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attributio
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@ -1,282 +0,0 @@
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/// Typst handout library, used for all documents in this repository.
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||||
|
||||
|
||||
/// If false, hide instructor info.
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///
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/// Compile with the following command to hide solutions:
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/// `typst compile main.typ --input show_solutions=false`
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///
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/// Solutions are shown by default. This behavior
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/// is less surprising than hiding content by default.
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#let show_solutions = {
|
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if "show_solutions" in sys.inputs {
|
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// Show solutions unless they're explicitly disabled
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not (
|
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sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
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||||
)
|
||||
} else {
|
||||
// Show solutions by default
|
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true
|
||||
}
|
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}
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|
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// Colors
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#let ored = rgb("D62121")
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#let ogrape = rgb("9C36B5")
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#let ocyan = rgb("2288BF")
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#let oteal = rgb("12B886")
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#let ogreen = rgb("37B26D")
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#let oblue = rgb("1C7ED6")
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|
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|
||||
|
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//
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// MARK: header
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//
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||||
|
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#let make_title(
|
||||
group,
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||||
quarter,
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||||
title,
|
||||
subtitle,
|
||||
) = {
|
||||
align(
|
||||
center,
|
||||
block(
|
||||
width: 60%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
align(
|
||||
center,
|
||||
stack(
|
||||
spacing: 7pt,
|
||||
(
|
||||
text(size: 10pt, group) + h(1fr) + text(size: 10pt, quarter)
|
||||
),
|
||||
line(length: 100%, stroke: 0.2mm),
|
||||
(
|
||||
text(size: 20pt, title) + linebreak() + text(size: 10pt, subtitle)
|
||||
),
|
||||
line(length: 100%, stroke: 0.2mm),
|
||||
),
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
|
||||
#let warn = {
|
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set text(ored)
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align(
|
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center,
|
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block(
|
||||
width: 60%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
fill: rgb(255, 255, 255),
|
||||
stroke: ored + 2pt,
|
||||
inset: 3mm,
|
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(
|
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align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
|
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+ parbreak()
|
||||
+ align(
|
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left,
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text(
|
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size: 10pt,
|
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[This handout contains solutions and notes.]
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+ linebreak()
|
||||
+ [Recompile without solutions before distributing.],
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),
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)
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),
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),
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)
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}
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|
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#let preparedby(name) = (
|
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text(
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size: 10pt,
|
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[Prepared by ]
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+ name
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||||
+ [ on ]
|
||||
+ datetime
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||||
.today()
|
||||
.display("[month repr:long] [day padding:none], [year]"),
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)
|
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)
|
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|
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//
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// MARK: Solutions
|
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//
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|
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#let solution(content) = {
|
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if show_solutions {
|
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align(
|
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center,
|
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stack(
|
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block(
|
||||
width: 100%,
|
||||
breakable: false,
|
||||
fill: ored,
|
||||
stroke: ored + 2pt,
|
||||
inset: 1.5mm,
|
||||
(
|
||||
align(left, text(fill: white, weight: "bold", [Solution:]))
|
||||
),
|
||||
),
|
||||
block(
|
||||
width: 100%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
fill: ored.lighten(80%).desaturate(10%),
|
||||
stroke: ored + 2pt,
|
||||
inset: 3mm,
|
||||
align(left, content),
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
}
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|
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#let notsolution(content) = {
|
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if not show_solutions { content }
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}
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|
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|
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//
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// MARK: Sections
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||||
//
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||||
|
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#let generic(t) = block(
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above: 8mm,
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below: 2mm,
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text(weight: "bold", t),
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)
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||||
|
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#let _generic_base(kind, ..args) = {
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counter("obj").step()
|
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if args.pos().len() == 0 {
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generic([
|
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#kind
|
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#context counter("obj").display():
|
||||
])
|
||||
} else {
|
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generic(
|
||||
[
|
||||
#kind
|
||||
#context counter("obj").display():
|
||||
]
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||||
+ " "
|
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+ args.pos().at(0),
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||||
)
|
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}
|
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}
|
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|
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#let problem(..args) = _generic_base("Problem", ..args)
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#let definition(..args) = _generic_base("Definition", ..args)
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#let theorem(..args) = _generic_base("Theorem", ..args)
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||||
|
||||
|
||||
//
|
||||
// MARK: Misc
|
||||
//
|
||||
|
||||
|
||||
#let hint(content) = {
|
||||
text(fill: rgb(100, 100, 100), style: "oblique", "Hint: ")
|
||||
text(fill: rgb(100, 100, 100), content)
|
||||
}
|
||||
|
||||
#let note(content) = {
|
||||
text(fill: rgb(100, 100, 100), content)
|
||||
}
|
||||
|
||||
#let examplesolution(content) = {
|
||||
let c = oblue
|
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|
||||
align(
|
||||
center,
|
||||
stack(
|
||||
block(
|
||||
width: 100%,
|
||||
breakable: false,
|
||||
fill: c,
|
||||
stroke: c + 2pt,
|
||||
inset: 1.5mm,
|
||||
(
|
||||
align(left, text(fill: white, weight: "bold", [Example solution:]))
|
||||
),
|
||||
),
|
||||
block(
|
||||
width: 100%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
fill: c.lighten(80%).desaturate(10%),
|
||||
stroke: c + 2pt,
|
||||
inset: 3mm,
|
||||
align(left, content),
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
|
||||
|
||||
//
|
||||
// MARK: wrapper
|
||||
//
|
||||
|
||||
#let handout(
|
||||
doc,
|
||||
group: none,
|
||||
quarter: none,
|
||||
title: none,
|
||||
by: none,
|
||||
subtitle: none,
|
||||
) = {
|
||||
set par(leading: 0.55em, first-line-indent: 0mm, justify: true)
|
||||
set text(font: "New Computer Modern")
|
||||
set par(spacing: 0.5em)
|
||||
show list: set block(spacing: 0.5em, below: 1em)
|
||||
set heading(numbering: (..nums) => nums.pos().at(0))
|
||||
|
||||
set page(
|
||||
margin: 20mm,
|
||||
width: 8in,
|
||||
height: 11.5in,
|
||||
footer: align(
|
||||
center,
|
||||
context counter(page).display(),
|
||||
),
|
||||
footer-descent: 5mm,
|
||||
)
|
||||
|
||||
|
||||
set list(
|
||||
tight: false,
|
||||
indent: 5mm,
|
||||
spacing: 3mm,
|
||||
)
|
||||
|
||||
show heading.where(level: 1): it => {
|
||||
set align(center)
|
||||
set text(weight: "bold")
|
||||
block[
|
||||
Section #counter(heading).display(): #text(it.body)
|
||||
]
|
||||
}
|
||||
|
||||
make_title(
|
||||
group,
|
||||
quarter,
|
||||
title,
|
||||
{
|
||||
if by == none { none } else { [#preparedby(by)\ ] }
|
||||
if subtitle == none { none } else { subtitle }
|
||||
},
|
||||
)
|
||||
|
||||
if show_solutions {
|
||||
warn
|
||||
}
|
||||
doc
|
||||
}
|
||||
|
||||
71
lib/typst/local/handout/0.1.0/header.typ
Normal file
71
lib/typst/local/handout/0.1.0/header.typ
Normal file
@ -0,0 +1,71 @@
|
||||
#import "misc.typ": ored
|
||||
|
||||
#let solution_warning() = {
|
||||
set text(ored)
|
||||
align(
|
||||
center,
|
||||
block(
|
||||
width: 60%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
fill: rgb(255, 255, 255),
|
||||
stroke: ored + 2pt,
|
||||
inset: 3mm,
|
||||
(
|
||||
align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
|
||||
+ parbreak()
|
||||
+ align(
|
||||
left,
|
||||
text(
|
||||
size: 10pt,
|
||||
[
|
||||
This handout contains solutions and notes. \
|
||||
Recompile without solutions before distributing.
|
||||
],
|
||||
),
|
||||
)
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
|
||||
#let make_header(
|
||||
title,
|
||||
subtitle: none,
|
||||
by: none,
|
||||
top_left: "",
|
||||
top_right: "",
|
||||
) = {
|
||||
let date = datetime
|
||||
.today()
|
||||
.display("[month repr:long] [day padding:none], [year]")
|
||||
|
||||
if (by != none) {
|
||||
by = text(size: 10pt, [Prepared by #by on #date])
|
||||
}
|
||||
|
||||
// Main title
|
||||
align(
|
||||
center,
|
||||
block(
|
||||
width: 60%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
align(
|
||||
center,
|
||||
stack(
|
||||
spacing: 7pt,
|
||||
// Top
|
||||
text(size: 10pt, top_left) + h(1fr) + text(size: 10pt, top_right),
|
||||
line(length: 100%, stroke: 0.2mm),
|
||||
// Title
|
||||
text(size: 20pt, title),
|
||||
// Subtitle
|
||||
if (by != none) { text(size: 10pt, by) },
|
||||
if (subtitle != none) { text(size: 10pt, subtitle) },
|
||||
line(length: 100%, stroke: 0.2mm),
|
||||
),
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
113
lib/typst/local/handout/0.1.0/lib.typ
Executable file
113
lib/typst/local/handout/0.1.0/lib.typ
Executable file
@ -0,0 +1,113 @@
|
||||
/// Typst handout library, used for all documents in this repository.
|
||||
|
||||
// Re-exports
|
||||
// All functions that maybe used by client code are listed here
|
||||
#import "misc.typ": *
|
||||
#import "object.typ": problem, definition, theorem
|
||||
#import "solution.typ": if_solutions, if_no_solutions, solution
|
||||
|
||||
|
||||
/// Main handout wrapper.
|
||||
/// Use this as follows:
|
||||
///
|
||||
/// ```
|
||||
/// #show: handout.with(
|
||||
/// group: "Advanced 2",
|
||||
/// title: [Handout Title],
|
||||
/// by: "author",
|
||||
/// subtitle: "optional",
|
||||
/// )
|
||||
///
|
||||
/// <rest of document>
|
||||
/// ```
|
||||
#let handout(
|
||||
doc,
|
||||
group: none,
|
||||
title: none,
|
||||
by: none,
|
||||
subtitle: none,
|
||||
) = {
|
||||
set page(
|
||||
margin: 20mm,
|
||||
width: 8.5in,
|
||||
height: 11in,
|
||||
footer: align(
|
||||
center,
|
||||
context counter(page).display(),
|
||||
),
|
||||
footer-descent: 5mm,
|
||||
)
|
||||
|
||||
//
|
||||
// Text style
|
||||
set text(font: "New Computer Modern")
|
||||
set par(
|
||||
leading: 0.55em,
|
||||
first-line-indent: 0mm,
|
||||
justify: true,
|
||||
spacing: 0.5em,
|
||||
)
|
||||
|
||||
//
|
||||
// List style
|
||||
show list: set block(spacing: 0.5em, below: 1em)
|
||||
set list(
|
||||
tight: false,
|
||||
indent: 5mm,
|
||||
spacing: 3mm,
|
||||
)
|
||||
|
||||
//
|
||||
// Heading style
|
||||
set heading(numbering: (..nums) => nums.pos().at(0))
|
||||
show heading.where(level: 1): it => {
|
||||
set align(center)
|
||||
set text(weight: "bold")
|
||||
block[
|
||||
Section #counter(heading).display(): #text(it.body)
|
||||
]
|
||||
}
|
||||
|
||||
//
|
||||
// Hack for custom references
|
||||
show ref: it => {
|
||||
import "object.typ": ref_obj
|
||||
|
||||
let x = ref_obj(it) // Custom impl for object references
|
||||
if (x != none) { return x }
|
||||
|
||||
return it // Use default `ref` implementation otherwise
|
||||
}
|
||||
|
||||
|
||||
//
|
||||
// Begin content
|
||||
//
|
||||
|
||||
// Make handout title
|
||||
{
|
||||
import "header.typ": make_header, solution_warning
|
||||
import "solution.typ": show_solutions
|
||||
|
||||
let url = link(
|
||||
"https://betalupi.com/handouts",
|
||||
`betalupi.com/handouts`,
|
||||
)
|
||||
|
||||
make_header(
|
||||
title,
|
||||
subtitle: subtitle,
|
||||
by: by,
|
||||
top_left: group,
|
||||
top_right: url,
|
||||
)
|
||||
|
||||
if show_solutions {
|
||||
solution_warning()
|
||||
}
|
||||
}
|
||||
|
||||
// Include rest of document
|
||||
doc
|
||||
}
|
||||
|
||||
49
lib/typst/local/handout/0.1.0/misc.typ
Executable file
49
lib/typst/local/handout/0.1.0/misc.typ
Executable file
@ -0,0 +1,49 @@
|
||||
/// Miscellaneous utilities
|
||||
|
||||
#let ored = rgb("D62121")
|
||||
#let oorange = rgb("#ffaa3b")
|
||||
#let ogrape = rgb("9C36B5")
|
||||
#let ocyan = rgb("2288BF")
|
||||
#let oteal = rgb("12B886")
|
||||
#let ogreen = rgb("37B26D")
|
||||
#let oblue = rgb("1C7ED6")
|
||||
|
||||
|
||||
#let note(content, type: none) = {
|
||||
set text(fill: rgb(100, 100, 100))
|
||||
if type != none {
|
||||
text(style: "oblique", [#type: ])
|
||||
}
|
||||
text(content)
|
||||
}
|
||||
|
||||
#let hint = note.with(type: "Hint")
|
||||
|
||||
#let examplesolution(content) = {
|
||||
let c = oblue
|
||||
|
||||
align(
|
||||
center,
|
||||
stack(
|
||||
block(
|
||||
width: 100%,
|
||||
breakable: false,
|
||||
fill: c,
|
||||
stroke: c + 2pt,
|
||||
inset: 1.5mm,
|
||||
(
|
||||
align(left, text(fill: white, weight: "bold", [Example solution:]))
|
||||
),
|
||||
),
|
||||
block(
|
||||
width: 100%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
fill: c.lighten(80%).desaturate(10%),
|
||||
stroke: c + 2pt,
|
||||
inset: 3mm,
|
||||
align(left, content),
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
100
lib/typst/local/handout/0.1.0/object.typ
Normal file
100
lib/typst/local/handout/0.1.0/object.typ
Normal file
@ -0,0 +1,100 @@
|
||||
|
||||
/// This module defines all handout "objects"
|
||||
/// (problems, theorems, definitions, etc)
|
||||
|
||||
/// Core render code for all objects (problems, theorems, etc)
|
||||
/// This should never be used directly by client code.
