RE: lazy comparison for equality ?
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:[EMAIL PROTECTED]] Sent: Wednesday, April 24, 2002 12:38 PM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] 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| [EMAIL PROTECTED] than astronomy is about telescopes. -Dijkstra | www.isi.edu/~hdaume On Wed, 24 Apr 2002 [EMAIL PROTECTED] 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 [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: finding ....
Moved to haskell café I do not have a code example, but how about a system that compiles different (and say independent) software projects in parallel, and which uses the same code to create temporary files needed for compilation? konst -Original Message- From: David Sankel [mailto:[EMAIL PROTECTED]] Sent: Tuesday, March 19, 2002 6:38 PM To: Pixel; Lennart Augustsson Cc: [EMAIL PROTECTED] Subject: Re: finding --- Pixel [EMAIL PROTECTED] wrote: Lennart Augustsson [EMAIL PROTECTED] writes: Diego Yanivello wrote: hi,is there (in Haskell) a function like existFile :: FilePath - IO (Bool) ? Thanks! Using such a function is generally a bad idea because of race conditions. however, real world programs use those tests since you don't need to care *everytime* about race conditions. (of course using this existFile before creating a temporary file is wrong, but existFile has *many* other applications) Could someone post an example of the creation of a temporary file where race conditions are important? Thanks, David J. Sankel __ Do You Yahoo!? Yahoo! Sports - live college hoops coverage http://sports.yahoo.com/ ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: Hiring Haskell programmers
I have always been wondering what exactly does quickly learn Haskell mean? Quickly learn Haskell syntax? Can one learn how to paint quickly? konst -Original Message- From: Eray Ozkural [mailto:[EMAIL PROTECTED]] Sent: Monday, March 11, 2002 4:24 PM To: Mark Carroll; [EMAIL PROTECTED] Subject: Re: Hiring Haskell programmers -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On Tuesday 12 March 2002 01:17, Mark Carroll wrote: How easy is it to hire reasonable Haskell programmers? Of course, this may mean, hiring people with the aptitude and interest to quickly learn Haskell. Has anyone any experience of this that they can share? I know it's easy. Try me. ;) Sincerely, - -- Eray Ozkural (exa) [EMAIL PROTECTED] Comp. Sci. Dept., Bilkent University, Ankara www: http://www.cs.bilkent.edu.tr/~erayo GPG public key fingerprint: 360C 852F 88B0 A745 F31B EA0F 7C07 AE16 874D 539C -BEGIN PGP SIGNATURE- Version: GnuPG v1.0.6 (GNU/Linux) Comment: For info see http://www.gnupg.org iD8DBQE8jUqcfAeuFodNU5wRAt1/AKCaN881FMXqzpx+xh1EpnFa/b6k9ACeNuPX aYrkL3CfiF2C6uBzH+3chPs= =U7Xm -END PGP SIGNATURE- ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
order of evaluation ?
Title: Message hello, below is the code that i wrote as an excercise for myself (I am still learning haskell). it implements a straighforward way to simplify boolean expressions, and should be self-explanatory. my question is, if i have an expression such as ((Const False) :: subexp), will subexp be reduced first (according to the definition 'simplify (x :: y) = simplify' ((simplify x) :: (simplify y))') or will laziness do the right thing, and emit (Const False) without looking into exp? i think the latter, but would appreciate a word from an expert. thanks konst PS: any coding suggestions, etc. are also welcome infixr 3 ::infixr 2 :|: data Exp = Const Bool | Sym String | Not Exp | Exp :: Exp | Exp :|: Exp instance Eq Exp where (Const x) == (Const y) = x==y (Sym x) == (Sym y) = x==y (Not x) == (Not y) = x==y (x :: y) == (u :: v) = x==u y==v || x==v y==u (x :|: y) == (u :|: v) = x==u y==v || x==v y==u _ == _ = False simplify (x :: y) = simplify' ((simplify x) :: (simplify y))simplify (x :|: y) = simplify' ((simplify x) :|: (simplify y))simplify (Not x) = simplify' (Not (simplify x))simplify x = x simplify' (Not (Const True)) = Const Falsesimplify' (Not (Const False)) = Const True simplify' (Not (Not x)) = x simplify' ((Not x) :: y) | x==y = Const Falsesimplify' (x :: (Not y)) | x==y = Const Falsesimplify' ((Not x) :|: y) | x==y = Const Truesimplify' (x :|: (Not y)) | x==y = Const True simplify' ((Const False) :: _) = Const Falsesimplify' (_ :: (Const False)) = Const Falsesimplify' ((Const True) :: x) = xsimplify' (x :: (Const True)) = x simplify' ((Const True) :|: _) = Const Truesimplify' (_ :|: (Const True)) = Const Truesimplify' ((Const False) :|: x) = xsimplify' (x :|: (Const False)) = x simplify' (x :: y) | x==y = xsimplify' (x :|: y) | x==y = x simplify' x = x
RE: efficiency question
