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

03 Jul '04

[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

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Newbie questions
by Crypt Master 03 Jul '04

03 Jul '04

Hi I consider myself a pretty good imperative programmer, been at it for decades, and am tryibng my hand at Haskell now. Unfrituantly I am not "getting" it all that quickly despite having the multimedia haskell book. To start ghetting some hands on, I thought iI owuld do something which i have done many times before in imperatives, a basic Genetic Algorithm Library. Unfortunatly I am comming unstuck straight away and was hoping some of you kind folk could point in the direction of dry land :-) iIf your not familiar with GAs its not too important, simply a GA searches over the whole solution space randomly. Well not quiet, its a guided search, its guided by evolution ie by the fitness of each indicidual of a random initial population. Evolution is simulated by 3 basic operators, selection, cross over and mutation. selection is which indiviuals get to breed, cross over is how they breed, and mutation is just things intetresting. So after reading the multimedia book, Iimmediately thought of a defining the solution space as a infinite list. I was hoping then I could do things like take 5 gaSolutionSpace to get 5 iterations or generations. My first attempt tied to use lis syntax [], but it wouldnt compile and after seeing "numsFrom" ina tutorial I redefined it as such. Here is what I have so far: -- gaSolutionSpace :: [a] -> [a] -- gaSolutionSpace [] = gaSolutionSpace createRandomPopulation -- recursive base case gaSolutionSpace x = x : gaSolutionSpace (evolvepopulation x) evolvepopulation :: a -> a evolvepopulation p = mutate cross select p -- createRandomPopulation :: [Num a] createRandomPopulation = [1,23,4,5,6] cross p = p mutate p = p select p = p ---- The take operator doesnt work on this. from hugs: HAGA> take 5 gaSolutionSpace [1,2,3,4,5] ERROR - Type error in application *** Expression : take 5 gaSolutionSpace [1,2,3,4,5] *** Term : take *** Type : Int -> [e] -> [e] *** Does not match : a -> b -> c -> d I am not sure I follow this. I assume its cause I didnt use the list notation in definaing gaSolutionSpace. Any ideas on how to do that. ? What about using map. It occurs to me that you can define gaSolutionspace as a map of the evolvolepopuloation function acrossan infinite solution space, but I dont know how to makethis infinte list to map over ...? There ar 2 papaers on GAs in haskell, but they use monads. I realise for performance I will probably have to use them too, but for now I would liek to do it without mondas even if performance isnt optimal. Sorry again for the newbie questions, but any help is appreciated. Stephen _________________________________________________________________ Help STOP SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail

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Audio signal processing in Haskell
by Henning Thielemann 03 Jul '04

03 Jul '04

As I can see a question similar to mine was already raised here: http://www.haskell.org/pipermail/glasgow-haskell-users/2004-February/006305… I also encountered the elegance that Haskell's laziness provides for signal processing. I use lists for the signals. With help of Haskore I put together a small song and it needed 5 minutes for rendering the quite simple music of about 30 seconds. Don't care about the exact numbers but I find it rather slow. What is the most promising strategy to speed up the computation without loosing the elegance of Haskell's lists? 1. Upgrading GHC from 6.0 to the current version? 2. Use of simplification rules? How can I find out what new rules may be useful? The several dumps that GHC can generate are a bit hard to understand. :-( Is there some more detailed information about what rules GHC already applies than what one can get with -ddump-simpl-stats and -ddump-rules? 3. More strictness? Is a custom list structure with strict entries and maybe with only one constructor (thus always infinite) a considerable alternative? This would disallow the use of lots of functions that deal with lists, including State monad functions. 4. A list of arrays? Then signal processes with feedback like an echo are much more difficult.