|
||||
///
|
||||
/// Args:
|
||||
/// - kind: the kind of object to make ("Problem", "Definition", etc)
|
||||
/// - label_name: a string. If provided, generate metadata for this object
|
||||
/// under the given label. Labels must be unique.
|
||||
/// This label can then be used to reference this object.
|
||||
///
|
||||
/// For example:
|
||||
/// ```
|
||||
/// #problem(label: "problem1")
|
||||
/// This is @problem1
|
||||
/// ```
|
||||
#let _obj_base(kind, ..args, label_name: none) = {
|
||||
counter("obj").step()
|
||||
let n = context counter("obj").get().first()
|
||||
|
||||
// The complete title text of this object,
|
||||
// like "Problem 5:" or "Theorem: "
|
||||
let obj_content = if args.pos().len() == 0 {
|
||||
[#kind #n:]
|
||||
} else {
|
||||
[#kind #n: #args.pos().at(0)]
|
||||
}
|
||||
|
||||
// Render the object
|
||||
block(
|
||||
above: 8mm,
|
||||
below: 2mm,
|
||||
text(weight: "bold", obj_content),
|
||||
)
|
||||
|
||||
// Generate labeled metadata for this object.
|
||||
//
|
||||
// This can be viewed directly with `#context query(<label>).first().value`,
|
||||
// Or referenced with `@label` (we define a custom renderer for this metadata later)
|
||||
if label_name != none {
|
||||
let meta = (
|
||||
"obj_meta_ref_kind": kind,
|
||||
// "obj_content": obj_content,
|
||||
"label": label(label_name),
|
||||
"counter": counter("obj"),
|
||||
)
|
||||
[ #metadata(meta) #label(label_name) ]
|
||||
}
|
||||
}
|
||||
|
||||
// `ref` implementation for object meta-references.
|
||||
// Returns `none` if `it` is not object metadata.
|
||||
#let ref_obj(it) = {
|
||||
let magic_key = "obj_meta_ref_kind"
|
||||
if not (
|
||||
it.element != none
|
||||
and it.element.has("value")
|
||||
and type(it.element.value) == "dictionary"
|
||||
and it.element.value.keys().contains(magic_key)
|
||||
) {
|
||||
// This label is not attached to object metadata
|
||||
return none
|
||||
}
|
||||
|
||||
let v = it.element.value
|
||||
let obj_type = v.at(magic_key)
|
||||
|
||||
// The value of this object's counter at its label
|
||||
let obj_count = v.counter.at(v.label).first()
|
||||
|
||||
// Produces text like "Problem 2",
|
||||
// which takes you to the referenced object when clicked.
|
||||
return link(v.label, [#obj_type #obj_count])
|
||||
}
|
||||
|
||||
/// Factory function for objects.
|
||||
/// Provided for convenience, lets us define objects in one line.
|
||||
#let _mkobj(kind) = {
|
||||
let out(..args, label: none) = _obj_base(
|
||||
kind,
|
||||
..args,
|
||||
label_name: label,
|
||||
)
|
||||
|
||||
return out
|
||||
}
|
||||
|
||||
|
||||
//
|
||||
// MARK: export
|
||||
//
|
||||
// Functions for client code are defined below
|
||||
|
||||
#let problem = _mkobj("Problem")
|
||||
#let definition = _mkobj("Definition")
|
||||
#let theorem = _mkobj("Theorem")
|
||||
#let example = _mkobj("Example")
|
||||
#let remark = _mkobj("Remark")
|
||||
56
lib/typst/local/handout/0.1.0/solution.typ
Executable file
56
lib/typst/local/handout/0.1.0/solution.typ
Executable file
@ -0,0 +1,56 @@
|
||||
#import "misc.typ": ored
|
||||
|
||||
/// If false, hide instructor info.
|
||||
///
|
||||
/// Compile with the following command to hide solutions:
|
||||
/// `typst compile main.typ --input show_solutions=false`
|
||||
///
|
||||
/// Solutions are shown by default. This behavior
|
||||
/// is less surprising than hiding content by default.
|
||||
#let show_solutions = {
|
||||
if "show_solutions" in sys.inputs {
|
||||
// Show solutions unless they're explicitly disabled
|
||||
not (
|
||||
sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
|
||||
)
|
||||
} else {
|
||||
// Show solutions by default
|
||||
true
|
||||
}
|
||||
}
|
||||
|
||||
#let if_solutions(content) = {
|
||||
if show_solutions { content }
|
||||
}
|
||||
|
||||
#let if_no_solutions(content) = {
|
||||
if not show_solutions { content }
|
||||
}
|
||||
|
||||
#let solution(content) = {
|
||||
if_solutions(
|
||||
align(
|
||||
center,
|
||||
stack(
|
||||
block(
|
||||
width: 100%,
|
||||
breakable: false,
|
||||
fill: ored,
|
||||
stroke: ored + 2pt,
|
||||
inset: 1.5mm,
|
||||
align(left, text(fill: white, weight: "bold", [Solution:])),
|
||||
),
|
||||
|
||||
block(
|
||||
width: 100%,
|
||||
height: auto,
|
||||
breakable: false,
|
||||
fill: ored.lighten(80%).desaturate(10%),
|
||||
stroke: ored + 2pt,
|
||||
inset: 3mm,
|
||||
align(left, content),
|
||||
),
|
||||
),
|
||||
),
|
||||
)
|
||||
}
|
||||
@ -1,6 +1,12 @@
|
||||
[package]
|
||||
name = "handout"
|
||||
description = "A library for math circle handouts"
|
||||
version = "0.1.0"
|
||||
entrypoint = "handout.typ"
|
||||
authors = []
|
||||
license = "GPL"
|
||||
entrypoint = "lib.typ"
|
||||
|
||||
homepage = "https://betalupi.com/handouts"
|
||||
repository = "https://git.betalupi.com/Mark/handouts"
|
||||
authors = ["Mark <mark@betalupi.com>"]
|
||||
license = "GPL-3.0-only "
|
||||
disciplines = ["education", "mathematics"]
|
||||
categories = ["layout", "components"]
|
||||
@ -1,13 +1,6 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: doc => handout(
|
||||
doc,
|
||||
group: "Advanced 2",
|
||||
quarter: link(
|
||||
"https://betalupi.com/handouts",
|
||||
"betalupi.com/handouts",
|
||||
),
|
||||
|
||||
#show: handout.with(
|
||||
title: [Tropical Polynomials],
|
||||
by: "Mark",
|
||||
subtitle: "Based on a handout by Bryant Mathews",
|
||||
|
||||
@ -3,5 +3,5 @@ title = "Tropical Polynomials"
|
||||
|
||||
|
||||
[publish]
|
||||
handout = true
|
||||
handout = false
|
||||
solutions = true
|
||||
|
||||
@ -104,7 +104,7 @@ Is there a tropical multiplicative identity? \
|
||||
Do tropical multiplicative inverses always exist? \
|
||||
#note([
|
||||
For every $x != #sym.infinity$, does there exist an inverse $y$ so that $x #tm y = i$, \
|
||||
where $i$ is the additive identity?
|
||||
where $i$ is the multiplicative identity?
|
||||
])
|
||||
|
||||
#solution([Yes, it is $-x$. For $x != 0$, $x #tm (-x) = 0$])
|
||||
@ -126,7 +126,7 @@ Fill the following tropical addition and multiplication tables
|
||||
|
||||
#let col = 10mm
|
||||
|
||||
#notsolution(
|
||||
#if_no_solutions(
|
||||
table(
|
||||
columns: (1fr, 1fr),
|
||||
align: center,
|
||||
|
||||
@ -63,7 +63,7 @@ where all exponents represent repeated tropical multiplication.
|
||||
Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
|
||||
#hint([$1x$ is not equal to $x$.])
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
$f(x) = min(2x , 1+x, 4)$, which looks like:
|
||||
@ -132,7 +132,7 @@ How can we use the graph to determine these roots?
|
||||
Graph $f(x) = -2x^2 #tp x #tp 8$. \
|
||||
#hint([Use half scale. 1 box = 2 units.])
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
#graphgrid({
|
||||
@ -140,7 +140,7 @@ Graph $f(x) = -2x^2 #tp x #tp 8$. \
|
||||
let step = 0.75
|
||||
|
||||
dotline((0, 0), (8 * step, 8 * step))
|
||||
dotline((0.5 * step, 0), (4 * step, 8 * step))
|
||||
dotline((0.5 * step, 0), (4.5 * step, 8 * step))
|
||||
dotline((0, 4 * step), (8 * step, 4 * step))
|
||||
|
||||
line(
|
||||
@ -210,7 +210,7 @@ and always produces $7$ for sufficiently large inputs.
|
||||
#problem()
|
||||
Graph $f(x) = 1x^2 #tp 3x #tp 5$.
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
The graphs of all three terms intersect at the same point:
|
||||
@ -261,7 +261,7 @@ How are the roots of $f$ related to its coefficients?
|
||||
#problem()
|
||||
Graph $f(x) = 2x^2 #tp 4x #tp 4$.
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution(
|
||||
graphgrid({
|
||||
|
||||
@ -5,12 +5,12 @@
|
||||
= Tropical Cubic Polynomials
|
||||
|
||||
#problem()
|
||||
Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
|
||||
Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 3x #tp 6$. \
|
||||
- sketch a graph of this polynomial
|
||||
- use this graph to find the roots of $f$
|
||||
- write (and expand) a product of linear factors with the same graph as $f$.
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
- Roots are 1, 2, and 3.
|
||||
@ -43,12 +43,12 @@ Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
|
||||
#pagebreak() // MARK: page
|
||||
|
||||
#problem()
|
||||
Consider the polynomial $f(x) = x^3 #tp x^2 #tp 6x #tp 6$. \
|
||||
Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 6x #tp 6$. \
|
||||
- sketch a graph of this polynomial
|
||||
- use this graph to find the roots of $f$
|
||||
- write (and expand) a product of linear factors with the same graph as $f$.
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
- Roots are 1, 2.5, and 2.5.
|
||||
@ -82,7 +82,7 @@ Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
|
||||
- use this graph to find the roots of $f$
|
||||
- write (and expand) a product of linear factors with the same graph as $f$.
|
||||
|
||||
#notsolution(graphgrid(none))
|
||||
#if_no_solutions(graphgrid(none))
|
||||
|
||||
#solution([
|
||||
- Roots are 2, 2, and 2.
|
||||
@ -118,10 +118,10 @@ Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b
|
||||
#solution([
|
||||
|
||||
$
|
||||
B = min(b, (a+c)/2, (2a+d)/2)
|
||||
B = min(b, (a+c)/2, (2a+d)/3)
|
||||
$
|
||||
$
|
||||
C = min(c, (b+d)/2, (a+2d)/2)
|
||||
C = min(c, (b+d)/2, (a+2d)/3)
|
||||
$
|
||||
])
|
||||
|
||||
@ -153,7 +153,7 @@ What are the roots of the following polynomial?
|
||||
#v(1fr)
|
||||
#pagebreak() // MARK: page
|
||||
|
||||
#problem()
|
||||
#problem(label: "findci")
|
||||
If
|
||||
$
|
||||
f(x) = c_0 #tp c_1 x #tp c_2 x^2 #tp ... #tp c_n x^n
|
||||
@ -183,7 +183,7 @@ Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
|
||||
|
||||
|
||||
#problem()
|
||||
With the same setup as the previous problem, \
|
||||
With the same setup as @findci, \
|
||||
find formulas for the roots $r_1, r_2, ..., r_n$.
|
||||
|
||||
#solution([
|
||||
|
||||
@ -1,35 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
|
||||
\title{Warm-Up: A Familiar Concept}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
|
||||
\problem{}<one>
|
||||
Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric. \par
|
||||
$$
|
||||
\sum_{\forall i} |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
|
||||
$$
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in \ref{one}.
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
What is this metric usually called?
|
||||
|
||||
\end{document}
|
||||
33
src/Warm-Ups/A Familiar Concept/main.typ
Normal file
33
src/Warm-Ups/A Familiar Concept/main.typ
Normal file
@ -0,0 +1,33 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: A Familiar Concept],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric:
|
||||
|
||||
#align(
|
||||
center,
|
||||
box(
|
||||
inset: 3mm,
|
||||
$
|
||||
sum_(#sym.forall i) |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
|
||||
$,
|
||||
),
|
||||
)
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in the previous problem.
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
What is this metric usually called?
|
||||
|
||||
|
||||
#v(0.25fr)
|
||||
@ -1,93 +0,0 @@
|
||||
\documentclass[
|
||||
nosolutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\title{Warm-Up: Adders}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
Fill the following binary addition table. \par
|
||||
\hint{s is \say{sum,} c is \say{carry}}
|
||||
|
||||
\begin{center}
|
||||
\begin{tabular}{ c c || c c }
|
||||
$a$ & $b$ & s & c \\
|
||||
\hline
|
||||
0 & 0 & ? & ? \\
|
||||
0 & 1 & ? & ? \\
|
||||
1 & 0 & ? & ? \\
|
||||
1 & 1 & ? & ?
