- Haskell-Cafe - Haskell.org

Working debuggers?
by Jim Apple 06 Jul '04

06 Jul '04

I downloaded and compiled buddha, but it apparently does not support Control.Monad.State. I downloaded Hat, but it requires hmake. Hmake fails to build (GHC 6.2.1). Any suggestions? Jim

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Question on Exercise from SOE
by Nathan Weston 05 Jul '04

05 Jul '04

I am learning haskell (and functional programming), from the School of Expression book. There's an exercise to rewrite the following function (for computing the area of a polygon) using map, fold, etc: data Shape = Polygon [Vertex] area (Polygon (v1:vs)) = polyArea vs where polyArea :: [Vertex] -> Float polyArea (v2:v3:vs') = triArea v1 v2 v3 + polyArea (v3:vs') polyArea _ = 0 My first thought is to use fold here, since this function accumulates a result as it traverses the list. But fold consumes one element of the list at a time, while area needs to examine the first three elements of the list, and consume two of them at a time. Is there a more general alternative to fold? Or is there some trick I'm missing here? Thanks, Nathan

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Pruning the tree
by paolo veronelli 05 Jul '04

05 Jul '04

Now I have my infinite tree inf t =Leaf "1" treeGrower :: Tree String-> Tree String treeGrower (Leaf a )= treeGrower (Fork (Leaf (a++"1")) (Leaf (a++"2"))) treeGrower (Fork l r) = Fork (treeGrower l) (treeGrower r) data Tree a = Fork (Tree a) (Tree a) | Leaf a deriving Show inf=treeGrower t I'd like to prune it to a level to have a show of a piece like piece= prune 5 inf How prune is to be coded ?? Thanks Paolino

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Question on Exercise from SOE
by Crypt Master 05 Jul '04

05 Jul '04

Hi Not looking to get flamed, just offering my opinion. I got the SOE first, and while I like the higher level of it, the way he thinks etc... I found it hard to learn haskell from it. I just recieved "Haskell: The Craft of cuntional Programming" this weekend, and have made huge leaps forward (for me) based on the information within. I would have prefered to start with this book in place of SOE. So my advice is SOE is great but back it up with The craft .. S _________________________________________________________________ Add photos to your e-mail with MSN 8. Get 2 months FREE*. http://join.msn.com/?page=features/featuredemail

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Using Product Algebraic types
by Crypt Master 05 Jul '04

05 Jul '04

Hi My question is how do I select or work with product Alhebraic types ? Let you give you some context: I need to be able to work with a list of items whos structure is onyl partially know. That is at the level of this module I dont care about what rest of it is. So I have this: type Fitness = Integer data Population a = Population [Fitness a] Hopefully this reads Population is constructed using the Population constructor and is a list who elements each conists a fitness value and some other value. Since I cant do poloymorphioc types with synonyms I went with the data type. My current task is to build a roulette wheel distribution of the fitness value. Basically I want to build and incremental summing of the fitness value so that each individual is paired with its upper range like so Population [10 x, 20 y, 30 z] New Population = [10 x, 20+10 y, 30+30 z] My first thought was a list comprehension: rwList :: Population -> Population rwList (Population popList) = Population [ (rwListSum i) | i <- popList] rwListSum :: [a] -> [a] rwListSum (fitness individual) = (fitness*2, individual) ingoring for a minute that this wont achieve the above goal, I immediatly ran into the issue of how do I represent in the inner portion of the Population List, i.e "Fitness a". "(fitness individual)" doesnt work ? I tried treating it as 2 normal arguments, but of course how can I return it then ... ? A function can only return one thing. If its not a tuple then it wont work. Hence my question of how you represent product types ? I did come up with one idea, that is of making the internal part a type on its own so now I have this: type Fitness = Integer data InternalIndividual a = InternalIndividual Fitness a data Population a = Population [InternalIndividual a] ... rwList :: Population -> Population rwList (Population popList) = Population [ (rwListSum i) | i <- popList] rwListSum :: InternalIndividual -> InternalIndividual rwListSum (InternalIndividual fitness individual) = (InternalIndividual fitness*2 individual) To my surpirse this compiles. I have been able to fully test it yet though. So is there a way to do this without a extra type ? Is there a better way to represent this ? Thanks for your time, PS Sorry for all the dumb questions, I always find asking better than just wondering for ages. _________________________________________________________________ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus

