Compare commits
4 Commits
typst-quan
...
main
| Author | SHA1 | Date | |
|---|---|---|---|
| 0b7acaf5ae | |||
| 69d835a2d2 | |||
| 81d6518553 | |||
| e5b0053465 |
@ -26,7 +26,7 @@ jobs:
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- name: "Download Typstyle"
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run: |
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wget -q "https://github.com/Enter-tainer/typstyle/releases/download/v0.12.14/typstyle-x86_64-unknown-linux-musl"
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wget -q "https://github.com/Enter-tainer/typstyle/releases/download/v0.13.17/typstyle-x86_64-unknown-linux-musl"
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chmod +x typstyle-x86_64-unknown-linux-musl
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- name: Check typst formatting
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@ -62,7 +62,7 @@ jobs:
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# more control anyway.
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- name: "Download Typst"
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run: |
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wget -q "https://github.com/typst/typst/releases/download/v0.12.0/typst-x86_64-unknown-linux-musl.tar.xz"
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wget -q "https://github.com/typst/typst/releases/download/v0.13.1/typst-x86_64-unknown-linux-musl.tar.xz"
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tar -xf "typst-x86_64-unknown-linux-musl.tar.xz"
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mv "typst-x86_64-unknown-linux-musl/typst" .
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rm "typst-x86_64-unknown-linux-musl.tar.xz"
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@ -3,13 +3,10 @@
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// Re-exports
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// All functions that maybe used by client code are listed here
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#import "misc.typ": *
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#import "object.typ": problem, definition, theorem, example, remark, generic
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#import "object.typ": definition, example, generic, problem, remark, theorem
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#import "solution.typ": (
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if_solutions,
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if_no_solutions,
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if_solutions_else,
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solution,
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instructornote,
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if_no_solutions, if_solutions, if_solutions_else, instructornote,
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sample_solution, solution,
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)
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@ -38,10 +35,7 @@
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margin: 20mm,
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width: 8.5in,
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height: 11in,
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footer: align(
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center,
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context counter(page).display(),
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),
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footer: align(center, context counter(page).display()),
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footer-descent: 5mm,
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)
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@ -102,8 +96,8 @@
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// Make handout title
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{
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import "header.typ": make_header, solution_warning, short_solution_warning
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import "solution.typ": solutions_state, reset_solutions
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import "header.typ": make_header, short_solution_warning, solution_warning
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import "solution.typ": reset_solutions, solutions_state
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reset_solutions()
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@ -29,11 +29,7 @@
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}
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// Render the object
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block(
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above: 8mm,
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below: 2mm,
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text(weight: "bold", obj_content),
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)
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block(above: 8mm, below: 2mm, text(weight: "bold", obj_content))
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// Generate labeled metadata for this object.
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//
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@ -57,7 +53,7 @@
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if not (
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it.element != none
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and it.element.has("value")
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and type(it.element.value) == "dictionary"
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and type(it.element.value) == dictionary
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and it.element.value.keys().contains(magic_key)
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) {
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// This label is not attached to object metadata
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@ -100,9 +96,5 @@
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#let remark = _mkobj("Remark")
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#let generic(obj_content) = {
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block(
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above: 8mm,
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below: 2mm,
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text(weight: "bold", obj_content),
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)
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block(above: 8mm, below: 2mm, text(weight: "bold", obj_content))
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}
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@ -1,4 +1,4 @@
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#import "misc.typ": ored, oblue
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#import "misc.typ": oblue, ored
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/// If false, hide instructor info.
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@ -61,57 +61,71 @@
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}
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#let solution(content) = {
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if_solutions(
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align(
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center,
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stack(
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block(
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width: 100%,
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breakable: false,
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fill: ored,
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stroke: ored + 2pt,
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inset: 1.5mm,
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align(left, text(fill: white, weight: "bold", [Solution:])),
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),
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block(
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width: 100%,
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height: auto,
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breakable: false,
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fill: ored.lighten(80%).desaturate(10%),
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stroke: ored + 2pt,
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inset: 3mm,
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align(left, content),
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),
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),
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if_solutions(align(center, stack(
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block(
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width: 100%,
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breakable: false,
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fill: ored,
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stroke: ored + 2pt,
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inset: 1.5mm,
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align(left, text(fill: white, weight: "bold", [Solution:])),
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),
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)
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block(
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width: 100%,
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height: auto,
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breakable: false,
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fill: ored.lighten(80%).desaturate(10%),
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stroke: ored + 2pt,
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inset: 3mm,
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align(left, content),
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),
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)))
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}
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#let sample_solution(content) = {
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align(center, stack(
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block(
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width: 100%,
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breakable: false,
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fill: oblue,
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stroke: oblue + 2pt,
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inset: 1.5mm,
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align(left, text(fill: white, weight: "bold", [Sample Solution:])),
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),
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block(
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width: 100%,
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height: auto,
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breakable: false,
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fill: oblue.lighten(80%).desaturate(10%),
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stroke: oblue + 2pt,
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inset: 3mm,
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align(left, content),
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),
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))
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}
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#let instructornote(content) = {
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if_solutions(
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align(
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center,
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stack(
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block(
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width: 100%,
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breakable: false,
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fill: oblue,
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stroke: oblue + 2pt,
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inset: 1.5mm,
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align(left, text(fill: white, weight: "bold", [Instructor note:])),
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),
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block(
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width: 100%,
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height: auto,
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breakable: false,
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fill: oblue.lighten(80%).desaturate(10%),
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stroke: oblue + 2pt,
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inset: 3mm,
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align(left, content),
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),
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),
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if_solutions(align(center, stack(
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block(
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width: 100%,
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breakable: false,
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fill: oblue,
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stroke: oblue + 2pt,
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inset: 1.5mm,
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align(left, text(fill: white, weight: "bold", [Instructor note:])),
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),
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)
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block(
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width: 100%,
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height: auto,
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breakable: false,
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fill: oblue.lighten(80%).desaturate(10%),
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stroke: oblue + 2pt,
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inset: 3mm,
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align(left, content),
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),
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)))
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}
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@ -1,5 +1,5 @@
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#import "@local/handout:0.1.0": *
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#import "@preview/cetz:0.3.1"
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#import "@preview/cetz:0.4.2"
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= Floats
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#definition()
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@ -31,75 +31,71 @@ Rewrite the following binary decimals in base 10: \
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#definition(label: "floatbits")
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Another way we can interpret a bit string is as a _signed floating-point decimal_, or a `float` for short. \
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Floats represent a subset of the real numbers, and are interpreted as follows: \
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#note([The following only applies to floats that consist of 32 bits. We won't encounter any others today.])
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#align(
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center,
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box(
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inset: 2mm,
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cetz.canvas({
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import cetz.draw: *
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let chars = (
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`0`,
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`b`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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)
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let x = 0
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for c in chars {
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content((x, 0), c)
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x += 0.25
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}
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let y = -0.4
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line((0.3, y), (0.65, y))
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content((0.45, y - 0.2), [s])
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line((0.85, y), (2.9, y))
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content((1.9, y - 0.2), [exponent])
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line((3.10, y), (9.4, y))
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content((6.3, y - 0.2), [fraction])
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}),
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),
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#note(
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[The following only applies to floats that consist of 32 bits. We won't encounter any others today.],
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)
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#align(center, box(inset: 2mm, cetz.canvas({
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import cetz.draw: *
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let chars = (
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`0`,
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`b`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`_`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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`0`,
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)
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let x = 0
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for c in chars {
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content((x, 0), c)
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x += 0.25
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}
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let y = -0.4
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line((0.3, y), (0.65, y))
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content((0.45, y - 0.2), [s])
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line((0.85, y), (2.9, y))
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content((1.9, y - 0.2), [exponent])
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line((3.10, y), (9.4, y))
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content((6.3, y - 0.2), [fraction])
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})))
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- The first bit denotes the sign of the float's value
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We'll label it $s$. \
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If $s = #text([`1`])$, this float is negative; if $s = #text([`0`])$, it is positive.
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@ -1,6 +1,6 @@
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#import "@local/handout:0.1.0": *
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#import "@preview/cetz:0.3.1"
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#import "@preview/cetz-plot:0.1.0": plot, chart
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#import "@preview/cetz:0.4.2"
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#import "@preview/cetz-plot:0.1.2": chart, plot
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= Integers and Floats
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@ -44,19 +44,11 @@ This allows us to improve the average error of our linear approximation:
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{
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let domain = (0, 1)
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plot.add(
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f1,
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domain: domain,
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label: $log(1+x)$,
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style: (stroke: ogrape),
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)
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plot.add(f1, domain: domain, label: $log(1+x)$, style: (
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stroke: ogrape,
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))
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plot.add(
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f2,
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domain: domain,
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label: $x$,
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style: (stroke: oblue),
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)
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plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
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},
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)
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})
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@ -90,19 +82,11 @@ This allows us to improve the average error of our linear approximation:
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{
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let domain = (0, 1)
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plot.add(
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f1,
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domain: domain,
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label: $log(1+x)$,
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style: (stroke: ogrape),
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)
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plot.add(f1, domain: domain, label: $log(1+x)$, style: (
|
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stroke: ogrape,
|
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))
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plot.add(
|
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f2,
|
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domain: domain,
|
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label: $x$,
|
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style: (stroke: oblue),
|
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)
|
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plot.add(f2, domain: domain, label: $x$, style: (stroke: oblue))
|
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},
|
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)
|
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})
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@ -120,16 +104,13 @@ We won't bother with this---we'll simply leave the correction term as an opaque
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|
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#v(1fr)
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|
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#note(
|
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type: "Note",
|
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[
|
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"Average error" above is simply the area of the region between the two graphs:
|
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$
|
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integral_0^1 abs( #v(1mm) log(1+x)_2 - (x+epsilon) #v(1mm))
|
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$
|
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Feel free to ignore this note, it isn't a critical part of this handout.
|
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],
|
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)
|
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#note(type: "Note", [
|
||||
"Average error" above is simply the area of the region between the two graphs:
|
||||
$
|
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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.
|
||||
])
|
||||
|
||||
|
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#pagebreak()
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@ -149,12 +130,11 @@ $
|
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Let $E$ and $F$ be the exponent and float bits of $x_f$. \
|
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We then have:
|
||||
$
|
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log_2(x_f)
|
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&= log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
|
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&= E - 127 + log_2(1 + F / (2^23)) \
|
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& approx E-127 + F / (2^23) + epsilon \
|
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&= 1 / (2^23)(2^23 E + F) - 127 + epsilon \
|
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&= 1 / (2^23)(x_i) - 127 + epsilon
|
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log_2(x_f) & = log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
|
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& = 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
|
||||
$
|
||||
])
|
||||
|
||||
|
||||
@ -156,9 +156,9 @@ float Q_rsqrt( float number ) {
|
||||
Using a calculator and some basic algebra, we can find the $epsilon$ this code uses: \
|
||||
#note[Remember, #text[`0x5f3759df`] is $6240089$ in hexadecimal.]
|
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$
|
||||
(3 times 2^22) (127 - epsilon) &= 6240089 \
|
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(127 - epsilon) &= 126.955 \
|
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epsilon &= 0.0450466
|
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(3 times 2^22) (127 - epsilon) & = 6240089 \
|
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(127 - epsilon) & = 126.955 \
|
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epsilon & = 0.0450466
|
||||
$
|
||||
|
||||
So, $0.045$ is the $epsilon$ used by Quake. \
|
||||
|
||||
134
src/Advanced/Symmetric Groups/macros.typ
Normal file
134
src/Advanced/Symmetric Groups/macros.typ
Normal file
@ -0,0 +1,134 @@
|
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#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#let markline(s, from, to, c: oblue) = {
|
||||
import cetz.draw: *
|
||||
let del = 0.4 // small line
|
||||
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: c,
|
||||
stroke: c,
|
||||
)
|
||||
|
||||
line(
|
||||
from + ".south",
|
||||
(v => cetz.vector.add(v, (0, -del * s)), from + ".south"),
|
||||
(v => cetz.vector.add(v, (0, (del + 0.4) * s)), to + ".north"),
|
||||
(v => cetz.vector.add(v, (0, 0.4 * s)), to + ".north"),
|
||||
stroke: c + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
to + ".north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
}
|
||||
|
||||
|
||||
#let marklinetop(s, from, to, del: 0.8, c: oblue) = {
|
||||
import cetz.draw: *
|
||||
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: c,
|
||||
stroke: c,
|
||||
)
|
||||
|
||||
line(
|
||||
from + ".north",
|
||||
(v => cetz.vector.add(v, (0, del * s)), from + ".north"),
|
||||
(v => cetz.vector.add(v, (0, del * s)), to + ".north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2 * s)), to + ".north"),
|
||||
stroke: c + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
to + ".north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
}
|
||||
|
||||
|
||||
|
||||
#let marklinebot(s, from, to, del: 0.8, c: oblue) = {
|
||||
import cetz.draw: *
|
||||
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: c,
|
||||
stroke: c,
|
||||
)
|
||||
|
||||
line(
|
||||
from + ".south",
|
||||
(v => cetz.vector.add(v, (0, -del * s)), from + ".south"),
|
||||
(v => cetz.vector.add(v, (0, -del * s)), to + ".south"),
|
||||
(v => cetz.vector.add(v, (0, -0.4 * s)), to + ".south"),
|
||||
stroke: c + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
to + ".south",
|
||||
90deg,
|
||||
..arrow,
|
||||
)
|
||||
}
|
||||
|
||||
|
||||
|
||||
#let marklinebotswap(s, from, to, del: 0.8, c: oblue) = {
|
||||
import cetz.draw: *
|
||||
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: c,
|
||||
stroke: c,
|
||||
)
|
||||
|
||||
line(
|
||||
from + ".south",
|
||||
(v => cetz.vector.add(v, (0, -del * s)), from + ".south"),
|
||||
(v => cetz.vector.add(v, (0.5 * s, -del * s)), to + ".south"),
|
||||
(v => cetz.vector.add(v, (0.5 * s, -del * s)), to + ".south"),
|
||||
(v => cetz.vector.add(v, (0.5 * s, del * s)), to + ".north"),
|
||||
(v => cetz.vector.add(v, (0, del * s)), to + ".north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2 * s)), to + ".north"),
|
||||
stroke: c + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
to + ".north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
}
|
||||
|
||||
#let marklinetopswap(s, from, to, del: 0.8, c: oblue) = {
|
||||
import cetz.draw: *
|
||||
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: c,
|
||||
stroke: c,
|
||||
)
|
||||
|
||||
line(
|
||||
from + ".north",
|
||||
(v => cetz.vector.add(v, (0, del * s)), from + ".north"),
|
||||
(v => cetz.vector.add(v, (0.5 * s, del * s)), to + ".north"),
|
||||
(v => cetz.vector.add(v, (0.5 * s, del * s)), to + ".north"),
|
||||
(v => cetz.vector.add(v, (0.5 * s, -del * s)), to + ".south"),
|
||||
(v => cetz.vector.add(v, (0, -del * s)), to + ".south"),
|
||||
(v => cetz.vector.add(v, (0, -0.2 * s)), to + ".south"),
|
||||
stroke: c + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
to + ".south",
|
||||
90deg,
|
||||
..arrow,
|
||||
)
|
||||
}
|
||||
@ -1,58 +0,0 @@
|
||||
% use [nosolutions] flag to hide solutions.
|
||||
% use [solutions] flag to show solutions.
