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Re: Read File
by Jorge Adriano 01 May '02

01 May '02

----------------- [moving to haskell-cafe] > I want to read some data from a text file. I accomplished it but I want to > read data from file as a Integer list for example text file has [1,2,3,5] ----------------- Wait you actually have "[1,2,3,5]" --------------------- > and when I read this data from file I handeled it as string. What can I do > to get it as integer list. Hint: map read ["1","2","3"] ::[Int] ------------------- this would be helpfull if you had "1 2 3 5". It's easier then Hint: read "[1,2,3,5]"::[Int] J.A.

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Re: Read File
by Jorge Adriano 01 May '02

01 May '02

[moving to haskell-cafe] > I want to read some data from a text file. I accomplished it but I want to > read data from file as a Integer list for example text file has [1,2,3,5] > and when I read this data from file I handeled it as string. What can I do > to get it as integer list. Hint: map read ["1","2","3"] ::[Int] J.A.

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(no subject)
by Ricky Kurniawan 30 Apr '02

30 Apr '02

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some algorithms are hard to code in O(n) in Haskell
by klusacek＠atrey.karlin.mff.cuni.cz 29 Apr '02

29 Apr '02

Hi - does anybody know how to do a term unification in Haskell in O(n), where n is length of the input terms? I know Montanari-Martelli algorithm for this, but I'm unable to code it in Haskell (in O(n)). I would need a destructive update of the directed graph representing terms (to do a substitution) and checking for cycles in directed graphs (these cycles emerge when unifying things like x=f(x) -- their detection is also called occurs-check). I've been trying hard but still have no solution. Maybe I'm not smart enough, maybe I don't know some trick, maybe it's impossible in Haskell. Thanks, _dave_

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Re: Binary search tree
by Rijk J. C. van Haaften 29 Apr '02

29 Apr '02

>Hi everybody, > >I studied haskell this semester at the university and I was required to >implement a binary search tree in haskell. >I would appreciate if anyone could send me an example code of this data >structure. Just read a standard textbook. Some useful course notes by Jeroen Fokker are available online at http://www.cs.uu.nl/people/jeroen/courses/fp-eng.pdf Enough to solve your problem. Rijk-Jan van Haaften

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Binary search tree
by Claudio Marcio de Souza Vicente 29 Apr '02

29 Apr '02

Hi everybody, I studied haskell this semester at the university and I was required to implement a binary search tree in haskell. I would appreciate if anyone could send me an example code of this data structure. thanks in advance, Claudio

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RE: defining (-> Bool) as a set
by Hal Daume III 26 Apr '02

26 Apr '02

Would there be any problems if there were an artificial "<-" type constructor in Haskell which was basically: type a <- b = b -> a but not a type synonym so you could define it to be instances of classes, but then those get applied "backwards" to ->? Okay, that made no sense. I'll try again. Type lambda = bad (undecidable). What about only type lambda on function types and not even type lambda; simply allow partial application to the second argument *only* on ->? This would solve my problem and I believe that of an earlier poster who wanted to define catamorphisms (I think; don't feel like checking), but couldn't because of this restriction. Is this still "too loose" to be made to work? -- Hal Daume III "Computer science is no more about computers | hdaume(a)isi.edu than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume On Mon, 22 Apr 2002, Simon Peyton-Jones wrote: > Hal, > > [I think this sort of question would be better on the haskell-cafe > list.] > > I don't think what you want can be done directly. It's the old > thing about not having lambdas at the type level. You want: > > instance Eq a => Coll (\x. x -> Bool) a where ... > > and you just can't do that. You *can* abstract the second argument > of (->): > > instance Eq a => Coll ((->) Bool) a where ... > > but not the first. It's a well known shortcoming in Haskell, that you > can > partially apply type constructors, but you can't do argument > permutation. > > I know of no good solution. Adding type lambdas in their full glory > makes type inference pretty much impossible. What we'd like is > a compromise. Maybe someone can invent one. But take care. > The ground is littered with corpses. > > Simon > > > | -----Original Message----- > | From: Hal Daume III [mailto:hdaume@ISI.EDU] > | Sent: 23 April 2002 01:30 > | To: Jorge Adriano > | Cc: Haskell Mailing List > | Subject: Re: defining (-> Bool) as a set > | > | > | Yeah, both options suggested are valid, of course. But I > | really don't want to have a constructor and I'm using Edison > | where Coll is defined something like: > | > | class Coll c e where > | empty :: c e > | insert :: c e -> e -> c e > | > | etc., which precludes the fun dep solution. > | > | - Hal > | > | -- > | Hal Daume III > | > | "Computer science is no more about computers | hdaume(a)isi.edu > | than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > | > | On Tue, 23 Apr 2002, Jorge Adriano wrote: > | > | > > | > > class Collection e ce | ce -> e where > | > > empty :: ce > | > > insert :: e -> ce -> ce > | > > member :: e -> ce -> Bool > | > > > | > > instance Eq a => Collection a (a -> Bool) where > | > > empty = (\x -> False) > | > > insert e f = (\x -> if x == e then True else f x) > | > > member e f = f e > | > > | > This is way better than my solution... > | > > | > I had never used multi-parameter classes before, so I forgot the > | > functional > | > dependency (right name? the "|ce->e"), and there was > | obviously no need for my > | > extra constructor. > | > > | > J.A. > | > > | > | _______________________________________________ > | Haskell mailing list > | Haskell(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell > | >

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RE: lazy comparison for equality ?
by Konst Sushenko 24 Apr '02

24 Apr '02

In addition to what Hal said, I think that even assuming that you want to write a function like that that is only supposed to be applied to finite lists (like, say, prelude's 'and' function, which does not terminate on an infinite list of True values) you still cannot do it because haskell lists are not like C lists. There is no notion, to my understanding, of list circularity. There are finite and infinite lists. Your example of a list that you call a circular list is just a definition of a recursive function that produces an infinite list. So what you are almost asking is how to write a function that checks if a list has repeated sub-sequences, which is not of course precisely the same that you asked (because even a finite list can have repeated sub-sequences), but close. And this is just one of many problems taken care of by various string search algorithms. konst > -----Original Message----- > From: Hal Daume III [mailto:hdaume@ISI.EDU] > Sent: Wednesday, April 24, 2002 12:38 PM > To: klusacek(a)atrey.karlin.mff.cuni.cz > Cc: haskell-cafe(a)haskell.org > Subject: Re: lazy comparison for equality ? > > > I don't think you can write such a function. For instance, > how would you > know whether [1..] is circular or not? In order to know that it's not > you'd need to evaluate it fully. > > -- > Hal Daume III > > "Computer science is no more about computers | hdaume(a)isi.edu > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > On Wed, 24 Apr 2002 klusacek(a)atrey.karlin.mff.cuni.cz wrote: > > > > > Hi - > > I'm a Haskell beginner and I have a problem. > > > > Let's have a list which may be normal list > > list1 = [1,2,3] > > or a circular list > > list2 = 1:2:list2 > > > > Now I'd like to have a function which tells me whether the > > given list is circular or not. This doesn't work: > > > > circ l = l l > > circ2 l [] = False > > circ2 l (_:as) | l==as = True > > | True = (circ2 l as) > > > > > > It seems that comparison l==as really compares element by > element thus > > falling into an infinite loop. I would need to compare > pointers instead of > > values. > > > > Does anybody know how this could be done ? > > > > Thanks. > > > > > > > > > > > > _______________________________________________ > > Haskell-Cafe mailing list > > Haskell-Cafe(a)haskell.org > > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >

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lazy comparison for equality ?
by klusacek＠atrey.karlin.mff.cuni.cz 24 Apr '02

24 Apr '02

Hi - I'm a Haskell beginner and I have a problem. Let's have a list which may be normal list list1 = [1,2,3] or a circular list list2 = 1:2:list2 Now I'd like to have a function which tells me whether the given list is circular or not. This doesn't work: circ l = l l circ2 l [] = False circ2 l (_:as) | l==as = True | True = (circ2 l as) It seems that comparison l==as really compares element by element thus falling into an infinite loop. I would need to compare pointers instead of values. Does anybody know how this could be done ? Thanks.

