00000000_00000000` represent $0.5 + 0.125 = 0.625$. \
+  We'll call the value of these bits as a binary integer $F$. \
+  Their value as a binary decimal is then $F div 2^23$. #note([(convince yourself of this)])
+
+
+#problem(label: "floata")
+Consider `0b01000001_10101000_00000000_00000000`. \
+#hint([The underscores here do _not_ match those in @floatbits])
+
+#v(2mm)
+
+Find the $s$, $E$, and $F$ we get if we interpret this bit string as a `float`. \
+#note([Leave $F$ as a sum of powers of two.])
+
+#solution([
+  $s = 0$ \
+  $E = 131$ \
+  $F = 2^21+2^19$
+])
+
+#v(1fr)
+
+
+#definition(label: "floatdef")
+The final value of a float with sign $s$, exponent $E$, and fraction $F$ is
+
+$
+  (-1)^s times 2^(E - 127) times (1 + F / (2^(23)))
+$
+
+Notice that this is very similar to base-10 scientific notation, which is written as
+
+$
+  (-1)^s times 10^(e) times (f)
+$
+
+#note[
+  We subtract 127 from $E$ so we can represent positive and negative numbers. \
+  $E$ is an eight bit binary integer, so $0 <= E <= 255$ and thus $-127 <= (E - 127) <= 127$.
+]
+
+#problem()
+Consider `0b01000001_10101000_00000000_00000000`. \
+This is the same bit string we used in @floata. \
+
+#v(2mm)
+
+What value do we get if we interpret this bit string as a float? \
+#hint([$21 div 16 = 1.3125$])
+
+#solution([
+  This is 21:
+  $
+    2^4 times (1 + (2^(21) + 2^(19)) / (2^(23)))
+    = 2^(4) times (1 + 2^(-2) + 2^(-4))
+    = 16 + 4 + 1
+    = 21
+  $
+])
+
+#v(1fr)
+#pagebreak()
+
+#problem()
+Encode $12.5$ as a float. \
+#hint([$12.5 div 8 = 1.5625$])
+
+#solution([
+  $
+    12.5
+    = 8 times 1.5625
+    = 2^(3) times (1 + (0.5 + 0.0625))
+    = 2^(130) times (1 + (2^(22) + 2^(19)) / (2^(23)))
+  $
+
+  which is `0b01000001_01001000_00000000_00000000`. \
+])
+
+
+#v(1fr)
+
+#definition()
+Say we have a bit string $x$. \
+We'll let $x_f$ denote the value we get if we interpret $x$ as a float, \
+and we'll let $x_i$ denote the value we get if we interpret $x$ an integer.
+
+#problem()
+Let $x = #text[`0b01000001_01001000_00000000_00000000`]$. \
+What are $x_f$ and $x_i$? #note([As always, you may leave big numbers as powers of two.])
+#solution([
+  $x_f = 12.5$
+
+  #v(2mm)
+
+  $x_i = 2^30 + 2^24 + 2^22 + 2^19 = 11,095,237,632$
+])
+
+#v(1fr)

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

@@ -0,0 +1,173 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+#import "@preview/cetz-plot:0.1.0": plot, chart
+
+= Integers and Floats
+
+#generic("Observation:")
+If $x$ is smaller than 1, $log_2(1 + x)$ is approximately equal to $x$. \
+Note that this equality is exact for $x = 0$ and $x = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
+
+#v(5mm)
+
+We'll add the _correction term_ $epsilon$ to our approximation: $log_2(1 + a) approx a + epsilon$. \
+This allows us to improve the average error of our linear approximation:
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr),
+  inset: 5mm,
+  [$log_2(1+x)$ and $x + 0$]
+    + cetz.canvas({
+      import cetz.draw: *
+
+      let f1(x) = calc.log(calc.abs(x + 1), base: 2)
+      let f2(x) = x
+
+      // Set-up a thin axis style
+      set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
+
+
+      plot.plot(
+        size: (7, 7),
+        x-tick-step: 0.2,
+        y-tick-step: 0.2,
+        y-min: 0,
+        y-max: 1,
+        x-min: 0,
+        x-max: 1,
+        legend: none,
+        axis-style: "scientific-auto",
+        x-label: none,
+        y-label: none,
+        {
+          let domain = (0, 1)
+
+          plot.add(
+            f1,
+            domain: domain,
+            label: $log(1+x)$,
+            style: (stroke: ogrape),
+          )
+
+          plot.add(
+            f2,
+            domain: domain,
+            label: $x$,
+            style: (stroke: oblue),
+          )
+        },
+      )
+    })
+    + [
+      Max error: 0.086 \
+      Average error: 0.0573
+    ],
+  [$log_2(1+x)$ and $x + 0.045$]
+    + cetz.canvas({
+      import cetz.draw: *
+
+      let f1(x) = calc.log(calc.abs(x + 1), base: 2)
+      let f2(x) = x + 0.0450466
+
+      // Set-up a thin axis style
+      set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
+
+
+      plot.plot(
+        size: (7, 7),
+        x-tick-step: 0.2,
+        y-tick-step: 0.2,
+        y-min: 0,
+        y-max: 1,
+        x-min: 0,
+        x-max: 1,
+        legend: none,
+        axis-style: "scientific-auto",
+        x-label: none,
+        y-label: none,
+        {
+          let domain = (0, 1)
+
+          plot.add(
+            f1,
+            domain: domain,
+            label: $log(1+x)$,
+            style: (stroke: ogrape),
+          )
+
+          plot.add(
+            f2,
+            domain: domain,
+            label: $x$,
+            style: (stroke: oblue),
+          )
+        },
+      )
+    })
+    + [
+      Max error: 0.041 \
+      Average error: 0.0254
+    ],
+)
+
+
+A suitiable value of $epsilon$ can be found using calculus or with computational trial-and-error. \
+We won't bother with this---we'll simply leave the correction term as an opaque constant $epsilon$.
+
+
+
+#v(1fr)
+
+#note(
+  type: "Note",
+  [
+    "Average error" above is simply the area of the region between the two graphs:
+    $
+      integral_0^1 abs( #v(1mm) log(1+x)_2 - (x+epsilon) #v(1mm))
+    $
+    Feel free to ignore this note, it isn't a critical part of this handout.
+  ],
+)
+
+
+#pagebreak()
+
+#problem(label: "convert")
+Use the fact that $log_2(1 + a) approx a + epsilon$ to approximate $log_2(x_f)$ in terms of $x_i$. \
+Namely, show that
+$
+  log_2(x_f) = (x_i) / (2^23) - 127 + epsilon
+$
+#note([
+  In other words, we're finding an expression for $x$ as a float
+  in terms of $x$ as an int.
+])
+
+#solution([
+  Let $E$ and $F$ be the exponent and float bits of $x_f$. \
+  We then have:
+  $
+    log_2(x_f)
+    &= log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
+    &= E - 127 + log_2(1 + F / (2^23)) \
+    & approx E-127 + F / (2^23) + epsilon \
+    &= 1 / (2^23)(2^23 E + F) - 127 + epsilon \
+    &= 1 / (2^23)(x_i) - 127 + epsilon
+  $
+])
+
+#v(1fr)
+
+
+#problem()
+Using basic log rules, rewrite $log_2(1 / sqrt(x))$ in terms of $log_2(x)$.
+
+#solution([
+  $
+    log_2(1 / sqrt(x)) = (-1) / (2)log_2(x)
+  $
+])
+
+#v(1fr)

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

@@ -0,0 +1,210 @@
+#import "@local/handout:0.1.0": *
+
+= The Fast Inverse Square Root
+
+A simplified version of the _Quake_ routine we are studying is reproduced below.
+
+#v(2mm)
+
+```c
+float Q_rsqrt( float number ) {
+    long i = * ( long * ) &number;
+    i = 0x5f3759df - ( i >> 1 );
+    return * ( float * ) &i;
+}
+```
+
+#v(2mm)
+
+This code defines a function `Q_rsqrt` that consumes a float `number` and approximates its inverse square root.
+If we rewrite this using notation we're familiar with, we get the following:
+$
+  #text[`Q_sqrt`] (n_f) =
+  6240089 - (n_i div 2)
+  #h(10mm)
+  approx 1 / sqrt(n_f)
+$
+
+#note[
+  `0x5f3759df` is $6240089$ in hexadecimal. \
+  Ask an instructor to explain if you don't know what this means. \
+  It is a magic number hard-coded into `Q_sqrt`.
+]
+
+#v(2mm)
+
+Our goal in this section is to understand why this works:
+- How does Quake approximate $1 / sqrt(x)$ by simply subtracting and dividing by two?
+- What's special about $6240089$?
+
+
+
+#v(1fr)
+
+#remark()
+For those that are interested, here are the details of the "code-to-math" translation:
+
+- "`long i = * (long *) &number`" is C magic that tells the compiler \
+  to set `i` to the `uint` value of the bits of `number`. \
+  #note[
+    "long" refers to a "long integer", which has 32 bits. \
+    Normal `int`s have 16 bits, `short int`s have 8.
+  ] \
+  In other words, `number` is $n_f$ and `i` is $n_i$.
+#v(2mm)
+
+
+- Notice the right-shift in the second line of the function. \
+  We translated `(i >> 1)` into $(n_i div 2)$.
+#v(2mm)
+
+- "`return * (float *) &i`" is again C magic. \
+  Much like before, it tells us to return the value of the bits of `i` as a float.
+#pagebreak()
+
+#generic("Setup:")
+We are now ready to show that $#text[`Q_sqrt`] (x)$ effectively approximates $1/sqrt(x)$. \
+For convenience, let's call the bit string of the inverse square root $r$. \
+In other words,
+$
+  r_f := 1 / (sqrt(n_f))
+$
+This is the value we want to approximate. \
+
+#problem(label: "finala")
+Find an approximation for $log_2(r_f)$ in terms of $n_i$ and $epsilon$ \
+#note[Remember, $epsilon$ is the correction constant in our approximation of $log_2(1 + x)$.]
+
+#solution[
+  $
+    log_2(r_f)
+    = log_2(1 / sqrt(n_f))
+    = (-1) / 2 log_2(n_f)
+    approx (-1) / 2 ( (n_i) / (2^23) + epsilon - 127 )
+  $
+]
+
+#v(1fr)
+
+#problem(label: "finalb")
+Let's call the "magic number" in the code above $kappa$, so that
+$
+  #text[`Q_sqrt`] (n_f) = kappa - (n_i div 2)
+$
+Use @convert and @finala to show that $#text[`Q_sqrt`] (n_f) approx r_i$ \
+#note(type: "Note")[
+  If we know $r_i$, we know $r_f$. \
+  We don't even need to convert between the two---the underlying bits are the same!
+]
+
+#solution[
+  From @convert, we know that
+  $
+    log_2(r_f) approx (r_i) / (2^23) + epsilon - 127
+  $
+
+  Combining this with the result from @finala, we get:
+  $
+    (r_i) / (2^23) + epsilon - 127
+    &approx (-1) / (2) ( (n_i) / (2^23) + epsilon - 127) \
+    (r_i) / (2^23)
+    &approx (-1) / (2) ( (n_i) / (2^23)) + 3 / 2 (127 - epsilon) \
+    r_i
+    &approx (-1) / 2 (n_i) + 2^23 3 / 2(127 - epsilon)
+    = 2^23 3 / 2 (127 - epsilon) - (n_i) / 2
+  $
+
+  #v(2mm)
+
+  This is exactly what we need! If we set $kappa$ to $(3 times 2^22) (127-epsilon)$, then
+  $
+    r_i approx kappa - (n_i div 2) = #text[`Q_sqrt`] (n_f)
+  $
+]
+
+#v(1fr)
+
+#problem(label: "finalc")
+What is the exact value of $kappa$ in terms of $epsilon$? \
+#hint[Look at @finalb. We already found it!]
+
+#solution[
+  This problem makes sure our students see that
+  $kappa = (3 times 2^22) (127 - epsilon)$. \
+  See the solution to @finalb.
+]
+
+#v(2cm)
+
+#pagebreak()
+
+#remark()
+In @finalc we saw that $kappa = (3 times 2^22) (127 - epsilon)$. \
+Looking at the code again, we see that $kappa = #text[`0x5f3759df`]$ in _Quake_:
+
+#v(2mm)
+
+```c
+float Q_rsqrt( float number ) {
+    long i = * ( long * ) &number;
+    i = 0x5f3759df - ( i >> 1 );
+    return * ( float * ) &i;
+}
+```
+
+#v(2mm)
+Using a calculator and some basic algebra, we can find the $epsilon$ this code uses: \
+#note[Remember, #text[`0x5f3759df`] is $6240089$ in hexadecimal.]
+$
+  (3 times 2^22) (127 - epsilon) &= 6240089 \
+  (127 - epsilon) &= 126.955 \
+  epsilon &= 0.0450466
+$
+
+So, $0.045$ is the $epsilon$ used by Quake. \
+Online sources state that this constant was generated by trial-and-error, \
+though it is fairly close to the ideal $epsilon$.
+
+#remark()
+And now, we're done! \
+We've shown that `Q_sqrt(x)` approximates $1/sqrt(x)$ fairly well. \
+
+#v(2mm)
+
+Notably, `Q_sqrt` uses _zero_ divisions or multiplications (`>>` doesn't count). \
+This makes it _very_ fast when compared to more traditional approximation techniques (i.e, Taylor series).
+
+#v(2mm)
+
+In the case of _Quake_, this is very important. 3D graphics require thousands of inverse-square-root calculations to render a single frame#footnote[e.g, to generate normal vectors], which is not an easy task for a Playstation running at 300MHz.
+
+#instructornote[
+  Let $x$ be a bit string. If we assume $x_f$ is positive and $E$ is even, then
+  $
+    (x #text[`>>`] 1)_f = 2^((E div 2) - 127) times (1 + (F div 2) / (2^(23)))
+  $
+  Notably: a right-shift divides the exponent of $x_f$ by two, \
+  which is, of course, a square root!
+
+  #v(2mm)
+
+  This intuition is hand-wavy, though: \
+  If $E$ is odd, its lowest-order bit becomes the highest-order bit of $F$ when we shift $x$ right. \
+  Also, a right shift doesn't divide the _entire_ exponent, skipping the $-127$ offset. \
+
+  #v(2mm)
+
+  Remarkably, this intuition is still somewhat correct. \
+  The bits align _just so_, and our approximation still works.
+
+  #v(8mm)
+
+  One can think of the fast inverse root as a "digital slide rule": \
+  The integer representation of $x_f$ already contains $log_2(x_f)$, offset and scaled. \
+  By subtracting and dividing in "log space", we effectively invert and root $x_f$!
+
+  After all,
+  $
+    - 1 / 2 log_2(n_f) = 1 / sqrt(n_f)
+  $
+]

src/Advanced/Fast Inverse Root/parts/05 bonus.typ

@@ -0,0 +1,36 @@
+#import "@local/handout:0.1.0": *
+
+= Bonus -- More about Floats
+
+#problem()
+Convince yourself that all numbers that can be represented as a float are rational.
+
+#problem()
+Find a rational number that cannot be represented as a float.
+
+#v(1fr)
+
+#problem()
+What is the smallest positive 32-bit float?
+
+#v(1fr)
+
+#problem()
+What is the largest positive 32-bit float?
+
+#v(1fr)
+
+#problem()
+How many floats are between $-1$ and $1$?
+
+#v(1fr)
+
+#problem()
+How many floats are between $1$ and $2$?
+
+#v(1fr)
+
+#problem()
+How many floats are between $1$ and $128$?
+
+#v(1fr)

src/Advanced/Introduction to Quantum/src/main.tex

@@ -1,18 +1,3 @@
-% Copyright (C) 2023 <Mark (mark@betalupi.com)>
-%
-% This program is free software: you can redistribute it and/or modify
-% it under the terms of the GNU General Public License as published by
-% the Free Software Foundation, either version 3 of the License, or
-% (at your option) any later version.
-%
-% You may have received a copy of the GNU General Public License
-% along with this program. If not, see <https://www.gnu.org/licenses/>.
-%
-%
-%
-% If you edit this, please give credit!
-% Quality handouts take time to make.
-
 % use the [nosolutions] flag to hide solutions,
 % use the [solutions] flag to show solutions.
 \documentclass[

src/Advanced/Logic Puzzles/main.typ

@@ -0,0 +1,17 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Bonus: Logic Puzzles],
+  by: "Mark",
+  subtitle: [Based on Raymond Smullyan's _To Mock a Mockingbird_.],
+)
+
+#set par(first-line-indent: 5mm)
+
+#include "parts/00 one.typ"
+#pagebreak()
+
+#include "parts/01 logician.typ"
+#pagebreak()
+
+#include "parts/02 gods.typ"

src/Advanced/Logic Puzzles/meta.toml

@@ -0,0 +1,7 @@
+[metadata]
+title = "Tropical Polynomials"
+
+
+[publish]
+handout = false
+solutions = true

src/Advanced/Logic Puzzles/parts/00 one.typ

@@ -0,0 +1,103 @@
+#import "@local/handout:0.1.0": *
+
+= Miscellaneous Puzzles
+
+#problem("The Flower Garden")
+In a certain flower garden, each flower was either red, yellow,
+or blue, and all three colors were represented. A statistician
+once visited the garden and made the observation that whatever three flowers you picked, at least one of them was bound
+to be red.
+A second statistician visited the garden and made
+the observation that whatever three flowers you picked, at
+least one was bound to be yellow.
+
+Two logic students heard about this and got into an argument. The first student said: "It therefore follows that whatever three flowers you pick, at least one is bound to be blue,
+doesn't it?" The second student said: "Of course not!"
+Which student was right, and why?
+
+#solution[
+  The first student was right, and here is why. From the
+  first statistician's report it follows that there cannot be more
+  than one yellow flower, because if there were two yellows,
+  you could pick two yellows and one blue, thus having a group
+  of three flowers that contained no red. This is contrary to the
+  report that every group of three is bound to contain at least
+  one red flower. Therefore there cannot be more than one yellow flower.
+
+  Similarly, there cannot be more than one blue
+  flower, because if there were two blues, you could pick two
+  blue flowers and one yellow and again have a group of three
+  that contained no red. And so from the first statistician's report
+  it follows that there is at most one yellow flower and one blue.
+  And it follows from the report of the second statistician that
+  there is at most one red flower, for if there were two reds,
+  you could pick two reds and one blue, thus obtaining a group
+  of three that contained no yellow. It also follows from the
+  second report that there cannot be more than one blue,
+  although we have already deduced this from the first report.
+
+  The upshot of all this is that there are only three flowers
+  in the entire garden-one red, one yellow, and one blue! And
+  so it is of course true that whatever three flowers you pick,
+  one of them must be blue.
+]
+
+#v(1fr)
+#pagebreak()
+
+
+#problem("What Question")
+There is a question I could ask you that has a definite correct
+answer---either yes or no---but it is logically impossible for
+you to give the correct answer. You might know what the
+correct answer is, but you cannot give it. Anybody other than
+you might possibly be able to give the correct answer, but
+you cannot!
+
+Can you figure out what question I could have in mind?
+
+#solution[
+  Suppose I ask you: "Is no your answer to this question?"
+  If you answer yes, then you are affirming that no is your answer to the question,
+  which is of course wrong. If you answer
+  no, then you are denying that no is your answer, although no
+  was your answer.
+
+  It is therefore impossible for you to answer
+  the question correctly even though the question does have a
+  correct answer: Either you answer no or you don't. If you do,
+  then yes is the correct answer; if you don't, then no is the
+  correct answer, but in neither case can you give the correct
+  answer.
+]
+
+#v(1fr)
+
+#problem("Which Way Would You Bet?")
+Choose your favorite baseball team and consider the scores it will make next
+season. \
+Which do you bet will be the larger number---the sum
+of these scores or the product of these scores?
+
+#solution[
+  The chances are that the sum will be the larger number,
+  because your team will probably score at least one zero, and
+  one zero makes the entire product zero.
+]
+
+#v(1fr)
+
+#generic("Aside:")
+Speaking of probability and statistics, there is the story of
+a statistician who told a friend that he never took airplanes: "I
+have computed the probability that there will be a bomb on
+the plane," he explained, "and although this probability is low,
+it is still too high for my comfort."
+
+Two weeks later, the friend
+met the statistician on a plane. "How come you changed your
+theory?" he asked. "Oh, I didn't change my theory; it's just
+that I subsequently computed the probability that there would
+simultaneously be two bombs on a plane. This probability is
+low enough for my comfort. \
+So now I simply carry my own bomb."

src/Advanced/Logic Puzzles/parts/01 logician.typ

@@ -0,0 +1,231 @@
+#import "@local/handout:0.1.0": *
+
+= The Absentminded Logician
+
+#problem("Only Three Words?")
+We are given three brothers named John, James, and William.
+John and James always lie, but William always
+tells the truth. The three are indistinguishable in appearance.
+You meet one of the three brothers on the street one day and
+wish to find out whether he is John (because John owes you
+money). You are allowed to ask him one question answerable
+by yes or no, but the question may not contain more than
+three words! What question would you ask?
+
+#solution[
+  The only three-word question I can think of that works
+  is: "Are you James?" If you are addressing John, he will answer
+  yes, since John lies, whereas both James and William would
+  answer no-James because he lies, and William because he tells
+  the truth. So a yes answer means that he is John and a no
+  answer means that he is not John.
+]
+
+#v(1fr)
+
+#problem("A Variant")
+Suppose we change the above conditions by making John and
+James both truthful and William a liar. Again you meet one
+of the three and wish to find out if he is John. Is there now a
+three-word yes/no question that can accomplish this?
+
+#solution[
+  The very same question---"Are you James?"---works,
+  only a yes answer now indicates that he isn't John and a no
+  answer indicates that he is John.
+]
+
+
+#v(1fr)
+#pagebreak()
+
+#problem("A More Subtle Puzzle")
+We now have only two brothers (identical twins). One of them
+is named Arthur and the other has a different name. One of
+the two always lies and the other always tells the truth, but
+we are not told whether Arthur is the liar or the truth-teller.
+One day you meet the two brothers together, and you wish
+to find out which one is Arthur. Note that you are not inter-
+ested in finding out which one lies and which one tells the
+truth, but only in finding out which one is Arthur. You are
+allowed to ask just one of them a question answerable by yes
+or no, and again the question may not contain more than three
+words. What question would you ask?
+
+#solution[
+  A common wrong guess is: "Are you Arthur?" This question is quite useless here; the answer you get could be the truth
+  or a lie, and you would still have no idea which one is really
+  Arthur.
+
+  A question that works is: "Is Arthur truthful?" Arthur will
+  surely answer yes to this question, because if Arthur is truthful,
+  he will truthfully claim that Arthur is truthful, and if Arthur
+  is not truthful, then he will falsely claim that Arthur is truthful.
+  So regardless of whether Arthur is truthful or whether he lies,
+  he will certainly claim that Arthur is truthful.
+
+  On the other hand, Arthur's brother---call him Henry---will claim that Arthur is not truthful, because if Henry is truthful, then Arthur is really not truthful and Henry will truthfully claim that Arthur is not. And if Henry lies, then Arthur really is truthful,inwhich case Henry will falsely claim that Arthur is not truthful. So whether Henry is truthful or not, he will surely claim that Arthur is not truthful.
+
+  In summary, Arthur will claim that
+  Arthur is truthful and Arthur's brother will claim that Arthur
+  is not truthful. So if you ask one of the brothers whether Arthur is truthful, and if you get yes for an answer, you will
+  know that you are speaking to Arthur; if you get no for an
+  answer, you will know that you are speaking to Arthur's
+  brother.
+
+  Incidentally, there is another three-word question that
+  works: "Does Arthur lie?" A yes answer to that question
+  would indicate that you are not speaking to Arthur, and a no
+  answer would indicate that you are speaking to Arthur. I leave
+  the verification of this to the reader.
+]
+
+#v(1fr)
+#pagebreak()
+
+#problem()
+Suppose that instead of wanting to find out which one is Arthur, you want to find out whether Arthur is the liar or the
+truth-teller. Again there is a three-word question that will do
+this. What three-word question will work? There is a pretty
+symmetry between the solutions of this and the last problem!
+
+#solution[
+  To find out whether Arthur is truthful, all you need to
+  ask is: "Are you Arthur?" Suppose you get the answer yes. If
+  it is a truthful answer, then the one addressed really is Arthur,
+  in which case Arthur is the truthful brother.
+
+  If the answer is
+  a lie, then the answerer is not really Arthur, in which case
+  Arthur must be the other one, again the truthful brother. So
+  regardless of whether the answer is truthful or a lie, a yes
+  answer indicates that Arthur-whichever one he is-must be
+  truthful. What if you get no for an answer?
+
+  Well, if it is a
+  truthful answer, then the speaker is not Arthur, but since he
+  is truthful, Arthur must be the brother who lies. On the other
+  hand, if the no answer was a lie, then the speaker really is
+  Arthur, in which case Arthur just told a lie. So a no answer,
+  whether it is the truth or a lie, indicates that Arthur is the liar.
+]
+
+#v(1fr)
+
+#problem()
+This time, all you are interested in finding out is which of the
+two brothers you meet is the liar and which is the truth-teller.
+You don't care which one is Arthur, or whether Arthur is the
+liar or the truth-teller. What three-word question will accomplish this?
+
+#solution[
+  Just ask him: "Do you exist?
+]
+
+#v(1fr)
+
+#problem()
+Next you are told to ask one of the brothers just one three-
+word question. If he answers yes, you will get a prize; if he
+answers no, then you get no prize. What question would you
+ask?
+
+#solution[
+  Just ask: "Are you truthful?" Both constant truth-tellers
+  and constant liars will answer yes to that question.
+]
+
+#v(1fr)
+#pagebreak()
+
+
+#problem("The Absentminded Logician")
+A certain logician, though absolutely brilliant in theoretical
+matters, was extremely unobservant and highly absent-
+minded. He met two beautiful identical-twin sisters named
+Teresa and Lenore. The two were indistinguishable ill ap-
+pearance, but Teresa always told the truth and Lenore always
+lied. The logician fell in love with one of them and married
+her, but unfortunately he forgot to find out her first name!
+The other sister didn't get married till a couple of years later.
+
+Quite shortly after the wedding, the logician had to go
+away for a logic conference. He returned a few days later. He
+then met one of the two sisters at a cocktail party and, of
+course, had no idea whether or not it was his wife. "I can find
+out in only one question," he thought proudly. "I'll simply
+use the Nelson Goodman principle and ask her if she is the
+type who could claim that she is my wife!" Then he had an
+even better idea: "I don't really have to be that elaborate and
+ask such a convoluted question. Why, I can find out if she is
+my wife by asking a much simpler question-in fact, one having only three words!"
+
+The logician was right! What three-word question answerable
+by yes or no should he ask to find out whether the lady he was addressing was his wife?
+
+#solution[
+  We recall that his wife's sister was not married at the time.
+  A three-word question that works is: "Is Teresa married?"
+  Suppose the lady answers yes. She is either Teresa or Lenore.
+  Suppose she is Teresa. Then the answer is truthful, hence Teresa is really married,
+  and the lady addressed is married and his
+  wife. If she is Lenore, the answer is a lie; Teresa is not really
+  married, so Lenore-who is the lady addressed-is married,
+  hence again the lady addressed is his wife. So a yes answer
+  indicates that he is speaking to his wife, regardless of whether
+  the answer is the truth or a lie. I leave it to the reader to verify
+  that a no answer indicates that he is speaking to his wife's sister.
+]
+
+#v(1fr)
+
+
+#problem()
+A few days later the logician again met one of the two sisters
+at another cocktail party. He again didn't know whether it was
+his wife or his sister-in-law. "It's high time I find out once
+and for all my wife's first name," he thought. "I can ask this
+lady just one three-word yes/no question, and then I'll know!"
+What three-word question could he ask?
+
+#solution[
+  The question to ask now is: "Are you married?" Suppose
+  she answers yes. Again, she is either Teresa or Lenore. Suppose
+  she is Teresa. Then the answer is truthful, hence the lady ad-
+  dressed is married, and since she is Teresa, he is married to
+  Teresa. But ,what if the lady addressed is Lenore? Then the
+  answer is a lie, hence the lady addressed is not really married,
+  and he is married to the other lady, again Teresa. So in either
+  case, a yes answer indicates that his wife's name is Teresa.
+  I again leave it to the reader to verify that a no answer
+  indicates that his wife's name is Lenore.
+]
+
+#v(1fr)
+#pagebreak()
+
+#problem()
+Suppose that in the last problem, the logician had wanted to
+know both the identity of the lady he met and the first name
+of his wife. He is again restricted to asking only one question
+answerable by yes or no, but this time there is no restriction
+on the number of words in the question.
+Can you find a question that will work?
+
+#solution[
+  No, because no such question exists!
+  You see, in each of the preceding problems, we were trying
+  to find out which of two possibilities holds, but in this problem,
+  we are trying to find out which ofJour possibilities holds.
+
+  (The four possibilities are that the lady addressed is Teresa,
+  his wife; that she is Lenore, his wife; that she is Teresa, his
+  sister-in-law; and that she is Lenore, his sister-in-law.)
+
+  However,
+  a yes/no question can elicit only two possible responses,
+  and with only two possible responses it is impossible to
+  determine which of four possibilities holds.
+]
+
+#v(1fr)

