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  • Is recursion preferred compare to iteration in multicore era?

    - by prM
    Or say, do multicore CPUs process recursion faster than iteration? Or it simply depends on how one language runs on the machine? like c executes function calls with large cost, comparing to doing simple iterations. I had this question because one day I told one of my friend that recursion isn't any amazing magic that can speed up programs, and he told me that with multicore CPUs recursion can be faster than iteration. EDIT: If we consider the most recursion-loved situation (data structure, function call), is it even possible for recursion to be faster?

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  • Explain Type Classes in Haskell

    - by Tsubasa Gomamoto
    Hi, I am a C++ / Java programmer and the main paradigm I happen to use in everyday programming is OOP. In some thread I read a comment that Type classes are more intuitive in nature than OOP. Can someone explain the concept of type classes in simple words so that an OOP guy like me can understand it?

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  • Haskell type classes and type families (cont'd)

    - by Giuseppe Maggiore
    I need some help in figuring a compiler error which is really driving me nuts... I have the following type class: infixl 7 --> class Selectable a s b where type Res a s b :: * (-->) :: (CNum n) => (Reference s a) -> (n,(a->b),(a->b->a)) -> Res a s b which I instance twice. First time goes like a charm: instance Selectable a s b where type Res a s b = Reference s b (-->) (Reference get set) (_,read,write) = (Reference (\s -> let (v,s') = get s in (read v,s')) (\s -> \x -> let (v,s') = get s v' = write v x (_,s'') = set s' v' in (x,s''))) since the type checker infers (-->) :: Reference s a -> (n,a->b,a->b->a) -> Reference s b and this signature matches with the class signature for (--) since Res a s b = Reference s b Now I add a second instance and everything breaks: instance (Recursive a, Rec a ~ reca) => Selectable a s (Method reca b c) where type Res a s (Method reca b c) = b -> Reference s c (-->) (Reference get set) (_,read,write) = \(x :: b) -> from_constant( Constant(\(s :: s)-> let (v,s') = get s :: (a,s) m = read v ry = m x :: Reference (reca) c (y,v') = getter ry (cons v) :: (c,reca) v'' = elim v' (_,s'') = set s' v'' in (y,s''))) :: Reference s c the compiler complains that Couldn't match expected type `Res a s (Method reca b c)' against inferred type `b -> Reference s c' The lambda expression `\ (x :: b) -> ...' has one argument, which does not match its type In the expression: \ (x :: b) -> from_constant (Constant (\ (s :: s) -> let ... in ...)) :: Reference s c In the definition of `-->': --> (Reference get set) (_, read, write) = \ (x :: b) -> from_constant (Constant (\ (s :: s) -> ...)) :: Reference s c reading carefully the compiler is telling me that it has inferred the type of (--) thusly: (-->) :: Reference s a -> (n,a->(Method reca b c),a->(Method reca b c)->a) -> (b -> Reference s c) which is correct since Res a s (Method reca b c) = b -> Reference s c but why can't it match the two definitions? Sorry for not offering a more succint and standalone example, but in this case I cannot figure how to do it...

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  • Using an element against an entire list in Haskell

    - by Snick
    I have an assignment and am currently caught in one section of what I'm trying to do. Without going in to specific detail here is the basic layout: I'm given a data element, f, that holds four different types inside (each with their own purpose): data F = F Float Int, Int a function: func :: F -> F-> Q Which takes two data elements and (by simple calculations) returns a type that is now an updated version of one of the types in the first f. I now have an entire list of these elements and need to run the given function using one data element and return the type's value (not the data element). My first analysis was to use a foldl function: myfunc :: F -> [F] -> Q myfunc y [] = func y y -- func deals with the same data element calls myfunc y (x:xs) = foldl func y (x:xs) however I keep getting the same error: "Couldn't match expected type 'F' against inferred type 'Q'. In the first argument of 'foldl', namely 'myfunc' In the expression: foldl func y (x:xs) I apologise for such an abstract analysis on my problem but could anyone give me an idea as to what I should do? Should I even use a fold function or is there recursion I'm not thinking about?

