23 Jun
2010
23 Jun
'10
12:41 p.m.
On Wed, Jun 23, 2010 at 4:35 PM, Daniel Lyons
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.
That was my first try too. If this answer did work, I don't think the question would be interesting. John