src/Advanced/Logic Puzzles/parts/02 gods.typ

@@ -0,0 +1,246 @@
+#import "@local/handout:0.1.0": *
+
+= Gods, Demons, and Mortals
+
+#definition()
+Shortly after Inspector Craig returned to London from his
+strange experience in Subterranea, he had a curious dream.
+He had been browsing that day in a library specializing in rare
+books on mythology, another of his many interests. His head
+was filled with gods and demons, and so his dream was perhaps not so surprising.
+Time sometimes passes in unusual ways in the course of a
+dream. Craig dreamed that he spent nine days in a region in
+which dwelled gods, demons, and mortals. The gods, of
+course, always told the truth, and the demons always lied. As
+to the mortals, half were knights and half were knaves. As
+usual, the knights told the truth and the knaves lied.
+
+#problem("The First Day")
+Craig dreamed that on the first day he met a dweller of the
+region who looked as ifhe might be a god, though Craig could
+not be sure. The dweller evidently guessed Craig's thoughts,
+smiled, and made a statement to reassure him. From this state-
+ment, Craig knew that he was in the presence of a god.
+Can you supply such a statement?
+
+#solution[
+  One statement that works is: "I am not a knight." If the
+  speaker were a knave or a demon, then it would be true that
+  he was not a knight, but knaves and demons don't make true
+  statements. Therefore the speaker was neither a knavtr nor a
+  demon, hence he was a knight or a god and his statement was
+  true. Since it was true, then he really is not a knight; hence
+  he must be a god.
+]
+
+
+#problem("The Second Day")
+In this episode of the dream, Craig met a terrifying-looking
+being who had every appearance of being a demon.
+"What sort of being are'you?" asked Craig, in some alarm.
+The being answered, and Craig then realized that he was confronting not a demon, but a knave. What could the being have
+answered?
+
+#solution[
+  A statement that works is: "I am a demon." Obviously
+  no demon can claim he is a demon, so the speaker is not a
+  demon. Therefore his statement was false and since he is not
+  a demon, he must be a knave.
+  Incidentally, this and the last puzzle are essentially the same
+  as Problems 4 and 5 of Chapter 1, the puzzles about the prizes.
+]
+
+#problem("The Third Day")
+In this episode, Craig met a totally nondescript-looking being
+who from appearances could have been anything at all. The
+being then made a statement from which Craig could deduce
+that he was either a god or a demon, but Craig could not tell
+which.
+Can you supply such a statement?
+
+#solution[
+  This is a bit more tricky: A statement that works is: "I
+  am either a god or a knave." That could be said by a god,
+  since a god is either a god or a knave; it could also be falsely
+  said by a demon. It couldn't be said by a knight, because a
+  knight would never lie and claim that he is either a god or a
+  knave, and it couldn't be said by a knave, because a knave
+  would never admit to the true fact that he is either a god or
+  a knave. And so the speaker must be either a god or a demon,
+  but there is no way to tell which.
+]
+
+#problem("The Fourth Day")
+Craig next met a being who made the following two statements:
+1. A god once claimed that I am a demon.
+2. No knight has ever claimed that I am a knave.
+What sort of being was he?
+
+
+#solution[
+  The speaker's first statement was obviously false, for if
+  it were true, a god would have once claimed that the speaker
+  was a demon, which would mean that the speaker really was
+  a demon, but no one who tells the truth can be a demon. Since
+  the first statement was false, so was the second statement, since
+  it was made by the same speaker. Therefore a knight did once
+  claim that the speaker was a knave, hence the speaker really
+  is a knave.
+]
+
+#problem("The Fifth Day")
+A being made the following two statements to Craig:
+1. I never claim to be a knave.
+2. I sometimes claim that I am a demon.
+What sort of being are we now dealing with?
+
+#solution[
+  The speaker's second statement was obviously a lie, be-
+  cause no truth-teller would ever say that he sometimes claims
+  to be a demon. Therefore the first statement was also a lie,
+  hence the speaker does sometimes claim to be a knave, hence
+  he must be a demon.
+]
+
+#problem("The Sixth Day")
+In this episode, Craig came across two beings, each of whom
+made a statement. Craig could then infer that at least one of
+them must be a god, but he could not tell which one. From
+neither statement alone could Craig have deduced this.
+What statements could the beings have made?
+
+#solution[
+  Many solutions are possible; here is one. Let us call the
+  two beings A and B. Now, suppose A and B make the fol-
+  lowing two statements:
+  A: B is a knight.
+  B: A is not a knight.
+  A is either telling the truth or lying.
+  Case 1-A is telling the truth: Then B really is a knight,
+  hence his statement is true, hence A is not a knight, therefore
+  A must be a god, since he is telling the truth.
+  Case 2-A is lying: Then B is not a knight, since A says
+  he is. Also, since A is lying, then A is certainly not a knight,
+  hence B's statement is true. Therefore B is telling the truth,
+  but is not a knight, hence B is a god.
+  So if Case 1 is true, A is a god; if Case 2 is true, then B is
+  a god. There is no way to tell whether A is telling the truth
+  or lying.
+]
+
+#problem("The Seventh Day")
+On the next day, Craig again met two beings each of whom
+made a statement. Craig could then infer that one of them was
+a knave and the other a demon, though he could not tell which
+was which. Again, from neither statement alone could Craig
+have inferred this. Can you supply two such statements?
+
+#solution[
+  gain let us call the two beings A and B. The following
+  statements would work:
+  A: Both of us are knaves.
+  B: Both of us are demons.
+  It is obvious that both are lying. Since A is lying, they are
+  not both knaves. Since B is lying, they are not both demons.
+  Therefore one is a knave and one is a demon, but there is no
+  way to tell which one is which.
+]
+
+#problem("Introducing Thor")
+On the eighth day, Craig met a being who had every ap-
+pearance of being the god Thor. The being made a statement,
+and Craig then knew he must be Thor.
+What statement could Thor have made?
+
+#solution[
+  A statement that works is: "I am either a knave or a demon
+  or the god Thor."
+  If the speaker were either a knave or a demon, then it would
+  be true that he is either a knave or a demon or the god Thor.
+  This would mean that a knave or a demon made a true state-
+  ment, which is not possible. Therefore the speaker is neither
+  a knave nor a demon, hence his statement is true. Hence he
+  must be the god Thor.
+]
+
+#problem("A Perplexity Resolved")
+Craig and Thor became fast friends. In fact, on the evening of
+the ninth day, Thor gave a magnificent banquet in Craig's
+honor. "I propose a toast to our illustrious guest!" said Thor,
+as he raised his glass of nectar.
+After a round of cheers, Craig was asked to speak.
+"I am very perplexed!" said Craig as he rose. "I wonder
+if this may not all be a dream!"
+"Why do you think you may be dreaming?" asked Thor.
+"Because," said Craig, "two incidents have occurred today
+that seem totally inexplicable. This morning I met someone
+who made a statement which no knight, knave, god, or demon
+could possibly make. Then this afternoon I met someone else
+who also made a statement which no dweller of this region
+could possibly make. That is why I suspect that I may be
+dreaming. "
+"Oh!" said Thor. "Be reassured; you are not dreaming.
+The two incidents have a perfectly rational explanation. You
+see, we have had two visitors here from another realm. Both
+of them are mortal. One is Cyrus, who always tells the truth,
+although he is not called a knight since he is not from this
+region. The other is Alexander, who sometimes tells the truth
+and sometimes lies. It must have been those two whom you
+met today. What statements did they make?"
+Craig then told the company what each had said.
+"That explains it perfectly!" said Thor. "Moreover, it fol-
+lows from their having said what they did that Cyrus was the
+one you met in the morning. And interestingly enough, if you
+hadn't met Alexander in the afternoon, you could never have
+known whether the one you met in the morning was Cyrus
+or Alexander."
+Craig thought the matter over and realized that Thor was
+right.
+What statements could these two outsiders have made
+which fulfill all of the above conditions?
+
+#solution[
+  Here is one possible solution.
+  MORNING SPEAKER: "I am neither a knight nor a god."
+  AFTERNOON SPEAKER: "I am either a knave or a demon."
+  No inhabitant of the region could make either of those
+  statements. No knight or god could claim that he is neither a
+  knight nor a god; no knave or demon could make the true
+  statement that he is neither a knight nor a god. As for the
+  second statement, obviously no knight or god would claim to
+  be either a knave or a demon and no knave or demon would
+  admit to being a knave or a demon. Therefore both were out-
+  siders; namely, Cyrus and Alexander. The statement of the
+  morning speaker was true and the statement of the afternoon
+  sp,eaker was false. Since Cyrus never makes false statements,
+  he couldn't have been the afternoon speaker. Thus he was the
+  morning speaker.
+]
+
+#problem("A Philosophical Puzzle")
+The next morning when Craig was wide awake and recalling
+his dream, he wondered whether he had been logically inconsistent in his sleep. "The trouble is this," thought Craig: "In
+my dream I believed that Thor was a god and that gods always
+tell the truth. Yet Thor told me that I wasn't dreaming. Now
+how could Thor, who tells the truth, say that I wasn't dreaming when in fact I was? Wasn't this an inconsistency on my
+part?"
+Would you say that Craig's dream was logically inconsistent?
+
+#solution[
+  As I see it, Craig's dream was not necessarily inconsistent. If
+  Craig had actually believed in the dream that he was dreaming,
+  then the set of his beliefs during his dream would have been
+  inconsistent, since the following propositions are indeed log-
+  ically contradictory: (1) Thor is a god; (2) Gods make only
+  true statements; (3) Thor stated that Craig was not dreaming;
+  (4) Craig was dreaming.
+  The contradiction is obvious. However, there is no evi-
+  dence that Craig at any time of his dream believed that he was
+  dreaming, although at one point he wondered whether he
+  might be dreaming. Craig presumably believed that he was
+  awake, and this belief, though false, was perfectly consistent
+  with the other beliefs of his dream.
+  Curiously enough, if Craig had formulated the belief that
+  he was dreaming, then this belief, though correct, would have
+  created a logical inconsistency!
+]

src/Advanced/Nonstandard Analysis/main.tex

@@ -1,19 +1,3 @@
-% Copyright (C) 2023 <Mark (mark@betalupi.com)>
-%
-% This program is free software: you can redistribute it and/or modify
-% it under the terms of the GNU General Public License as published by
-% the Free Software Foundation, either version 3 of the License, or
-% (at your option) any later version.
-%
-% You may have received a copy of the GNU General Public License
-% along with this program. If not, see <https://www.gnu.org/licenses/>.
-%
-%
-%
-% If you edit this, please give credit!
-% Quality handouts take time to make.
-
-
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[

src/Advanced/Nonstandard Analysis/parts/dual.tex

@@ -1,19 +1,3 @@
-% Copyright (C) 2023 <Mark (mark@betalupi.com)>
-%
-% This program is free software: you can redistribute it and/or modify
-% it under the terms of the GNU General Public License as published by
-% the Free Software Foundation, either version 3 of the License, or
-% (at your option) any later version.
-%
-% You may have received a copy of the GNU General Public License
-% along with this program. If not, see <https://www.gnu.org/licenses/>.
-%
-%
-%
-% If you edit this, please give credit!
-% Quality handouts take time to make.
-
-
 \section{Dual Numbers}
 
 \definition{}

src/Advanced/Nonstandard Analysis/parts/extensions.tex

@@ -1,18 +1,3 @@
-% Copyright (C) 2023 <Mark (mark@betalupi.com)>
-%
-% This program is free software: you can redistribute it and/or modify
-% it under the terms of the GNU General Public License as published by
-% the Free Software Foundation, either version 3 of the License, or
-% (at your option) any later version.
-%
-% You may have received a copy of the GNU General Public License
-% along with this program. If not, see <https://www.gnu.org/licenses/>.
-%
-%
-%
-% If you edit this, please give credit!
-% Quality handouts take time to make.
-
 \section{Extensions of $\mathbb{R}$}
 
 \definition{}

src/Advanced/Nonstandard Analysis/parts/supremum.tex

@@ -1,19 +1,3 @@
-% Copyright (C) 2023 <Mark (mark@betalupi.com)>
-%
-% This program is free software: you can redistribute it and/or modify
-% it under the terms of the GNU General Public License as published by
-% the Free Software Foundation, either version 3 of the License, or
-% (at your option) any later version.
-%
-% You may have received a copy of the GNU General Public License
-% along with this program. If not, see <https://www.gnu.org/licenses/>.
-%
-%
-%
-% If you edit this, please give credit!
-% Quality handouts take time to make.
-
-
 \section*{The supremum \& the infimum}
 
 \definition{}

src/Advanced/Relativity/diagram.tex

@@ -0,0 +1,249 @@
+% Diagrams adapted from https://tikz.net/relativity_minkowski_diagram/
+
+%
+% MARK: Matt's tweaks
+%
+
+\newcommand{\emptydiagram}[1]{
+	\begin{center}
+		\begin{tikzpicture}[scale=2]
+			\def\xmax{2}
+			\def\xmaxp{2.2} % maximum of rotated axis
+			\def\Nlines{6} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+			\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+			\coordinate (O) at (0,0);
+			\coordinate (X) at (\xmax+0.2,0);
+			\coordinate (T) at (0,\xmax+0.2);
+			\coordinate (X') at (\ang:\xmaxp+0.2);
+			\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+				\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+				\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+				\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+			}
+
+			% AXES
+			\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {#1: $t$};
+			\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+		\end{tikzpicture}
+	\end{center}
+}
+
+\newcommand{\emptydiagramc}[1]{
+	\begin{center}
+		\begin{tikzpicture}[scale=2]
+			\def\xmax{2}
+			\def\xmaxp{2.2} % maximum of rotated axis
+			\def\Nlines{6} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+			\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+			\coordinate (O) at (0,0);
+			\coordinate (X) at (\xmax+0.2,0);
+			\coordinate (T) at (0,\xmax+0.2);
+			\coordinate (X') at (\ang:\xmaxp+0.2);
+			\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+				\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+				\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+				\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+			}
+
+			% AXES
+			\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {#1: $ct$};
+			\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+		\end{tikzpicture}
+	\end{center}
+}
+
+\newcommand{\halfdiagramc}[1]{
+	\begin{center}
+		\begin{tikzpicture}[scale=2]
+			\def\xmax{2}
+			\def\xmaxp{2.2} % maximum of rotated axis
+			\def\Nlines{6} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+			\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+			\coordinate (O) at (0,0);
+			\coordinate (X) at (\xmax+0.2,0);
+			\coordinate (T) at (0,\xmax+0.2);
+			\coordinate (X') at (\ang:\xmaxp+0.2);
+			\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+			% WORLD LINE GRID
+			\message{  Making world lines...^^J}
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+			\draw[world line]   (-\x,-0.2) -- (-\x,\xmax);
+			\draw[world line]   ( \x,-0.2) -- ( \x,\xmax);
+			%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+			\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+			}
+
+			% AXES
+			\draw[->,thick] (0,-0.2) -- (T) node[left=-1] {#1: $ct$};
+			\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+		\end{tikzpicture}
+	\end{center}
+}
+
+\newcommand{\halfdiagramcwide}[1]{
+	\begin{center}
+		\begin{tikzpicture}[scale=2]
+			\def\xmax{3.5}
+			\def\xmaxp{2.2} % maximum of rotated axis
+			\def\Nlines{10} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+			\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+			\coordinate (O) at (0,0);
+			\coordinate (X) at (\xmax+0.2,0);
+			\coordinate (T) at (0,1.95);
+			\coordinate (X') at (\ang:\xmaxp+0.2);
+			\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\draw[world line]   (-\x,-0.2) -- (-\x,1.95);
+				\draw[world line]   ( \x,-0.2) -- ( \x,1.95);
+				%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+				%\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+			}
+
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,5}{
+				% \draw[world line]   (-\x,-0.2) -- (-\x,\xmax);
+				% \draw[world line]   ( \x,-0.2) -- ( \x,\xmax);
+				%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+				\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+			}
+
+			% AXES
+			\draw[->,thick] (0,-0.2) -- (T) node[left=-1] {#1: $ct$};
+			\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+		\end{tikzpicture}
+	\end{center}
+}
+
+
+%
+% MARK: tikz setup
+%
+
+\usepackage{tikz}
+\usetikzlibrary{patterns}
+
+\usepackage{hyperref}
+\usepackage{graphicx}
+\usepackage{caption}
+
+\usepackage{etoolbox} % ifthen
+\usepackage[outline]{contour} % glow around text
+\usetikzlibrary{calc} % for adding up coordinates
+\usetikzlibrary{decorations.markings,decorations.pathmorphing}
+\usetikzlibrary{angles,quotes} % for pic (angle labels)
+\usetikzlibrary{arrows.meta} % for arrow size
+\usepackage{xfp} % higher precision (16 digits?)
+\contourlength{1.1pt}
+
+\tikzset{>=latex} % for LaTeX arrow head
+\colorlet{myred}{red!85!black}
+\colorlet{mydarkred}{red!55!black}
+\colorlet{mylightred}{red!85!black!12}
+\colorlet{myfieldred}{mydarkred!5} % for S' background
+\colorlet{myredhighlight}{myred!20} % highlights simultaneity in ladder paradox
+\colorlet{myblue}{blue!80!black}
+\colorlet{mydarkblue}{blue!50!black}
+\colorlet{mylightblue}{blue!50!black!30}
+\colorlet{mylightblue2}{myblue!10}
+\colorlet{mygreen}{green!80!black}
+\colorlet{mypurple}{blue!40!red!80!black}
+\colorlet{mydarkgreen}{green!50!black}
+\colorlet{mydarkpurple}{blue!40!red!50!black}
+\colorlet{myorange}{orange!40!yellow!95!black}
+\colorlet{mydarkorange}{orange!40!yellow!85!black}
+\colorlet{mybrown}{brown!20!orange!90!black}
+\colorlet{mydarkbrown}{brown!20!orange!55!black}
+\colorlet{mypurplehighlight}{mydarkpurple!20} % highlights simultaneity in ladder paradox
+\tikzstyle{world line}=[myblue!40,line width=0.3]
+\tikzstyle{world line t}=[mypurple!50!myblue!40,line width=0.3]
+\tikzstyle{world line'}=[mydarkred!40,line width=0.3]
+\tikzstyle{mysmallarr}=[-{Latex[length=3,width=2]},thin]
+\tikzstyle{mydashed}=[dash pattern=on 3 off 3]
+\tikzstyle{rod}=[mydarkbrown,draw=mydarkbrown,double=mybrown,double distance=2pt,
+                 line width=0.2,line cap=round,shorten >=1pt,shorten <=1pt]
+%\tikzstyle{rod'}=[rod,draw=mydarkbrown!80!red!85,double=mybrown!80!red!85]
+\tikzstyle{vector}=[->,line width=1,line cap=round]
+\tikzstyle{vector'}=[vector,shorten >=1.2]
+\tikzstyle{particle}=[mygreen,line width=0.9]
+\tikzstyle{photon}=[-{Latex[length=5,width=4]},myorange,line width=0.8,decorate,
+                    decoration={snake,amplitude=1.0,segment length=5,post length=5}]
+
+\def\tick#1#2{\draw[thick] (#1) ++ (#2:0.06) --++ (#2-180:0.12)}
+\def\tickp#1#2{\draw[thick,mydarkred] (#1) ++ (#2:0.06) --++ (#2-180:0.12)}
+\def\Nsamples{100} % number samples in plot
+
+% COMMON AXES
+\pgfdeclarelayer{back} % to draw on background
+\pgfsetlayers{back,main} % set order
+\def\xmin{0.23}
+\def\xmax{2}
+\def\Nlines{6} % number of world lines (at constant x/t)
+\def\DNxp{0}   % difference in number of world lines of x' axis
+\def\DNyp{0}   % difference in number of world lines of ct' axis
+\def\DNy{0}    % difference in number of world lines of ct axis
+\def\ang{20}   % angle between x and x' axes
+\def\xplabelang{180} % anchor angle of x' axis label
+%\pgfmathsetmacro\ang{atan(0.44)} % angle between x and x' axes
+\def\axes{
+	\pgfmathsetmacro\d{\xmax/(\Nlines+0.4)} % grid size
+	\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+	\pgfmathsetmacro\ymax{\xmax+\DNy*\d} % maximum of y = ct axis
+	\pgfmathsetmacro\xmaxp{(\xmax/\d+\DNxp)*\D} % maximum of x' axis
+	\pgfmathsetmacro\ymaxp{(\xmax/\d+\DNyp)*\D} % maximum of y' = ct' axis
+	\pgfmathsetmacro\Nylines{\Nlines+\DNy} % number of world lines at constant ct'
+	\pgfmathsetmacro\Nxplines{\Nlines+\DNxp} % number of world lines at constant x'
+	\pgfmathsetmacro\Nyplines{\Nlines+\DNyp} % number of world lines at constant ct'
+	\coordinate (O) at (0,0);
+	\coordinate (X) at (\xmax+0.15,0);
+	\coordinate (T) at (0,\ymax+0.15);
+	\coordinate (X') at (\ang:\xmaxp+0.2);
+	\coordinate (T') at (90-\ang:\ymaxp+0.2);
+
+	% FILL
+	\begin{pgfonlayer}{back} % draw on back
+	\fill[myfieldred]
+		(\ang:-\xmin) -- (\ang:\xmaxp) --++ (90-\ang:\ymaxp) --++ (\ang:-\xmaxp)
+		-- (90-\ang:-\xmin) -- cycle;
+	\end{pgfonlayer}
+
+	% WORLD LINE GRID
+	\message{  Making world lines...^^J}
+	\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+	\draw[world line]   (\x,0) -- (\x,\ymax);
+	}
+	\foreach \i [evaluate={\t=\i*\d;}] in {1,...,\Nylines}{
+	\draw[world line t] (0,\t) -- (\xmax,\t);
+	}
+
+	% BOOSTED WORLD LINE GRID
+	\foreach \i [evaluate={\x=\i*\D;}] in {1,...,\Nxplines}{
+		\draw[world line'] (\ang:\x) --++ (90-\ang:\ymaxp);
+	}
+	\foreach \i [evaluate={\t=\i*\D;}] in {1,...,\Nyplines}{
+		\draw[world line'] (90-\ang:\t) --++ (\ang:\xmaxp);
+	}
+
+	% AXES
+	\draw[->,thick] (0,-\xmin) -- (T) node[left=-1] {$ct$};
+	\draw[->,thick] (-\xmin,0) -- (X) node[below=0] {$x$};
+	\draw[->,thick,mydarkred] (90-\ang:-\xmin) -- (T')
+	node[right=5,above=-1] {$ct'$};
+	\draw[->,thick,mydarkred] (\ang:-\xmin) -- (X')
+	node[anchor=\xplabelang,inner sep=2] {$x'$};
+}

src/Advanced/Relativity/main.tex

@@ -0,0 +1,32 @@
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	singlenumbering,
+	shortwarning
+]{../../../lib/tex/ormc_handout}
+\usepackage{../../../lib/tex/macros}
+\input{diagram}
+
+
+
+\uptitlel{Advanced 2}
+\uptitler{\smallurl{}}
+\title{Special Relativity}
+\subtitle{
+	Prepared by Matthew Kowalski on \today{} \\
+}
+
+\begin{document}
+	\maketitle
+
+	%\input{parts/relative velocity}
+	\input{parts/01 spacetime diagrams}
+	\input{parts/02 galilean}
+	\input{parts/03 special}
+	%\input{parts/relative velocity}
+	\input{parts/04 simultaneity}
+	\input{parts/05 proper}
+	\input{parts/06 contraction}
+
+\end{document}

src/Advanced/Relativity/meta.toml

@@ -0,0 +1,8 @@
+[metadata]
+title = "Special Relativity"
+
+# Matthew Kowalski, 2025-02-02
+
+[publish]
+handout = false
+solutions = true

src/Advanced/Relativity/parts/01 spacetime diagrams.tex

@@ -0,0 +1,196 @@
+\section{Spacetime Diagrams}
+
+We are going to derive the principles and consequences of special relativity using basic geometry. \\
+o help with our visualization, we will be using spacetime diagrams (called {\em Minkowski diagrams}).
+
+To make our models simpler, we will only be considering {\em one spatial dimension}.
+
+We plot space, which we denote by $x$, as the horizontal axis, and time, which we denote by $t$ as the vertical axis. For a given object, we can then plot its position at any given time. \\
+This will give a (potentially curvy) line that we call the object's {\em world line}.
+
+\example{}
+Suppose that at time $t = 0$, you are standing still with your cat at your feet.
+Your cat walks away from you at speed $1$. We can represent this with a spacetime diagram:
+
+\begin{center}
+	% SPACETIME DIAGRAM with WORLD LINES
+	\begin{tikzpicture}[scale=2.0]
+		%\message{Worldlines^^J}
+		\def\ymin{0.2}
+		\def\xmin{1.6}
+		\def\xmax{2}
+		\def\Nlines{4} % number of world lines (at constant x/t)
+		\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+		\coordinate (O) at (0,0);
+		\coordinate (T) at (0,\xmax+0.2);
+
+		% WORLD LINES GRID
+		%\message{  Making world lines...^^J}
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+			\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+		}
+		\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+		\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+		\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+
+		% AXES
+		\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
+		\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
+
+		% VECTORS
+		\draw[vector,myred, very thick] (O) -- (4*\d,4*\d)
+		node[mydarkred,right=10,above] {\contour{white}{cat: $x(t)=t$}};
+		\draw[vector,myblue, very thick] (O) -- (0,4*\d)
+		node[mydarkblue,below left=0] {\contour{white}{you: $x(t)=0$}};
+		%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+	\end{tikzpicture}
+\end{center}
+
+\note[Note]{
+	The horizontal axis is space and the vertical axis is time. \par
+	We are only working with one spatial dimension.
+}
+
+\vfill
+\pagebreak
+
+\problem{}<pets scatter>
+Suppose that you are standing still at time $t = 0$ and your many pets lie at your feet.
+
+\begin{itemize}
+	\item Your cat, unhappy that she is not fed, begins walking away to your right at speed $2$.
+	\item Your dog, distracted by a squirrel, walks away to your left at speed $1$.
+	\item Your hamster, just wanting to feel included, waits a second and then follows the dog at speed $2$.
+	After reaching your dog, your hamster turns around and sprints after the cat at speed $3$.
+\end{itemize}
+
+Draw this situation in the provided spacetime diagram.
+
+\makeatletter
+\if@solutions\else
+	\emptydiagram{Alice}
+\fi
+\makeatother
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=2.0]
+			\def\ymin{0.2}
+			\def\xmin{2}
+			\def\xmax{2}
+			\def\Nlines{4} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+			\coordinate (O) at (0,0);
+			\coordinate (T) at (0,\xmax+0.2);
+
+			% WORLD LINES GRID
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+				\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+				\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+			}
+			\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+			\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+			\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+			\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
+
+			% AXES
+			\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
+			\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
+
+			% VECTORS
+			\draw[vector,myred] (O) -- (0,4*\d)
+			node[mydarkred,below left=0] {\contour{white}{you: $x(t)=0$}};
+			\draw[vector,myblue] (O) -- (4*\d,2*\d)
+			node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=2t$}};
+			\draw[vector,mygreen] (O) -- (-4*\d,4*\d)
+			node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=-t$}};
+			\draw[vector,black] (O) -- (0,\d) -- (-2*\d, 2*\d) -- (4*\d, 4*\d)
+			node[black,below right=0] {\contour{white}{hamster}};
+			% \draw[vector,myblue]
+			%   (O) to[out=35,in=-100] (O)
+			%   to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
+			%   node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
+			%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+		\end{tikzpicture}
+	\end{center}
+\end{solution}
+
+
+\vfill
+
+\problem{Event}
+Any single point $(x,t)$ on a spacetime diagram is considered an {\em event}
+because it describes a time and place. For instance, what is the event that
+corresponds to your hamster catching up to your dog?
+
+\begin{solution}
+	$(-2,2)$
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}<pets scatter train>
+Suppose that the situation of \ref{pets scatter} occurred while you were riding on
+a train moving to the right at speed $1$.
+Everything occurs relative to you in the same way.
+Draw the same diagram in this new situation.
+Are any of your pets staying still in this new situation?
+
+\makeatletter
+\if@solutions\else
+	\emptydiagram{Train}
+\fi
+\makeatother
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=2.0]
+			\def\ymin{0.2}
+			\def\xmin{2}
+			\def\xmax{2}
+			\def\Nlines{4} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+			\coordinate (O) at (0,0);
+			\coordinate (T) at (0,\xmax+0.2);
+
+			% WORLD LINES GRID
+			\message{  Making world lines...^^J}
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+			\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+			\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+			}
+			\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+			\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+			\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+			\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
+
+			% AXES
+			\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
+			\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
+
+			% VECTORS
+			\draw[vector,myred] (O) -- (4*\d,4*\d)
+			node[mydarkred,below left=0] {\contour{white}{you: $x(t)=t$}};
+			\draw[vector,myblue] (O) -- (6*\d,2*\d)
+			node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=3t$}};
+			\draw[vector,mygreen] (O) -- (0,4*\d)
+			node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=0$}};
+			\draw[vector,black] (O) -- (\d,\d) -- (0, 2*\d) -- (8*\d, 4*\d)
+			node[black,below right=0] {\contour{white}{hamster}};
+			% \draw[vector,myblue]
+			%   (O) to[out=35,in=-100] (O)
+			%   to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
+			%   node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
+			%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+		\end{tikzpicture}
+	\end{center}
+
+	The dog remains stationary in this reference frame.
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Relativity/parts/02 galilean.tex