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RE: "do" notation and ">>"
by Simon Marlow 24 Apr '02

24 Apr '02

> I was not suggesting that the Haskell'98 report should > change or even give a warning -- I was giving a warning to > compiler implementors, that this simple change might have > disastrous effects. > > BTW, I remember a similar discussion along these lines on > the Haskell mailing list that happened in 1997 I think, but > I cannot find the archives. > > Another comment one can make here is the following: if > Haskell does not care about sharing, why is the monomorphism > restriction there? One difference is that when the compiler gives you *less* sharing than you were expecting, then it is in some sense worse than getting more sharing. More sharing might result in space leaks, but in practice (in my experience anyway) this isn't as noticeable as the time leaks you get when sharing is lost. GHC has implemented almost-full-laziness for a long time, and I don't recall anyone ever complaining about space leaks introduced as a result. The monomorphism restriction, on the other hand, is concerned with the user getting no less than the amount of sharing he/she was expecting (personally I think this should be a compiler warning rather than a language restriction, but I'm aware there are other reasons for wanting the MR too). Cheers, Simon

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how to call Fortran procedures in Haskell program?
by Zhe Fu 21 Apr '02

21 Apr '02

Does anybody know how to call Fortran procedures in Haskell program? I tried Green Card, but seems it only works with C codes. Thanks.

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�𾴵��Ͽͻ��Ǻ�!
by web 20 Apr '02

20 Apr '02

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Hi.
by wooil 20 Apr '02

20 Apr '02

How I can register for the mailing list? Philip

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Re: [HOpenGL] ANNOUNCE: FunGEn-0.1 released!
by Sven Panne 19 Apr '02

19 Apr '02

[ Moved from haskell to haskell-cafe ] Cool! :-) I just had a quick test on my Solaris box, and most things went fine. Just some small hints: * The Makefile contains WinDoze CR/LF, which *nix-makes don't really like. * There were uppercase/lowercase mismatches in the file names of the bitmaps for the worms game. Well, WinDoze doesn't care, but "real" OSes do. :-) * The easiest way to compile FunGEn and the examples is probably: ghc --make -O -fglasgow-exts -package HOpenGL -o game game.hs ( cd pong ; ghc --make -O -package HOpenGL -i.. -o pong pong.hs ) ( cd worms ; ghc --make -O -package HOpenGL -i.. -o worms worms.hs ) At least if you have a recent GHC + an installed HOpenGL package... Cheers, S.

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Question on style - Implicit parameters
by Jorge Adriano 18 Apr '02

18 Apr '02

Hi all, Lately I've found implicit parameters to be extremely usefull in many situations. The most simple and obvious is when you need to calculate some values that depend on specific parameters, in some context where these same parameters constant. There is no poblem if all the calculus are defined in the scope of those 'constant' parameters, but if you need to define functions outside their scope then you have to pass them as arguments, and sometimes that complicates type signatures. Now my question. I've end up in situations where I need to write something like --------- step :: Inst-> Train->Inst step inst train = changeInst deltaW deltaV inst where deltaW = someSemiConst1 with ?inst=inst; ?train train deltaV = someSemiConst2 with ?inst=inst; ?train train --------- Ok the problem is that both deltaV and deltaW are assigned some constants that implicitly depend on the arguments 'inst' and 'train'. Now imagine that someSemiConst1 = f n someSemiConst2 = g n for some n that also depends implicitly on 'inst' and 'train'. In this case 'n' would (could? haven't actually tried it) be calculated twice, one fro the first semiConst, another for the second. It would be nice to have something like with ?inst=inst; ?train train { deltaW = someSemiConst1 deltaV = someSemiConst2 } My solution was to define some function (deltaW, deltaV)=calculateDeltas with ?inst=inst; ?train=train And in this function I simple does (deltaW, deltaV)= (someSemiConst1, someSemiConst2) In case you're wondering, why not just call the 'step' function bounding it's context to 'inst' and 'train'. It doesn't makes much sense to me in this case, becouse the whole purpose of this function is to change a 'inst' acording to some 'train' (in fact it is beeing used in a foldl). How do you usually handle this situations? Is there an elegant solution for this? J.A.

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Your message to Cvs-nofib awaits moderator approval
by cvs-nofib-admin＠haskell.org 18 Apr '02

18 Apr '02

Your mail to 'Cvs-nofib' with the subject How are you Is being held until the list moderator can review it for approval. The reason it is being held: Message body is too big: 134653 bytes but there's a limit of 40 KB Either the message will get posted to the list, or you will receive notification of the moderator's decision.

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ASSISTANCE NEEDED.
by DR DAN MOMOH 15 Apr '02

15 Apr '02

DR DAN MOMOH FED.MINISTRY OF PETROLUEM REOURCES(F.M.P.R) LAGOS NIGERIA. ATTN:SIR, REQUEST FOR ASSISTANCE- STRICTLY CONFIDENTIAL I amDR. DAN MOMOH, an accountant in the Ministry of petroleum Resources (MPR) and a member of a seven-man Tender Board in charge of contract review and payment approvals. I came to know of you in my search for a reliable person to handle a very confidential transaction that involves the transfer of a huge sum of money to a foreign account. It may sound strange but exercise patience and read on. There were series of contracts executed by a Consortium Multinational in the Oil industry in favor of Ministry of Petroleum Resources among which were: 1. The extension of pipeline network within Nigeria for crude oil, Down stream products distribution and subsequent evacuation- US$ 195 Million. 2. Contract for the Turn Around Maintenance (TAM) of the various refineries in the country~ US$ 152 Million. 3. The construction of storage tanks for petroleum products (Depots)~ US$240 Million. The original values of this contract were deliberately over invoiced to the sum of US $35.5million American Dollars which has now been approved and is now ready to be transferred being that the companies that actually executed these contracts has been fully paid and project officially commissioned. Consequently, my colleagues and I are willing to transfer the total amount to your account for subsequent disbursement since we are Civil Servant and are prohibited by the Code of Conduct Bureau ( Civil Servant Laws) from opening/ operating foreign account in our names. Needless to say, the trust posed on you at this juncture is enormous. In return, we have agreed to offer you 20% of the transferred sum while 10% shall be set aside for accidental expenses (Internal & External) between the parties in the course of the transfer. We will mandate you to draw up an investment plan for us of which you will control since we cannot bring our shares back to Nigeria. We are very interested in investing in real estate in your country. You must however note that this transaction is subjected to the following terms and conditions: 1. Our conviction of your transparent honesty and diligence. 2. That you would treat this transaction with utmost secrecy and confidentiality because of the circumstances we now find ourselves. 3. That the funds would be transferred to an account over which you have absolute control. Modalities have been worked out at the highest level of the presidency at the Central Bank of Nigeria for the immediate transfer of the funds within 14 working days, subject to your satisfaction of the above stated terms. Our assurance is that your role is 100% risk free. To accord this transaction, the legality it deserves and for mutual security of the parties involved and the funds, the whole approval procedures will be officially and legally processed with your name or the name of your company you may nominate as the bonafide beneficiary. Also, be informed that our main reason of contacting you is for the obvious reason that we will want to invest these funds in the real estate. It is our believe that you can be of immense help in this regards. Kindly expedite action so as to enable us include this transaction into this batch as payment to foreign contractors is usually carried out on quarterly basis. Please, contact me as soon as possible on my private e-mail address:almoore@hotvoice.com Best regards, DR. DAN MOMOH

