Trees indexed by a Cayley Monoid

Posted on December 27, 2020
Tags: Haskell

The Cayley monoid is well-known in Haskell (difference lists, for instance, are a specific instance of the Cayley monoid), because it gives us O(1)O(1) <>. What’s less well known is that it’s also important in dependently typed programming, because it gives us definitional associativity. In other words, the type x . (y . z) is definitionally equal to (x . y) . z in the Cayley monoid.

Some helpers and extra code

data Nat = Z | S Nat

type family (+) (n :: Nat) (m :: Nat) :: Nat where
  Z   + m = m
  S n + m = S (n + m)

I used a form of the type-level Cayley monoid in a previous post to type vector reverse without proofs. I figured out the other day another way to use it to type tree flattening.

Say we have a size-indexed tree and vector:

data Tree (a :: Type) (n :: Nat) :: Type where
  Leaf  :: a -> Tree a (S Z)
  (:*:) :: Tree a n -> Tree a m -> Tree a (n + m)

data Vec (a :: Type) (n :: Nat) :: Type where
  Nil  :: Vec a Z
  (:-) :: a -> Vec a n -> Vec a (S n)

And we want to flatten it to a list in O(n)O(n) time:

treeToList :: Tree a n -> Vec a n
treeToList xs = go xs Nil
    go :: Tree a n -> Vec a m -> Vec a (n + m)
    go (Leaf    x) ks = x :- ks
    go (xs :*: ys) ks = go xs (go ys ks)

Haskell would complain specifically that you hadn’t proven the monoid laws:

• Couldn't match type ‘n’ with ‘n + 'Z’
• Could not deduce: (n2 + (m1 + m)) ~ ((n2 + m1) + m)

But it seems difficult at first to figure out how we can apply the same trick as we used for vector reverse: there’s no real way for the Tree type to hold a function from Nat to Nat.

To solve this problem we can borrow a trick that Haskellers had to use in the good old days before type families to represent type-level functions: types (or more usually classes) with multiple parameters.

data Tree' (a :: Type) (n :: Nat) (m :: Nat) :: Type where
  Leaf  :: a -> Tree' a n (S n)
  (:*:) :: Tree' a n2 n3
        -> Tree' a n1 n2
        -> Tree' a n1 n3

The Tree' type here has three parameters: we’re interested in the last two. The first of these is actually an argument to a function in disguise; the second is its result. To make it back into a normal size-indexed tree, we apply that function to zero:

type Tree a = Tree' a Z

three :: Tree Int (S (S (S Z)))
three = (Leaf 1 :*: Leaf 2) :*: Leaf 3

This makes the treeToList function typecheck without complaint:

treeToList :: Tree a n -> Vec a n
treeToList xs = go xs Nil
    go :: Tree' a x y -> Vec a x -> Vec a y
    go (Leaf    x) ks = x :- ks
    go (xs :*: ys) ks = go xs (go ys ks)