John Lato wrote:
How would you implement bfnum? (If you've already read the paper, what was your first answer?)
My first idea was something similar to what is described in appendix A. However, after reading the paper, I wrote the following code: data Tree a = E | T a (Tree a) (Tree a) deriving Show bfnum :: Tree a -> Tree Int bfnum = num . bf bf :: Tree a -> [Tree a] bf root = output where children = go 1 output output = root : children go 0 _ = [] go n (E : rest) = go (pred n) rest go n (T _ a b : rest) = a : b : go (succ n) rest num :: [Tree a] -> Tree Int num input = root where root : children = go 1 input children go k (E : input) children = E : go k input children go k (T _ _ _ : input) ~(a : ~(b : children)) = T k a b : go (succ k) input children Maybe one could try to fuse bf and num. Tillmann