Re: [Haskell-cafe] What is this applicative functor?

These instances are quite similar to https://github.com/ambiata/disorder.hs/blob/ce9ffc1139e32eaa9b82a1a6e2cfeb91... (I wish `disorder-jack` were already on Hackage!) - Oleg On 01.02.2017 07:28, David Menendez wrote:

On Tue, Jan 31, 2017 at 11:55 PM, Joachim Breitner mailto: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 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 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 mailto:dave@zednenem.com> <http://www.eyrie.org/~zednenem/ http://www.eyrie.org/%7Ezednenem/>

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