Sized Types

Note

This is a stub.

Sizes help the termination checker by tracking the depth of data structures across definition boundaries.

The built-in combinators for sizes are described in Sized types.

Example for coinduction: finite languages

See Abel 2017 and Traytel 2017.

Decidable languages can be represented as infinite trees. Each node has as many children as the number of characters in the alphabet A. Each path from the root of the tree to a node determines a possible word in the language. Each node has a boolean label, which is true if and only if the word corresponding to that node is in the language. In particular, the root node of the tree is labelled true if and only if the word ε belongs to the language.

These infinite trees can be represented as the following coinductive data-type:

record Lang (i : Size) (A : Set) : Set where
  coinductive
  field
    ν : Bool
    δ : ∀{j : Size< i} → A → Lang j A

open Lang

As we said before, given a language a : Lang A, ν a ≡ true iff ε ∈ a. On the other hand, the language δ a x : Lang A is the Brzozowski derivative of a with respect to the character x, that is, w ∈ δ a x iff xw ∈ a.

With this data type, we can define some regular languages. The first one, the empty language, contains no words; so all the nodes are labelled false:

∅ : ∀ {i A}  → Lang i A
ν ∅   = false
δ ∅ _ = ∅

The second one is the language containing a single word; the empty word. The root node is labelled true, and all the others are labelled false:

ε : ∀ {i A} → Lang i A
ν ε   = true
δ ε _ = ∅

To compute the union (or sum) of two languages, we do a point-wise or operation on the labels of their nodes:

_+_ : ∀ {i A} → Lang i A → Lang i A → Lang i A
ν (a + b)   = ν a   ∨ ν b
δ (a + b) x = δ a x + δ b x

infixl 10 _+_

Now, lets define concatenation. The base case (ν) is straightforward: ε ∈ a · b iff ε ∈ a and ε ∈ b.

For the derivative (δ), assume that we have a word w, w ∈ δ (a · b) x. This means that xw = αβ, with α ∈ a and β ∈ b.

We have to consider two cases:

  1. ε ∈ a. Then, either:
    • α = ε, and β = xw, where w ∈ δ b x.
    • α = xα’, with α’ ∈ δ a x, and w = α’β ∈ δ a x · b.
  2. ε ∉ a. Then, only the second case above is possible:
    • α = xα’, with α’ ∈ δ a x, and w = α’β ∈ δ a x · b.
_·_ : ∀ {i A} → Lang i A → Lang i A → Lang i A
ν (a · b)   = ν a ∧ ν b
δ (a · b) x = if ν a then δ a x · b + δ b x else δ a x · b

infixl 20 _·_

Here is where sized types really shine. Without sized types, the termination checker would not be able to recognize that _+_ or if_then_else are not inspecting the tree, which could render the definition non-productive. By contrast, with sized types, we know that the a + b is defined to the same depth as a and b are.

In a similar spirit, we can define the Kleene star:

_* : ∀ {i A} → Lang i A → Lang i A
ν (a *)   = true
δ (a *) x = δ a x · a *

infixl 30 _*

Again, because the types tell us that _·_ preserves the size of its inputs, we can have the recursive call to a * under a function call to _·_.

Testing

First, we want to give a precise notion of membership in a language. We consider a word as a List of characters.

_∈_ : ∀ {i} {A} → List i A → Lang i A → Bool
[]      ∈ a = ν a
(x ∷ w) ∈ a = w ∈ δ a x

Note how the size of the word we test for membership cannot be larger than the depth to which the language tree is defined.

If we want to use regular, non-sized lists, we need to ask for the language to have size ∞.

_∈_ : ∀ {A} → List A → Lang ∞ A → Bool
[]      ∈ a = ν a
(x ∷ w) ∈ a = w ∈ δ a x

Intuitively, ∞ is a Size larger than the size of any term than one could possibly define in Agda.

Now, let’s consider binary strings as words. First, we define the languages ⟦ x ⟧ containing the single word “x” of length 1, for alphabet A = Bool:

⟦_⟧ : ∀ {i} → Bool → Lang i Bool
ν ⟦ _     ⟧       = false

δ ⟦ false ⟧ false = ε
δ ⟦ true  ⟧ true  = ε
δ ⟦ false ⟧ true  = ∅
δ ⟦ true  ⟧ false = ∅

Now we can define the bip-bop language, consisting of strings of even length alternating letters “true” and “false”.

bip-bop = (⟦ true ⟧ · ⟦ false ⟧)*

Let’s test a few words for membership in the language bip-bop!

test₁ : (true ∷ false ∷ true ∷ false ∷ true ∷ false ∷ []) ∈ bip-bop ≡ true
test₁ = refl

test₂ : (true ∷ false ∷ true ∷ false ∷ true ∷ []) ∈ bip-bop ≡ false
test₂ = refl

test₃ : (true ∷ true ∷ false ∷ []) ∈ bip-bop ≡ false
test₃ = refl