A terminal view of testing polymorphic functions

Posted on June 29, 2017

Parametricity restricts the behavior of a polymorphic function $$\phi : \forall \alpha. T \alpha \to U \alpha$$, so that, picking one type $$\tau$$ and one input value $$x : T \tau$$ and inspecting the result $$\phi~x : U\tau$$ often tells us about what happens on many other inputs.

To test polymorphic functions, Bernardy et al.1 (2010) have a monomorphization technique using initial algebras (the “initial view”). They also conjectured the existence of a dual method based on terminal coalgebras (the “terminal view”), which I will show in this post.

Initial view

We assume that $$T\alpha$$ has the form $$(F\alpha \to \alpha)\times(G\alpha\to X)$$, for some functors $$F$$ and $$G$$ and a non-parametric type $$X$$ (i.e., in which $$\alpha$$ does not occur), and that $$U$$ (the result type) is also a functor. Then, to test a function of type $$\forall\alpha.(F\alpha\to\alpha)\times(G\alpha\to X)\to U\alpha$$, it suffices to only try inputs where the first argument is the initial $$F$$-algebra, $$f_0 : F\mu\to\mu$$ (and $$\mu$$ is the “least fixed point” of the functor $$F$$). The second argument may vary freely in the type $$G\mu\to X$$.

The intuition is that the first argument, of type $$F\alpha\to\alpha$$, is used by the polymorphic function to construct values of abstract type $$\alpha$$, while the second argument provides ways of “observing” these constructed values. The initial algebra $$f_0 : F\mu\to\mu$$ corresponds to an injective constructor, so that a tester can inspect values of type $$\mu$$ in the output $$U\mu$$ to know exactly how they were constructed by $$\phi$$. By parametricity, the polymorphic function $$\phi$$ must construct its output “uniformly” over all types, so looking at the result on one input allows us to deduce its behavior on many other inputs.

This effectively establishes an isomorphism:

$(\forall \alpha. (F\alpha\to\alpha)\times(G\alpha\to X)\to U\alpha) \quad\simeq\quad (G\mu\to X)\to U\mu$

Testing a polymorphic function is equivalent to testing a corresponding monomorphic function, which is more straightforward to do.

Definitions

Given $$\phi : \forall\alpha. (F\alpha\to\alpha)\times(G\alpha\to\alpha)\to U\alpha.$$ and the initial algebra $$f_0 : F\mu\to\mu$$, the monomorphization of $$\phi$$ is defined by:

\DeclareMathOperator\mono{mono} \DeclareMathOperator\poly{poly} \DeclareMathOperator\id{id} \begin{aligned} & \mono{\phi} : (G\mu\to\mu)\to U\mu \\ & \mono{\phi}~g_0 = \phi~(f_0, g_0) \end{aligned}

To define the inverse, we recall the definition of the initial algebra $$f_0$$: every algebra $$f : F\alpha\to\alpha$$ induces a catamorphism $$\eta_{f} : \mu\to\alpha$$, which is the unique function making the following square commute:

$\require{AMScd} \begin{CD} F\mu @>{f_0}>> \mu \\ @V{F\eta_{f}}VV @VV{\eta_{f}}V \\ F\alpha @>{f}>> \alpha \end{CD}$

Given a monomorphic $$\psi : (G\mu\to\mu)\to U\mu$$, its polymorphization is:

\begin{aligned} & \poly{\psi} : \forall\alpha. (F\alpha\to\alpha)\times(G\alpha\to\alpha)\to U\alpha \\ & \poly{\psi}~(f, g) = U\eta_{f}~(\psi~(g \circ G\eta_{f})) \end{aligned}

Monopoly theorem

Polymorphization is the inverse of monomorphization.

1. $$\poly$$ is a right-inverse for $$\mono$$:

$\mono\circ\poly = \id$

Unfolding the definitions, we need to check the following equation:

$U\eta_{f_0}~(\psi~(g_0\circ G\eta_{f_0})) = \psi~g_0$

Indeed, the catamorphism $$\eta_{f_0}$$ of the initial algebra must be the identity function. We may thus simplify the left hand side by functoriality of $$U$$ and $$G$$.

2. $$\poly$$ is a left-inverse for $$\mono$$:

$\poly \circ \mono = \id$

Equivalently,

$U\eta_{f}~(\phi~(f_0,g\circ G\eta_{f})) = \phi~(f, g)$

That is a consequence of the free theorem of $$\phi$$, with the catamorphism $$\eta_{f}$$ as a functional relation between $$\mu$$ and $$\alpha$$ (details omitted).