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Draw a logic circuit that atisfies the above table. \par
|
||||
This is called a \textit{half adder}. \par
|
||||
\hint{You should need exactly two gates.}
|
||||
|
||||
\begin{solution}
|
||||
$s = a \texttt{ xor } b$ \par
|
||||
$c = a \texttt{ and } b$
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
\definition{}
|
||||
A \textit{full adder} is similar to a half adder, but it has an extra input: \par
|
||||
a full adder takes $a$, $b$, and $c_\text{in}$, and produces $s$ and $c_\text{out}$. \par
|
||||
\hint{$c_\text{in}$ is \say{carry in}}
|
||||
|
||||
\problem{}
|
||||
Use two half adders to construct a full adder.
|
||||
|
||||
\begin{solution}
|
||||
$s_1, c_1 = \texttt{HA}(a, b)$ \par
|
||||
$s_2, c_2 = \texttt{HA}(s_1, c_\text{in})$ \par
|
||||
$s_\text{out} = s_2$ \par
|
||||
$c_\text{out} = \texttt{OR}(c_1, c_2)$
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Of course, the class should just draw the circuit.
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}<rippleadder>
|
||||
How can we add two four-bit binary numbers using the full adder? \par
|
||||
We want a four-bit output sum and a one-bit $c_\text{out}$.
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Say that all basic logic gates need $1u$ of time to fully switch states. \par
|
||||
\note[Note]{This is called \textit{gate delay}}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
How much time does a full adder need to fully switch states? \par
|
||||
How about your circuit from \ref{rippleadder}?
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{Bonus}
|
||||
Design a faster solution to \ref{rippleadder}.
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
85
src/Warm-Ups/Adders/main.typ
Normal file
85
src/Warm-Ups/Adders/main.typ
Normal file
@ -0,0 +1,85 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Adders],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Fill the following binary addition table. \
|
||||
#hint([s is "sum," c is "carry"])
|
||||
|
||||
#align(
|
||||
center,
|
||||
table(
|
||||
columns: (9mm, 9mm, 9mm, 9mm),
|
||||
align: center,
|
||||
$a$, $b$, $s$, $c$,
|
||||
[0], [0], [?], [?],
|
||||
[0], [1], [?], [?],
|
||||
[1], [0], [?], [?],
|
||||
[1], [1], [?], [?],
|
||||
),
|
||||
)
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Draw a logic circuit that atisfies the above table. \
|
||||
This is called a _half adder_. \
|
||||
#hint([You should need exactly two gates.])
|
||||
|
||||
#solution([
|
||||
$s = a #text([`xor`]) b$ \
|
||||
$c = a #text([`and`]) b$
|
||||
])
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#definition()
|
||||
A _full adder_ is similar to a half adder, but it has an extra input: \
|
||||
a full adder takes $a$, $b$, and $c_"in"$, and produces $s$ and $c_"out"$. \
|
||||
#hint([$c_"in"$ is "carry in"])
|
||||
|
||||
#problem()
|
||||
Use two half adders to construct a full adder.
|
||||
|
||||
#solution([
|
||||
$
|
||||
s_1, c_1 &= "HA"(a, b) \
|
||||
s_2, c_2 &= "HA"(s_1, c_"in") \
|
||||
s_"out" &= s_2 \
|
||||
c_"out" &= "OR"(c_1, c_2)
|
||||
$
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Of course, the class should just draw the circuit.
|
||||
])
|
||||
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
#problem(label: "ripple-adder")
|
||||
How can we add two four-bit binary numbers using the full adder? \
|
||||
We want a four-bit output sum and a one-bit $c_"out"$.
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Say that all basic logic gates need $1u$ of time to fully switch states. \
|
||||
#note([This is called _gate delay_], type: "Note")
|
||||
|
||||
#v(2mm)
|
||||
|
||||
How much time does a full adder need to fully switch states? \
|
||||
How about your circuit from @ripple-adder?
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem("Bonus")
|
||||
Design a faster solution to @ripple-adder.
|
||||
|
||||
#v(1fr)
|
||||
@ -1,187 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
% x,y,scale,style
|
||||
\def\ttt#1#2#3#4{
|
||||
\draw[#4] (#1, #2+#3) -- (#1+#3+#3+#3, #2+#3);
|
||||
\draw[#4] (#1, #2+#3+#3) -- (#1+#3+#3+#3, #2+#3+#3);
|
||||
\draw[#4] (#1+#3, #2) -- (#1+#3, #2+#3+#3+#3);
|
||||
\draw[#4] (#1+#3+#3, #2) -- (#1+#3+#3, #2+#3+#3+#3);
|
||||
}
|
||||
|
||||
|
||||
\geometry{
|
||||
paper = letterpaper,
|
||||
top = 25mm,
|
||||
bottom = 30mm,
|
||||
left = 20mm,
|
||||
right = 20mm,
|
||||
headheight = 75mm,
|
||||
footskip = 15mm
|
||||
}
|
||||
|
||||
|
||||
% misere ttt
|
||||
|
||||
% Numerical Tic Tac Toe is a variation invented by the mathematician Ronald Graham.
|
||||
% The numbers 1 to 9 are used in this game. The first player plays with the odd numbers,
|
||||
% the second player plays with the even numbers. All numbers can be used only once.
|
||||
% The player who puts down 15 points in a line wins (sum of 3 numbers).
|
||||
% This game can be generalized to a n × n board.
|
||||
|
||||
|
||||
% In Treblecross, both players play with the same symbol.
|
||||
% The game is played on a 1-by-n board with k equal to 3.
|
||||
% The player who makes a three in a row of Xs (or black chips) wins the game
|
||||
|
||||
\title{Warm-Up: Big-Tac-Toe}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
|
||||
|
||||
We have large tic-tac-toe grid, each cell of which contains another.
|
||||
On each turn, one player puts their glyph into a cell of a small grid. When the next player goes,
|
||||
they must make their move in the small grid in the same position as the previous player's move.
|
||||
\begin{itemize}
|
||||
\item The first player to move may pick any small grid to start in.
|
||||
\item If a player is directed to a grid that is \textit{full}, that player may go anywhere. \par
|
||||
A sub-grid that is \say{won} but not full may still be played in.
|
||||
\end{itemize}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
The first player to complete a line of three \say{won} subgrids wins the game.
|
||||
|
||||
\vfill\null\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.65]
|
||||
\ttt{0}{0}{4}{line width=0.5mm}
|
||||
\ttt{0.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{8.5}{1}{line width=0.25mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.65]
|
||||
\ttt{0}{0}{4}{line width=0.5mm}
|
||||
\ttt{0.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{8.5}{1}{line width=0.25mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill\null\vfill
|
||||
|
||||
\problem{}
|
||||
Play a few rounds of this game with someone nearby. \par
|
||||
Can either player force a win?
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Modify the rules of this game to disallow play in won subgrids. \par
|
||||
How does your strategy change?
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
\null\vfill\null\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.65]
|
||||
\ttt{0}{0}{4}{line width=0.5mm}
|
||||
\ttt{0.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{8.5}{1}{line width=0.25mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.65]
|
||||
\ttt{0}{0}{4}{line width=0.5mm}
|
||||
\ttt{0.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{8.5}{1}{line width=0.25mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill\null\vfill
|
||||
|
||||
\vfill\null\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.65]
|
||||
\ttt{0}{0}{4}{line width=0.5mm}
|
||||
\ttt{0.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{8.5}{1}{line width=0.25mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.65]
|
||||
\ttt{0}{0}{4}{line width=0.5mm}
|
||||
\ttt{0.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{0.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{4.5}{8.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{0.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{4.5}{1}{line width=0.25mm}
|
||||
\ttt{8.5}{8.5}{1}{line width=0.25mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill\null\vfill
|
||||
|
||||
\end{document}
|
||||
90
src/Warm-Ups/Big-Tac-Toe/main.typ
Normal file
90
src/Warm-Ups/Big-Tac-Toe/main.typ
Normal file
@ -0,0 +1,90 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Big-Tac-Toe],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#let extra_boards = false;
|
||||
|
||||
#let ttt(s, p) = {
|
||||
// s: scale,
|
||||
// p: position
|
||||
let x = p.at(0) * s
|
||||
let y = p.at(1) * s
|
||||
|
||||
cetz.draw.line((-1 * s + x, 3 * s + y), (-1 * s + x, -3 * s + y))
|
||||
cetz.draw.line((1 * s + x, 3 * s + y), (1 * s + x, -3 * s + y))
|
||||
cetz.draw.line((3 * s + x, -1 * s + y), (-3 * s + x, -1 * s + y))
|
||||
cetz.draw.line((3 * s + x, 1 * s + y), (-3 * s + x, 1 * s + y))
|
||||
}
|
||||
|
||||
#let btt(s) = cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
set-style(stroke: (thickness: 0.5mm * s))
|
||||
|
||||
ttt(s, (-7, -7))
|
||||
ttt(s, (-7, 0))
|
||||
ttt(s, (-7, 7))
|
||||
ttt(s, (0, -7))
|
||||
ttt(s, (0, 0))
|
||||
ttt(s, (0, 7))
|
||||
ttt(s, (7, -7))
|
||||
ttt(s, (7, 0))
|
||||
ttt(s, (7, 7))
|
||||
|
||||
set-style(stroke: (thickness: 2mm * s))
|
||||
ttt(s * 3.5, (0, 0))
|
||||
})
|
||||
|
||||
|
||||
#problem()
|
||||
Consider a large tic-tac-toe grid, each cell of which contains another.
|
||||
On each turn, one player puts their glyph into a cell of a small grid. When the next player goes,
|
||||
they must make their move in the small grid in the same position as the previous player's move.
|
||||
- The first player to move may pick any small grid to start in.
|
||||
- If a player is directed to a grid that is _full_, that player may go anywhere. \
|
||||
A sub-grid that is "won" but not full may still be played in.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
The first player to complete a line of three "won" subgrids wins the game.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#problem()
|
||||
Play a few rounds of this game with someone nearby. \
|
||||
Can either player force a win?
|
||||
|
||||
#table(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr),
|
||||
btt(0.35), btt(0.35),
|
||||
);
|
||||
|
||||
#problem()
|
||||
Modify the rules of this game to disallow play in won subgrids. \
|
||||
How does your strategy change? \
|
||||
#if extra_boards { note([Additional boards are available on the next page.]) }
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#if extra_boards {
|
||||
pagebreak()
|
||||
|
||||
align(
|
||||
center,
|
||||
grid(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr),
|
||||
rows: (1fr, 1fr, 1fr),
|
||||
btt(0.35), btt(0.35),
|
||||
btt(0.35), btt(0.35),
|
||||
btt(0.35), btt(0.35),
|
||||
),
|
||||
)
|
||||
}
|
||||
@ -1,31 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
|
||||
\title{Warm-Up: Fuse Timers}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
|
||||
\problem{}
|
||||
Suppose we have two strings and a lighter. Each string takes an hour to fully burn. \par
|
||||
However, we do not know how fast each part of the string burns:
|
||||
half might burn in 1 minute, and the rest could take 59.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
How would we measure exactly 45 minutes using these strings?
|
||||
|
||||
\vfill
|
||||
|
||||
\end{document}
|
||||
15
src/Warm-Ups/Fuse Timers/main.typ
Normal file
15
src/Warm-Ups/Fuse Timers/main.typ
Normal file
@ -0,0 +1,15 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Fuse Timers],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Suppose we have two strings and a lighter. Each string takes exactly an hour to fully burn. \
|
||||
However, we do not know how fast each part of the string burns:
|
||||
half might burn in 1 minute, and the rest could take 59.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
How can we measure exactly 45 minutes using these two strings?
|
||||
@ -1,121 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\makeatletter
|
||||
\newcommand{\thisone}{
|
||||
\if@solutions
|
||||
{\color{red} $\Leftarrow$ \texttt{this one}}
|
||||
\else\fi
|
||||
}
|
||||
|
||||
\title{Warm-Up: The Gallery}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
A museum curator is arranging seven photographs on a gallery wall in accordance with the photographer's requirements.
|
||||
They are titled as follows: Fence, Gardenias, Hibiscus, Irises, Katydid, Lotus, and Magnolia.
|
||||
|
||||
The photograph's requirements are as follows:
|
||||
\begin{itemize}
|
||||
\item Gardenias must be immediately before Katydid.
|
||||
\item Hibiscus must be somewhere before Katydid but cannot be the first photograph.
|
||||
\item Irises and Lotus must be next to one another.
|
||||
\item Magnolia must be one of the first three photographs.
|
||||
\item Fence must be either first or seventh.
|
||||
\end{itemize}
|
||||
|
||||
\problem{}
|
||||
Which of the below could be a valid ordering? \par
|
||||
\note[Note]{We denote each painting by the first letter of its title.}
|
||||
\begin{itemize}
|
||||
\item \texttt{FHGMKIL}
|
||||
\item \texttt{HMGKILF}
|
||||
\item \texttt{ILMHGKF} \thisone{}
|
||||
\item \texttt{LMIHGKF}
|
||||
\item \texttt{MFHGKLI}
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
If Irises is immediately before Gardenias, which of the following could be true?
|
||||
\begin{itemize}
|
||||
\item Gardenias is fourth
|
||||
\item Hibiscus is fourth
|
||||
\item Irises is third
|
||||
\item Lotus is second
|
||||
\item Magnolia is third \thisone{}
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
The ordering of the photographs is fully determined if...
|
||||
\begin{itemize}
|
||||
\item Gardenias is fourth
|
||||
\item Hibiscus is second
|
||||
\item Irises is second
|
||||
\item Lotus is first \thisone{}
|
||||
\item Magnolia is third
|
||||
\end{itemize}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}
|
||||
If Magnolia is second, what CANNOT be true?
|
||||
\begin{itemize}
|
||||
\item Hibiscus is third
|
||||
\item Hibiscus is fourth \thisone{}
|
||||
\item Hibiscus is fifth
|
||||
\item Gardenias is fourth
|
||||
\item Gardenias is sixth
|
||||
\end{itemize}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Katydid cannot be in which position?
