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

04 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 04 Jul '04

04 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 03 Jul '04

03 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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Re: [Haskell-cafe] Newbie questions
by Crypt Master 03 Jul '04

03 Jul '04

Hi Thanks for responding and taking the time to respond so fully. Its much appreciated! >From what I can see, a key difficulty you're having is with the >"evolvepopulation" function. You've given it the type a -> a, which >pretty much requires it to be an identity function, because it's not >allowed to make any assumptions about the values it takes. I was going trying for generic as possible, I am keen to expore how much more expressive one can be in functional lanaguages. >To make things easier to read, try defining types for Population and >Individual (they can always be generalized later). > type Individual = Int > type Population = [Individual] A use fill tip, thanks. [Snip] >It looks like you're defining evolution as the composition of mutate, >cross, and select, but got tripped up by the order of evaluation. >Specifically, "mutate cross select p" is the same as "((mutate cross) >select) p", which probably isn't what you want. It only recently dawned on me that functions are left asscoaitive and your absolutly right its not what i had in mind. >If you have: > mutate :: Population -> Population > cross :: Population -> Population > select :: Population -> Population >Then you could define > evolve p = mutate (cross (select p)) > > -- alternative: > -- evolve = mutate . cross . select I now get what the f.g is as well :-) Thanks, the alternate format with the original has just clicked something else into place. >Starting from an initial population, you want to generate a sequence by >iterating evolve, starting from an initial population. Haskell provides >a function "iterate" for this purpose, or you can define your own. Yeah other posters suggested this as well. I now have the the same function as you mentioned which is very nice: gaSolutionSpaceFrom :: a -> [a] gaSolutionSpaceFrom = iterate evolvePopulation [snip] >I found it helpful to remember that you can think of a function f :: a >-> b -> c as taking two arguments of types a and b and returning a >result c, or as taking a single argument of type a and returning a >result of type b -> c. Once you become comfortable with partial >application, a lot of Haskell code starts to make more sense. I understood the partial application when explained in simple terms, but I find I have a hard time when it used in more coimplicated examples. The currying in the "The Haskell School of Expression" left with a big "Huh", even though a simple application makes sense to me. "The craft of Functional Programming" it seesm thicker more methdoical, so I may get some more of the basics down in the next few weeks. Thanks again. ------------------------------ Message: 8 Date: Sat, 03 Jul 2004 08:38:55 +0000 From: "Crypt Master" <cryptmaster(a)hotmail.com> Subject: Re: [Haskell-cafe] Newbie questions To: duncan.coutts(a)worcester.oxford.ac.uk Cc: haskell-cafe(a)haskell.org Message-ID: <BAY18-F13KHmj9n1xxW00056ac5(a)hotmail.com> Content-Type: text/plain; format=flowed [Snip] People say function application in Haskell is written without brackets but this can be misleading, here you do need brackets to indicate that 'gaSolutionSpace [1,2,3,4,5]' is one argument and not two. So you should write: take 5 (gaSolutionSpace [1,2,3,4,5]) [Snip] Thanks alot, this was very helpfull. It also makes more sense now that I looked up the associativity of functions and found it to be left assoc. For some reason I assumed it would automaically bracket form the right as such (take (5 (gaSolutionSpace [1,2,3,4,5]))) but its actually this (((take 5) gaSolutionSpace) [1,2,3,4,5]) Thanks again. _________________________________________________________________ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail ------------------------------ Message: 9 Date: Sat, 03 Jul 2004 08:46:24 +0000 From: "Crypt Master" <cryptmaster(a)hotmail.com> Subject: [Haskell-cafe] Re: Newbie questions To: wferi(a)afavant.elte.hu Cc: haskell-cafe(a)haskell.org Message-ID: <BAY18-F16gyf2M0KFHR000262c6(a)hotmail.com> Content-Type: text/plain; format=flowed >-- gaSolutionSpace :: [a] -> [a] >>gaSolutionSpace x = x : gaSolutionSpace (evolvepopulation x) -Stop deceiving yourself until it's too late. -Why did you comment out the type annotation? *Sheepish Grin* its historical, my original thought and attempt was that you would recieve a list of populations and evolve it to a bigger list of populations. Hence the [a] -> [a]. It didnt work out too well as this is what I came up with: gaSolSpace [x:xs] = gaSolutionSpace [x : evolePopulation x] Eventually i realised that I needed to evolve a single population, not a list, which let to a -> [a] and thanks to Keith I now have this: gaSolutionSpaceFrom :: a -> [a] gaSolutionSpaceFrom = iterate evolvePopulation Thanks _________________________________________________________________ Add photos to your e-mail with MSN 8. Get 2 months FREE*. http://join.msn.com/?page=features/featuredemail ------------------------------ _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe End of Haskell-Cafe Digest, Vol 11, Issue 3 ******************************************* _________________________________________________________________ Help STOP SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail

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Re: [Haskell-cafe] Newbie questions
by Crypt Master 03 Jul '04

03 Jul '04

Hi Very very helpful, thanks. I wasnt sure about the "From" on the end, but I get the subtle change in way you see it. Much better than mine. Using you other advice I got: gaSolutionSpaceFrom :: a -> [a] gaSolutionSpaceFrom p = iterate evolvePopulation p and even managed to curry it :-) : gaSolutionSpaceFrom :: a -> [a] gaSolutionSpaceFrom = iterate evolvePopulation Thanks again. -------------- Kieth: [Snip] I think you mean: gaSolutionSpaceFrom :: a -> [a] gaSolutionSpaceFrom x = x : gaSolutionSpaceFrom (evolvepopulation x) gaSolutionSpace = gaSolutionSpaceFrom createRandomPopulation Note that "a" above should be replaced with your population type. Also note the "iterate" function in the standard library does just this. _________________________________________________________________ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail

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Re: Newbie questions
by Crypt Master 03 Jul '04

03 Jul '04

>-- gaSolutionSpace :: [a] -> [a] >>gaSolutionSpace x = x : gaSolutionSpace (evolvepopulation x) -Stop deceiving yourself until it's too late. -Why did you comment out the type annotation? *Sheepish Grin* its historical, my original thought and attempt was that you would recieve a list of populations and evolve it to a bigger list of populations. Hence the [a] -> [a]. It didnt work out too well as this is what I came up with: gaSolSpace [x:xs] = gaSolutionSpace [x : evolePopulation x] Eventually i realised that I needed to evolve a single population, not a list, which let to a -> [a] and thanks to Keith I now have this: gaSolutionSpaceFrom :: a -> [a] gaSolutionSpaceFrom = iterate evolvePopulation Thanks _________________________________________________________________ Add photos to your e-mail with MSN 8. Get 2 months FREE*. http://join.msn.com/?page=features/featuredemail

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