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
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