Hello Conor,

The following seems to work:

On 31 Jan 2007, at 11:46, Conor McBride wrote:


Hi folks

A puzzle for you (and for me too). I just tried to translate an old Epigram favourite (which predates Epigram by some time) into GADT- speak. It went wrong. I had a feeling it might. I wonder how to fix it: I imagine it's possible to fix it, but that some cunning may be necessary. Here it is:

Type-level Nat

> data Z = Z
> data S x = S x


Lets also have naturals reflected at the value level:

> data Nat :: * -> * where
>    Zero :: Nat Z
>    Succ :: Nat n -> Nat (S n)



Finite sets: Fin Z is empty, Fin (S n) has one more element than Fin n

> data Fin :: * -> * where
>   Fz :: Fin (S n)
>   Fs :: Fin n -> Fin (S n)

The "thinning" function: thin i is the order-preserving embedding from Fin n to Fin (S n) which never returns i.


Now we modify thin to take an extra Nat n (not needed, but just to show the duality with thickening):

> thin :: Nat n -> Fin (S n) -> Fin n -> Fin (S n)
> thin n         Fz      i       = Fs i
> thin (Succ n)  (Fs i)  (Fz)    = Fz
> thin (Succ n)  (Fs i)  (Fs j)  = Fs (thin n i j)

The thing to note here is that only:

> thin (Succ n)  (Fs i)  (Fz)    = Fz

is well typed; if you had the case:

> thin Zero  (Fs i)  (Fz)    = Fz

you get:

Conor.lhs:28:21:
    Inaccessible case alternative: Can't match types `S n' and `Z'
    In the pattern: Fz
    In the definition of `thin': thin Zero (Fs i) (Fz) = Fz
Failed, modules loaded: none.




Its partial inverse, "thickening" has the spec

thicken i i = Nothing
thicken i (thin i j) = Just j



Now lets go for the thickening function:

> thicken :: Nat n -> Fin (S n) -> Fin (S n) -> Maybe (Fin n)
> thicken n                Fz      Fz      = Nothing
> thicken n                Fz      (Fs j)  = Just j
> thicken Zero             (Fs i)  Fz      = Nothing
> thicken (Succ _)         (Fs i)  Fz      = Just Fz
> thicken (Succ n)         (Fs i)  (Fs j)  = fmap Fs (thicken n i j)

Note that this version of thicken is more precise than the version you had -- your third case splits into two
different cases now. The problem is that when you had:

> thicken (Fs i) Fz = ...

you need to account for two possibilities: the first possibility is (Fz :: Fin (S Zero)), to which you want to return Nothing
and (Fz :: Fin (S (S n))) for which you can safely return Just Fz.

Let me know if that helps.

Cheers,

Bruno

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