Lambda edits

Advanced/Lambda Calculus/main.tex

@@ -2,7 +2,8 @@
 % use [solutions] flag to show solutions.
 \documentclass[
 	solutions,
-	singlenumbering
+	singlenumbering,
+	shortwarning
 ]{../../resources/ormc_handout}
 
 \usepackage{url}
@@ -26,7 +27,13 @@
 
 \newcommand{\lm}{\lambda}
 
-% Notice that I a b = 1 a b!
+
+
+% TODO:
+% Lazy evaluation (alternate Y)
+% Add a few theorems
+% Better ending -> applications?
+% - nix, comparison to imperitive
 
 
 \uptitlel{Advanced 2}
@@ -39,7 +46,7 @@
 	\maketitle
 
 	\begin{minipage}{8cm}
-		Beware of the Turing tar-pit in which everything is possible but nothing of interest is easy.
+		Beware of the Turing tar pit, in which everything is possible but nothing of interest is easy.
 
 		\vspace{2ex}
 

Advanced/Lambda Calculus/parts/00 intro.tex

@@ -6,11 +6,9 @@ Consider the following statement:
 $$
 	I = \lm a . a
 $$
-
 This tells us that $I$ is a function that takes its input, $a$, to itself. We'll call this the \textit{identity function}.
 
 To apply functions, put them next to their inputs. We'll omit the usual parentheses to save space.
-
 $$
 (I~\star) =
 (\lm \tzm{b}a. a)~\tzmr{a}\star =
@@ -26,13 +24,10 @@ $$
 		(a.center) to (b.center);
 \end{tikzpicture}
 $$
-
 Functions are left-associative: If $A$ and $B$ are functions, $(A~B~\star)$ is equivalent to $((A~B)~\star)$.
 As usual, we'll use parentheses to group terms if we want to override this order: $(A~(B~\star)) \neq (A~B~\star)$ \par
 In this handout, all types of parentheses ( $(), [~],$ etc ) are equivalent.
 
-\vfill
-
 \generic{$\beta$-Reduction:}
 
 $\beta$-reduction is a fancy name for \say{simplifying an expression.} We've already done it once above.
@@ -61,12 +56,8 @@ $$
 			(c.center) to (d.center);
 	\end{tikzpicture}
 $$
-
 We cannot reduce this any further, so we stop. Our expression is now in \textit{$\beta$-normal form}.
 
-\vfill
-\pagebreak
-
 \problem{}
 Reduce the following expressions:
 \begin{itemize}

Advanced/Lambda Calculus/parts/01 combinators.tex

@@ -6,6 +6,8 @@ For example, $b$ is a free variable in $\lm a. b$. The same is true of $\star$ i
 
 A \textit{combinator} is a function with no free variables.
 
+
+
 \definition{The Kestrel}
 
 Notable combinators are often named after birds.\hspace{-0.5ex}\footnotemark{} We've already met a few: \par
@@ -21,6 +23,9 @@ Another notable combinator is $K$, the \textit{Kestrel}:
 $$
 	K = \lm ab . a
 $$
+
+
+
 \problem{}
 What does the Kestrel do? Explain in plain English. \par
 \hint{What is $(K~\heartsuit~\star)$?}

Advanced/Lambda Calculus/parts/03 numbers.tex

@@ -49,13 +49,9 @@ How about $(8~NOT~F)$?
 \pagebreak
 
 \problem{}
-This handout may remind you of Professor Oleg's handout on Peano's axioms. Good. \par
+Peano's axioms state that we only need a zero element and a \say{successor} operation to
-Recall the tools we used to build the natural numbers: \par
+build the natural numbers. We've already defined zero.
-We had a zero element and a \say{successor} operation so that $1 \coloneqq S(0)$, $2 \coloneqq S(1)$, and so on.
+Now, create a successor operation so that $1 \coloneqq S(0)$, $2 \coloneqq S(1)$, and so on. \par
-
-\vspace{1ex}
-
-Create a successor operation for the Church numerals. \par
 \hint{A good signature for this function is $\lm nfa$, or more clearly $\lm n.\lm fa$. Do you see why?}
 
 \begin{solution}
@@ -175,36 +171,5 @@ $D(1) = 0$, $D(2) = 1$, etc. $D(0)$ should be zero. \par
 	$D = \lm n . \Bigl[(~n~H~\langle 0, 0 \rangle~)~T\Bigr]$
 \end{solution}
 
-\begin{solution}
-	Here's a different solution. \par
-	Can you figure out how it works?
-
-	\vspace{1ex}
-
-	$
-	D_0 =
-	\lm p . \Bigl[p~T\Bigr]
-	\Bigl\langle
-		F ~,~ p~F
-	\Bigr\rangle
-	\Bigl\langle
-		F
-		~,~
-		\bigl\langle
-			p~F~T ~,~ ( (p~F~T)~(P~F~F) )
-		\bigr\rangle
-	\Bigr\rangle
-	$
-
-	\vspace{1ex}
-
-	$
-	D = \lm nfa .
-	\Bigl(
-		n D_0 \Bigl\langle T, \langle f, a \rangle \Bigr\rangle
-	\Bigr)~F~F
-	$
-\end{solution}
-
 \vfill
 \pagebreak

Advanced/Lambda Calculus/parts/04 recursion.tex

@@ -35,21 +35,24 @@ Write an expression that resolves to itself. \par
 \vspace{1ex}
 
 This expression is often called $\Omega$, after the last letter of the Greek alphabet. \par
-$\Omega$ useless on its own, but it gives us a starting point for recursion.
+$\Omega$ useless on its own, but it gives us a starting point for recursion. \par
 
 \begin{solution}
 	$\Omega = M~M = (\lm x . xx) (\lm x . xx)$
 
-	\vspace{1ex}
+	\vspace{1mm}
-
+	
 	An uninspired mathematician might call the Mockingbird $\omega$, \say{little omega}.
 \end{solution}
 
+
 \vfill
+\pagebreak
+
+
 
 \definition{}
-This is the \textit{Y-combinator}, easily the most famous $\lm$ expression. \par
+This is the \textit{Y-combinator}. You may notice that it's just $\Omega$ put to work.
-You may notice that it's just $\Omega$, put to work.
 $$
 	Y = \lm f . (\lm x . f(x~x))(\lm x . f(x~x))
 $$
@@ -58,5 +61,27 @@ $$
 What does this thing do? \par
 Evaluate $Y f$.
 