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Re: ASSISTANCE NEEDED.
by Brian Tawney 15 Apr '02

15 Apr '02

>REQUEST FOR ASSISTANCE- STRICTLY CONFIDENTIAL This is a common scam. Everyone should ignore it and move on. If you're interested, here's a URL to a site about this particular con-game: http://www.egnet.co.uk/nigeria/index.asp _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp.

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Re: Help need "Cannot use type synonym in instance head"
by Jon Cast 14 Apr '02

14 Apr '02

Please keep this on the list, so everyone can benefit. Benny <neon_2001(a)hotmail.com> wrote: > Hello Jon > Thanks for the help. But it still didn't work after I changed the > codes from: > > type Poly = [Float] > > addPoly::Poly->Poly->Poly > > addPoly x y = zipWith (+) x y > > instance (Float a) => Num (Poly a) where > > (+) = addPoly > to your suggestion: > type Poly = [Float] > addPoly::Poly->Poly->Poly > addPoly x y = zipWith (+) x y > instance Num [Float] where > (+) = addPoly > I received this erreor message "Syntax error in instance head > (variable expected)". I sent this suggestion without actually trying it; unfortunately, as you discovered, it violates the rules of the Haskell 98 language. Please disregard this suggestion. > Then I tried your second suggestion, so I changed to codes to: > type Poly = [Float] > newtype Poly = Poly [Float] > addPoly::Poly->Poly->Poly > addPoly (Poly x) (Poly y) = Poly (zipWith (+) x y) > instance (Float a) => Num (Poly a) where > (+) = addPoly > Then I received this error message " Repeated definition of type > constructor "Poly"". What I meant was the following: > newtype Poly = Poly [Float] > addPoly::Poly->Poly->Poly > addPoly (Poly x) (Poly y) = Poly (zipWith (+) x y) > instance Num Poly where > (+) = addPoly I.e., delete the type declaration as well as the extraneous information from the instance. <snip> Btw., you'll also want instances of Eq and Show to get Num to go through. > Thank you very much You're welcome > Cheers > Benny Jon Cast

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Help need "Cannot use type synonym in instance head"
by Benny 14 Apr '02

14 Apr '02

type Poly = [Float] addPoly::Poly->Poly->Poly addPoly x y = zipWith (+) x y instance (Float a) => Num (Poly a) where (+) = addPoly I got the error message "Cannot use type synonym in instance head" when I was trying to compile the codes above. Can anyone tell me why and how to solve it? Cheers Benny

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Re: Help need "Cannot use type synonym in instance head"
by Ashley Yakeley 14 Apr '02

14 Apr '02

At 2002-04-13 21:22, Benny wrote: >type Poly = [Float] > >addPoly::Poly->Poly->Poly >addPoly x y = zipWith (+) x y > >instance (Float a) => Num (Poly a) where > (+) = addPoly > > >I got the error message "Cannot use type synonym in instance head" when I >was trying to compile the codes above. Can anyone tell me why and how to >solve it? There are at least two errors here. Firstly the context of your instance "(Float a) =>" is wrong. "Float" is not a class. Secondly, "(Poly a)" is meaningless, since Poly is a type and does not take any parameter. -- Ashley Yakeley, Seattle WA

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Theorems for free.
by Long Goodbye 13 Apr '02

13 Apr '02

I have heard that Philip Wadlers article "Theorems for free" has been written in a "CT terms" version in addition to the one that can be found at Nec. If this is the case then I would very much like to have a pointer to where this file can be found. Thank you, Henrik Ljung _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp.

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writeFile and buffering
by Jorge Adriano 13 Apr '02

13 Apr '02

Hi, Is there any way to use writeFile with no buffering? I know that with handles I can do, > hSetBuffering hdl NoBuffering Since all I do is write a string to a file, I don't see much need for handles here though. I wanted no buffer to be able to check the status of the program by accessing the file - I calculate thousands of values, but just write few hundred so it usually ends up just writing the string in the very end. Thanks in advance J.A.

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Re: Bug in IO libraries when sending data through a pipe?
by Manuel M. T. Chakravarty 13 Apr '02

13 Apr '02

Jens Petersen <juhp(a)01.246.ne.jp> wrote, > Jens Petersen <petersen(a)redhat.com> writes: > > Also I just rediscovered Manuel Chakravarty's HPL (Haskell > Ports Library), which provides a rather elegant, > sophisticated approach. > > http://www.cse.unsw.edu.au/~chak/haskell/ports/ > > It compiles fine under ghc-5.02.2, and using the BufferMode > patch included at the end output seems to be ok, Why LineBuffering? Doesn't really change anything for me (same GHC version). > but input > of more than 2048 bytes doesn't seem to be being handled > reliably. [..] > mostly gives no output, but occasionally I see > > Warning: Ports.listenToPort: Attempted to listen to a closed port! > > Needs some debugging I guess. :) I guess, it needs more documentation. Your test program is wrong in two places. The function `listenToPort' starts to listens at the given port exactly when `listenToPort' is executed; ie, any data that goes over the port earlier will not be received in the stream that this call to `listenToPort' returns. You wrote > outpt <- newPort ' ' > errpt <- newPort ' ' > let p = proc cmd args > p inpt outpt errpt > putStrLn "output:" > out <- listenToPort outpt As the processes `cmd' is fork()ed in a separate thread, it may already have finshed its business until you get to `listenToPort outpt', which means you have missed all the data. For similar reasons > errclosed <- isClosedPort errpt > unless errclosed $ > do > err <- listenToPort errpt isn't recommended. I changed your program to what I have appended. It then works for me for the % cat 4096 | test-processes cat test...except that I sometimes get a "Broken Pipe". So, I guess, I have to do some debugging after all (plus improve the documentation). Cheers, Manuel PS: I'll add your program to the ports/tests/ directory if you don't mind. -=- module Main where import Processes import Ports import IO (openFile, hGetContents, IOMode(..), hSetBuffering, BufferMode(..)) import Monad (unless) main :: IO() main = do inpt <- getContents withPorts [] $ \ (cmd:args) -> do outpt <- newPort ' ' errpt <- newPort ' ' out <- listenToPort outpt err <- listenToPort errpt let p = proc cmd args p inpt outpt errpt putStrLn "output:" mapM_ putStrLn $ lines out putStrLn "error:" putStr err putStrLn "test finished"

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recursive function
by excellent2002 11 Apr '02