@@ -0,0 +1,195 @@
+\section{Galilean Relativity}
+
+Much like you can watch your pets scatter from the perspective of a train, we can watch the world from anyone's perspective. When we shift perspective like this, just using our normal intuition, we call this {\em Galilean relativity}.
+
+\example{}
+Consider the situation of Example $1$ again, but now from the perspective of your cat.
+From your cat's perspective, she's the one staying still and you're the one walking away,
+only now you're walking away at speed $1$ to the left. We'll denote our new spatial variable with $x'$.
+
+\begin{center}
+	% SPACETIME DIAGRAM with WORLD LINES
+	\begin{tikzpicture}[scale=2.0]
+		\def\ymin{0.2}
+		\def\xmin{2}
+		\def\xmax{2}
+		\def\Nlines{4} % number of world lines (at constant x/t)
+		\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+		\coordinate (O) at (0,0);
+		\coordinate (T) at (0,\xmax+0.2);
+
+		% WORLD LINES GRID
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+			\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+		}
+		\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+		\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+		\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+		\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
+
+		% AXES
+		\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
+		\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x'$};
+
+		% VECTORS
+		\draw[vector,myred, very thick] (O) -- (0,4*\d)
+		node[mydarkred,above right] {\contour{white}{cat: $x'(t)=0$}};
+		\draw[vector,myblue, very thick] (O) -- (-4*\d,4*\d)
+		node[mydarkblue,below left=0] {\contour{white}{you: $x'(t)=-t$}};
+		%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+	\end{tikzpicture}
+\end{center}
+
+\problem{}
+Draw the situation from \ref{pets scatter} in your cat's perspective. \par
+What if we drew the situation from \ref{pets scatter train} in your cat's perspective? \par
+Would there be any change when the cat is on the train? Why or why not?
+
+
+\makeatletter
+\if@solutions\else
+	\emptydiagram{Cat}
+\fi
+\makeatother
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=1.8]
+		\def\ymin{0.2}
+		\def\xmin{4}
+		\def\xmax{2}
+		\def\Nlines{4} % number of world lines (at constant x/t)
+		\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+		\coordinate (O) at (0,0);
+		\coordinate (T) at (0,\xmax+0.2);
+
+		% WORLD LINES GRID
+		\message{  Making world lines...^^J}
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+			\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+		}
+		\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+		\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+		\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+		\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
+		\draw[world line] (-5*\d,-\ymin) -- (-5*\d,\xmax);
+		\draw[world line] (-6*\d,-\ymin) -- (-6*\d,\xmax);
+		\draw[world line] (-7*\d,-\ymin) -- (-7*\d,\xmax);
+		\draw[world line] (-8*\d,-\ymin) -- (-8*\d,\xmax);
+
+		% AXES
+		\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
+		\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
+
+		% VECTORS
+		\draw[vector,myred] (O) -- (-8*\d,4*\d)
+		node[mydarkred,below left=0] {\contour{white}{you: $x(t)=-2t$}};
+		\draw[vector,myblue] (O) -- (0,4*\d)
+		node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=0$}};
+		\draw[vector,mygreen] (O) -- (-9*\d,3*\d)
+		node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=-3t$}};
+		\draw[vector,black] (O) -- (-2*\d,1*\d) -- (-6*\d, 2*\d) -- (-4*\d, 4*\d)
+		node[black,below right=0] {\contour{white}{hamster}};
+		% \draw[vector,myblue]
+		%   (O) to[out=35,in=-100] (O)
+		%   to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
+		%   node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
+		%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+		\end{tikzpicture}
+	\end{center}
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+\definition{Reference Frame}
+When we view the world from the perspective of different objects,
+we say that we are working in different {\em reference frames}.
+The original Example 1, where you are stationary, is your reference frame.
+The new plot in Example 5, where your cat is stationary, is your cat's reference frame.
+
+If we want to compare what is happening in multiple reference frames at once, we can graph
+multiple spacetime grids on one plot. If we overlay the cat's reference frame onto your reference frame,
+we can visualize everything in Example 1 as:
+
+% SPACETIME DIAGRAM - GALILEAN TRANSFORMATION
+\begin{center}
+	\begin{tikzpicture}[scale=1.8]
+		\def\xmax{2}
+		\def\xmaxp{2.1} % maximum of rotated axis
+		\def\Nlines{4} % number of world lines (at constant x/t)
+		\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+		\pgfmathsetmacro\ang{atan(1)} % angle
+		\coordinate (O) at (0,0);
+		\coordinate (X) at (\xmax+0.2,0);
+		\coordinate (T) at (0,\xmax+0.2);
+		\coordinate (X') at (\ang:\xmaxp+0.2);
+		\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+		% WORLD LINES GRID
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+			\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+			\draw[world line t] ({-\xmax-tan(\ang)*\x},-\x) -- (\xmax,-\x);
+			\draw[world line t] (-\xmax,\x) -- ({\xmax+tan(\ang)*\x},\x);
+		}
+
+		% AXES
+		\draw[->,thick] (0,-\xmax) -- (T) node[left=0] {$t$};
+		\draw[->,thick] (-\xmax,0) -- (X) node[right=6,below=-1] {$x={\color{mydarkred}x'}$};
+		\draw[->,thick,mydarkred, very thick] (90-\ang:-\xmaxp) -- (T')
+		node[left=-1] {$t'$}
+		node[right=2,below right=-2] {cat: $x(t) = t$};
+		% VECTORS
+		\draw[vector,myblue, very thick] (O) -- (0,4*\d)
+		node[mydarkblue,below left=0] {\contour{white}{you: $x(t)=0$}};
+
+		% WORLD LINES GRID - BOOSTED
+		\message{  Making world lines, boosted...^^J}
+		\fill[mydarkred,opacity=0.05]
+		(O) --++ (90-\ang:\xmax) --++ (\xmax,0) --++ (90-\ang:-\xmax) -- cycle;
+		\fill[mydarkred,opacity=0.05]
+		(O) --++ (90-\ang:-\xmax) --++ (-\xmax,0) --++ (90-\ang:\xmax) -- cycle;
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line'] (\x,0) --++ (90-\ang:\xmax);
+			\draw[world line'] (-\x,0) --++ (90-\ang:-\xmax);
+		}
+
+		\node[right] at (5.6*\d, \d) {$x=\color{mydarkred}x' + 1$};
+		%\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
+	\end{tikzpicture}
+\end{center}
+
+Here $x,t$ are the spacetimes coordinates in your perspective and $x',t'$ are
+the spacetime coordinates in your cat's perspective. Note that $t = t'$ for any point while $x = x' + t$.
+
+
+
+\problem{}
+What does it mean for two events to lie on the same vertical (blue) line from your perspective? \par
+What does it mean for two events to lie on the same slanted (red) line from your cat's perspective?
+\vfill
+
+\begin{solution}
+	Two events lie on the same vertical line if they occur at the same location in your reference frame. \par
+	Two events lie on the same slanted line if they occur at the same location in your cat's reference frame.
+\end{solution}
+
+
+% \remark{}
+%     Here we are forcing time in the cat's reference frame to behave the same as in our reference frame. As in, one second for the cat is one second for us and vice versa. However, looking at the plot and measuring distances, it almost looks like one second for us is longer than one second for the cat... Suspicious...
+
+
+\problem{}
+In the situation from Problem 2, when will your hamster catch up to your cat? \\
+Choose the most convenient reference frame to work in, you shouldn't have to do much math.
+
+\begin{solution}
+	Using the cat's reference frame drawn in Problem 6, the hamster will catch up to the cat at $t = 8$.
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Relativity/parts/03 special.tex

@@ -0,0 +1,301 @@
+\section{Special Relativity}
+Galilean relativity is nice until we start going really, {\em really} fast.
+Since most of us are terribly slow, we can use it without any issues.
+However, in reality, things are much weirder. In particular, there is a maximum speed: the speed of light,
+$$c = 299,792,458 \tfrac{m}{s}.$$
+Nothing can move faster than the speed of light and {\bf in every reference frame, light will move at this speed}.
+
+Let's see if this is consistent with Galilean relativity.
+We are going to making things easier for ourselves now and change units. Instead of measuring time $t$, we are now going to measure $ct$.
+
+\problem{}<photon diagram>
+Suppose you are sitting still and you send one photon to your right. Draw this photon on a spacetime diagram, with horizontal axis $x$ and vertical axis $ct$.
+
+
+\makeatletter
+\if@solutions\else
+	\halfdiagramc{You}
+\fi
+\makeatother
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=2.0]
+			\def\ymin{0.2}
+			\def\xmin{1.6}
+			\def\xmax{2}
+			\def\Nlines{4} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+			\coordinate (O) at (0,0);
+			\coordinate (T) at (0,\xmax+0.2);
+
+			% WORLD LINES GRID
+			\message{  Making world lines...^^J}
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+				\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+			}
+			\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+			\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+			\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+
+			% AXES
+			\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
+			\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
+
+			% VECTORS
+			\draw[vector,myred] (O) -- (0,4*\d)
+			node[mydarkred,below left=0] {\contour{white}{you: $x(t)=0$}};
+			\draw[photon,shorten >=2] (O) -- (4*\d, 4*\d)
+			node[black, above right] {\contour{white}{photon}};
+			%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+		\end{tikzpicture}
+	\end{center}
+\end{solution}
+
+\vfill
+
+\problem{}
+Suppose you are now sitting on a train that is moving to the right at $c/2$
+and again send one photon to your right.
+Draw this diagram in the reference frame of the ground.
+
+Draw your (Galilean) reference frame on top of this diagram
+What is the speed of the photon in your reference frame? Is that a problem?
+
+\makeatletter
+\if@solutions\else
+	\halfdiagramc{Train}
+\fi
+\makeatother
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=1.5]
+			\pgfmathsetmacro\ang{atan(1/2)} % angle
+			\def\ymin{0.2}
+			\def\xmin{1.6}
+			\def\xmax{2}
+			\def\Nlines{4} % number of world lines (at constant x/t)
+			\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+			\coordinate (O) at (0,0);
+			\coordinate (T) at (0,\xmax+0.2);
+
+			% WORLD LINES GRID
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\draw[world line]   ( \x,-\ymin) -- ( \x,\xmax);
+				\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+			}
+			\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
+			\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
+			\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
+
+			% AXES
+			\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
+			\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x=\color{mydarkred}x'$};
+
+			% VECTORS
+			\draw[vector,myred] (O) -- (2*\d,4*\d)
+			node[mydarkred,below left=0] {\contour{white}{you: $x(t)=ct/2$}};
+			\draw[photon,shorten >=2] (O) -- (4*\d, 4*\d)
+			node[black, above right] {\contour{white}{photon}};
+			%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
+
+			% WORLD LINES GRID - BOOSTED
+			\fill[mydarkred,opacity=0.05]
+			(O) --++ (90-\ang:\xmax) --++ (\xmax,0) --++ (90-\ang:-\xmax) -- cycle;
+			% \fill[mydarkred,opacity=0.05]
+			%   (O) --++ (90-\ang:-\xmax) --++ (-\xmax,0) --++ (90-\ang:\xmax) -- cycle;
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line'] (\x,0) --++ (90-\ang:\xmax);
+			% \draw[world line'] (-\x,0) --++ (90-\ang:-\xmax);
+			}
+		\end{tikzpicture}
+	\end{center}
+
+	In your reference frame, the light is only moving at speed $c/2$. Uh oh.
+\end{solution}
+
+
+\vfill
+\pagebreak
+
+
+\problem{}
+Clearly, Galilean relativity and the absolute speed of light do not mix well together. Having noticed this, you are now in the same boat as early 20th century physicists.
+Can you brainstorm any ways to fix Galilean relativity to account for this absolute speed of light?
+
+{\em Hint 1}: Try different methods of drawing the axes of your reference frame that would maintain the speed of light in both the rest frame and in your reference frame.
+
+{\em Hint 2}: The photon worldline always bisects the angle between the space and time axes. \\ Is there a way that you can make that happen in your reference frame?
+
+{\em \color{gray} Don't worry if you don't have any ideas! It took physicists a while to figure this out. Whenever you want to move on, we have the solution on the new page.}
+
+\makeatletter
+\if@solutions\else
+	\emptydiagramc{Train}
+\fi
+\makeatother
+
+
+
+\vfill
+\pagebreak
+
+
+\definition{Lorentz Boost}
+Looking at our spacetime diagram from \ref{photon diagram}, we see that the photon worldline bisects the angle between the $ct$ axis and $x$ axis. So if we want to maintain this speed in all reference frames, we just need to make sure that photons bisect our new time axis $ct'$ and our new space axis $x'$.
+
+In order to do this, we're going to to rotate our space axis $x'$ by the same angle that our $ct'$ axis is rotated. Rotating both axes like this is called a {\em Lorentz boost} and is best visualized in the following diagram:
+
+% SPACETIME DIAGRAM - LORENTZ BOOST
+\begin{center}
+	\begin{tikzpicture}[scale=1.8]
+		\def\xmax{2}
+		\def\xmaxp{2.2} % maximum of rotated axis
+		\def\Nlines{5} % number of world lines (at constant x/t)
+		\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+		\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+		\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+		\coordinate (O) at (0,0);
+		\coordinate (X) at (\xmax+0.2,0);
+		\coordinate (T) at (0,\xmax+0.2);
+		\coordinate (X') at (\ang:\xmaxp+0.2);
+		\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+		% WORLD LINE GRID
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+			\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+			\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+			\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+		}
+
+		% BOOSTED WORLD LINE GRID
+		\message{  Making world lines for boosted frame...^^J}
+		\fill[mydarkred,opacity=0.05]
+		(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
+		\fill[mydarkred,opacity=0.05]
+		(O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
+		\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
+			\draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
+			\draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
+			\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
+			\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
+		}
+
+		% AXES
+		\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
+		\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+		\draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
+		node[right=5,above=-1] {you: $ct'$};
+		\draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
+
+		% ANGLES
+		\draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
+		\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
+
+		% PHOTON
+		\draw[photon] (O) --++ (5*\d, 5*\d)
+		node[black, above right] {\contour{white}{photon}};
+	\end{tikzpicture}
+\end{center}
+
+As before, the slanted (red) axes are your reference frame as you're moving. \\
+Note that you (and anything moving the same speed as you) are stationary in this reference frame.
+
+\remark{}
+In this diagram, we've not only rotated your space axis ($x'$), we've also adjusted the scale of $ct'$ and $x'$ relative to the rest frame. This scaling comes from physical experiments which we will conduct later.
+
+\problem{}
+Please verify that in the diagram above, if you shoot a photon behind you, it still moves at speed $c$ in both your reference frame and the ground's reference frame. You can do this by drawing directly on the diagram.
+
+\begin{solution}
+	This follows by just extending the boosted axes to the second quadrant and drawing the photon's worldline.
+\end{solution}
+
+
+\vfill
+\pagebreak
+
+% \begin{solution}\begin{center}\begin{tikzpicture}[scale=1.8]
+%   \message{Lorentz boost^^J}
+
+%   \def\xmax{2}
+%   \def\xmaxp{2.2} % maximum of rotated axis
+%   \def\Nlines{5} % number of world lines (at constant x/t)
+%   \pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+%   \pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+%   \pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+%   \coordinate (O) at (0,0);
+%   \coordinate (X) at (\xmax+0.2,0);
+%   \coordinate (T) at (0,\xmax+0.2);
+%   \coordinate (X') at (\ang:\xmaxp+0.2);
+%   \coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+%   % WORLD LINE GRID
+%   \message{  Making world lines...^^J}
+%   \foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+%     \message{  Running i/N=\i/\Nlines, x=\x...^^J}
+%     \draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+%     \draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+%     \draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+%     \draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+%   }
+
+%   % BOOSTED WORLD LINE GRID
+%   \message{  Making world lines for boosted frame...^^J}
+%   \fill[mydarkred,opacity=0.05]
+%     (O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
+%   \fill[mydarkred,opacity=0.05]
+%     (O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
+%   \foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
+%     \message{  Running i/N=\i/\Nlines, x=\x...^^J}
+%     \draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
+%     \draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
+%     \draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
+%     \draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
+%   }
+
+%   % AXES
+%   \draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
+%   \draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+%   \draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
+%     node[right=5,above=-1] {you: $ct'$};
+%   \draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
+
+%   % ANGLES
+%   \draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
+%   \draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
+
+%   % PHOTON
+%   \draw[photon] (O) --++ (4*\d,4*\d);
+
+% \end{tikzpicture}\end{center}\end{solution}
+
+\problem{}
+A caveat to Lorentz boosts is that we cannot boost to reference frames which are at the speed of light or
+faster. Based on the diagram given, why can't we do that? \\
+{\em "If my calculations are correct, when this baby hits $c$, you're gonna see some serious stuff."}
+
+\begin{solution}
+	If boosted to $c$, our axes would overlap, compressing time and space into one.
+	If we boosted past $c$, our axes would flip, making the past the future and the future the past.
+\end{solution}
+\vfill
+
+\problem{}
+This diagram implies some strange things. We'll spend the next sections discussing
+some consequences of this, but take a minute to note anything weird that you notice.
+
+In particular, look at a unit of time (or a unit of length) in your frame vs the rest frame.
+
+Which is longer, one unit of time in your reference frame or in the rest frame?
+
+Which is longer, one unit of distance in your reference frame or in the rest frame?
+
+Consider the implications of a slanted space line.
+What does it mean if two events both lie on this line?
+
+\vfill
+\pagebreak

src/Advanced/Relativity/parts/04 simultaneity.tex

@@ -0,0 +1,257 @@
+\section{Simultaneity}
+Let's look now at the consequences of a rotated spatial coordinate.
+To help ourselves switch between different perspectives,
+we'll bring in some professionals: Alice and Bob.
+
+\ref{simultaneity setup} to \ref{Bob overreacted} can all be completed on one spacetime diagram. Feel free to use the one provided below, or draw your own if it gets too crowded.
+
+\makeatletter
+\if@solutions\else
+	\emptydiagramc{Alice}
+\fi
+\makeatother
+
+\problem{}<simultaneity setup>
+Draw a spacetime diagram from Alice's reference frame or use the one provided.
+What do the horizontal gridlines represent?
+
+If two events (remember, points in spacetime) lie on the same horizontal line,
+what does that imply about the events?
+
+\begin{solution}
+	Horizontal lines have constant $t$ value.
+	Therefore, if two events lie on the same horizontal
+	line then they occur at the same time.
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}
+Suppose that Bob is walking to the right at speed $c/3$, relative to Alice.
+Add Bob's worldline to your spacetime diagram. Superimpose Bob's reference frame onto the diagram
+
+What do the "horizontal" lines in Bob's reference frame represent to Bob?
+
+
+\begin{solution}
+	Same solution, horizontal lines in Bob's reference frame represent events at a particular time.
+\end{solution}
+
+
+\vfill
+
+
+
+\problem{}
+Suppose that Alice is passing time, snapping to some music with her arms out.\\
+Both of her arms are length $1$ and she snaps both hands every unit time.
+So there is a snap at location $x = \pm 1$ at $ct = 1$, $ct = 2$, and so on.
+
+Draw this situation on your spacetime diagram.
+
+Consider this from Bob's perspective. What does Bob hear?
+(We'll assume that sound propagates instantly, so if a sound occurs at $ct' = 3.5$ then Bob hears it instantly at $ct' = 3.5$.)
+\begin{solution}
+	The snaps will be out of time in Bob's reference frame.
+	He will hear Alice's right hand first and then her left hand after.
+\end{solution}
+\vfill
+
+\problem{}<Bob overreacted>
+Bob is deeply annoyed at Alice because she {keeps}.~{\em snapping}.~{\em out}.~{\bf\em of}.~{\bf\em TIME}.
+
+In a fit of frustration he starts running faster and faster to the right.
+Eventually, he notices that the snaps are changing.
+What happens to the timing of the snaps? Do the snaps ever come back into time?
+At what speed is Bob running if they do?
+
+\begin{solution}
+	We can line up new sets of snaps with lines of slope $1/2$.
+	So if Bob is running at speed $c/2$then he will hear the snaps in time again.
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}
+Done with music, Alice and Bob decide to have a race and they agree to race from $x = 0$ to $x = 2$.
+Suppose that Alice can run at an impressive speed of $c/2$ while Bob can only run at a measly speed of $c/4$.
+\begin{enumerate}
+	\item Who wins the race?
+	\begin{solution}
+		Alice
+	\end{solution}
+	\vspace{20pt}
+
+	\item Is there any reference frame in which Bob wins?
+
+	\makeatletter
+	\if@solutions\else
+		\halfdiagramcwide{Ground}
+	\fi
+	\makeatother
+
+	\begin{solution}
+		No
+	\end{solution}
+	\vfill
+
+	\item Suppose instead that Alice starts at $x = 0$ and is racing to $x = 2$, while Bob starts at $x = 8$ and is racing to $x = 10$. Is there now a reference frame where Bob wins? Why?\\
+	{\em Hint: Plot these events on a spacetime diagram.}
+	\begin{solution}
+		Bob finishes at $(2, 8)$ and Alice finishes at $(10,4)$.
+		If you consider a reference frame moving to the left at speed $c/2$, then Alice and Bob will tie. Anything faster, Bob will win. Anything slower, Alice will win.
+	\end{solution}
+\end{enumerate}
+
+
+\makeatletter
+\if@solutions\else
+	\halfdiagramcwide{Ground}
+\fi
+\makeatother
+
+
+\vfill
+\pagebreak
+
+
+
+% \problem{}
+%     We've seen that by changing your reference frame, events that were simultaneous may now occur at different times.
+
+%     Which of the following pairs of events (points in spacetime $(x,ct)$) are simultaneous in {\em some} inertial reference frame?
+
+%     % Remember, we can only consider reference frames which are moving less than the speed of light.
+%     \begin{enumerate}
+%         \item $(1,1)$ and $(0,1)$
+%         \item $(2,1)$ and $(0,0)$
+%         \item $(2,3)$ and $(0,1)$
+%         \item $(1,2)$ and $(0,0)$
+%         \item $(1,1)$ and $(1,3)$
+%     \end{enumerate}
+%     \begin{solution}
+%     \begin{enumerate}
+%         \item $(1,1)$ and $(0,1)$ - yes
+%         \item $(2,1)$ and $(0,0)$ - yes
+%         \item $(2,3)$ and $(0,1)$ - no, would require a reference frame at the speed of light
+%         \item $(1,2)$ and $(0,0)$ - no, would require a reference frame moving faster than the speed of light
+%         \item $(1,1)$ and $(1,3)$ - definitely not
+%     \end{enumerate}
+%     \end{solution}
+
+\problem{}
+This weirdness with simultaneity might make us question what the past, present, and future are.
+To make sense of this, suppose that we want to define the {\em present} to be every event which is simultaneous to you, right here, right now --- a.k.a.~the spacetime point $(0,0)$ --- in at least one reference frame.
+
+On the given spacetime diagram, draw the region which represents the present.
+
+Draw a region which represents the future: events that are later in time in every reference frame.
+
+Draw a region which represents the past: events that are earlier in time in every reference frame.
+
+\begin{center}
+	\begin{tikzpicture}[scale=2]
+	\def\xmax{2}
+	\def\xmaxp{2.2} % maximum of rotated axis
+	\def\Nlines{5} % number of world lines (at constant x/t)
+	\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+	\pgfmathsetmacro\ang{atan(1/3)} % angle between x and x' axes
+	\coordinate (O) at (0,0);
+	\coordinate (X) at (\xmax+0.2,0);
+	\coordinate (T) at (0,\xmax+0.2);
+
+	% WORLD LINE GRID
+	\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+		\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+		\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+		\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+		\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+	}
+
+	% AXES
+	\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
+	\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+
+	% LABELS
+	%\draw pic[->,"$45^\circ$",draw=black,angle radius=23,angle eccentricity=1.38] {angle = X--O--C};
+	% \node[mydarkorange,above right] at (0.1*\xmax,\xmax) {future light cone};
+	% \node[mydarkorange,below] at (0,-\xmax) {past light cone};
+
+	% FILLS
+	% \fill[myblue,opacity=0.05] % SPACELIKE
+	%   (\xmax,\xmax) -- (-\xmax,-\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- cycle;
+	% \fill[myorange,opacity=0.05] % TIMELIKE
+	%   (\xmax,\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- (-\xmax,-\xmax) -- cycle;
+	% \node[mydarkblue,right,align=center] at (-\xmax,0.18*\xmax)
+	%   {\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
+	% \node[mydarkblue,left,align=center] at (\xmax,0.18*\xmax)
+	%   {\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
+	% \node[mydarkorange,align=center] at (-0.22*\xmax,0.67*\xmax)
+	%   {\contour{myorange!5}{future}\\[-2]\contour{myorange!5}{}};
+	% \node[mydarkorange,align=center] at (0.22*\xmax,-0.67*\xmax)
+	%   {\contour{myorange!5}{past}\\[-2]\contour{myorange!5}{}};
+
+	% PHOTON
+	\draw[photon] ( \xmax,-\xmax) -- ( 0.02*\xmax,-0.02*\xmax);
+	\draw[photon] (-\xmax,-\xmax) -- (-0.02*\xmax,-0.02*\xmax);
+	\draw[photon] ( 0.02*\xmax,0.02*\xmax) -- ( \xmax,\xmax)
+	node[mydarkorange,above right] {$x=ct$};
+	\draw[photon] (-0.02*\xmax,0.02*\xmax) -- (-\xmax,\xmax);
+
+	% % PARTICLE WORLDLINE
+	% \draw[particle,decoration={markings,mark=at position 0.27 with {\arrow{latex}},
+	%                                     mark=at position 0.76 with {\arrow{latex}}},postaction={decorate}]
+	%     (-0.5*\xmax,-\xmax) to[out=80,in=-110] (O) to[out=70,in=-100] (0.45*\xmax,\xmax);
+	% \fill[mydarkgreen] (O) circle(0.04); % event
+	\end{tikzpicture}
+\end{center}
+Another way to think of this is thinking of causality. In a reference frame, the {\em future} is every event that can be affected by an event at $(0,0)$. The {\em past} is every event that could have affected an event at $(0,0)$. The {\em present} is every event that is causally independent of $(0,0)$.
+
+
+\begin{solution}
+	% SPACETIME DIAGRAM - LIGHT CONE
+	The photons split the diagram into four regions.
+	The top region is the future, the bottom region is the past, and the left/right regions are the present.
+\end{solution}
+
+
+\vfill
+\pagebreak
+
+
+\problem{Bell's Spaceship Paradox}
+Suppose that we have two spaceships, distance $L$ apart, tied together with floss of length $L$. The floss is so weak that any stretching at all will cause it to disintegrate.
+
+The spaceships are at rest and then simultaneously accelerate to speed $c/2$.
+
+Draw the spacetime diagram for this situation. Include the reference frame of the spaceships {\em after} they start moving (i.e. the reference frame moving at speed $c/2$).\\
+What happens to the floss in the boosted frame? Does it break? Why?
+
+\makeatletter
+\if@solutions\else
+	\emptydiagramc{Rest}
+\fi
+\makeatother
+\vfill
+
+\problem{Bell's Spaceship Paradox (continued)}
+The same outcome has to occur in all reference frames, so we know that the floss breaks in the stationary reference frame. However, in the stationary reference frame, the two spaceships accelerate at the same time, so our explanation no longer seems accurate.
+
+Can you come up with a hypothesis for why the floss breaks in the stationary reference frame?
+
+{\em Hint: The floss breaks if the ships are farther apart than the length of the floss.}
+
+\begin{solution}
+	The floss will break as soon as the spaceships accelerate. \par
+	In the boosted frame, the spaceship on the right accelerates first,
+	stretching the floss, causing it to break. \par
+	In the rest frame, this paradox motivates length contraction.
+	Either the ships get farther apart (doesn't happen) or the floss gets shorter.
+	This implies that the length of moving objects must get smaller.
+\end{solution}
+
+
+\vfill
+\pagebreak

