Dominic Steinitz wrote:
I've been trying to re-label nodes in a rose tree without re-inventing wheels (although I'm beginning to wish I had). I've got as far as this but haven't yet cracked the general case for Traversable.
Solution 1) Data.Tree is already an instance of Traversable. :) 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)
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
*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)) Regards, apfelmus