The form of $$T\alpha$$ as $$(F\alpha\to\alpha)\times(G\alpha\to X)$$ above may seem restrictive, but the paper also shows how a large class of types can actually be transformed to make that result applicable. Basically, the type $$T\alpha$$ may be a product of types of the form $$C_i\alpha\to D_i\alpha$$ where $$C_i, D_i$$ are functors, and $$D_i$$ is made of sums, products, and fixpoints.

To motivate the dual approach, let’s look at an example where that is not the case.

Example

Let $$\mathbb{B}$$ be the type of booleans, with constructors $$0,1 : \mathbb{B}$$ and destructor $$\mathrm{if}:\forall\beta.\mathbb{B}\to\beta\to\beta\to\beta$$. Our example will be:

$\forall \alpha. ((\alpha\to\mathbb{B}) \times ((\alpha\to\mathbb{B})\to\mathbb{B})) \to \mathbb{B}$

Given arguments $$f : \alpha\to\mathbb{B}$$ and $$g : (\alpha\to\mathbb{B})\to\mathbb{B}$$, the result will be a boolean combination of applications of $$f$$ and $$g$$. Since there is no $$\alpha$$ in the immediate context to apply $$f$$ to, we can only apply $$g$$ at first. Here is a typical inhabitant of the above type:

$\phi = \lambda(f, g). g~(\lambda a. \mathrm{if}~(f~a)~0~1)$

The only way a polymorphic function $$\phi$$ of that type can inspect a value of type $$\alpha$$ is to use the first parameter. Hence, from the point of view of $$\phi$$, a value of type $$\alpha$$ contains only as much information as a boolean. To test $$\phi$$, it is actually sufficient to instantiate $$\alpha$$ with $$\mathbb{B}$$ and set the first argument to the identity function $$f_0 = \lambda x.x$$.

In fact, the above type is isomorphic to the following finite type:

$((\mathbb{B}\to\mathbb{B})\to\mathbb{B})\to\mathbb{B}$

Terminal view

With the initial view, we examined how a polymorphic function uses constructors of abstract type $$\alpha$$, represented by algebras $$F\alpha\to\alpha$$. With an initial algebra $$f_0 : F\mu\to\mu$$, the least fixed point $$\mu$$ was as large as the sum of ways of constructing its inhabitants using $$f_0$$.

Here, we shall look at destructors: coalgebras $$\alpha \to F\alpha$$. (Above, $$F\alpha=\mathbb{B}$$.) With a terminal coalgebra $$f_1 : \nu\to F\nu$$, the “greatest fixed point” $$\nu$$ will be as large as the “product” of observations we can make about an inhabitant using $$f_1$$.

We now consider the following type, where $$F$$ is a functor and we’ll see later what $$U$$ can be:

$\forall \alpha. (\alpha\to F\alpha)\to U\alpha$

What happens if we take the terminal coalgebra, $$f_1:\nu\to F\nu$$?

By definition, every coalgebra $$f : \alpha\to F\alpha$$ induces an anamorphism $$\epsilon_{f} : \alpha\to\nu$$, which is the unique function with the commutative square:

$\require{AMScd} \begin{CD} \alpha @>{f}>> F\alpha \\ @V{\epsilon_{f}}VV @VV{F\epsilon_{f}}V \\ \nu @>{f_1}>> F\nu \end{CD}$

Under the extra assumption that $$U$$ is a contravariant functor, we show this isomorphism:

$(\forall\alpha. (\alpha\to F\alpha)\to U\alpha) \quad\simeq\quad U\nu$

Definitions

The monomorphization of $$\phi : \forall\alpha. (\alpha\to F\alpha)\to U\alpha$$ is defined by:

\begin{aligned} & \mono\phi : U\nu \\ & \mono\phi = \phi~f_1 \end{aligned}

The polymorphization of $$\psi : U\nu$$ is defined by:

\begin{aligned} & \poly\psi : \forall\alpha. (\alpha\to F\alpha)\to U\alpha \\ & \poly\psi~f = U\epsilon_{f}~\psi \end{aligned}

Note that the contravariant functor $$U$$ lifts $$\epsilon_{f} : \alpha\to\nu$$ to $$U\epsilon_{f} : U\nu\to U\alpha$$.

Theorem

Polymorphization is the inverse of monomorphization.

1. $$\poly$$ is a right-inverse for $$\mono$$:

$\mono\circ\poly = \id$

Equivalently, $$U\epsilon_{f_1}~\psi = \psi$$.

Indeed, the anamorphism $$\epsilon_{f_1}$$ of the terminal coalgebra must be the identity function.