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering
|
||||
]{../../../lib/tex/handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
\usetikzlibrary{calc}
|
||||
|
||||
\uptitlel{Advanced 2}
|
||||
\uptitler{\smallurl{}}
|
||||
\title{Symmetric Groups}
|
||||
\subtitle{Prepared by Mark on \today{}}
|
||||
|
||||
|
||||
|
||||
\def\line#1#2{
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(#1)
|
||||
-- ($(#1) + (0, -1)$)
|
||||
-- ($(#2) + (0,1)$)
|
||||
-- (#2);
|
||||
}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\input{parts/0 intro}
|
||||
\input{parts/1 cycle}
|
||||
\input{parts/2 groups}
|
||||
\input{parts/3 subgroup}
|
||||
|
||||
|
||||
\section{Bonus problems}
|
||||
|
||||
\problem{}
|
||||
Show that $x \in \mathbb{Z}^+$ has a multiplicative inverse mod $n$ iff $\text{gcd}(x, n) = 1$
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Let $\sigma = (\sigma_1 \sigma_2 ... \sigma_k)$ be a $k$-cycle in $S_n$, and let $\tau$ be an arbitrary element of $S_n$. \par
|
||||
Show that $\tau \sigma \tau^{-1}$ = $\bigl(\tau(\sigma_1), \tau(\sigma_2), ..., \tau(\sigma_k)\bigr)$ \par
|
||||
\hint{As usual, $\tau$ is a permutation. Thus, $\tau(x)$ is the value at position $x$ after applying $\tau$.}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Show that the set $\Bigl\{ (1, 2),~ (1,2,...,n) \Bigr\}$ generates $S_n$.
|
||||
\vfill
|
||||
|
||||
% TODO: (a second day?)
|
||||
% alternating group
|
||||
% type and sign and conjugation
|
||||
% isomorphisms & automorphisms
|
||||
% automorphism groups
|
||||
\end{document}
|
||||
35
src/Advanced/Symmetric Groups/main.typ
Normal file
35
src/Advanced/Symmetric Groups/main.typ
Normal file
@ -0,0 +1,35 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Symmetric Groups],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#include "parts/00 intro.typ"
|
||||
#pagebreak()
|
||||
|
||||
#include "parts/01 cycle.typ"
|
||||
#pagebreak()
|
||||
|
||||
#include "parts/02 groups.typ"
|
||||
#pagebreak()
|
||||
|
||||
#include "parts/03 subgroup.typ"
|
||||
#pagebreak()
|
||||
|
||||
= Bonus problems
|
||||
|
||||
#problem()
|
||||
Show that $x in ZZ^+$ has a multiplicative inverse mod $n$ iff $gcd(x, n) = 1$
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Let $sigma = (sigma_1 sigma_2 ... sigma_k)$ be a $k$-cycle in $S_n$, and let $tau$ be an arbitrary element of $S_n$. \
|
||||
Show that $tau sigma tau^(-1)$ = $(tau(sigma_1), tau(sigma_2), ..., tau(sigma_k))$ \
|
||||
#hint[$tau$ is a permutation, so $tau(x)$ is the value at position $x$ after applying $tau$.]
|
||||
|
||||
#v(1fr)
|
||||
#problem()
|
||||
Show that the set ${ (1, 2), (1,2,...,n)}$ generates $S_n$.
|
||||
#v(1fr)
|
||||
@ -1,199 +0,0 @@
|
||||
\section{Introduction}
|
||||
|
||||
|
||||
\definition{}
|
||||
Informally, a \textit{permutation} of a collection of $n$ objects is an ordering of these $n$ objects. \par
|
||||
For example, a few permutations of $\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}$ are $\texttt{ABCD}$,
|
||||
$\texttt{BCDA}$, and $\texttt{DACB}$. \par
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
This, however, isn't the definition we'll use today. Instead of defining permutations as \say{ordered lists,}
|
||||
(as we do above), we'll define them as functions. Our first goal today is to make sense of this definition.
|
||||
|
||||
|
||||
|
||||
\definition{Permutations}
|
||||
Let $\Omega$ be an arbitrary set of $n$ objects. \par
|
||||
A \textit{permutation} on $\Omega$ is a map from $\Omega$ to itself that produces a \textit{unique} output for each input. \par
|
||||
\note{In other words, if $a$ and $b$ are different, $f(a)$ and $f(b)$ must also be different.}
|
||||
|
||||
|
||||
\footnotetext{The words \say{function} and \say{map} are equivalent.}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
For example, consider $\{1, 2, 3\}$. \par
|
||||
One permutation on this set can be defined as follows: \par
|
||||
\begin{itemize}
|
||||
\item $f(1) = 3$
|
||||
\item $f(2) = 1$
|
||||
\item $f(3) = 2$
|
||||
\end{itemize}
|
||||
|
||||
If we take the array $123$ and apply
|
||||
|
||||
|
||||
\problem{}
|
||||
List all permutations on three objects. \par
|
||||
How many permutations of $n$ objects are there?
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
What map corresponds to the permutation $[321]$?
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
What map corresponds to the \say{do-nothing} permutation? \par
|
||||
Write it as a function and in square-bracket notation. \par
|
||||
\note[Note]{We usually call this the \textit{trivial permutation}}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
We can visualize permutations with a \textit{string diagram}, shown below. \par
|
||||
The arrows in this diagram denote the image of $f$ for each possible input.
|
||||
Two examples are below:
|
||||
|
||||
\vspace{2mm}
|
||||
\hfill
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1a) at (0, 0.5) {1};
|
||||
\node (2a) at (1, 0.5) {2};
|
||||
\node (3a) at (2, 0.5) {3};
|
||||
\node (4a) at (3, 0.5) {4};
|
||||
|
||||
\node (1b) at (0, -2) {1};
|
||||
\node (3b) at (1, -2) {3};
|
||||
\node (4b) at (2, -2) {4};
|
||||
\node (2b) at (3, -2) {2};
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
\line{4a}{4b}
|
||||
\end{tikzpicture}
|
||||
\hfill
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1a) at (0, 0.5) {1};
|
||||
\node (2a) at (1, 0.5) {2};
|
||||
\node (3a) at (2, 0.5) {3};
|
||||
\node (4a) at (3, 0.5) {4};
|
||||
|
||||
\node (2b) at (0, -2) {2};
|
||||
\node (1b) at (1, -2) {1};
|
||||
\node (3b) at (2, -2) {3};
|
||||
\node (4b) at (3, -2) {4};
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
\line{4a}{4b}
|
||||
\end{tikzpicture}
|
||||
\hfill\null
|
||||
\vspace{2mm}
|
||||
|
||||
|
||||
Note that in all our examples thus far, the objects in our set have an implicit order.
|
||||
This is only for convenience. The elements of $\Omega$ are not ordered (it is a \textit{set}, after all),
|
||||
and we may present them however we wish.
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\vspace{1cm}
|
||||
|
||||
For example, consider the diagrams below. \par
|
||||
On the left, 1234 are ordered as usual. In the middle, they are ordered alphabetically. \par
|
||||
The rightmost diagram uses arbitrary, meaningless labels.
|
||||
|
||||
\vspace{2mm}
|
||||
\hfill
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1a) at (0, 0.5) {1};
|
||||
\node (2a) at (1, 0.5) {2};
|
||||
\node (3a) at (2, 0.5) {3};
|
||||
\node (4a) at (3, 0.5) {4};
|
||||
|
||||
\node (2b) at (0, -2) {2};
|
||||
\node (1b) at (1, -2) {1};
|
||||
\node (3b) at (2, -2) {3};
|
||||
\node (4b) at (3, -2) {4};
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
\line{4a}{4b}
|
||||
\end{tikzpicture}
|
||||
\hfill
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (4a) at (0, 0.5) {4};
|
||||
\node (1a) at (1, 0.5) {1};
|
||||
\node (3a) at (2, 0.5) {3};
|
||||
\node (2a) at (3, 0.5) {2};
|
||||
|
||||
\node (1b) at (0, -2) {1};
|
||||
\node (4b) at (1, -2) {4};
|
||||
\node (3b) at (2, -2) {3};
|
||||
\node (2b) at (3, -2) {2};
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
\line{4a}{4b}
|
||||
\end{tikzpicture}
|
||||
\hfill
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1a) at (0, 0.5) {$\triangle$};
|
||||
\node (2a) at (1, 0.5) {$\divideontimes$};
|
||||
\node (3a) at (2, 0.5) {$\circledcirc$};
|
||||
\node (4a) at (3, 0.5) {$\boxdot$};
|
||||
|
||||
\node (2b) at (0, -2) {$\divideontimes$};
|
||||
\node (1b) at (1, -2) {$\triangle$};
|
||||
\node (3b) at (2, -2) {$\circledcirc$};
|
||||
\node (4b) at (3, -2) {$\boxdot$};
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
\line{4a}{4b}
|
||||
\end{tikzpicture}
|
||||
\hfill\null
|
||||
\vspace{2mm}
|
||||
|
||||
|
||||
It shouldn't be hard to see that despite the different \say{output} order (2134 and 1432), \par
|
||||
the same permutation is depicted in all three diagrams. This example demonstrates two things:
|
||||
\begin{itemize}[itemsep=2mm]
|
||||
\item First, the names of the items in our set do not have any meaning. \par
|
||||
$\Omega$ is just a set of $n$ arbitrary things, which we may label however we like.
|
||||
|
||||
\item Second, permutations are verbs. We do not care about the \say{output} of a certain permutation,
|
||||
we care about what it \textit{does}. We could, for example, describe the permutation above as
|
||||
\say{swap the first two of four elements.}
|
||||
\end{itemize}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
|
||||
Why, then, do we order our elements when we talk about permutations? As noted before, this is for convenience.
|
||||
If we assign a natural order to the elements of $\Omega$ (say, 1234), we can identify permutations by simply listing
|
||||
their output:
|
||||
Clearly, $[1234]$ represents the trivial permutation, $[2134]$ represents \say{swap first two,}
|
||||
and $[4123]$ represents \say{cycle right.}
|
||||
|
||||
\problem{}
|
||||
Draw string diagrams for $[4123]$ and $[2341]$.
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
378
src/Advanced/Symmetric Groups/parts/00 intro.typ
Normal file
378
src/Advanced/Symmetric Groups/parts/00 intro.typ
Normal file
@ -0,0 +1,378 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.4.2"
|
||||
#import "../macros.typ": *
|
||||
|
||||
= Introduction
|
||||
|
||||
#definition()
|
||||
Informally, a _permutation_ on a collection of $n$ objects is an ordering of these $n$ objects.
|
||||
|
||||
For example, a few permutations of `A`, `B`, `C`, and `D` are `ABCD`, `BCDA`, and `DACB`.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
This, however, isn't the definition we'll use today.
|
||||
Instead of defining permutations as "ordered lists" \
|
||||
(like we do above), we'll define them as _functions_ on finite sets. \
|
||||
Our first goal today is to make sense of this definition.
|
||||
|
||||
|
||||
#definition("Permutations")
|
||||
Let $Omega$ be a set of $n$ arbitrary objects.
|
||||
|
||||
A _permutation_ $f$ on $Omega$ is a map#footnote[The words "function" and "map" are equivalent.]
|
||||
from $Omega$ to itself that produces a _unique_ output for each input.
|
||||
|
||||
#note[This means that if $a$ and $b$ are different, $f(a)$ and $f(b)$ must also be different.]
|
||||
|
||||
|
||||
|
||||
#v(2mm)
|
||||
|
||||
For example, consider ${1, 2, 3}$. \
|
||||
One permutation on this set can be defined as follows:
|
||||
- $f(1) = 3$
|
||||
- $f(2) = 1$
|
||||
- $f(3) = 2$
|
||||
|
||||
If we take the array $123$ and apply $f$, we get $312$.
|
||||
|
||||
#problem()
|
||||
List all permutations on three objects. \
|
||||
How many permutations of $n$ objects are there?
|
||||
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
What map corresponds to the permutation that produces the array `312` from the array `123`?