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  • Haskell Write Computation result to file

    - by peterwkc
    Hello to all, i have function which create a tuple after computation but i would like to write it to file. I know how to write file using writeFile but did not know how to combine computation and monads IO together in the type signature This is my code. invest :: ([Char]->Int->Int->([Char], Int) ) -> [Char]->Int->Int->([Char], Int) invest myinvest x y = myinvest x y myinvest :: [Char]->Int->Int->([Char], Int) myinvest w x y | y > 0 = (w, x + y) | otherwise = error "Invest amount must greater than zero" where I have a function which computes the maximum value from list but i want to these function receive input from file then perform the computation of maximum value. maximuminvest :: (Ord a) => [a] -> a maximuminvest [] = error "Empty Invest Amount List" maximuminvest [x] = x maximuminvest (x:xs) | x > maxTail = x | otherwise = maxTail where maxTail = maximuminvest xs Please help. Thanks.

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  • Haskell: writing the result of a computation to file

    - by peterwkc
    I have a function which creates a tuple after computation, but I would like to write it to file. I know how to write to a file using writeFile, but do not know how to combine the computation and monads IO together in the type signature. This is my code. invest :: ([Char]->Int->Int->([Char], Int) ) -> [Char]->Int->Int->([Char], Int) invest myinvest x y = myinvest x y myinvest :: [Char]->Int->Int->([Char], Int) myinvest w x y | y > 0 = (w, x + y) | otherwise = error "Invest amount must greater than zero" where I have a function which computes the maximum value from a list, but I want this function to receive input from a file, and then perform the computation of maximum value. maximuminvest :: (Ord a) => [a] -> a maximuminvest [] = error "Empty Invest Amount List" maximuminvest [x] = x maximuminvest (x:xs) | x > maxTail = x | otherwise = maxTail where maxTail = maximuminvest xs Please help. Thanks.

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  • List of divisors of an integer n (Haskell)

    - by Code-Guru
    I currently have the following function to get the divisors of an integer: -- All divisors of a number divisors :: Integer -> [Integer] divisors 1 = [1] divisors n = firstHalf ++ secondHalf where firstHalf = filter (divides n) (candidates n) secondHalf = filter (\d -> n `div` d /= d) (map (n `div`) (reverse firstHalf)) candidates n = takeWhile (\d -> d * d <= n) [1..n] I ended up adding the filter to secondHalf because a divisor was repeating when n is a square of a prime number. This seems like a very inefficient way to solve this problem. So I have two questions: How do I measure if this really is a bottle neck in my algorithm? And if it is, how do I go about finding a better way to avoid repetitions when n is a square of a prime?

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  • Haskell: Gluing a char and a list together?

    - by Vincent
    So I have this code here: toWords :: String - [a] toWords "" = [] toWords (nr1 : rest) | nr1 == ' ' = toWords rest | otherwise = nr1 : toWords rest The "toWords" function should simply remove all spaces and return a list with all the words. But I keep getting this error: test.hs:5:18: Couldn't match expected type a' against inferred typeChar' `a' is a rigid type variable bound by the type signature for `toWords' at test.hs:1:22 In the first argument of `(:)', namely `nr1' In the expression: nr1 : toWords rest In the definition of `toWords': toWords (nr1 : rest) | nr1 == ' ' = toWords rest | otherwise = nr1 : toWords rest Failed, modules loaded: none.

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  • How and why is ap defined as liftM2 id in Haskell

    - by luke_randall
    Whilst trying to better understand Applicative, I looked at the definition of <*, which tends to be defined as ap, which in turn is defined as: ap :: (Monad m) => m (a -> b) -> m a -> m b ap = liftM2 id Looking at the type signatures for liftM2 and id, namely: liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r id :: a -> a I fail to understand how just by passing in id, the relevant part of the type signature seems to transform from (a1 -> a2 -> r) -> m a1 to m (a -> b). What am I missing here?