(moved to haskell-café) I ran Hal's code on my computer, and with test2 I get a stack overflow (so I had to use +RTS option for it to finish). test1 does not overflow stack (of standard size, I mean without +RTS). Which implies that test2 uses more stack space than test1. why would it use more stack if not because of laziness? konst -Original Message- From: Hal Daume III [mailto:[EMAIL PROTECTED]] Sent: Friday, February 08, 2002 4:35 PM To: Jorge Adriano Cc: Konst Sushenko; [EMAIL PROTECTED] Subject: Re: efficiency question I agree that it's the overhead of (,), but I don't see why there would be any overhead for doing this. -- Hal Daume III Computer science is no more about computers| [EMAIL PROTECTED] than astronomy is about telescopes. -Dijkstra | www.isi.edu/~hdaume On Sat, 9 Feb 2002, Jorge Adriano wrote: On Friday 08 February 2002 23:52, Hal Daume III wrote: I've tried using a strict fold: foldl' f a [] = a foldl' f a (x:xs) = (foldl' f $! f a x) xs but that has no effect (or minimal effect). That wouldn't work even if if laziness is the problem because that would only cause the elements of the list to be evaluated to head normal form, the elements of the pair would not be evaluated so you'd have a 'suspension of (minus and plus) operations'. instead of (\x (a,b) - (x+a,x-b)) try (\x (a,b) - (((,) $! x-a)$! x-b) ) I just noticed that you were the one who sent me the DeepSeq module. This is the kind of place where I want to use it. Instead of $!, try $!!. And Konst Sushenko wrote: My guess is that it is due to the laziness of the addition/subtraction in (,) Seems to me like lazyness is not the right guess because both functions Hall first posted were lazy. So I think it's just the overhead of applying (,) besides (+) and (-) in each step. Do I make sense or am I missing something? J.A. ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
examples using built-in state monad
hello, in my program i used my own parameterised state transformer monad, which iswell described in literature: newtype State s m a = ST (s - m (a,s)) ghc and hugs contain built in implementation of state monad ST. is it the same thing? the documentation is not clear on that. if it is the same, is it faster? also, could someoneplease recommend any samples that use the built in ST monad? thanks konst
RE: newbie: running a state transformer in context of a state reader
Marcin,
thanks for your help.
to implement the lift functionality i added these well
known definitions:
class (Monad m, Monad (t m)) = TransMonad t m where
lift :: m a - t m a
instance (Monad m, Monad (State s m)) = TransMonad (State s) m where
lift m = ST (\s - m = (\a - return (a,s)))
but my lookahead function
lookahead p = do { s - fetch
; lift (evalState p s)
}
is typed as
lookahead :: State MyState Maybe a - State MyState Maybe (a,MyState)
but i need
lookahead :: State MyState Maybe a - State MyState Maybe a
apparently, the (=) and return used in the definition of lift above are
for the monad (State s m), and not monad m...
everything works if i do not use the TransMonad class, but define lift
manually as:
lift :: Parser a - Parser a
lift m = ST (\s - unST m s = (\(a,_) - return (a,s)))
but this looks like a special case of the lift above, except the right hand
side of
'bind' is executed in the right context.
i am still missing something
konst
-Original Message-
From: Marcin 'Qrczak' Kowalczyk [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, February 20, 2001 10:17 AM
To: [EMAIL PROTECTED]
Subject: Re: newbie: running a state transformer in context of a state
reader
Mon, 19 Feb 2001 18:07:17 -0800, Konst Sushenko [EMAIL PROTECTED] pisze:
now i am curious if it is possible to run the given parser (state
transformer) in a context of a state reader somehow, so as the state
gets preserved automatically. something that would let me omit the
calls to fetch and set methods.
It should be possible to do something like this:
lookahead:: Parser a - Parser a
lookahead p = do { s - fetch
; lift (evalState p s)
}
where evalState :: Monad m = State s m a - s - m a
lift :: Monad m = m a - State s m a
are functions which should be available or implementable in a monad
transformer framework. I don't have the Hutton/Meijer's paper at hand
so I don't know if they provided them and under which names. Such
functions are provided e.g. in the framework provided with ghc (by
Andy Gill, inspired by Mark P Jones' paper "Functional Programming
with Overloading and Higher-Order Polymorphism").
This definition of lookahead uses a separate state transformer thread
instead of making changes in place and undoing them later. I don't
think that it could make sense to convert a state transformer to
a state reader by replacing its internals, because p does want to
transform the state locally; a value of type Parser a represents
a state transformation. The changes must be isolated from the main
parser, but they must happen in some context.
--
__(" Marcin Kowalczyk * [EMAIL PROTECTED] http://qrczak.ids.net.pl/
\__/
^^ SYGNATURA ZASTEPCZA
QRCZAK
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need help w/ monad comprehension syntax
hello, i am having troublegetting my program below to work. i think i implemented the monad methods correctly, but the function 'g' does not type as i would expect. Hugs thinks that it is just a list (if i remove the explicit typing). i want it to be functionally identical to the function 'h'. what am i missing? thanks konst newtype State s a = ST (s - (a,s)) unST (ST m) = m instance Functor (State s) where fmap f m = ST (\s - let (a,s') = unST m s in (f a, s')) instance Monad (State s) where return a = ST (\s - (a,s)) m = f = ST (\s - let (a,s') = unST m s in unST (f a) s') --g :: State String Char g = [ x | x - return 'a' ] h :: State String Char h = return 'a'
