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haskelltreemonadsapplicativefree-monad

Are free monads also zippily applicative?

I think I've come up with an interesting "zippy" Applicative instance for Free.

data FreeMonad f a = Free (f (FreeMonad f a))
                   | Return a

instance Functor f => Functor (FreeMonad f) where
    fmap f (Return x) = Return (f x)
    fmap f (Free xs) = Free (fmap (fmap f) xs)

instance Applicative f => Applicative (FreeMonad f) where
    pure = Return

    Return f <*> xs = fmap f xs
    fs <*> Return x = fmap ($x) fs
    Free fs <*> Free xs = Free $ liftA2 (<*>) fs xs

It's sort of a zip-longest strategy. For example, using data Pair r = Pair r r as the functor (so FreeMonad Pair is an externally labelled binary tree):

    +---+---+    +---+---+               +-----+-----+
    |       |    |       |      <*>      |           |
 +--+--+    h    x    +--+--+   -->   +--+--+     +--+--+
 |     |              |     |         |     |     |     |
 f     g              y     z        f x   g x   h y   h z

I haven't seen anyone mention this instance before. Does it break any Applicative laws? (It doesn't agree with the usual Monad instance of course, which is "substitutey" rather than "zippy".)

Solution

  • Yes, it looks like this is a lawful Applicative. Weird!

    As @JosephSible points out, you can read off the identity, homomorphism and interchange laws immediately from the definitions. The only tricky one is the composition law.

    pure (.) <*> u <*> v <*> w = u <*> (v <*> w)

    There are eight cases to check, so strap in.

    • One case with three Returns: pure (.) <*> Return f <*> Return g <*> Return z
      • Follows trivially from associativity of (.).
    • Three cases with one Free:
      • pure (.) <*> Free u <*> Return g <*> Return z
        • Working backwards from Free u <*> (Return g <*> Return z) you get fmap (\f -> f (g z)) (Free u), so this follows from the functor law.
      • pure (.) <*> Return f <*> Free v <*> Return z
fmap ($z) $ fmap f (Free v)
fmap (\g -> f (g z)) (Free v)                  -- functor law
fmap (f . ($z)) (Free v)
fmap f (fmap ($z) (Free v))                    -- functor law
Return f <$> (Free v <*> Return z)             -- RHS of `<*>` (first and second cases)
QED
      • pure (.) <*> Return f <*> Return g <*> Free w
        • Reduces immediately to fmap (f . g) (Free w), so follows from the functor law.
    • Three cases with one Return: 
      • pure (.) <*> Return f <*> Free v <*> Free w
Free $ fmap (<*>) (fmap (fmap (f.)) v) <*> w
Free $ fmap (\y z -> fmap (f.) y <*> z) v <*> w                  -- functor law
Free $ fmap (\y z -> fmap (.) <*> Return f <*> y <*> z) v <*> w  -- definition of fmap, twice
Free $ fmap (\y z -> Return f <*> (y <*> z)) v <*> w             -- composition
Free $ fmap (\y z -> fmap f (y <*> z)) v <*> w                   -- RHS of fmap, definition of liftA2
Free $ fmap (fmap f) $ fmap (<*>) v <*> w                        -- functor law, eta reduce
fmap f $ Free $ liftA2 (<*>) v w                                 -- RHS of fmap
Return f <*> Free v <*> Free w                                   -- RHS of <*>
QED.

      • pure (.) <*> Free u <*> Return g <*> Free w
Free ((fmap (fmap ($g))) (fmap (fmap (.)) u)) <*> Free w
Free (fmap (fmap (\f -> f . g) u)) <*> Free w                    -- functor law, twice
Free $ fmap (<*>) (fmap (fmap (\f -> f . g)) u) <*> w
Free $ fmap (\x z -> fmap (\f -> f . g) x <*> z) u <*> w         -- functor law
Free $ fmap (\x z -> pure (.) <*> x <*> Return g <*> z) u <*> w
Free $ fmap (\x z -> x <*> (Return g <*> z)) u <*> w             -- composition
Free $ fmap (<*>) u <*> fmap (Return g <*>) w                    -- https://gist.github.com/benjamin-hodgson/5b36259986055d32adea56d0a7fa688f
Free u <*> fmap g w                                              -- RHS of <*> and fmap
Free u <*> (Return g <*> w)
QED.

      • pure (.) <*> Free u <*> Free v <*> Return z
Free (fmap (<*>) (fmap (fmap (.)) u) <*> v) <*> Return z
Free (fmap (\x y -> fmap (.) x <*> y) u <*> v) <*> Return z        -- functor law
Free $ fmap (fmap ($z)) (fmap (\x y -> fmap (.) x <*> y) u <*> v)
Free $ liftA2 (\x y -> (fmap ($z)) (fmap (.) x <*> y)) u v         -- see Lemma, with f = fmap ($z) and g x y = fmap (.) x <*> y
Free $ liftA2 (\x y -> fmap (.) x <*> y <*> Return z) u v          -- interchange
Free $ liftA2 (\x y -> x <*> (y <*> Return z)) u v                 -- composition
Free $ liftA2 (\f g -> f <*> fmap ($z) g) u v                      -- interchange
Free $ fmap (<*>) u <*> (fmap (fmap ($z)) v)                       -- https://gist.github.com/benjamin-hodgson/5b36259986055d32adea56d0a7fa688f
Free u <*> Free (fmap (fmap ($z)) v)
Free u <*> (Free v <*> Return z)
QED.
    • Three Frees: pure (.) <*> Free u <*> Free v <*> Free w
      • This case only exercises the Free/Free case of <*>, whose right-hand side is identical to that of Compose's <*>. So this one follows from the correctness of Compose's instance.

    For the pure (.) <*> Free u <*> Free v <*> Return z case I used a lemma:

    Lemma: fmap f (fmap g u <*> v) = liftA2 (\x y -> f (g x y)) u v.

    fmap f (fmap g u <*> v)
pure (.) <*> pure f <*> fmap g u <*> v  -- composition
fmap (f .) (fmap g u) <*> v             -- homomorphism
fmap ((f .) . g) u <*> v                -- functor law
liftA2 (\x y -> f (g x y)) u v          -- eta expand
QED.

    Variously I'm using functor and applicative laws under the induction hypothesis.

    This was pretty fun to prove! I'd love to see a formal proof in Coq or Agda (though I suspect the termination/positivity checker might mess it up).