11 Apr '02

Hi,I want to write a recursive function to change list as this form ["whantever","\n","it is","\n","I","don;t","care"] to ["whatever","it is","I \n don't \n care"], but I failed to work it out. Anybody can help me ? Thank you very much If there is no "\n",between the strings, add "\n" beteen them and merge these string. Regards Alan ----------------------------------------- ÒÚÌÆÉú»îÆ·ÅÆµÚÒ»Ö÷Žò http://bag.etang.com/payment_1.htm ÒÚÌÆ¶ÌÐÅ£¬ÄúÌùÉíµÄÊ±ÉÐ¹ËÎÊ http://sms.etang.com Ã¿ÌìÈÃÄãÇáËÉŒÇµ¥ŽÊ http://www.etang.com/texthanburg/index.htm ÃÀ¹úŽóÑ§Ö®ÃÅÎªÄã¶ø¿ª °Ù·Ö°ÙÄÃœ±Ñ§œð http://campus.etang.com/html/academic/usa/scho/scho017.htm

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subscribe
by Leif Eriksen 09 Apr '02

09 Apr '02

Leif Eriksen Software Engineer AusRegistry Pty Ltd Level 12, 10 Queens Rd. Melbourne, Victoria. Australia 3004. Ph: +61 3 9866 1990 Fax: +61 3 9866 1970 Email: leif(a)ausregistry.com.au Web: www.ausregistry.com.au Registrars: www.ausregistry.net.au

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subscribe
by Leif Eriksen 09 Apr '02

09 Apr '02

Leif Eriksen Software Engineer AusRegistry Pty Ltd Level 12, 10 Queens Rd. Melbourne, Victoria. Australia 3004. Ph: +61 3 9866 1990 Fax: +61 3 9866 1970 Email: leif(a)ausregistry.com.au Web: www.ausregistry.com.au Registrars: www.ausregistry.net.au

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RE: if (++) were left associative ?
by Konst Sushenko 08 Apr '02

08 Apr '02

> > I don't think so. I think it only takes linear time to get the head. > But once you've gotten the head, it takes linear time again to get the > head of the tail, .... You get (I think...) a progression like > n+(n-1)+(n-2)+...+1, which is in O(n^2). > What does 'n' denote? The get the head, it takes time proportional to the number of (++) invocations. Is that what you mean? konst

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RE: if (++) were left associative ?
by Konst Sushenko 07 Apr '02

07 Apr '02

This one helped. Thanks. 'reduction' was the key word that made it clear for me. It is not that left associated (++) reconstructs the list once and again (although one can say that) it is just that to return the head from the recursive invocations takes quadratic number of reductions. Right? konst > -----Original Message----- > From: Jay Cox [mailto:sqrtofone@yahoo.com] > Sent: Sunday, April 07, 2002 11:09 AM > To: Konst Sushenko > Cc: haskell-cafe(a)haskell.org; Jon Fairbairn > Subject: Re: if (++) were left associative ? > > > On Sun, 7 Apr 2002, Jon Fairbairn wrote: > > > The answer is that you have to evaluate ((([] ++ [1]) ++ > > [2]) ++ [3]) before you can find it, and to get that, you > > have to evaluate (([] ++ [1]) ++ [2]) and so on. > > > > > > In the other case, you can get the head of (a:...)++ (b ++ > > c) by evaluating only the first steps of the first (++). > > > > > > Does that help? > > > > Jón > > > > > Just to add a bit. "evaluate" is such a fuzzy word, > especially in a lazy > language. You must know the context of the expression in > order to know > how much of it is actually evaluated*. > > (* Eventually, the context of all expressions in haskell will be the > evaluation of something in the IO monad, but generally you don't > have to go that far in your reasoning about contexts and stuff.) > > > Ok. I shall now attempt to explain a bit more deeply the differences > between the evaluation expressions which use a left associative append > versis expressions uing a right associative append. > > > > > > In the following,"tail" means any arbitrary tail of a list. > Also, for the > sake of argument, assume : binds closer than ++ and function > application > so we dont have to mess with so many damn parentheses. (in > reality I don't > think anything binds closer than function application in Haskell.) > > the definition of ++ is like > > [] ++ b = b > (l:ls) ++ b = l:(ls++b) > > > applist = ((w:tail ++ x:tail) ++ y:tail) ++ z:tail > > This can be rewritten as the following, so we can litterally see > the leftmost-outermost-reduction. > > applist = (++) ((++) ((++) w:tail x:tail) y:tail) z:tail > > ick! lets say f=(++) (or that f is the same function as the append > operator (++)) > > applist = f (f (f w:tail x:tail) y:tail) z:tail > > > > Alright. Now say we want the head of this applist. So lets use > the head function. > > head patern matches against that thing, causing > a reduction of the outmost f. Since f > pattern matches against its first arguement, > it must cause a reduction in the second f > in that expression, and likewise the third f. > The third f will succeed in its pattern match, > namely (w:tail). > > Now, f, which was (++), will get unfolded. > using the original definition of (++) > (l:ls) ++ b = l:(ls++b) > > this becomes > (w:tail) ++ (x:tail) = w:(tail ++ x:tail) > > (which in our notation is w:(f tail x:tail)) > > substuting our unfolding back into applist and continuing with > reductions: > > f (f (f w:tail x:tail) y:tail) z:tail > -> f (f w:(f tail x:tail) y:tail) z:tail > -> f w:(f (f tail x:tail) y:tail) z:tail > -> w:(f(f(f tail x:tail) y:tail) z:tail) > > so its not like it will take length (w:tail) + length (x:tail) + > length(y:tail) reductions or whatever, but it does take > 3 reductions to get to the head, AND WILL BY NECESSITY take 3 > reductions to get each element of the rest of the tail of w:tail. > > I >think< in the end, if you need all of applist, you will need > something like > > 3*length (w:tail) + 2*length(x:tail) + length(y:tail) > > reductions of f. > > That is definitely NOT linear with > (length x:tail) + (length y:tail) + (length z:tail) > > > > > Now. Lets try go get the head of > > w:tail ++ (x:tail ++ (y:tail ++ z:tail)) > > doing like before with (++), that is > > f w:tail (f x:tail (f y:tail z:tail)) > > and now f pattern matches, etc. > > > ->w:(f tail (f x:tail (f y:tail z:tail))) > > > and thats the ONLY reduction necessary. > > getting all of applist should only require something close to > > (length x:tail) + (length y:tail) + (length z:tail) > > reductions of f. > > > > I'm sure I'm not %100 accurate here, (and It's not worth the effort > for this message!) but I hope this gives more insight and not > more confusion! > > Cheers, > > > Jay Cox > > >

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RE: if (++) were left associative ?
by Konst Sushenko 07 Apr '02

07 Apr '02

Thanks, but it still does not help... > Well, you've removed the parentheses that give you the > information you want. > Yes, I overlooked the parentheses. I meant what you said below: > foldl (++) [] [[1],[2],[3],[4]] ... > -> (((([] ++ [1]) ++ [2]) ++ [3]) ++ [4]) > > now you have to ask what has to be evaluated in order to get > the head of the result. > > The answer is that you have to evaluate ((([] ++ [1]) ++ > [2]) ++ [3]) before you can find it, and to get that, you > have to evaluate (([] ++ [1]) ++ [2]) and so on. > I understand that I have to evaluate all that to get the head of the result, I just do not see how this ends up being quadratic. I have no problem seeing that if each function's argument had to be evaluated before the function is called but this is not the case here, right? To simplify, lets have (([] ++ [1]) ++ [2]) ++ [3]. To get the head of result (++) is called recursively three times. The last invocation returns [1] because its left argument is []. The invocation before it (... ++ [2]) receives [1], and can return (1:..) now, so it does. It does not finish constructing its whole result before something is returned to (... ++ [3]). Here is the root of my misunderstanding. The way I see it is that every time an element is output by (++) at the bottom of the invocation chain, it is propagated up the chain. No? konst