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  • Where can I learn advanced Haskell?

    - by FredOverflow
    In a comment to one of my answers, SO user sdcwc essentially pointed out that the following code: comb 0 = [[]] comb n = let rest = comb (n-1) in map ('0':) rest ++ map ('1':) rest could be replaced by: comb n = replicateM n "01" which had me completely stunned. Now I am looking for a tutorial, book or PDF that teaches these advanced concepts. I am not looking for a "what's a monad" tutorial aimed at beginners or online references explaining the type of replicateM. I want to learn how to think in monads and use them effectively, monadic "patterns" if you will.

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  • Haskell: Defaulting constraints to type

    - by yairchu
    Consider this example: applyKTimes :: Integral i => i -> (a -> a) -> a -> a applyKTimes 0 _ x = x applyKTimes k f x = applyKTimes (k-1) f (f x) applyThrice :: (a -> a) -> a -> a applyThrice = applyKTimes 3 The 3 in applyThrice is defaulted by GHC to an Integer as shown when compiling with -Wall: Warning: Defaulting the following constraint(s) to type 'Integer' 'Integral t' arising from a use of 'applyKTimes' So I guess that Integer is the default Integral a => a. Is there a way to define "default types" for other constraints too? Is using default types bad practice? (it does complain when using -Wall..)

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  • Haskell map function with predicate

    - by Paul
    I feel like this should be fairly obvious, or easy, but I just can't get it. What I want to do is apply a function to a list (using map) but only if a condition is held. Imagine you only wanted to divide the numbers which were even: map (`div` 2) (even) [1,2,3,4] And that would give out [1,1,3,2] since only the even numbers would have the function applied to them. Obviously this doesn't work, but is there a way to make this work without having to write a seperate function that you can give to map? Filter is almost there, except I also want to keep the elements which the condition doesn't hold for, and just not apply the function to them. Thanks

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  • lists searches in SYB or uniplate haskell

    - by Chris
    I have been using uniplate and SYB and I am trying to transform a list For instance type Tree = [DataA] data DataA = DataA1 [DataB] | DataA2 String | DataA3 String [DataA] deriving Show data DataB = DataB1 [DataA] | DataB2 String | DataB3 String [DataB] deriving Show For instance, I would like to traverse my tree and append a value to all [DataB] So my first thought was to do this: changeDataB:: Tree -> Tree changeDataB = everywhere(mkT changeDataB') chanegDataB'::[DataB] -> [DataB] changeDataB' <add changes here> or if I was using uniplate changeDataB:: Tree -> Tree changeDataB = transformBi changeDataB' chanegDataB'::[DataB] -> [DataB] changeDataB' <add changes here> The problem is that I only want to search on the full list. Doing either of these searches will cause a search on the full list and all of the sub-lists (including the empty list) The other problem is that a value in [DataB] may generate a [DataB], so I don't know if this is the same kind of solution as not searching chars in a string. I could pattern match on DataA1 and DataB3, but in my real application there are a bunch of [DataB]. Pattern matching on the parents would be extensive. The other thought that I had was to create a data DataBs = [DataB] and use that to transform on. That seems kind of lame, there must be a better solution.

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  • Problem when trying to define Show for my Point3D type in Haskell

    - by devoured elysium
    I am trying to define Show for my Point3D type: type Point3D = (Integer, Integer, Integer) instance Show Point3D where show (x,y,z) = "<" ++ (show x) ++ "," ++ (show y) ++ "," ++ (show z) ++ ">" yet I must be missing something in the sintax, as I am always getting an error: Illegal instance declaration for `Show Point3D' (All instance types must be of the form (T t1 ... tn) where T is not a synonym. Use -XTypeSynonymInstances if you want to disable this.) In the instance declaration for `Show Point3D' What am I doing wrong?