|
||||
\begin{itemize}
|
||||
\item Third \thisone{}
|
||||
\item Fourth
|
||||
\item Fifth
|
||||
\item Sixth
|
||||
\item Seventh
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
If Gardenias is fourth, what must be true?
|
||||
\begin{itemize}
|
||||
\item Fence is first \thisone{}
|
||||
\item Hibiscus is third
|
||||
\item Irises is seventh
|
||||
\item Magnolia is first
|
||||
\item Magnolia is second
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Which one of the following,
|
||||
if substituted for the second condition,
|
||||
would have the same effect in determining the
|
||||
arrangement of the photographs?
|
||||
|
||||
\begin{itemize}
|
||||
\item If Fence is seventh, Hibiscus is second
|
||||
\item Gardenias is somewhere after Hibiscus, and either Fence or Magnolia is first
|
||||
\item Hibiscus must be somewhere between the first and sixth photographs
|
||||
\item Unless Hibiscus is second, it must be somewhere between Magnolia and Gardenias \thisone{}
|
||||
\item Katydid is somewhere after Hibiscus, which must be after Fence.
|
||||
\end{itemize}
|
||||
\vfill
|
||||
\pagebreak
|
||||
\end{document}
|
||||
103
src/Warm-Ups/Gallery/main.typ
Normal file
103
src/Warm-Ups/Gallery/main.typ
Normal file
@ -0,0 +1,103 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: The Gallery],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#let thisone = if_solutions(
|
||||
text(fill: ored, [#sym.arrow.l.double.long `this one`]),
|
||||
)
|
||||
|
||||
A museum curator is arranging seven photographs on a gallery wall in accordance with the photographer's requirements.
|
||||
They are titled as follows: Fence, Gardenias, Hibiscus, Irises, Katydid, Lotus, and Magnolia.
|
||||
|
||||
The photograph's requirements are as follows:
|
||||
#v(2mm)
|
||||
- Gardenias must be immediately before Katydid.
|
||||
- Hibiscus must be somewhere before Katydid but cannot be the first photograph.
|
||||
- Irises and Lotus must be next to one another.
|
||||
- Magnolia must be one of the first three photographs.
|
||||
- Fence must be either first or seventh.
|
||||
|
||||
|
||||
#problem()
|
||||
Which of the below could be a valid ordering? \
|
||||
#note([We denote each painting by the first letter of its title.], type: "Note")
|
||||
- `FHGMKIL`
|
||||
- `HMGKILF`
|
||||
- `ILMHGKF` #thisone
|
||||
- `LMIHGKF`
|
||||
- `MFHGKLI`
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
If Irises is immediately before Gardenias, which of the following could be true?
|
||||
- Gardenias is fourth
|
||||
- Hibiscus is fourth
|
||||
- Irises is third
|
||||
- Lotus is second
|
||||
- Magnolia is third #thisone
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
The ordering of the photographs is fully determined if...
|
||||
- Gardenias is fourth
|
||||
- Hibiscus is second
|
||||
- Irises is second
|
||||
- Lotus is first #thisone
|
||||
- Magnolia is third
|
||||
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
If Magnolia is second, what CANNOT be true?
|
||||
- Hibiscus is third
|
||||
- Hibiscus is fourth #thisone
|
||||
- Hibiscus is fifth
|
||||
- Gardenias is fourth
|
||||
- Gardenias is sixth
|
||||
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Katydid cannot be in which position?
|
||||
- Third #thisone
|
||||
- Fourth
|
||||
- Fifth
|
||||
- Sixth
|
||||
- Seventh
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
If Gardenias is fourth, what must be true?
|
||||
- Fence is first #thisone
|
||||
- Hibiscus is third
|
||||
- Irises is seventh
|
||||
- Magnolia is first
|
||||
- Magnolia is second
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Which one of the following,
|
||||
if substituted for the second condition,
|
||||
would have the same effect in determining the
|
||||
arrangement of the photographs?
|
||||
|
||||
- If Fence is seventh, Hibiscus is second
|
||||
- Gardenias is somewhere after Hibiscus, and either Fence or Magnolia is first
|
||||
- Hibiscus must be somewhere between the first and sixth photographs
|
||||
- Unless Hibiscus is second, it must be somewhere between Magnolia and Gardenias \
|
||||
#if_solutions(text(fill: ored, [#sym.arrow.t.double `this one`]))
|
||||
- Katydid is somewhere after Hibiscus, which must be after Fence.
|
||||
|
||||
#v(1fr)
|
||||
@ -1,54 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\title{Warm-Up: Mario Kart}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
A standard Mario Kart cup consists of 12 players and four races. \par
|
||||
Each race is scored as follows:
|
||||
\begin{itemize}
|
||||
\item 15 points are awarded for first place;
|
||||
\item 12 for second;
|
||||
\item and $(13 - \text{place})$ otherwise.
|
||||
\end{itemize}
|
||||
In any one race, no players may tie.
|
||||
A player's score at the end of a cup is the sum of their scores for each of the four races.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
An $n$-way tie occurs when the top $n$ players have the same score at the end of a round. \par
|
||||
What is the largest possible $n$, and how is it achieved?
|
||||
|
||||
\begin{solution}
|
||||
A 12-way tie is impossible, since the total number of point is not divisible by 12.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
A 11-way tie is possible, with a top score of 28:
|
||||
\begin{itemize}
|
||||
\item Four players finish $1^\text{st}$, $3^\text{ed}$, $11^\text{th}$, and $12^\text{th}$;
|
||||
|
||||
% spell:off
|
||||
\item Four players finish $2^\text{nd}$, $4^\text{th}$, $9^\text{th}$, and $10^\text{th}$;
|
||||
% spell:on
|
||||
|
||||
\item Two players finish fifth twice and seventh twice,
|
||||
\item One player finishes sixth in each race.
|
||||
\end{itemize}
|
||||
The final player always finishes eighth, with a non-tie score of 20.
|
||||
\end{solution}
|
||||
|
||||
\end{document}
|
||||
35
src/Warm-Ups/Mario Kart/main.typ
Normal file
35
src/Warm-Ups/Mario Kart/main.typ
Normal file
@ -0,0 +1,35 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Mario Kart],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
A standard Mario Kart cup consists of 12 players and four races. \
|
||||
Each race is scored as follows:
|
||||
- 15 points are awarded for first place;
|
||||
- 12 for second;
|
||||
- and $(13 - #text("place"))$ otherwise.
|
||||
|
||||
In any one race, no players may tie. \
|
||||
A player's score at the end of a cup is the sum of their scores for each of the four races.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
An $n$-way tie occurs when the top $n$ players have the same score at the end of a round. \
|
||||
What is the largest possible $n$, and how is it achieved?
|
||||
|
||||
#solution([
|
||||
A 12-way tie is impossible, since the total number of point is not divisible by 12.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
A 11-way tie is possible, with a top score of 28:
|
||||
- Four players finish $1^#text("st")$, $3^#text("ed")$, $11^#text("th")$, and $12^#text("th")$;
|
||||
- Four players finish $2^#text("nd")$, $4^#text("th")$, $9^#text("th")$, and $10^#text("th")$; // spell:disable-line
|
||||
- Two players finish fifth twice and seventh twice,
|
||||
- One player finishes sixth in each race.
|
||||
The final player always finishes eighth, with a non-tie score of 20.
|
||||
|
||||
])
|
||||
@ -1,132 +0,0 @@
|
||||
\documentclass[
|
||||
nosolutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{arrows.meta}
|
||||
\usetikzlibrary{shapes.geometric}
|
||||
|
||||
% We put nodes in a separate layer, so we can
|
||||
% slightly overlap with paths for a perfect fit
|
||||
\pgfdeclarelayer{nodes}
|
||||
\pgfdeclarelayer{path}
|
||||
\pgfsetlayers{main,nodes}
|
||||
|
||||
% Layer settings
|
||||
\tikzset{
|
||||
% Layer hack, lets us write
|
||||
% later = * in scopes.
|
||||
layer/.style = {
|
||||
execute at begin scope={\pgfonlayer{#1}},
|
||||
execute at end scope={\endpgfonlayer}
|
||||
},
|
||||
%
|
||||
% Arrowhead tweak
|
||||
>={Latex[ width=2mm, length=2mm ]},
|
||||
%
|
||||
% Nodes
|
||||
main/.style = {
|
||||
draw,
|
||||
circle,
|
||||
fill = white,
|
||||
line width = 0.35mm
|
||||
}
|
||||
}
|
||||
|
||||
\title{Warm Up: Odd Dice}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
|
||||
We say a set of dice $\{A, B, C\}$ is \textit{nontransitive}
|
||||
if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
|
||||
In other words, we get a counterintuitive \say{rock - paper - scissors} effect.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Create a set of nontransitive six-sided dice. \par
|
||||
\hint{All sides should be numbered with positive integers less than 10.}
|
||||
|
||||
\begin{solution}
|
||||
One possible set can be numbered as follows:
|
||||
\begin{itemize}
|
||||
\item Die $A$: $2, 2, 4, 4, 9, 9$
|
||||
\item Die $B$: $1, 1, 6, 6, 8, 8$
|
||||
\item Die $C$: $3, 3, 5, 5, 7, 7$
|
||||
\end{itemize}
|
||||
|
||||
\vspace{4mm}
|
||||
|
||||
Another solution is below:
|
||||
\begin{itemize}
|
||||
\item Die $A$: $3, 3, 3, 3, 3, 6$
|
||||
\item Die $B$: $2, 2, 2, 5, 5, 5$
|
||||
\item Die $C$: $1, 4, 4, 4, 4, 4$
|
||||
\end{itemize}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Now, consider the set of six-sided dice below:
|
||||
\begin{itemize}
|
||||
\item Die $A$: $4, 4, 4, 4, 4, 9$
|
||||
\item Die $B$: $3, 3, 3, 3, 8, 8$
|
||||
\item Die $C$: $2, 2, 2, 7, 7, 7$
|
||||
\item Die $D$: $1, 1, 6, 6, 6, 6$
|
||||
\item Die $E$: $0, 5, 5, 5, 5, 5$
|
||||
\end{itemize}
|
||||
On average, which die beats each of the others? Draw a graph. \par
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale = 0.5]
|
||||
\begin{scope}[layer = nodes]
|
||||
\node[main] (a) at (-2, 0.2) {$a$};
|
||||
\node[main] (b) at (0, 2) {$b$};
|
||||
\node[main] (c) at (2, 0.2) {$c$};
|
||||
\node[main] (d) at (1, -2) {$d$};
|
||||
\node[main] (e) at (-1, -2) {$e$};
|
||||
\end{scope}
|
||||
|
||||
\draw[->]
|
||||
(a) edge (b)
|
||||
(b) edge (c)
|
||||
(c) edge (d)
|
||||
(d) edge (e)
|
||||
(e) edge (a)
|
||||
|
||||
(a) edge (c)
|
||||
(b) edge (d)
|
||||
(c) edge (e)
|
||||
(d) edge (a)
|
||||
(e) edge (b)
|
||||
;
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
Now, say we roll each die twice. What happens to the graph above?
|
||||
|
||||
\begin{solution}
|
||||
The direction of each edge is reversed!
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
111
src/Warm-Ups/Odd Dice/main.typ
Normal file
111
src/Warm-Ups/Odd Dice/main.typ
Normal file
@ -0,0 +1,111 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Odd Dice],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
We say a set of dice ${A, B, C}$ is _nontransitive_
|
||||
if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
|
||||
In other words, we get a counterintuitive "rock - paper - scissors" effect.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Create a set of nontransitive six-sided dice. \
|
||||
#hint([All sides should be numbered with positive integers less than 10.])
|
||||
|
||||
#solution([
|
||||
One possible set can be numbered as follows:
|
||||
- Die $A$: $2, 2, 4, 4, 9, 9$
|
||||
- Die $B$: $1, 1, 6, 6, 8, 8$
|
||||
- Die $C$: $3, 3, 5, 5, 7, 7$
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Another solution is below:
|
||||
- Die $A$: $3, 3, 3, 3, 3, 6$
|
||||
- Die $B$: $2, 2, 2, 5, 5, 5$
|
||||
- Die $C$: $1, 4, 4, 4, 4, 4$
|
||||
|
||||
])
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Now, consider the set of six-sided dice below:
|
||||
- Die $A$: $4, 4, 4, 4, 4, 9$
|
||||
- Die $B$: $3, 3, 3, 3, 8, 8$
|
||||
- Die $C$: $2, 2, 2, 7, 7, 7$
|
||||
- Die $D$: $1, 1, 6, 6, 6, 6$
|
||||
- Die $E$: $0, 5, 5, 5, 5, 5$
|
||||
On average, which die beats each of the others? Draw a diagram.
|
||||
|
||||
#solution(
|
||||
align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let s = 0.8 // Scale
|
||||
let t = 13pt * s // text size
|
||||
let radius = 0.3 * s
|
||||
|
||||
// Points
|
||||
let a = (-2 * s, 0.2 * s)
|
||||
let b = (0 * s, 2 * s)
|
||||
let c = (2 * s, 0.2 * s)
|
||||
let d = (1.2 * s, -2.1 * s)
|
||||
let e = (-1.2 * s, -2.1 * s)
|
||||
|
||||
set-style(
|
||||
stroke: (thickness: 0.6mm * s),
|
||||
mark: (
|
||||
end: (
|
||||
symbol: ">",
|
||||
fill: black,
|
||||
offset: radius + (0.025 * s),
|
||||
width: 1.2mm * s,
|
||||
length: 1.2mm * s,
|
||||
),
|
||||
),
|
||||
)
|
||||
|
||||
line(a, b)
|
||||
line(b, c)
|
||||
line(c, d)
|
||||
line(d, e)
|
||||
line(e, a)
|
||||
line(a, c)
|
||||
line(b, d)
|
||||
line(c, e)
|
||||
line(d, a)
|
||||
line(e, b)
|
||||
|
||||
circle(a, radius: radius, fill: oblue, stroke: none)
|
||||
circle(b, radius: radius, fill: oblue, stroke: none)
|
||||
circle(c, radius: radius, fill: oblue, stroke: none)
|
||||
circle(d, radius: radius, fill: oblue, stroke: none)
|
||||
circle(e, radius: radius, fill: oblue, stroke: none)
|
||||
|
||||
content(a, text(fill: white, size: t, [*A*]))
|
||||
content(b, text(fill: white, size: t, [*B*]))
|
||||
content(c, text(fill: white, size: t, [*C*]))
|
||||
content(d, text(fill: white, size: t, [*D*]))
|
||||
content(e, text(fill: white, size: t, [*E*]))
|
||||
}),
|
||||
),
|
||||
)
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Now, say we roll each die twice. What happens to the graph from the previous problem?
