Langton's Ant Redux, and Unrestricted Types

Fri Sep 26, 2014

So this past Wednesday was the monthly Toronto Haskell User Group meeting, and I used the opportunity to A) show off my Langton's Ant implementation to people that might be able to offer pointers and B) ask a question about types that I'd been thinking about. We'll handle those in reverse order, with a bit of exposition on the question.

Splay Trees

Here's a Haskell implementation of a Splay Tree not interesting enough to dwell on:

module SplayTree (Tree(..), fromList, SplayTree.elem, insert, remove) where

data Tree a = Node { value :: a
                   , left :: Tree a
                   , right :: Tree a }
            | Empty deriving (Eq, Ord, Show, Read)

rotateR Empty = Empty
rotateR tree@(Node _ Empty _) = tree
rotateR (Node val (Node child ll lr) right) = Node child ll (Node val lr right)

rotateL Empty = Empty
rotateL tree@(Node _ _ Empty) = tree
rotateL (Node val left (Node child rl rr)) = Node child (Node val left rl) rr

insertRaw :: Ord a => Tree a -> a -> Tree a
insertRaw Empty a = Node a Empty Empty
insertRaw tree@(Node val left right) a
    | a == val = tree
    | a > val = Node val left $ insertRaw right a
    | otherwise = Node val (insertRaw left a) right

fromList :: Ord a => [a] -> Tree a
fromList lst = foldl (\memo a -> splayTo (insertRaw memo a) a) Empty lst

splayTo :: Ord a => Tree a -> a -> Tree a
splayTo Empty _ = Empty
splayTo tree@(Node val left right) a
    | a == val = tree
    | a > val = rotateL $ Node val left $ splayTo right a
    | otherwise = rotateR $ Node val (splayTo left a) right

elem :: Ord a => Tree a -> a -> Maybe (a, Tree a)
elem tree a = case a == value splayed of
                True -> Just (a, splayed)
                _ -> Nothing
    where splayed = splayTo tree a

insert :: Ord a => Tree a -> a -> Tree a
insert tree elem = splayTo (insertRaw tree elem) elem

remove :: Ord a => Tree a -> a -> Tree a
remove tree elem = if root == elem
                   then deleteRoot splayed
                   else tree
    where splayed@(Node root _ _) = splayTo tree elem

deleteRoot :: Tree a -> Tree a
deleteRoot (Node val Empty Empty) = Empty
deleteRoot tree@(Node val left Empty) = Node (rightLeaf left) (trim left) Empty
    where rightLeaf (Node val _ Empty) = val
          rightLeaf (Node _ _ right) = rightLeaf right
          trim (Node _ Empty Empty) = Empty
          trim (Node _ left Empty) = left
          trim (Node val left right) = Node val left $ trim right
deleteRoot tree@(Node val left right) = Node (leftLeaf right) left (trim right)
    where leftLeaf (Node val Empty _) = val
          leftLeaf (Node _ left _) = leftLeaf left
          trim (Node _ Empty Empty) = Empty
          trim (Node _ Empty right) = right
          trim (Node val left right) = Node val (trim left) right

Notice in particular that the Tree type has an unrestricted a in it.

--        v
data Tree a = Nod...
--        ^

but most of the manipulation functions that act on a Tree a do have a restriction

insertRaw :: Ord a => Tree a -> a -> Tree a
fromList :: Ord a => [a] -> Tree a
splayTo :: Ord a => Tree a -> a -> Tree a
elem :: Ord a => Tree a -> a -> Maybe (a, Tree a)
insert :: Ord a => Tree a -> a -> Tree a
remove :: Ord a => Tree a -> a -> Tree a

They all need an a that's a member of the Ord typeclass. That is, you need to be able to run compare on the elements of a Splay Tree in order for it to do you any good.

So from my perspective, the unrestricted a in my data declaration feels like lying. I'm giving the impression that you can make a tree out of anything, when in fact you can only make a tree out of anything you can sort. My question was, "Is there a way to restrict this structure at the type level?". What I wanted to write was something like

data Ord a => Tree a = Nod...

To paraphrase Ben Darwin and Albert Lai,

You Don't Want That

As it happens, they managed to convince me of this. Here's a paraphrase of the argument they used; maybe it'll work on you too.

There are only two ways you'll ever want to export your type. You either want to export it as a black box, or you want to export it in a non-opaque way so that your users can see some or all of its implementation details. If you're going the black box route

```haskell module Foo (YourTypeConstructor, apiFn, apiFn', apiFn''...) where ... ```

then your users are only going to be interacting with your type through your API functions, which are already suitably annotated, and restricted to valid inputs, so there's no need to redundantly restrict the type. If you're going the non-opaque route|1|,

```haskell module Foo (YourTypeConstructor(..), apiFn, apiFn', apiFn''...) where ... ```

then your users might ignore some or all of your API, and just use your structure. In doing so, they may find a purpose for it that doesn't require its elements to be members of Ord. In this case, you would be doing them a disservice by restricting your type declaration.

And I'll buy that.

So I don't really want a restricted Tree, I just want restricted manipulation functions, which I have. Case closed.

Langton's Haskelly Ant Redux

I also ended up showing off my ants implementation. I'm near the bottom in terms of Haskell skill-level at the group, so the explanation went quite quickly. The only part there was any confusion about was the admittedly over-complicated animate function. I mentioned that ideally, I'd want to write something like

iterateM_ (\w -> do { printWorld w ; wait 10 ; step w }) test

instead of the code I had actually written, but didn't have iterateM_. The group considered that no excuse, and promptly found a definition from Control.Monad.Loop|2| via Hayoo. It turned out to be a bit more baroque than necessary.

iterateM_ :: Monad m => (a -> m a) -> a -> m b
iterateM_ f = g
    where g x = f x >>= g

I was feeling a bit intimidated by that, but the group pushed me to write my own, starting with just the type signature. "It's extremely simple", they said, "and should really use do-notation", unless I shared the Control.Monad.Loop writers' apparently bizarre taste for point-free style. They were right:

loop :: Monad m => (a -> m a) -> a -> m b
loop f a = do res <- f a
              loop f res


With that, we can basically write animate out of existence entirely by re-defining main as

main :: IO ()
main = concurrent $ loop nextFrame (test, 0)
    where nextFrame (w, ct) = do liftIO $ setContent "world" $ showWorld w
                                 liftIO $ setContent "generations" $ show ct
                                 wait 50
                                 return $ (step w, succ ct)

which is at once easier for me to read, and apparently more idiomatic than the setTimeout recursion I had set up last time. The concurrent call at the top, and the liftIOs in nextFrame are there because wait|3| is a monadic function introduced in Haste, but it's in a monad called CIO, whose goal is to be used for client-side concurrency, while setContent, and more importantly main, are both plain IO () functions.

The complete revised code, and recompiled .js are both up at the same CodeRetreat github.


1 - |back| - Incidentally, the example below exports all internals of YourTypeConstructor, but the argument works whenever you expose any internal implementation detail to your users, not just at the transparent extreme.

2 - |back| - This would be a hackage link, but it's down as I write this. It probably won't be by the time I get around to the next draft, but it happens with irritating regularity when I'm collecting links for these pieces.

3 - |back| - Scroll down to the bottom for the actual documentation of wait; it doesn't have a direct link target.

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