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Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>

src/Advanced/Lambda Calculus/parts/05 challenges.tex

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+\section{Challenges}
+
+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.}
+
+\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