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\documentclass[../main.tex]{subfiles}
\begin{document}
\section{Challenges}
Do \ref{Yfac} first, then finish the rest in any order. \\
Have fun!
\problem{}<Yfac>
Design a recursive factorial function using $Y$. \\
\vfill
\problem{}
Design a non-recursive factorial function. \\
\note{This one is a lot 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}. \\
Lists have a defined length, so you should be able to tell when you're on the last element.
\problem{}
Write POW $a$ $b$, which raises $b$ to the $a$th power.
\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]
~,~
\Bigl[\text{MULT}~(p~T)~(p~F)\Bigr]
~~\Bigr\rangle
$
\vspace{2ex}
$
\text{FAC} = \lm n .
\bigl( n~\text{FAC}_0~\langle n, 1 \rangle \bigr)
$
\linehack{}
\textbf{Lists:}
\vspace{3ex}
One possible implementation is
$\Bigl\langle
\langle \text{is last} ~,~ \text{item} \rangle
~,~
\text{next}...
\Bigr\rangle$, where:
\vspace{1ex}
\say{is last} is a boolean, true iff this is the last item in the list. \\
\say{item} is the thing you're storing \\
\say{next...} is another one of these list fragments. \\
It doesn't matter what \say{next} is in the last list fragment. A dedicated \say{is last} slot allows us to store ANY function in this list.
\vspace{1ex}
Here, $\text{GET} = \lm nL.[(n~L~F)~T~F$] \\
This will break if $n$ is out of range.
\end{solution}
\vfill
\problem{Bonus}
Play with \textit{Lamb}, an automatic lambda expression evaluator. \\
\url{https://git.betalupi.com/Mark/lamb}
\end{document}