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
What map corresponds to the "do-nothing" permutation?
|
||||
|
||||
Write it as a function and in square-bracket notation.
|
||||
|
||||
#note([We will call this the _trivial permutation_])
|
||||
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
We can visualize a permutation using a _string diagram_.
|
||||
The arrows in this diagram denote \
|
||||
the output of $f$ for each possible input.
|
||||
Two examples are below:
|
||||
|
||||
#table(
|
||||
columns: (1fr, 1fr),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 0.5 // scale
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
content((0 * s, 0 * s), $1$, name: "1b")
|
||||
content((1 * s, 0 * s), $3$, name: "3b")
|
||||
content((2 * s, 0 * s), $4$, name: "4b")
|
||||
content((3 * s, 0 * s), $2$, name: "2b")
|
||||
|
||||
markline(s, "1a", "1b")
|
||||
markline(s, "2a", "2b")
|
||||
markline(s, "3a", "3b")
|
||||
markline(s, "4a", "4b")
|
||||
})),
|
||||
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 0.5 // scale
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
content((0 * s, 0 * s), $2$, name: "2b")
|
||||
content((1 * s, 0 * s), $1$, name: "1b")
|
||||
content((2 * s, 0 * s), $3$, name: "3b")
|
||||
content((3 * s, 0 * s), $4$, name: "4b")
|
||||
|
||||
markline(s, "1a", "1b")
|
||||
markline(s, "2a", "2b")
|
||||
markline(s, "3a", "3b")
|
||||
markline(s, "4a", "4b")
|
||||
})),
|
||||
)
|
||||
|
||||
Note that the elements of the set we are permuting are not ordered. (it is a _set_, after all!) \
|
||||
For example, consider the diagrams below.
|
||||
On the left, 1234 are ordered as usual. \
|
||||
In the middle, they are ordered alphabetically. \
|
||||
The rightmost diagram uses arbitrary, meaningless labels.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#table(
|
||||
columns: (1fr, 1fr, 1fr),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.5 // scale
|
||||
let del = 0.2 // small line
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
content((0 * s, 0 * s), $2$, name: "2b")
|
||||
content((1 * s, 0 * s), $1$, name: "1b")
|
||||
content((2 * s, 0 * s), $3$, name: "3b")
|
||||
content((3 * s, 0 * s), $4$, name: "4b")
|
||||
|
||||
line(
|
||||
"1a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "1a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "1b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "1b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"1b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"2a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "2a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "2b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "2b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"2b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"3a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "3a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "3b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "3b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"3b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"4a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "4a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "4b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "4b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"4b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
})),
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.5 // scale
|
||||
let del = 0.2 // small line
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $4$, name: "4a")
|
||||
content((1 * s, 3 * s), $1$, name: "1a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $2$, name: "2a")
|
||||
|
||||
content((0 * s, 0 * s), $1$, name: "1b")
|
||||
content((1 * s, 0 * s), $4$, name: "4b")
|
||||
content((2 * s, 0 * s), $3$, name: "3b")
|
||||
content((3 * s, 0 * s), $2$, name: "2b")
|
||||
|
||||
line(
|
||||
"1a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "1a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "1b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "1b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"1b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"2a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "2a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "2b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "2b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"2b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"3a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "3a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "3b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "3b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"3b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"4a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "4a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "4b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "4b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"4b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
})),
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.5 // scale
|
||||
let del = 0.2 // small line
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $ast.circle$, name: "4a")
|
||||
content((1 * s, 3 * s), $hexa.stroked$, name: "1a")
|
||||
content((2 * s, 3 * s), $triangle.stroked.b$, name: "3a")
|
||||
content((3 * s, 3 * s), $\#$, name: "2a")
|
||||
|
||||
content((0 * s, 0 * s), $hexa.stroked$, name: "1b")
|
||||
content((1 * s, 0 * s), $ast.circle$, name: "4b")
|
||||
content((2 * s, 0 * s), $triangle.stroked.b$, name: "3b")
|
||||
content((3 * s, 0 * s), $\#$, name: "2b")
|
||||
|
||||
line(
|
||||
"1a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "1a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "1b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "1b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"1b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"2a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "2a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "2b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "2b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"2b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"3a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "3a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "3b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "3b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"3b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
|
||||
line(
|
||||
"4a.south",
|
||||
(v => cetz.vector.add(v, (0, -del)), "4a.south"),
|
||||
(v => cetz.vector.add(v, (0, del + 0.2)), "4b.north"),
|
||||
(v => cetz.vector.add(v, (0, 0.2)), "4b.north"),
|
||||
stroke: oblue + s * 1mm,
|
||||
)
|
||||
mark(
|
||||
"4b.north",
|
||||
270deg,
|
||||
..arrow,
|
||||
)
|
||||
})),
|
||||
)
|
||||
|
||||
#v(2mm)
|
||||
|
||||
It shouldn't be hard to see that despite the different "output" each diagram displays \
|
||||
($2134$, $1432$, and $hexa.stroked ast.circle triangle.stroked.b \#$), the same permutation ("swap first two") is shown in each.
|
||||
|
||||
Observe the following:
|
||||
- The "names" of the items in our set do not have any meaning. \
|
||||
We are interested in sets of $n$ arbitrary things, which we may label however we like.
|
||||
- Permutations are _verbs_. \
|
||||
We do not care about the "output" of a certain permutation. Rather, we care about what it _does_. \
|
||||
We could, for example, describe the permutation in the above three diagrams as "swap the first two elements."
|
||||
|
||||
|
||||
#definition("Square Brackets")
|
||||
However, elements with an implicit order (1, 2, 3, ...) are convenient. \
|
||||
Such sets let us denote a permutation by writing the array it produces \
|
||||
after transforming the "reference order" $123...n$.
|
||||
|
||||
We will call this _square-bracket notation_. \
|
||||
$[312]$ denotes the permutation that produces $312$ when applied to $123$.
|
||||
|
||||
#problem()
|
||||
Draw string diagrams for $[4123]$ and $[2341]$.
|
||||
|
||||
#v(1fr)
|
||||
635
src/Advanced/Symmetric Groups/parts/01 cycle.typ
Executable file
635
src/Advanced/Symmetric Groups/parts/01 cycle.typ
Executable file
@ -0,0 +1,635 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.4.2"
|
||||
#import "../macros.typ": *
|
||||
|
||||
= Cycle Notation
|
||||
|
||||
#definition("Order")
|
||||
The _order_ of a permutation $f$ is the smallest positive $n$ where $f^n (x) = x$ for all $x$. \
|
||||
In other words, if we repeatedly apply a permutation with order $n$, \
|
||||
we will get back to where we started after $n$ steps. \
|
||||
|
||||
#v(2mm)
|
||||
|
||||
For example, consider $[2134]$. This permutation has order $2$, as we can see below:
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.5 // scale
|
||||
let del = 0.2 // small line
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
content((0 * s, 0 * s), $2$, name: "2b")
|
||||
content((1 * s, 0 * s), $1$, name: "1b")
|
||||
content((2 * s, 0 * s), $3$, name: "3b")
|
||||
content((3 * s, 0 * s), $4$, name: "4b")
|
||||
|
||||
content((0 * s, -3 * s), $1$, name: "1c")
|
||||
content((1 * s, -3 * s), $2$, name: "2c")
|
||||
content((2 * s, -3 * s), $3$, name: "3c")
|
||||
content((3 * s, -3 * s), $4$, name: "4c")
|
||||
|
||||
markline(s, "1a", "1b")
|
||||
markline(s, "2a", "2b")
|
||||
markline(s, "3a", "3b")
|
||||
markline(s, "4a", "4b")
|
||||
|
||||
markline(s, "1b", "1c")
|
||||
markline(s, "2b", "2c")
|
||||
markline(s, "3b", "3c")
|
||||
markline(s, "4b", "4c")
|
||||
})),
|
||||
)
|
||||
|
||||
Swapping the first two elements of a list twice changes nothing. \
|
||||
Thus, $[2134]$ has an order of two.
|
||||
|
||||
|
||||
#problem()
|
||||
What is the order of $[2314]$? \
|
||||
How about $[4321]$? \
|
||||
#note(type: "Note")[Try to solve this problem without drawing any strings!]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Find a permutation on five elements with order 4.
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem(label: "finiteorder")
|
||||
Show that all permutations on a finite set have a well-defined order. \
|
||||
In other words, show that there must always be an integer $n$ where $f^n (x) = x$.
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
#definition("Composition", label: "compdef")
|
||||
The _composition_ of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \
|
||||
We'll denote this by simply writing the permutations we're composing next to each other, like $f g$. \
|
||||
Note that $g$ is applied _before_ $f$ in $f g$.
|
||||
|
||||
#problem()
|
||||
Show that function composition is associative. \
|
||||
That is, show that $f(g h) = (f g)h$.
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
What is $[1324][4321]$? \
|
||||
How about $[321][213][231]$? \
|
||||
Rewrite these compositions as one permutation in square brackets.
|
||||
|
||||
#solution([
|
||||
- $[1324][4321]$ is $[4321]$
|
||||
- $[321][213][231]$ is $[123]$
|
||||
])
|
||||
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
|
||||
Permutations are verbs---but we've been referring to them using a noun (i.e, their output).
|
||||
|
||||
Square-bracket notation fails to capture the structure of the permutation it identifies.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Is the permutation $[1234]$ different than the permutation $[12345]$? \
|
||||
These permutations operate on different sets---but they are both the identity! \
|
||||
Are $[5342761]$ and $[1342567][5234761]$ similar? What are their orders?
|
||||
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Good notation should help us understand the objects we are studying. \
|
||||
We need something better than square brackets.
|
||||
|
||||
|
||||
#pagebreak()
|
||||
|
||||
|
||||
#remark("Cycles")
|
||||
Any permutation is composed of a number of _cycles_. \
|
||||
Reread @finiteorder to convince yourself of this fact.
|
||||
|
||||
|
||||
#example()
|
||||
Consider the permutation $[2134]$. \
|
||||
It consists of one two-cycle: $1 arrow.r 2 arrow.r 1$, which we can see in the diagram below. \
|
||||
#note(
|
||||
type: "Note",
|
||||
)[
|
||||
$3 arrow.r 3$ and $4 arrow.r 4$ are also cycles, but we'll ignore them.
|
||||
One-cycles aren't interesting.
|
||||
]
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.6 // scale
|
||||
let del = 0.4 // small line
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
marklinetop(s, "1a", "2a")
|
||||
marklinebot(s, "2a", "1a")
|
||||
})),
|
||||
)
|
||||
|
||||
#v(4mm)
|
||||
|
||||
The permutation $[431265]$ is a bit more interesting---it contains two cycles: \
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.6 // scale
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
content((4 * s, 3 * s), $5$, name: "5a")
|
||||
content((5 * s, 3 * s), $6$, name: "6a")
|
||||
|
||||
|
||||
marklinetop(s, "3a", "2a", del: 0.8)
|
||||
marklinebot(s, "2a", "4a", del: 1.3)
|
||||
marklinetop(s, "4a", "1a", del: 1.3)
|
||||
marklinebot(s, "1a", "3a", del: 0.8)
|
||||
marklinebot(s, "5a", "6a", del: 0.8, c: ogreen)
|
||||
marklinetop(s, "6a", "5a", del: 0.8, c: ogreen)
|
||||
})),
|
||||
)
|
||||
|
||||
#remark()
|
||||
Two-cycles may also be called _transpositions_. \
|
||||
Any permutation that swaps two elements is a transposition.
|
||||
|
||||
#problem()
|
||||
Find all cycles in $[5342761]$.
|
||||
|
||||
#solution[
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.5 // scale
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (
|
||||
frame: "rect",
|
||||
stroke: none,
|
||||
padding: .1,
|
||||
))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
content((4 * s, 3 * s), $5$, name: "5a")
|
||||
content((5 * s, 3 * s), $6$, name: "6a")
|
||||
content((6 * s, 3 * s), $7$, name: "7a")
|
||||
|
||||
|
||||
marklinetop(s, "1a", "7a", del: 1.6)
|
||||
marklinebot(s, "7a", "5a", del: 1.2)
|
||||
marklinetopswap(s, "5a", "1a", del: 1.2)
|
||||
|
||||
|
||||
marklinebot(s, "2a", "4a", del: 1.2, c: ogreen)
|
||||
marklinetop(s, "4a", "3a", del: 0.8, c: ogreen)
|
||||
marklinebotswap(s, "3a", "2a", del: 0.8, c: ogreen)
|
||||
})),
|
||||
)
|
||||
|
||||
There are two non-trivial cycles:
|
||||
- $4 arrow.r 3 arrow.r 2 arrow.r 4$
|
||||
- $1 arrow.r 7 arrow.r 5 arrow.r 1$
|
||||
]
|
||||
|
||||
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
What permutation on five objects is formed by the cycles $3 arrow.r 5 arrow.r 3$ and $1 arrow.r 2 arrow.r 4 arrow.r 1$? \
|
||||
Write it in square-bracket notation.
|
||||
|
||||
#solution[
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.6 // scale
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (
|
||||
frame: "rect",
|
||||
stroke: none,
|
||||
padding: .1,
|
||||
))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
content((4 * s, 3 * s), $5$, name: "5a")
|
||||
|
||||
marklinetop(s, "3a", "5a", del: 0.8, c: ogreen)
|
||||
marklinebot(s, "5a", "3a", del: 0.8, c: ogreen)
|
||||
|
||||
marklinebot(s, "1a", "2a", del: 0.8)
|
||||
marklinetop(s, "2a", "4a", del: 1.2)
|
||||
|
||||
marklinebotswap(s, "4a", "1a", del: 1.2)
|
||||
})),
|
||||
)
|
||||
|
||||
|
||||
This is $[41523]$.
|
||||
]
|
||||
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
#definition("Cycle Notation")
|
||||
We can use cycles to develop better notation: \
|
||||
Instead of identifying permutations using their output, we'll identify them using their _cycles_.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
For example, we'll write $[2134]$ is $(12)$ in cycle notation, \
|
||||
since it consists only of the cycle $1 arrow.r 2 arrow.r 1$:
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.6 // scale
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (
|
||||
frame: "rect",
|
||||
stroke: none,
|
||||
padding: .1,
|
||||
))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
marklinebot(s, "1a", "2a", del: 0.8)
|
||||
marklinetop(s, "2a", "1a", del: 0.8)
|
||||
})),
|
||||
)
|
||||
|
||||
#v(2mm)
|
||||
|
||||
|
||||
Permutations that consist of more than one cycle are written as a composition. \
|
||||
$[2143]$ is written as $(12)(34)$. Applying the permutation $[2143]$ has the same effect as applying $(34)$, then applying $(12)$.
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.6 // scale
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (
|
||||
frame: "rect",
|
||||
stroke: none,
|
||||
padding: .1,
|
||||
))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
|
||||
marklinetop(s, "1a", "2a", del: 0.8)
|
||||
marklinebot(s, "2a", "1a", del: 0.8)
|
||||
marklinetop(s, "3a", "4a", del: 0.8, c: ogreen)
|
||||
marklinebot(s, "4a", "3a", del: 0.8, c: ogreen)
|
||||
})),
|
||||
)
|
||||
|
||||
|
||||
|
||||
#remark()
|
||||
According to @finiteorder, any permutation may be written as a composition of disjoint cycles. \
|
||||
Convince yourself of this fact.
|
||||
|
||||
#problem()
|
||||
Rewrite $[431265]$ in cycle notation.
|
||||
|
||||
#solution[
|
||||
$[431265]$ is $(1324)(56)$:
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.6 // scale
|
||||
let arrow = (
|
||||
symbol: ")>",
|
||||
scale: s * 2.2,
|
||||
fill: oblue,
|
||||
stroke: oblue,
|
||||
)
|
||||
|
||||
set-style(content: (
|
||||
frame: "rect",
|
||||
stroke: none,
|
||||
padding: .1,
|
||||
))
|
||||
content((0 * s, 3 * s), $1$, name: "1a")
|
||||
content((1 * s, 3 * s), $2$, name: "2a")
|
||||
content((2 * s, 3 * s), $3$, name: "3a")
|
||||
content((3 * s, 3 * s), $4$, name: "4a")
|
||||
content((4 * s, 3 * s), $5$, name: "5a")
|
||||
content((5 * s, 3 * s), $6$, name: "6a")
|
||||
|
||||
marklinetop(s, "1a", "3a", del: 0.8, c: ogreen)
|
||||
marklinebot(s, "4a", "1a", del: 1.3, c: ogreen)
|
||||
marklinebot(s, "3a", "2a", del: 0.8, c: ogreen)
|
||||
marklinetop(s, "2a", "4a", del: 1.3, c: ogreen)
|
||||
|
||||
marklinetop(s, "5a", "6a", del: 0.8)
|
||||
marklinebot(s, "6a", "5a", del: 0.8)
|
||||
})),
|
||||
)
|
||||
]
|
||||
|
||||
|
||||
|
||||
|
||||
#remark()
|
||||
The identity permutation $f(x) = x$ is written as $()$ in cycle notation.
