handouts/Advanced/Lambda Calculus/parts/04 recursion.tex

\section{Recursion}


%\iftrue
\iffalse

	You now have a choice. Choose wisely --- there's no going back.

	\begin{tcolorbox}[
		breakable,
		colback=white,
		colframe=gray,
		arc=0pt, outer arc=0pt
	]
		\raggedright
		\textbf{Take the red pill:} You stay on this page and try to solve \ref{thechallenge}. \par
		This will take a while, and it's very unlikely you'll finish before class ends.


		\vspace{4mm}

		I strongly prefer this option. It's not easy, but you'll be very happy if you solve it yourself.
		This is a chance to build your own solution to a fundamental problem in this field, just as
		Curry, Church, and Turing did when first developing the theory of lambda calculus.

		- Mark
	\end{tcolorbox}

	\begin{tcolorbox}[
		breakable,
		colback=white,
		colframe=gray,
		arc=0pt, outer arc=0pt
	]
		\textbf{Take the blue pill:} You skip this problem and turn the page.
		Half of the answer to \ref{thechallenge} will be free, and the rest will be
		broken into smaller steps. This is how we usually learn out about interesting
		mathematics, both in high school and in university.

		\vspace{2mm}

		This path isn't as rewarding as the one above, but it is well-paved
		and easier to traverse.
	\end{tcolorbox}

	\problem{}<thechallenge>
	Can you find a way to recursively call functions in lambda calculus? \par
	Find a way to define a recursive factorial function. \par

	\note{$A = (\lm a. A~a)$ doesn't count. You can't use a macro inside itself.}

	\vfill
	\pagebreak
\fi





















Say we want a function that computes the factorial of a positive integer. Here's one way we could define it:
$$
	x! = \begin{cases}
		x \times (x-1)! & x \neq 0 \\
		1 & x = 0
	\end{cases}
$$

We cannot re-create this in lambda calculus, since we aren't given a way to recursively call functions.

\vspace{2mm}

One could think that $A = \lm a. A~a$ is a recursive function. In fact, it is not. \par
Remember that such \say{definitions} aren't formal structures in lambda calculus. \par
They're just shorthand that simplifies notation.

\begin{instructornote}
	We're talking about recursion, and \textit{computability} isn't far away. At one point or another, it may be good to give the class a precise definition of \say{computable by lambda calculus:}

	\vspace{4ex}

	Say we have a device that reduces a $\lm$ expression to $\beta$-normal form. We give it an expression, and the machine simplifies it as much as it can and spits out the result.

	\vspace{1ex}

	An algorithm is \say{computable by lambda calculus} if we can encode its input in an expression that resolves to the algorithm's output.
\end{instructornote}

\problem{}
Write an expression that resolves to itself. \par
\hint{Your answer should be quite short.}

\vspace{1ex}

This expression is often called $\Omega$, after the last letter of the Greek alphabet. \par
$\Omega$ useless on its own, but it gives us a starting point for recursion. \par

\begin{solution}
	$\Omega = M~M = (\lm x . xx) (\lm x . xx)$

	\vspace{1mm}
	
	An uninspired mathematician might call the Mockingbird $\omega$, \say{little omega}.
\end{solution}

\vfill
\pagebreak





\definition{}
This is the \textit{Y-combinator}. You may notice that it's just $\Omega$ put to work.
$$
	Y = \lm f . (\lm x . f(x~x))(\lm x . f(x~x))
$$

\problem{}
What does this thing do? \par
Evaluate $Y f$.



%\vfill

%\definition{}
%We say $x$ is a \textit{fixed point} of a function $f$ if $f(x) = x$.

%\problem{}
%Show that $Y F$ is a fixed point of $F$.

%\vfill

%\problem{}
%Let $b = (\lm xy . y(xxy))$ and $B = b ~ b$. \par
%Let $N = B F$ for an arbitrary lambda expression $F$. \par

%Show that $F N = N$.

%\vfill

%\problem{Bonus}
%Find a fixed-point combinator that isn't $Y$ or $\Theta$.

\vfill
\pagebreak