src/Advanced/Relativity/parts/05 proper.tex

@@ -0,0 +1,391 @@
+\section{Proper Time and Proper Length}
+
+We mentioned that the space and time axes also get rescaled during a Lorentz boost. Let's figure out by how much! To do this, we're going to use the only consistent tool at our disposal, the speed of light.
+
+\problem{Time dilation}
+Suppose that Alice remains stationary.
+In her reference frame, Bob moves to the right at speed $v$.
+Every quantity in Alice's reference frame will be denoted normally,
+i.e.~$x,t$. Every quantity in Bob's reference frame will be given a prime, i.e.~$x',t'$.
+
+Bob holds onto a very special clock.
+The clock consists of a light bulb, a mirror distance $L$ away from the light
+(perpendicular to the direction Bob is moving) and a photosensor.
+
+\begin{enumerate}
+	\item Draw the experimental setup.
+	Note again that the light bulb and mirror are separated in the $y$ direction, not the $x$.
+	\vfill
+
+	\item In Bob's reference frame (where the clock isn't moving), how long does it take the light to be emitted, travel to the mirror, bounce back, and be reabsorbed? Write this as $t'$.
+	\begin{solution}
+		$ct' = 2L \implies t' = 2L/c$
+	\end{solution}
+	\vfill\pagebreak
+
+	\item In Alice's reference frame, she still sees the light move, reflect off the mirror, then come back. How long does this take?
+	{\em Note: The mirror and photosensor are moving at speed $v$ during this process. Also, the light is now moving at an angle, but it still moves at $c$.}
+	\begin{solution}
+		Pythagorean theorem:
+		$$(ct)^2 = (vt)^2 + 4L^2 \implies t^2 = \frac{4L^2}{c^2 - v^2} \implies t = \frac{2L}{\sqrt{c^2 - v^2}}$$
+	\end{solution}
+	\vfill
+
+	\item Combine these two formulas to find a relationship between $t$ and $t'$.
+
+	If one unit of time elapses in Bob's reference frame, does more or less time elapse in Alice's reference frame? Who ages faster?
+	\begin{solution}
+		$$ct' = \sqrt{c^2 - v^2}t \implies t' = \sqrt{1 - v^2/c^2} t$$
+
+		Since $v < c$, $0 \leq v^2/c^2 <1$. Then $t' < t$.
+
+		Alice ages faster. The time that Bob experiences is less than Alice.
+	\end{solution}
+\end{enumerate}
+
+
+\vfill
+\pagebreak
+
+\problem{Space dilation}
+Since the speed of light is always constant, we can use time dilation to see what happens to our notion of distance. Suppose that a photon is emitted from one location and absorbed in another.
+
+\begin{enumerate}
+		\item Suppose that in Alice's reference frame, the photon takes time $t$ to be emitted and absorbed. In this time, what distance $x$ does the photon travel?
+	\begin{solution}
+		$x = ct$
+	\end{solution}
+	\vfill
+
+	\item Suppose that Bob is traveling at speed $v$ relative to Alice. Using time dilation, how long does the photon take to be emitted and absorbed in Bob's reference frame? What distance $x'$ does the photon travel in that time?
+	{\em This will give you a formula relating distance $x'$ in Bob's reference frame to distance $x$ in Alice's reference frame.}
+	\begin{solution}
+		$x' = ct' = ct \sqrt{1 - v^2/c^2} = x \sqrt{1 - v^2/c^2}$
+	\end{solution}
+	\vfill
+\end{enumerate}
+\vfill \pagebreak
+
+\remark{}
+Switching back and forth between reference frames gets confusing quickly,
+especially when you are consideration time and space dilation. To avoid this,
+we will define a new notion of time and space that doesn't care about which reference frame we're in.
+Since this will be universal, we'll call these {\em proper time}, denoted by $\tau$, and {\em proper distance},
+denoted by $\chi$.
+
+Recall our notion of past and future from earlier:
+% SPACETIME DIAGRAM - LIGHT CONE
+\begin{center}
+	\begin{tikzpicture}[scale=1.8]
+		\message{Light cone^^J}
+
+		\def\xmax{2}
+		\def\xmaxp{2.2} % maximum of rotated axis
+		\def\Nlines{5} % number of world lines (at constant x/t)
+		\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
+		\pgfmathsetmacro\ang{atan(1/3)} % angle between x and x' axes
+		\coordinate (O) at (0,0);
+		\coordinate (X) at (\xmax+0.2,0);
+		\coordinate (T) at (0,\xmax+0.2);
+
+		% WORLD LINE GRID
+		\message{  Making world lines...^^J}
+		\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+			\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+			\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+			\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+			\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+			\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+		}
+
+		% AXES
+		\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
+		\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+
+		% LABELS
+		%\draw pic[->,"$45^\circ$",draw=black,angle radius=23,angle eccentricity=1.38] {angle = X--O--C};
+		\node[mydarkorange,above right] at (0.1*\xmax,\xmax) {future light cone};
+		\node[mydarkorange,below] at (0,-\xmax) {past light cone};
+
+		% FILLS
+		\fill[myblue,opacity=0.05] % SPACELIKE
+			(\xmax,\xmax) -- (-\xmax,-\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- cycle;
+		\fill[myorange,opacity=0.05] % TIMELIKE
+			(\xmax,\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- (-\xmax,-\xmax) -- cycle;
+		\node[mydarkblue,right,align=center] at (-\xmax,0.18*\xmax)
+			{\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
+		\node[mydarkblue,left,align=center] at (\xmax,0.18*\xmax)
+			{\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
+		\node[mydarkorange,align=center] at (-0.22*\xmax,0.67*\xmax)
+			{\contour{myorange!5}{future}\\[-2]\contour{myorange!5}{}};
+		\node[mydarkorange,align=center] at (0.22*\xmax,-0.67*\xmax)
+			{\contour{myorange!5}{past}\\[-2]\contour{myorange!5}{}};
+
+		\node at (2*\d, 4*\d){\textbullet};
+		\node[black, above] at (2*\d, 4*\d){future event};
+
+		\node at (4*\d, 2*\d){\textbullet};
+		\node[black, above right] at (4*\d, 2*\d){present event};
+
+		% PHOTON
+		\draw[photon] ( \xmax,-\xmax) -- ( 0.02*\xmax,-0.02*\xmax);
+		\draw[photon] (-\xmax,-\xmax) -- (-0.02*\xmax,-0.02*\xmax);
+		\draw[photon] ( 0.02*\xmax,0.02*\xmax) -- ( \xmax,\xmax)
+			node[mydarkorange,above right] {$x=ct$};
+		\draw[photon] (-0.02*\xmax,0.02*\xmax) -- (-\xmax,\xmax);
+
+		% % PARTICLE WORLDLINE
+		% \draw[particle,decoration={markings,mark=at position 0.27 with {\arrow{latex}},
+		%                                     mark=at position 0.76 with {\arrow{latex}}},postaction={decorate}]
+		%     (-0.5*\xmax,-\xmax) to[out=80,in=-110] (O) to[out=70,in=-100] (0.45*\xmax,\xmax);
+		% \fill[mydarkgreen] (O) circle(0.04); % event
+	\end{tikzpicture}
+\end{center}
+
+
+\problem{Proper Time}
+For any event $(x,ct)$ in the past or future, we can choose a reference frame to make it occur at $x' = 0$.
+\begin{enumerate}
+	\item What velocity do we need to boost to so that the event $(x,ct)$ now occurs at $x' = 0$?
+	\begin{solution}
+		We need to boost to $v = x/t$.
+	\end{solution}
+	\vfill
+
+	\item After boosting to this reference frame, what time $ct'$ does the event occur at? \\
+	This will be our proper time $c\tau$.
+	\begin{solution}
+		Using time dilation, the event now occurs at $$c\tau = ct' = ct\sqrt{1 - x^2/c^2 t^2} = \sqrt{c^2 t^2 - x^2}$$
+	\end{solution}
+	\vfill
+\end{enumerate}
+
+\pagebreak
+
+
+\problem{Proper Distance}
+
+Similarly, for any event $(x,ct)$ in the present, we can choose a reference frame to make it occur at $t' = 0$.
+\begin{enumerate}
+	\item What velocity do we need to boost to so that the event $(x,ct)$ now occurs at $t' = 0$?
+	\begin{solution}
+		We need to boost to $v = c^2t/x$.
+	\end{solution}
+	\vfill
+
+	\item After boosting to that velocity, what spatial position $x'$ does the event occur at? This will be our proper distance $\chi$.
+	\begin{solution}
+	Using space dilation, the event now occurs at $$\chi = x' = x\sqrt{1 - c^2t^2/x^2} = \sqrt{x^2 - c^2t^2}$$.
+	\end{solution}
+\end{enumerate}
+\vfill
+
+Defined in this way, proper time and proper distance give a quick formula to
+figure out how old an object is, or how far an object traveled, in its own reference frame!
+
+We'll now use this proper time and proper space to quickly solve one of
+the most famous paradoxes in special relativity.
+\pagebreak
+
+% \remark{}
+% To visualize the consequences of this change, let's see what lines of constant proper time and constant proper space look like.
+% Lines in green are constant proper time and lines in blue are constant proper space.
+
+% % SPACETIME DIAGRAM - MULTIPLE INVARIANT HYPERBOLOIDS
+% % Inspiration: https://commons.wikimedia.org/wiki/File:Spacelike_and_Timelike_Invariant_Hyperbolas.png
+% \begin{center}\begin{tikzpicture}[scale=1.8]
+%   \message{Multiple invariant hyperboloids^^J}
+
+%   \def\xmax{2}
+%   \def\Nlines{4} % number of world lines (at constant x/t)
+%   \pgfmathsetmacro\w{\xmax/(\Nlines+1)}
+
+%   % AXES
+%   \draw[->,thick] (0,-\xmax) -- (0,\xmax+0.2) node[left=-1] {$ct$};
+%   \draw[->,thick] (-\xmax,0) -- (\xmax+0.2,0) node[right=-1] {$x$};
+
+%   % LIGHTCONE
+%   \draw[myorange,thick] (-\xmax,-\xmax) -- (\xmax, \xmax);
+%   \draw[myorange,thick] (-\xmax, \xmax) -- (\xmax,-\xmax);
+
+%   \foreach \i [evaluate={\s=\xmax*\i/(\Nlines+1); \xm=sqrt(\xmax^2-\s^2);}] in {1,...,\Nlines}{
+
+%     % SPACELIKE HYPERBOLOIDS
+%     \draw[mygreen,thick,samples=\Nsamples,smooth,variable=\x,domain=-\xm:\xm]
+%       plot(\x,-{sqrt(\s^2+(\x)^2)})
+%       plot(\x,{sqrt(\s^2+(\x)^2)});
+
+%     % TIMELIKE HYPERBOLOIDS
+%     \draw[myblue,thick,samples=\Nsamples,smooth,variable=\y,domain=-\xm:\xm]
+%       plot(-{sqrt(\s^2+(\y)^2)},\y)
+%       plot({sqrt(\s^2+(\y)^2)},\y);
+
+%   }
+
+%   % LABELS
+%   \node[mydarkgreen,above left=2,align=center] at (-0.2*\xmax,\xmax)
+%     {timelike separation\\[-1]$s^2 = c^2t^2 - x^2 > 0$};
+%   \node[mydarkorange,left=2,above right=-2,align=center] at (\xmax,\xmax)
+%     {lightlike separation\\[-1]$s^2 = c^2t^2 - x^2 = 0$};
+%   \node[mydarkblue,right=0,align=center] at (0.88*\xmax,-0.24*\xmax)
+%     {spacelike separation\\[-1]$s^2 = c^2t^2 - x^2 < 0$};
+
+%   % % VECTORS
+%   % \def\xa{0.5}
+%   % \def\xb{2.7}
+%   % \def\ta{-0.7}
+%   % \def\tb{1.7}
+%   % \draw[mydarkgreen,very thick,decoration={markings,mark=at position 0.55 with {\arrow{latex}}},
+%   %       postaction={decorate},samples=20,variable=\x,domain=\xa:\xb]
+%   %   plot({\w*\x},{\w*sqrt((\x)^2+3^2)});
+%   % \draw[mydarkblue,very thick,decoration={markings,mark=at position 0.6 with {\arrow{latex}}},
+%   %       postaction={decorate},samples=20,variable=\x,domain=\ta:\tb]
+%   %   plot({\w*sqrt((\x)^2+3^2)},{\w*\x});
+%   % \fill[mydarkgreen] ({\w*\xa},{\w*sqrt(\xa^2+3^2)}) coordinate (A) circle(0.03);
+%   % \fill[mydarkgreen] ({\w*\xb},{\w*sqrt(\xb^2+3^2)}) coordinate (A') circle(0.03);
+%   % \fill[mydarkblue] ({(\w*sqrt((\ta)^2+3^2)},{\w*\ta}) coordinate (B) circle(0.03);
+%   % \fill[mydarkblue] ({(\w*sqrt((\tb)^2+3^2)},{\w*\tb}) coordinate (B') circle(0.03);
+%   % \draw[vector',mydarkgreen] (0,0) -- (A)
+%   %   node[pos=0.53,right=-2] {$s$};
+%   % \draw[vector',mydarkgreen] (0,0) -- (A')
+%   %   node[pos=0.57,right=-2] {$s$};
+%   % \draw[vector',mydarkblue] (0,0) -- (B)
+%   %   node[pos=0.53,below=-1] {$s$};
+%   % \draw[vector',mydarkblue] (0,0) -- (B')
+%   %   node[pos=0.53,above=-1] {$s$};
+
+% \end{tikzpicture}\end{center}
+
+\problem{Twin paradox}
+Suppose that you have two twins, Mark and Lucas, born at the exact same instant and location. \\
+One day, Mark sends Lucas off in a spaceship traveling at $c/2$ to a planet that is one lightyear away ($c*(\text{1 year})$). Upon reaching the planet, Lucas realizes that he's all alone! Saddened, he turns around immediately, traveling back at speed $c/2$.
+
+The question is: Which twin is older? We'll break this down step by step.
+\begin{enumerate}
+	\item From Mark's perspective, what time does Lucas reach the planet? What time does he return?
+	\begin{solution}
+		Lucas reaches the planet at $t = 2$ years.
+
+		Lucas reaches home at $t = 4$ years.
+	\end{solution}
+	\vfill
+
+	\item From Lucas' perspective, how long does it take him to reach the planet? Hint: use proper time.
+	\begin{solution}
+		The proper time of Lucas arriving at the planet is $$c\tau = \sqrt{4 - 1}\text{ lightyears} \implies \tau = \sqrt{3}\text{ years}.$$
+	\end{solution}
+	\vfill
+
+	\item How long does it take him to return?
+	\begin{solution}
+		$\tau = \sqrt{3}\text{ years}.$
+	\end{solution}
+	\vfill
+
+	\item Which twin is older?
+	\begin{solution}
+		Mark
+	\end{solution}
+\end{enumerate}
+
+\vfill
+\pagebreak
+
+\problem{Twin paradox (continued)}
+The weird part is, from Lucas' perspective, it was Mark who flew off and then came back! So why is Mark not younger than Lucas?
+Let's break this down step by step.
+
+\begin{enumerate}
+	\item[\bf E:]
+	To help answer this, draw a spacetime diagram of this situation. In Lucas' reference frames (one for going out to the planet and one for coming back), draw lines of constant time. What happens to Mark from Lucas' perspective?\\
+	{\em Hint: changing speed is weird. Focus on where Lucas changes speed.}
+
+	\makeatletter
+	\if@solutions\else
+		\halfdiagramc{Mark}
+	\fi
+	\makeatother
+
+
+	\begin{solution}
+		% SPACETIME DIAGRAM of TWIN PARADOX
+		\begin{tikzpicture}[scale=2.0]
+			\def\xmin{0.2}
+			\def\xmax{2}
+			\def\ymax{2.3}
+			\def\Nlines{5} % number of world lines (at constant x/t)
+			\def\ang{60} % angle between ct and ct' axes
+			\pgfmathsetmacro\d{0.94*\xmax/\Nlines} % grid size
+			\pgfmathsetmacro\dt{3*\d} % time of half trip
+			\pgfmathsetmacro\D{\dt/tan(\ang)} % distance between observers
+			\pgfmathsetmacro\h{\dt-\D/tan(\ang)} % half time gap of return
+			\coordinate (A) at (0,0); % observer A at t=0
+			\coordinate (B) at (\D,0); % observer B at t=0
+			\coordinate (C) at (\D,\dt); % point of return
+			\coordinate (T1) at (0,\dt); % time of return
+			\coordinate (T2) at (0,2*\dt); % time of arrival
+
+			% WORLD LINES GRID
+			\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+				\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+				\draw[world line]   ( \x,-\xmin) -- ( \x,\ymax);
+				\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
+			}
+			\draw[world line t] (-\xmin,{(\Nlines+1)*\d}) -- (\xmax,{(\Nlines+1)*\d});
+
+			% AXES
+			\draw[->,thick] (0,-\xmin) -- (0,\ymax+0.2) node[above left=-2] {$ct$};
+			\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
+
+			% VECTORS
+			\draw[vector,myred,shorten >=1] (A) -- (T2);
+			\draw[vector,mygreen,shorten >=2] (A) -- (C);
+			\draw[vector,mygreen,shorten >=1] (C) -- (T2);
+
+			% PLANES OF SIMULTANEITY
+			\fill[mydarkred,opacity=0.06]
+				(0,\h) -- (C) -- (0,2*\dt-\h) -- cycle;
+			\pgfmathsetmacro\ystep{\h/3}
+			\foreach \i [evaluate={\dy=(\i-1)*\ystep; \ya=\i*\ystep; \yb=2*\dt-\i*\ystep;}] in {1,...,3}{
+				\draw[mydarkred,dashed,line width=0.6]
+				(0,\ya)++(90-\ang:-0.8*\xmin) --++ (90-\ang:{1.2*\xmin+\D/sin(\ang)});
+				\draw[mydarkblue,dashed,line width=0.6]
+				(0,\yb)++(\ang-90:-0.8*\xmin) --++ (\ang-90:{1.2*\xmin+\D/sin(\ang)});
+				\fill[mydarkred]  (0,\ya) circle(0.02);
+				\fill[mydarkblue] (0,\yb) circle(0.02);
+				%\fill[mydarkblue] ({\D-\dy*cot(\ang)},\dt+\dy) circle(0.02);
+				%\fill[mydarkred]  ({\D-\dy*cot(\ang)},\dt-\dy) circle(0.02);
+				\fill[mydarkblue] (C)++(-\ang:{\dy*sin(\ang)/cos(2*\ang)}) circle(0.02);
+				\fill[mydarkred]  (C)++( \ang:{\dy*sin(\ang)/cos(2*\ang)}) circle(0.02);
+			}
+			\fill[mydarkred] (A) circle(0.04) node[below left=-1] {Mark}; % observer A
+			\fill[mydarkgreen] (C) circle(0.04)
+				node[right=4] {\contour{white}{Lucas turns around}}; % observer B returns
+			\node[mydarkblue,above right=0,align=left] at (2*\d,1.15*\dt)
+				{\contour{white}{planes of}\\[-2]\contour{white}{simultaneity}};
+			\node[mydarkred,below right=0,align=left] at (2*\d,0.85*\dt)
+				{\contour{white}{planes of}\\[-2]\contour{white}{simultaneity}};
+
+			% TICKS
+			\node[fill=white,inner sep=1,above=1,left=3] at (T1) {$\dfrac{ct_2}{2}=ct_1$};
+			\node[fill=white,inner sep=1,above=2,left=3] at (T2) {$ct_2$};
+			\tick{T1}{0};
+			\tick{T2}{0};
+		\end{tikzpicture}
+
+	\end{solution}
+	\vfill
+
+	\item[\bf F:] We said that when Lucas landed on the planet, he immediately took off back towards Earth. From Mark's perspective, how long was Lucas on that planet?
+	\begin{solution}
+		$4 - 2 \sqrt{3}$ years.
+	\end{solution}
+	\vfill
+
+	\item[\bf G:] What is physically different between Lucas' perspective and Mark's perspective?
+	\begin{solution}
+		Lucas is the one who accelerates. His reference frame is non-inertial (not defined in this packet) and would require additional care.
+	\end{solution}
+	\vfill
+\end{enumerate}
+\pagebreak

