23 Jun
2010
23 Jun
'10
11:35 a.m.
On Wed, Jun 23, 2010 at 03:41:29PM +0100, John Lato wrote:
How would you implement bfnum? (If you've already read the paper, what was your first answer?)
This was my first answer, and it is wrong, but I thought it was slightly clever, so here it is: bfnum :: Tree a -> Tree Int bfnum tree = bfnum' tree 1 where bfnum' E _ = E bfnum' (T _ l r) i = T i (bfnum' l (i*2)) (bfnum' r ((i*2)+1)) If you have an incomplete tree, it will skip though. I didn't realize it was wrong until I finished reading the paper though, so I don't have a better solution that actually works. -- Daniel