|
||||
|
||||
#solution([
|
||||
The direction of each edge is reversed!
|
||||
])
|
||||
|
||||
#v(1fr)
|
||||
@ -1,80 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
\title{Warm-Up: The Painting}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
You have a painting on a string. \par
|
||||
Hang the painting on two nails so that if either is removed, the painting falls. \par
|
||||
|
||||
\vspace{2mm}
|
||||
You may detach the string as you hang the painting, but it must be re-attached once you're done.
|
||||
|
||||
\note{The solution to this problem isn't a \say{think outside the box} trick, it's a clever wrapping of the string.}
|
||||
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=2.0]
|
||||
\draw[line width = 0.5mm]
|
||||
(0, 1) --
|
||||
(2, 1) --
|
||||
(2, 0) --
|
||||
(0, 0) --
|
||||
cycle
|
||||
;
|
||||
|
||||
\draw[line width = 0.5mm, dotted]
|
||||
(0.1, 1) --
|
||||
(1, 1.5) --
|
||||
(1.9, 1)
|
||||
;
|
||||
|
||||
\fill (1, 1.5) circle[radius = 0.25mm];
|
||||
|
||||
|
||||
\draw[line width = 0.2mm]
|
||||
(0.66, 0.66) --
|
||||
(0.66, 0.35) --
|
||||
(0.60, 0.1)
|
||||
(0.72, 0.1)--(0.66, 0.35)
|
||||
;
|
||||
|
||||
\draw[line width = 0.2mm]
|
||||
(0.66, 0.575) --
|
||||
(0.6, 0.475) --
|
||||
(0.525, 0.575)
|
||||
(0.66, 0.575) --
|
||||
(0.72, 0.475) --
|
||||
(0.795, 0.575)
|
||||
;
|
||||
|
||||
\fill[color=white] (0.66, 0.66) circle[radius = 0.8mm];
|
||||
\draw (0.66, 0.66) circle[radius = 0.8mm];
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
\begin{solution}
|
||||
|
||||
Say we have a left nail and a right nail. The path of the string is as follows:
|
||||
\begin{itemize}
|
||||
\item Start on the left
|
||||
\item Move over both nails, wrap right nail cw
|
||||
\item Wrap left nail ccw
|
||||
\item Wrap right nail ccw
|
||||
\item Exit downwards, in between both nails
|
||||
\end{itemize}
|
||||
|
||||
\end{solution}
|
||||
|
||||
\end{document}
|
||||
82
src/Warm-Ups/Painting/main.typ
Normal file
82
src/Warm-Ups/Painting/main.typ
Normal file
@ -0,0 +1,82 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: What's an AST?],
|
||||
by: "Mark",
|
||||
subtitle: "Based on a true story.",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Say we have a painting on a string. \
|
||||
Hang the painting on two nails so that if either is removed, the painting falls. \
|
||||
|
||||
#v(2mm)
|
||||
You may detach the string as you hang the painting, but it must be re-attached once you're done. \
|
||||
#hint[The solution to this problem isn't a "think outside the box" trick, it's a clever wrapping of the string.]
|
||||
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 2.5
|
||||
|
||||
|
||||
line(
|
||||
(0 * s, 1 * s),
|
||||
(2 * s, 1 * s),
|
||||
(2 * s, 0 * s),
|
||||
(0 * s, 0 * s),
|
||||
close: true,
|
||||
stroke: (thickness: 0.8mm),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.1 * s, 1 * s),
|
||||
(0.5 * s, 1.5 * s),
|
||||
(1.5 * s, 1.5 * s),
|
||||
(1.9 * s, 1 * s),
|
||||
stroke: (thickness: 0.5mm, dash: "dotted"),
|
||||
)
|
||||
|
||||
circle((0.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
|
||||
circle((1.5 * s, 1.5 * s), radius: 0.04 * s, fill: black, stroke: none)
|
||||
|
||||
line(
|
||||
(0.66 * s, 0.66 * s),
|
||||
(0.66 * s, 0.35 * s),
|
||||
(0.60 * s, 0.1 * s),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.72 * s, 0.1 * s),
|
||||
(0.66 * s, 0.35 * s),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.66 * s, 0.575 * s),
|
||||
(0.6 * s, 0.475 * s),
|
||||
(0.525 * s, 0.575 * s),
|
||||
)
|
||||
|
||||
line(
|
||||
(0.66 * s, 0.575 * s),
|
||||
(0.72 * s, 0.475 * s),
|
||||
(0.795 * s, 0.575 * s),
|
||||
)
|
||||
|
||||
circle((0.66 * s, 0.66 * s), radius: 0.07 * s, fill: white)
|
||||
}),
|
||||
)
|
||||
|
||||
#solution([
|
||||
Say we have a left nail and a right nail. The path of the string is as follows:
|
||||
- Start on the left
|
||||
- Move over both nails, wrap right nail cw
|
||||
- Wrap left nail ccw
|
||||
- Wrap right nail ccw
|
||||
- Exit downwards, in between both nails
|
||||
])
|
||||
@ -1,57 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\title{Warm-Up: Partition Products}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
Take any positive integer $n$. \par
|
||||
Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... + a_k$ \par
|
||||
Maximize the product $a_1 \times a_2 \times ... \times a_k$
|
||||
|
||||
|
||||
|
||||
\begin{solution}
|
||||
|
||||
\textbf{Interesting Solution:}
|
||||
|
||||
Of course, all $a_i$ should be greater than $1$. \par
|
||||
Also, all $a_i$ should be smaller than four, since $x \leq x(x-2)$ if $x \geq 4$. \par
|
||||
Thus, we're left with sequences that only contain 2 and 3. \par
|
||||
\note{Note that two twos are the same as one four, but we exclude fours for simplicity.}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \par
|
||||
The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
|
||||
|
||||
\linehack{}
|
||||
|
||||
|
||||
|
||||
\textbf{Calculus Solution:}
|
||||
|
||||
First, solve this problem for equal, non-integer $a_i$:
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
We know $n = \prod{a_i}$, thus $\ln(n) = \sum{\ln(a_i)}$. \par
|
||||
If all $a_i$ are equal, we get $\ln(n) = k \times \ln(n / k)$. \par
|
||||
Derive wrt $k$ and set to zero to get $\ln(n / k) = 1$ \par
|
||||
So $k = n / e$ and $n / k = e \approx 2.7$
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
If we try to approximate this with integers, we get the same solution as above.
|
||||
\end{solution}
|
||||
\end{document}
|
||||
41
src/Warm-Ups/Partition Products/main.typ
Normal file
41
src/Warm-Ups/Partition Products/main.typ
Normal file
@ -0,0 +1,41 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Partition Products],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Take any positive integer $n$. \
|
||||
Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... a_k$ \
|
||||
Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
|
||||
|
||||
|
||||
#solution([
|
||||
*Interesting Solution:*
|
||||
|
||||
Of course, all $a_i$ should be greater than $1$. \
|
||||
Also, all $a_i$ should be smaller than four, since $x <= x(x-2)$ if $x >= 4$. \
|
||||
Thus, we're left with sequences that only contain 2 and 3. \
|
||||
#note([Note that two twos are the same as one four, but we exclude fours for simplicity.])
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \
|
||||
The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
|
||||
|
||||
#v(8mm)
|
||||
|
||||
*Calculus Solution:*
|
||||
|
||||
First, solve this problem for equal, real $a_i$:
|
||||
#v(2mm)
|
||||
We know $n = product(a_i)$, thus $ln(n) = sum(ln(a_i))$. \
|
||||
If all $a_i$ are equal, we get $ln(n) = k #sym.times ln(n / k)$. \
|
||||
Derive wrt $k$ and set to zero to get $ln(n / k) = 1$ \
|
||||
So $k = n / e$ and $n / k = e #sym.approx 2.7$
|
||||
|
||||
#v(2mm)
|
||||
|
||||
If we try to approximate this with integers, we get the same solution as above.
|
||||
])
|
||||
@ -1,47 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
\usepackage{graphicx}
|
||||
|
||||
\title{Warm-Up: Passing Balls}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
|
||||
Twelve people are standing in a circle. Each is assigned a number between 1 and 12. \par
|
||||
Participants numbered 1, 2, 3, and 4 hold red, green, yellow, and black balls, respectively. \par
|
||||
Everyone else is empty-handed.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Each participant can pass their ball to any student that is exactly 5 positions away. \par
|
||||
Balls cannot be passed to someone who has one in hand.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
After a number of passes, the first four participants again hold all the balls. \par
|
||||
Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4?
|
||||
|
||||
\begin{solution}
|
||||
\begin{itemize}
|
||||
\item the graph of possible moves is isomorphic to a circle (since 5 and 12 are coprime),
|
||||
\item but the balls get passed around, so swapping the place of any two balls is not allowed. \\
|
||||
Therefore, the balls will stay in their initial (cyclic) order:
|
||||
\end{itemize}
|
||||
|
||||
\begin{center}
|
||||
\includegraphics[width=\textwidth]{pass-sol.png}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
|
||||
\end{document}
|
||||
192
src/Warm-Ups/Passing Balls/main.typ
Normal file
192
src/Warm-Ups/Passing Balls/main.typ
Normal file
@ -0,0 +1,192 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Passing Balls],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Twelve people are standing in a circle. Each is assigned a number between 1 and 12. \
|
||||
Participants numbered 1, 2, 3, and 4 hold red, green, yellow, and black balls, respectively. \
|
||||
Everyone else is empty-handed.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Each participant can pass their ball to any student that is exactly 5 positions away. \
|
||||
Balls cannot be passed to someone who has one in hand.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
After a number of passes, the first four participants again hold all the balls. \
|
||||
Participant 1 has a black ball. Which balls are held by participants 2, 3, and 4?
|
||||
|
||||
#solution([
|
||||
The graph of possible moves is isomorphic to a circle (since 5 and 12 are coprime), \
|
||||
so the order of the balls cannot be changed as they are passed around.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Therefore, the balls will stay in their initial (cyclic) order:
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#{
|
||||
let s = 0.7 // scale
|
||||
let t = 12pt * s // text size
|
||||
let radius = 0.35
|
||||
|
||||
let pts = (
|
||||
(0 * s, 3 * s),
|
||||
(1 * s, 2 * s),
|
||||
(2 * s, 1 * s),
|
||||
(3 * s, 0 * s),
|
||||
(2 * s, -1 * s),
|
||||
(1 * s, -2 * s),
|
||||
(0 * s, -3 * s),
|
||||
(-1 * s, -2 * s),
|
||||
(-2 * s, -1 * s),
|
||||
(-3 * s, 0 * s),
|
||||
(-2 * s, 1 * s),
|
||||
(-1 * s, 2 * s),
|
||||
)
|
||||
|
||||
let pts_shuf = (
|
||||
(0 * s, 3 * s), // 1
|
||||
(1 * s, -2 * s), // 6
|
||||
(-2 * s, 1 * s), // 11
|
||||
(3 * s, 0 * s), // 4
|
||||
(-2 * s, -1 * s), // 9
|
||||
(1 * s, 2 * s), // 2
|
||||
(0 * s, -3 * s), // 7
|
||||
(-1 * s, 2 * s), // 12
|
||||
(2 * s, -1 * s), // 5
|
||||
(-3 * s, 0 * s), // 10
|
||||
(2 * s, 1 * s), // 3
|
||||
(-1 * s, -2 * s), // 8
|
||||
)
|
||||
|
||||
table(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr, 1fr),
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
set-style(stroke: (thickness: 0.4mm, paint: black))
|
||||
line(..pts_shuf, close: true)
|
||||
|
||||
let i = 1
|
||||
for p in pts {
|
||||
circle(
|
||||
p,
|
||||
radius: radius * s,
|
||||
fill: if i == 1 {
|
||||
ored
|
||||
} else if i == 2 {
|
||||
ogreen
|
||||
} else if i == 3 {
|
||||
oorange
|
||||
} else if i == 4 {
|
||||
oblue
|
||||
} else { white },
|
||||
)
|
||||
|
||||
content(
|
||||
p,
|
||||
text(
|
||||
fill: if i <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#i*],
|
||||
),
|
||||
)
|
||||
i = i + 1
|
||||
}
|
||||
}),
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
set-style(stroke: (thickness: 0.4mm, paint: black))
|
||||
line(..pts, close: true)
|
||||
|
||||
let i = 1
|
||||
for p in pts {
|
||||
let l = calc.rem(((i - 1) * 5), 12) + 1
|
||||
|
||||
|
||||
circle(
|
||||
p,
|
||||
radius: radius * s,
|
||||
fill: if l == 1 {
|
||||
ored
|
||||
} else if l == 2 {
|
||||
ogreen
|
||||
} else if l == 3 {
|
||||
oorange
|
||||
} else if l == 4 {
|
||||
oblue
|
||||
} else { white },
|
||||
)
|
||||
|
||||
content(
|
||||
p,
|
||||
text(
|
||||
fill: if l <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#l*],
|
||||
),
|
||||
)
|
||||
i = i + 1
|
||||
}
|
||||
}),
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
set-style(stroke: (thickness: 0.4mm, paint: black))
|
||||
line(..pts, close: true)
|
||||
|
||||
let i = 1
|
||||
for p in pts {
|
||||
let l = calc.rem(((i - 1) * 5), 12) + 1
|
||||
|
||||
|
||||
circle(
|
||||
p,
|
||||
radius: radius * s,
|
||||
fill: if l == 1 {
|
||||
oblue
|
||||
} else if l == 2 {
|
||||
oorange
|
||||
} else if l == 3 {
|
||||
ored
|
||||
} else if l == 4 {
|
||||
ogreen
|
||||
} else { white },
|
||||
)
|
||||
|
||||
content(
|
||||
p,
|
||||
text(
|
||||
fill: if l <= 4 {
|
||||
white
|
||||
} else {
|
||||
black
|
||||
},
|
||||
size: t,
|
||||
[*#l*],
|
||||
),
|
||||
)
|
||||
i = i + 1
|
||||
}
|
||||
}),
|
||||
)
|
||||
}
|
||||
])
|
||||
Binary file not shown.