2. $$\poly$$ is a left-inverse for $$\mono$$:

$\poly \circ \mono = \id$

Equivalently, $$U\epsilon_{f}~(\phi~f_1) = \phi~f$$.

That is a consequence of the free theorem of $$\phi$$, with the catamorphism $$\epsilon_{f}$$ as a functional relation between $$\nu$$ and $$\alpha$$ (details omitted).

Application

This technique actually applies to our example; with $$\alpha\to F\alpha=\alpha\to\mathbb{B}$$ and $$U\alpha=((\alpha\to\mathbb{B})\to\mathbb{B})\to\mathbb{B}$$, we obtain the same monomorphization:

\begin{aligned} \nu &= \mathbb{B} \\ f_1 &= \lambda x. x \end{aligned}

Notice that simple trick of pushing the extra argument type $$(\alpha\to\mathbb{B})\to\mathbb{B}$$ into the result type $$U\alpha$$. This happens to work whenever the argument type is covariant in $$\alpha$$ (this includes types like $$\alpha$$, $$X\to\alpha$$, $$X$$, where $$X$$ is non-parametric).

Dually, in the initial view, we separated the algebra $$F\alpha\to\alpha$$ from an “observation function” $$G\alpha\to X$$; we can simplify that assumption by shoving that type (contravariant in $$\alpha$$, since $$G$$ is covariant) into the result type $$U\alpha$$, which remains covariant.

To summarize, we have two dual methods of monomorphizing polymorphic functions, of type $$\forall\alpha.T\alpha\to U\alpha$$, in the following situations:

• $$\forall\alpha. (F\alpha\to\alpha)\to U\alpha$$, where $$F$$ and $$U$$ are covariant—in particular, $$U\alpha$$ may be a function type whose arguments $$G\alpha\to X$$ correspond to “observation functions”;
• $$\forall\alpha. (\alpha\to F\alpha)\to U\alpha$$, where $$F$$ is covariant and $$U$$ is contravariant—$$U\alpha$$ may be a function type with “constructors” $$Y\to H\alpha$$ as arguments for example.

Overlapping views

There are cases where both techniques apply. We should get equivalent results since monomorphizations are isomorphisms. For instance:

$\forall\alpha. (X\to \alpha) \to (\alpha\to Y) \to Z$

The initial view yields $$\alpha = X$$, with the first argument set to the identity function; the second argument, which may vary freely, has type $$X\to Y$$.

The terminal view yields $$\alpha = Y$$, with the second argument set to the identity function; the first argument has type $$X\to Y$$.

Here is another example:

$\forall\alpha. (\alpha\to\alpha\times X)\to\alpha\to Y$

The coalgebra $$\alpha\to\alpha\times X$$ views $$\alpha$$ as an infinite stream of $$X$$, i.e., the type $$X^\omega$$. We fix the first argument as the stream destructor $$X^\omega \to X^\omega\times X$$, and the second argument may be any stream $$X^\omega$$.

Before taking the initial view, we rewrite that type a bit, first by commutativity $$A \to B \to C \simeq B \to A \to C$$ and second by distributivity of exponentials over products $$A \to B\times C \simeq (A\to B)\times(A\to C)$$.

\begin{aligned} & (\alpha\to\alpha\times X)\to\alpha\to Y \\ \quad\simeq\quad& \alpha\to(\alpha\to\alpha\times X)\to Y \\ \quad\simeq\quad& \alpha\to(\alpha\to\alpha)\to(\alpha\to X)\to Y \end{aligned}

The algebra $$\alpha \times (\alpha\to\alpha)$$ (isomorphic to $$(\unicode{x1D7D9} + \alpha) \to \alpha$$) views $$\alpha$$ as a natural number. With $$\alpha = \mathbb{N}$$, we fix the first two arguments to the Peano constructors (zero and successor), and the last argument varies over $$\mathbb{N}\to X$$, which is isomorphic to streams $$X^\omega$$.

Questions

The initial view can be adapted to other types $$\forall\alpha.T\alpha\to U\alpha$$ when there is an embedding of $$T\alpha$$ in some $$(F\alpha\to\alpha)\times C\alpha$$ for a covariant $$F$$ and contravariant $$C$$.

Dually, are there interesting types to embed in $$(\alpha\to F\alpha)\times D\alpha$$, for $$F$$ and $$D$$ both covariant?

Is there a unified view that generalizes the above?

What can we do for a type like $$\forall\alpha.((\alpha\to\alpha)\to\alpha)\to\alpha$$, for which neither the initial nor terminal view are applicable?

1. Testing polymorphic properties. Jean-Philippe Bernardy, Patrik Jansson, Koen Claessen. ESOP 2010.