[Haskell-cafe] dependent types, singleton types....

Wed Apr 29 12:54:38 UTC 2015

Accchhhh

Thats another level of pain, i'm not a big haskeller so i'll have to wrestle with it to get all the knobs and dials working, leave it with me for the moment and i'll start wrestling with cabal

Excuse the spelling, sent from a phone with itty bitty keys, it like trying to sow a button on a shirt with a sausage.

On 29 Apr 2015, at 12:43, Richard Eisenberg <eir at cis.upenn.edu<mailto:eir at cis.upenn.edu>> wrote:

Hello Mark,

Your suspicion that your singleton tree type is wrong is well-founded.

The problem is that, in my opinion, that exercise is mentioned too early in the tutorial. To properly implement a singleton type for a parameterized type, like a binary tree, you will need `data family Sing (a :: k)`, as explained just a little bit further down in the post. You'll need to rewrite your definition for singleton numbers and booleans to work with `Sing` as well.

Your code except the definition for SBranch is all correct. The problem with your definition is that you don't get the right information when pattern-matching. For example, say you have x with type `SBTree a`. If you successfully pattern match against `SBranch SZ SLeaf SLeaf`, you would want to learn `a ~ Branch Z Leaf Leaf`. But that's not what you'll get in your implementation: you'll get a type error saying that we don't know that `a0` is an `SNat`, where `a ~ Branch a0 Leaf Leaf`, or something like that. The type-level information is simply encoded in the wrong place for this to work out.

Write back and I'll give you the full answer if this isn't enough to get you moving in the right direction!

Richard

On Apr 28, 2015, at 10:45 AM, "Nicholls, Mark" <nicholls.mark at vimn.com<mailto:nicholls.mark at vimn.com>> wrote:

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/dependent-types-in-haskell

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/dependent-types-in-haskell

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

Mark Nicholls | Senior Technical Director, Programmes & Development - Viacom International Media Networks
A: 17-29 Hawley Crescent London NW1 8TT | e: Nicholls.Mark at vimn.com<mailto:mik at vimn.com> T: +44 (0)203 580 2223

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