|
Before Width: | Height: | Size: 47 KiB |
@ -1,34 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber,
|
||||
hidewarning
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
\title{Warm-Up: Prime Factors}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
What proportion of integers have $2$ as their smallest prime factor?
|
||||
% 1^2
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
What proportion of integers have $3$ as their second-smallest prime factor?
|
||||
% 1/6
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
What is the median second-smallest prime factor?
|
||||
% 37
|
||||
\vfill
|
||||
|
||||
\end{document}
|
||||
23
src/Warm-Ups/Prime Factors/main.typ
Normal file
23
src/Warm-Ups/Prime Factors/main.typ
Normal file
@ -0,0 +1,23 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Prime Factors],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
What proportion of integers have $2$ as their smallest prime factor?
|
||||
#solution([$1 div 2$])
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
What proportion of integers have $3$ as their second-smallest prime factor?
|
||||
#solution([$1 div 6$])
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
What is the median second-smallest prime factor?
|
||||
#solution([37])
|
||||
#v(1fr)
|
||||
@ -1,153 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\usepackage{xcolor}
|
||||
\usepackage{soul}
|
||||
\usepackage{hyperref}
|
||||
|
||||
\definecolor{Light}{gray}{.90}
|
||||
\sethlcolor{Light}
|
||||
\newcommand{\htexttt}[1]{\texttt{\hl{#1}}}
|
||||
|
||||
|
||||
\title{The Regex Warm-Up}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
|
||||
Last time, we discussed Deterministic Finite Automata. One interesting application of these mathematical objects is found in computer science: Regular Expressions. \par
|
||||
This is often abbreviated \say{regex}, which is pronounced like \say{gif.}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Regex is a language used to specify patterns in a string. You can think of it as a concise way to define a DFA, using text instead of a huge graph. \par
|
||||
|
||||
Often enough, a clever regex pattern can do the work of a few hundred lines of code.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Like the DFAs we've studied, a regex pattern \textit{accepts} or \textit{rejects} a string. However, we don't usually use this terminology with regex, and instead say that a string \textit{matches} or \textit{doesn't match} a pattern.
|
||||
|
||||
\vspace{5mm}
|
||||
|
||||
Regex strings consist of characters, quantifiers, sets, and groups.
|
||||
|
||||
\vspace{5mm}
|
||||
|
||||
\textbf{Quantifiers} \par
|
||||
Quantifiers specify how many of a character to match. \par
|
||||
There are four of these: \htexttt{+}, \htexttt{*}, \htexttt{?}, and \htexttt{\{ \}}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
\htexttt{+} means \say{match one or more of the preceding token} \par
|
||||
\htexttt{*} means \say{match zero or more of the preceding token}
|
||||
|
||||
For example, the pattern \htexttt{ca+t} will match the following strings:
|
||||
\begin{itemize}
|
||||
\item \texttt{cat}
|
||||
\item \texttt{caat}
|
||||
\item \texttt{caaaaaaaat}
|
||||
\end{itemize}
|
||||
\htexttt{ca+t} will \textbf{not} match the string \texttt{ct}. \par
|
||||
The pattern \htexttt{ca*t} will match all the strings above, including \texttt{ct}.
|
||||
\vspace{2mm}
|
||||
|
||||
|
||||
\htexttt{?} means \say{match one or none of the preceding token} \par
|
||||
The pattern \htexttt{linea?r} will match only \texttt{linear} and \texttt{liner}.
|
||||
\vspace{2mm}
|
||||
|
||||
Brackets \htexttt{\{min, max\}} are the most flexible quantifier. \par
|
||||
They specify exactly how many tokens to match: \par
|
||||
\htexttt{ab\{2\}a} will match only \texttt{abba}. \par
|
||||
\htexttt{ab\{1,3\}a} will match only \texttt{aba}, \texttt{abba}, and \texttt{abbba}. \par
|
||||
% spell:off
|
||||
\htexttt{ab\{2,\}a} will match any \texttt{ab...ba} with at least two \texttt{b}s.
|
||||
% spell:on
|
||||
|
||||
\vspace{5mm}
|
||||
|
||||
\problem{}
|
||||
Write the patterns \htexttt{a*} and \htexttt{a+} using only \htexttt{\{ \}}.
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Draw a DFA equivalent to the regex pattern \htexttt{01*0}.
|
||||
\vfill
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\textbf{Characters, Sets, and Groups} \par
|
||||
In the previous section, we saw how we can specify characters literally: \par
|
||||
\texttt{a+} means \say{one or more \texttt{a} character}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
There are, of course, other ways we can specify characters.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
The first such way is the \textit{set}, denoted \htexttt{[ ]}. A set can pretend to be any character inside it. \par
|
||||
For example, \htexttt{m[aoy]th} will match \texttt{math}, \texttt{moth}, or \texttt{myth}. \par
|
||||
\htexttt{a[01]+b} will match \texttt{a0b}, \texttt{a111b}, \texttt{a1100110b}, and any other similar string. \par
|
||||
You may negate a set with a \htexttt{\textasciicircum}. \par
|
||||
\htexttt{[\textasciicircum abc]} will match any character except \texttt{a}, \texttt{b}, or \texttt{c}, including symbols and spaces.
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
If we want to keep characters together, we can use the \textit{group}, denoted \htexttt{( )}. \par
|
||||
|
||||
Groups work exactly as you'd expect, representing an atomic\footnotemark{} group of characters. \par
|
||||
\htexttt{a(01)+b} will match \texttt{a01b} and \texttt{a010101b}, but will \textbf{not} match \texttt{a0b}, \texttt{a1b}, or \texttt{a1100110b}.
|
||||
|
||||
\footnotetext{In other words, \say{unbreakable}}
|
||||
|
||||
|
||||
\problem{}<regex>
|
||||
You are now familiar with most of the tools regex has to offer. \par
|
||||
Write patterns that match the following strings:
|
||||
\begin{enumerate}[itemsep=1mm]
|
||||
\item An ISO-8601 date, like \texttt{2022-10-29}. \par
|
||||
\hint{Invalid dates like \texttt{2022-13-29} should also be matched.}
|
||||
|
||||
\item An email address. \par
|
||||
\hint{Don't forget about subdomains, like \texttt{math.ucla.edu}.}
|
||||
|
||||
\item A UCLA room number, like \texttt{MS 5118} or \texttt{Kinsey 1220B}.
|
||||
|
||||
\item Any ISBN-10 of the form \texttt{0-316-00395-7}. \par
|
||||
\hint{Remember that the check digit may be an \texttt{X}. Dashes are optional.}
|
||||
|
||||
\item A word of even length. \par
|
||||
\hint{The set \texttt{[A-z]} contains every english letter, capitalized and lowercase. \\
|
||||
\texttt{[a-z]} will only match lowercase letters.}
|
||||
|
||||
\item A word with exactly 3 vowels. \par
|
||||
\hint{The special token \texttt{\textbackslash w} will match any word character. It is equivalent to \texttt{[A-z0-9\_]} \\ \texttt{\_} stands for a literal underscore.}
|
||||
|
||||
\item A word that has even length and exactly 3 vowels.
|
||||
|
||||
\item A sentence that does not start with a capital letter.
|
||||
\end{enumerate}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
If you'd like to know more, check out \url{https://regexr.com}. It offers an interactive regex prompt, as well as a cheatsheet that explains every other regex token there is. \par
|
||||
You will find a nice set of challenges at \url{https://alf.nu/RegexGolf}.
|
||||
I especially encourage you to look into this if you are interested in computer science.
|
||||
\end{document}
|
||||
135
src/Warm-Ups/Regex/main.typ
Normal file
135
src/Warm-Ups/Regex/main.typ
Normal file
@ -0,0 +1,135 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [The Regex Warm-Up],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
|
||||
Last time, we discussed Deterministic Finite Automata. One interesting application of these mathematical objects is found in computer science: Regular Expressions. \
|
||||
This is often abbreviated "regex," which is pronounced like "gif."
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Regex is a language used to specify patterns in a string. You can think of it as a concise way to define a DFA, using text instead of a huge graph. \
|
||||
|
||||
Often enough, a clever regex pattern can do the work of a few hundred lines of code.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Like the DFAs we've studied, a regex pattern _accepts_ or _rejects_ a string. However, we don't usually use this terminology with regex, and instead say that a string _matches_ or _doesn't match_ a pattern.
|
||||
|
||||
#v(5mm)
|
||||
|
||||
Regex strings consist of characters, quantifiers, sets, and groups.
|
||||
|
||||
#v(5mm)
|
||||
|
||||
|
||||
|
||||
*Quantifiers* \
|
||||
Quantifiers specify how many of a character to match. \
|
||||
There are four of these: `+`, `*`, `?`, and `{ }`.
|
||||
|
||||
#v(4mm)
|
||||
|
||||
`+` means "match one or more of the preceding token" \
|
||||
`*` means "match zero or more of the preceding token"
|
||||
|
||||
For example, the pattern `ca+t` will match the following strings:
|
||||
- `cat`
|
||||
- `caat`
|
||||
- `caaaaaaaat`
|
||||
`ca+t` will *not* match the string `ct`. \
|
||||
The pattern `ca*t` will match all the strings above, including `ct`.
|
||||
|
||||
|
||||
#v(4mm)
|
||||
|
||||
|
||||
`?` means "match one or none of the preceding token" \
|
||||
The pattern `linea?r` will match only `linear` and `liner`.
|
||||
|
||||
#v(4mm)
|
||||
|
||||
Brackets `{min, max}` are the most flexible quantifier. \
|
||||
They specify exactly how many tokens to match: \
|
||||
`ab{2}a` will match only `abba`. \
|
||||
`ab{1,3}a` will match only `aba`, `abba`, and `abbba`. \
|
||||
`ab{2,}a` will match any `ab...ba` with at least two `b`s. // spell:disable-line
|
||||
|
||||
#problem()
|
||||
Write the patterns `a*` and `a+` using only `{ }`.
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Draw a DFA equivalent to the regex pattern `01*0`.
|
||||
#v(1fr)
|
||||
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
*Characters, Sets, and Groups* \
|
||||
In the previous section, we saw how we can specify characters literally: \
|
||||
`a+` means "one or more `a` characters" \
|
||||
There are, of course, other ways we can specify characters.
|
||||
|
||||
#v(4mm)
|
||||
|
||||
The first such way is the _set_, denoted `[ ]`. A set can pretend to be any character inside it. \
|
||||
For example, `m[aoy]th` will match `math`, `moth`, or `myth`. \
|
||||
`a[01]+b` will match `a0b`, `a111b`, `a1100110b`, and any other similar string. \
|
||||
|
||||
#v(4mm)
|
||||
|
||||
We can negate a set with a `^`. \
|
||||
`[^abc]` will match any single character except `a`, `b`, or `c`, including symbols and spaces.
|
||||
|
||||
#v(4mm)
|
||||
|
||||
If we want to keep characters together, we can use the _group_, denoted `( )`. \
|
||||
|
||||
Groups work exactly as you'd expect, representing an atomic#footnote([In other words, "unbreakable"]) group of characters. \
|
||||
`a(01)+b` will match `a01b` and `a010101b`, but will *not* match `a0b`, `a1b`, or `a1100110b`.
|
||||
|
||||
#problem()
|
||||
You are now familiar with most of the tools regex has to offer. \
|
||||
Write patterns that match the following strings:
|
||||
|
||||
- An ISO-8601 date, like `2022-10-29`. \
|
||||
#hint([Invalid dates like `2022-13-29` should also be matched.])
|
||||
|
||||
- An email address. \
|
||||
#hint([Don't forget about subdomains, like `math.ucla.edu`.])
|
||||
|
||||
- A UCLA room number, like `MS 5118` or `Kinsey 1220B`.
|
||||
|
||||
- Any ISBN-10 of the form `0-316-00395-7`. \
|
||||
#hint([Remember that the check digit may be an `X`. Dashes are optional.])
|
||||
|
||||
- A word of even length. \
|
||||
#hint([
|
||||
The set `[A-z]` contains every english letter, capitalized and lowercase. \
|
||||
`[a-z]` will only match lowercase letters.
|
||||
])
|
||||
|
||||
- A word with exactly 3 vowels. \
|
||||
#hint([
|
||||
The special token `\w` will match any word character. \
|
||||
It is equivalent to `[A-z0-9_]`. `_` represents a literal underscore.
|
||||
])
|
||||
|
||||
- A word that has even length and exactly 3 vowels.
|
||||
|
||||
- A sentence that does not start with a capital letter.