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Re: Parsec: No newline after last line and 'many'
by Iavor Diatchki 07 Apr '02

07 Apr '02

hello, does this work? > pLine = many (noneOf "\n") > pLines = pLine `sepEndBy` char '\n' bye iavor On Sun, Apr 07, 2002 at 07:44:42PM +0200, Till Doerges wrote: > Hi folks, > > when trying to write a parser for unix-style textfiles, I stumbled > across the problem having to distinguish between lines w/ a newline > character at the end and lines w/o a newline at the end (i.e. the last > line in a file doesn't have to have it). > > Eventually I want to be able to use my parser 'linep' with 'many' on > no matter which text-files. > > The first version I tried, was > > --- snip --- > skip1 :: Parser() > skip1 = do > anyChar > return() > > linep :: Parser(String) > linep = do { cs <- many (noneOf "\n") > ; skip1 > ; return cs > } > --- snap --- > > Well, this only works, if the last line in the file is also terminated > by '\n'. So, I tried to be smarter using: > > --- snip --- > lineWith :: Parser(String) > lineWith = try (do { cs <- many (noneOf "\n") > ; skip1 > ; return cs > }) > > lineWithOut :: Parser(String) > lineWithOut = many anyChar > > linep :: Parser(String) > linep = lineWith > <|> lineWithOut > <?> "line" > --- snap --- > > But this doesn't do the trick either, instead it'll overflow my stack > if used in conjunction w/ 'many': > > --- snip --- > parse (do {many linep}) "stdin" (unlines ["hallo","ballo"]) > ^CInterrupted. > ReadOrig> parse (do {many linep}) "stdin" ("hallo\nballo") > ^CInterrupted. > --- snap --- > > Avoiding many OTOH works fine: > --- snip --- > parse (do {linep;linep}) "stdin" (unlines ["hallo","ballo"]) > Right "ballo" > ReadOrig> parse (do {linep;linep}) "stdin" ("hallo\nballo") > Right "ballo" > --- snap --- > > Besides these two attempts explicitly spelled out I've made quite a > few others, but none of them did the trick. > > The documentation for 'try' says that it'll only consume input upon > success, but why doesn't it work w/ 'many'? > > I know about the Prelude-functions unlines, lines etc., but they don't > really help, since linep should be used within another parser of mine. > > I am using Parsec as of Jan 25, 2001 (the version included w/ ghc 5.02). > > Any hints/solutions/... would be greatly appreciated! > > Bye -- Till > _______________________________________________ > Haskell mailing list > Haskell(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell -- ---------------------------------------+---------------------------------------- Iavor S. Diatchki | email: diatchki(a)cse.ogi.edu Dept. of Computer Science | web: http://www.cse.ogi.edu/~diatchki OGI School of Science and Engineering | work phone: 5037481631 OHSU | home phone: 5036434085 ---------------------------------------+----------------------------------------

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RE: if (++) were left associative ?
by Konst Sushenko 07 Apr '02

07 Apr '02

Thanks, but you are assuming that foldl evaluates the intermediate result right away. If it were, then I have no problem seeing why time is quadratic. I am concerned about laziness though... My understanding of foldl is thus: foldl (++) [] [[1],[2],[3],[4]] -> foldl (++) ([] ++ [1]) [[2],[3],[4]] -> foldl (++) ([] ++ [1] ++ [2]) [[3],[4]] -> foldl (++) ([] ++ [1] ++ [2] ++ [3]) [[4]] -> foldl (++) ([] ++ [1] ++ [2] ++ [3] ++ [4]) [] -> [] ++ [1] ++ [2] ++ [3] ++ [4] Which brings me back to my original question... > -----Original Message----- > From: David Feuer [mailto:dfeuer@cs.brown.edu] > Sent: Sunday, April 07, 2002 3:21 AM > To: haskell-cafe(a)haskell.org > Subject: Re: if (++) were left associative ? > > > On Sun, Apr 07, 2002, Konst Sushenko wrote: > > Hello, > > > > this is probably embarrassing, but I just realised that I do not > > understand why list concatenation operation takes quadratic > time if it > > is associated from left to right (see e.g. 'The Haskell School of > > Expression' by Hudak, p. 335): > > > > cat1 = (++) > > cat2 = (++) > > > > ( [1,2] `cat1` [3,4] ) `cat2` [5,6] > > > > > > My understanding is that cat2 is invoked before cat1, and > as soon as the > > first element of cat1 is available, cat2 starts building > its own result. > > cat2 does not need to wait until the whole list is emitted > by cat1, does > > it? Where is quadratic time complexity? > > Suppose we define > > listseq [] x = x > listseq (a:b) x = listseq b x > > l = [1..n] > > inter = foldl (++) [] (map (:[]) foo) > > final = listseq inter inter > > > Then inter will be the result of concatenating [1],[2],[3],... from > left to right. One meaningful question is: how long will it take to > calculate "final"? This is (I believe) called the _shared cost_ of > inter (while inter does not actually do any significant work, it > produces a chain of suspensions that together do quite a > bit.... I'm not > sure if the way they are chained still allows this to be > considered the > shared cost, but I'm not sure). > > 1. How long does it take to calculate final ? > Well, let's try it for n = 4: > > inter = foldl (++) [] [[1],[2],[3],[4]] > = foldl (++) ([]++[1]) [[2],[3],[4]] = foldl (++) [1] > [[2],[3],[4]] > = foldl (++) ([1]++[2]) [[3],[4]] = foldl (++) [1,2] [[3],[4]] > = foldl (++) ([1,2]++[3]) [[4]] = foldl (++) [1,2,3] [[4]] > = foldl (++) ([1,2,3]++[4]) [] = foldl (++) [1,2,3,4] [] > = [1,2,3,4] > > Note that in successive iterations, the following are calculated: > > []++[1] > [1]++[2] > [1,2]++[3] > [1,2,3]++[4] > > Since l1++l2 takes order length(l1), each concatenation takes > more time > than the previous one (linearly), so the total time for all of them is > quadratic. >

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Re: Lambda over types.
by anatoli 07 Apr '02

