# reperiendi

## Another piece of the stone

Posted in Category theory, Math, Programming by Mike Stay on 2007 December 28

A few days ago, I thought that I had understood pi calculus in terms of category theory, and I did, in a way.

 $\lambda-$calculus model in Set type set of values term function (deterministic outcome) rewrite rule identity functional
 $\pi-$calculus model in Set process set of states reduction rule state update rule (deterministic outcome)

To make lambda calculus into a category, we mod out by the rewrite rules and consider equivalence classes of terms instead of actual terms. A model of this category (i.e. a functor from the category to Set) picks out a set of values for each datatype and a function for each program. Given a value from the input set, we get a single value from the output set.

Similarly, a model of the pi calculus assigns to each process a set of states and to each reduction rule a function that updates the state. A morphism in this way of thinking is a certain number of reductions to perform. The reductions are deterministic in the sense that we can model “A or B” as $A \sqcup B.$ Given an object’s state at a certain point, we can tell what set of states it can be in after the system reduces n messages.

However, what we really want is to know the set of states it can be in after all the messages have been processed: what is its normal form? This is far more like the rewrite rules in lambda calculus. It suggests that we should be treating the reduction rules like 2-morphisms instead of 1-morphisms. There’s one important difference from lambda calculus, however: the 2-morphisms of pi calculus are not confluent! It matters very much in which order they are evaluated. Thus processes can’t map to anything but trivial functions.

It looks like a better fit for models of the pi calculus is Rel, the category of sets and relations. A morphism in Rel can relate a single input state to multiple output states. This suggests we have a single object * in the pi calculus that gets mapped to a set of possible states for the system, while each process gets mapped to a relation that encodes all its possible reductions.

I’m rather embarrassed that it took me so long to notice this, since my advisor has been talking about replacing Set by Rel for years.

 category lambda calculus pi calculus objects types only one type *, a system of processes a morphism an equivalence class of terms a structural congruence class of processes dinatural transformation from the constant functor (mapping to the terminal set and its identity) to a functor generated by hom, products, projections, and exponentials (if they’re there) combinator: template for programs mapping between isomorphic types (usually) since there’s only one type, this is trivial Model of the category in Rel (usually taken in Set, a subcategory of Rel) a set of values for each data type and a function for each morphism between them * maps to a set S of states for the system, and each process gets mapped to a relation that relates each element of S to its possible reductions in S

## The Yoda embedding

Posted in Category theory by Mike Stay on 2007 December 21

$\overline{\mbox{Contravariant}}\,(\mbox{it is}).$

## Continuation passing as a reflection

Posted in Category theory, Math, Programming by Mike Stay on 2007 December 21

We can write any expression like $f(g(x,y))$ as a full binary tree where the nodes denote evaluation of the left child at the right, and the leaves are values:

$\begin{array}{ccccccc}&&ev\\&\swarrow&&\searrow\\f&&&&ev\\&&&\swarrow&&\searrow\\&&ev&&&&y\\&\swarrow&&\searrow\\g&&&&x\end{array}$

Figure 1: $f(g(x)(y))$

[In the caption of figure 1, the expression is slightly different; when using trees, it’s more convenient to curry all the functions—that is, replace every comma “,” by back-to-back parens: “)(” .]

The continuation passing transform (Yoneda embedding) first reflects the tree across the vertical axis and then replaces the root and all the left children with their continuized form—a value $x$ gets replaced with a function value $\overline{x} = (f \mapsto f(x)):$

$\begin{array}{ccccccc}&&&&\overline{ev}\\&&&\swarrow&&\searrow\\&&\overline{ev}&&&&f\\&\swarrow&&\searrow\\\overline{y}&&&&ev\\&&&\swarrow&&\searrow\\&&\overline{x}&&&&g\end{array}$

Figure 2: $\overline{\overline{ \overline{y}(\overline{x}(g)) } (f)}$

What does this evaluate to? Well,

$\begin{array}{rl} & \overline{\overline{ \overline{y}(\overline{x}(g)) } (f) }\\ = & \overline{f(\overline{y}(\overline{x}(g)))} \\ = & \overline{f(\overline{x}(g)(y))} \\ = & \overline{f(g(x)(y))} \end{array}$

As we hoped, it’s the continuization (Yoneda embedding) of the original expression. Iterating, we get

$\begin{array}{ccccccc}&&\overline{\overline{ev}}\\&\swarrow&&\searrow\\\overline{f}&&&&\overline{ev}\\&&&\swarrow&&\searrow\\&&\overline{ev}&&&&\overline{y}\\&\swarrow&&\searrow\\\overline{g}&&&&\overline{x}\end{array}$

Figure 3: $\overline{\overline{\overline{f}\left(\overline{\overline{\overline{g}\left(\overline{x}\right)}\left(\overline{y}\right)}\right)}}$

At this point, we get an enormous simplification: we can get rid of overlines whenever the left and right branch both have them. For instance,

$\overline{g}(\overline{x}) = \overline{x}(g) = g(x).$

Actually working out the whole expression would mean lots of epicycular reductions like this one, but taking the shortcut, we just get rid of all the lines except at the root. That means that we end up with

$\overline{\overline{f(g(x)(y))}}$

for our final expression.

