<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\section{Challenges}

Do \ref{Yfac} first, then finish the rest in any order.

\problem{}&lt;Yfac&gt;
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.}

\begin{solution}
	$\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)
	$
\end{solution}

\vfill



\problem{}
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

	\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
	$\Bigl\langle
		\langle \text{is last} ~,~ \text{item} \rangle
		~,~
		\text{next}...
	\Bigr\rangle$, where:

	\vspace{1ex}

	\say{is last} is a boolean, true iff this is the last item in the list. \par
	\say{item} is the thing you're storing \par
	\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 arbitrary functions 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
\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 function MOD so that $(\text{MOD}~a~b)$ 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}

\pagebreak</pre></body></html>