07 Apr '02

(Redirected from haskell to haskell-cafe.) Well, I tried to fix the problems you mention. Results attached. I have managed to do without de Brujin notation. The evaluator is (modulo bugs!) normal order. One can easily do an applicative order evaluator as well. Evaluating to WHNF is probably easy too (this is what the previous version supposed to do). I'll try to come up with a regression test suite for this stuff as my (copious :) free time permits, but on the surface it looks ok. --- oleg(a)pobox.com wrote: > > > anatoli <anatoli at yahoo> wrote: > > > This is all, of course, of purely academical interest. The notation > > is extremely inconvenient to do any real work. I'd rather prefer > > a real, language-supported lambda over types. > > > Or... wait a minute! You did find all those problems; does it mean > > you tried to *use* this stuff for something? Just curious. > > I must start with a profuse apology, because what follows is perhaps > of little relevance to the list. I also propose to re-direct the > follow-ups to the Haskell Cafe. > > I have examined your code and then tried a few examples, some of which > from my code's regression tests. > > I have implemented a compile-time lambda-calculator, in a different > language. I should say, in a meta-language. The output of the > evaluator is a term that can then be compiled and evaluated. The > lambda-calculator acts as a partial evaluator. The calculator > normalizes a term in a pure untyped lambda calculus. The result is > compiled and evaluated in a call-by-value lambda-calculus with > constants. > > Haskell typechecker (augmented with multi-parameter classes with > functional dependencies and let on loose) may do something similar. > > BTW, the meta-language lambda-evaluator code (with the regression tests) > can be found at > http://pobox.com/~oleg/ftp/Computation/rewriting-rule-lambda.txt > The calculator is implemented in CPS, in some sort of extended lambda > calculus. Therefore, the code has three kinds of lambdas: of the source > language, of the transformer meta-language, and of the target > language. The first and the third lambdas are spelled the same and > are the same. > > > _______________________________________________ > Haskell mailing list > Haskell(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell __________________________________________________ Do You Yahoo!? Yahoo! Tax Center - online filing with TurboTax http://taxes.yahoo.com/

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if (++) were left associative ?
by Konst Sushenko 07 Apr '02

07 Apr '02

Hello, this is probably embarrassing, but I just realised that I do not understand why list concatenation operation takes quadratic time if it is associated from left to right (see e.g. 'The Haskell School of Expression' by Hudak, p. 335): cat1 = (++) cat2 = (++) ( [1,2] `cat1` [3,4] ) `cat2` [5,6] My understanding is that cat2 is invoked before cat1, and as soon as the first element of cat1 is available, cat2 starts building its own result. cat2 does not need to wait until the whole list is emitted by cat1, does it? Where is quadratic time complexity? konst

2 1

please help me
by Eric 07 Apr '02

07 Apr '02

Hi there, I am a beginner a haskell,and I have some difficulty in dealing with an assignment that is to translate a string representation of a list of appointments into a list of appointments. For example: strToApps "! 10-11 lecture 12-13 lunch at the refec :(" Would yield: [((10,11),"lecture",True), ((12,13),"lunch at the refec :(",False)] "!" is used to determine that appointment is important or not I have done some work,those are: > type Date = (Day,Month,Year) -- e.g. (1,1,2000) = 1/1/2000 > type Year = Int > type Month= Int > type Day = Int > type Hour = Int > type Minute = Int > type Time = Hour -- only times on the hour for now > type Appointment = (Duration, Note, Important) > type Duration = (Time,Time) -- e.g. (13,14) = 1-2pm, forall (s,f) in Duration: s < f > type Note = String > type Important = Bool > strToApps :: String -> [Appointment] > strToApps s = How can I finish the requirement make use of library functions such as words, unwords, lines and break. I am eager to receive your help, thank you very much. Best Wishes Leo =================================================================== ÐÂÀËÃâ·Ñµç×ÓÓÊÏä (http://mail.sina.com.cn) ÐÂÀË·ÖÀàÐÅÏ¢£º¶þÊÖÊÐ³¡×ßÒ»×ß£¬žÃ³öÊÖÊ±ŸÍ³öÊÖ£¡ (http://classad.sina.com.cn/2shou/)

3 3

Ex 9.9 in Paul Hudak's book
by Ludovic Kuty 05 Apr '02

05 Apr '02

The exercise is short. First, he defines a "fix" function: fix f = f (fix f) Which has the type (a -> a) -> a Then the function "remainder": remainder :: Integer -> Integer -> Integer remainder a b = if a < b then a else remainder (a - b) b The function fix, has far as i understand the matter, has no base case. So if someone wants to evaluate it, it will indefinitely apply the function f to itself, leading the hugs interpreter to its end. That's ok. But next he asks the reader to rewrite the function remainder using fix and i cannot find out how. I tried something like: remainder2 a b = fix (\x -> x - b) but there is no base case too. As soon as the lambda expression wants to evaluate its argument, the interpreter crashes. Can someone help me to solve the problem ? TIA Ludovic Kuty

5 8

newbie:: getting random Ints
by Peter Rooney 04 Apr '02

04 Apr '02

hello all, total newbie to haskell, armed with "the craft of functional programming". i can't seem to generate random numbers. i have read this: http://www.haskell.org/onlinelibrary/random.html and searched the web and archives, and read thru Random.hs (mostly over my head), but have been unable to get any combination of getStdRandom randomR, etc. to work. even the example in the 98 report, import Random rollDice :: IO Int rollDice = getStdRandom (randomR (1,6)) gets me: Main> rollDice Main> after loading the file, which makes me think i'm missing something! TCFP has an example of how to do it yourself, but i can see that my needs are more than met by Random.hs. my various attempts all look a lot like this: Main> getStdGen (random 5) ERROR - Type error in application *** Expression : getStdGen (random 5) *** Term : getStdGen *** Type : IO StdGen *** Does not match : a -> b could someone post an example of how to generate random integers within a range? tia, peter

5 5

Re: hashmap withdrawal and poor haskell style
by Long Goodbye 04 Apr '02

04 Apr '02

module Main where main = do content <- getContents let starstat = oneline content alphabet = ['a'..'z'] count ch = length . filter (==ch) oneline str ch = [ch] ++ " " ++ stars (count ch str) stars x = take x ['*','*'..] sequence $ map (putStrLn . starstat) alphabet Just another example ... I thought you might like an example close to your original code _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp.

1 0

hashmap withdrawal and poor haskell style
by Michal Wallace 04 Apr '02

04 Apr '02

Hello everyone, I just wrote my first haskell program. I started with a simple python program and tried to see if I could port it to haskell. The program reads text from stdin and prints out a histogram of all the letters: """ alphabet = 'abcdefghjiklmnopqrstuvwxyz' def letter_count(lines): res = {} for line in lines: for char in line.lower(): if char not in alphabet: continue if res.has_key(char): res[char] += 1 else: res[char] = 1 return res if __name__=="__main__": import sys hist = letter_count(sys.stdin) for letter in alphabet: print letter, "*" * hist.get(letter, 0) """ Kid stuff, but it took me about 10 hours to port it... :) Here's what I came up with: """ module Main where alphabet = "abcdefghijklmnopqrstuvwxyz" count ch str = length [c | c <- str , c == ch] hist str = [count letter str | letter <- alphabet] oneline ch str = [ch] ++ " " ++ stars (count ch str) stars x = if x == 0 then "" else "*" ++ stars ( x - 1 ) report str ch = do putStrLn ( oneline ch str ) loop f (h:t) = if t == [] then f h else do f h loop f t main = do content <- getContents let rpt letter = report content letter loop rpt alphabet """ Other than ignoring upper case letters, and being really really slow, it seems to work fine in hugs.... One thing I really missed was a hash / dictionary. I tried for about an hour to use Assoc following the examples from PLEAC: http://pleac.sourceforge.net/pleac_haskell/hashes.html ... But I never got it working: :> module Main where :> import Assoc (empty) :> main :: IO() :> main = do line <- getContents :> let w = length line :> count:: AssocDefault String Int :> count = w.foldl (\a s -> a.update s (+1)) empty :> print x -> ERROR "alphahist.hs":6 - Undefined type constructor "AssocDefault" (I couldn't get Assoc to work either, even with "Import Assoc (Assoc)" though I looked in the Assoc.hs file and could see it in there.. This was the most frustrating part of the experiment) Can anyone point me in the right direction here? Also, I'd really like to here anyone's thoughts on the code I have above, especially concercing what I could have done better. :) Thanks! Cheers, - Michal http://www.sabren.net/ sabren(a)manifestation.com ------------------------------------------------------------ Give your ideas the perfect home: http://www.cornerhost.com/ cvs - weblogs - php - linux shell - perl/python/cgi - java ------------------------------------------------------------

