Can someone check my answer (no I'm not doing an assessment...I'm actually learning stuff out of interest!) working through https://www.fpcomplete.com/user/konn/prove-your-haskell-for-great-safety/dep... still there is a section about singleton types and the exercise is "Exercise: Define the binary tree type and implement its singleton type." Ok, I think I'm probably wrong....a binary tree is something like...
data BTree a = Leaf | Branch a (BTree a) (BTree a)
With DataKind My logic goes... Leaf is an uninhabited type, so I need a value isomorphic to it.... Easy?
data SBTree a where SLeaf :: SBTree Leaf
Things like Branch Integer Leaf (Branch String Leaf Leaf) Are uninhabited...so I need to add
SBranch :: (a :: *) -> (SBTree (b :: BTree *)) -> (SBTree (c :: BTree *)) -> SBTree (Branch a b c)
? It compiles...but....is it actually correct? Things like
y = SBranch (SS (SS SZ)) SLeaf SLeaf z = SBranch (SS (SS SZ)) (SBranch SZ SLeaf SLeaf) SLeaf
Seem to make sense ish. From: Nicholls, Mark Sent: 28 April 2015 9:33 AM To: Nicholls, Mark Subject: sds Hello, working through https://www.fpcomplete.com/user/konn/prove-your-haskell-for-great-safety/dep... but a bit stuck...with an error...
{-# LANGUAGE DataKinds, TypeFamilies, TypeOperators, UndecidableInstances, GADTs, StandaloneDeriving #-}
data Nat = Z | S Nat
data Vector a n where Nil :: Vector a Z (:-) :: a -> Vector a n -> Vector a (S n) infixr 5 :-
I assume init...is a bit like tail but take n - 1 elements from the front....but...
init' :: Vector a ('S n) -> Vector a n init' (x :- Nil) = Nil init' (x :- xs@(_ :- _)) = x :- (init' xs)
zipWithSame :: (a -> b -> c) -> Vector a n -> Vector b n -> Vector c n zipWithSame f Nil Nil = Nil zipWithSame f (x :- xs) (y :- xs@(_ :- _)) = Nil
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