[Haskell-cafe] What is this applicative functor?

[David Menendez]

On Tue, Jan 31, 2017 at 11:55 PM, Joachim Breitner <mail at 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.

-- 
Dave Menendez <dave at zednenem.com>
<http://www.eyrie.org/~zednenem/>
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