8 10

Re: hashmap withdrawal and poor haskell style
by Yoann Padioleau 03 Apr '02

03 Apr '02

Remi Turk <remi(a)abcweb.nl> writes: > > -------------------- > module Main where > > import Char > import FiniteMap > > printCount :: (Char, Integer) -> IO () > printCount (letter, count) > =3D putStrLn $ letter : replicate count '*' > > countLetters :: String -> [(Char, Integer)] > countLetters letters > =3D fmToList $ addListToFM_C (+) emptyFM $ zip letters (repeat 1) > > main =3D getContents >>=3D mapM_ printCount . countLetters . filter isAl= > pha this is not exactly the same semantic than the python version, for example if 'z' is not in the contents, then you will not display 'z' (without star) and you will not respect also the order of the letter (i think) > -------------------- > > Happy Hacking > > Remi > > --=20 > See the light and feel my warm desire, > Run through your veins like the evening sun > It will live but no eyes will see it, > I'll bless your name before I die. > > Key fingerprint =3D CC90 A1BA CF6D 891C 5B88 C543 6C5F C469 8F20 70F4 > > --wRRV7LY7NUeQGEoC > Content-Type: application/pgp-signature > Content-Disposition: inline > > -----BEGIN PGP SIGNATURE----- > Version: GnuPG v1.0.6 (GNU/Linux) > Comment: For info see http://www.gnupg.org > > iD8DBQE8qzpFbF/EaY8gcPQRAm8DAJ44i0avXmnavtDu+OOy1mdrBL88HwCfWsx+ > GFJUUd1zm1wV8CwfqU2TN0k= > =X3DP > -----END PGP SIGNATURE----- > > --wRRV7LY7NUeQGEoC-- > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > -- Yoann Padioleau, INSA de Rennes, France, Opinions expressed here are only mine. Je n'écris qu'à titre personnel. **____ Get Free. Be Smart. Simply use Linux and Free Software. ____**

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03 Apr '02

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Re: hashmap withdrawal and poor haskell style
by Yoann Padioleau 03 Apr '02

03 Apr '02

Michal Wallace <sabren(a)manifestation.com> writes: > module Main where > alphabet = "abcdefghijklmnopqrstuvwxyz" in haskell you can do alphabet = ['a'..'z'] > count ch str = length [c | c <- str , c == ch] can do count c s = length (filter (c ==) s) or more cryptic: count c = length . (filter (c==)) > hist str = [count letter str | letter <- alphabet] > oneline ch str = [ch] ++ " " ++ stars (count ch str) > stars x = if x == 0 > then "" > else "*" ++ stars ( x - 1 ) stars n = replicate n '*' > report str ch = do putStrLn ( oneline ch str ) > loop f (h:t) = if t == [] > then f h > else do f h > loop f t you can use mapM_ main = do content <- getContents mapM_ (\a -> report content a) alphabet > main = do content <- getContents > let rpt letter = report content letter > loop rpt alphabet > """ you dont choose the more efficient strategy cos you parse 26 times the contents of the file, it is better to do (as you did in python) parse 1 and remember the number of occurence. > > Other than ignoring upper case letters, and being really > really slow, it seems to work fine in hugs.... > > One thing I really missed was a hash / dictionary. I > tried for about an hour to use Assoc following the > examples from PLEAC: > > http://pleac.sourceforge.net/pleac_haskell/hashes.html > > > ... But I never got it working: that's because i dont really use standard haskell library, i redefine a library in a file named Common.hs where there is a constructor AssocDefault. So if you want use AssocDefault, they you have to do: module Main where import Prelude hiding (($),(^),(.),(!!),map,take,lookup,.... AS in the example in the pleac section* import Common Put your code here. so it becomes: #!/usr/bin/runhugs -98 module Main where import Prelude hiding (($),(^),(.),(!!),map,take,lookup,drop,splitAt,reverse,filter,takeWhile,dropWhile,null,foldl,length) import Common main = do s <- getContents s.downcase .foldl(\h c -> if isAlpha c then h.update c (+1) else h) (empty::AssocDefault Char Int) .(\h -> ['a'..'z'].each (\c -> putStrLn (c^replicate (h!c) '*'))) or in more standard haskell, i will do: #!/usr/bin/runhugs -98 module Main where import Prelude hiding ((.)) (.) o f = f o update k f xs = xs.map (\ (k2,v) -> if k == k2 then (k,f v) else (k2,v)) each:: (e -> IO ()) -> [e] -> IO () each = mapM_ main :: IO () main = do s <- getContents s.map toLower .foldl(\h c -> if isAlpha c then h.update c (+1) else h) [(c, 0::Int) | c <- ['a'..'z']] .each (\ (c,n) -> putStrLn (show c ++ (replicate n '*' ))) > > > :> module Main where > :> import Assoc (empty) > :> main :: IO() > :> main = do line <- getContents > :> let w = length line > :> count:: AssocDefault String Int > :> count = w.foldl (\a s -> a.update s (+1)) empty > :> print x > > -> ERROR "alphahist.hs":6 - Undefined type constructor "AssocDefault" that's because AssocDefault is a constructor that standard haskell does not define. > > Also, I'd really like to here anyone's thoughts on the code > I have above, especially concercing what I could have done > better. :) > > Thanks! > > Cheers, > > - Michal http://www.sabren.net/ sabren(a)manifestation.com > ------------------------------------------------------------ > Give your ideas the perfect home: http://www.cornerhost.com/ > cvs - weblogs - php - linux shell - perl/python/cgi - java > ------------------------------------------------------------ > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > -- Yoann Padioleau, INSA de Rennes, France, Opinions expressed here are only mine. Je n'écris qu'à titre personnel. **____ Get Free. Be Smart. Simply use Linux and Free Software. ____**

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Re: Ex 9.9 in Paul Hudak's book
by Lennart Augustsson 03 Apr '02

03 Apr '02

> What I would like to know is how the 'fix' function could be used > to find the fixed point of a function like ( \n -> 2*n ). The 'fix' function does compute a (the least) fixed point of ( \n -> 2*n ). The least fixed point is bottom. -- Lennart

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Re: sort inefficiency
by Dylan Thurston 03 Apr '02

03 Apr '02

On Wed, Apr 03, 2002 at 09:35:51AM +0400, Serge D. Mechveliani wrote: > The Standard library specifies only the map related to the name > `sort'. This map can be described, for example, via sort-by-insertion > program. > And the algorithm choice is a matter of each particular > implementation. Implementation has right to change the algorithm. Reading this, it occurred to me that if you're very picky the implementation probably isn't allowed to pick the algorithm: you need to assume that '<' is actually a total order to have much leeway at all. (Suppose, e.g., that comparing two particular elements yields an exception.) It seems to me this is a problem with providing code as specification: you probably fix the details more than you want. Best, Dylan Thurston

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guards
by Cagdas Ozgenc 29 Mar '02

29 Mar '02

Greetings. Is it requried that the type used in guards always Bool? It seems to me that the Haskell language is dependent to Bool type, and Bool type is depending to the Haskell language causing a circular dependency problem. Of course this is not a practical issue. Is there a theoretical explanation? Thanks for taking time.

