Lambda edits
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@ -2,7 +2,8 @@
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions,
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singlenumbering
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singlenumbering,
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shortwarning
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]{../../resources/ormc_handout}
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\usepackage{url}
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@ -26,7 +27,13 @@
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\newcommand{\lm}{\lambda}
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% Notice that I a b = 1 a b!
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% TODO:
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% Lazy evaluation (alternate Y)
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% Add a few theorems
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% Better ending -> applications?
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% - nix, comparison to imperitive
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\uptitlel{Advanced 2}
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@ -39,7 +46,7 @@
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\maketitle
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\begin{minipage}{8cm}
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Beware of the Turing tar-pit in which everything is possible but nothing of interest is easy.
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Beware of the Turing tar pit, in which everything is possible but nothing of interest is easy.
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\vspace{2ex}
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@ -6,11 +6,9 @@ Consider the following statement:
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$$
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I = \lm a . a
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$$
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This tells us that $I$ is a function that takes its input, $a$, to itself. We'll call this the \textit{identity function}.
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To apply functions, put them next to their inputs. We'll omit the usual parentheses to save space.
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$$
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(I~\star) =
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(\lm \tzm{b}a. a)~\tzmr{a}\star =
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@ -26,13 +24,10 @@ $$
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(a.center) to (b.center);
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\end{tikzpicture}
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$$
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Functions are left-associative: If $A$ and $B$ are functions, $(A~B~\star)$ is equivalent to $((A~B)~\star)$.
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As usual, we'll use parentheses to group terms if we want to override this order: $(A~(B~\star)) \neq (A~B~\star)$ \par
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In this handout, all types of parentheses ( $(), [~],$ etc ) are equivalent.
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\vfill
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\generic{$\beta$-Reduction:}
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$\beta$-reduction is a fancy name for \say{simplifying an expression.} We've already done it once above.
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@ -61,12 +56,8 @@ $$
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(c.center) to (d.center);
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\end{tikzpicture}
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$$
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We cannot reduce this any further, so we stop. Our expression is now in \textit{$\beta$-normal form}.
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\vfill
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\pagebreak
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\problem{}
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Reduce the following expressions:
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\begin{itemize}
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@ -6,6 +6,8 @@ For example, $b$ is a free variable in $\lm a. b$. The same is true of $\star$ i
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A \textit{combinator} is a function with no free variables.
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\definition{The Kestrel}
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Notable combinators are often named after birds.\hspace{-0.5ex}\footnotemark{} We've already met a few: \par
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@ -21,6 +23,9 @@ Another notable combinator is $K$, the \textit{Kestrel}:
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$$
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K = \lm ab . a
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$$
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\problem{}
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What does the Kestrel do? Explain in plain English. \par
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\hint{What is $(K~\heartsuit~\star)$?}
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@ -49,13 +49,9 @@ How about $(8~NOT~F)$?
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\pagebreak
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\problem{}
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This handout may remind you of Professor Oleg's handout on Peano's axioms. Good. \par
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Recall the tools we used to build the natural numbers: \par
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We had a zero element and a \say{successor} operation so that $1 \coloneqq S(0)$, $2 \coloneqq S(1)$, and so on.
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\vspace{1ex}
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Create a successor operation for the Church numerals. \par
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Peano's axioms state that we only need a zero element and a \say{successor} operation to
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build the natural numbers. We've already defined zero.
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Now, create a successor operation so that $1 \coloneqq S(0)$, $2 \coloneqq S(1)$, and so on. \par
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\hint{A good signature for this function is $\lm nfa$, or more clearly $\lm n.\lm fa$. Do you see why?}
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\begin{solution}
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@ -175,36 +171,5 @@ $D(1) = 0$, $D(2) = 1$, etc. $D(0)$ should be zero. \par
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$D = \lm n . \Bigl[(~n~H~\langle 0, 0 \rangle~)~T\Bigr]$
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\end{solution}
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\begin{solution}
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Here's a different solution. \par
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Can you figure out how it works?
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\vspace{1ex}
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$
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D_0 =
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\lm p . \Bigl[p~T\Bigr]
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\Bigl\langle
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F ~,~ p~F
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\Bigr\rangle
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\Bigl\langle
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F
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~,~
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\bigl\langle
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p~F~T ~,~ ( (p~F~T)~(P~F~F) )
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\bigr\rangle
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\Bigr\rangle
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$
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\vspace{1ex}
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$
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D = \lm nfa .
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\Bigl(
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n D_0 \Bigl\langle T, \langle f, a \rangle \Bigr\rangle
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\Bigr)~F~F
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$
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\end{solution}
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\vfill
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\pagebreak
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@ -35,21 +35,24 @@ Write an expression that resolves to itself. \par
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\vspace{1ex}
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This expression is often called $\Omega$, after the last letter of the Greek alphabet. \par
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$\Omega$ useless on its own, but it gives us a starting point for recursion.
