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Questions About Rose Trees Analysis
by Lorenzo Isella 20 Dec '15

20 Dec '15

Dear All, I am trying to learn Haskell (for the second time), but I have to admit I am still struggling with relative simple stuff, so I ask if anybody can give me a hand. I am dealing with the issues of extracting info out of some rose trees and probably I am also having some troubles with the notation. The rose tree is defined with the following notation data Rose a = a :> [Rose a] deriving (Eq, Show) and the root can be detected simply as root (a :> rs) = a I would like to have the expression (with the ":>" notation for the Node) of the function to find the children of the root. Example of expected behavior of this function children children (1 :> [2 :> [], 3 :> []]) = [2 :> [], 3 :> []] On top of that, I am trying to get the functions to have the functions size :: Rose a -> Int leaves :: Rose a -> Int that count the number of nodes in a rose tree, respectively the number of leaves (nodes without any children). For the children function, I have tried stuff like children (a :> rs) = rs quite unsuccessfully. Any suggestion is appreciated. Cheers Lorenzo

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Learning Haskell: Algebraic Data Types
by Manuel M T Chakravarty 18 Dec '15

18 Dec '15

The latest chapter of our ”Learning Haskell” tutorial takes a first stab at algebraic data types — there will be a second chapter continuing the topic sometime soon: http://blog.haskellformac.com/blog/algebraic-data-types-demystified Happy Coding! Manuel

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Extend instance for List
by Graham Gill 16 Dec '15

16 Dec '15

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

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Monad and bind operator interpretation
by Raja 15 Dec '15

15 Dec '15

The type signature of bind (>>=) is as follows: (>>=) :: m a -> (a -> m b) -> m b One interpretation of this could be as follows: bind takes two parameters (m a & f) and returns m b (the same type returned by f) So extending this interpretation - can I swap the two parameters (?) Now my new hypothetical interpretation becomes: (>>=) :: (a -> m b) -> m a -> m b If i further add parens: (>>=) (a -> m b) -> (m a -> m b) This allows me to slightly tweak my interpretation: bind takes one param f (of type a -> m b) and returns another param f (of type m a -> m b) This feels like a more intuitive way to think about Monads - am I on the right track? (not that I want to switch the params permanently - just trying to get a feel for monads)

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IO a, IO b
by Imants Cekusins 15 Dec '15

15 Dec '15

Hello all, This looks like another potential 1+ day problem for me. Could you please help? Problem: a few (... -> IO a) functions are called as part of the same "do" block of container function these functions may return various IO a: "a" can be of different types I need the return values to proceed i.e. would call val <- (... -> IO a) container function returns IO b specifically, let's consider 2 functions called inside do block: ... -> IO Bool ... -> IO (Either String a) container function returns IO (Either String a) Questions: 1) what is a beginner-friendly way of doing this? 2) how would you do this? links, ideas or snippets will all be appreciated.

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basic State Monad
by Imants Cekusins 15 Dec '15

15 Dec '15

This snippet increments integer n times using state monad. How to call: main 10 5 module BasicState where import Control.Monad.State.Strict import Debug.Trace type St a = State a a -- caller is not aware that main uses state -- mai is a pure function main :: Int -> Int -> Int main start0 repeat0 = evalState (repeatN repeat0) start0 -- state-passing computation repeatN :: Int -> St Int repeatN n0 -- repeat n times | n0 < 1 = get -- current state | otherwise = do withState pureStateModifier get -- update state repeatN $ n0 - 1 -- recurse -- state unaware modifier function pureStateModifier :: Int -> Int pureStateModifier = (+ 1)

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Understanding Haskell Map.lookup Example in LYH
by Olumide 14 Dec '15

14 Dec '15

Hello Haskellers! I'd appreciate help understanding the origin of the extra 'map' in 'Map.lookup lockerNumber map', from the following example taken from chapter 8 of "Learn You a Haskell for Great Good" http://learnyouahaskell.com/making-our-own-types-and-typeclasses#type-synon… lockerLookup :: Int -> LockerMap -> Either String Code lockerLookup lockerNumber map = case Map.lookup lockerNumber map of Nothing -> Left $ "Locker number " ++ show lockerNumber ++ " doesn't exist!" Just (state, code) -> if state /= Taken then Right code else Left $ "Locker " ++ show lockerNumber ++ " is already taken!" Regards, - Olumide

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Haskell code optimisation
by Abhishek Kumar 13 Dec '15

13 Dec '15

I was trying to write below program for ackerman function but it fails (waits too long) for ack(4,1) whereas a recursive C program gives result in 37secs.Can someone pls explain this behaviour and recomend some optimisation. ------haskell code f m n | m==0 =n+1 | n==0 = f (m-1) 1 | otherwise = f (m-1) (f m (n-1)) Thanks Abhishek Kumar

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Trouble understanding the type of sequence [Just, Just]
by Lim H. 12 Dec '15

12 Dec '15

Hi everyone, Sorry if this email disturbs you. I haven't used a developer's mailing list before so I'm not sure if I'm violating any etiquette. If I do, please excuse me. I'm trying to understand the type of sequence [Just, Just]. I can understand sequence [Just 1, Just 2] :: Num a => Maybe [a] because when looking at the type of sequence sequence :: (Monad m, Traversable t) => t (m a) -> m (t a) it is clear that this function takes a collection of monadic values and return a single monadic value of the collection. Thus, when we call sequence [Just 1, Just 2] we should get back a Just of [1,2]. Following that train of thoughts, shouldn't sequence [Just, Just] return a single Just? Here is the corresponding SO question http://stackoverflow.com/questions/34244574/trouble-understanding-the-type-… Lim

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foldr, Foldable and right-side folding
by Raja 11 Dec '15

11 Dec '15

foldr is supposed to start folding from the right side (as the name suggests). and this is why it is synonymous to "list construction" as I'm told for e.g: > foldr (:) [ ] [1,2,3,4,5] [1,2,3,4,5] In the same spirit I'm trying to construct a Foldable instance of my own type: data Li a = Nil | Cons a (Li a) deriving (Show) instance Foldable Li where foldr f b Nil = b foldr f b (Cons a y) = foldr f (f a b) y So I'm trying out foldr for my type: > foldr Cons Nil (Cons 1 (Cons 2 Nil)) Cons 2 (Cons 1 Nil) This shows my foldr implementation i'm not folding from right side, but how can I possibly do that - the data could have been an infinite stream. It feels like I will never be able to truly write a foldr implementation with "right" folding mechanism. Any thoughts?

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