[Haskell-cafe] Re: Foldable Rose Trees
Dominic Steinitz
dominic.steinitz at blueyonder.co.uk
Tue Dec 18 13:09:41 EST 2007
> Solution 1) Data.Tree is already an instance of Traversable. :)
>
Yes it's all there but I would have missed the fun of trying to do it
myself ;-) Plus the data structure I actually want to re-label isn't
quite a rose tree.
> Solution 2) The key observation is that you the instances for rose trees
> can/should be bootstrapped from corresponding instances for lists [].
> With this, we have
>
>>> instance Functor Rose' where
>>> fmap f (Rose' x rs) = Rose' (f x) (map (fmap f) rs)
>
> fmap f (Rose' x rs) = Rose' (f x) (fmap (fmap f) rs)
>
> (fmap instead of map to highlight the general structure)
>
>>> instance Foldable Rose' where
>>> foldMap f (Rose' x rs) = f x `mappend` (mconcat (map (foldMap f) rs))
>
> foldMap f (Rose' x rs) = f x `mappend` (foldMap (foldMap f) rs)
Interesting - I hadn't twigged that they were the same modulo
instantiation for [].
> ((.).(.)) mconcat map :: forall a b. (Monoid b) => (a -> b) -> [a] -> b
> *Main> :t foldMap
> foldMap :: forall a m (t :: * -> *).
> (Monoid m, Foldable t) =>
> (a -> m) -> t a -> m
>
>>> instance Traversable Rose' where
>>> traverse f (Rose' x []) = Rose' <$> f x <*> pure []
>>> traverse f (Rose' x [x0]) = Rose' <$> f x <*> (pure (\x -> [x]) <*> traverse f x0)
>>> traverse f (Rose' x [x0,x1]) = Rose' <$> f x <*> (pure (\x y -> x:y:[]) <*> traverse f x0 <*> traverse f x1)
>>> traverse f (Rose' x [x0,x1,x2]) = Rose' <$> f x <*> (pure (\x y z -> x:y:z:[]) <*> traverse f x0 <*> traverse f x1 <*> traverse f x2)
>
> traverse f (Rose' x xs) = Rose' <$> f x <*> traverse (traverse f) xs
>
And then this becomes something you might guess.
>
>
>>> *Main> let (p,_) = runState (unwrapMonad (traverse (\x -> WrapMonad update) (Rose' 3 [Rose' 5 [Rose' 11 [Rose' 19 []], Rose' 13 [], Rose' 17[]], Rose' 7 []]))) 0 in p
>>> Rose' 0 [Rose' 1 [Rose' 2 [Rose' 3 []],Rose' 4 [],Rose' 5 []],Rose' 6 []]
>
> This can be made shorter:
>
> Data.Traversable.mapM m = unwrapMonad . traverse . (\x -> WrapMonad (m x))
>
>
Your help as ever is excellent.
Many thanks, Dominic.
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