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
If you'd like to know more, check out `https://regexr.com`.
|
||||
It offers an interactive regex prompt,
|
||||
as well as a cheatsheet that explains every other regex token there is. \
|
||||
You can find a nice set of challenges at `https://alf.nu/RegexGolf`.
|
||||
75
src/Warm-Ups/Somewhat Random Numbers/main.typ
Normal file
75
src/Warm-Ups/Somewhat Random Numbers/main.typ
Normal file
@ -0,0 +1,75 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Somewhat Random Numbers],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Alice generates 100 random numbers uniformly from $[0,1]$. \
|
||||
Bob generates 101 random numbers from $[0, 1]$, but deletes the lowest result.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Say we have both of the resulting arrays, but do not know who generated each one. \
|
||||
We would like to guess which of the two was generated by Bob. \
|
||||
What is the optimal strategy, and what is its probability of guessing correctly?
|
||||
|
||||
|
||||
#solution([
|
||||
Looking at the mean seems like a good idea, but there's a better way: \
|
||||
Assign the array with the smaller _minimum_ to Alice.
|
||||
|
||||
#v(3mm)
|
||||
|
||||
To compute the probability, generate 201 numbers. \
|
||||
Assign the first 100 to Alice and the rest to Bob. \
|
||||
Look at the lowest two numbers (of these 201, *before* Bob drops his lowest).
|
||||
|
||||
#v(8mm)
|
||||
|
||||
We'll use the following notation: \
|
||||
`AB` means the lowest was owned by Alice, and the second-lowest, by Bob.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Probabilities are as follows: \
|
||||
- `AA`: $100\/201 times 99\/200 approx 0.246$
|
||||
- `AB`: $100\/201 times 101\/200 approx 0.251$
|
||||
- `BA`: $101\/201 times 100\/200 approx 0.251$ // spell:disable-line
|
||||
- `BB`: $101\/201 times 100\/200 approx 0.251$
|
||||
|
||||
#v(4mm)
|
||||
Now, Bob drops his lowest number. \
|
||||
We'll cross out the number he drops and box the new lowest number (i.e, the one we observe):
|
||||
- #{
|
||||
(
|
||||
box(`A`, stroke: ored, inset: 1pt)
|
||||
+ box(`A`, inset: 1pt)
|
||||
+ box([: $approx 0.246$], inset: (top: 1pt, bottom: 1pt))
|
||||
)
|
||||
}
|
||||
- #{
|
||||
(
|
||||
box(`A`, stroke: ored, inset: 1pt)
|
||||
+ box(strike(`B`), inset: 1pt)
|
||||
+ box([: $approx 0.251$], inset: (top: 1pt, bottom: 1pt))
|
||||
)
|
||||
}
|
||||
- #{
|
||||
(
|
||||
box(strike(`B`), inset: 1pt)
|
||||
+ box(`A`, stroke: ored, inset: 1pt)
|
||||
+ box([: $approx 0.251$], inset: (top: 1pt, bottom: 1pt))
|
||||
)
|
||||
}
|
||||
- #{
|
||||
(
|
||||
box(strike(`B`), inset: 1pt)
|
||||
+ box(`B`, stroke: ored, inset: 1pt)
|
||||
+ box([: $approx 0.251$], inset: (top: 1pt, bottom: 1pt))
|
||||
)
|
||||
}
|
||||
#v(8mm)
|
||||
Alice has the smallest number in 3 of 4 cases, which have a total probability of $approx 0.749$.
|
||||
])
|
||||
6
src/Warm-Ups/Somewhat Random Numbers/meta.toml
Normal file
6
src/Warm-Ups/Somewhat Random Numbers/meta.toml
Normal file
@ -0,0 +1,6 @@
|
||||
[metadata]
|
||||
title = "Somewhat Random Numbers"
|
||||
|
||||
[publish]
|
||||
handout = true
|
||||
solutions = true
|
||||
@ -1,89 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
|
||||
\usepackage{tikz}
|
||||
|
||||
|
||||
\title{The Sysadmin's Warm-Up}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
|
||||
Most of you have seen a hard drive. Many have touched one, and a lucky few have poked around inside one. These devices have two interesting properties:
|
||||
|
||||
\begin{enumerate}
|
||||
\item They hold valuable data
|
||||
\item They eventually fail
|
||||
\end{enumerate}
|
||||
|
||||
Needless to say, this is a problem. \par
|
||||
We need to design a system that allows hard drives to fail without data loss.
|
||||
|
||||
\definition{}
|
||||
You can think of a hard drive as a long string of bits. \par
|
||||
Assume all hard drives can store 1 TiB of data.
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}
|
||||
|
||||
\node[above] at (1/2, 0) {Drive A};
|
||||
\draw (0cm, 0cm) -- (0cm, -3cm);
|
||||
\draw (1cm, 0cm) -- (1cm, -3cm);
|
||||
\foreach \i in {0,...,-6} {
|
||||
\draw (0cm,\i cm / 2) -- (1cm ,\i cm / 2);
|
||||
}
|
||||
|
||||
\node at (1/2, - 1 / 4) {1};
|
||||
\node at (1/2, - 3 / 4) {1};
|
||||
\node at (1/2, - 5 / 4) {0};
|
||||
\node at (1/2, - 7 / 4) {...};
|
||||
\node at (1/2, - 9 / 4) {1};
|
||||
\node at (1/2, -11 / 4) {0};
|
||||
|
||||
|
||||
\node[above] at (5/2, 0) {Drive B};
|
||||
\draw (2cm, 0cm) -- (2cm, -3cm);
|
||||
\draw (3cm, 0cm) -- (3cm, -3cm);
|
||||
\foreach \i in {0,...,-6} {
|
||||
\draw (2cm,\i cm / 2) -- (3cm ,\i cm / 2);
|
||||
}
|
||||
|
||||
\node at (5/2, - 1 / 4) {0};
|
||||
\node at (5/2, - 3 / 4) {1};
|
||||
\node at (5/2, - 5 / 4) {0};
|
||||
\node at (5/2, - 7 / 4) {...};
|
||||
\node at (5/2, - 9 / 4) {0};
|
||||
\node at (5/2, -11 / 4) {1};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\problem{}
|
||||
Suppose we have two hard drives. How can we arrange our data so that...
|
||||
\begin{enumerate}
|
||||
\item We get 1 TiB of usable storage
|
||||
\item We lose no data if any one drive fails
|
||||
\end{enumerate}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Suppose we have three hard drives. How can we arrange our data so that...
|
||||
\begin{enumerate}
|
||||
\item We get 2 TiB of usable storage
|
||||
\item We lose no data if any one drive fails
|
||||
\end{enumerate}
|
||||
|
||||
\vfill
|
||||
\end{document}
|
||||
35
src/Warm-Ups/Sysadmin/main.typ
Normal file
35
src/Warm-Ups/Sysadmin/main.typ
Normal file
@ -0,0 +1,35 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [The Sysadmin's Warm-Up],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
Most of you have seen a hard drive. \
|
||||
Many have touched one, and a lucky few have taken one apart. \
|
||||
These devices have two interesting properties:
|
||||
- They hold valuable data
|
||||
- They inevitably fail
|
||||
|
||||
Needless to say, this is a problem. \
|
||||
We would like to design a system that tolerates hard drive failures without data loss.
|
||||
|
||||
#definition()
|
||||
You can think of a hard drive as a long string of bits. \
|
||||
Assume all hard drives in the following problems have the same size. \
|
||||
If a hard drive "fails", all data on it is instantly lost.
|
||||
|
||||
#problem()
|
||||
Suppose we have two hard drives. How can we arrange our data so that...
|
||||
- We get 1 TiB of usable storage
|
||||
- We lose no data if any one drive fails
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Suppose we have three hard drives. How can we arrange our data so that...
|
||||
- We get 2 TiB of usable storage
|
||||
- We lose no data if any one drive fails
|
||||
|
||||
#v(1fr)
|
||||
|
||||
@ -1,30 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
\title{Warm-Up: Travellers}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
Four travellers are on a plane, each moving along a straight line at an arbitrary constant speed. \par
|
||||
No two of their paths are parallel, and no three intersect at the same point. \par
|
||||
We know that traveller A has met travelers B, C, and D, \par
|
||||
and that traveller B has met C and D (and A). Show that C and D must also have met. \par
|
||||
|
||||
\begin{solution}
|
||||
When a body travels at a constant speed, its graph with respect to time is a straight line. \par
|
||||
So, we add time axis in the third dimension, perpendicular to our plane. \par
|
||||
Naturally, the projection of each of these onto the plane corresponds to a road.
|
||||
|
||||
Now, note that two intersecting lines define a plane and use the conditions in the problem to show that no two lines are parallel.
|
||||
\end{solution}
|
||||
|
||||
\end{document}
|
||||
20
src/Warm-Ups/Travellers/main.typ
Normal file
20
src/Warm-Ups/Travellers/main.typ
Normal file
@ -0,0 +1,20 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Travellers],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Four travellers are on a plane, each moving along a straight line at an arbitrary constant speed. \
|
||||
No two of their paths are parallel, and no three intersect at the same point. \
|
||||
We know that traveller A has met travelers B, C, and D, \
|
||||
and that traveller B has met C and D (and A). Show that C and D must also have met.
|
||||
|
||||
#solution([
|
||||
When a body travels at a constant speed, its graph with respect to time is a straight line. \
|
||||
So, we add time axis in the third dimension, perpendicular to our plane. \
|
||||
Naturally, the projection of each of these onto the plane corresponds to a road.
|
||||
|
||||
Now, note that two intersecting lines define a plane and use the conditions in the problem to show that no two lines are parallel.
|
||||
])
|
||||
26
src/Warm-Ups/Tuesday/main.typ
Normal file
26
src/Warm-Ups/Tuesday/main.typ
Normal file
@ -0,0 +1,26 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Tuesday],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Charlie has two children. \
|
||||
One of them is a boy born on a Tuesday. \
|
||||
What is the probability that Charlie has two boys?
|
||||
|
||||
#solution([
|
||||
$13 / 27$, which notably isn't $1/2$.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Draw a $14 times 14$ square, highlight the possible cases, and you'll see it. \
|
||||
|
||||
#v(2mm)
|
||||
|
||||
Order is very important in this problem. \
|
||||
If we knew that the _first_ child was a boy born on Tuesday,
|
||||
the probability of two boys would be $1/2$. \
|
||||
Also consider a smaller case where a week has two days.
|
||||
])
|
||||
6
src/Warm-Ups/Tuesday/meta.toml
Normal file
6
src/Warm-Ups/Tuesday/meta.toml
Normal file
@ -0,0 +1,6 @@
|
||||
[metadata]
|
||||
title = "Tuesday"
|
||||
|
||||
[publish]
|
||||
handout = true
|
||||
solutions = true
|
||||
@ -1,43 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\usepackage[linguistics]{forest}
|
||||
|
||||
|
||||
|
||||
\title{Warm-Up: What's an AST?}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today. \\ Based on a true story.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
Say you have a valid string of simple arithmetic that contains no unary operators (like $3!$ or $-4$) and no parenthesis:
|
||||
$$
|
||||
3 + 9 \times 8 \div 5 \land 6
|
||||
$$
|
||||
|
||||
You may assume that all numbers and operators in this string consist of exactly one character. \\
|
||||
|
||||
Devise an algorithm that turns this string into a tree (as shown below), respecting the order of operations $[\land, \times, \div, +, -]$.
|
||||
|
||||
\begin{center}
|
||||
\begin{forest}
|
||||
[$+$
|
||||
[3]
|
||||
[$\div$
|
||||
[$\times$[9][8]]
|
||||
[$\land$[5][6]]
|
||||
]
|
||||
]
|
||||
\end{forest}
|
||||
\end{center}
|
||||
|
||||
\end{document}
|
||||
74
src/Warm-Ups/What's an AST/main.typ
Normal file
74
src/Warm-Ups/What's an AST/main.typ
Normal file
@ -0,0 +1,74 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: What's an AST?],
|
||||
by: "Mark",
|
||||
subtitle: "Based on a true story.",
|
||||
)
|
||||
|
||||
#problem()
|
||||
Say we have a valid string of simple arithmetic that contains \
|
||||
no unary operators (like $3!$ or $-4$) and no parenthesis:
|
||||
|
||||
#v(2mm)
|
||||
$
|
||||
3 + 9 times 8 div 5 and 6
|
||||
$
|
||||
#v(2mm)
|
||||
|
||||
You may assume that all numbers and operators in this string consist of exactly one character. \
|
||||
|
||||
Devise an algorithm that turns such strings into a tree (as shown below), \
|
||||
respecting the order of operations $[and, times, div, +, -]$.
|
||||
|
||||
#v(2mm)
|
||||
|
||||
#align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
// spell:off
|
||||
content((0, 0), $+$, name: "r")
|
||||
content((-0.5, -1), $3$, name: "a")
|
||||
content((0.5, -1), $div$, name: "b")
|
||||
content((-0.3, -2), $times$, name: "ba")
|
||||
content((1.3, -2), $and$, name: "bb")
|
||||
content((-0.8, -3), $9$, name: "baa")
|
||||
content((0.2, -3), $8$, name: "bab")
|
||||
content((0.8, -3), $5$, name: "bba")
|
||||
content((1.8, -3), $6$, name: "bbb")
|
||||
// spell:on
|
||||
|
||||
// Zero-sized arrows are a hack for offset.
|
||||
set-style(
|
||||
stroke: (thickness: 0.3mm),
|
||||
mark: (
|
||||
start: (
|
||||
symbol: "|",
|
||||
offset: 0.25,
|
||||
width: 0mm,
|
||||
length: 0mm,
|
||||
),
|
||||
end: (
|
||||
symbol: "|",
|
||||
offset: 0.25,
|
||||
width: 0mm,
|
||||
length: 0mm,
|
||||
),
|
||||
),
|
||||
)
|
||||
|
||||
// spell:off
|
||||
line("r", "a")
|
||||
line("r", "b")
|
||||
line("b", "ba")
|
||||
line("b", "bb")
|
||||
line("ba", "baa")
|
||||
line("ba", "bab")
|
||||
line("bb", "bba")
|
||||
line("bb", "bbb")
|
||||
// spell:on
|
||||
}),
|
||||
)
|
||||
@ -1,66 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
% x,y,scale,style
|
||||
\def\ttt#1#2#3#4{
|
||||
\draw[#4] (#1, #2+#3) -- (#1+#3+#3+#3, #2+#3);
|
||||
\draw[#4] (#1, #2+#3+#3) -- (#1+#3+#3+#3, #2+#3+#3);
|
||||
\draw[#4] (#1+#3, #2) -- (#1+#3, #2+#3+#3+#3);
|
||||
\draw[#4] (#1+#3+#3, #2) -- (#1+#3+#3, #2+#3+#3+#3);
|
||||
}
|
||||
|
||||
|
||||
\title{Warm-Up: Wild Tic-Tac-Toe}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
|
||||
\problem{}
|
||||
In wild tic-tac-toe, players may place either an X or O on each move. The player that first completes a
|
||||
row of any three symbols wins. Show that the first player always has a winning strategy.
