Final lambda edits

Advanced/Lambda Calculus/main.tex

@@ -2,12 +2,11 @@
 % use [solutions] flag to show solutions.
 \documentclass[
 	solutions,
-	singlenumbering,
+	singlenumbering
-	shortwarning
 ]{../../resources/ormc_handout}
 
-\usepackage{url}
 \usepackage{mathtools} % for \coloneqq
+\usepackage{hyperref}
 
 % An invisible marker, used to
 % draw arrows in equations.
@@ -37,7 +36,7 @@
 
 
 \uptitlel{Advanced 2}
-\uptitler{Fall 2022}
+\uptitler{Fall 2023}
 \title{Lambda Calculus}
 \subtitle{Prepared by Mark on \today{}}
 

Advanced/Lambda Calculus/parts/00 intro.tex

@@ -114,7 +114,13 @@ Reduce the following expressions:
 \generic{Currying:}
 
 In lambda calculus, functions are only allowed to take one argument.
-However, we can emulate multivariable functions through \textit{currying}.
+However, we can emulate multivariable functions through \textit{currying}\footnotemark{}\hspace{-1ex}.
+
+\footnotetext{After Haskell Brooks Curry\footnotemark{}\hspace{-1ex}, a logician that contributed to the theory of functional computation.}
+\footnotetext{
+	There are three programming languages named after him: Haskell, Brook, and Curry. \par
+	Two of these are functional, and one is an oddball GPU language last released in 2007.
+}
 
 \vspace{1ex}
 

Advanced/Lambda Calculus/parts/02 boolean.tex

@@ -1,6 +1,7 @@
 \section{Boolean Algebra}
 
-The Kestrel selects its first argument, and the Kite selects its second. This \say{choosing} behavior is similar to something you may have already seen...
+The Kestrel selects its first argument, and the Kite selects its second. \par
+Maybe we can somehow put this \say{choosing} behavior to work...
 
 \vspace{1ex}
 
@@ -18,6 +19,14 @@ Write a function $\text{NOT}$ so that $(\text{NOT} ~ T) = F$ and $(\text{NOT}~F)
 
 \vfill
 
+\problem{}
+How would \say{if} statements work in this model of boolean logic? \par
+Say we have a boolean $B$ and two expressions $E_T$ and $E_F$.
+Can we write a function that evaluates to $E_T$ if $B$ is true, and to $E_F$ otherwise?
+
+\vfill
+\pagebreak
+
 \problem{}
 Write functions $\text{AND}$, $\text{OR}$, and $\text{XOR}$ that satisfy the following table.
 
@@ -62,5 +71,7 @@ What inputs should it take? What outputs should it produce?
 	$\text{EQ} = \lm ab . [\text{NOT}~(\text{XOR}~a~b)]$
 \end{solution}
 
+
 \vfill
+
 \pagebreak

Advanced/Lambda Calculus/parts/04 recursion.tex

@@ -1,4 +1,77 @@
 \section{Recursion}
+
+
+%\iftrue
+\iffalse
+
+	You now have a choice. Choose wisely --- there's no going back.
+
+	\begin{tcolorbox}[
+		breakable,
+		colback=white,
+		colframe=gray,
+		arc=0pt, outer arc=0pt
+	]
+		\raggedright
+		\textbf{Take the red pill:} You stay on this page and try to solve \ref{thechallenge}. \par
+		This will take a while, and it's very unlikely you'll finish before class ends.
+
+
+		\vspace{4mm}
+
+		I strongly prefer this option. It's not easy, but you'll be very happy if you solve it yourself.
+		This is a chance to build your own solution to a fundamental problem in this field, just as
+		Curry, Church, and Turing did when first developing the theory of lambda calculus.
+
+		- Mark
+	\end{tcolorbox}
+
+	\begin{tcolorbox}[
+		breakable,
+		colback=white,
+		colframe=gray,
+		arc=0pt, outer arc=0pt
+	]
+		\textbf{Take the blue pill:} You skip this problem and turn the page.
+		Half of the answer to \ref{thechallenge} will be free, and the rest will be
+		broken into smaller steps. This is how we usually learn out about interesting
+		mathematics, both in high school and in university.
+
+		\vspace{2mm}
+
+		This path isn't as rewarding as the one above, but it is well-paved
+		and easier to traverse.
+	\end{tcolorbox}
+
+	\problem{}<thechallenge>
+	Can you find a way to recursively call functions in lambda calculus? \par
+	Find a way to define a recursive factorial function. \par
+
+	\note{$A = (\lm a. A~a)$ doesn't count. You can't use a macro inside itself.}
+
+	\vfill
+	\pagebreak
+\fi
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
 
 Say we want a function that computes the factorial of a positive integer. Here's one way we could define it:
 $$
@@ -30,7 +103,7 @@ They're just shorthand that simplifies notation.
 
 \problem{}
 Write an expression that resolves to itself. \par
-\note{Your answer should be quite short.}
+\hint{Your answer should be quite short.}
 
 \vspace{1ex}
 
@@ -45,12 +118,13 @@ $\Omega$ useless on its own, but it gives us a starting point for recursion. \pa
 	An uninspired mathematician might call the Mockingbird $\omega$, \say{little omega}.
 \end{solution}
 
-
 \vfill
 \pagebreak
 
 
 
+
+
 \definition{}
 This is the \textit{Y-combinator}. You may notice that it's just $\Omega$ put to work.
 $$
@@ -62,26 +136,27 @@ What does this thing do? \par
 Evaluate $Y f$.
 
 
-\vfill
 
-\definition{}
+%\vfill
-We say $x$ is a \textit{fixed point} of a function $f$ if $f(x) = x$.
 
-\problem{}
+%\definition{}
-Show that $Y F$ is a fixed point of $F$.
+%We say $x$ is a \textit{fixed point} of a function $f$ if $f(x) = x$.
 
-\vfill
+%\problem{}
+%Show that $Y F$ is a fixed point of $F$.
 
-\problem{}
+%\vfill
-Let $\theta = (\lm xy . y(xxy))$ and $\Theta = \theta \theta$. \par
-Let $N = \Theta F$ for an arbitrary lambda expression $F$. \par
 
-Show that $F N = N$.
+%\problem{}
+%Let $b = (\lm xy . y(xxy))$ and $B = b ~ b$. \par
+%Let $N = B F$ for an arbitrary lambda expression $F$. \par
 
-\vfill
+%Show that $F N = N$.
 
-\problem{Bonus}
+%\vfill
-Find a fixed-point combinator that isn't $Y$ or $\Theta$.
+
+%\problem{Bonus}
+%Find a fixed-point combinator that isn't $Y$ or $\Theta$.
 
 \vfill
 \pagebreak

Advanced/Lambda Calculus/parts/05 challenges.tex

@@ -39,7 +39,8 @@ Design a non-recursive factorial function. \par
 
 
 \problem{}
-Solve \ref{decrement} without using $H$.
+Solve \ref{decrement} without using $H$. \par
+In \ref{decrement}, we created the \say{decrement} function.
 
 \begin{solution}
 	One solution is below. Can you figure out how it works? \par
@@ -118,7 +119,7 @@ Write a lambda expression that evaluates to $T$ if a number $n$ is prime, and to
 \vfill
 
 \problem{}
-Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
+Write a function MOD so that $(\text{MOD}~a~b)$ reduces to the remainder of $a \div b$.
 
 \vfill