src/Advanced/Relativity/parts/06 contraction.tex

@@ -0,0 +1,366 @@
+\section{Length Contraction}
+With proper time and distance done, we can now tackle length contraction easily! Length contraction is weird because different parts of our object will now be experiencing different times.
+
+\problem{}
+Suppose that you (at rest) see a rod of length $L$ moving at speed $v$.
+
+\makeatletter
+\if@solutions\else
+	\halfdiagramc{you}
+\fi
+\makeatother
+
+\begin{enumerate}
+	\item Using the provided grid, draw a spacetime diagram where the left side of the rod is at $x = 0$ at $t = 0$.
+
+	\item When we switch to the rod's reference frame, space gets rotated. Draw a line in the rod's reference frame which represents the rod at time $t' = 0$, when the left side is at $x' = 0$. Call the right side of the rod at this time $P$.
+
+	\item Switching back to your reference frame, what are the spacetime coordinates $(ct,x)$ of $P$?
+	\vfill
+
+	\item Use the formula for proper distance to compute the length $L'$ of the rod in its own reference frame.
+	\vfill
+
+	\item Which is larger, $L'$ or $L$? Do moving objects shrink?
+	\vspace{30pt}
+\end{enumerate}
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=1.5]
+			%\def\R{2*\xmax} % radius of clip
+			%\clip (-\xmin,\R) |- (\R,-\xmin) arc(0:90:\xmin+\R);
+
+			% AXES
+			\def\xmin{0.2}
+			\def\xmax{2.6}
+			\def\ang{33.5}  % angle between x and x' axes
+			\def\Nlines{12} % number of world lines (at constant x/t)
+			\def\DNy{1}     % difference in number of world lines of y axis (lengthen)
+			\def\DNxp{-6}   % difference in number of world lines of x' axis (shorten)
+			\def\DNyp{-2}   % difference in number of world lines of y' axis (shorten)
+			\def\xplabelang{170} % anchor angle of x' axis label
+			\axes
+
+			% SETTINGS
+			\pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
+			\pgfmathsetmacro\L{cos(2*\ang)/cos(\ang)*\Lz} % contracted length L in S
+			\pgfmathsetmacro\yminb{-0.7*\xmin} % ymin of barn in S
+			\pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
+			\pgfmathsetmacro\wb{3.08*\d} % width of barn in S
+			\pgfmathsetmacro\yA{(\xb+0.04*\d)/tan(\ang)} % y = ct coordinate when ladder is fully in barn in S
+			\coordinate (L) at (\L,0); % ladder end in S
+			\coordinate (L') at (\ang:\Lz); % ladder end in S'
+			\coordinate (A) at (90-\ang:{(\xb+0.04*\d)/sin(\ang)}); % left end of ladder when fully in barn
+			\coordinate (B) at ($(A)+(\L,0)$); % right end of ladder when fully in barn
+			\coordinate (C) at (90-\ang:{(\xb+\wb+0.08*\d)/sin(\ang)}); % left end of ladder when fully passed through barn
+
+			% FILL
+			\begin{pgfonlayer}{back} % draw on back (behind axes)
+			% \fill[mydarkblue!22] % barn frame
+			%   (\xb,\yminb) rectangle (\xb+\wb,\ymax);
+			\fill[mylightred] % ladder frame
+			(90-\ang:-\xmin) -- (90-\ang:\ymaxp) --++ (\ang:\Lz) -- (L) --++ (90-\ang:-\xmin) -- cycle;
+			% \begin{scope}
+			%   \clip (0,0) rectangle(1.3*\xmax,\ymax+0.2);
+			%   \draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
+			%     (A)++(\ang:{-\xb/cos(\ang)-0.05}) --++ (\ang:\xmax+3.6*\D)
+			%     (B)++(\ang:{-\xb/cos(\ang)-0.05-\Lz}) --++ (\ang:\xmax+3.8*\D);
+			%   \draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
+			%     (0,\yA) --++ (\xmax+0.8*\d,0);
+			% \end{scope}
+			\end{pgfonlayer}
+			% \draw[->,thick,mydarkblue] % barn left door
+			%   (\xb,\yminb) -- (\xb,\ymax+0.15);
+			% \draw[->,thick,mydarkblue] % barn right door
+			%   (\xb+\wb,\yminb) -- (\xb+\wb,\ymax+0.15);
+			\draw[->,thick,mydarkbrown] % rod left end
+			(L)++(90-\ang:-\xmin) -- (L) -- (L') --++ (90-\ang:\ymaxp+0.2);
+
+			% LADDER
+			\draw[rod] (O) -- (L')
+			node[pos=0.55,above=2,scale=0.8] {\contour{mylightred}{$L'$}};
+			\draw[rod] (O) -- (L)
+			node[pos=0.46,below=1,scale=0.8] {$L$};
+
+			% % LADDER IN BARN
+			% \draw[rod] (A) --++ (L');
+			% \draw[rod] (B) --++ (\ang:-\Lz);
+			% \draw[rod] (A) --++ (L);
+
+			% % LADDER RIGHT OF BARN
+			% \draw[rod] (C) --++ (L');
+			% \draw[rod] (C) --++ (L);
+
+			% LABELS
+			% \node[mydarkblue,below=0,align=center,scale=0.8,yshift=1] at (\xb+\wb/2,0)
+			%   {barn\\$L$};
+			% \node[mydarkpurple,right,align=left,scale=0.65,yshift=1.2] at (\xb+3.6*\d,\yA)
+			%   {both doors closed in S};
+			%   %{both doors\\[-3]close in S};
+			% \node[mydarkred,right,scale=0.65,yshift=1.8,rotate=\ang] at ($(A)+(\ang:\Lz+0.8*\D)$)
+			%   {left door closed in S$'$};
+			% \node[mydarkred,right,scale=0.65,yshift=0.7,rotate=\ang] at ($(B)+(\ang:0.8*\D)$)
+			%   {right door closed in S$'$};
+		\end{tikzpicture}
+	\end{center}
+
+	We want to calculate the spacetime position of the right side side of the rod when the left side is at the origin.
+
+	In the rest frame, the equation for the right edge of the rod is $x = vt + L$.
+	The equation for the spatial axis of the rod's rest frame is $c^2t = vx$.
+
+	This implies that they intersect at $ct = \frac{cvL}{c^2 - v^2}$ and $x = \frac{L}{1-v^2/c^2}$.
+
+	Calculating the proper length between the left and right side of the rod, we find that
+	\begin{align*}
+		L' = \chi = \sqrt{\frac{c^4L^2}{(c^2 - v^2)^2} - \frac{c^2 v^2 L^2}{(c^2 - v^2)}}
+		= L\sqrt{\frac{c^2(c^2 - v^2)}{(c^2 - v^2)^2}}
+		= \frac{L}{\sqrt{1 - v^2/c^2}}
+	\end{align*}
+
+	We note that this implies $L' > L$. i.e., moving objects get shorter
+\end{solution}
+
+
+\pagebreak
+
+\problem{Ladder paradox}
+Aiden and Matt have a ladder of length $2L$ that they are trying to squeeze into a barn of length $L$. Suppose that the barn has a front door and back door which can be open/closed simultaneously.
+
+Now, Aiden is particular smart, so he gives Matt the ladder and has Matt run at the barn at speed $v = \sqrt{3}c/2$. Does the ladder fit in the barn? We'll analyze this in the new few questions.
+
+\begin{enumerate}
+	\item
+	What is the length of the ladder from Aiden's perspective when Matt is running? Does the ladder fit in the barn?
+	\begin{solution}
+		$L$
+	\end{solution}
+	\vfill
+
+	\item As soon as Matt and the ladder are inside the barn, Aiden quickly closes and opens the doors of the barn. Success!
+
+	However, consider this from Matt's perspective. From Matt's perspective, he's holding a ladder of length $2L$ and a barn is flying at him at speed $v = \sqrt{3}c/2$. By length contraction, what is the length of the barn?
+	\begin{solution}
+		$L/2$
+	\end{solution}
+	\vfill\pagebreak
+
+	\item Despite the barn being too short, we know that the ladder has to fit! Using the provided grid, draw a spacetime diagram of the situation. Include Matt's reference frame on your diagram.
+	\makeatletter
+	\if@solutions\else
+		\emptydiagramc{Aiden}
+	\fi
+	\makeatother
+
+
+	\vfill
+	\item From Matt's perspective, why don't the doors of the barn crush the ladder?
+\end{enumerate}
+
+
+\begin{solution}
+	In short, from Matt's perspective, the barn doors do not close at the same time. The back door closes right when the front of the ladder reaches it, then opens again. Later, the front door closes right when the back of the ladder passes it, then opens again.
+	% SPACETIME DIAGRAM - LADDER PARADOX
+	\begin{tikzpicture}[scale=2.5]
+		\message{Ladder paradox^^J}
+		%\def\R{2*\xmax} % radius of clip
+		%\clip (-\xmin,\R) |- (\R,-\xmin) arc(0:90:\xmin+\R);
+
+		% AXES
+		\def\xmin{0.2}
+		\def\xmax{2.6}
+		\def\ang{33.5}  % angle between x and x' axes
+		\def\Nlines{12} % number of world lines (at constant x/t)
+		\def\DNy{1}     % difference in number of world lines of y axis (lengthen)
+		\def\DNxp{-6}   % difference in number of world lines of x' axis (shorten)
+		\def\DNyp{-2}   % difference in number of world lines of y' axis (shorten)
+		\def\xplabelang{170} % anchor angle of x' axis label
+		\axes
+
+		% SETTINGS
+		\pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
+		\pgfmathsetmacro\L{cos(2*\ang)/cos(\ang)*\Lz} % contracted length L in S
+		\pgfmathsetmacro\yminb{-0.7*\xmin} % ymin of barn in S
+		\pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
+		\pgfmathsetmacro\wb{3.08*\d} % width of barn in S
+		\pgfmathsetmacro\yA{(\xb+0.04*\d)/tan(\ang)} % y = ct coordinate when ladder is fully in barn in S
+		\coordinate (L) at (\L,0); % ladder end in S
+		\coordinate (L') at (\ang:\Lz); % ladder end in S'
+		\coordinate (A) at (90-\ang:{(\xb+0.04*\d)/sin(\ang)}); % left end of ladder when fully in barn
+		\coordinate (B) at ($(A)+(\L,0)$); % right end of ladder when fully in barn
+		\coordinate (C) at (90-\ang:{(\xb+\wb+0.08*\d)/sin(\ang)}); % left end of ladder when fully passed through barn
+
+		% FILL
+		\begin{pgfonlayer}{back} % draw on back (behind axes)
+		\fill[mydarkblue!22] % barn frame
+		(\xb,\yminb) rectangle (\xb+\wb,\ymax);
+		\fill[mylightred] % ladder frame
+		(90-\ang:-\xmin) -- (90-\ang:\ymaxp) --++ (\ang:\Lz) -- (L) --++ (90-\ang:-\xmin) -- cycle;
+		\begin{scope}
+		\clip (0,0) rectangle(1.3*\xmax,\ymax+0.2);
+		\draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
+		(A)++(\ang:{-\xb/cos(\ang)-0.05}) --++ (\ang:\xmax+3.6*\D)
+		(B)++(\ang:{-\xb/cos(\ang)-0.05-\Lz}) --++ (\ang:\xmax+3.8*\D);
+		\draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
+		(0,\yA) --++ (\xmax+0.8*\d,0);
+		\end{scope}
+		\end{pgfonlayer}
+		\draw[->,thick,mydarkblue] % barn left door
+		(\xb,\yminb) -- (\xb,\ymax+0.15);
+		\draw[->,thick,mydarkblue] % barn right door
+		(\xb+\wb,\yminb) -- (\xb+\wb,\ymax+0.15);
+		\draw[->,thick,mydarkbrown] % rod left end
+		(L)++(90-\ang:-\xmin) -- (L) -- (L') --++ (90-\ang:\ymaxp+0.2);
+
+		% LADDER
+		\draw[rod] (O) -- (L')
+		node[pos=0.55,above=2,scale=0.8] {\contour{mylightred}{$2L$}};
+		\draw[rod] (O) -- (L)
+		node[pos=0.46,below=1,scale=0.8] {$L$};
+
+		% LADDER IN BARN
+		\draw[rod] (A) --++ (L');
+		\draw[rod] (B) --++ (\ang:-\Lz);
+		\draw[rod] (A) --++ (L);
+
+		% LADDER RIGHT OF BARN
+		\draw[rod] (C) --++ (L');
+		\draw[rod] (C) --++ (L);
+
+		% LABELS
+		\node[mydarkblue,below=0,align=center,scale=0.8,yshift=1] at (\xb+\wb/2,0)
+		{barn\\$L$};
+		\node[mydarkpurple,right,align=left,scale=0.65,yshift=1.2] at (\xb+3.6*\d,\yA)
+		{both doors closed in S};
+		%{both doors\\[-3]close in S};
+		\node[mydarkred,right,scale=0.65,yshift=1.8,rotate=\ang] at ($(A)+(\ang:\Lz+0.8*\D)$)
+		{left door closed in S$'$};
+		\node[mydarkred,right,scale=0.65,yshift=0.7,rotate=\ang] at ($(B)+(\ang:0.8*\D)$)
+		{right door closed in S$'$};
+	\end{tikzpicture}
+
+
+	% SPACETIME DIAGRAM - LADDER PARADOX from perspective of S' (i.e. in the S' frame)
+	% \begin{tikzpicture}[scale=2.5]
+	%   \message{Ladder paradox from the perspective of S'^^J}
+
+	%   % SETTINGS
+	%   \def\ang{-33.5}   % angle between x and x' axes
+	%   \def\Nxlines{9}   % number of world lines (at constant x)
+	%   \def\Nylines{13}  % number of world lines (at constant t)
+	%   \def\Nxplines{6}  % number of world lines (at constant x')
+	%   \def\Nyplines{10} % number of world lines (at constant t')
+	%   \def\xmin{0.2}
+	%   \pgfmathsetmacro\D{2.6/13} % grid size
+	%   \pgfmathsetmacro\d{\D/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+	%   \pgfmathsetmacro\xmax{(\Nxlines+0.4)*\d}   % maximum of x axis in S
+	%   \pgfmathsetmacro\ymax{(\Nylines+0.4)*\d}   % maximum of y = ct axis in S
+	%   \pgfmathsetmacro\xmaxp{(\Nxplines+0.4)*\D} % maximum of x' axis in S'
+	%   \pgfmathsetmacro\ymaxp{(\Nyplines+0.4)*\D} % maximum of y' = ct' axis in S'
+	%   \pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
+	%   \pgfmathsetmacro\L{\Lz/cos(\ang)} % contracted length L in S
+	%   \pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
+	%   \pgfmathsetmacro\wb{3.08*\d} % width of barn in S
+	%   \pgfmathsetmacro\yAp{(\xb+0.04*\d)*sin(\ang)*(1-cot(\ang)^2)} % y' = ct' coordinate when ladder is fully in barn in S
+	%   \pgfmathsetmacro\yBp{\yAp+\L*sin(\ang)} % y' = ct' coordinate when ladder is fully in barn in S
+	%   \coordinate (O) at (0,0);
+	%   \coordinate (X) at (\ang:\xmax+0.05);
+	%   \coordinate (T) at (90-\ang:\ymax+0.05);
+	%   \coordinate (X') at (\xmaxp+0.15,0);
+	%   \coordinate (T') at (0,\ymaxp+0.15);
+	%   \coordinate (L) at (\ang:\L); % ladder end in S
+	%   \coordinate (L') at (\Lz,0); % ladder end in S'
+	%   \coordinate (A) at (0,\yAp); % left end of ladder when fully in barn in S
+	%   \coordinate (B) at ($(A)+(L)$); % right end of ladder when fully in barn in S
+	%   \coordinate (C) at (0,{(\xb+\wb+0.08*\d)*sin(\ang)*(1-cot(\ang)^2)}); % left end of ladder when fully passed through barn
+
+	%   % FILL
+	%   \fill[myfieldred]
+	%     (-\xmin,0) -| (\xmaxp,\ymaxp) -| (0,-\xmin) -| cycle;
+	%   \fill[mylightred] % ladder frame
+	%     (0,-\xmin) |- (\Lz,\ymaxp) -- ($(L)+(0,-\xmin)$) -- cycle;
+	%   \fill[mydarkblue!22] % barn frame
+	%     (\ang:\xb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax)
+	%     --++ (\ang:\wb) --++ (90-\ang:-\xmin-\ymax) -- cycle;
+
+	%   % HIGHLIGHT DOORS OPEN/CLOSED
+	%   \begin{scope}
+	%     \clip (0,0) --++ (90-\ang:\ymax) -- (\xmaxp+1.8*\d,\ymaxp) --++ (0,-1.1*\ymaxp) -- cycle;
+	%     \draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
+	%       ({\yAp*tan(\ang)-0.1},\yAp) -- (\xmaxp+1.6*\d,\yAp)
+	%       ({\yBp*tan(\ang)-0.1},\yBp) -- (\xmaxp+1.8*\d,\yBp);
+	%     \draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
+	%       (A)++(\ang:-\xb-0.1) --++ (\ang:{\xb+\L+3.75*\d});
+	%   \end{scope}
+
+	%   % BOOSTED WORLD LINE GRID
+	%   \message{  Making world lines for boosted frame...^^J}
+	%   \foreach \i [evaluate={\x=\i*\D;}] in {1,...,\Nxplines}{
+	%     \draw[world line]   (\x,0) -- (\x,\ymaxp);
+	%   }
+	%   \foreach \i [evaluate={\t=\i*\D;}] in {1,...,\Nyplines}{
+	%     \draw[world line t] (0,\t) -- (\xmaxp,\t);
+	%   }
+
+	%   % WORLD LINE GRID
+	%   \message{  Making world lines...^^J}
+	%   \foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nxlines}{
+	%     \draw[world line'] (\ang:\x) --++ (90-\ang:\ymax);
+	%   }
+	%   \foreach \i [evaluate={\t=\i*\d;}] in {1,...,\Nylines}{
+	%     \draw[world line'] (90-\ang:\t) --++ (\ang:\xmax);
+	%   }
+
+	%   % WORLD LINES BARN & ROD
+	%   \draw[->,thick,mydarkblue] % barn left door
+	%     (\ang:\xb+\wb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax+0.2);
+	%   \draw[->,thick,mydarkblue] % barn right door
+	%     (\ang:\xb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax+0.2);
+	%   \draw[->,thick,mydarkbrown] % rod right end
+	%     (L)++(0,-\xmin) -- (\Lz,\ymaxp+0.15);
+
+	%   % AXES
+	%   \draw[->,thick] (90-\ang:-\xmin) -- (T) node[below left=-1] {$ct$};
+	%   \draw[->,thick] (\ang:-\xmin) -- (X) node[below left=0] {$x$};
+	%   \draw[->,thick,mydarkred] (0,-\xmin) -- (T')
+	%     node[right=3,above=-1] {$ct'$};
+	%   \draw[->,thick,mydarkred] (-\xmin,0) -- (X')
+	%     node[anchor=140,inner sep=0.5] {$x'$};
+
+	%   % LADDER
+	%   \draw[rod] (O) -- (L)
+	%     node[pos=0.45,below=2,scale=0.8] {$L$};
+	%   \draw[rod] (O) -- (L')
+	%     node[pos=0.45,above=0.6,scale=0.8] {\contour{mylightred}{$2L$}};
+	%   \draw[rod] (O) -- (L');
+
+	%   % LADDER IN BARN
+	%   \draw[rod] (A) --++ (L);
+	%   \draw[rod] (A) --++ (L');
+	%   \draw[rod] (B) --++ (-\Lz,0);
+
+	%   % LADDER RIGHT OF BARN
+	%   \draw[rod] (C) --++ (L');
+	%   \draw[rod] (C) --++ (L);r
+
+	%   % LABELS
+	%   \node[mydarkbrown,above=1,scale=0.8] at (\Lz/2,\ymaxp)
+	%     {rod};
+	%   \node[mydarkblue,below=0,scale=0.8]
+	%     at ({(\xb+\wb/2)*cos(\ang)*(1-tan(\ang)^2)+0.07},0)
+	%     {\contour{mydarkblue!22}{barn}};
+	%   %\node[mydarkblue,anchor=90-\ang,inner sep=2,scale=0.8,rotate=\ang] at (\ang:\xb+\L/2)
+	%   %  {barn}; %{barn\\$L$};
+	%   \node[mydarkpurple,right,align=left,scale=0.65,yshift=1.6,rotate=\ang]
+	%     at ($(B)+(\ang:0.4*\d)$) {both doors closed in S};
+	%   \node[mydarkred,right,scale=0.65,yshift=1.8] at ($(A)+(\Lz+0.3*\d,0)$)
+	%     {left door closed in S$'$};
+	%   \node[mydarkred,right,scale=0.65,yshift=0.7] at ($(B)+(0.3*\d,0)$)
+	%     {right door closed in S$'$};
+
+	% \end{tikzpicture}
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Relativity/parts/relative velocity.tex

@@ -0,0 +1,79 @@
+\section{Relative Velocity}
+The first weirdness occurs if we consider multiple people running in multiple directions and speeds. Let's explore this.
+
+You shouldn't have to do any math here besides thinking about angles.
+
+\problem{}
+Using the given diagram, verify that if, in Bob's reference frame, Alice is running away from Bob at speed $v$, then in Alice's reference frame, Bob is running away at speed $v$.
+
+{\em Hint: slopes}
+
+% SPACETIME DIAGRAM - LORENTZ BOOST
+\begin{center}\begin{tikzpicture}[scale=1.8]
+\message{Lorentz boost^^J}
+
+\def\xmax{2}
+\def\xmaxp{2.2} % maximum of rotated axis
+\def\Nlines{5} % number of world lines (at constant x/t)
+\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
+\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
+\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
+\coordinate (O) at (0,0);
+\coordinate (X) at (\xmax+0.2,0);
+\coordinate (T) at (0,\xmax+0.2);
+\coordinate (X') at (\ang:\xmaxp+0.2);
+\coordinate (T') at (90-\ang:\xmaxp+0.2);
+
+% WORLD LINE GRID
+\message{  Making world lines...^^J}
+\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
+\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+\draw[world line]   (-\x,-\xmax) -- (-\x,\xmax);
+\draw[world line]   ( \x,-\xmax) -- ( \x,\xmax);
+\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
+\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
+}
+
+% BOOSTED WORLD LINE GRID
+\message{  Making world lines for boosted frame...^^J}
+\fill[mydarkred,opacity=0.05]
+(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
+\fill[mydarkred,opacity=0.05]
+(O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
+\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
+\message{  Running i/N=\i/\Nlines, x=\x...^^J}
+\draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
+\draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
+\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
+\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
+}
+
+% AXES
+\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {Alice: $ct$};
+\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
+\draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
+node[right=5,above=-1] {Bob: $ct'$};
+\draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
+
+% ANGLES
+\draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
+\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
+
+% % PHOTON
+% \draw[photon] (O) --++ (4*\d, 4*\d);
+\end{tikzpicture}\end{center}
+\begin{solution}
+Extend the graph to the left and look at the slope of Alice's world line in Bob's reference frame. You'll find that Alice's slope in Bob's reference frame is the opposite of Bob's slope in Alice's reference frame.
+\end{solution}
+
+\problem{}
+What if Alice runs to the left at speed $v$, Bob stays still, and Charlie runs to the right at speed $v$? 
+
+From Charlie's reference frame, is Alice running away at speed $2v$? Why or why not? 
+{\em Think about the consequences.}
+
+\begin{solution}
+If $v > c/2$, then this poses an immediate problem. Alice has to move at a speed below $2v$. So velocities do not just add. 
+\end{solution}
+\vfill
+\pagebreak

src/Advanced/Tropical Polynomials/main.typ

@@ -1,13 +1,6 @@
 #import "@local/handout:0.1.0": *
 
-#show: doc => handout(
-  doc,
-  group: "Advanced 2",
-  quarter: link(
-    "https://betalupi.com/handouts",
-    "betalupi.com/handouts",
-  ),
-
+#show: handout.with(
   title: [Tropical Polynomials],
   by: "Mark",
   subtitle: "Based on a handout by Bryant Mathews",

src/Advanced/Tropical Polynomials/meta.toml

@@ -3,5 +3,5 @@ title = "Tropical Polynomials"
 
 
 [publish]
-handout = true
+handout = false
 solutions = true

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

@@ -104,7 +104,7 @@ Is there a tropical multiplicative identity? \
 Do tropical multiplicative inverses always exist? \
 #note([
   For every $x != #sym.infinity$, does there exist an inverse $y$ so that $x #tm y = i$, \
-  where $i$ is the additive identity?
+  where $i$ is the multiplicative identity?
 ])
 
 #solution([Yes, it is $-x$. For $x != 0$, $x #tm (-x) = 0$])
@@ -126,7 +126,7 @@ Fill the following tropical addition and multiplication tables
 
 #let col = 10mm
 
-#notsolution(
+#if_no_solutions(
   table(
     columns: (1fr, 1fr),
     align: center,

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

@@ -63,7 +63,7 @@ where all exponents represent repeated tropical multiplication.
 Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
 #hint([$1x$ is not equal to $x$.])
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution([
   $f(x) = min(2x , 1+x, 4)$, which looks like:
@@ -132,7 +132,7 @@ How can we use the graph to determine these roots?
 Graph $f(x) = -2x^2 #tp x #tp 8$. \
 #hint([Use half scale. 1 box = 2 units.])
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution([
   #graphgrid({
@@ -140,7 +140,7 @@ Graph $f(x) = -2x^2 #tp x #tp 8$. \
     let step = 0.75
 
     dotline((0, 0), (8 * step, 8 * step))
-    dotline((0.5 * step, 0), (4 * step, 8 * step))
+    dotline((0.5 * step, 0), (4.5 * step, 8 * step))
     dotline((0, 4 * step), (8 * step, 4 * step))
 
     line(
@@ -210,7 +210,7 @@ and always produces $7$ for sufficiently large inputs.
 #problem()
 Graph $f(x) = 1x^2 #tp 3x #tp 5$.
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution([
   The graphs of all three terms intersect at the same point:
@@ -261,7 +261,7 @@ How are the roots of $f$ related to its coefficients?
 #problem()
 Graph $f(x) = 2x^2 #tp 4x #tp 4$.
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution(
   graphgrid({

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

@@ -5,12 +5,12 @@
 = Tropical Cubic Polynomials
 
 #problem()
-Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
+Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 3x #tp 6$. \
 - sketch a graph of this polynomial
 - use this graph to find the roots of $f$
 - write (and expand) a product of linear factors with the same graph as $f$.
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution([
   - Roots are 1, 2, and 3.
@@ -43,12 +43,12 @@ Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
 #pagebreak() // MARK: page
 
 #problem()
-Consider the polynomial $f(x) = x^3 #tp x^2 #tp 6x #tp 6$. \
+Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 6x #tp 6$. \
 - sketch a graph of this polynomial
 - use this graph to find the roots of $f$
 - write (and expand) a product of linear factors with the same graph as $f$.
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution([
   - Roots are 1, 2.5, and 2.5.
@@ -82,7 +82,7 @@ Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
 - use this graph to find the roots of $f$
 - write (and expand) a product of linear factors with the same graph as $f$.
 
-#notsolution(graphgrid(none))
+#if_no_solutions(graphgrid(none))
 
 #solution([
   - Roots are 2, 2, and 2.
@@ -118,10 +118,10 @@ Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b
 #solution([
 
   $
-    B = min(b, (a+c)/2, (2a+d)/2)
+    B = min(b, (a+c)/2, (2a+d)/3)
   $
   $
-    C = min(c, (b+d)/2, (a+2d)/2)
+    C = min(c, (b+d)/2, (a+2d)/3)
   $
 ])
 
@@ -153,7 +153,7 @@ What are the roots of the following polynomial?
 #v(1fr)
 #pagebreak() // MARK: page
 
-#problem()
+#problem(label: "findci")
 If
 $
   f(x) = c_0 #tp c_1 x #tp c_2 x^2 #tp ... #tp c_n x^n
@@ -183,7 +183,7 @@ Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
 
 
 #problem()
-With the same setup as the previous problem, \
+With the same setup as @findci, \
 find formulas for the roots $r_1, r_2, ..., r_n$.
 
