On December 14, 2015 6:57:27 AM GMT+02:00, Graham Gill
The NICTA course https://github.com/NICTA/course includes exercises on the type class Extend, in Course/Extend.hs. Extend is a superclass of Comonad. Here's the class definition:
-- | All instances of the `Extend` type-class must satisfy one law. This law -- is not checked by the compiler. This law is given as: -- -- * The law of associativity -- `∀f g. (f <<=) . (g <<=) ≅ (<<=) (f . (g <<=))` class Functor f => Extend f where -- Pronounced, extend. (<<=) :: (f a -> b) -> f a -> f b
infixr 1 <<=
Could someone please motivate the Extend instance for List? (Its implementation is left as an exercise. In the course, type List a is isomorphic to [a].) Some of the tests (<<=) is expected to pass are shown, and make clear what ought to happen.
-- | Implement the @Extend@ instance for @List@. -- -- >>> length <<= ('a' :. 'b' :. 'c' :. Nil) -- [3,2,1] -- -- >>> id <<= (1 :. 2 :. 3 :. 4 :. Nil) -- [[1,2,3,4],[2,3,4],[3,4],[4]] -- -- >>> reverse <<= ((1 :. 2 :. 3 :. Nil) :. (4 :. 5 :. 6 :. Nil) :. Nil) -- [[[4,5,6],[1,2,3]],[[4,5,6]]]
The following (wrong, according to the tests) Extend instance for List nevertheless obeys the types and obeys the Extend law of associativity.
instance Extend List where (<<=) :: (List a -> b) -> List a -> List b (<<=) f = (:. Nil) . f (:. Nil) is analogous to (: []), create a singleton list.
I can't find a good reference on the Extend type class to convince me why the correct Extend instance for List in the course is the desirable
one. (I'm not saying my version is desirable, in fact it seems fairly useless, but it works.)
Graham
------------------------------------------------------------------------
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Indeed, your implementation is a valid Extend instance. However, it cannot be extended into a valid Comonad, whereas the supplied instance can. Can you see why? Hint: In order to be a Comonad, an Extend instance must also supply a function extract:: f a -> a which must be an identity for extend. Is this possible for your instance? The intuition behind Extend is that given a computation that takes into account the context of the values in your container, it applies that computation everywhere, passing it the appropriate context. Thus, elements of List may be considered as having the remainder of the list as their context, and that is indeed what is passed to the computation, as evidence by extend id. Indeed, this function is so essential to the essence of a Comonad that it is given its own name - duplicate - and forms the building block for an equivalent set of laws for Comonad, namely: - duplicate . duplicate = fmap duplicate . duplicate - extract . duplicate = id = fmap extract . duplicate (If you've heard of join in the context of Monad, this is precisely the dual set of laws it satisfies) Indeed, it may be easier to prove your Extend instance doesn't extend to a Comonad instance by using this formulation of the laws. HTH, Gesh