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Relating functors in Category Theory to Functor
by David Menendez 03 Jul '04

03 Jul '04

Until recently, I've had difficulty relating the concept of a "functor" from Category Theory to the class Functor in Haskell. I've figured out a relationship that seems to make sense, but I wanted to check with the list to make sure I'm not on the wrong track. In Category Theory, a functor F between categories C and D has two parts: (1) a mapping F[ob] from the objects of C to the objects of D (2) a mapping F[mor] from the morphisms of C to the morphisms of D For any morphism f: a -> b in C, there is a morphism F(f): F(a) -> F(b) in D. For two morphisms f and g in C, F(f . g) = F(f) . F(g) in D. We can interpret Haskell as a category with types as objects and functions as morphisms. For example, isDigit :: Char -> Bool. So, for some Functor f, the type constructor f corresponds to F[ob], and fmap to F[mor]. For any function g :: a -> b, fmap g :: f a -> f b, and fmap (g . h) = fmap g . fmap h. -- For two functors F, G: C -> D, there is a natural transformation tau from F to G, if for every object a in C, there is morphism tau[a]: F(a) -> G(a) in D. In Haskell, natural transformations are polymorphic functions, tau :: f a -> g a. For example, maybeToList :: Maybe a -> [a]. -- A monad is a functor, F: C -> C, plus two natural transformations mu: F^2 -> F and eta: I -> F (I being the identity functor for C). In Haskell, F[ob] becomes the type constructor, F[mor] becomes fmap (or liftM), mu becomes join :: f (f a) -> f a, and eta becomes return :: a -> f a. So, am I on the right track here? -- David Menendez <zednenem(a)psualum.com> <http://www.eyrie.org/~zednenem/>

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readMVar and the devils
by Conor T McBride 03 Jul '04

03 Jul '04

Hi folks I'm having a play with Concurrent Haskell, and came across this in the library. readMVar :: MVar a -> IO a This is a combination of takeMVar and putMVar; ie. it takes the value from the MVar, puts it back, and also returns it. I was just wondering if I understand things correctly. Suppose myMVar is nonempty and I have one thread executing do x <- takeMVar myMVar -- (A) putMVar myMVar x -- (B) return x and another executing putMVar myMVar y -- (C) Questions (1) Is it possible that an evil scheduler will execute (A) then (C) and block on (B)? (2) Is it the case that if myMVar is nonempty and readMVar is chosen for execution, that readMVar myMVar takes the value from myMVar, guaranteed that the _same_ value will be put back immediately, and also returned? (3) If yes to both, what combination of takeMVar and putMVar manages to maintain the lock in between the two? I'm only asking, because I'm trying to cook up some kind of partial evaluator for programs which are being simultaneously edited by fiercely argumentative devils. Kind of `demonic laziness', where the computations decide when _they_ feel like running (eg sometime after the program turns up), rather than the `angelic laziness' we're used to. I'd like to use MVars in a write-once-read-many style, so that once someone fills in a bit of program, it stays put. I'd hate for another devil to sneak in a replacement program just in the twinkling of an eye between taking something out to read and putting it back to be read again. I guess I could use some kind of extra semaphore MVar to ensure that the reader has the lock on the program MVar, but that's more like hard work. Is readMVar what I want? Another more general question would be `is this an utterly stupid thing to do in practice?'. I mean, just consider an evaluator for some simple language (eg arithmetic expressions), implemented by forking a thread for every subcomputation, and using MVars to transmit the values. Is that diabolically slow compared to a normal evaluator? I guess I'll try it, but I was wondering if anyone had any useful experience to share? Cheers Conor

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some newbie FFI questions
by John Kozak 02 Jul '04