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A binary tree
by paolo veronelli 04 Jul '04

04 Jul '04

I want to build a binary tree where each leaf is a string of "L" and "R" defining their position from the top This should be done without context isn't it? data Tree a = Fork (Tree a) (Tree a) | Leaf a deriving Show t =Leaf "" treeGrower :: Tree a -> Tree a treeGrower (Leaf a )= treeGrower (Fork (Leaf (a++"1")) (Leaf (a++"2"))) treeGrower (Fork l r) = Fork (treeGrower l) (treeGrower r) ghci says: Cannot unify the type-signature variable `a' with the type `[a1]' Expected type: [a1] Inferred type: a In the first argument of `(++)', namely `a' In the first argument of `Leaf', namely `(a ++ "1")' I don't get it.........

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Re: [Haskell-cafe] newbie problems
by MR K P SCHUPKE 04 Jul '04

04 Jul '04

The zipper should work on n-ary trees. all the zipper does is store the tree as (Context,Subtree) so you just need to adapt that to a Rose-Tree. you could do something like (and I am thinking out loud here as I havent written zipper code for rose trees)... data Context x = Root | Parent x (Context x) [Rose x] [Rose x] data Rose x = Node x [Rose x] so at any given point in the tree the context is either Root, or the Parent (with the value at that node, the parents context, then the left siblings and right siblings as lists) As to the difference between types and values, a type limits the permitted values. For example a :: [Int] a is a list of integers. now data IntList = Value Int | List [Int] IntList can be a single int or a list of ints... data IntList = Value Int | List [IntList] Now intlist can be a value or a list of IntLists... we are dealing with the type of permitted data not the data itself. when we define a value for example: a :: IntList a = Value 3 This must obey the construction rules given in the data type "IntList" So you see when a data declaration refers to itself it is not a "pointer" but an inclusion of the type for example: a :: IntList a = List [Value 3,List [Value 2,Value 1],Value 4] is a valid instance of IntList, but notice that although the 'type' refers to itself the value is not self-referential. Thats not to say self referential values are not possible - but lets avoid confusing the issue with such things. Keean.

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Running Total or Incremental Sum as a higher order type ?
by Crypt Master 03 Jul '04

03 Jul '04

Hi I was attempting to get a running a total of a list i.e for [1,2,3] the result should be [1,3,6] -- [1, 1+2, 2+3] I build this function: incrementalSum [] x = [] incrementalSum (x:xs) runningTotal = currentSum : incrementalSum xs currentSum where currentSum = runningTotal + x I was trying to see how I could higher order functions for this but came up empty. Surely this is a pattern which has been abstracted ? I feel I have missed the obvious here. Thanks, S _________________________________________________________________ MSN 8 helps eliminate e-mail viruses. Get 2 months FREE*. http://join.msn.com/?page=features/virus

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RE: [Haskell-cafe] List syntax -- [] vs ()
by Crypt Master 03 Jul '04

03 Jul '04

Hi Thanks. A light bulb just went on :). I am still in awe of how much "special syntax" is made of such simple base. Haskell certainly is an interesting thing to learn, its like fractals, complexity and beuty from simplicity. S ----Original Message Follows---- Crypt Master, CM> I have noticed that lists seem to swtich between CM> using [] and using (). for example: CM> CM> listSum [] = 0 CM> listSum (x:xs) = x + listsum xs The parentheses are just 'normal' parentheses that are needed because application binds stronger than (:). Without the parentheses, you would get listSum x : xs which is the same as (listSum x) : xs . HTH, Stefan _________________________________________________________________ MSN 8 helps eliminate e-mail viruses. Get 2 months FREE*. http://join.msn.com/?page=features/virus

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List syntax -- [] vs ()
by Crypt Master 03 Jul '04

03 Jul '04

Hi I have noticed that lists seem to swtich between using [] and using (). for example: listSum [] = 0 listSum (x:xs) = x + listsum xs but when specificy lists you use [] as in [1,2,3]. or type signatures are [a] -> [a] It also seems when they mentioned on the right hand side it also always []. Is it just for pattern matching that you use the "tuple" syntax ? How does haskell know we dont mean a tuple ? Or do we mean a tuple when we say (x:xs) ? Thanks _________________________________________________________________ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus

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