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Re: Little problem
by Jorge Adriano 27 Mar '02

27 Mar '02

[moving to haskell-cafe] > Hello > I have little problem. I want to solve this equation > > f(n-1) = f(n)+n > f(6)=6 This is a simple recursive definition.. By the way, are you sure you didn't meant: f(n-1)=f(n)+(n-1) > => f(2)=? Not exactly sure what you're after here. You could simply do: f(2) =f(3)+3 =f(4)+4+3 =f(5)+5+4+3 =f(6)+6+5+4+3 =6+6+5+4+3 =24 anyway it's simple to see, or even prove by induction, that f(n) = (\sum_{i=n+1}^{7} i) -1 = (n+8)(7-n)/2 - 1 If I'm rigth and you're formula is suposed to be f(n-1) = f(n)+n it is even easier. J.A.

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Re: Is there a name for this structure?
by dominic.j.steinitz＠britishairways.com 27 Mar '02

27 Mar '02

The name escapes me but any category theory book should supply the answer. Dominic. Tom Pledger <Tom.Pledger(a)peace.com>@haskell.org on 27/03/2002 04:22:29 Sent by: haskell-cafe-admin(a)haskell.org To: haskell-cafe cc: bcc: Subject: Is there a name for this structure? Joe English writes: : | Suppose you have two morphisms f : A -> B and g : B -> A | such that neither (f . g) nor (g . f) is the identity, | but satisfying (f . g . f) = f. Is there a conventional name | for this? Is it equivalent to saying that (f . g) is the identity on the range of f? That's shorter, though still not a snappy single word term. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ------------------------------------------------------------------------------------------------- Save time by using an eTicket and our Self-Service Check-in Kiosks. For more information go to http://www.britishairways.com/eservice1

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Is there a name for this structure?
by Joe English 27 Mar '02

27 Mar '02

Not really a Haskell question, but someone here might know the answer... Suppose you have two morphisms f : A -> B and g : B -> A such that neither (f . g) nor (g . f) is the identity, but satisfying (f . g . f) = f. Is there a conventional name for this? Alternately, same question, but f and g are functors and A and B categories. In some cases (g . f . g) is also equal to g; is there a name for this as well? I find myself running into pairs of functions with this property over and over again, and am looking for a short way to describe the property... Thanks, --Joe English jenglish(a)flightlab.com

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Re: I need some help
by Jorge Adriano 26 Mar '02

26 Mar '02

> > However the user of the program give some argument then program is > > starting. And I want to use this arguments for example at the end of the > > program. What will I do. > > Good question. I've had a bad time with this, some in this list proably got > tired of me saying the same thing over and over again... those of you might > want to skip the rest of my mail :-) huh? what is this line doing here.. I thought I had cut/pasted it below, but instead I copy/pasted it (then changed it). Most of the is pretty readble, I just complain about stuff in the last part of the e-mail :-) J.A.

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Re: I need some help
by Jorge Adriano 26 Mar '02

26 Mar '02

[Obs: moving to haskell cafe] [beware, huge answer follows] > Hello > I am writing a whole program which is analyz a elegtromagnetic wave. This > is ok but in the program I use a lot of constant for example pi,epsilon > etc. I want to ask that > how I use a constant in Haskell ? Defining constants it's easy, examples. -- with no type signature epsilon = 3.1 -- with explicit type signature epsilon2 :: Double epsilon2 = 3.1 -- specifying the type but no type signature epsilon3 = 3.1 :: Double (by the way, wouldn't it make sense not to get a warning about missing type signature in this last case. I know there is no signature there but still...) > However the user of the program give some argument then program is > starting. And I want to use this arguments for example at the end of the > program. What will I do. Good question. I've had a bad time with this, some in this list proably got tired of me saying the same thing over and over again... those of you might want to skip the rest of my mail :-) If you want the user to be able to chose those values, then they are NOT constants, period. Yes their values won't chnage during the rest of the algorithm, but that doesn't makes it constants, so you can't declare them that way. What *can* you do. 1. One *awfull* option is to pass thos values as arguments to all your functions. Yes signatures will get monstruous, it will complicate the developing process etc, 2. Instead of passing all those variables, group them in a record - data type with named fields - and pass that record arround. (http://www.haskell.org/onlinereport/exps.html#sect3.15) Signatures will get better. It's kind of weird though, to always have to pass around that record. Things can get quite messy, IMO. -- foo f :: Double -> Double -> Double f x y = eta*x - epsilon^2*y -- foo2 f :: Options -> Double -> Double -> Double f opt x y = (eta opt)* (epsilon opt)^2*y And this is just a very simple example, it can get much worst. 3. Implicit parameters. (At least in Ghc, what about others?) Now this is the closest to an elegant solution. *Read both*: http://citeseer.nj.nec.com/246042.html http://www.cs.chalmers.se/~rjmh/Globals.ps f :: (?eta::Double, ?epsilon::Double)=> Double -> Double -> Double f x y = ?eta*x - ?epsilon^2*y Now you don't have to specify eta and epsilon in the type signature itself, but in the context - no passing around extra (not natural) parameters. It makes an huge diference IMO when developing your algorithms. Still signatures wil get even bigger than if you just passed them as parameters, you can't even type the context like: (?eta, ?epsilon::Double) So that is just awfull! Once again records will help. f :: (?opt::Options)=> Double -> Double -> Double f x y = (eta ?opt) *x - (epsilon ?opt)^2*y It still lokes quite bad. I *hate* the field projection functions, and I hate the '?'!! So you could simply rename you data opt fields and declare, eta :: (?opt:: Options) => Double eta = etaF ?opt epsilon :: (?opt:: Options) => Double epsilon = epsilonF ?opt Now f could be written like: f :: (?opt::Options)=> Double -> Double -> Double f x y = eta*x - epsilon^2*y Which differs from the first f function only in the context. It is *almost* good enough for me :-) "Why the almost?" you ask... Well, implicit parameters are not only non-standard features, but also experimental features (right?), so you they can change anytime. -------------------------------------------------------------------------------------- NOTE: some of you might wanna quit reading here, I keep complaining about this stuff, you are probably tired of it :-) -------------------------------------------------------------------------------------- In a couple of days you will also want to use more than one record, because grouping certain constants together doesn't feel right, and contexts will get bigger. Also, you will want to use global variables and instead of a record you will use an STRef to a record, and very small signatures will have in they're context something like: (?opt :: STRef s Opt). And then you will need to update some option fields, and you will want to use modifySTRef, and you will need to create an applyField function to every damn field of the record... So you will start asking 1. (when) will we have implicit contexts available? 2. wouldn't it be good to have, besides the data projection functions, some update and apply field functions, automaticly derived when declaring the data type? [yes am actually *asking* if *anyone agrees* with me? haven't had any feedback on this one, so I don't no wether it is a good or just plain dumb idea] IMVHO, it is of extreme importance and *urgent* to get implicit parameters right. This question about how to use this 'user difined constants' is a *simple* question and should have a *very simple* and standard answer. The same goes to global variables. We need them in certain situations and it is supposed to be, and it can be from my experience, *simple*. J.A.

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