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$\Omega$ useless on its own, but it gives us a starting point for recursion. \par
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\begin{solution}
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$\Omega = M~M = (\lm x . xx) (\lm x . xx)$
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\vspace{1ex}
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\vspace{1mm}
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An uninspired mathematician might call the Mockingbird $\omega$, \say{little omega}.
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\end{solution}
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\vfill
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\pagebreak
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\definition{}
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This is the \textit{Y-combinator}, easily the most famous $\lm$ expression. \par
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You may notice that it's just $\Omega$, put to work.
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This is the \textit{Y-combinator}. You may notice that it's just $\Omega$ put to work.
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$$
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Y = \lm f . (\lm x . f(x~x))(\lm x . f(x~x))
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$$
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@ -58,5 +61,27 @@ $$
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What does this thing do? \par
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Evaluate $Y f$.
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\vfill
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\definition{}
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We say $x$ is a \textit{fixed point} of a function $f$ if $f(x) = x$.
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\problem{}
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Show that $Y F$ is a fixed point of $F$.
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\vfill
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\problem{}
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Let $\theta = (\lm xy . y(xxy))$ and $\Theta = \theta \theta$. \par
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Let $N = \Theta F$ for an arbitrary lambda expression $F$. \par
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Show that $F N = N$.
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\vfill
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\problem{Bonus}
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Find a fixed-point combinator that isn't $Y$ or $\Theta$.
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\vfill
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\pagebreak
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@ -5,34 +5,19 @@ Do \ref{Yfac} first, then finish the rest in any order.
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\problem{}<Yfac>
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Design a recursive factorial function using $Y$.
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\begin{solution}
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$\text{FAC} = \lm yn.[Z~n][1][\text{MULT}~n~(y~(\text{D}~n))]$
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To compute the factorial of 5, evaluate $(\text{Y}~\text{FAC}~5)$.
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\end{solution}
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\vfill
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\problem{}
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Design a non-recursive factorial function. \par
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\note{This one is easier than \ref{Yfac}, but I don't think it will help you solve it.}
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\problem{}
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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.
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$\text{GET}~L~0$ should give the first item in the list, and $\text{GET}~L~1$, the \textit{second}. \par
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Lists have a defined length, so you should be able to tell when you're on the last element.
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\problem{}
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Solve \ref{decrement} without using $H$.
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\problem{}
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Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
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\begin{solution}
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\textbf{Factorial with recursion:}
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\vspace{3ex}
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$\text{FAC} = \lm yn.[Z~n][1][\text{MULT}~n~(y~(\text{D}~n))]$
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\linehack{}
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\textbf{Factorial without recursion:}
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\vspace{3ex}
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$\text{FAC}_0 = \lm p .
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\Bigl\langle~~
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\Bigl[D~(p~t)\Bigr]
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\text{FAC} = \lm n .
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\bigl( n~\text{FAC}_0~\langle n, 1 \rangle \bigr)
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$
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\end{solution}
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\linehack{}
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\vfill
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\textbf{Lists:}
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\vspace{3ex}
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\problem{}
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Solve \ref{decrement} without using $H$.
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\begin{solution}
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One solution is below. Can you figure out how it works? \par
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\vspace{1ex}
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$
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D_0 =
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\lm p . \Bigl[p~T\Bigr]
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\Bigl\langle
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F ~,~ p~F
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\Bigr\rangle
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\Bigl\langle
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F
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~,~
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\bigl\langle
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p~F~T ~,~ ( (p~F~T)~(P~F~F) )
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\bigr\rangle
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\Bigr\rangle
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$
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\vspace{1ex}
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$
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D = \lm nfa .
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\Bigl(
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n D_0 \Bigl\langle T, \langle f, a \rangle \Bigr\rangle
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\Bigr)~F~F
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$
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\end{solution}
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\vfill
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\problem{}
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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.
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$\text{GET}~L~0$ should give the first item in the list, and $\text{GET}~L~1$, the \textit{second}. \par
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Lists have a defined length, so you should be able to tell when you're on the last element.
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\begin{solution}
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One possible implementation is
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$\Bigl\langle
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\langle \text{is last} ~,~ \text{item} \rangle
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@ -75,8 +103,27 @@ Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
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This will break if $n$ is out of range.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Write a lambda expression that represents the Fibonacci function: \par
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$f(0) = 1$, $f(1) = 1$, $f(n + 2) = f(n + 1) + f(n)$.
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\vfill
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\problem{}
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Write a lambda expression that evaluates to $T$ if a number $n$ is prime, and to $F$ otherwise.
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\vfill
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\problem{}
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Write a MOD $a$ $b$ function that reduces to the remainder of $a \div b$.
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\vfill
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\problem{Bonus}
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Play with \textit{Lamb}, an automatic lambda expression evaluator. \par
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\url{https://git.betalupi.com/Mark/lamb}
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\pagebreak
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