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
Convince yourself that disjoint cycles commute. \
|
||||
That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \
|
||||
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
#problem(label: "insquare")
|
||||
Write the following in square-bracket notation.
|
||||
|
||||
- $(12)$ on a set of 2 elements
|
||||
- $(12)(435)$ on a set of 5 elements
|
||||
#v(2mm)
|
||||
- $(321)$ on a set of 3 elements
|
||||
- $(321)$ on a set of 6 elements
|
||||
#v(2mm)
|
||||
- $(1234)$ on a set of 4 elements
|
||||
- $(3412)$ on a set of 4 elements
|
||||
|
||||
#note[
|
||||
Note that $(12)$ refers the "swap first two" permutation on a set of _any_ size. \
|
||||
We can use consistent notation for the same action on two different sets! \
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
Write the following in square-bracket notation.
|
||||
Pay attention!
|
||||
- $(13)(243)$ on a set of 4 elements
|
||||
- $(243)(13)$ on a set of 4 elements
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
Consider the last two permutations in @insquare, $(1234)$ and $(3412)$. \
|
||||
These are _identical_---they are the same cycle written in two different ways. \
|
||||
List all other ways to write this cycle. \
|
||||
#hint[There are two more.]
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
#definition("Inverse")
|
||||
The _inverse_ of a permitation $f$ is a permutation $g$ that "un-does" $f$. \
|
||||
This means that $g(f(x)) = x$ for all $x$.
|
||||
|
||||
#problem()
|
||||
What is the inverse of $(12)$? \
|
||||
How about $(123)$? And $(4231)$? \
|
||||
#note[
|
||||
Note we do not need to know the size of the set we are operating on. \
|
||||
The inverse of $(12)$ is the same in sets of all sizes!
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
Let $sigma$ be a permutation composed of disjoint cycles $sigma_1sigma_2...sigma_k$. \
|
||||
Say we know the order of all $sigma_i$. What is the order of $sigma$?
|
||||
|
||||
#solution[
|
||||
$
|
||||
#text[lcm]\(#text[ord]\(sigma_1),#h(0.5em) #text[ord]\(sigma_2),#h(0.5em) ...,#h(0.5em) #text[ord]\(sigma_k))
|
||||
$
|
||||
]
|
||||
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem(label: "cycletrans")
|
||||
Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
|
||||
|
||||
#solution[
|
||||
*Intuition:*\
|
||||
$(123...n)$ is a right-shift. Swapping all pairs from right to left achieves the same effect.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
*Complete solution:* \
|
||||
Consider $n-1$. After applying $(123...n)$, it takes the position of $n$.
|
||||
|
||||
After applying $(n-1, n)$, $n-1$ moves to the same position _and is never moved again!_ \
|
||||
Repeat this argument for all other $n$.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Write $(7126453)$ as a product of transpositions. \
|
||||
|
||||
#solution[
|
||||
Move elements one at a time, and using the last position as temporary storage.
|
||||
|
||||
We get $(71)(72)(76)(74)(75)(73)$.
|
||||
Other solutions are possible. \
|
||||
|
||||
#v(2mm)
|
||||
|
||||
*Bonus:* How can we do this in the fewest number of transpositions?
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
#problem(label: "simpletrans")
|
||||
Show that any permutation is a product of transpositions.
|
||||
|
||||
#solution[
|
||||
Re-use the argument in @cycletrans. \
|
||||
Pick an arbitrary "working slot," and re-build all cycles. \
|
||||
Use the "not touched again" argument for a proper proof.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem(label: "onetrans")
|
||||
Show that any permutation is a product of transpositions of the form $(1, k)$. \
|
||||
|
||||
#solution[
|
||||
Use @simpletrans to rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$. \
|
||||
Showing that $(a, b) = (1, a)(1, b)(1, a)$ is fairly easy.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
#problem(label: "oneplustrans")
|
||||
Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.
|
||||
|
||||
#solution[
|
||||
This is the same as @onetrans,
|
||||
but we use $a + 1$ as a "working slot" instead of $1$.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
Show that any permutation is a product of adjacent transpositions. \
|
||||
An _adjacent transposition_ swaps two adjacent elements, and thus looks like $(n, n+1)$.
|
||||
|
||||
#solution[
|
||||
As before, we will use @simpletrans and rewrite the transpositions it produces in a convenient fashion.
|
||||
To do this, we must show that every transposition $(a, b)$ is a product of adjacent transpositions.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
In the proof below, assume that $a < b$ and perform induction on $b - a$. \
|
||||
|
||||
#v(4mm)
|
||||
|
||||
|
||||
*Base Case:*\
|
||||
If $b - a = 1$, $(a, b)$ is a product of adjacent transpositions. \
|
||||
In fact, it _is_ an adjacent transposition.
|
||||
|
||||
#v(4mm)
|
||||
|
||||
*Induction:*\
|
||||
Now, say $b - a = n + 1$. \
|
||||
Assume that all $(a, b)$ where $b - a <= n$ are products of adjacent transpositions.\
|
||||
By @oneplustrans, $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
$(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \
|
||||
Thus, $(a, b)$ is a product of adjacent transpositions.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
139
src/Advanced/Symmetric Groups/parts/02 groups.typ
Executable file
139
src/Advanced/Symmetric Groups/parts/02 groups.typ
Executable file
@ -0,0 +1,139 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.4.2"
|
||||
#import "../macros.typ": *
|
||||
|
||||
= Groups (review)
|
||||
|
||||
#definition()
|
||||
Before we continue, we must introduce a bit of notation:
|
||||
- $S_n$ is the set of permutations on $n$ objects.
|
||||
- $ZZ_n$ is the set of integers mod $n$.
|
||||
|
||||
- $ZZ_n^times$ is the set of integers mod $n$ with multiplicative inverses. \
|
||||
In other words, it is the set of integers smaller than $n$ and coprime to $n$.#footnote[We proved this in another handout, but you may take it as fact here.] \
|
||||
For example, $ZZ_12^times = {1, 5, 7, 11}$.
|
||||
|
||||
#problem()
|
||||
What are the elements of $S_3$? #hint[Use cycle notation] \
|
||||
How about $ZZ_17^times$?
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#definition()
|
||||
A _group_ $(G, *)$ consists of a set $G$ and an operator $*$. \
|
||||
Groups always have the following properties:
|
||||
|
||||
+ $G$ is closed under $*$. In other words, $a, b in G => a * b in G$.
|
||||
+ $*$ is _associative_: $(a * b) * c = a * (b * c)$ for all $a,b,c in G$
|
||||
+ There is an _identity_ $e in G$, so that $a * e = e * a = a$ for all $a in G$.
|
||||
+ For any $a in G$, there exists a $b in G$ so that $a * b = b * a = e$. $b$ is called the _inverse_ of $a$. \
|
||||
This element is written as $-a$ if our operator is addition and $a^(-1)$ otherwise.
|
||||
|
||||
Any pair $(G, *)$ that satisfies these properties is a group.
|
||||
|
||||
#problem()
|
||||
Is $(ZZ_5, +)$ a group? \
|
||||
Is $(ZZ_5, -)$ a group? \
|
||||
#note[$+$ and $-$ refer to the usual operations in modular arithmetic.]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
What is the group with the fewest number of elements?
|
||||
|
||||
#solution[
|
||||
Let $(G, star)$ be our group, where $G = {x}$ and $star$ is defined by $x star x = x$
|
||||
|
||||
Verifying that the trivial group is a group is trivial.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
Show that function composition is associative
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Show that $S_n$ is a group under composition.
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Let $(G, *)$ be a group with finitely many elements, and let $a in G$. \
|
||||
Show that there is an $n$ in $in ZZ^+$ so that $a^n = e$ \
|
||||
#hint[$a^n = a * a * ... * a$ repeated $n$ times.]
|
||||
|
||||
#v(2mm)
|
||||
|
||||
The smallest such $n$ defines the _order_ of $g$.
|
||||
|
||||
#hint[
|
||||
We've already done a special case of this problem! \
|
||||
Find it in this handout, then rewrite your proof for an arbitrary (finite) group.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
What is the order of 5 in $(ZZ_25, +)$? \
|
||||
What is the order of 2 in $(ZZ_17^times, times)$? \
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
|
||||
#definition("Generator", label: "gendef")
|
||||
Let $G$ be a group, and let $g$ be an element of $G$. \
|
||||
We say $g$ is a _generator_ if every other element of $G$ may be written as a power of $g$. \
|
||||
|
||||
#problem()
|
||||
Let $G$ be a group of $n$ elements. \
|
||||
If $g$ is a generator, what is its order? \
|
||||
Provide a proof.
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Find the two generators in $(ZZ, +)$ \
|
||||
Then, find all generators of $(ZZ_5, +)$
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
How many groups have only one generator?
|
||||
|
||||
#solution[
|
||||
Only one: the trivial group. The inverse of a generator is also a generator!
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#definition()
|
||||
Let $S$ be a subset of the elements in $G$. \
|
||||
We say that $S$ _generates_ $G$ if every element of $G$ may be written as a product of elements in $S$. \
|
||||
#note(type: "Note")[This is an extension of @gendef.]
|
||||
|
||||
#problem()
|
||||
We've already found a few generating sets of $S_n$. What are they?
|
||||
|
||||
#solution[
|
||||
The following sets generate $S_n$:
|
||||
- All transpositions
|
||||
- All transpositions of the form $(1, k)$
|
||||
- All adjacent transpositions
|
||||
|
||||
#v(2mm)
|
||||
|
||||
The smallest generating set of $S_n$ consists of the transposition $(12)$ and the $n$-cycle $(1,2,...,n)$. \
|
||||
The proof of this is a bonus problem later in the handout.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
172
src/Advanced/Symmetric Groups/parts/03 subgroup.typ
Normal file
172
src/Advanced/Symmetric Groups/parts/03 subgroup.typ
Normal file
@ -0,0 +1,172 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.4.2"
|
||||
#import "../macros.typ": *
|
||||
|
||||
= Subgroups
|
||||
|
||||
#problem(label: "s2s3share")
|
||||
What elements do $S_2$ and $S_3$ share?
|
||||
|
||||
#v(2cm)
|
||||
|
||||
|
||||
Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} subset \{1, 2, 3\}$. \
|
||||
Can we say something similar about $S_2$ and $S_3$?
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Looking at @s2s3share, we may want to say that $S_2 subset S_3$ since every element of $S_2$ is in $S_3$. \
|
||||
This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are _groups_, not _sets_: \
|
||||
their elements come with structure, which the "subset" relation does not capture.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
To account for this, we'll define a similar relation: subgroups.
|
||||
|
||||
#definition("Subgroup")
|
||||
Let $G$ and $G'$ be groups. We say $G'$ is a _subgroup_ of $G$ (and write $G' subset G$) if the following are true:\
|
||||
(Note that $x, y$ are elements of $G$, and $x y$ is multiplication in $G$)
|
||||
- the set of elements in $G'$ is a subset of the set of elements in $G$.
|
||||
- the identity of $G$ is in $G'$
|
||||
- $x,y in G' => x y in G'$
|
||||
- $x in G' => x^(-1) in G'$
|
||||
|
||||
The above definition may look fairly scary, but the idea behind a subgroup is simple. \
|
||||
Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Say we have a set of four elements and only look at the first three. \
|
||||
$S_3$ fully describes all the ways we can arrange those three elements:
|
||||
|
||||
#table(
|
||||
columns: (1fr,),
|
||||
align: center,
|
||||
stroke: none,
|
||||
align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.7
|
||||
|
||||
set-style(content: (frame: "rect", stroke: none, fill: white, padding: .1))
|
||||
content((0 * s, 0.5 * s), $1$, name: "1a")
|
||||
content((1 * s, 0.5 * s), $2$, name: "2a")
|
||||
content((2 * s, 0.5 * s), $3$, name: "3a")
|
||||
content((3 * s, 0.5 * s), $4$, name: "4a")
|
||||
|
||||
content((0 * s, -2 * s), $2$, name: "2b")
|
||||
content((1 * s, -2 * s), $3$, name: "3b")
|
||||
content((2 * s, -2 * s), $1$, name: "1b")
|
||||
content((3 * s, -2 * s), $4$, name: "4b")
|
||||
|
||||
|
||||
// These arrows are wrong,
|
||||
// but create a symmetric picture
|
||||
markline(s, "1a", "1b")
|
||||
markline(s, "2a", "3b")
|
||||
markline(s, "3a", "2b")
|
||||
markline(s, "4a", "4b", c: ogreen)
|
||||
|
||||
content(
|
||||
(1 * s, -0.55 * s),
|
||||
$S_3$,
|
||||
fill: white,
|
||||
stroke: oblue + 0.6mm,
|
||||
padding: 1.3mm,
|
||||
)
|
||||
})),
|
||||
)
|
||||
|
||||
|
||||
#problem()
|
||||
Show that $S_3$ is a subgroup of $S_4$.
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
#definition("Isomorphism")
|
||||
Let $G$ and $H$ be groups. We say that $G$ and $H$ are _isomorphic_ (and write $G tilde.equiv H$) \
|
||||
if there is a bijection $f: G -> H$ with the following properties:
|
||||
- $f(e_G) = e_H$, where $e_G$ is the identity in $G$
|
||||
- $f(x^(-1)) = f(x)^(-1)$ for all $x$ in $G$
|
||||
- $f(x y) = f(x) f(y)$ for all $x, y$ in $G$
|
||||
|
||||
Intuitively, you can think of isomorphism as a form of equivalence. \
|
||||
If two groups are isomorphic, they only differ by the names of their elements. \
|
||||
The function $f$ above tells us how to map one set of labels to the other.
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
Show that $ZZ_7^times$ and $ZZ_9^times$ are isomorphic.
|
||||
#hint[
|
||||
Build a bijection with the above properties. \
|
||||
Remember that a group is fully defined by its multiplication table.
|
||||
]
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
Show that $ZZ_10^times$, $ZZ_5^times$, and $ZZ_4$ are isomorphic.
|
||||
#hint[
|
||||
Build a bijection with the above properties. \
|
||||
Remember that a group is fully defined by its multiplication table.
|
||||
]
|
||||
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Show that isomorphism is transitive. \
|
||||
That is, if $A tilde.equiv B$ and $B tilde.equiv C$, then $A tilde.equiv C$.
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
#problem(label: "firstindex")
|
||||
How many subgroups of $S_4$ are isomorphic to $S_3$? \
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
What are the orders of $S_3$ and $S_4$? \
|
||||
How is this related to @firstindex?
|
||||
|
||||
#solution[
|
||||
$|S_4| = |S_3| times [S_4 : S_3]$
|
||||
|
||||
#v(2mm)
|
||||
|
||||
This solution is written using index notation, \
|
||||
but the class doesn't need to know what it means yet.
