- Haskell-Cafe - Haskell.org

hackage, gtk2hs
by Cetin Sert 01 Jan '09

01 Jan '09

Hi, A package I want to upload only builds with the unreleased gtk2hs version from the darcs repository and not the latest released version 0.9.13 or any lesser. But the repository version seems to not have changed, so if I say in my cabal file: gtk >= 0.9.14 it does not even build with the gtk2hs version from the repos, on the other hand if I say: gtk >= 0.9.13 people using 0.9.13 may be surprised to see the package does not build. What should I do o__O? Wait for the next gtk2hs release before releasing my package? Or is there a way to notify _potential users/those interested_ of the need to use the latest gtk2hs they can get from the darcs repos. Best Regards, Cetin Sert

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definition of data
by Max.cs 01 Jan '09

01 Jan '09

hi all, I want to define a data type Tree a that can either be a or Branch (Tree a) (Tree a)? I tried data Tree a = a | Branch (Tree a) (Tree a) deriving Show but it seems not accpetable in haskell ? any way I could achieve this ? Thanks max

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ANN: gitit-0.4.1, recaptcha-0.1
by John MacFarlane 01 Jan '09

01 Jan '09

I'm pleased to announce the release of gitit-0.4.1, which I've just uploaded to HackageDB. Gitit is a wiki program that stores pages in a git repostory. Gitit now has support for (optional) captchas, using the reCAPTCHA service. I've packaged up the reCAPTCHA code as a separate library on HackageDB, recaptcha. Upgrading from older versions of gitit: The format of gitit's user database has changed. (gitit now generates a new random salt for each user, instead of using a single static salt for all users.) Unfortunately, this means that current users of gitit (yes, all seven of you) will have to delete your gitit-users file and have your users create new accounts. Sorry about the breaking change, but better now than later. When you upgrade, you'll also need to delete the _local directory, since changes have been made to the data structure that holds the application state. This shouldn't have any ill effects, since everything of lasting importance (users, pages) is stored elsewhere. If you use a configuration file, or if you want to start using reCAPTCHA (which requires a configuration file), you will also have to add fields for the captcha system. See data/SampleConfig.hs for the format. John

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Re: [Haskell-cafe] bottom case in proof by induction
by raeck＠msn.com 01 Jan '09

01 Jan '09

I am afraid I am still confused. > foo [] = ... > foo (x:xs) = ... > There is an implied: > foo _|_ = _|_ > The right side cannot be anything but _|_. If it could, then that would imply we could solve the halting problem: in a proof, how I could say the right side must be _|_ without defining foo _|_ = _|_ ? and in the case of > bad () = _|_ > bad _|_ = () mean not every function with a _|_ input will issue a _|_ output, so we have to say what result will be issued by a _|_ input in the definitions of the functions if we want to prove the equvalence between them? However, in the case of map f _|_ , I do believe the result will be _|_ since it can not be anything else, but how I could prove this? any clue? ps, the definition of map does not mention anything about _|_ . Thanks Raeck From: Luke Palmer Sent: Wednesday, December 31, 2008 10:43 PM To: Max.cs ; raeck(a)msn.com Subject: Re: [Haskell-cafe] bottom case in proof by induction On Wed, Dec 31, 2008 at 3:28 PM, Max.cs <max.cs.2009(a)googlemail.com> wrote: thanks Luke, so you mean the _|_ is necessary only when I have defined the pattern _|_ such as foo [] = [] foo _|_ = _|_ foo (x:xs) = x( foo xs ) -- consider non-termination case That is illegal Haskell. But another way of putting that is that whenever you do any pattern matching, eg.: foo [] = ... foo (x:xs) = ... There is an implied: foo _|_ = _|_ The right side cannot be anything but _|_. If it could, then that would imply we could solve the halting problem: halts () = True halts _|_ = False Because _|_ is the denotation of a program which never halts. To do it a bit more domain-theoretically, I'll first cite the result that every function has a fixed point. That is, for every f, there is a function fix f, where fix f = f (fix f). (The fix function is actually available in Haskell from the module Data.Function). Then let's consider this bad function: bad () = _|_ -- you can't write _|_ in Haskell, but "undefined" or "let x = x in x" mean the same thing bad _|_ = () Then what is fix f? Well, it either terminates or it doesn't, right? I.e. fix f = () or fix f = _|_. Taking into account that fix f = f (fix f): If it does: fix f = () = f () = _|_, a contradiction. If it doesn't: fix f = _|_ = f _|_ = (), another contradiction. >From a mathematical perspective, that's why you can't pattern match on _|_. >From an operational perspective, it's just that _|_ means "never terminates", and we can't determine that, because we would try to run it "until it doesn't terminate", which is meaningless... Luke

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bottom case in proof by induction
by raeck＠msn.com 01 Jan '09

01 Jan '09

Dear all, Happy New Year! I am learning the Induction Proof over Haskell, I saw some proofs for the equivalence of two functions will have a case called 'bottom' but some of them do no have. What kind of situation we should also include the bottom case to the proof? How about the functions do not have the 'bottom' input such as: foo [] = [] foo (x:xs) = x : (foo xs) thank you, Raeck

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ANN: monte-carlo-0.2, gsl-random-0.2.3
by Patrick Perry 01 Jan '09