However, if this expression is just part of a larger one—if what we’re calling the “root” is really the child of some other node—then the cancellation of lines on siblings applies to our “root” and its sibling, and we really do get back to where we started!

## A piece of the Rosetta stone

Posted in Category theory, Math, Programming by Mike Stay on 2007 December 20
 category lambda calculus pi calculus Turing machine objects types structural congruence classes of processes $\mathbb{N}\times S^*,$ where $\mathbb{N}$ is the natural numbers and $S^*$ is all binary sequences with finitely many ones. a morphism an equivalence class of terms a specific reduction from one process state to the next a specific transition from one state and position of the machine to the next dinatural transformation from the constant functor (mapping to the terminal set and its identity) to a functor generated by hom, products, projections, and exponentials (if they’re there) combinator reduction rule (covers all reductions of a particular form) tape-head update rule (covers all transitions with the current cell and state in common) products product types parallel processes multiple tapes internal hom exponential types all types are exponentials? ? Model of the category in Set A set of values for each data type and a function for each morphism between them A set of states for each process and a single evolution function out of each set. ?

This won’t be appearing in our Rosetta stone paper, but I wanted to write it down. What flavor of logic corresponds to the pi calculus? To the Turing machine?

## The continuation passing transform and the Yoneda embedding

Posted in Category theory, Math, Programming by Mike Stay on 2007 December 19

They’re the same thing! Why doesn’t anyone ever say so?

Assume A and B are types; the continuation passing transform takes a function (here I’m using C++ notation)

B f(A a);

and produces a function

 CPT_f( (*k)(B), A a) {
return k(f(a));
}

where X is any type. In CPT_f, instead of returning the value f(a) directly, it reads in a continuation function k and “passes” the result to it. Many compilers use this transform to optimize the memory usage of recursive functions; continuations are also used for exception handling, backtracking, coroutines, and even show up in English.

The Yoneda embedding takes a category $C^{\rm op}$ and produces a category $\mbox{HOM}(C, \mbox{Set})$:

$\begin{array}{ccccccc}\mbox{CPT}: & C^{\rm op} & \to & \mbox{HOM}(C, \mbox{Set}) \\ & A & \mapsto & \mbox{hom}(A, -) \\ & f:B\leftarrow A & \mapsto & \mbox{CPT}(f): & \mbox{hom}(B, -) & \to & \mbox{hom}(A, -) \\ & & & & k & \mapsto & k \circ f \end{array}$

We get the transformation above by uncurrying to get $\mbox{CPT}(f):\mbox{hom}(B, -) \times A \to -.$

In Java, a (cartesian closed) category $C$ is an interface C with a bunch of internal interfaces and methods mapping between them. A functor $F:C \to {\rm Set}$ is written

class F implements C.

Then each internal interface C.A gets instantiated as a set F.A of values and each method C.f() becomes instantiated as a function F.f() between the sets.

The continuation passing transform can be seen as a parameterized functor ${\rm CPT\langle X\rangle}:C \to {\rm Set}$. We’d write

class CPT<X> implements C.

Then each internal interface C.A gets instantiated as a set CPT.A of methods mapping from C.A to X—i.e. continuations that accept an input of type C.A—and each method C.f maps to the continuized function CPT.f described above.

Then the Yoneda lemma says that for every model of $C$—that is, for every class F implementing the interface C—there’s a natural isomorphism between the set $F(A)$ and the set of natural transformations ${\rm hom}({\rm hom}(A, -), F).$

A natural transformation $\alpha$ between $F:C \to {\rm Set}$ and ${\rm CPT\langle X \rangle}:C \to {\rm Set}$ is a way to cast the class F to the class CPT<X> such that for any method of C, you can either

• invoke its implementation directly (as a method of F) and then continuize the result (using the type cast), or
• continuize first (using the type cast) and then invoke the continuized function (as a method of CPT<X>) on the result

and you’ll get the same answer. Because it’s a natural isomorphism, the cast has an inverse.

The power of the Yoneda lemma is taking a continuized form (which apparently turns up in lots of places) and replacing it with the direct form. The trick to using it is recognizing a continuation when you see one.