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
$S_4$ also has $S_2$ and the trivial group as subgroups. \
|
||||
How many instances of each does $S_4$ contain?
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
$(ZZ_4, +)$ is also a subgroup of $S_4$. Find it! \
|
||||
How many subgroups of $ZZ_4$ are isomorphic to $S_4$?
|
||||
|
||||
#solution[
|
||||
A good hint is "look at generators."
|
||||
|
||||
#v(4mm)
|
||||
|
||||
There are four instances of $ZZ_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \
|
||||
(i.e, the group generated by $(1234)$ is isomorphic to $ZZ_4$)
|
||||
]
|
||||
|
||||
#v(1fr)
|
||||
@ -1,536 +0,0 @@
|
||||
|
||||
\section{Cycle Notation}
|
||||
|
||||
\definition{Order}
|
||||
The \textit{order} of a permutation $f$ is the \textbf{smallest} positive $n$ so that $f^n(x) = x$ for all $x$. \par
|
||||
If we repeatedly apply a permutation with order $n$, we will get back to where we started after $n$ steps. \par
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
For example, consider $[2134]$. This permutation has order $2$, as we clearly see below:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1a) at (0, 0.5) {1};
|
||||
\node (2a) at (1, 0.5) {2};
|
||||
\node (3a) at (2, 0.5) {3};
|
||||
\node (4a) at (3, 0.5) {4};
|
||||
|
||||
\node (2b) at (0, -2) {2};
|
||||
\node (1b) at (1, -2) {1};
|
||||
\node (3b) at (2, -2) {3};
|
||||
\node (4b) at (3, -2) {4};
|
||||
|
||||
\node (1c) at (0, -4.5) {1};
|
||||
\node (2c) at (1, -4.5) {2};
|
||||
\node (3c) at (2, -4.5) {3};
|
||||
\node (4c) at (3, -4.5) {4};
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
\line{4a}{4b}
|
||||
\line{1b}{1c}
|
||||
\line{2b}{2c}
|
||||
\line{3b}{3c}
|
||||
\line{4b}{4c}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
Swapping the first two elements of a list twice changes nothing. \par
|
||||
Thus, $[2134]$ has an order of two.
|
||||
|
||||
|
||||
\problem{}
|
||||
What is the order of $[2314]$? \par
|
||||
How about $[4321]$? \par
|
||||
\note[Note]{You shouldn't need to draw any strings to solve this problem.}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Show that all permutations (on a finite set) have a well-defined order. \par
|
||||
In other words, show that there is always an integer $n$ so that $f^n(x) = x$.
|
||||
|
||||
\vfill
|
||||
|
||||
\definition{Composition}<compdef>
|
||||
The \textit{composition} of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \par
|
||||
We'll denote this as $fg$---that is, by simply writing the permutations we're composing next to each other.
|
||||
|
||||
\problem{}
|
||||
Show that function composition is associative. \par
|
||||
That is, show that $f(gh) = (fg)h$.
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
What is $[1324][4321]$? \par
|
||||
How about $[321][213][231]$? \par
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
|
||||
Permutations are verbs---but we've been referring to them using a noun (namely, their output when
|
||||
applied to an ordered sequence of numbers). Our notation fails to capture the meaning of the
|
||||
underlying object.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Think about it: is the permutation $[1234]$ different than the permutation $[12345]$? \par
|
||||
Indeed, these permutations operate on different sets---but they are both the identity! \par
|
||||
What should we do if we want to talk about the identity on $\{1, 2, ..., 10\}$?
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
We need something better.
|
||||
|
||||
|
||||
|
||||
|
||||
\definition{Cycles}
|
||||
Any permutation is composed of a number of \textit{cycles}. \par
|
||||
|
||||
For example, consider the permutation $[2134]$, which consists of one two-cycle: $1 \to 2 \to 1$ \par
|
||||
\note[Note]{$3 \to 3$ and $4 \to 4$ are also cycles, but we'll ignore them. One-cycles aren't aren't interesting.}
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1) at (0, 0) {1};
|
||||
\node (2) at (1, 0) {2};
|
||||
\node (3) at (2, 0) {3};
|
||||
\node (4) at (3, 0) {4};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(1)
|
||||
-- ($(1) + (0,-1)$)
|
||||
-- ($(2) + (0,-1)$)
|
||||
-- (2);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(2)
|
||||
-- ($(2) + (0, 1)$)
|
||||
-- ($(1) + (0, 1)$)
|
||||
-- (1);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
The permutation $[431265]$ is a bit more interesting---it contains two cycles: \par
|
||||
($1 \to 3 \to 2 \to 4 \to 1$ and $5 \to 6 \to 5$)
|
||||
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1) at (0, 0) {1};
|
||||
\node (2) at (1, 0) {2};
|
||||
\node (3) at (2, 0) {3};
|
||||
\node (4) at (3, 0) {4};
|
||||
\node (5) at (4, 0) {5};
|
||||
\node (6) at (5, 0) {6};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(3)
|
||||
-- ($(3) + (0,-1)$)
|
||||
-- ($(2) + (0,-1)$)
|
||||
-- (2);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(2)
|
||||
-- ($(2) + (0,1.5)$)
|
||||
-- ($(4) + (0,1.5)$)
|
||||
-- (4);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(4)
|
||||
-- ($(4) + (0,-1.5)$)
|
||||
-- ($(1) + (0,-1.5)$)
|
||||
-- (1);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(1)
|
||||
-- ($(1) + (0,1)$)
|
||||
-- ($(3) + (0,1)$)
|
||||
-- (3);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(5)
|
||||
-- ($(5) + (0,-1)$)
|
||||
-- ($(6) + (0,-1)$)
|
||||
-- (6);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(6)
|
||||
-- ($(6) + (0,1)$)
|
||||
-- ($(5) + (0,1)$)
|
||||
-- (5);
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
Another name we'll often use for two-cycles is \textit{transposition}. \par
|
||||
Any permutation that swaps two adjacent elements is called a transposition. \par
|
||||
|
||||
|
||||
\problem{}
|
||||
Find all cycles in $[5342761]$.
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1) at (0, 0) {1};
|
||||
\node (2) at (1, 0) {2};
|
||||
\node (3) at (2, 0) {3};
|
||||
\node (4) at (3, 0) {4};
|
||||
\node (5) at (4, 0) {5};
|
||||
\node (6) at (5, 0) {6};
|
||||
\node (7) at (6, 0) {7};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(1)
|
||||
-- ($(1) + (0,2)$)
|
||||
-- ($(7) + (0,2)$)
|
||||
-- (7);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(7)
|
||||
-- ($(7) + (0,-1.5)$)
|
||||
-- ($(5) + (0,-1.5)$)
|
||||
-- (5);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(5)
|
||||
-- ($(5) + (0,1.5)$)
|
||||
-- ($(1) + (0.5,1.5)$)
|
||||
-- ($(1) + (0.5,-1)$)
|
||||
-- ($(1) + (0,-1)$)
|
||||
-- (1);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(2)
|
||||
-- ($(2) + (0,-1.5)$)
|
||||
-- ($(4) + (0,-1.5)$)
|
||||
-- (4);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(4)
|
||||
-- ($(4) + (0,1)$)
|
||||
-- ($(3) + (0,1)$)
|
||||
-- (3);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(3)
|
||||
-- ($(3) + (0,-1)$)
|
||||
-- ($(2) + (0.5,-1)$)
|
||||
-- ($(2) + (0.5,1)$)
|
||||
-- ($(2) + (0,1)$)
|
||||
-- (2);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
What permutation (on five objects) is formed by the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?
|
||||
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1) at (0, 0) {1};
|
||||
\node (2) at (1, 0) {2};
|
||||
\node (3) at (2, 0) {3};
|
||||
\node (4) at (3, 0) {4};
|
||||
\node (5) at (4, 0) {5};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(3)
|
||||
-- ($(3) + (0,1)$)
|
||||
-- ($(5) + (0,1)$)
|
||||
-- (5);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(5)
|
||||
-- ($(5) + (0,-1)$)
|
||||
-- ($(3) + (0,-1)$)
|
||||
-- (3);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(1)
|
||||
-- ($(1) + (0,-1)$)
|
||||
-- ($(2) + (0,-1)$)
|
||||
-- (2);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(2)
|
||||
-- ($(2) + (0,1.5)$)
|
||||
-- ($(4) + (0,1.5)$)
|
||||
-- (4);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(4)
|
||||
-- ($(4) + (0,-1.5)$)
|
||||
-- ($(1) + (0.5,-1.5)$)
|
||||
-- ($(1) + (0.5,1)$)
|
||||
-- ($(1) + (0,1)$)
|
||||
-- (1);
|
||||
\end{tikzpicture}
|
||||
|
||||
This is $[41523]$
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\definition{Cycle Notation}
|
||||
We now have a solution to our problem of notation.
|
||||
Instead of referring to permutations using their output, we will refer to them using their \textit{cycles}.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
For example, we'll write $[2134]$ as $(12)$, which denotes the cycle $1 \to 2 \to 1$:
|
||||
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1) at (0, 0) {1};
|
||||
\node (2) at (1, 0) {2};
|
||||
\node (3) at (2, 0) {3};
|
||||
\node (4) at (3, 0) {4};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(1)
|
||||
-- ($(1) + (0,-1)$)
|
||||
-- ($(2) + (0,-1)$)
|
||||
-- (2);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(2)
|
||||
-- ($(2) + (0, 1)$)
|
||||
-- ($(1) + (0, 1)$)
|
||||
-- (1);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
|
||||
As another example, $[431265]$ is $(1324)(56)$ in cycle notation. \par
|
||||
Note that we write $[431265]$ as a \textit{composition} of two cycles: \par
|
||||
applying the permutation $[431265]$ is the same as applying $(1324)$, then applying $(56)$.
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1) at (0, 0) {1};
|
||||
\node (2) at (1, 0) {2};
|
||||
\node (3) at (2, 0) {3};
|
||||
\node (4) at (3, 0) {4};
|
||||
\node (5) at (4, 0) {5};
|
||||
\node (6) at (5, 0) {6};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(3)
|
||||
-- ($(3) + (0,-1)$)
|
||||
-- ($(2) + (0,-1)$)
|
||||
-- (2);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(2)
|
||||
-- ($(2) + (0,1.5)$)
|
||||
-- ($(4) + (0,1.5)$)
|
||||
-- (4);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(4)
|
||||
-- ($(4) + (0,-1.5)$)
|
||||
-- ($(1) + (0,-1.5)$)
|
||||
-- (1);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ocyan]
|
||||
(1)
|
||||
-- ($(1) + (0,1)$)
|
||||
-- ($(3) + (0,1)$)
|
||||
-- (3);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(5)
|
||||
-- ($(5) + (0,-1)$)
|
||||
-- ($(6) + (0,-1)$)
|
||||
-- (6);
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(6)
|
||||
-- ($(6) + (0,1)$)
|
||||
-- ($(5) + (0,1)$)
|
||||
-- (5);
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Any permutation $\sigma$ may be written as a product (i.e, composition) of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
|
||||
Make sure you believe this fact. If you don't, ask an instructor. \par
|
||||
Also, the identity $f(x) = x$ is written as $()$ in cycle notation.
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Convince yourself that disjoint cycles commute. \par
|
||||
That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \par
|
||||
|
||||
|
||||
|
||||
|
||||
\problem{}<insquare>
|
||||
Write the following in square-bracket notation.
|
||||
\begin{itemize}
|
||||
\item $(12)$ \tab~\tab on a set of 2 elements
|
||||
\item $(12)(435)$ \tab on a set of 5 elements
|
||||
\vspace{2mm}
|
||||
\item $(321)$ \tab~\tab on a set of 3 elements
|
||||
\item $(321)$ \tab~\tab on a set of 6 elements
|
||||
\vspace{2mm}
|
||||
\item $(1234)$ \tab on a set of 4 elements
|
||||
\item $(3412)$ \tab on a set of 4 elements
|
||||
\end{itemize}
|
||||
\note{
|
||||
Note that $(12)$ refers the \say{swap first two} permutation on a set of \textit{any} size. \\
|
||||
We can now use the same name for the same permutation on two different sets! \\
|
||||
}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Write the following in square-bracket notation.
|
||||
Be careful.
|
||||
\begin{itemize}
|
||||
\item $(13)(243)$ \tab on a set of 4 elements
|
||||
\item $(243)(13)$ \tab on a set of 4 elements
|
||||
\end{itemize}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Look at the last two permutations in \ref{insquare}, $(1234)$ and $(3412)$. \par
|
||||
These are \textit{identical}---they are the same cycle written in two different ways. \par
|
||||
List all other ways to write this cycle. \hint{There are two more.} \par
|
||||
\note{Also, note that the last two permutations in \ref{insquare} are the same.}
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
\problem{}
|
||||
What is the inverse of $(12)$? \par
|
||||
How about $(123)$? And $(4231)$? \par
|
||||
\note{
|
||||
Note that again, we don't need to know how big our set is. \\
|
||||
The inverse of $(12)$ is the same in all sets.
|
||||
}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
Say $\sigma$ is a permutation composed of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
|
||||
Say we know the order of all $\sigma_i$. What is the order of $\sigma$?
|
||||
|
||||
\begin{solution}
|
||||
$\text{lcm}\Bigl(\text{ord}(\sigma_1),~ \text{ord}(\sigma_2),~ ..., ~ \text{ord}(\sigma_k)\Bigr)$
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}<cycletrans>
|
||||
Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
|
||||
|
||||
\begin{solution}
|
||||
TODO
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Write $(7126453)$ as a product of transpositions. \par
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}<simpletrans>
|
||||
Show that any permutation is a product of transpositions.
|
||||
|
||||
\begin{solution}
|
||||
Use \ref{cycletrans}.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that any permutation is a product of transpositions of the form $(1, k)$. \par
|
||||
|
||||
\begin{solution}
|
||||
Use \ref{simpletrans} and rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$. \par
|
||||
Showing that $(a, b) = (1, a)(1, b)(1, a)$ is fairly easy.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.
|
||||
|
||||
\begin{solution}
|
||||
This is the same as the $(1, a)(1, b)(1, a)$ case above, but we use $a + 1$
|
||||
as a \say{working slot} instead of $1$.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that any permutation is a product of adjacent transpositions. \par
|
||||
(An \textit{adjacent transposition} swaps two adjacent elements, and thus looks like $(n, n+1)$)
|
||||
|
||||
\begin{solution}
|
||||
As before, we will use \ref{simpletrans} and rewrite the transpositions it produces in a form that fits the problem.
|
||||
We thus need to show that every transposition $(a, b)$ is a product of adjacent transpositions.
|
||||
|
||||
\vspace{8mm}
|
||||
|
||||
In the proof below, assume that $a < b$ and perform induction on $b - a$. \par
|
||||
|
||||
\textbf{Base Case:}\par
|
||||
If $b - a = 1$, we clearly see that $(a, b)$ is a product of adjacent. \par
|
||||
In fact, it \textit{is} an adjacent transposition.
|
||||
|
||||
\vspace{4mm}
|
||||
|
||||
\textbf{Induction:}\par
|
||||
Now, say $b - a = n + 1$. \par
|
||||
Assume that all $(a, b)$ where $b - a \leq n$ are products of adjacent transpositions.\par
|
||||
Note that $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
$(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \par
|
||||
Thus, $(a, b)$ is a product of adjacent transpositions.
