On Tue, Jan 31, 2017 at 11:55 PM, Joachim Breitner <mail@joachim-breitner.de> wrote:
Hi,

Am Dienstag, den 31.01.2017, 15:22 -0500 schrieb Joachim Breitner:
> I recently wrote this applicative functor:
>
>     data OneStep a = OneStep a [a]
>
>     instance Functor OneStep where
>         fmap f (OneStep o s) = OneStep (f o) (map f s)
>
>     instance Applicative OneStep where
>         pure x = OneStep x []
>         OneStep f fs <*> OneStep x xs = OneStep (f x) (map ($x) fs ++
> map f xs)
>
>     takeOneStep :: OneStep t -> [t]
>     takeOneStep (OneStep _ xs) = xs
>
> This was useful in the context of writing a shrink for QuickCheck, as
> discussed at http://stackoverflow.com/a/41944525/946226.
>
> Now I wonder: Does this functor have a proper name? Does it already
> exist in the libraries somewhere? Should it?

I guess it does not exist, so I am preparing a package for it here:
https://github.com/nomeata/haskell-successors

The source code contains (in comments) a proof of the Applicative laws.

My gut feeling says that this does not have a Monad instance that is
compatible with the given Applicative instance, but it is too late
today to substantiate this feeling. If anyone feels like puzzling: Can
you come up with a Monad instance, or (more likely) give good reasons
why there cannot be one?

How about this?

hd (OneStep x xs) = x

instance Monad OneStep where
    OneStep x xs >>= f = OneStep y (map (hd . f) xs ++ ys)
        where
        OneStep y ys = f x

Not sure if it’s good for anything, but it seems valid and consistent based on a preliminary investigation.

--