Dr. David J. Pearce

Iso-Recursive versus Equi-Recursive Types

An important component of the Whiley language is the use of recursive data types.  Whilst these are similar to the algebraic data types found in languages like Haskell, they are also more powerful since Whiley employs a structural type system. So, what does all this mean? Well, let’s look at an example:

define IntList as null | {int data, IntList next}
define AnyList as null | {any data, AnyList next}

AnyList f(IntList ls):
    return ls

Here, we’ve got two recursive data types which describe something akin to a linked list. For example, an IntList describes a recursive structure made up of zero or more nodes. Each node contains an int data field, and a next field to access either the next node, or null if the end is reached. An AnyList is very similar, except that its payload consists of arbitrary data, rather than just integer data (as for IntList).

In the above example, we see that the parameter ls is returned without an explicit cast. In other words, we know that IntList is a subtype of AnyList. This is a key difference from standard algebraic data types, where IntList and AnyList would always be considered distinct (i.e. unrelated) types.

Subtyping Recursive Types

The ability to have implicit subtyping relationships between recursive data types is a key strength compared with algebraic data types.  At the same time, it also presents a complex algorithmic challenge and numerous approaches have been proposed in the literature.  Previously, I have written extensively on this subject (see e.g. here, here and here).  In fact, there are two broad approaches taken to subtyping recursive data types: iso-recursive and equi-recursive.  In Whiley, and my previous writings on this topic, I have strictly followed the equi-recursive approach and I would strongly recommend this to anyone developing a recursive type system.

A good account of the iso- versus equi-recursive approaches can be found in Pierce’s excellent book [1].  The key difference between the two approaches is whether the recursion is “implicit” (equi-recursive) or “explicit” (iso-recursive).  In the equi-recursive approach, types are implemented under-the-hood as directed graphs where recursion corresponds to a cycle in the graph (see here for an example).  In the iso-recursive approach, special types of the form μX.T are used (the so-called “mu” types). In such a type, X is a recursive variable used within the body T. For example, a mu type corresponding to our IntList example is: μX.(null | {int data, X next}).

Mu types can be “folded” and “unfolded”. To unfold a type μX.T we generate the type T[X/μX.T] (that’s T with X replaced by μX.T). For example, unfolding μX.(null | {int data, X next}) gives null | {int data, μX.(null | {int data, X next}) next}. Explicit operators for unfolding and folding are provided for manipulating types where fold(unfold(T)) = T holds for any type T. In the iso-recursive scheme (and unlike the equi-recursive scheme), a type and its unfolding are distinct and unrelated. To show that one mu type subtypes another, we must first fold/unfold them to have the same recursive structure, after which we can establish the subtyping relation via the so-called “Amber Rule” (see e.g. [1,2] for more on this).

Subtyping Iso-Recursive Types

Having considered how subtyping is performed for iso-recursive types, the question is: for two types which should be related, can we always fold/unfold them to reach a matching recursive structure? Well, I believe the answer is no. Here’s my informal proof:

define LTree as null | { int data, LTree left, LTree right}
define RTree as null | { int data, RTree_b left, RTree right}
define RTree_b as null | { int data, RTree left, RTree right}

Now, we have to ask ourselves the question: is there a sequence of fold/unfold operations that will transform LTree into RTree (i.e. to show that they are equivalent)? To see why this is impossible, we consider the notion of “balance” (as in balanced tree). After any number of fold / unfold operations the LTree type will remain balanced; but, for the RTree type, it will never be balanced.

For a more detailed investigation into the expressiveness of iso-recursive types, I’d suggest looking at this recent paper [2].


  1. Types and Programming Languages, Benjamin C. Pierce. The MIT Press, ISBN 0-262-16209-1.

  2. Completely Subtyping Iso-recursive Types, Technical Report CSE-071012, Jeremy Blackburn Ivory Hernandez Jay Ligatti Michael Nachtigal, University of South Florida.