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
@ -1,165 +0,0 @@
|
||||
\section{Groups (review)}
|
||||
|
||||
\definition{}
|
||||
Before we continue, we must introduce a bit of notation:
|
||||
\begin{itemize}
|
||||
\item $S_n$ is the set of permutations on $n$ objects.
|
||||
\item $\mathbb{Z}_n$ is the set of integers mod $n$.
|
||||
|
||||
\item $\mathbb{Z}_n^\times$ is the set of integers mod $n$ with multiplicative inverses. \par
|
||||
In other words, it is the set of integers smaller than $n$ and coprime to $n$.\footnotemark{} \par
|
||||
For example, $\mathbb{Z}_{12}^\times = \{1, 5, 7, 11\}$.
|
||||
|
||||
\footnotetext{We proved this in another handout, but you may take it as fact here.}
|
||||
\end{itemize}
|
||||
|
||||
\problem{}
|
||||
What are the elements of $S_3$? \tab\hint{Use cycle notation}\par
|
||||
How about $\mathbb{Z}_{17}^\times$?
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\definition{}
|
||||
A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \par
|
||||
Groups always have the following properties:
|
||||
|
||||
\begin{enumerate}
|
||||
\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
|
||||
\item $\ast$ is \textit{associative}: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
|
||||
\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
|
||||
\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = b \ast a = e$. $b$ is called the \textit{inverse} of $a$. \par
|
||||
This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
|
||||
\end{enumerate}
|
||||
|
||||
Any pair $(G, \ast)$ that satisfies these properties is a group.
|
||||
|
||||
\problem{}
|
||||
Is $(\mathbb{Z}_5, +)$ a group? \par
|
||||
Is $(\mathbb{Z}_5, -)$ a group? \par
|
||||
\note[Note]{$+$ and $-$ refer to the usual operations in modular arithmetic.}
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
What is the group with the fewest elements?
|
||||
|
||||
\begin{solution}
|
||||
Let $(G, \star)$ be our group, where $G = \{x\}$ and $\star$ is defined by $x \star x = x$
|
||||
|
||||
Verifying that the trivial group is a group is trivial.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that function composition is associative
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that $S_n$ is a group under composition.
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Let $(G, \ast)$ be a group with finitely many elements, and let $a \in G$. \par
|
||||
Show that $\exists n \in \mathbb{Z}^+$ so that $a^n = e$ \par
|
||||
\hint{$a^n = a \ast a \ast ... \ast a$ repeated $n$ times.}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
The smallest such $n$ defines the \textit{order} of $g$.
|
||||
|
||||
\begin{examplesolution}
|
||||
We've already done a special case of this problem! \par
|
||||
Find it in this handout, then rewrite your proof for an arbitrary (finite) group.
|
||||
\end{examplesolution}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
What is the order of 5 in $(\mathbb{Z}_{25}, +)$? \par
|
||||
What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \par
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\definition{}<gendef>
|
||||
Let $G$ be a group, and let $g$ be an element of $G$. \par
|
||||
We say $g$ is a \textit{generator} if every other element of $G$ may be written as a power of $g$. \par
|
||||
|
||||
\problem{}
|
||||
Say the size of a group $G$ is $n$. \par
|
||||
If $g$ is a generator, what is its order? \par
|
||||
Provide a proof.
|
||||
\vfill
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Find the two generators in $(\mathbb{Z}, +)$ \par
|
||||
Then, find all generators of $(\mathbb{Z}_5, +)$
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
How many groups have only one generator?
|
||||
|
||||
\begin{solution}
|
||||
Only one: the trivial group. The inverse of a generator is also a generator!
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\definition{}
|
||||
Let $S$ be a subset of the elements in $G$. \par
|
||||
We say that $S$ \textit{generates} $G$ if every element of $G$ may be written as a product of elements in $S$. \par
|
||||
\note{Note that this is an extension of \ref{gendef}.}
|
||||
|
||||
\problem{}
|
||||
We've already found a few generating sets of $S_n$. What are they?
|
||||
|
||||
\begin{solution}
|
||||
The following sets generate $S_n$:
|
||||
\begin{itemize}
|
||||
\item All transpositions
|
||||
\item All transpositions of the form $(1, k)$
|
||||
\item All adjacent transpositions
|
||||
\end{itemize}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
The smallest generating set of $S_n$ consists of the transposition $(12)$ and the $n$-cycle $(1,2,...,n)$. \par
|
||||
The proof of this is a bonus problem later in the handout.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
@ -1,163 +0,0 @@
|
||||
\section{Subgroups}
|
||||
|
||||
\problem{}<s2s3share>
|
||||
What elements do $S_2$ and $S_3$ share?
|
||||
\vspace{2cm}
|
||||
|
||||
|
||||
|
||||
Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} \subset \{1, 2, 3\}$. \par
|
||||
Can we say something similar about $S_2$ and $S_3$?
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Looking at \ref{s2s3share}, we may want to say that $S_2 \subset S_3$ since every element of $S_2$ is in $S_3$. \par
|
||||
This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are \textit{groups}, not \textit{sets}: \par
|
||||
their elements come with structure, which the \say{subset} relation does not capture.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
To account for this, we'll define a similar relation: subgroups.
|
||||
|
||||
\definition{}
|
||||
Let $G$ and $G'$ be groups. We say $G'$ is a \textit{subgroup} of $G$ (and write $G' \subset G$) if the following are true:\par
|
||||
(Note that $x, y$ are elements of $G$, and $xy$ is multiplication in $G$)
|
||||
\begin{itemize}
|
||||
\item the set of elements in $G'$ is a subset of the set of elements in $G$.
|
||||
\item the identity of $G$ is in $G'$
|
||||
\item $x,y \in G' \implies xy \in G'$
|
||||
\item $x \in G' \implies x^{-1} \in G'$
|
||||
\end{itemize}
|
||||
|
||||
The above definition may look faily scary, but the idea behind a subgroup is simple. \par
|
||||
Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \par
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Say we have a set of four elements and only look at the first three. \par
|
||||
$S_3$ fully describes all the ways we can arrange those three elements:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.5]
|
||||
\node (1a) at (0, 0.5) {1};
|
||||
\node (2a) at (1, 0.5) {2};
|
||||
\node (3a) at (2, 0.5) {3};
|
||||
\node (4a) at (3, 0.5) {4};
|
||||
|
||||
\node (2b) at (0, -2) {2};
|
||||
\node (3b) at (1, -2) {3};
|
||||
\node (1b) at (2, -2) {1};
|
||||
\node (4b) at (3, -2) {4};
|
||||
|
||||
\draw[line width = 0.3mm, ->, ogreen]
|
||||
(4a)
|
||||
-- ($(4a) + (0, -1)$)
|
||||
-- ($(4b) + (0,1)$)
|
||||
-- (4b);
|
||||
|
||||
\line{1a}{1b}
|
||||
\line{2a}{2b}
|
||||
\line{3a}{3b}
|
||||
|
||||
\node[fill=white,draw=oblue,line width=0.3mm] at (1, -0.75) {$S_3$};
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that $S_3$ is a subgroup of $S_4$.
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
\definition{}
|
||||
Let $G$ and $H$ be groups. We say that $G$ and $H$ are \textit{isomorphic} (and write $A \simeq B$) \par
|
||||
if there is a bijection $f: G \to H$ with the following properties:
|
||||
\begin{itemize}
|
||||
\item $f(e_G) = e_H$, where $e_G$ is the identity in $G$
|
||||
\item $f(x^{-1}) = f(x)^{-1}$ for all $x$ in $G$
|
||||
\item $f(xy) = f(x)f(y)$ for all $x, y$ in $G$
|
||||
\end{itemize}
|
||||
|
||||
Intuitively, you can think of isomorphism as a form of equivalence. \par
|
||||
If two groups are isomorphic, they only differ by the names of their elements. \par
|
||||
The function $f$ above tells us how to map one set of labels to the other.
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that $\mathbb{Z}_7^\times$ and $\mathbb{Z}_9^\times$ are isomorphic.
|
||||
\hint{
|
||||
Build a bijection with the above properties. \\
|
||||
Remember that a group is fully defined by its multiplication table.
|
||||
}
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
Show that $\mathbb{Z}_{10}^\times$ and $\mathbb{Z}_5^\times$, and $\mathbb{Z}_4$ are isomorphic.
|
||||
\hint{
|
||||
Build a bijection with the above properties. \\
|
||||
Remember that a group is fully defined by its multiplication table.
|
||||
}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Show that isomorphism is transitive. \par
|
||||
That is, if $A \simeq B$ and $B \simeq C$, then $A \simeq C$.
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
\problem{}<firstindex>
|
||||
How many subgroups of $S_4$ are isomorphic to $S_3$? \par
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
What are the orders of $S_3$ and $S_4$? \par
|
||||
How is this related to \ref{firstindex}?
|
||||
|
||||
\begin{solution}
|
||||
$|S_4| = |S_3| \times [S_4 : S_3]$
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
This solution is written using index notation, \par
|
||||
but the class doesn't need to know what it means yet.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
$S_4$ also has $S_2$ and the trivial group as subgroups. \par
|
||||
How many instances of each does $S_4$ contain?
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
$(\mathbb{Z}_4, +)$ is also a subgroup of $S_4$. Find it! \par
|
||||
How many subgroups of $\mathbb{Z}_4$ are isomorphic to $S_4$?.
|
||||
|
||||
\begin{solution}
|
||||
A good hint is \say{look at generators.}
|
||||
|
||||
\vspace{4mm}
|
||||
|
||||
There are four instances of $\mathbb{Z}_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \par
|
||||
(i.e, the group generated by $(1234)$ is isomorphic to $\mathbb{Z}_4$)
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
|
||||
// Shorthand, we'll be using these a lot.
|
||||
@ -7,35 +7,31 @@
|
||||
#let tm = sym.times.circle
|
||||
|
||||
#let graphgrid(inner_content) = {
|
||||
align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let x = 5.25
|
||||
align(center, box(inset: 3mm, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let x = 5.25
|
||||
|
||||
grid(
|
||||
(0, 0), (x, x), step: 0.75,
|
||||
stroke: luma(100) + 0.3mm
|
||||
)
|
||||
grid(
|
||||
(0, 0),
|
||||
(x, x),
|
||||
step: 0.75,
|
||||
stroke: luma(100) + 0.3mm,
|
||||
)
|
||||
|
||||
if (inner_content != none) {
|
||||
inner_content
|
||||
}
|
||||
if (inner_content != none) {
|
||||
inner_content
|
||||
}
|
||||
|
||||
mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
|
||||
mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
|
||||
mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
|
||||
mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
|
||||
|
||||
line(
|
||||
(0, x + 0.25),
|
||||
(0, 0),
|
||||
(x + 0.25, 0),
|
||||
stroke: 0.75mm + black,
|
||||
)
|
||||
}),
|
||||
),
|
||||
)
|
||||
line(
|
||||
(0, x + 0.25),
|
||||
(0, 0),
|
||||
(x + 0.25, 0),
|
||||
stroke: 0.75mm + black,
|
||||
)
|
||||
})))
|
||||
}
|
||||
|
||||
/// Adds extra padding to an equation.
|
||||
@ -48,23 +44,16 @@
|
||||
/// Note that there are newlines between the $ and content,
|
||||
/// this gives us display math (which is what we want when using this macro)
|
||||
#let eqnbox(eqn) = {
|
||||
align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
eqn,
|
||||
),
|
||||
)
|
||||
align(center, box(
|
||||
inset: 3mm,
|
||||
eqn,
|
||||
))
|
||||
}
|
||||
|
||||
#let dotline(a, b) = {
|
||||
cetz.draw.line(
|
||||
a,
|
||||
b,
|
||||
stroke: (
|
||||
dash: "dashed",
|
||||
thickness: 0.5mm,
|
||||
paint: ored,
|
||||
),
|
||||
)
|
||||
cetz.draw.line(a, b, stroke: (
|
||||
dash: "dashed",
|
||||
thickness: 0.5mm,
|
||||
paint: ored,
|
||||
))
|
||||
}
|
||||
|
||||
@ -26,9 +26,9 @@ $
|
||||
|
||||
#solution([
|
||||
- Is tropical addition commutative?\
|
||||
Yes, $min(min(x,y),z) = min(x,y,z) = min(x,min(y,z))$
|
||||
Yes, $min(min(x, y), z) = min(x, y, z) = min(x, min(y, z))$
|
||||
- Is tropical addition associative? \
|
||||
Yes, $min(x,y) = min(y,x)$
|
||||
Yes, $min(x, y) = min(y, x)$
|
||||
- Is there a tropical additive identity? \
|
||||
No. There is no $n$ where $x <= n$ for all real $x$
|
||||
])
|
||||
@ -117,7 +117,7 @@ Do tropical multiplicative inverses always exist? \
|
||||
Is tropical multiplication distributive over addition? \
|
||||
#note([Does $x #tm (y #tp z) = x #tm y #tp x #tm z$?])
