\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}