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  • Haskell: Constrain function on type Double to only work with Integers

    - by thurn
    Suppose I'm writing a function that takes a list of integers and returns only those integers in the list that are less than 5.2. I might do something like this: belowThreshold = filter (< 5.2) Easy enough, right? But now I want to constrain this function to only work with input lists of type [Int] for design reasons of my own. This seems like a reasonable request. Alas, no. A declaration that constraints the types as so: belowThreshold :: [Integer] -> [Integer] belowThreshold = filter (< 5.2) Causes a type error. So what's the story here? Why does doing filter (< 5.2) seem to convert my input list into Doubles? How can I make a version of this function that only accepts integer lists and only returns integer lists? Why does the type system hate me?

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  • Haskell: Problems with overloading: Interpreter can´t tell which + to use

    - by Ben
    Hi, I want to make functions Double - Double an instance of the Num typeclass. I want to define the sum of two functions as sum of their images. So I wrote instance Num Function where f + g = (\ x - (f x) + (g x)) Here the compiler complains he can´t tell whether I´m using Prelude.+ or Module.+ in the lambda expression. So I imported Prelude qualified as P and wrote instance Num Function where f + g = (\ x - (f x) P.+ (g x)) This compiles just fine, but when I try to add two functions in GHCi the interpreter complains again he can´t tell whether I´m using Prelude.+ or Module.+. Is there any way I can solve this problem?

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  • Haskell Add Function Return to List Until Certain Length

    - by kienjakenobi
    I want to write a function which takes a list and constructs a subset of that list of a certain length based on the output of a function. If I were simply interested in the first 50 elements of the sorted list xs, then I would use fst (splitAt 50 (sort xs)). However, the problem is that elements in my list rely on other elements in the same list. If I choose element p, then I MUST also choose elements q and r, even if they are not in the first 50 elements of my list. I am using a function finderFunc which takes an element a from the list xs and returns a list with the element a and all of its required elements. finderFunc works fine. Now, the challenge is to write a function which builds a list whose total length is 50 based on multiple outputs of finderFunc. Here is my attempt at this: finish :: [a] -> [a] -> [a] --This is the base case, which adds nothing to the final list finish [] fs = [] --The function is recursive, so the fs variable is necessary so that finish -- can forward the incomplete list to itself. finish ps fs -- If the final list fs is too small, add elements to it | length fs < 50 && length (fs ++ newrs) <= 50 = fs ++ finish newps newrs -- If the length is met, then add nothing to the list and quit | length fs >= 50 = finish [] fs -- These guard statements are currently lacking, not the main problem | otherwise = finish [] fs where --Sort the candidate list sortedps = sort ps --(finderFunc a) returns a list of type [a] containing a and all the -- elements which are required to go with it. This is the interesting -- bit. rs is also a subset of the candidate list ps. rs = finderFunc (head sortedps) --Remove those elements which are already in the final list, because -- there can be overlap newrs = filter (`notElem` fs) rs --Remove the elements we will add to the list from the new list -- of candidates newps = filter (`notElem` rs) ps I realize that the above if statements will, in some cases, not give me a list of exactly 50 elements. This is not the main problem, right now. The problem is that my function finish does not work at all as I would expect it to. Not only does it produce duplicate elements in the output list, but it sometimes goes far above the total number of elements I want to have in the list. The way this is written, I usually call it with an empty list, such as: finish xs [], so that the list it builds on starts as an empty list.

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  • Linking/Combining Type Classes in Haskell

    - by thegravian
    Say I have two type classes defined as follows that are identical in function but different in names: class Monad m where (>>=) :: m a -> (a -> m b) -> m b return :: a -> m a class PhantomMonad p where pbind :: p a -> (a -> p b) -> p b preturn :: a -> p b Is there a way to tie these two classes together so something that is an instance of PhantomMonad will automatically be an instance of Monad, or will instances for each class have to be explicitly written? Any insight would be most appreciated, thanks!

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  • Point-free in Haskell

    - by wwrob
    I have this code that I want to make point-free; (\k t -> chr $ a + flip mod 26 (ord k + ord t -2*a)) How do I do that? Also are there some general rules for point free style other than "think about this amd come up with something"?

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