|
||||
|
||||
#solution([Yes, $x + min(y,z) = min(x+y, x+z)$])
|
||||
#solution([Yes, $x + min(y, z) = min(x+y, x+z)$])
|
||||
|
||||
#v(1fr)
|
||||
|
||||
@ -134,14 +134,7 @@ Fill the following tropical addition and multiplication tables
|
||||
table(
|
||||
columns: (col, col, col, col, col, col),
|
||||
align: center,
|
||||
table.header(
|
||||
[$#tp$],
|
||||
[$1$],
|
||||
[$2$],
|
||||
[$3$],
|
||||
[$4$],
|
||||
[$#sym.infinity$],
|
||||
),
|
||||
table.header([$#tp$], [$1$], [$2$], [$3$], [$4$], [$#sym.infinity$]),
|
||||
|
||||
box(inset: 3pt, $1$), [], [], [], [], [],
|
||||
box(inset: 3pt, $2$), [], [], [], [], [],
|
||||
@ -152,14 +145,7 @@ Fill the following tropical addition and multiplication tables
|
||||
table(
|
||||
columns: (col, col, col, col, col, col),
|
||||
align: center,
|
||||
table.header(
|
||||
[$#tm$],
|
||||
[$0$],
|
||||
[$1$],
|
||||
[$2$],
|
||||
[$3$],
|
||||
[$4$],
|
||||
),
|
||||
table.header([$#tm$], [$0$], [$1$], [$2$], [$3$], [$4$]),
|
||||
|
||||
box(inset: 3pt, $0$), [], [], [], [], [],
|
||||
box(inset: 3pt, $1$), [], [], [], [], [],
|
||||
@ -178,14 +164,7 @@ Fill the following tropical addition and multiplication tables
|
||||
table(
|
||||
columns: (col, col, col, col, col, col),
|
||||
align: center,
|
||||
table.header(
|
||||
[$#tp$],
|
||||
[$1$],
|
||||
[$2$],
|
||||
[$3$],
|
||||
[$4$],
|
||||
[$#sym.infinity$],
|
||||
),
|
||||
table.header([$#tp$], [$1$], [$2$], [$3$], [$4$], [$#sym.infinity$]),
|
||||
|
||||
box(inset: 3pt, $1$),
|
||||
box(inset: 3pt, $1$),
|
||||
@ -225,14 +204,7 @@ Fill the following tropical addition and multiplication tables
|
||||
table(
|
||||
columns: (col, col, col, col, col, col),
|
||||
align: center,
|
||||
table.header(
|
||||
[$#tm$],
|
||||
[$0$],
|
||||
[$1$],
|
||||
[$2$],
|
||||
[$3$],
|
||||
[$4$],
|
||||
),
|
||||
table.header([$#tm$], [$0$], [$1$], [$2$], [$3$], [$4$]),
|
||||
|
||||
box(inset: 3pt, $0$),
|
||||
box(inset: 3pt, $0$),
|
||||
@ -281,10 +253,9 @@ Adjacent parenthesis imply tropical multiplication
|
||||
|
||||
#solution([
|
||||
$
|
||||
(x #tp 2)(x #tp 3)
|
||||
&= x^2 #tp 2x #tp 3x #tp (2 #tm 3) \
|
||||
&= x^2 #tp (2 #tp 3)x #tp (2 #tm 3) \
|
||||
&= x^2 #tp 2x #tp 5
|
||||
(x #tp 2)(x #tp 3) & = x^2 #tp 2x #tp 3x #tp (2 #tm 3) \
|
||||
& = x^2 #tp (2 #tp 3)x #tp (2 #tm 3) \
|
||||
& = x^2 #tp 2x #tp 5
|
||||
$
|
||||
|
||||
Also, $f(1) = 2$ and $f(4) = 5$.
|
||||
|
||||
@ -1,21 +1,18 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "../macros.typ": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
= Tropical Polynomials
|
||||
|
||||
#definition()
|
||||
A _polynomial_ is an expression formed by adding and multiplying numbers and a variable $x$. \
|
||||
Every polynomial can be written as
|
||||
#align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
$
|
||||
c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
|
||||
$,
|
||||
),
|
||||
)
|
||||
#align(center, box(
|
||||
inset: 3mm,
|
||||
$
|
||||
c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
|
||||
$,
|
||||
))
|
||||
for some nonnegative integer $n$ and coefficients $c_0, c_1, ..., c_n$. \
|
||||
The _degree_ of a polynomial is the largest $n$ for which $c_n$ is nonzero.
|
||||
|
||||
@ -43,15 +40,12 @@ In this section, we will analyze tropical polynomials:
|
||||
#definition()
|
||||
A _tropical_ polynomial is a polynomial that uses tropical addition and multiplication. \
|
||||
In other words, it is an expression of the form
|
||||
#align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
$
|
||||
c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
|
||||
$,
|
||||
),
|
||||
)
|
||||
#align(center, box(
|
||||
inset: 3mm,
|
||||
$
|
||||
c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
|
||||
$,
|
||||
))
|
||||
where all exponents represent repeated tropical multiplication.
|
||||
|
||||
#pagebreak() // MARK: page
|
||||
@ -66,7 +60,7 @@ Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
$f(x) = min(2x , 1+x, 4)$, which looks like:
|
||||
$f(x) = min(2x, 1+x, 4)$, which looks like:
|
||||
|
||||
#graphgrid({
|
||||
import cetz.draw: *
|
||||
@ -90,15 +84,12 @@ Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
|
||||
#problem()
|
||||
Now, factor $f(x) = x^2 #tp 1x #tp 4$ into two polynomials with degree 1. \
|
||||
In other words, find $r$ and $s$ so that
|
||||
#align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
$
|
||||
x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
|
||||
$,
|
||||
),
|
||||
)
|
||||
#align(center, box(
|
||||
inset: 3mm,
|
||||
$
|
||||
x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
|
||||
$,
|
||||
))
|
||||
|
||||
we will call $r$ and $s$ the _roots_ of $f$.
|
||||
|
||||
@ -159,15 +150,19 @@ Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
|
||||
#solution([
|
||||
We (tropically) factor out $-2$ to get
|
||||
|
||||
#eqnbox($
|
||||
f(x) = -2(x^2 #tp 2x #tp 10)
|
||||
$)
|
||||
#eqnbox(
|
||||
$
|
||||
f(x) = -2(x^2 #tp 2x #tp 10)
|
||||
$,
|
||||
)
|
||||
|
||||
|
||||
by the same process as the previous problem, we get
|
||||
#eqnbox($
|
||||
f(x) = -2(x #tp 2)(x #tp 8)
|
||||
$)
|
||||
#eqnbox(
|
||||
$
|
||||
f(x) = -2(x #tp 2)(x #tp 8)
|
||||
$,
|
||||
)
|
||||
])
|
||||
|
||||
#v(1fr)
|
||||
@ -236,11 +231,11 @@ Graph $f(x) = 1x^2 #tp 3x #tp 5$.
|
||||
#problem()
|
||||
Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
|
||||
|
||||
#solution(
|
||||
eqnbox($
|
||||
#solution(eqnbox(
|
||||
$
|
||||
f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2
|
||||
$),
|
||||
)
|
||||
$,
|
||||
))
|
||||
|
||||
#v(1fr)
|
||||
|
||||
@ -263,23 +258,21 @@ Graph $f(x) = 2x^2 #tp 4x #tp 4$.
|
||||
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution(
|
||||
graphgrid({
|
||||
import cetz.draw: *
|
||||
let step = 0.75
|
||||
#solution(graphgrid({
|
||||
import cetz.draw: *
|
||||
let step = 0.75
|
||||
|
||||
dotline((0, 2 * step), (3 * step, 8 * step))
|
||||
dotline((0, 4 * step), (5 * step, 8 * step))
|
||||
dotline((0, 4 * step), (8 * step, 4 * step))
|
||||
dotline((0, 2 * step), (3 * step, 8 * step))
|
||||
dotline((0, 4 * step), (5 * step, 8 * step))
|
||||
dotline((0, 4 * step), (8 * step, 4 * step))
|
||||
|
||||
line(
|
||||
(0, 2 * step),
|
||||
(1 * step, 4 * step),
|
||||
(7.5 * step, 4 * step),
|
||||
stroke: 1mm + oblue,
|
||||
)
|
||||
}),
|
||||
)
|
||||
line(
|
||||
(0, 2 * step),
|
||||
(1 * step, 4 * step),
|
||||
(7.5 * step, 4 * step),
|
||||
stroke: 1mm + oblue,
|
||||
)
|
||||
}))
|
||||
|
||||
|
||||
#problem()
|
||||
@ -325,7 +318,7 @@ Find a formula for $B$ in terms of $a$, $b$, and $c$. \
|
||||
|
||||
#solution([
|
||||
If we want to factor $a(x^2 #tp (b-a)x #tp (c-a))$, we need to find $r$ and $s$ so that
|
||||
- $min(r,s) = b-a$, and
|
||||
- $min(r, s) = b-a$, and
|
||||
- $r + s = c - a$
|
||||
|
||||
#v(2mm)
|
||||
@ -341,9 +334,8 @@ Find a formula for $B$ in terms of $a$, $b$, and $c$. \
|
||||
|
||||
*Case 2:* If $b > (a + c #sym.div) 2$, then
|
||||
$
|
||||
accent(f, macron)(x)
|
||||
&= a x^2 #tp ((a+c)/2)x #tp c \
|
||||
&= a(x #tp (c-a)/2)^2
|
||||
accent(f, macron)(x) & = a x^2 #tp ((a+c)/2)x #tp c \
|
||||
& = a(x #tp (c-a)/2)^2
|
||||
$
|
||||
has the same graph as $f$, and thus $B = (a+c) #sym.div 2$
|
||||
|
||||
|
||||
@ -1,6 +1,6 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "../macros.typ": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
= Tropical Cubic Polynomials
|
||||
|
||||
@ -131,15 +131,12 @@ Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b
|
||||
#problem()
|
||||
What are the roots of the following polynomial?
|
||||
|
||||
#align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
$
|
||||
3 x^6 #tp 4 x^5 #tp 2 x^4 #tp x^3 #tp x^2 #tp 4 x #tp 5
|
||||
$,
|
||||
),
|
||||
)
|
||||
#align(center, box(
|
||||
inset: 3mm,
|
||||
$
|
||||
3 x^6 #tp 4 x^5 #tp 2 x^4 #tp x^3 #tp x^2 #tp 4 x #tp 5
|
||||
$,
|
||||
))
|
||||
|
||||
#solution([
|
||||
We have
|
||||
@ -169,9 +166,8 @@ Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
|
||||
|
||||
#solution([
|
||||
$
|
||||
A_j
|
||||
&= min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
|
||||
&= min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
|
||||
A_j & = min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
|
||||
& = min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
|
||||
$
|
||||
|
||||
#v(2mm)
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
= Wallpaper Symmetries
|
||||
|
||||
@ -12,7 +12,9 @@ There are four classes of Euclidean isometries:
|
||||
- reflections
|
||||
- rotations
|
||||
- glide reflections
|
||||
#note([We can prove there are no others, but this is beyond the scope of this handout.]) \
|
||||
#note(
|
||||
[We can prove there are no others, but this is beyond the scope of this handout.],
|
||||
) \
|
||||
A simple example of each isometry is shown below:
|
||||
|
||||
#let demo(c) = {
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
= Mirror Symmetry
|
||||
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
= Rotational Symmetry
|
||||
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#let pat(img, sol) = {
|
||||
problem()
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
= The Signature-Cost Theorem
|
||||
|
||||
@ -7,23 +7,20 @@
|
||||
First, we'll associate a _cost_ to each type of symmetry in orbifold notation:
|
||||
|
||||
#v(4mm)
|
||||
#align(
|
||||
center,
|
||||
table(
|
||||
stroke: (1pt, 1pt),
|
||||
align: center,
|
||||
columns: (auto, auto, auto, auto),
|
||||
[*Symbol*], [*Cost*], [*Symbol*], [*Cost*],
|
||||
[#sym.circle.small], [2], [#sym.times or #sym.convolve], [1],
|
||||
[#sym.diamond.stroked.small`2`], [1/2], [#sym.convolve`2`], [1/4],
|
||||
[#sym.diamond.stroked.small`3`], [2/3], [#sym.convolve`3`], [1/3],
|
||||
[#sym.dots], [#sym.dots], [#sym.dots], [#sym.dots],
|
||||
[#sym.diamond.stroked.small`n`],
|
||||
[$(n-1) / n$],
|
||||
[#sym.convolve`n`],
|
||||
[$(n-1) / (2n)$],
|
||||
),
|
||||
)
|
||||
#align(center, table(
|
||||
stroke: (1pt, 1pt),
|
||||
align: center,
|
||||
columns: (auto, auto, auto, auto),
|
||||
[*Symbol*], [*Cost*], [*Symbol*], [*Cost*],
|
||||
[#sym.circle.small], [2], [#sym.times or #sym.convolve], [1],
|
||||
[#sym.diamond.stroked.small`2`], [1/2], [#sym.convolve`2`], [1/4],
|
||||
[#sym.diamond.stroked.small`3`], [2/3], [#sym.convolve`3`], [1/3],
|
||||
[#sym.dots], [#sym.dots], [#sym.dots], [#sym.dots],
|
||||
[#sym.diamond.stroked.small`n`],
|
||||
[$(n-1) / n$],
|
||||
[#sym.convolve`n`],
|
||||
[$(n-1) / (2n)$],
|
||||
))
|
||||
|
||||
|
||||
We then calculate the total "cost" of a signature by adding up the costs of each component.
|
||||
|
||||
@ -46,10 +46,10 @@ Use two half adders to construct a full adder.
|
||||
|
||||
#solution([
|
||||
$
|
||||
s_1, c_1 &= "HA"(a, b) \
|
||||
s_2, c_2 &= "HA"(s_1, c_"in") \
|
||||
s_"out" &= s_2 \
|
||||
c_"out" &= "OR"(c_1, c_2)
|
||||
s_1, c_1 & = "HA"(a, b) \
|
||||
s_2, c_2 & = "HA"(s_1, c_"in") \
|
||||
s_"out" & = s_2 \
|
||||
c_"out" & = "OR"(c_1, c_2)
|
||||
$
|
||||
|
||||
#v(2mm)
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Big-Tac-Toe],
|
||||
@ -75,16 +75,13 @@ How does your strategy change? \
|
||||
#if extra_boards {
|
||||
pagebreak()
|
||||
|
||||
align(
|
||||
center,
|
||||
grid(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr),
|
||||
rows: (1fr, 1fr, 1fr),
|
||||
btt(0.35), btt(0.35),
|
||||
btt(0.35), btt(0.35),
|
||||
btt(0.35), btt(0.35),
|
||||
),
|
||||
)
|
||||
align(center, grid(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr),
|
||||
rows: (1fr, 1fr, 1fr),
|
||||
btt(0.35), btt(0.35),
|
||||
btt(0.35), btt(0.35),
|
||||
btt(0.35), btt(0.35),
|
||||
))
|
||||
}
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
|
||||
#show: handout.with(
|
||||
@ -43,61 +43,56 @@ Now, consider the set of six-sided dice below:
|
||||
- Die $E$: $0, 5, 5, 5, 5, 5$
|
||||
On average, which die beats each of the others? Draw a diagram.