02 Jul '04

I'm trying to experiment with image processing in haskell (with which I haven't much experience). I've written some FFI code to talk to the ImageMagick library which provokes a few questions (environment is ghc 6.2.1 on debian): 1. Speed: I'm reading in a 2000x1500 pixel image, and have defined a Pixel type like this: data Pixel a = Pixel !a !a !a deriving Show I use ImageMagick to load the image, then build an Array of Pixel Floats. Building the array takes 45 seconds on a 2.5Ghz P4 with code compiled -O2, which seems slow to me - are my expectations unrealistic? I've tried various UNPACK things which didn't make much difference. 2. How do I convert a CFloat into a Float? 3. I get the wrong answer ;-) I expect the C and haskell code below to produce the same pixel data, but they don't (the C code is right). C code: #include <stdio.h> #include <wand/magick_wand.h> int main(int argc,char *argv[]) { MagickWand *wand = NewMagickWand(); int width,height; float *pixels; int i,j; if (argc!=2) { fprintf(stderr,"bad args\n"); exit(100); } if (!MagickReadImage(wand,argv[1])) { fprintf(stderr,"can't read file\n"); exit(1); } height = MagickGetImageHeight(wand); width = MagickGetImageWidth(wand); printf("dimension: %dx%d\n",width,height); pixels = (float*) malloc(3*width*height*sizeof(float)); if (!MagickGetImagePixels(wand,0,0,width,height,"RGB",FloatPixel,(unsigned char*) pixels)) { fprintf(stderr,"can't get pixels\n"); exit(1); } printf("%f %f %f\n",pixels[0],pixels[1],pixels[2]); printf("%f %f %f\n",pixels[4],pixels[5],pixels[6]); printf("%f %f %f\n",pixels[7],pixels[8],pixels[9]); } haskell code: -- requires flags: -fffi -fglasgow-exts import Foreign import Foreign.C import MarshalArray import Char import List import Array import Bits data Pixel a = Pixel !a !a !a deriving Show data MagickWand -- opaque foreign import ccall "wand/magick_wand.h NewMagickWand" newMagickWand :: IO (Ptr MagickWand) foreign import ccall "wand/magick_wand.h DestroyMagickWand" destroyMagickWand :: (Ptr MagickWand) -> IO () foreign import ccall "wand/magick_wand.h MagickSetFilename" setFilename :: Ptr MagickWand -> CString -> IO () foreign import ccall "wand/magick_wand.h MagickReadImage" readImage :: Ptr MagickWand -> CString -> IO CInt foreign import ccall "wand/magick_wand.h MagickGetImageWidth" getWidth :: Ptr MagickWand -> CULong foreign import ccall "wand/magick_wand.h MagickGetImageHeight" getHeight :: Ptr MagickWand -> CULong foreign import ccall "wand/magick_wand.h MagickGetImagePixels" getPixels :: Ptr MagickWand -> CULong -> CULong -> CULong -> CULong -> CString -> CInt -> (Ptr a) -> IO CInt -- StorageType: enum Char=1,Short,Integer,Long,Float=5,Double type Frame a = Array (Int,Int) (Pixel a) getPixel :: ([a],Int) -> Pixel a getPixel (floats,i) = Pixel (floats!!(i*3)) (floats!!(i*3+1)) (floats!!(i*3+2)) getFrame :: (Storable a) => (Ptr a) -> Int -> Int -> IO (Frame a) getFrame arr h w = do putStr ("getFrame: "++(show (w,h))++"\n") floats <- peekArray (3*h*w) arr return (array ((1,1),(h,w)) [((i,j),getPixel (floats,((w*i)+j))) | i<-[1..h],j<-[1..w]]) loadFrameFloat :: FilePath -> IO (Frame CFloat) loadFrameFloat filename = do wand <- newMagickWand f <- (newCString filename) ok <- readImage wand f let w = (fromInteger.toInteger) (getWidth wand) h = (fromInteger.toInteger) (getHeight wand) in do arr <- mallocArray (w*h*3) :: (IO (Ptr CFloat)) rgb <- (newCString "RGB") ok1 <- getPixels wand 0 0 (getHeight wand) (getWidth wand) rgb 5 arr -- PixelFloat==5 frame <- getFrame arr h w free arr destroyMagickWand wand return frame thanks, John

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My old chess player
by paolo veronelli 02 Jul '04

02 Jul '04

Ok Crypt Master give the hint to pronounce something ,because I feel a complete stupid in front of Functional Programming' though I'm a trained imperative one. I've read the nice paper www.cs.chalmers.se/~rjmh/Papers/whyfp.html and remember may years ago the hard job done to make my chess engine work ,but that was imperative C that's why it was hard ,isn't it? John Huges say the way for trees is data Albero position=Foglia position |Ramo position [Albero position] This is like Quantum-Mechanics for my head but it works ,I've tried. figliatore creatore (Foglia position) = Ramo position (map (figliatore creatore) (creatore (Foglia Position)) This I don't know how to try it but I'm faithfully blind........ Now I would like to make a change to fit my player,but the wall is too high. My function 'creatore' in fact would take a list of moves and a starting position as argument, not a nice ready position. My imperative engine stored leaves with a pointer to their mothers so that it was easy to track the list of moves. Now maybe I have to change Albero definition in data Albero move mother=Foglia move mother |Ramo move mother [Albero move mother] Is this the way? the road tracker is going to be pather (Foglia move mother)= if mother==() = () else = (pather madre ):madre This doesn't compile!! And I don't write the new 'figliatore' not to insult anyone Thanks For Comments Paolino Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/

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