Table of Contents

Implementing the Y-Combinator

Curry defines the $Y$-combinator as:

\begin{eqnarray*} Y &=& \lambda f .(\lambda x.f(xx))(\lambda x.f(xx)) \end{eqnarray*}

which, when applied to a lambda term $g$, $\beta$-reduces in two steps to:

\begin{eqnarray*} Y g &=& g (Y g) \end{eqnarray*}

An equivalent $Y$-combinator definition is the following:

\begin{eqnarray*} Y &=& \lambda f.(\lambda x.xx)(\lambda x.f(xx)) \end{eqnarray*}

which consists of part of the $\Omega$ combinator $\Omega = (\lambda x.xx)(\lambda x.xx)$ named $\omega = \lambda x.xx$ and the term $\lambda x.f(xx)$.

If we $\beta$-reduce the equivalent form once:

\begin{eqnarray*} Y &=& \lambda f.(\lambda x.xx)(\lambda x.f(xx)) \end{eqnarray*}

we would obtain Curry's form of the $Y$-combinator.

Whilst the Curry form should work perfectly fine in Scheme and untyped lambda calculus, a statically typed language will not accept a function application such as $x x$ since the type system will forbid it. In order to do that, we start by applying an additional $\eta$-abstraction to obtain the form:

\begin{eqnarray*} Y &=& \lambda f.(\lambda x.xx)(\lambda x.(f (\lambda y.((x x) y)))) \end{eqnarray*}

which can be re-written as:

\begin{eqnarray*} Y &=& \lambda f.(\lambda x.xx)(f (\lambda y.((\lambda x.xx) y))) \end{eqnarray*}

We notice that the $\omega$ combinator ($\lambda x.xx$) appears twice in the resulting $Y$-combinator which lets us express the $Y$-combinator using the $\omega$ combinator:

\begin{eqnarray*} \omega &=& \lambda x.xx \\ Y' &=& \lambda f.\omega(f (\lambda y.\omega y)) \end{eqnarray*}

where $Y'$ and $Y$ are $\beta$-equivalent.

Expressing the $Y$ combinator in terms of $\omega$ allows us to implement the $Y$ combinator in statically typed programming languages.

Implementations

Exponentiation with Church Numerals

You may have seen the definition of exponentiation as the combinator $\mathbbold{EXP}$ (or $\mathbbold{POW}$) given by the lambda abstraction:

\begin{eqnarray*} \mathbbold{EXP} &\equiv& \lambda m.\lambda n.mn \end{eqnarray*}

where $n$ and $m$ are both Church numerals. $\mathbbold{EXP}$ will perform the mathematical equivalent of $m^{n}$ using Church numerals.

Therefore the application of a Church numeral $n$ to another Church numeral $\mathbbold{m}$ will yield the Church numeral equivalent to $m^n$. For instance, applying the Church numeral $\mathbbold{1}$ to Church numeral $\mathbbold{2}$ should yield $m^{n} = 2^{1} \equiv \mathbbold{2}$. While all the numerals work out, the application of the Church numeral zero to any other Church numeral $\beta$-reduces to the identity function $\mathbbold{I}$. For instance, applying the Church numeral $\mathbbold{0}$ to $\mathbbold{2}$ we have:

\begin{eqnarray*} \mathbbold{0}(\mathbbold{2}) &\equiv& (\lambda f.\lambda x.x)(\lambda f.\lambda x.f(f(x))) \\ &\xrightarrow{\beta-reduce: \lambda f}& \lambda x.x \\ &\equiv& \mathbbold{I} \end{eqnarray*}

Since the identity function $\mathbbold{I}$ is not a Church numeral, it would seem that the application is not defined. However, by the rules of exponentiation, any number to the power zero should yield one; in other words: $2^{0} = 1$.

The trick here is that the identity function $\mathbbold{I}$ is equivalent with the Church numeral $\mathbbold{I}$ under the $\eta$-expansion.

The $\eta$-expansion is given by:

\begin{eqnarray*} f &\xrightarrow{\eta-expand: f}& \lambda x.f(x) \end{eqnarray*}

so we can re-work the application of the Church numeral zero to the Church numeral two:

\begin{eqnarray*} \mathbbold{0}(\mathbbold{2}) &\equiv& (\lambda f.\lambda x.x)(\lambda f.\lambda x.f(f(x))) \\ &\xrightarrow{\beta-reduce: \lambda f}& \lambda x.x \\ &\xrightarrow{\alpha-convert: x \mapsto f}& \lambda f.f \\ &\xrightarrow{\eta-expand: f \mapsto \lambda x.f(x)}& \lambda f.\lambda x.f(x) \\ &\equiv& \mathbbold{1} \end{eqnarray*}

which yields the expected result.

η-Reduction and η-Expansion

$\eta$-reduction is given by:

\begin{eqnarray*} \lambda x.f(x) &\xrightarrow{\eta-reduce: \lambda x.f(x)}& f \end{eqnarray*}

$\eta$-expansion is given by:

\begin{eqnarray*} f &\xrightarrow{\eta-expand: f}& \lambda x.f(x) \end{eqnarray*}

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