|
||||
|
||||
#solution(
|
||||
align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
#solution(align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.8 // Scale
|
||||
let t = 13pt * s // text size
|
||||
let radius = 0.3 * s
|
||||
let s = 0.8 // Scale
|
||||
let t = 13pt * s // text size
|
||||
let radius = 0.3 * s
|
||||
|
||||
// Points
|
||||
let a = (-2 * s, 0.2 * s)
|
||||
let b = (0 * s, 2 * s)
|
||||
let c = (2 * s, 0.2 * s)
|
||||
let d = (1.2 * s, -2.1 * s)
|
||||
let e = (-1.2 * s, -2.1 * s)
|
||||
// Points
|
||||
let a = (-2 * s, 0.2 * s)
|
||||
let b = (0 * s, 2 * s)
|
||||
let c = (2 * s, 0.2 * s)
|
||||
let d = (1.2 * s, -2.1 * s)
|
||||
let e = (-1.2 * s, -2.1 * s)
|
||||
|
||||
set-style(
|
||||
stroke: (thickness: 0.6mm * s),
|
||||
mark: (
|
||||
end: (
|
||||
symbol: ">",
|
||||
fill: black,
|
||||
offset: radius + (0.025 * s),
|
||||
width: 1.2mm * s,
|
||||
length: 1.2mm * s,
|
||||
),
|
||||
),
|
||||
)
|
||||
set-style(
|
||||
stroke: (thickness: 0.6mm * s),
|
||||
mark: (
|
||||
end: (
|
||||
symbol: ">",
|
||||
fill: black,
|
||||
offset: radius + (0.025 * s),
|
||||
width: 1.2mm * s,
|
||||
length: 1.2mm * s,
|
||||
),
|
||||
),
|
||||
)
|
||||
|
||||
line(a, b)
|
||||
line(b, c)
|
||||
line(c, d)
|
||||
line(d, e)
|
||||
line(e, a)
|
||||
line(a, c)
|
||||
line(b, d)
|
||||
line(c, e)
|
||||
line(d, a)
|
||||
line(e, b)
|
||||
line(a, b)
|
||||
line(b, c)
|
||||
line(c, d)
|
||||
line(d, e)
|
||||
line(e, a)
|
||||
line(a, c)
|
||||
line(b, d)
|
||||
line(c, e)
|
||||
line(d, a)
|
||||
line(e, b)
|
||||
|
||||
circle(a, radius: radius, fill: oblue, stroke: none)
|
||||
circle(b, radius: radius, fill: oblue, stroke: none)
|
||||
circle(c, radius: radius, fill: oblue, stroke: none)
|
||||
circle(d, radius: radius, fill: oblue, stroke: none)
|
||||
circle(e, radius: radius, fill: oblue, stroke: none)
|
||||
circle(a, radius: radius, fill: oblue, stroke: none)
|
||||
circle(b, radius: radius, fill: oblue, stroke: none)
|
||||
circle(c, radius: radius, fill: oblue, stroke: none)
|
||||
circle(d, radius: radius, fill: oblue, stroke: none)
|
||||
circle(e, radius: radius, fill: oblue, stroke: none)
|
||||
|
||||
content(a, text(fill: white, size: t, [*A*]))
|
||||
content(b, text(fill: white, size: t, [*B*]))
|
||||
content(c, text(fill: white, size: t, [*C*]))
|
||||
content(d, text(fill: white, size: t, [*D*]))
|
||||
content(e, text(fill: white, size: t, [*E*]))
|
||||
}),
|
||||
),
|
||||
)
|
||||
content(a, text(fill: white, size: t, [*A*]))
|
||||
content(b, text(fill: white, size: t, [*B*]))
|
||||
content(c, text(fill: white, size: t, [*C*]))
|
||||
content(d, text(fill: white, size: t, [*D*]))
|
||||
content(e, text(fill: white, size: t, [*E*]))
|
||||
})))
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: What's an AST?],
|
||||
@ -18,59 +18,56 @@ You may detach the string as you hang the painting, but it must be re-attached o
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 2.5
|
||||
#align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 2.5
|
||||
|
||||
|
||||
line(
|
||||
(0 * s, 1 * s),
|
||||
(2 * s, 1 * s),
|
||||
(2 * s, 0 * s),
|
||||
(0 * s, 0 * s),
|
||||
close: true,
|
||||
stroke: (thickness: 0.8mm),
|
||||
)
|
||||
line(
|
||||
(0 * s, 1 * s),
|
||||
(2 * s, 1 * s),
|
||||
(2 * s, 0 * s),
|
||||
(0 * s, 0 * s),
|
||||
close: true,
|
||||
stroke: (thickness: 0.8mm),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.1 * s, 1 * s),
|
||||
(0.5 * s, 1.5 * s),
|
||||
(1.5 * s, 1.5 * s),
|
||||
(1.9 * s, 1 * s),
|
||||
stroke: (thickness: 0.5mm, dash: "dotted"),
|
||||
)
|
||||
line(
|
||||
(0.1 * s, 1 * s),
|
||||
(0.5 * s, 1.5 * s),
|
||||
(1.5 * s, 1.5 * s),
|
||||
(1.9 * s, 1 * s),
|
||||
stroke: (thickness: 0.5mm, dash: "dotted"),
|
||||
)
|
||||
|
||||
circle((0.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
|
||||
circle((1.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
|
||||
circle((0.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
|
||||
circle((1.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
|
||||
|
||||
line(
|
||||
(0.66 * s, 0.66 * s),
|
||||
(0.66 * s, 0.35 * s),
|
||||
(0.60 * s, 0.1 * s),
|
||||
)
|
||||
line(
|
||||
(0.66 * s, 0.66 * s),
|
||||
(0.66 * s, 0.35 * s),
|
||||
(0.60 * s, 0.1 * s),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.72 * s, 0.1 * s),
|
||||
(0.66 * s, 0.35 * s),
|
||||
)
|
||||
line(
|
||||
(0.72 * s, 0.1 * s),
|
||||
(0.66 * s, 0.35 * s),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.66 * s, 0.575 * s),
|
||||
(0.6 * s, 0.475 * s),
|
||||
(0.525 * s, 0.575 * s),
|
||||
)
|
||||
line(
|
||||
(0.66 * s, 0.575 * s),
|
||||
(0.6 * s, 0.475 * s),
|
||||
(0.525 * s, 0.575 * s),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.66 * s, 0.575 * s),
|
||||
(0.72 * s, 0.475 * s),
|
||||
(0.795 * s, 0.575 * s),
|
||||
)
|
||||
line(
|
||||
(0.66 * s, 0.575 * s),
|
||||
(0.72 * s, 0.475 * s),
|
||||
(0.795 * s, 0.575 * s),
|
||||
)
|
||||
|
||||
circle((0.66 * s, 0.66 * s), radius: 0.07 * s, fill: white)
|
||||
}),
|
||||
)
|
||||
circle((0.66 * s, 0.66 * s), radius: 0.07 * s, fill: white)
|
||||
}))
|
||||
|
||||
#solution([
|
||||
Say we have a left nail and a right nail. The path of the string is as follows:
|
||||
|
||||
@ -7,7 +7,7 @@
|
||||
|
||||
#problem()
|
||||
Take any positive integer $n$. \
|
||||
Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... a_k$ \
|
||||
Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... a_k$ \
|
||||
Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
|
||||
|
||||
|
||||
@ -17,7 +17,9 @@ Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
|
||||
Of course, all $a_i$ should be greater than $1$. \
|
||||
Also, all $a_i$ should be smaller than four, since $x <= x(x-2)$ if $x >= 4$. \
|
||||
Thus, we're left with sequences that only contain 2 and 3. \
|
||||
#note([Note that two twos are the same as one four, but we exclude fours for simplicity.])
|
||||
#note(
|
||||
[Note that two twos are the same as one four, but we exclude fours for simplicity.],
|
||||
)
|
||||
|
||||
#v(2mm)
|
||||
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Passing Balls],
|
||||
@ -78,32 +78,25 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
|
||||
|
||||
let i = 1
|
||||
for p in pts {
|
||||
circle(
|
||||
p,
|
||||
radius: radius * s,
|
||||
fill: if i == 1 {
|
||||
ored
|
||||
} else if i == 2 {
|
||||
ogreen
|
||||
} else if i == 3 {
|
||||
oorange
|
||||
} else if i == 4 {
|
||||
oblue
|
||||
} else { white },
|
||||
)
|
||||
circle(p, radius: radius * s, fill: if i == 1 {
|
||||
ored
|
||||
} else if i == 2 {
|
||||
ogreen
|
||||
} else if i == 3 {
|
||||
oorange
|
||||
} else if i == 4 {
|
||||
oblue
|
||||
} else { white })
|
||||
|
||||
content(
|
||||
p,
|
||||
text(
|
||||
fill: if i <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#i*],
|
||||
),
|
||||
)
|
||||
content(p, text(
|
||||
fill: if i <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#i*],
|
||||
))
|
||||
i = i + 1
|
||||
}
|
||||
}),
|
||||
@ -118,32 +111,25 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
|
||||
let l = calc.rem(((i - 1) * 5), 12) + 1
|
||||
|
||||
|
||||
circle(
|
||||
p,
|
||||
radius: radius * s,
|
||||
fill: if l == 1 {
|
||||
ored
|
||||
} else if l == 2 {
|
||||
ogreen
|
||||
} else if l == 3 {
|
||||
oorange
|
||||
} else if l == 4 {
|
||||
oblue
|
||||
} else { white },
|
||||
)
|
||||
circle(p, radius: radius * s, fill: if l == 1 {
|
||||
ored
|
||||
} else if l == 2 {
|
||||
ogreen
|
||||
} else if l == 3 {
|
||||
oorange
|
||||
} else if l == 4 {
|
||||
oblue
|
||||
} else { white })
|
||||
|
||||
content(
|
||||
p,
|
||||
text(
|
||||
fill: if l <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#l*],
|
||||
),
|
||||
)
|
||||
content(p, text(
|
||||
fill: if l <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#l*],
|
||||
))
|
||||
i = i + 1
|
||||
}
|
||||
}),
|
||||
@ -158,32 +144,25 @@ Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4
|
||||
let l = calc.rem(((i - 1) * 5), 12) + 1
|
||||
|
||||
|
||||
circle(
|
||||
p,
|
||||
radius: radius * s,
|
||||
fill: if l == 1 {
|
||||
oblue
|
||||
} else if l == 2 {
|
||||
oorange
|
||||
} else if l == 3 {
|
||||
ored
|
||||
} else if l == 4 {
|
||||
ogreen
|
||||
} else { white },
|
||||
)
|
||||
circle(p, radius: radius * s, fill: if l == 1 {
|
||||
oblue
|
||||
} else if l == 2 {
|
||||
oorange
|
||||
} else if l == 3 {
|
||||
ored
|
||||
} else if l == 4 {
|
||||
ogreen
|
||||
} else { white })
|
||||
|
||||
content(
|
||||
p,
|
||||
text(
|
||||
fill: if l <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#l*],
|
||||
),
|
||||
)
|
||||
content(p, text(
|
||||
fill: if l <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#l*],
|
||||
))
|
||||
i = i + 1
|
||||
}
|
||||
}),
|
||||
|
||||
@ -1,5 +1,5 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: What's an AST?],
|
||||
@ -24,51 +24,48 @@ respecting the order of operations $[and, times, div, +, -]$.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
#align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
// spell:off
|
||||
content((0, 0), $+$, name: "r")
|
||||
content((-0.5, -1), $3$, name: "a")
|
||||
content((0.5, -1), $div$, name: "b")
|
||||
content((-0.3, -2), $times$, name: "ba")
|
||||
content((1.3, -2), $and$, name: "bb")
|
||||
content((-0.8, -3), $9$, name: "baa")
|
||||
content((0.2, -3), $8$, name: "bab")
|
||||
content((0.8, -3), $5$, name: "bba")
|
||||
content((1.8, -3), $6$, name: "bbb")
|
||||
// spell:on
|
||||
// spell:off
|
||||
content((0, 0), $+$, name: "r")
|
||||
content((-0.5, -1), $3$, name: "a")
|
||||
content((0.5, -1), $div$, name: "b")
|
||||
content((-0.3, -2), $times$, name: "ba")
|
||||
content((1.3, -2), $and$, name: "bb")
|
||||
content((-0.8, -3), $9$, name: "baa")
|
||||
content((0.2, -3), $8$, name: "bab")
|
||||
content((0.8, -3), $5$, name: "bba")
|
||||
content((1.8, -3), $6$, name: "bbb")
|
||||
// spell:on
|
||||
|
||||
// Zero-sized arrows are a hack for offset.
|
||||
set-style(
|
||||
stroke: (thickness: 0.3mm),
|
||||
mark: (
|
||||
start: (
|
||||
symbol: "|",
|
||||
offset: 0.25,
|
||||
width: 0mm,
|
||||
length: 0mm,
|
||||
),
|
||||
end: (
|
||||
symbol: "|",
|
||||
offset: 0.25,
|
||||
width: 0mm,
|
||||
length: 0mm,
|
||||
),
|
||||
// Zero-sized arrows are a hack for offset.
|
||||
set-style(
|
||||
stroke: (thickness: 0.3mm),
|
||||
mark: (
|
||||
start: (
|
||||
symbol: "|",
|
||||
offset: 0.25,
|
||||
width: 0mm,
|
||||
length: 0mm,
|
||||
),
|
||||
)
|
||||
end: (
|
||||
symbol: "|",
|
||||
offset: 0.25,
|
||||
width: 0mm,
|
||||
length: 0mm,
|
||||
),
|
||||
),
|
||||
)
|
||||
|
||||
// spell:off
|
||||
line("r", "a")
|
||||
line("r", "b")
|
||||
line("b", "ba")
|
||||
line("b", "bb")
|
||||
line("ba", "baa")
|
||||
line("ba", "bab")
|
||||
line("bb", "bba")
|
||||
line("bb", "bbb")
|
||||
// spell:on
|
||||
}),
|
||||
)
|
||||
// spell:off
|
||||
line("r", "a")
|
||||
line("r", "b")
|
||||
line("b", "ba")
|
||||
line("b", "bb")
|
||||
line("ba", "baa")
|
||||
line("ba", "bab")
|
||||
line("bb", "bba")
|
||||
line("bb", "bbb")
|
||||
// spell:on
|
||||
}))
|
||||
|
||||
@ -1,24 +1,21 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
#import "@preview/cetz:0.4.2"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Wild Tic-Tac-Toe],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#let ttt = align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 0.7 // scale
|
||||
#let ttt = align(center, cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 0.7 // scale
|
||||
|
||||
set-style(stroke: (thickness: 0.5mm * s))
|
||||
line((-1 * s, 3 * s), (-1 * s, -3 * s))
|
||||
line((1 * s, 3 * s), (1 * s, -3 * s))
|
||||
line((3 * s, -1 * s), (-3 * s, -1 * s))
|
||||
line((3 * s, 1 * s), (-3 * s, 1 * s))
|
||||
}),
|
||||
)
|
||||
set-style(stroke: (thickness: 0.5mm * s))
|
||||
line((-1 * s, 3 * s), (-1 * s, -3 * s))
|
||||
line((1 * s, 3 * s), (1 * s, -3 * s))
|
||||
line((3 * s, -1 * s), (-3 * s, -1 * s))
|
||||
line((3 * s, 1 * s), (-3 * s, 1 * s))
|
||||
}))
|
||||
|
||||
|
||||
#problem()
|
||||
|
||||
Reference in New Issue
Block a user