|
||||
|
||||
\vfill
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.60]
|
||||
\ttt{0}{0}{2}{line width=0.3mm}
|
||||
\ttt{7}{0}{2}{line width=0.3mm}
|
||||
\ttt{14}{0}{2}{line width=0.3mm}
|
||||
|
||||
\ttt{0}{7}{2}{line width=0.3mm}
|
||||
\ttt{7}{7}{2}{line width=0.3mm}
|
||||
\ttt{14}{7}{2}{line width=0.3mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Now, say the first player to complete a row loses. Who has a winning strategy now?
|
||||
|
||||
\vfill
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.60]
|
||||
\ttt{0}{0}{2}{line width=0.3mm}
|
||||
\ttt{7}{0}{2}{line width=0.3mm}
|
||||
\ttt{14}{0}{2}{line width=0.3mm}
|
||||
|
||||
\ttt{0}{7}{2}{line width=0.3mm}
|
||||
\ttt{7}{7}{2}{line width=0.3mm}
|
||||
\ttt{14}{7}{2}{line width=0.3mm}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
47
src/Warm-Ups/Wild Tic-Tac-Toe/main.typ
Normal file
47
src/Warm-Ups/Wild Tic-Tac-Toe/main.typ
Normal file
@ -0,0 +1,47 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
#import "@preview/cetz:0.3.1"
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Wild Tic-Tac-Toe],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#let ttt = align(
|
||||
center,
|
||||
cetz.canvas({
|
||||
import cetz.draw: *
|
||||
let s = 0.7 // scale
|
||||
|
||||
set-style(stroke: (thickness: 0.5mm * s))
|
||||
line((-1 * s, 3 * s), (-1 * s, -3 * s))
|
||||
line((1 * s, 3 * s), (1 * s, -3 * s))
|
||||
line((3 * s, -1 * s), (-3 * s, -1 * s))
|
||||
line((3 * s, 1 * s), (-3 * s, 1 * s))
|
||||
}),
|
||||
)
|
||||
|
||||
|
||||
#problem()
|
||||
In wild tic-tac-toe, players may place either an X or O on each move. The player that first completes a
|
||||
row of any three symbols wins. Show that the first player always has a winning strategy.
|
||||
#v(4mm)
|
||||
|
||||
#table(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr, 1fr),
|
||||
ttt, ttt, ttt,
|
||||
);
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Now, say the first player to complete a row loses. Who has a winning strategy now?
|
||||
#v(4mm)
|
||||
|
||||
#table(
|
||||
stroke: none,
|
||||
align: center,
|
||||
columns: (1fr, 1fr, 1fr),
|
||||
ttt, ttt, ttt,
|
||||
);
|
||||
#v(1fr)
|
||||
@ -1,150 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\makeatletter
|
||||
\newcommand{\thisone}{
|
||||
\if@solutions
|
||||
{\color{red} $\Leftarrow$ \texttt{this one}}
|
||||
\else\fi
|
||||
}
|
||||
|
||||
\title{Zeno's Furniture}
|
||||
\uptitlel{Warm Ups}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
Zeno Furniture sells exactly five types of furniture:
|
||||
\begin{itemize}
|
||||
\item Footstools
|
||||
\item Hutches
|
||||
\item Sideboards
|
||||
\item Tables
|
||||
\item Vanities
|
||||
\end{itemize}
|
||||
Irene buys four items, each of a different type,
|
||||
and each made of exactly one kind of wood:
|
||||
\begin{itemize}
|
||||
\item Maple
|
||||
\item Oak
|
||||
\item Pine
|
||||
\item Rosewood
|
||||
\end{itemize}
|
||||
|
||||
The following conditions govern Irene's purchases:
|
||||
\begin{itemize}
|
||||
\item Any vanity she buys is Maple.
|
||||
\item Any rosewood item she buys is a sideboard.
|
||||
\item If she buys a vanity, she does not buy a footstool.
|
||||
\item If she buys a footstool, she also buys a table made of the same material.
|
||||
\item Irene does not buy an oak table.
|
||||
\item Exactly two of the items she buys are made of the same kind of wood.
|
||||
\end{itemize}
|
||||
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
Which one of the following could be an accurate
|
||||
list of the items Irene buys? \par
|
||||
\begin{itemize}
|
||||
\item maple footstool, maple hutch, rosewood sideboard, maple table
|
||||
\item oak hutch, rosewood sideboard, pine table, oak vanity
|
||||
\item rosewood hutch, maple sideboard, oak table, maple vanity
|
||||
\item pine footstool, rosewood sideboard, pine table, maple vanity
|
||||
\item maple footstool, pine hutch, oak sideboard, maple table \thisone{}
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
If Irene buys one item made of rosewood and two items made
|
||||
of maple, then which one of the following pairs could be two
|
||||
of the items she buys?
|
||||
\begin{itemize}
|
||||
\item a rosewood sideboard and an oak footstool
|
||||
\item an oak hutch and a pine sideboard
|
||||
\item an oak hutch and a maple table \thisone{}
|
||||
\item a maple sideboard and a maple vanity
|
||||
\item a maple hutch and a maple table
|
||||
\end{itemize}
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
\problem{}
|
||||
Which one of the following is a complete and accurate list
|
||||
of all the woods any footstool that Irene buys could be made of?
|
||||
\begin{itemize}
|
||||
\item maple, oak
|
||||
\item maple, pine \thisone{}
|
||||
\item maple, rosewood
|
||||
\item maple, oak, pine
|
||||
\item maple, oak, pine, rosewood
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
Suppose Irene buys a footstool. Then which one of the following
|
||||
is a complete and accurate list of items and any one of which she
|
||||
could buy in maple?
|
||||
\begin{itemize}
|
||||
\item footstool, hutch, sideboard, table, vanity
|
||||
\item footstool, hutch, sideboard, table \thisone{}
|
||||
\item footstool, hutch, sideboard
|
||||
\item footstool, hutch
|
||||
\item footstool
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Which one of the following cannot be the two items Irene
|
||||
buys that are made of the same wood as each other?
|
||||
\begin{itemize}
|
||||
\item footstool, hutch \thisone{}
|
||||
\item hutch, sideboard
|
||||
\item hutch, table
|
||||
\item sideboard, vanity
|
||||
\item table, vanity
|
||||
\end{itemize}
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}
|
||||
If Irene does not buy an item made of maple, then each of the
|
||||
following must be true except...
|
||||
\begin{itemize}
|
||||
\item Irene buys a footstool
|
||||
\item Irene buys a pine hutch \thisone{}
|
||||
\item Irene buys a rosewood sideboard
|
||||
\item Irene buys exactly one item made of oak
|
||||
\item Irene buys exactly two items made of pine
|
||||
\end{itemize}
|
||||
\vfill
|
||||
|
||||
|
||||
\problem{}
|
||||
Suppose the condition that Irene does not buy an oak table is
|
||||
replaced with the condition that she does not buy a pine table.
|
||||
If all the other conditions hold as originally given, which of the
|
||||
following cannot be true?
|
||||
\begin{itemize}
|
||||
\item Irene buys an oak footstool.
|
||||
\item Irene buys a hutch and a table made of the same wood.
|
||||
\item Irene buys a vanity, but she does not buy an oak table.
|
||||
\item Irene buys a maple table and an oak hutch.
|
||||
\item Irene buys a rosewood sideboard and exactly two items made of pine. \thisone{}
|
||||
\end{itemize}
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
125
src/Warm-Ups/Zeno's Furniture/main.typ
Normal file
125
src/Warm-Ups/Zeno's Furniture/main.typ
Normal file
@ -0,0 +1,125 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: Zeno's Furniture],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#let thisone = if_solutions(
|
||||
text(fill: ored, [#sym.arrow.l.double.long `this one`]),
|
||||
)
|
||||
|
||||
|
||||
Zeno's Furniture sells exactly five types of furniture: \
|
||||
Footstools, Hutches, Sideboards, Tables, and Vanities.
|
||||
|
||||
#v(3mm)
|
||||
|
||||
Each can be made of exactly one kind of wood: \
|
||||
Maple, Oak, Pine, or Rosewood
|
||||
|
||||
#v(3mm)
|
||||
|
||||
Irene buys four items, each of a different type. \
|
||||
The following conditions govern Irene's purchases:
|
||||
- Any vanity she buys is Maple.
|
||||
- Any rosewood item she buys is a sideboard.
|
||||
- If she buys a vanity, she does not buy a footstool.
|
||||
- If she buys a footstool, she also buys a table made of the same material.
|
||||
- Irene does not buy an oak table.
|
||||
- Exactly two of the items she buys are made of the same kind of wood.
|
||||
|
||||
|
||||
|
||||
#v(5mm)
|
||||
|
||||
|
||||
#problem()
|
||||
Which one of the following could be an accurate
|
||||
list of the items Irene buys? \
|
||||
- maple footstool, maple hutch, rosewood sideboard, maple table
|
||||
- oak hutch, rosewood sideboard, pine table, oak vanity
|
||||
- rosewood hutch, maple sideboard, oak table, maple vanity
|
||||
- pine footstool, rosewood sideboard, pine table, maple vanity
|
||||
- maple footstool, pine hutch, oak sideboard, maple table #thisone
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
If Irene buys one item made of rosewood and two items made
|
||||
of maple, then which one of the following pairs could be two
|
||||
of the items she buys?
|
||||
- a rosewood sideboard and an oak footstool
|
||||
- an oak hutch and a pine sideboard
|
||||
- an oak hutch and a maple table #thisone
|
||||
- a maple sideboard and a maple vanity
|
||||
- a maple hutch and a maple table
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
#problem()
|
||||
Which one of the following is a complete and accurate list
|
||||
of all the woods any footstool that Irene buys could be made of?
|
||||
- maple, oak
|
||||
- maple, pine #thisone
|
||||
- maple, rosewood
|
||||
- maple, oak, pine
|
||||
- maple, oak, pine, rosewood
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
Suppose Irene buys a footstool. Then which one of the following
|
||||
is a complete and accurate list of items and any one of which she
|
||||
could buy in maple?
|
||||
- footstool, hutch, sideboard, table, vanity
|
||||
- footstool, hutch, sideboard, table #thisone
|
||||
- footstool, hutch, sideboard
|
||||
- footstool, hutch
|
||||
- footstool
|
||||
|
||||
#v(1fr)
|
||||
|
||||
#problem()
|
||||
Which one of the following cannot be the two items Irene
|
||||
buys that are made of the same wood as each other?
|
||||
- footstool, hutch #thisone
|
||||
- hutch, sideboard
|
||||
- hutch, table
|
||||
- sideboard, vanity
|
||||
- table, vanity
|
||||
|
||||
#v(1fr)
|
||||
#pagebreak()
|
||||
|
||||
|
||||
|
||||
|
||||
#problem()
|
||||
If Irene does not buy an item made of maple, then each of the
|
||||
following must be true except...
|
||||
- Irene buys a footstool
|
||||
- Irene buys a pine hutch #thisone
|
||||
- Irene buys a rosewood sideboard
|
||||
- Irene buys exactly one item made of oak
|
||||
- Irene buys exactly two items made of pine
|
||||
|
||||
#v(1fr)
|
||||
|
||||
|
||||
#problem()
|
||||
Suppose the condition that Irene does not buy an oak table is
|
||||
replaced with the condition that she does not buy a pine table.
|
||||
If all the other conditions hold as originally given, which of the
|
||||
following cannot be true?
|
||||
- Irene buys an oak footstool.
|
||||
- Irene buys a hutch and a table made of the same wood.
|
||||
- Irene buys a vanity, but she does not buy an oak table.
|
||||
- Irene buys a maple table and an oak hutch.
|
||||
- Irene buys a rosewood sideboard and exactly two items made of pine. #thisone
|
||||
|
||||
#v(1fr)
|
||||
@ -1,22 +0,0 @@
|
||||
\documentclass[
|
||||
solutions,
|
||||
hidewarning,
|
||||
singlenumbering,
|
||||
nopagenumber
|
||||
]{../../../lib/tex/ormc_handout}
|
||||
\usepackage{../../../lib/tex/macros}
|
||||
|
||||
|
||||
\title{Warm-Up: \texttt{fmod}}
|
||||
\uptitler{\smallurl{}}
|
||||
\subtitle{Prepared by Mark on \today.}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\problem{}
|
||||
I'm sure you're all familiar with how \texttt{mod(a, b)} and \texttt{remainder(a, b)} work with integers. \par
|
||||
Devise an equivalent for floats (i.e, real numbers).
|
||||
|
||||
\end{document}
|
||||
10
src/Warm-Ups/fmod/main.typ
Normal file
10
src/Warm-Ups/fmod/main.typ
Normal file
@ -0,0 +1,10 @@
|
||||
#import "@local/handout:0.1.0": *
|
||||
|
||||
#show: handout.with(
|
||||
title: [Warm-Up: `fmod`],
|
||||
by: "Mark",
|
||||
)
|
||||
|
||||
#problem()
|
||||
I'm sure you're all familiar with how `mod(a, b)` and `remainder(a, b)` \ work when `a` and `b` are integers.
|
||||
Devise an equivalent for floats (i.e, real numbers).
|
||||
Loading…
x
Reference in New Issue
Block a user