Functor instance for FunPB
Hey Everyone, ich have a algebraic data type: data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) } I got a Monoid Instance instance Monoid (FunPB a b) where mempty = FunPB $ \k -> (k,mempty) mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2] Now I need to write a functor instance instance Functor (FunPB a) where fmap f b = ? How can I map over a->(a,[b])? Lg Alex
mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2]
This cracks me up. Well played. Well played indeed. Well all the things you need are right here in this line, right? I mean you sent it because you want to show us what you know, right? There's higher order functions, there's mapping and folding and currying of different kinds and everything. So what's missing?
Hey, No, I do not want to show anything. The monoid instance is from my teacher. It is an exercise. Sry for the misunderstanding. LG Alex Am 06.05.2017 um 20:25 schrieb MarLinn:
mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2]
This cracks me up. Well played. Well played indeed.
Well all the things you need are right here in this line, right? I mean you sent it because you want to show us what you know, right? There's higher order functions, there's mapping and folding and currying of different kinds and everything. So what's missing?
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Hi, Am Samstag, den 06.05.2017, 20:32 +0200 schrieb Alex Wede:
The monoid instance is from my teacher.
it is hard to tell with confidence, given that we do not know what FunFB is supposed to “mean”, but data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) } instance Monoid (FunPB a b) where mempty = FunPB $ \k -> (k,mempty) mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2] looks very fishy, as the return value of of type a from pb1 and pb2 are ignored in mappend The latter means that this is not a lawful monoid, because for pb = (\n -> (n+1, []) we have mempty `mappend` pb == mempty /= pb Did you maybe mean mappend pb1 pb2 = FunPB $ \n0 -> let (n1,xs1) = runFunFB n1 (n2,xs2) = runFunFB n2 in (n2, xs1 ++ xs2) Greetings, Joachim -- Joachim Breitner mail@joachim-breitner.de http://www.joachim-breitner.de/
On 06/05/17 14:50, Alex Wede wrote:
Now I need to write a functor instance
instance Functor (FunPB a) where
fmap f b = ?
It will help to see the specific types of the instance instance Functor (FunPB a) where fmap :: (c -> d) -> FunPB a c -> FunPB a d ie we need to change only the second argument of FunPB a c. By you definition, that would be the [b] in data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) } so you can remove the constructor to get the inner function, create a new function that uses the inner one and fmap the second part of the tuple instance Functor (FunPB a) where fmap f (FunPB lambda) = FunPB $ \a1 -> let (a2, bs) = lambda a1 in (a2 , fmap f bs) see that we have a `fmap f bs` in the second part, we use that we already have a Functor instance for [b], making our job easier. -- -- Ruben
The only lawful instance I can think of is: instance Functor (FunPB a) where fmap f (FunPB g) = FunPB $ \k -> let (a,b) = g k in (a, fmap f b) which is fairly straightforward IMO. On 2017-05-06 19:50, Alex Wede wrote:
Hey Everyone,
ich have a algebraic data type:
data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) }
I got a Monoid Instance
instance Monoid (FunPB a b) where mempty = FunPB $ \k -> (k,mempty) mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2]
Now I need to write a functor instance
instance Functor (FunPB a) where
fmap f b = ?
How can I map over a->(a,[b])?
Lg
Alex
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On 2017-05-06 20:38, David Kraeutmann wrote:
The only lawful instance I can think of is: instance Functor (FunPB a) where fmap f (FunPB g) = FunPB $ \k -> let (a,b) = g k in (a, fmap f b)
which is fairly straightforward IMO. And because it's closer to the original style:
fmap f = FunPB .((.)((<$>).(<$>)$f)). runFunPB Now it's STARRING INTO YOU SOUL. And it's also so much easier to read.
Hallo. You can try reusing fmap.
data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) fmap f (FunPB g) = FunPB $ fmap (fmap (fmap f)) g
This works because (a->), (a,), and [] are all functors.
Freundliche Grüße,
Daniel.
On Sat, May 6, 2017 at 9:04 PM, MarLinn
On 2017-05-06 20:38, David Kraeutmann wrote:
The only lawful instance I can think of is: instance Functor (FunPB a) where fmap f (FunPB g) = FunPB $ \k -> let (a,b) = g k in (a, fmap f b)
which is fairly straightforward IMO.
And because it's closer to the original style:
fmap f = FunPB .((.)((<$>).(<$>)$f)). runFunPB
Now it's STARRING INTO YOU SOUL. And it's also so much easier to read.
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participants (6)
-
Alex Wede -
Daniel Díaz Casanueva -
David Kraeutmann -
Joachim Breitner -
MarLinn -
Ruben Astudillo