+
 \vfill
-\pagebreak
+
+\definition{}
+We say $x$ is a \textit{fixed point} of a function $f$ if $f(x) = x$.
+
+\problem{}
+Show that $Y F$ is a fixed point of $F$.
+
+\vfill
+
+\problem{}
+Let $\theta = (\lm xy . y(xxy))$ and $\Theta = \theta \theta$. \par
+Let $N = \Theta F$ for an arbitrary lambda expression $F$. \par
+
+Show that $F N = N$.
+
+\vfill
+
+\problem{Bonus}
+Find a fixed-point combinator that isn't $Y$ or $\Theta$.
+
+\vfill
+\pagebreak

Advanced/Lambda Calculus/parts/05 challenges.tex

@@ -5,34 +5,19 @@ Do \ref{Yfac} first, then finish the rest in any order.
 \problem{}<Yfac>
 Design a recursive factorial function using $Y$.
 
+\begin{solution}
+	$\text{FAC} = \lm yn.[Z~n][1][\text{MULT}~n~(y~(\text{D}~n))]$
+
+	To compute the factorial of 5, evaluate $(\text{Y}~\text{FAC}~5)$.
+\end{solution}
+
 \vfill
 
 \problem{}
 Design a non-recursive factorial function. \par
 \note{This one is easier than \ref{Yfac}, but I don't think it will help you solve it.}
 
-\problem{}
-Using pairs, make a \say{list} data structure. Define a GET function, so that $\text{GET}~L~n$ reduces to the nth item in the list.
-$\text{GET}~L~0$ should give the first item in the list, and $\text{GET}~L~1$, the \textit{second}. \par
-Lists have a defined length, so you should be able to tell when you're on the last element.
-
-\problem{}
-Solve \ref{decrement} without using $H$.
-
-\problem{}
-Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
-
 \begin{solution}
-	\textbf{Factorial with recursion:}
-	\vspace{3ex}
-
-	$\text{FAC} = \lm yn.[Z~n][1][\text{MULT}~n~(y~(\text{D}~n))]$
-
-	\linehack{}
-
-	\textbf{Factorial without recursion:}
-	\vspace{3ex}
-
 	$\text{FAC}_0 = \lm p .
 	\Bigl\langle~~
 	\Bigl[D~(p~t)\Bigr]
@@ -47,12 +32,55 @@ Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
 	\text{FAC} = \lm n .
 	\bigl( n~\text{FAC}_0~\langle n, 1 \rangle \bigr)
 	$
+\end{solution}
 
-	\linehack{}
+\vfill
 
-	\textbf{Lists:}
-	\vspace{3ex}
 
+
+\problem{}
+Solve \ref{decrement} without using $H$.
+
+\begin{solution}
+	One solution is below. Can you figure out how it works? \par
+
+	\vspace{1ex}
+
+	$
+	D_0 =
+	\lm p . \Bigl[p~T\Bigr]
+	\Bigl\langle
+		F ~,~ p~F
+	\Bigr\rangle
+	\Bigl\langle
+		F
+		~,~
+		\bigl\langle
+			p~F~T ~,~ ( (p~F~T)~(P~F~F) )
+		\bigr\rangle
+	\Bigr\rangle
+	$
+
+	\vspace{1ex}
+
+	$
+	D = \lm nfa .
+	\Bigl(
+		n D_0 \Bigl\langle T, \langle f, a \rangle \Bigr\rangle
+	\Bigr)~F~F
+	$
+\end{solution}
+
+\vfill
+
+
+
+\problem{}
+Using pairs, make a \say{list} data structure. Define a GET function, so that $\text{GET}~L~n$ reduces to the nth item in the list.
+$\text{GET}~L~0$ should give the first item in the list, and $\text{GET}~L~1$, the \textit{second}. \par
+Lists have a defined length, so you should be able to tell when you're on the last element.
+
+\begin{solution}
 	One possible implementation is
 	$\Bigl\langle
 		\langle \text{is last} ~,~ \text{item} \rangle
@@ -75,8 +103,27 @@ Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
 	This will break if $n$ is out of range.
 \end{solution}
 
+\vfill
+\pagebreak
+
+\problem{}
+Write a lambda expression that represents the Fibonacci function: \par
+$f(0) = 1$, $f(1) = 1$, $f(n + 2) = f(n + 1) + f(n)$.
+
+\vfill
+
+\problem{}
+Write a lambda expression that evaluates to $T$ if a number $n$ is prime, and to $F$ otherwise.
+
+\vfill
+
+\problem{}
+Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
+
 \vfill
 
 \problem{Bonus}
 Play with \textit{Lamb}, an automatic lambda expression evaluator. \par
-\url{https://git.betalupi.com/Mark/lamb}
+\url{https://git.betalupi.com/Mark/lamb}
+
+\pagebreak