 #solution([

src/Warm-Ups/A Familiar Concept/main.tex

@@ -1,35 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\title{Warm-Up: A Familiar Concept}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-
-	\problem{}<one>
-	Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric. \par
-	$$
-		\sum_{\forall i} |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
-	$$
-	\vfill
-
-	\problem{}
-	Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in \ref{one}.
-	\vfill
-
-	\problem{}
-	What is this metric usually called?
-
-\end{document}

src/Warm-Ups/A Familiar Concept/main.typ

@@ -0,0 +1,33 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: A Familiar Concept],
+  by: "Mark",
+)
+
+#problem()
+Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric:
+
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      sum_(#sym.forall i) |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
+    $,
+  ),
+)
+
+#v(1fr)
+
+#problem()
+Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in the previous problem.
+
+#v(1fr)
+
+
+#problem()
+What is this metric usually called?
+
+
+#v(0.25fr)

src/Warm-Ups/Adders/main.tex

@@ -1,93 +0,0 @@
-\documentclass[
-	nosolutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: Adders}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	Fill the following binary addition table. \par
-	\hint{s is \say{sum,} c is \say{carry}}
-
-	\begin{center}
-		\begin{tabular}{ c c || c c }
-			$a$ & $b$ & s & c \\
-			\hline
-			0 & 0 & ? & ? \\
-			0 & 1 & ? & ? \\
-			1 & 0 & ? & ? \\
-			1 & 1 & ? & ?
-		\end{tabular}
-	\end{center}
-
-	\vfill
-
-	\problem{}
-	Draw a logic circuit that atisfies the above table. \par
-	This is called a \textit{half adder}. \par
-	\hint{You should need exactly two gates.}
-
-	\begin{solution}
-		$s = a \texttt{ xor } b$ \par
-		$c = a \texttt{ and } b$
-	\end{solution}
-
-	\vfill
-
-	\definition{}
-	A \textit{full adder} is similar to a half adder, but it has an extra input: \par
-	a full adder takes $a$, $b$, and $c_\text{in}$, and produces $s$ and $c_\text{out}$. \par
-	\hint{$c_\text{in}$ is \say{carry in}}
-
-	\problem{}
-	Use two half adders to construct a full adder.
-
-	\begin{solution}
-		$s_1, c_1 = \texttt{HA}(a, b)$ \par
-		$s_2, c_2 = \texttt{HA}(s_1, c_\text{in})$ \par
-		$s_\text{out} = s_2$ \par
-		$c_\text{out} = \texttt{OR}(c_1, c_2)$
-
-		\vspace{2mm}
-
-		Of course, the class should just draw the circuit.
-	\end{solution}
-
-
-	\vfill
-	\pagebreak
-
-	\problem{}<rippleadder>
-	How can we add two four-bit binary numbers using the full adder? \par
-	We want a four-bit output sum and a one-bit $c_\text{out}$.
-
-	\vfill
-
-	\problem{}
-	Say that all basic logic gates need $1u$ of time to fully switch states. \par
-	\note[Note]{This is called \textit{gate delay}}
-
-	\vspace{2mm}
-
-	How much time does a full adder need to fully switch states? \par
-	How about your circuit from \ref{rippleadder}?
-
-	\vfill
-
-	\problem{Bonus}
-	Design a faster solution to \ref{rippleadder}.
-
-
-	\vfill
-	\pagebreak
-
-\end{document}

src/Warm-Ups/Adders/main.typ

@@ -0,0 +1,85 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Adders],
+  by: "Mark",
+)
+
+#problem()
+Fill the following binary addition table. \
+#hint([s is "sum," c is "carry"])
+
+#align(
+  center,
+  table(
+    columns: (9mm, 9mm, 9mm, 9mm),
+    align: center,
+    $a$, $b$, $s$, $c$,
+    [0], [0], [?], [?],
+    [0], [1], [?], [?],
+    [1], [0], [?], [?],
+    [1], [1], [?], [?],
+  ),
+)
+
+#v(1fr)
+
+#problem()
+Draw a logic circuit that atisfies the above table. \
+This is called a _half adder_. \
+#hint([You should need exactly two gates.])
+
+#solution([
+  $s = a #text([`xor`]) b$ \
+  $c = a #text([`and`]) b$
+])
+
+#v(1fr)
+
+#definition()
+A _full adder_ is similar to a half adder, but it has an extra input: \
+a full adder takes $a$, $b$, and $c_"in"$, and produces $s$ and $c_"out"$. \
+#hint([$c_"in"$ is "carry in"])
+
+#problem()
+Use two half adders to construct a full adder.
+
+#solution([
+  $
+    s_1, c_1 &= "HA"(a, b) \
+    s_2, c_2 &= "HA"(s_1, c_"in") \
+    s_"out" &= s_2 \
+    c_"out" &= "OR"(c_1, c_2)
+  $
+
+  #v(2mm)
+
+  Of course, the class should just draw the circuit.
+])
+
+
+#v(1fr)
+#pagebreak()
+
+
+
+#problem(label: "ripple-adder")
+How can we add two four-bit binary numbers using the full adder? \
+We want a four-bit output sum and a one-bit $c_"out"$.
+#v(1fr)
+
+#problem()
+Say that all basic logic gates need $1u$ of time to fully switch states. \
+#note([This is called _gate delay_], type: "Note")
+
+#v(2mm)
+
+How much time does a full adder need to fully switch states? \
+How about your circuit from @ripple-adder?
+
+#v(1fr)
+
+#problem("Bonus")
+Design a faster solution to @ripple-adder.
+
+#v(1fr)

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

@@ -1,187 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-% x,y,scale,style
-\def\ttt#1#2#3#4{
-	\draw[#4] (#1, #2+#3) -- (#1+#3+#3+#3, #2+#3);
-	\draw[#4] (#1, #2+#3+#3) -- (#1+#3+#3+#3, #2+#3+#3);
-	\draw[#4] (#1+#3, #2) -- (#1+#3, #2+#3+#3+#3);
-	\draw[#4] (#1+#3+#3, #2) -- (#1+#3+#3, #2+#3+#3+#3);
-}
-
-
-\geometry{
-	paper      = letterpaper,
-	top        = 25mm,
-	bottom     = 30mm,
-	left       = 20mm,
-	right      = 20mm,
-	headheight = 75mm,
-	footskip   = 15mm
-}
-
-
-% misere ttt
-
-% Numerical Tic Tac Toe is a variation invented by the mathematician Ronald Graham.
-% The numbers 1 to 9 are used in this game. The first player plays with the odd numbers,
-% the second player plays with the even numbers. All numbers can be used only once.
-% The player who puts down 15 points in a line wins (sum of 3 numbers).
-% This game can be generalized to a n × n board.
-
-
-% In Treblecross, both players play with the same symbol.
-% The game is played on a 1-by-n board with k equal to 3.
-% The player who makes a three in a row of Xs (or black chips) wins the game
-
-\title{Warm-Up: Big-Tac-Toe}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-
-
-	We have large tic-tac-toe grid, each cell of which contains another.
-	On each turn, one player puts their glyph into a cell of a small grid. When the next player goes,
-	they must make their move in the small grid in the same position as the previous player's move.
-	\begin{itemize}
-		\item The first player to move may pick any small grid to start in.
-		\item If a player is directed to a grid that is \textit{full}, that player may go anywhere. \par
-		A sub-grid that is \say{won} but not full may still be played in.
-	\end{itemize}
-
-	\vspace{2mm}
-
-	The first player to complete a line of three \say{won} subgrids wins the game.
-
-	\vfill\null\hfill
-	\begin{minipage}{0.48\textwidth}
-		\begin{center}
-			\begin{tikzpicture}[scale=0.65]
-				\ttt{0}{0}{4}{line width=0.5mm}
-				\ttt{0.5}{0.5}{1}{line width=0.25mm}
-				\ttt{0.5}{4.5}{1}{line width=0.25mm}
-				\ttt{0.5}{8.5}{1}{line width=0.25mm}
-				\ttt{4.5}{0.5}{1}{line width=0.25mm}
-				\ttt{4.5}{4.5}{1}{line width=0.25mm}
-				\ttt{4.5}{8.5}{1}{line width=0.25mm}
-				\ttt{8.5}{0.5}{1}{line width=0.25mm}
-				\ttt{8.5}{4.5}{1}{line width=0.25mm}
-				\ttt{8.5}{8.5}{1}{line width=0.25mm}
-			\end{tikzpicture}
-		\end{center}
-	\end{minipage}
-	\hfill
-	\begin{minipage}{0.48\textwidth}
-		\begin{center}
-			\begin{tikzpicture}[scale=0.65]
-				\ttt{0}{0}{4}{line width=0.5mm}
-				\ttt{0.5}{0.5}{1}{line width=0.25mm}
-				\ttt{0.5}{4.5}{1}{line width=0.25mm}
-				\ttt{0.5}{8.5}{1}{line width=0.25mm}
-				\ttt{4.5}{0.5}{1}{line width=0.25mm}
-				\ttt{4.5}{4.5}{1}{line width=0.25mm}
-				\ttt{4.5}{8.5}{1}{line width=0.25mm}
-				\ttt{8.5}{0.5}{1}{line width=0.25mm}
-				\ttt{8.5}{4.5}{1}{line width=0.25mm}
-				\ttt{8.5}{8.5}{1}{line width=0.25mm}
-			\end{tikzpicture}
-		\end{center}
-	\end{minipage}
-	\hfill\null\vfill
-
-	\problem{}
-	Play a few rounds of this game with someone nearby. \par
-	Can either player force a win?
-
-	\vfill
-
-	\problem{}
-	Modify the rules of this game to disallow play in won subgrids. \par
-	How does your strategy change?
-
-	\vfill
-	\pagebreak
-
-
-	\null\vfill\null\hfill
-	\begin{minipage}{0.48\textwidth}
-		\begin{center}
-			\begin{tikzpicture}[scale=0.65]
-				\ttt{0}{0}{4}{line width=0.5mm}
-				\ttt{0.5}{0.5}{1}{line width=0.25mm}
-				\ttt{0.5}{4.5}{1}{line width=0.25mm}
-				\ttt{0.5}{8.5}{1}{line width=0.25mm}
-				\ttt{4.5}{0.5}{1}{line width=0.25mm}
-				\ttt{4.5}{4.5}{1}{line width=0.25mm}
-				\ttt{4.5}{8.5}{1}{line width=0.25mm}
-				\ttt{8.5}{0.5}{1}{line width=0.25mm}
-				\ttt{8.5}{4.5}{1}{line width=0.25mm}
-				\ttt{8.5}{8.5}{1}{line width=0.25mm}
-			\end{tikzpicture}
-		\end{center}
-	\end{minipage}
-	\hfill
-	\begin{minipage}{0.48\textwidth}
-		\begin{center}
-			\begin{tikzpicture}[scale=0.65]
-				\ttt{0}{0}{4}{line width=0.5mm}
-				\ttt{0.5}{0.5}{1}{line width=0.25mm}
-				\ttt{0.5}{4.5}{1}{line width=0.25mm}
-				\ttt{0.5}{8.5}{1}{line width=0.25mm}
-				\ttt{4.5}{0.5}{1}{line width=0.25mm}
-				\ttt{4.5}{4.5}{1}{line width=0.25mm}
-				\ttt{4.5}{8.5}{1}{line width=0.25mm}
-				\ttt{8.5}{0.5}{1}{line width=0.25mm}
-				\ttt{8.5}{4.5}{1}{line width=0.25mm}
-				\ttt{8.5}{8.5}{1}{line width=0.25mm}
-			\end{tikzpicture}
-		\end{center}
-	\end{minipage}
-	\hfill\null\vfill
-
-	\vfill\null\hfill
-	\begin{minipage}{0.48\textwidth}
-		\begin{center}
-			\begin{tikzpicture}[scale=0.65]
-				\ttt{0}{0}{4}{line width=0.5mm}
-				\ttt{0.5}{0.5}{1}{line width=0.25mm}
-				\ttt{0.5}{4.5}{1}{line width=0.25mm}
-				\ttt{0.5}{8.5}{1}{line width=0.25mm}
-				\ttt{4.5}{0.5}{1}{line width=0.25mm}
-				\ttt{4.5}{4.5}{1}{line width=0.25mm}
-				\ttt{4.5}{8.5}{1}{line width=0.25mm}
-				\ttt{8.5}{0.5}{1}{line width=0.25mm}
-				\ttt{8.5}{4.5}{1}{line width=0.25mm}
-				\ttt{8.5}{8.5}{1}{line width=0.25mm}
-			\end{tikzpicture}
-		\end{center}
-	\end{minipage}
-	\hfill
-	\begin{minipage}{0.48\textwidth}
-		\begin{center}
-			\begin{tikzpicture}[scale=0.65]
-				\ttt{0}{0}{4}{line width=0.5mm}
-				\ttt{0.5}{0.5}{1}{line width=0.25mm}
-				\ttt{0.5}{4.5}{1}{line width=0.25mm}
-				\ttt{0.5}{8.5}{1}{line width=0.25mm}
-				\ttt{4.5}{0.5}{1}{line width=0.25mm}
-				\ttt{4.5}{4.5}{1}{line width=0.25mm}
-				\ttt{4.5}{8.5}{1}{line width=0.25mm}
-				\ttt{8.5}{0.5}{1}{line width=0.25mm}
-				\ttt{8.5}{4.5}{1}{line width=0.25mm}
-				\ttt{8.5}{8.5}{1}{line width=0.25mm}
-			\end{tikzpicture}
-		\end{center}
-	\end{minipage}
-	\hfill\null\vfill
-
-\end{document}

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

@@ -0,0 +1,90 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+#show: handout.with(
+  title: [Warm-Up: Big-Tac-Toe],
+  by: "Mark",
+)
+
+#let extra_boards = false;
+
+#let ttt(s, p) = {
+  // s: scale,
+  // p: position
+  let x = p.at(0) * s
+  let y = p.at(1) * s
+
+  cetz.draw.line((-1 * s + x, 3 * s + y), (-1 * s + x, -3 * s + y))
+  cetz.draw.line((1 * s + x, 3 * s + y), (1 * s + x, -3 * s + y))
+  cetz.draw.line((3 * s + x, -1 * s + y), (-3 * s + x, -1 * s + y))
+  cetz.draw.line((3 * s + x, 1 * s + y), (-3 * s + x, 1 * s + y))
+}
+
+#let btt(s) = cetz.canvas({
+  import cetz.draw: *
+
+  set-style(stroke: (thickness: 0.5mm * s))
+
+  ttt(s, (-7, -7))
+  ttt(s, (-7, 0))
+  ttt(s, (-7, 7))
+  ttt(s, (0, -7))
+  ttt(s, (0, 0))
+  ttt(s, (0, 7))
+  ttt(s, (7, -7))
+  ttt(s, (7, 0))
+  ttt(s, (7, 7))
+
+  set-style(stroke: (thickness: 2mm * s))
+  ttt(s * 3.5, (0, 0))
+})
+
+
+#problem()
+Consider a large tic-tac-toe grid, each cell of which contains another.
+On each turn, one player puts their glyph into a cell of a small grid. When the next player goes,
+they must make their move in the small grid in the same position as the previous player's move.
+- The first player to move may pick any small grid to start in.
+- If a player is directed to a grid that is _full_, that player may go anywhere. \
+  A sub-grid that is "won" but not full may still be played in.
+
+#v(2mm)
+
+The first player to complete a line of three "won" subgrids wins the game.
+
+#v(2mm)
+
+#problem()
+Play a few rounds of this game with someone nearby. \
+Can either player force a win?
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr),
+  btt(0.35), btt(0.35),
+);
+
+#problem()
+Modify the rules of this game to disallow play in won subgrids. \
+How does your strategy change? \
+#if extra_boards { note([Additional boards are available on the next page.]) }
+
+#v(1fr)
+
+#if extra_boards {
+  pagebreak()
+
+  align(
+    center,
+    grid(
+      stroke: none,
+      align: center,
+      columns: (1fr, 1fr),
+      rows: (1fr, 1fr, 1fr),
+      btt(0.35), btt(0.35),
+      btt(0.35), btt(0.35),
+      btt(0.35), btt(0.35),
+    ),
+  )
+}

src/Warm-Ups/Fuse Timers/main.tex

@@ -1,31 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\title{Warm-Up: Fuse Timers}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-
-	\problem{}
-	Suppose we have two strings and a lighter. Each string takes an hour to fully burn. \par
-	However, we do not know how fast each part of the string burns:
-	half might burn in 1 minute, and the rest could take 59.
-
-	\vspace{2mm}
-
-	How would we measure exactly 45 minutes using these strings?
-
-	\vfill
-
-\end{document}

src/Warm-Ups/Fuse Timers/main.typ

@@ -0,0 +1,15 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Fuse Timers],
+  by: "Mark",
+)
+
+#problem()
+Suppose we have two strings and a lighter. Each string takes exactly an hour to fully burn. \
+However, we do not know how fast each part of the string burns:
+half might burn in 1 minute, and the rest could take 59.
+
+#v(2mm)
+
+How can we measure exactly 45 minutes using these two strings?

src/Warm-Ups/Gallery/main.tex

@@ -1,121 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\makeatletter
-\newcommand{\thisone}{
-	\if@solutions
-	{\color{red} $\Leftarrow$ \texttt{this one}}
-	\else\fi
-}
-
-\title{Warm-Up: The Gallery}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-	A museum curator is arranging seven photographs on a gallery wall in accordance with the photographer's requirements.
-	They are titled as follows: Fence, Gardenias, Hibiscus, Irises, Katydid, Lotus, and Magnolia.
-
-	The photograph's requirements are as follows:
-	\begin{itemize}
-		\item Gardenias must be immediately before Katydid.
-		\item Hibiscus must be somewhere before Katydid but cannot be the first photograph.
-		\item Irises and Lotus must be next to one another.
-		\item Magnolia must be one of the first three photographs.
-		\item Fence must be either first or seventh.
-	\end{itemize}
-
-	\problem{}
-	Which of the below could be a valid ordering? \par
-	\note[Note]{We denote each painting by the first letter of its title.}
-	\begin{itemize}
-		\item \texttt{FHGMKIL}
-		\item \texttt{HMGKILF}
-		\item \texttt{ILMHGKF} \thisone{}
-		\item \texttt{LMIHGKF}
-		\item \texttt{MFHGKLI}
-	\end{itemize}
-	\vfill
-
-	\problem{}
-	If Irises is immediately before Gardenias, which of the following could be true?
-	\begin{itemize}
-		\item Gardenias is fourth
-		\item Hibiscus is fourth
-		\item Irises is third
-		\item Lotus is second
-		\item Magnolia is third \thisone{}
-	\end{itemize}
-	\vfill
-
-	\problem{}
-	The ordering of the photographs is fully determined if...
-	\begin{itemize}
-		\item Gardenias is fourth
-		\item Hibiscus is second
-		\item Irises is second
-		\item Lotus is first \thisone{}
-		\item Magnolia is third
-	\end{itemize}
-
-	\vfill
-	\pagebreak
-
-	\problem{}
-	If Magnolia is second, what CANNOT be true?
-	\begin{itemize}
-		\item Hibiscus is third
-		\item Hibiscus is fourth \thisone{}
-		\item Hibiscus is fifth
-		\item Gardenias is fourth
-		\item Gardenias is sixth
-	\end{itemize}
-
-	\vfill
-
-	\problem{}
-	Katydid cannot be in which position?
-	\begin{itemize}
-		\item Third \thisone{}
-		\item Fourth
-		\item Fifth
-		\item Sixth
-		\item Seventh
-	\end{itemize}
-	\vfill
-
-	\problem{}
-	If Gardenias is fourth, what must be true?
-	\begin{itemize}
-		\item Fence is first \thisone{}
-		\item Hibiscus is third
-		\item Irises is seventh
-		\item Magnolia is first
-		\item Magnolia is second
-	\end{itemize}
-	\vfill
-
-	\problem{}
-	Which one of the following,
-	if substituted for the second condition,
-	would have the same effect in determining the
-	arrangement of the photographs?
-
-	\begin{itemize}
-		\item If Fence is seventh, Hibiscus is second
-		\item Gardenias is somewhere after Hibiscus, and either Fence or Magnolia is first
-		\item Hibiscus must be somewhere between the first and sixth photographs
-		\item Unless Hibiscus is second, it must be somewhere between Magnolia and Gardenias \thisone{}
-		\item Katydid is somewhere after Hibiscus, which must be after Fence.
-	\end{itemize}
-	\vfill
-	\pagebreak
-\end{document}

src/Warm-Ups/Gallery/main.typ

@@ -0,0 +1,103 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: The Gallery],
+  by: "Mark",
+)
+
+#let thisone = if_solutions(
+  text(fill: ored, [#sym.arrow.l.double.long `this one`]),
+)
+
+A museum curator is arranging seven photographs on a gallery wall in accordance with the photographer's requirements.
+They are titled as follows: Fence, Gardenias, Hibiscus, Irises, Katydid, Lotus, and Magnolia.
+
+The photograph's requirements are as follows:
+#v(2mm)
+- Gardenias must be immediately before Katydid.
+- Hibiscus must be somewhere before Katydid but cannot be the first photograph.
+- Irises and Lotus must be next to one another.
+- Magnolia must be one of the first three photographs.
+- Fence must be either first or seventh.
+
+
+#problem()
+Which of the below could be a valid ordering? \
+#note([We denote each painting by the first letter of its title.], type: "Note")
+- `FHGMKIL`
+- `HMGKILF`
+- `ILMHGKF` #thisone
+- `LMIHGKF`
+- `MFHGKLI`
+
+#v(1fr)
+
+#problem()
+If Irises is immediately before Gardenias, which of the following could be true?
+- Gardenias is fourth
+- Hibiscus is fourth
+- Irises is third
+- Lotus is second
+- Magnolia is third #thisone
+
+#v(1fr)
+
+#problem()
+The ordering of the photographs is fully determined if...
+- Gardenias is fourth
+- Hibiscus is second
+- Irises is second
+- Lotus is first #thisone
+- Magnolia is third
+
+
+#v(1fr)
+#pagebreak()
+
+
+
+#problem()
+If Magnolia is second, what CANNOT be true?
+- Hibiscus is third
+- Hibiscus is fourth #thisone
+- Hibiscus is fifth
+- Gardenias is fourth
+- Gardenias is sixth
+
+
+#v(1fr)
+
+#problem()
+Katydid cannot be in which position?
+- Third #thisone
+- Fourth
+- Fifth
+- Sixth
+- Seventh
+
+#v(1fr)
+
+#problem()
+If Gardenias is fourth, what must be true?
+- Fence is first #thisone
+- Hibiscus is third
+- Irises is seventh
+- Magnolia is first
+- Magnolia is second
+
+#v(1fr)
+
+#problem()
+Which one of the following,
+if substituted for the second condition,
+would have the same effect in determining the
+arrangement of the photographs?
+
+- If Fence is seventh, Hibiscus is second
+- Gardenias is somewhere after Hibiscus, and either Fence or Magnolia is first
+- Hibiscus must be somewhere between the first and sixth photographs
+- Unless Hibiscus is second, it must be somewhere between Magnolia and Gardenias \
+  #if_solutions(text(fill: ored, [#sym.arrow.t.double `this one`]))
+- Katydid is somewhere after Hibiscus, which must be after Fence.
+
+#v(1fr)

src/Warm-Ups/Mario Kart/main.tex

@@ -1,54 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: Mario Kart}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	A standard Mario Kart cup consists of 12 players and four races. \par
-	Each race is scored as follows:
-	\begin{itemize}
-		\item 15 points are awarded for first place;
-		\item 12 for second;
-		\item and $(13 - \text{place})$ otherwise.
-	\end{itemize}
-	In any one race, no players may tie.
-	A player's score at the end of a cup is the sum of their scores for each of the four races.
-
-	\vspace{2mm}
-
-	An $n$-way tie occurs when the top $n$ players have the same score at the end of a round. \par
-	What is the largest possible $n$, and how is it achieved?
-
-	\begin{solution}
-		A 12-way tie is impossible, since the total number of point is not divisible by 12.
-
-		\vspace{2mm}
-
-		A 11-way tie is possible, with a top score of 28:
-		\begin{itemize}
-			\item Four players finish $1^\text{st}$, $3^\text{ed}$, $11^\text{th}$, and $12^\text{th}$;
-
-			% spell:off
-			\item Four players finish $2^\text{nd}$, $4^\text{th}$, $9^\text{th}$, and $10^\text{th}$;
-			% spell:on
-
-			\item Two players finish fifth twice and seventh twice,
-			\item One player finishes sixth in each race.
-		\end{itemize}
-		The final player always finishes eighth, with a non-tie score of 20.
-	\end{solution}
-
-\end{document}

src/Warm-Ups/Mario Kart/main.typ

@@ -0,0 +1,35 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Mario Kart],
+  by: "Mark",
+)
+
+#problem()
+A standard Mario Kart cup consists of 12 players and four races. \
+Each race is scored as follows:
+- 15 points are awarded for first place;
+- 12 for second;
+- and $(13 - #text("place"))$ otherwise.
+
+In any one race, no players may tie. \
+A player's score at the end of a cup is the sum of their scores for each of the four races.
+
+#v(2mm)
+
+An $n$-way tie occurs when the top $n$ players have the same score at the end of a round. \
+What is the largest possible $n$, and how is it achieved?
+
+#solution([
+  A 12-way tie is impossible, since the total number of point is not divisible by 12.
+
+  #v(2mm)
+
+  A 11-way tie is possible, with a top score of 28:
+  - Four players finish $1^#text("st")$, $3^#text("ed")$, $11^#text("th")$, and $12^#text("th")$;
+  - Four players finish $2^#text("nd")$, $4^#text("th")$, $9^#text("th")$, and $10^#text("th")$; // spell:disable-line
+  - Two players finish fifth twice and seventh twice,
+  - One player finishes sixth in each race.
+  The final player always finishes eighth, with a non-tie score of 20.
+
+])

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 +1,111 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+
+#show: handout.with(
+  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(2mm)
+
+  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(
+  align(
+    center,
+    cetz.canvas({
+      import cetz.draw: *
+
+      let s = 0.8 // Scale
+      let t = 13pt * s // text size
+      let radius = 0.3 * s
+
+      // Points
+      let a = (-2 * s, 0.2 * s)
+      let b = (0 * s, 2 * s)
+      let c = (2 * s, 0.2 * s)
+      let d = (1.2 * s, -2.1 * s)
+      let e = (-1.2 * s, -2.1 * s)
+
+      set-style(
+        stroke: (thickness: 0.6mm * s),
+        mark: (
+          end: (
+            symbol: ">",
+            fill: black,
+            offset: radius + (0.025 * s),
+            width: 1.2mm * s,
+            length: 1.2mm * s,
+          ),
+        ),
+      )
+
+      line(a, b)
+      line(b, c)
+      line(c, d)
+      line(d, e)
+      line(e, a)
+      line(a, c)
+      line(b, d)
+      line(c, e)
+      line(d, a)
+      line(e, b)
+
+      circle(a, radius: radius, fill: oblue, stroke: none)
+      circle(b, radius: radius, fill: oblue, stroke: none)
+      circle(c, radius: radius, fill: oblue, stroke: none)
+      circle(d, radius: radius, fill: oblue, stroke: none)
+      circle(e, radius: radius, fill: oblue, stroke: none)
+
+      content(a, text(fill: white, size: t, [*A*]))
+      content(b, text(fill: white, size: t, [*B*]))
+      content(c, text(fill: white, size: t, [*C*]))
+      content(d, text(fill: white, size: t, [*D*]))
+      content(e, text(fill: white, size: t, [*E*]))
+    }),
+  ),
+)
+
+#v(1fr)
+
+#problem()
+Now, say we roll each die twice. What happens to the graph from the previous problem?
+
+#solution([
+  The direction of each edge is reversed!
+])
+
+#v(1fr)

src/Warm-Ups/Odd Dice/meta.toml

@@ -4,8 +4,3 @@ title = "Odd Dice"
 [publish]
 handout = true
 solutions = true
-
-[[attribution]]
-who = "mark"
-when = 2024-02-13
-what = "Initial version of handout"

src/Warm-Ups/Painting/main.tex

@@ -1,80 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-\title{Warm-Up: The Painting}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	You have a painting on a string. \par
-	Hang the painting on two nails so that if either is removed, the painting falls. \par
-
-	\vspace{2mm}
-	You may detach the string as you hang the painting, but it must be re-attached once you're done.
-
-	\note{The solution to this problem isn't a \say{think outside the box} trick, it's a clever wrapping of the string.}
-
-
-	\begin{center}
-	\begin{tikzpicture}[scale=2.0]
-		\draw[line width = 0.5mm]
-			(0, 1) --
-			(2, 1) --
-			(2, 0) --
-			(0, 0) --
-			cycle
-		;
-
-		\draw[line width = 0.5mm, dotted]
-			(0.1, 1) --
-			(1, 1.5) --
-			(1.9, 1)
-		;
-
-		\fill (1, 1.5) circle[radius = 0.25mm];
-
-
-		\draw[line width = 0.2mm]
-			(0.66, 0.66) --
-			(0.66, 0.35) --
-			(0.60, 0.1)
-			(0.72, 0.1)--(0.66, 0.35)
-		;
-
-		\draw[line width = 0.2mm]
-			(0.66, 0.575) --
-			(0.6, 0.475) --
-			(0.525, 0.575)
-			(0.66, 0.575) --
-			(0.72, 0.475) --
-			(0.795, 0.575)
-		;
-
-		\fill[color=white] (0.66, 0.66) circle[radius = 0.8mm];
-		\draw (0.66, 0.66) circle[radius = 0.8mm];
-	\end{tikzpicture}
-	\end{center}
-
-
-	\begin{solution}
-
-		Say we have a left nail and a right nail. The path of the string is as follows:
-		\begin{itemize}
-			\item Start on the left
-			\item Move over both nails, wrap right nail cw
-			\item Wrap left nail ccw
-			\item Wrap right nail ccw
-			\item Exit downwards, in between both nails
-		\end{itemize}
-
-	\end{solution}
-
-\end{document}

src/Warm-Ups/Painting/main.typ

@@ -0,0 +1,82 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+#show: handout.with(
+  title: [Warm-Up: What's an AST?],
+  by: "Mark",
+  subtitle: "Based on a true story.",
+)
+
+#problem()
+Say we have a painting on a string. \
+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.]
+
+
+#v(2mm)
+
+#align(
+  center,
+  cetz.canvas({
+    import cetz.draw: *
+    let s = 2.5
+
+
+    line(
+      (0 * s, 1 * s),
+      (2 * s, 1 * s),
+      (2 * s, 0 * s),
+      (0 * s, 0 * s),
+      close: true,
+      stroke: (thickness: 0.8mm),
+    )
+
+    line(
+      (0.1 * s, 1 * s),
+      (0.5 * s, 1.5 * s),
+      (1.5 * s, 1.5 * s),
+      (1.9 * s, 1 * s),
+      stroke: (thickness: 0.5mm, dash: "dotted"),
+    )
+
+    circle((0.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
+    circle((1.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
+
+    line(
+      (0.66 * s, 0.66 * s),
+      (0.66 * s, 0.35 * s),
+      (0.60 * s, 0.1 * s),
+    )
+
+    line(
+      (0.72 * s, 0.1 * s),
+      (0.66 * s, 0.35 * s),
+    )
+
+    line(
+      (0.66 * s, 0.575 * s),
+      (0.6 * s, 0.475 * s),
+      (0.525 * s, 0.575 * s),
+    )
+
+    line(
+      (0.66 * s, 0.575 * s),
+      (0.72 * s, 0.475 * s),
+      (0.795 * s, 0.575 * s),
+    )
+
+    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:
+  - Start on the left
+  - Move over both nails, wrap right nail cw
+  - Wrap left nail ccw
+  - Wrap right nail ccw
+  - Exit downwards, in between both nails
+])

