Hmm ...
How about:
Perfect :: * -> * = Fix (L :: * -> *) . /\ A . (A + L (A,A))
unfold Perfect = [L := Fix L . t] t where t = /\ A . (A + L (A,A))
= /\ A . (A + (Fix L . /\ B . (B + L (B,B))) (A,A))
assuming alpha-equality. Because L and (Fix L . t) are of kind (* ->
*), the substitution should be okay. Am I missing something, again?
-Antoine
On Jan 24, 2008 10:31 AM, Edsko de Vries
On Thu, Jan 24, 2008 at 10:06:04AM -0600, Antoine Latter wrote:
Can "Fix" be made to work with higher-kinded types? If so, would the following work:
Perfect = /\ A . Fix (L :: * -> *) . (A + L (A,A))
Hi,
Thanks for your quick reply. Unfortunately, your solution does not work. For
Fix X. t
to be well-defined, we must have that if 'X' has kind 'k', then 't' must also have kind 'k' (compare the type of 'fix' in Haskell: (a -> a) -> a).
Keep in mind I have no idea what the "Perfect" data structure is supposed to look like.
The Haskell equivalent would be
data Perfect a = Leaf a | Branch (Perfect (a, a))
and models perfect binary trees (I admit, slightly headwrecking datatype! :)
Edsko