Converting Data.Reify explicit sharing graph to AST with de Bruijn indices

Question

I\'m trying to recover sharing (in the Type-Safe Observable Sharing in Haskell sense) for a simple AST, using Data.Reify:

{-# LANGUAGE DeriveFol         


        

Answer 1

We'll divide this problem into 3 parts. The first part is to use the data-reify library to recover the graph of the AstF. The second part will create an abstract syntax tree with Let bindings represented with de Bruijn indices. Finally, we will remove all of the unnecessary let bindings.

These are all the toys we will use along the way. StandaloneDeriving and UndecidableInstances are only needed to provide Eq and Show instances for things like Fix.

{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE DeriveFoldable #-}
{-# LANGUAGE DeriveTraversable #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE UndecidableInstances #-}

import Data.Foldable
import Data.Reify
import Data.Traversable
import qualified Data.List as List

import Data.IntMap ((!))
import qualified Data.IntMap as IntMap

import Prelude hiding (any)

Use data-reify

You have almost all of the pieces in place to use the data-reify library.

data AstF f =
      LitF Int
    | AddF f f
    deriving (Eq, Show, Functor, Foldable, Traversable)


newtype Fix f = In { out :: f (Fix f) }

deriving instance Eq (f (Fix f)) => Eq (Fix f)
deriving instance Show (f (Fix f)) => Show (Fix f)

instance Traversable a => MuRef (Fix a) where
    type DeRef (Fix a) = a
    mapDeRef f = traverse f . out

All that's missing is the call to reifyGraph. Let's try a small example

do
    let example = In (AddF (In (AddF (In (LitF 1)) (In (LitF 2)))) example)
    graph <- reifyGraph example
    print graph

This outputs

let [(1,AddF 2 1),(2,AddF 3 4),(4,LitF 2),(3,LitF 1)] in 1

graph has the type Graph AstF, and is constructed by the constructor Graph [(Unique, AstF Unique)] Unique. The first argument to the constructor is the list of nodes with their new unique keys. Each edge in the structure has been replaced with the new unique key of the node at the edge's head. The second argument to the constructor is the unique key of the node of the root of the tree.

Convert graph to Let representation

We will convert the Graph from data-reify into a de Bruijn indexed abstract syntax tree with Let bindings. We will represent the AST using the following type. This type doesn't need to know anything about the internal representation of the AST.

type Index = Int

-- This can be rewritten in terms of Fix and Functor composition
data Indexed f
    = Var Index
    | Let (Indexed f) (Indexed f)
    | Exp (f (Indexed f))

deriving instance Eq (f (Indexed f)) => Eq (Indexed f)
deriving instance Show (f (Indexed f)) => Show (Indexed f)

The Indexes represent the number of Lets between where the variable is used and the Let where it was declared. You should read Let a b as let (Var 0)=a in b

Our strategy to convert the graph into an Indexed AST is to traverse the graph starting at the root node. At every node, we will introduce a Let binding for that node. For every edge we will check to see if the node it refers to is already in an introduced Let binding that is in scope. If it is, we will replace the edge with the variable for that Let binding. If it is not already introduced by a Let binding, we will traverse it. The only thing we need to know about the AST we are operating on is that it is a Functor.

index :: Functor f => Graph (DeRef (Fix f)) -> Indexed f
index (Graph edges root) = go [root]
    where
        go keys@(key:_) =
            Let (Exp (fmap lookup (map ! key))) (Var 0)
                where
                    lookup unique = 
                        case List.elemIndex unique keys of
                            Just n -> Var n
                            Nothing -> go (unique:keys)
        map = IntMap.fromList edges

We will define the following for convenience.

reifyLet :: Traversable f => Fix f -> IO (Indexed f)
reifyLet = fmap index . reifyGraph

We'll try the same example as before

do
    let example = In (AddF (In (AddF (In (LitF 1)) (In (LitF 2)))) example)
    lets <- reifyLet example
    print lets

This outputs

Let (Exp (AddF (Let (Exp (AddF (Let (Exp (LitF 1)) (Var 0)) (Let (Exp (LitF 2)) (Var 0)))) (Var 0)) (Var 0))) (Var 0)

We only had 1 let binding in example but this has 4 Lets. We will remove the unnecessary Let binding in the next step.

Remove unnecessary `Let` bindings

To remove Let bindings that introduce unused variables, we need a notion of what a used variable is. We will define it for any Foldable AST.

used :: (Foldable f) => Index -> Indexed f -> Bool
used x (Var y) = x == y
used x (Let a b) = used (x+1) a || used (x+1) b
used x (Exp a)  = any (used x) a

When we remove a Let bindings, the number of intervening Let bindings, and thus the de Bruijn indices for variables, will change. We will need to be able to remove a variable from an Indexed AST

remove x :: (Functor f) => Index -> Indexed f -> Indexed f
remove x (Var y) =
    case y `compare` x of
        EQ -> error "Removed variable that's being used`
        LT -> Var y
        GT -> Var (y-1)
remove x (Let a b) = Let (remove (x+1) a) (remove (x+1) b)
remove x (Exp a) = Exp (fmap (remove x) a)

There are two ways a Let binding can introduce an unused variable. The variable can be completely unused, for example let a = 1 in 2, or it can be trivially used, as in let a = 1 in a. The first can be replaced by 2 and the second can be replaced by 1. When we remove the Let binding, we also need to adjust all of the remaining variables in the AST with remove. Things that aren't Let don't introduce unused variables, and have nothing to replace.

removeUnusedLet :: (Functor f, Foldable f) => Indexed f -> Indexed f
removeUnusedLet (Let a b) =
    if (used 0 b) 
    then
        case b of
            Var 0 ->
                if (used 0 a)
                then (Let a b)
                else remove 0 a
            _     -> (Let a b)
    else remove 0 b
removeUnusedLet x = x

We'd like to be able to apply removeUnusedLet everywhere in the Indexed AST. We could use something more generic for this, but we'll just define for ourselves how to apply a function everywhere in an Indexed AST

mapIndexed :: (Functor f) => (Indexed f -> Indexed f) -> Indexed f -> Indexed f
mapIndexed f (Let a b) = Let (f a) (f b)
mapIndexed f (Exp a)   = Exp (fmap f a)
mapIndexed f x         = x

postMap :: (Functor f) => (Indexed f -> Indexed f) -> Indexed f -> Indexed f
postMap f = go
    where
        go = f . mapIndexed go

Then we can remove all the unused lets with

removeUnusedLets = postMap removeUnusedLet

We'll try our example again

do
    let example = In (AddF (In (AddF (In (LitF 1)) (In (LitF 2)))) example)
    lets <- reifyLet example
    let simplified = removeUnusedLets lets
    print simplified

This introduces only a single Let

   Let (Exp (AddF (Exp (AddF (Exp (LitF 1)) (Exp (LitF 2)))) (Var 0))) (Var 0)

Limitations

Mutually recursive definitions don't result in mutually recursive Let bindings. For example

do
    let
        left   =  In (AddF (In (LitF 1)) right       )
        right   = In (AddF left         (In (LitF 2)))
        example = In (AddF left          right       )
    lets <- reifyLet example
    let simplified = removeUnusedLets lets
    print simplified

Results in

Exp (AddF
    (Let (Exp (AddF
        (Exp (LitF 1))
        (Exp (AddF (Var 0) (Exp (LitF 2))))
    )) (Var 0))
    (Let (Exp (AddF
        (Exp (AddF (Exp (LitF 1)) (Var 0)))
        (Exp (LitF 2))
    )) (Var 0)))

I don't believe there is a mutually recursive representation for these in Indexed without using a negative Index.