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

@@ -1,57 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: Partition Products}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	Take any positive integer $n$. \par
-	Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... + a_k$ \par
-	Maximize the product $a_1 \times a_2 \times ... \times a_k$
-
-
-
-	\begin{solution}
-
-		\textbf{Interesting Solution:}
-
-		Of course, all $a_i$ should be greater than $1$. \par
-		Also, all $a_i$ should be smaller than four, since $x \leq x(x-2)$ if $x \geq 4$. \par
-		Thus, we're left with sequences that only contain 2 and 3. \par
-		\note{Note that two twos are the same as one four, but we exclude fours for simplicity.}
-
-		\vspace{2mm}
-
-		Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \par
-		The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
-
-		\linehack{}
-
-
-
-		\textbf{Calculus Solution:}
-
-		First, solve this problem for equal, non-integer $a_i$:
-
-		\vspace{2mm}
-
-		We know $n = \prod{a_i}$, thus $\ln(n) = \sum{\ln(a_i)}$. \par
-		If all $a_i$ are equal, we get $\ln(n) = k \times \ln(n / k)$. \par
-		Derive wrt $k$ and set to zero to get $\ln(n / k) = 1$ \par
-		So $k = n / e$ and $n / k = e \approx 2.7$
-
-		\vspace{2mm}
-
-		If we try to approximate this with integers, we get the same solution as above.
-	\end{solution}
-\end{document}

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

@@ -0,0 +1,41 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Partition Products],
+  by: "Mark",
+)
+
+#problem()
+Take any positive integer $n$. \
+Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ...  a_k$ \
+Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
+
+
+#solution([
+  *Interesting Solution:*
+
+  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.])
+
+  #v(2mm)
+
+  Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \
+  The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
+
+  #v(8mm)
+
+  *Calculus Solution:*
+
+  First, solve this problem for equal, real $a_i$:
+  #v(2mm)
+  We know $n = product(a_i)$, thus $ln(n) = sum(ln(a_i))$. \
+  If all $a_i$ are equal, we get $ln(n) = k #sym.times ln(n / k)$. \
+  Derive wrt $k$ and set to zero to get $ln(n / k) = 1$ \
+  So $k = n / e$ and $n / k = e #sym.approx 2.7$
+
+  #v(2mm)
+
+  If we try to approximate this with integers, we get the same solution as above.
+])

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

@@ -1,47 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-\usepackage{graphicx}
-
-\title{Warm-Up: Passing Balls}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-
-	Twelve people are standing in a circle. Each is assigned a number between 1 and 12. \par
-	Participants numbered 1, 2, 3, and 4 hold red, green, yellow, and black balls, respectively. \par
-	Everyone else is empty-handed.
-
-	\vspace{2mm}
-
-	Each participant can pass their ball to any student that is exactly 5 positions away. \par
-	Balls cannot be passed to someone who has one in hand.
-
-	\vspace{2mm}
-
-	After a number of passes, the first four participants again hold all the balls. \par
-	Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4?
-
-	\begin{solution}
-		\begin{itemize}
-			\item the graph of possible moves is isomorphic to a circle (since 5 and 12 are coprime),
-			\item but the balls get passed around, so swapping the place of any two balls is not allowed. \\
-			Therefore, the balls will stay in their initial (cyclic) order:
-		\end{itemize}
-
-		\begin{center}
-			\includegraphics[width=\textwidth]{pass-sol.png}
-		\end{center}
-	\end{solution}
-
-
-\end{document}

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

@@ -0,0 +1,192 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+#show: handout.with(
+  title: [Warm-Up: Passing Balls],
+  by: "Mark",
+)
+
+#problem()
+Twelve people are standing in a circle. Each is assigned a number between 1 and 12. \
+Participants numbered 1, 2, 3, and 4 hold red, green, yellow, and black balls, respectively. \
+Everyone else is empty-handed.
+
+#v(2mm)
+
+Each participant can pass their ball to any student that is exactly 5 positions away. \
+Balls cannot be passed to someone who has one in hand.
+
+#v(2mm)
+
+After a number of passes, the first four participants again hold all the balls. \
+Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4?
+
+#solution([
+  The graph of possible moves is isomorphic to a circle (since 5 and 12 are coprime), \
+  so the order of the balls cannot be changed as they are passed around.
+
+  #v(2mm)
+
+  Therefore, the balls will stay in their initial (cyclic) order:
+
+  #v(2mm)
+
+  #{
+    let s = 0.7 // scale
+    let t = 12pt * s // text size
+    let radius = 0.35
+
+    let pts = (
+      (0 * s, 3 * s),
+      (1 * s, 2 * s),
+      (2 * s, 1 * s),
+      (3 * s, 0 * s),
+      (2 * s, -1 * s),
+      (1 * s, -2 * s),
+      (0 * s, -3 * s),
+      (-1 * s, -2 * s),
+      (-2 * s, -1 * s),
+      (-3 * s, 0 * s),
+      (-2 * s, 1 * s),
+      (-1 * s, 2 * s),
+    )
+
+    let pts_shuf = (
+      (0 * s, 3 * s), // 1
+      (1 * s, -2 * s), // 6
+      (-2 * s, 1 * s), // 11
+      (3 * s, 0 * s), // 4
+      (-2 * s, -1 * s), // 9
+      (1 * s, 2 * s), // 2
+      (0 * s, -3 * s), // 7
+      (-1 * s, 2 * s), // 12
+      (2 * s, -1 * s), // 5
+      (-3 * s, 0 * s), // 10
+      (2 * s, 1 * s), // 3
+      (-1 * s, -2 * s), // 8
+    )
+
+    table(
+      stroke: none,
+      align: center,
+      columns: (1fr, 1fr, 1fr),
+      cetz.canvas({
+        import cetz.draw: *
+
+        set-style(stroke: (thickness: 0.4mm, paint: black))
+        line(..pts_shuf, close: true)
+
+        let i = 1
+        for p in pts {
+          circle(
+            p,
+            radius: radius * s,
+            fill: if i == 1 {
+              ored
+            } else if i == 2 {
+              ogreen
+            } else if i == 3 {
+              oorange
+            } else if i == 4 {
+              oblue
+            } else { white },
+          )
+
+          content(
+            p,
+            text(
+              fill: if i <= 4 {
+                white
+              } else {
+                black
+              },
+              size: t,
+              [*#i*],
+            ),
+          )
+          i = i + 1
+        }
+      }),
+      cetz.canvas({
+        import cetz.draw: *
+
+        set-style(stroke: (thickness: 0.4mm, paint: black))
+        line(..pts, close: true)
+
+        let i = 1
+        for p in pts {
+          let l = calc.rem(((i - 1) * 5), 12) + 1
+
+
+          circle(
+            p,
+            radius: radius * s,
+            fill: if l == 1 {
+              ored
+            } else if l == 2 {
+              ogreen
+            } else if l == 3 {
+              oorange
+            } else if l == 4 {
+              oblue
+            } else { white },
+          )
+
+          content(
+            p,
+            text(
+              fill: if l <= 4 {
+                white
+              } else {
+                black
+              },
+              size: t,
+              [*#l*],
+            ),
+          )
+          i = i + 1
+        }
+      }),
+      cetz.canvas({
+        import cetz.draw: *
+
+        set-style(stroke: (thickness: 0.4mm, paint: black))
+        line(..pts, close: true)
+
+        let i = 1
+        for p in pts {
+          let l = calc.rem(((i - 1) * 5), 12) + 1
+
+
+          circle(
+            p,
+            radius: radius * s,
+            fill: if l == 1 {
+              oblue
+            } else if l == 2 {
+              oorange
+            } else if l == 3 {
+              ored
+            } else if l == 4 {
+              ogreen
+            } else { white },
+          )
+
+          content(
+            p,
+            text(
+              fill: if l <= 4 {
+                white
+              } else {
+                black
+              },
+              size: t,
+              [*#l*],
+            ),
+          )
+          i = i + 1
+        }
+      }),
+    )
+  }
+])

src/Warm-Ups/Prime Factors/main.tex

@@ -1,34 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber,
-	hidewarning
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-\title{Warm-Up: Prime Factors}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	What proportion of integers have $2$ as their smallest prime factor?
-	% 1^2
-	\vfill
-
-
-	\problem{}
-	What proportion of integers have $3$ as their second-smallest prime factor?
-	% 1/6
-	\vfill
-
-
-	\problem{}
-	What is the median second-smallest prime factor?
-	% 37
-	\vfill
-
-\end{document}

src/Warm-Ups/Prime Factors/main.typ

@@ -0,0 +1,23 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Prime Factors],
+  by: "Mark",
+)
+
+#problem()
+What proportion of integers have $2$ as their smallest prime factor?
+#solution([$1 div 2$])
+#v(1fr)
+
+
+#problem()
+What proportion of integers have $3$ as their second-smallest prime factor?
+#solution([$1 div 6$])
+#v(1fr)
+
+
+#problem()
+What is the median second-smallest prime factor?
+#solution([37])
+#v(1fr)

src/Warm-Ups/Regex/main.tex

@@ -1,153 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\usepackage{xcolor}
-\usepackage{soul}
-\usepackage{hyperref}
-
-\definecolor{Light}{gray}{.90}
-\sethlcolor{Light}
-\newcommand{\htexttt}[1]{\texttt{\hl{#1}}}
-
-
-\title{The Regex Warm-Up}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-
-	Last time, we discussed Deterministic Finite Automata. One interesting application of these mathematical objects is found in computer science: Regular Expressions. \par
-	This is often abbreviated \say{regex}, which is pronounced like \say{gif.}
-
-	\vspace{2mm}
-
-	Regex is a language used to specify patterns in a string. You can think of it as a concise way to define a DFA, using text instead of a huge graph. \par
-
-	Often enough, a clever regex pattern can do the work of a few hundred lines of code.
-
-	\vspace{2mm}
-
-	Like the DFAs we've studied, a regex pattern \textit{accepts} or \textit{rejects} a string. However, we don't usually use this terminology with regex, and instead say that a string \textit{matches} or \textit{doesn't match} a pattern.
-
-	\vspace{5mm}
-
-	Regex strings consist of characters, quantifiers, sets, and groups.
-
-	\vspace{5mm}
-
-	\textbf{Quantifiers} \par
-	Quantifiers specify how many of a character to match. \par
-	There are four of these: \htexttt{+}, \htexttt{*}, \htexttt{?}, and \htexttt{\{ \}}
-
-	\vspace{2mm}
-
-	\htexttt{+} means \say{match one or more of the preceding token} \par
-	\htexttt{*} means \say{match zero or more of the preceding token}
-
-	For example, the pattern \htexttt{ca+t} will match the following strings:
-	\begin{itemize}
-		\item \texttt{cat}
-		\item \texttt{caat}
-		\item \texttt{caaaaaaaat}
-	\end{itemize}
-	\htexttt{ca+t} will \textbf{not} match the string \texttt{ct}. \par
-	The pattern \htexttt{ca*t} will match all the strings above, including \texttt{ct}.
-	\vspace{2mm}
-
-
-	\htexttt{?} means \say{match one or none of the preceding token} \par
-	The pattern \htexttt{linea?r} will match only \texttt{linear} and \texttt{liner}.
-	\vspace{2mm}
-
-	Brackets \htexttt{\{min, max\}} are the most flexible quantifier. \par
-	They specify exactly how many tokens to match: \par
-	\htexttt{ab\{2\}a} will match only \texttt{abba}. \par
-	\htexttt{ab\{1,3\}a} will match only \texttt{aba}, \texttt{abba}, and \texttt{abbba}. \par
-	% spell:off
-	\htexttt{ab\{2,\}a} will match any \texttt{ab...ba} with at least two \texttt{b}s.
-	% spell:on
-
-	\vspace{5mm}
-
-	\problem{}
-	Write the patterns \htexttt{a*} and \htexttt{a+} using only \htexttt{\{ \}}.
-	\vfill
-
-	\problem{}
-	Draw a DFA equivalent to the regex pattern \htexttt{01*0}.
-	\vfill
-
-	\pagebreak
-
-
-
-
-
-
-	\textbf{Characters, Sets, and Groups} \par
-	In the previous section, we saw how we can specify characters literally: \par
-	\texttt{a+} means \say{one or more \texttt{a} character}
-
-	\vspace{2mm}
-
-	There are, of course, other ways we can specify characters.
-
-	\vspace{2mm}
-
-	The first such way is the \textit{set}, denoted \htexttt{[ ]}. A set can pretend to be any character inside it. \par
-	For example, \htexttt{m[aoy]th} will match \texttt{math}, \texttt{moth}, or \texttt{myth}. \par
-	\htexttt{a[01]+b} will match \texttt{a0b}, \texttt{a111b}, \texttt{a1100110b}, and any other similar string. \par
-	You may negate a set with a \htexttt{\textasciicircum}. \par
-	\htexttt{[\textasciicircum abc]} will match any character except \texttt{a}, \texttt{b}, or \texttt{c}, including symbols and spaces.
-
-	\vspace{2mm}
-
-	If we want to keep characters together, we can use the \textit{group}, denoted \htexttt{( )}. \par
-
-	Groups work exactly as you'd expect, representing an atomic\footnotemark{} group of characters. \par
-	\htexttt{a(01)+b} will match \texttt{a01b} and \texttt{a010101b}, but will \textbf{not} match \texttt{a0b}, \texttt{a1b}, or \texttt{a1100110b}.
-
-	\footnotetext{In other words, \say{unbreakable}}
-
-
-	\problem{}<regex>
-	You are now familiar with most of the tools regex has to offer. \par
-	Write patterns that match the following strings:
-	\begin{enumerate}[itemsep=1mm]
-		\item An ISO-8601 date, like \texttt{2022-10-29}. \par
-		\hint{Invalid dates like \texttt{2022-13-29} should also be matched.}
-
-		\item An email address. \par
-		\hint{Don't forget about subdomains, like \texttt{math.ucla.edu}.}
-
-		\item A UCLA room number, like \texttt{MS 5118} or \texttt{Kinsey 1220B}.
-
-		\item Any ISBN-10 of the form \texttt{0-316-00395-7}. \par
-		\hint{Remember that the check digit may be an \texttt{X}. Dashes are optional.}
-
-		\item A word of even length. \par
-		\hint{The set \texttt{[A-z]} contains every english letter, capitalized and lowercase. \\
-		\texttt{[a-z]} will only match lowercase letters.}
-
-		\item A word with exactly 3 vowels. \par
-		\hint{The special token \texttt{\textbackslash w} will match any word character. It is equivalent to \texttt{[A-z0-9\_]} \\ \texttt{\_} stands for a literal underscore.}
-
-		\item A word that has even length and exactly 3 vowels.
-
-		\item A sentence that does not start with a capital letter.
-	\end{enumerate}
-
-	\vfill
-
-	\problem{}
-	If you'd like to know more, check out \url{https://regexr.com}. It offers an interactive regex prompt, as well as a cheatsheet that explains every other regex token there is. \par
-	You will find a nice set of challenges at \url{https://alf.nu/RegexGolf}.
-	I especially encourage you to look into this if you are interested in computer science.
-\end{document}

src/Warm-Ups/Regex/main.typ

@@ -0,0 +1,135 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [The Regex Warm-Up],
+  by: "Mark",
+)
+
+
+Last time, we discussed Deterministic Finite Automata. One interesting application of these mathematical objects is found in computer science: Regular Expressions. \
+This is often abbreviated "regex," which is pronounced like "gif."
+
+#v(2mm)
+
+Regex is a language used to specify patterns in a string. You can think of it as a concise way to define a DFA, using text instead of a huge graph. \
+
+Often enough, a clever regex pattern can do the work of a few hundred lines of code.
+
+#v(2mm)
+
+Like the DFAs we've studied, a regex pattern _accepts_ or _rejects_ a string. However, we don't usually use this terminology with regex, and instead say that a string _matches_ or _doesn't match_ a pattern.
+
+#v(5mm)
+
+Regex strings consist of characters, quantifiers, sets, and groups.
+
+#v(5mm)
+
+
+
+*Quantifiers* \
+Quantifiers specify how many of a character to match. \
+There are four of these: `+`, `*`, `?`, and `{ }`.
+
+#v(4mm)
+
+`+` means "match one or more of the preceding token" \
+`*` means "match zero or more of the preceding token"
+
+For example, the pattern `ca+t` will match the following strings:
+- `cat`
+- `caat`
+- `caaaaaaaat`
+`ca+t` will *not* match the string `ct`. \
+The pattern `ca*t` will match all the strings above, including `ct`.
+
+
+#v(4mm)
+
+
+`?` means "match one or none of the preceding token" \
+The pattern `linea?r` will match only `linear` and `liner`.
+
+#v(4mm)
+
+Brackets `{min, max}` are the most flexible quantifier. \
+They specify exactly how many tokens to match: \
+`ab{2}a` will match only `abba`. \
+`ab{1,3}a` will match only `aba`, `abba`, and `abbba`. \
+`ab{2,}a` will match any `ab...ba` with at least two `b`s. // spell:disable-line
+
+#problem()
+Write the patterns `a*` and `a+` using only `{ }`.
+#v(1fr)
+
+#problem()
+Draw a DFA equivalent to the regex pattern `01*0`.
+#v(1fr)
+
+#pagebreak()
+
+
+
+
+
+
+*Characters, Sets, and Groups* \
+In the previous section, we saw how we can specify characters literally: \
+`a+` means "one or more `a` characters" \
+There are, of course, other ways we can specify characters.
+
+#v(4mm)
+
+The first such way is the _set_, denoted `[ ]`. A set can pretend to be any character inside it. \
+For example, `m[aoy]th` will match `math`, `moth`, or `myth`. \
+`a[01]+b` will match `a0b`, `a111b`, `a1100110b`, and any other similar string. \
+
+#v(4mm)
+
+We can negate a set with a `^`. \
+`[^abc]` will match any single character except `a`, `b`, or `c`, including symbols and spaces.
+
+#v(4mm)
+
+If we want to keep characters together, we can use the _group_, denoted `( )`. \
+
+Groups work exactly as you'd expect, representing an atomic#footnote([In other words, "unbreakable"]) group of characters. \
+`a(01)+b` will match `a01b` and `a010101b`, but will *not* match `a0b`, `a1b`, or `a1100110b`.
+
+#problem()
+You are now familiar with most of the tools regex has to offer. \
+Write patterns that match the following strings:
+
+- An ISO-8601 date, like `2022-10-29`. \
+  #hint([Invalid dates like `2022-13-29` should also be matched.])
+
+- An email address. \
+  #hint([Don't forget about subdomains, like `math.ucla.edu`.])
+
+- A UCLA room number, like `MS 5118` or `Kinsey 1220B`.
+
+- Any ISBN-10 of the form `0-316-00395-7`. \
+  #hint([Remember that the check digit may be an `X`. Dashes are optional.])
+
+- A word of even length. \
+  #hint([
+    The set `[A-z]` contains every english letter, capitalized and lowercase. \
+    `[a-z]` will only match lowercase letters.
+  ])
+
+- A word with exactly 3 vowels. \
+  #hint([
+    The special token `\w` will match any word character. \
+    It is equivalent to `[A-z0-9_]`. `_` represents a literal underscore.
+  ])
+
+- A word that has even length and exactly 3 vowels.
+
+- A sentence that does not start with a capital letter.
+#v(1fr)
+
+#problem()
+If you'd like to know more, check out `https://regexr.com`.
+It offers an interactive regex prompt,
+as well as a cheatsheet that explains every other regex token there is. \
+You can find a nice set of challenges at `https://alf.nu/RegexGolf`.

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

@@ -0,0 +1,78 @@
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	shortwarning
+]{../../../lib/tex/ormc_handout}
+\usepackage{../../../lib/tex/macros}
+
+
+\usepackage{pdfpages}
+\usepackage{sliderule}
+\usepackage{changepage}
+
+% Args:
+% x, top scale y, label
+\newcommand{\slideruleind}[3]{
+	\draw[
+		line width=1mm,
+		draw=black,
+		opacity=0.3,
+		text opacity=1
+	]
+		({#1}, {#2 + 1})
+		--
+		({#1}, {#2 - 1.1})
+		node [below] {#3};
+}
+
+
+\uptitlel{Advanced}
+\uptitler{\smallurl{}}
+\title{Warm-Up: Slide Rules}
+\subtitle{Prepared by Mark on \today}
+
+\begin{document}
+
+	\maketitle
+
+
+	\begin{center}
+	\begin{minipage}{6cm}
+		Dad says that anyone who can't use
+		a slide rule is a cultural illiterate
+		and should not be allowed to vote.
+
+		\vspace{1ex}
+
+		\textit{Have Space Suit --- Will Travel, 1958}
+	\end{minipage}
+	\end{center}
+	\hfill
+
+	\input{parts/0 logarithms.tex}
+	\input{parts/1 intro.tex}
+	\input{parts/2 multiplication.tex}
+
+	% Make sure the slide rule is on an odd page,
+	% so that double-sided printing won't require
+	% students to tear off problems.
+	\checkoddpage
+	\ifoddpage\else
+		\vspace*{\fill}
+		\begin{center}
+		{
+			\Large
+			\textbf{This page unintentionally left blank.}
+		}
+		\end{center}
+		\vspace{\fill}
+		\pagebreak
+	\fi
+
+	\includepdf[
+		pages=1,
+		fitpaper=true
+	]{resources/rule.pdf}
+
+\end{document}

src/Warm-Ups/Slide Rules/meta.toml

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

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

@@ -0,0 +1,63 @@
+\section{Logarithms}
+
+\definition{}<logdef>
+The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\
+In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\
+
+\problem{}
+Evaluate the following by hand:
+
+\begin{enumerate}
+	\item $\log_{10}{(1000)}$
+	\vfill
+	\item $\log_2{(64)}$
+	\vfill
+	\item $\log_2{(\frac{1}{4})}$
+	\vfill
+	\item $\log_x{(x)}$ for any $x$
+	\vfill
+	\item $log_x{(1)}$ for any $x$
+	\vfill
+\end{enumerate}
+
+\problem{}<logids>
+Prove the following:
+
+\begin{enumerate}[itemsep=2mm]
+	\item $\log_b{(b^x)} = x$
+	\vfill
+	\item $b^{\log_b{x}} = x$
+	\vfill
+	\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
+	\vfill
+	\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
+	\vfill
+	\item $\log_b{(x^y)} = y \log_b{(x)}$
+	\vfill
+\end{enumerate}
+
+
+\begin{instructornote}
+	A good intro to the following sections is the linear slide rule:
+	\note{(note that these rules start at 0)}
+	\begin{center}
+		\begin{tikzpicture}[scale=0.5]
+			\linearscale{2}{1}{}
+			\linearscale{0}{0}{}
+
+			\slideruleind
+				{5}
+				{1}
+				{2 + 3 = 5}
+	\end{tikzpicture}
+	\end{center}
+
+	Take two linear rules, offset one, and you add.
+	Do the same with a log scale, and you multiply! \\
+
+	\linehack{}
+
+	After assembling the paper slide rule, you can make a visor with some transparent tape.
+\end{instructornote}
+
+\pagebreak

src/Warm-Ups/Slide Rules/parts/1 intro.tex

@@ -0,0 +1,43 @@
+\section{Introduction}
+
+Mathematicians, physicists, and engineers needed to quickly compute products long before computers conquered the world.
+
+\medskip
+
+The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
+
+\medskip
+
+There are four scales on your slide rule, each labeled with a letter on the left side:
+
+\def\sliderulewidth{13}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\tscale{0}{9}{T}
+	\kscale{0}{8}{K}
+	\abscale{0}{7}{A}
+
+	\abscale{0}{5.5}{B}
+	\ciscale{0}{4.5}{CI}
+	\cdscale{0}{3.5}{C}
+
+	\cdscale{0}{2}{D}
+	\lscale{0}{1}{L}
+	\sscale{0}{0}{S}
+\end{tikzpicture}
+\end{center}
+
+Each scale's ``generating function'' is on the right:
+\begin{itemize}
+	\item T: $\tan$
+	\item K: $x^3$
+	\item A,B: $x^2$
+	\item CI: $\frac{1}{x}$
+	\item C, D: $x$
+	\item L: $\log_{10}(x)$
+	\item S: $\sin$
+\end{itemize}
+
+Once you understand the layout of your slide rule, move on to the next page.
+
+\pagebreak

src/Warm-Ups/Slide Rules/parts/2 multiplication.tex

@@ -0,0 +1,299 @@
+\section{Multiplication}
+
+We'll use the C and D scales of your slide rule to multiply. \\
+
+Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$:
+
+\def\sliderulewidth{10}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+\end{tikzpicture}
+\end{center}
+
+Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{1}
+		{6}
+
+\end{tikzpicture}
+\end{center}
+
+Of course, our answer is 6.
+
+\problem{}
+What is $1.15 \times 2.1$? \\
+Use your slide rule.
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=1]
+			\cdscale{\cdscalefn(1.15)}{1}{C}
+			\cdscale{0}{0}{D}
+
+			\slideruleind
+				{\cdscalefn(1.15)}
+				{1}
+				{1.15}
+
+			\slideruleind
+				{\cdscalefn(1.15) + \cdscalefn(2.1)}
+				{1}
+				{2.415}
+
+		\end{tikzpicture}
+		\end{center}
+\end{solution}
+
+\vfill
+
+Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within two decimal places is close enough for most practical applications. \\
+
+\pagebreak
+
+Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
+What should we do if we want to calculate $32 \times 210$? \\
+
+\problem{}
+Using your slide rule, calculate $32 \times 210$. \\
+%\hint{$32 = 3.2 \times 10^1$}
+
+\begin{solution}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(2.1)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(2.1)}
+			{1}
+			{2.1}
+
+		\slideruleind
+			{\cdscalefn(2.1) + \cdscalefn(3.2)}
+			{1}
+			{6.72}
+
+	\end{tikzpicture}
+	\end{center}
+
+		Placing the decimal point correctly is your job. \\
+		$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
+\end{solution}
+
+\vfill
+
+%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
+
+%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
+
+
+\problem{}
+Compute the following:
+\begin{enumerate}
+	\item $1.44 \times 52$
+	\item $0.38 \times 1.24$
+	\item $\pi \times 2.35$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $1.44 \times 52 = 74.88$
+		\item $0.38 \times 1.24 = 0.4712$
+		\item $\pi \times 2.35 = 7.382$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}<provemult>
+Note that the numbers on your C and D scales are logarithmically spaced.
+
+\def\sliderulewidth{13}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{0}{1}{C}
+	\cdscale{0}{0}{D}
+\end{tikzpicture}
+\end{center}
+
+Why does our multiplication procedure work? \\
+%\hint{See \ref{logids}}
+
+\vfill
+\pagebreak
+
+Now we want to compute $7.2 \times 5.5$:
+
+\def\sliderulewidth{10}
+\begin{center}
+\begin{tikzpicture}[scale=0.8]
+	\cdscale{\cdscalefn(5.5)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(5.5)}
+		{1}
+		{5.5}
+
+	\slideruleind
+		{\cdscalefn(5.5) + \cdscalefn(7.2)}
+		{1}
+		{???}
+
+\end{tikzpicture}
+\end{center}
+
+No matter what order we go in, the answer ends up off the scale. There must be another way. \\
+
+\medskip
+
+Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}. Move it over the \textit{larger} number, $7.2$:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(7.2)}
+		{1}
+		{7.2}
+
+\end{tikzpicture}
+\end{center}
+
+Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+
+	\slideruleind
+		{\cdscalefn(7.2)}
+		{1}
+		{7.2}
+
+	\slideruleind
+		{\cdscalefn(3.96)}
+		{1}
+		{3.96}
+
+\end{tikzpicture}
+\end{center}
+
+Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$. We can do this by hand to verify our answer. \\
+
+\medskip
+
+\problem{}
+Why does this work? \par
+\hint{Add a second $D$ scale.}
+
+\begin{solution}
+	Consider the following picture, where I've put two D scales next to each other:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=0.7]
+		\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+		\cdscale{0}{0}{}
+		\cdscale{-10}{0}{}
+
+		\draw[
+			draw=black,
+		]
+			(0, 0)
+			--
+			(0, -0.3)
+			node [below] {D};
+
+		\draw[
+			draw=black,
+		]
+			(-10, 0)
+			--
+			(-10, -0.3)
+			node [below] {D};
+
+		\slideruleind
+			{-10 + \cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(3.96)}
+			{1}
+			{3.96}
+
+	\end{tikzpicture}
+	\end{center}
+
+	\medskip
+
+	The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$ smaller than it should be.
+
+	\medskip
+
+	\medskip
+	In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
+	This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}
+Compute the following using your slide rule:
+\begin{enumerate}
+	\item $9 \times 8$
+	\item $15 \times 35$
+	\item $42.1 \times 7.65$
+	\item $6.5^2$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $9 \times 8 = 72$
+		\item $15 \times 35 = 525$
+		\item $42.1 \times 7.65 = 322.065$
+		\item $6.5^2 = 42.25$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+
+\problem{}
+Compute the following using your slide rule. \\
+
+\begin{enumerate}
+	\item $135 \div 15$
+	\item $68.2 \div 0.575$
+	\item $(118 \times 0.51) \div 6.6$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $135 \div 15 = 9$
+		\item $68.2 \div 0.575 = 118.609$
+		\item $(118 \times 0.51) \div 6.6 = 9.118$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+\pagebreak

src/Warm-Ups/Slide Rules/sliderule.sty

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

src/Warm-Ups/Somewhat Random Numbers/main.typ

@@ -0,0 +1,78 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Somewhat Random Numbers],
+  by: "Mark",
+)
+
+#problem()
+Alice generates 100 random numbers uniformly from $[0,1]$. \
+Bob generates 101 random numbers from $[0, 1]$, but deletes the lowest result.
+
+#v(2mm)
+
+Say we have both of the resulting arrays, but do not know who generated each one. \
+We would like to guess which of the two was generated by Bob. \
+- Say we assign the array with the smallest mean to Alice. \
+  What is our probability of guessing correctly?
+- How often do we guess correctly if we instead assign \
+  the array with the smaller _minimum_ to Alice?
+
+
+#solution([
+  Looking at the mean seems like a good idea, but there's a better way: \
+  Assign the array with the smaller _minimum_ to Alice.
+
+  #v(3mm)
+
+  To compute the probability, generate 201 numbers. \
+  Assign the first 100 to Alice and the rest to Bob. \
+  Look at the lowest two numbers (of these 201, *before* Bob drops his lowest).
+
+  #v(8mm)
+
+  We'll use the following notation: \
+  `AB` means the lowest was owned by Alice, and the second-lowest, by Bob.
+
+  #v(2mm)
+
+  Probabilities are as follows: \
+  - `AA`: $100\/201 times 99\/200 approx 0.246$
+  - `AB`: $100\/201 times 101\/200 approx 0.251$
+  - `BA`: $101\/201 times 100\/200 approx 0.251$ // spell:disable-line
+  - `BB`: $101\/201 times 100\/200 approx 0.251$
+
+  #v(4mm)
+  Now, Bob drops his lowest number. \
+  We'll cross out the number he drops and box the new lowest number (i.e, the one we observe):
+  - #{
+      (
+        box(`A`, stroke: ored, inset: 1pt)
+          + box(`A`, inset: 1pt)
+          + box([: $approx 0.246$], inset: (top: 1pt, bottom: 1pt))
+      )
+    }
+  - #{
+      (
+        box(`A`, stroke: ored, inset: 1pt)
+          + box(strike(`B`), inset: 1pt)
+          + box([: $approx 0.251$], inset: (top: 1pt, bottom: 1pt))
+      )
+    }
+  - #{
+      (
+        box(strike(`B`), inset: 1pt)
+          + box(`A`, stroke: ored, inset: 1pt)
+          + box([: $approx 0.251$], inset: (top: 1pt, bottom: 1pt))
+      )
+    }
+  - #{
+      (
+        box(strike(`B`), inset: 1pt)
+          + box(`B`, stroke: ored, inset: 1pt)
+          + box([: $approx 0.251$], inset: (top: 1pt, bottom: 1pt))
+      )
+    }
+  #v(8mm)
+  Alice has the smallest number in 3 of 4 cases, which have a total probability of $approx 0.749$.
+])

src/Warm-Ups/Somewhat Random Numbers/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Somewhat Random Numbers"
+
+[publish]
+handout = true
+solutions = true

src/Warm-Ups/Sysadmin/main.tex

@@ -1,89 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\usepackage{tikz}
-
-
-\title{The Sysadmin's Warm-Up}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-
-	Most of you have seen a hard drive. Many have touched one, and a lucky few have poked around inside one. These devices have two interesting properties:
-
-	\begin{enumerate}
-		\item They hold valuable data
-		\item They eventually fail
-	\end{enumerate}
-
-	Needless to say, this is a problem. \par
-	We need to design a system that allows hard drives to fail without data loss.
-
-	\definition{}
-	You can think of a hard drive as a long string of bits. \par
-	Assume all hard drives can store 1 TiB of data.
-
-	\begin{center}
-	\begin{tikzpicture}
-
-		\node[above] at (1/2, 0) {Drive A};
-		\draw (0cm, 0cm) -- (0cm, -3cm);
-		\draw (1cm, 0cm) -- (1cm, -3cm);
-		\foreach \i in {0,...,-6} {
-			\draw (0cm,\i cm / 2) -- (1cm ,\i cm / 2);
-		}
-
-		\node at (1/2, - 1 / 4) {1};
-		\node at (1/2, - 3 / 4) {1};
-		\node at (1/2, - 5 / 4) {0};
-		\node at (1/2, - 7 / 4) {...};
-		\node at (1/2, - 9 / 4) {1};
-		\node at (1/2, -11 / 4) {0};
-
-
-		\node[above] at (5/2, 0) {Drive B};
-		\draw (2cm, 0cm) -- (2cm, -3cm);
-		\draw (3cm, 0cm) -- (3cm, -3cm);
-		\foreach \i in {0,...,-6} {
-			\draw (2cm,\i cm / 2) -- (3cm ,\i cm / 2);
-		}
-
-		\node at (5/2, - 1 / 4) {0};
-		\node at (5/2, - 3 / 4) {1};
-		\node at (5/2, - 5 / 4) {0};
-		\node at (5/2, - 7 / 4) {...};
-		\node at (5/2, - 9 / 4) {0};
-		\node at (5/2, -11 / 4) {1};
-	\end{tikzpicture}
-	\end{center}
-
-	\problem{}
-	Suppose we have two hard drives. How can we arrange our data so that...
-	\begin{enumerate}
-		\item We get 1 TiB of usable storage
-		\item We lose no data if any one drive fails
-	\end{enumerate}
-
-	\vfill
-
-	\problem{}
-	Suppose we have three hard drives. How can we arrange our data so that...
-	\begin{enumerate}
-		\item We get 2 TiB of usable storage
-		\item We lose no data if any one drive fails
-	\end{enumerate}
-
-	\vfill
-\end{document}

src/Warm-Ups/Sysadmin/main.typ

@@ -0,0 +1,35 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [The Sysadmin's Warm-Up],
+  by: "Mark",
+)
+
+Most of you have seen a hard drive. \
+Many have touched one, and a lucky few have taken one apart. \
+These devices have two interesting properties:
+- They hold valuable data
+- They inevitably fail
+
+Needless to say, this is a problem. \
+We would like to design a system that tolerates hard drive failures without data loss.
+
+#definition()
+You can think of a hard drive as a long string of bits. \
+Assume all hard drives in the following problems have the same size. \
+If a hard drive "fails", all data on it is instantly lost.
+
+#problem()
+Suppose we have two hard drives. How can we arrange our data so that...
+- We get 1 TiB of usable storage
+- We lose no data if any one drive fails
+
+#v(1fr)
+
+#problem()
+Suppose we have three hard drives. How can we arrange our data so that...
+- We get 2 TiB of usable storage
+- We lose no data if any one drive fails
+
+#v(1fr)
+

src/Warm-Ups/Travellers/main.tex

@@ -1,30 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-\title{Warm-Up: Travellers}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	Four travellers are on a plane, each moving along a straight line at an arbitrary constant speed. \par
-	No two of their paths are parallel, and no three intersect at the same point. \par
-	We know that traveller A has met travelers B, C, and D, \par
-	and that traveller B has met C and D (and A). Show that C and D must also have met. \par
-
-	\begin{solution}
-		When a body travels at a constant speed, its graph with respect to time is a straight line. \par
-		So, we add time axis in the third dimension, perpendicular to our plane. \par
-		Naturally, the projection of each of these onto the plane corresponds to a road.
-
-		Now, note that two intersecting lines define a plane and use the conditions in the problem to show that no two lines are parallel.
-	\end{solution}
-
-\end{document}

src/Warm-Ups/Travellers/main.typ

@@ -0,0 +1,20 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Travellers],
+  by: "Mark",
+)
+
+#problem()
+Four travellers are on a plane, each moving along a straight line at an arbitrary constant speed. \
+No two of their paths are parallel, and no three intersect at the same point. \
+We know that traveller A has met travelers B, C, and D, \
+and that traveller B has met C and D (and A). Show that C and D must also have met.
+
+#solution([
+  When a body travels at a constant speed, its graph with respect to time is a straight line. \
+  So, we add time axis in the third dimension, perpendicular to our plane. \
+  Naturally, the projection of each of these onto the plane corresponds to a road.
+
+  Now, note that two intersecting lines define a plane and use the conditions in the problem to show that no two lines are parallel.
+])

src/Warm-Ups/Tuesday/main.typ

@@ -0,0 +1,26 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Tuesday],
+  by: "Mark",
+)
+
+#problem()
+Charlie has two children. \
+One of them is a boy born on a Tuesday. \
+What is the probability that Charlie has two boys?
+
+#solution([
+  $13 / 27$, which notably isn't $1/2$.
+
+  #v(2mm)
+
+  Draw a $14 times 14$ square, highlight the possible cases, and you'll see it. \
+
+  #v(2mm)
+
+  Order is very important in this problem. \
+  If we knew that the _first_ child was a boy born on Tuesday,
+  the probability of two boys would be $1/2$. \
+  Also consider a smaller case where a week has two days.
+])

src/Warm-Ups/Tuesday/meta.toml

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

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

@@ -1,43 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\usepackage[linguistics]{forest}
-
-
-
-\title{Warm-Up: What's an AST?}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today. \\ Based on a true story.}
-
-\begin{document}
-
-	\maketitle
-
-	Say you have a valid string of simple arithmetic that contains no unary operators (like $3!$ or $-4$) and no parenthesis:
-	$$
-	3 + 9 \times 8 \div 5 \land 6
-	$$
-
-	You may assume that all numbers and operators in this string consist of exactly one character. \\
-
-	Devise an algorithm that turns this string into a tree (as shown below), respecting the order of operations $[\land, \times, \div, +, -]$.
-
-	\begin{center}
-	\begin{forest}
-		[$+$
-			[3]
-			[$\div$
-				[$\times$[9][8]]
-				[$\land$[5][6]]
-			]
-		]
-	\end{forest}
-	\end{center}
-
-\end{document}

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

@@ -0,0 +1,74 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+#show: handout.with(
+  title: [Warm-Up: What's an AST?],
+  by: "Mark",
+  subtitle: "Based on a true story.",
+)
+
+#problem()
+Say we have a valid string of simple arithmetic that contains \
+no unary operators (like $3!$ or $-4$) and no parenthesis:
+
+#v(2mm)
+$
+  3 + 9 times 8 div 5 and 6
+$
+#v(2mm)
+
+You may assume that all numbers and operators in this string consist of exactly one character. \
+
+Devise an algorithm that turns such strings into a tree (as shown below), \
+respecting the order of operations $[and, times, div, +, -]$.
+
+#v(2mm)
+
+#align(
+  center,
+  cetz.canvas({
+    import cetz.draw: *
+
+    // spell:off
+    content((0, 0), $+$, name: "r")
+    content((-0.5, -1), $3$, name: "a")
+    content((0.5, -1), $div$, name: "b")
+    content((-0.3, -2), $times$, name: "ba")
+    content((1.3, -2), $and$, name: "bb")
+    content((-0.8, -3), $9$, name: "baa")
+    content((0.2, -3), $8$, name: "bab")
+    content((0.8, -3), $5$, name: "bba")
+    content((1.8, -3), $6$, name: "bbb")
+    // spell:on
+
+    // Zero-sized arrows are a hack for offset.
+    set-style(
+      stroke: (thickness: 0.3mm),
+      mark: (
+        start: (
+          symbol: "|",
+          offset: 0.25,
+          width: 0mm,
+          length: 0mm,
+        ),
+        end: (
+          symbol: "|",
+          offset: 0.25,
+          width: 0mm,
+          length: 0mm,
+        ),
+      ),
+    )
+
+    // spell:off
+    line("r", "a")
+    line("r", "b")
+    line("b", "ba")
+    line("b", "bb")
+    line("ba", "baa")
+    line("ba", "bab")
+    line("bb", "bba")
+    line("bb", "bbb")
+    // spell:on
+  }),
+)

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

@@ -1,66 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-% x,y,scale,style
-\def\ttt#1#2#3#4{
-	\draw[#4] (#1, #2+#3) -- (#1+#3+#3+#3, #2+#3);
-	\draw[#4] (#1, #2+#3+#3) -- (#1+#3+#3+#3, #2+#3+#3);
-	\draw[#4] (#1+#3, #2) -- (#1+#3, #2+#3+#3+#3);
-	\draw[#4] (#1+#3+#3, #2) -- (#1+#3+#3, #2+#3+#3+#3);
-}
-
-
-\title{Warm-Up: Wild Tic-Tac-Toe}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-
-	\problem{}
-	In wild tic-tac-toe, players may place either an X or O on each move. The player that first completes a
-	row of any three symbols wins. Show that the first player always has a winning strategy.
-
-	\vfill
-	\begin{center}
-		\begin{tikzpicture}[scale=0.60]
-			\ttt{0}{0}{2}{line width=0.3mm}
-			\ttt{7}{0}{2}{line width=0.3mm}
-			\ttt{14}{0}{2}{line width=0.3mm}
-
-			\ttt{0}{7}{2}{line width=0.3mm}
-			\ttt{7}{7}{2}{line width=0.3mm}
-			\ttt{14}{7}{2}{line width=0.3mm}
-		\end{tikzpicture}
-	\end{center}
-
-	\vfill
-
-	\problem{}
-	Now, say the first player to complete a row loses. Who has a winning strategy now?
-
-	\vfill
-	\begin{center}
-		\begin{tikzpicture}[scale=0.60]
-			\ttt{0}{0}{2}{line width=0.3mm}
-			\ttt{7}{0}{2}{line width=0.3mm}
-			\ttt{14}{0}{2}{line width=0.3mm}
-
-			\ttt{0}{7}{2}{line width=0.3mm}
-			\ttt{7}{7}{2}{line width=0.3mm}
-			\ttt{14}{7}{2}{line width=0.3mm}
-		\end{tikzpicture}
-	\end{center}
-
-
-	\vfill
-	\pagebreak
-
-\end{document}

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

@@ -0,0 +1,47 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+#show: handout.with(
+  title: [Warm-Up: Wild Tic-Tac-Toe],
+  by: "Mark",
+)
+
+#let ttt = align(
+  center,
+  cetz.canvas({
+    import cetz.draw: *
+    let s = 0.7 // scale
+
+    set-style(stroke: (thickness: 0.5mm * s))
+    line((-1 * s, 3 * s), (-1 * s, -3 * s))
+    line((1 * s, 3 * s), (1 * s, -3 * s))
+    line((3 * s, -1 * s), (-3 * s, -1 * s))
+    line((3 * s, 1 * s), (-3 * s, 1 * s))
+  }),
+)
+
+
+#problem()
+In wild tic-tac-toe, players may place either an X or O on each move. The player that first completes a
+row of any three symbols wins. Show that the first player always has a winning strategy.
+#v(4mm)
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr, 1fr),
+  ttt, ttt, ttt,
+);
+#v(1fr)
+
+#problem()
+Now, say the first player to complete a row loses. Who has a winning strategy now?
+#v(4mm)
+
+#table(
+  stroke: none,
+  align: center,
+  columns: (1fr, 1fr, 1fr),
+  ttt, ttt, ttt,
+);
+#v(1fr)

src/Warm-Ups/Zeno's Furniture/main.tex

@@ -1,150 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\makeatletter
-\newcommand{\thisone}{
-	\if@solutions
-	{\color{red} $\Leftarrow$ \texttt{this one}}
-	\else\fi
-}
-
-\title{Zeno's Furniture}
-\uptitlel{Warm Ups}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-	Zeno Furniture sells exactly five types of furniture:
-	\begin{itemize}
-		\item Footstools
-		\item Hutches
-		\item Sideboards
-		\item Tables
-		\item Vanities
-	\end{itemize}
-	Irene buys four items, each of a different type,
-	and each made of exactly one kind of wood:
-	\begin{itemize}
-		\item Maple
-		\item Oak
-		\item Pine
-		\item Rosewood
-	\end{itemize}
-
-	The following conditions govern Irene's purchases:
-	\begin{itemize}
-		\item Any vanity she buys is Maple.
-		\item Any rosewood item she buys is a sideboard.
-		\item If she buys a vanity, she does not buy a footstool.
-		\item If she buys a footstool, she also buys a table made of the same material.
-		\item Irene does not buy an oak table.
-		\item Exactly two of the items she buys are made of the same kind of wood.
-	\end{itemize}
-
-
-
-
-	\problem{}
-	Which one of the following could be an accurate
-	list of the items Irene buys? \par
-	\begin{itemize}
-		\item maple footstool, maple hutch, rosewood sideboard, maple table
-		\item oak hutch, rosewood sideboard, pine table, oak vanity
-		\item rosewood hutch, maple sideboard, oak table, maple vanity
-		\item pine footstool, rosewood sideboard, pine table, maple vanity
-		\item maple footstool, pine hutch, oak sideboard, maple table \thisone{}
-	\end{itemize}
-	\vfill
-
-
-	\problem{}
-	If Irene buys one item made of rosewood and two items made
-	of maple, then which one of the following pairs could be two
-	of the items she buys?
-	\begin{itemize}
-		\item a rosewood sideboard and an oak footstool
-		\item an oak hutch and a pine sideboard
-		\item an oak hutch and a maple table \thisone{}
-		\item a maple sideboard and a maple vanity
-		\item a maple hutch and a maple table
-	\end{itemize}
-	\vfill
-	\pagebreak
-
-
-	\problem{}
-	Which one of the following is a complete and accurate list
-	of all the woods any footstool that Irene buys could be made of?
-	\begin{itemize}
-		\item maple, oak
-		\item maple, pine \thisone{}
-		\item maple, rosewood
-		\item maple, oak, pine
-		\item maple, oak, pine, rosewood
-	\end{itemize}
-	\vfill
-
-
-	\problem{}
-	Suppose Irene buys a footstool. Then which one of the following
-	is a complete and accurate list of items and any one of which she
-	could buy in maple?
-	\begin{itemize}
-		\item footstool, hutch, sideboard, table, vanity
-		\item footstool, hutch, sideboard, table \thisone{}
-		\item footstool, hutch, sideboard
-		\item footstool, hutch
-		\item footstool
-	\end{itemize}
-	\vfill
-
-	\problem{}
-	Which one of the following cannot be the two items Irene
-	buys that are made of the same wood as each other?
-	\begin{itemize}
-		\item footstool, hutch \thisone{}
-		\item hutch, sideboard
-		\item hutch, table
-		\item sideboard, vanity
-		\item table, vanity
-	\end{itemize}
-	\vfill
-	\pagebreak
-
-	\problem{}
-	If Irene does not buy an item made of maple, then each of the
-	following must be true except...
-	\begin{itemize}
-		\item Irene buys a footstool
-		\item Irene buys a pine hutch \thisone{}
-		\item Irene buys a rosewood sideboard
-		\item Irene buys exactly one item made of oak
-		\item Irene buys exactly two items made of pine
-	\end{itemize}
-	\vfill
-
-
-	\problem{}
-	Suppose the condition that Irene does not buy an oak table is
-	replaced with the condition that she does not buy a pine table.
-	If all the other conditions hold as originally given, which of the
-	following cannot be true?
-	\begin{itemize}
-		\item Irene buys an oak footstool.
-		\item Irene buys a hutch and a table made of the same wood.
-		\item Irene buys a vanity, but she does not buy an oak table.
-		\item Irene buys a maple table and an oak hutch.
-		\item Irene buys a rosewood sideboard and exactly two items made of pine. \thisone{}
-	\end{itemize}
-	\vfill
-	\pagebreak
-
-\end{document}

src/Warm-Ups/Zeno's Furniture/main.typ

@@ -0,0 +1,125 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Zeno's Furniture],
+  by: "Mark",
+)
+
+#let thisone = if_solutions(
+  text(fill: ored, [#sym.arrow.l.double.long `this one`]),
+)
+
+
+Zeno's Furniture sells exactly five types of furniture: \
+Footstools, Hutches, Sideboards, Tables, and Vanities.
+
+#v(3mm)
+
+Each can be made of exactly one kind of wood: \
+Maple, Oak, Pine, or Rosewood
+
+#v(3mm)
+
+Irene buys four items, each of a different type. \
+The following conditions govern Irene's purchases:
+- Any vanity she buys is Maple.
+- Any rosewood item she buys is a sideboard.
+- If she buys a vanity, she does not buy a footstool.
+- If she buys a footstool, she also buys a table made of the same material.
+- Irene does not buy an oak table.
+- Exactly two of the items she buys are made of the same kind of wood.
+
+
+
+#v(5mm)
+
+
+#problem()
+Which one of the following could be an accurate
+list of the items Irene buys? \
+- maple footstool, maple hutch, rosewood sideboard, maple table
+- oak hutch, rosewood sideboard, pine table, oak vanity
+- rosewood hutch, maple sideboard, oak table, maple vanity
+- pine footstool, rosewood sideboard, pine table, maple vanity
+- maple footstool, pine hutch, oak sideboard, maple table #thisone
+
+#v(1fr)
+
+
+#problem()
+If Irene buys one item made of rosewood and two items made
+of maple, then which one of the following pairs could be two
+of the items she buys?
+- a rosewood sideboard and an oak footstool
+- an oak hutch and a pine sideboard
+- an oak hutch and a maple table #thisone
+- a maple sideboard and a maple vanity
+- a maple hutch and a maple table
+
+#v(1fr)
+#pagebreak()
+
+
+#problem()
+Which one of the following is a complete and accurate list
+of all the woods any footstool that Irene buys could be made of?
+- maple, oak
+- maple, pine #thisone
+- maple, rosewood
+- maple, oak, pine
+- maple, oak, pine, rosewood
+
+#v(1fr)
+
+
+#problem()
+Suppose Irene buys a footstool. Then which one of the following
+is a complete and accurate list of items and any one of which she
+could buy in maple?
+- footstool, hutch, sideboard, table, vanity
+- footstool, hutch, sideboard, table #thisone
+- footstool, hutch, sideboard
+- footstool, hutch
+- footstool
+
+#v(1fr)
+
+#problem()
+Which one of the following cannot be the two items Irene
+buys that are made of the same wood as each other?
+- footstool, hutch #thisone
+- hutch, sideboard
+- hutch, table
+- sideboard, vanity
+- table, vanity
+
+#v(1fr)
+#pagebreak()
+
+
+
+
+#problem()
+If Irene does not buy an item made of maple, then each of the
+following must be true except...
+- Irene buys a footstool
+- Irene buys a pine hutch #thisone
+- Irene buys a rosewood sideboard
+- Irene buys exactly one item made of oak
+- Irene buys exactly two items made of pine
+
+#v(1fr)
+
+
+#problem()
+Suppose the condition that Irene does not buy an oak table is
+replaced with the condition that she does not buy a pine table.
+If all the other conditions hold as originally given, which of the
+following cannot be true?
+- Irene buys an oak footstool.
+- Irene buys a hutch and a table made of the same wood.
+- Irene buys a vanity, but she does not buy an oak table.
+- Irene buys a maple table and an oak hutch.
+- Irene buys a rosewood sideboard and exactly two items made of pine. #thisone
+
+#v(1fr)

src/Warm-Ups/fmod/main.tex

@@ -1,22 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: \texttt{fmod}}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	I'm sure you're all familiar with how \texttt{mod(a, b)} and \texttt{remainder(a, b)} work with integers. \par
-	Devise an equivalent for floats (i.e, real numbers).
-
-\end{document}

src/Warm-Ups/fmod/main.typ

@@ -0,0 +1,10 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: `fmod`],
+  by: "Mark",
+)
+
+#problem()
+I'm sure you're all familiar with how `mod(a, b)` and `remainder(a, b)` \ work when `a` and `b` are integers.
+Devise an equivalent for floats (i.e, real numbers).