01 Jan '09

I've released a new version of the monte-carlo packages for haskell. Here are the highlights for monte-carlo: Changes in 0.2: * More general type class, MonadMC, which allows all the functions to work in both MC and MCT monads. * Functions to sample from discrete distributions. * Functions to sample subsets For a quick tutorial, see my blog post at http://quantile95.com/2008/12/31/monte-carlo-poker-odds/ Happy New Year, everyone! Patrick

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Re: [Haskell-cafe] about the concatenation on a tree
by Henk-Jan van Tuyl 31 Dec '08

31 Dec '08

On Wed, 31 Dec 2008 17:19:09 +0100, Max cs <max.cs.2009(a)googlemail.com> wrote: > Hi Henk-Jan van Tuyl, > > Thank you very much for your reply! > > I think the concatenation should be different to thhe > > treeConcat :: Tree a -> Tree a -> Tree a > > the above is a combination of two trees instead of a concatenation, so > I think the type of treeConcat should be: > > treeConcat :: Tree (Tree a) -> Tree a > > instead. How do you think? : ) I tried to implement it .. but it seems > confusing.. to me > > Thanks > > Max Hello Max, The function treeConcat :: Tree (Tree a) -> Tree a cannot be created, as it has an infinite type; you can however, define a function that replaces leafs with trees, for example treeConcat' in the following module, that replaces all leaves that contains a one with a given tree: > module TreeConcat where > data Tree a = Leaf a | Branch (Tree a) (Tree a) > deriving Show > treeConcat' :: Num a => Tree a -> Tree a -> Tree a > treeConcat' (Leaf 1) tree = tree > treeConcat' (Leaf x) _ = Leaf x > treeConcat' (Branch x y) tree = Branch (treeConcat' x tree) (treeConcat' > y tree) > main :: IO () > main = print $ treeConcat' (Branch (Leaf 1) (Leaf 2)) (Branch (Leaf 3) > (Leaf 4)) This displays: Branch (Branch (Leaf 3) (Leaf 4)) (Leaf 2) If this doen't help you either, I need to know more about what you are trying to do. Regards, Henk-Jan van Tuyl -- http://functor.bamikanarie.com http://Van.Tuyl.eu/ -- > > > On Wed, Dec 31, 2008 at 3:33 PM, Henk-Jan van Tuyl > <hjgtuyl(a)chello.nl>wrote: > >> Hi Max, >> >> A simple way to do this: >> >> module TreeConcat where >>> >> >> data Tree a = Leaf a | Branch (Tree a) (Tree a) >>> deriving Show >>> >> >> treeConcat :: Tree a -> Tree a -> Tree atreeConcat xs ys = Branch xs ys >>> >> >> main :: IO () >>> main = print $ treeConcat (Leaf 1) (Leaf 2) >>> >> >> But perhaps you want a certain ordering? Have a look at: >> >> http://hackage.haskell.org/packages/archive/AvlTree/4.2/doc/html/Data-Tree-… >> >> >> -- >> Regards, >> Henk-Jan van Tuyl >> >> >> -- >> http://functor.bamikanarie.com >> http://Van.Tuyl.eu/ >> -- >> >> >> --

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about the concatenation on a tree
by Max cs 31 Dec '08

31 Dec '08

hi all, not sure if there is someone still working during holiday like me : ) I got a little problem in implementing some operations on tree. suppose we have a tree date type defined: data Tree a = Leaf a | Branch (Tree a) (Tree a) I want to do a concatenation on these tree just like the concat on list. Anyone has idea on it? or there are some existing implementation? Thank you and Happy New Year! regards, Max

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Re: [Haskell-beginners] about the concatenation on a tree
by Thomas Davie 31 Dec '08

31 Dec '08

On 31 Dec 2008, at 16:02, Max cs wrote: > hi all, not sure if there is someone still working during holiday > like me : ) > > I got a little problem in implementing some operations on tree. > > suppose we have a tree date type defined: > > data Tree a = Leaf a | Branch (Tree a) (Tree a) > > I want to do a concatenation on these tree just like the concat on > list. > Anyone has idea on it? or there are some existing implementation? > > Thank you and Happy New Year! > How would you like to concatenate them? Concatonation on lists is easy because there's only one end point to attach the next list to, on a tree though, there are many leaves to attach things to. Here's a few examples though: Attaching to the right most point on the tree (tree data structure modified to store data in branches not leaves here) data Tree a = Leaf | Branch (Tree a) a (Tree a) concatT :: [Tree a] -> Tree a concatT = foldr1 appendT appendT :: Tree a -> Tree a -> Tree a appendT Leaf t = t appendT (Branch l x r) t = Branch l x (appendT r t) Attaching to *all* the leaves on the tree (same modification to the data structure) concatT :: [Tree a] -> Tree a concatT = foldr1 appendT appendT :: Tree a -> Tree a -> Tree a appendT Leaf t = t appendT (Branch l x r) t = Branch (appendT l t) x (appendT r t) merging a list of trees maintaining them as ordered binary trees concatT :: Ord a => [Tree a] -> Tree a concatT = foldr1 unionT unionT :: Ord a => Tree a -> Tree a -> Tree a unionT t = foldrT insertT t foldrT :: (a -> b -> b) -> b -> Tree a -> b foldrT f z Leaf = z foldrT f z (Branch l x r) = f x (foldrT f (foldrT f z r) l) insertT :: Ord a => a -> Tree a -> Tree a insertT x Leaf = Branch Leaf x Leaf insertT x (Branch l y r) | x <= y = Branch (insertT x l) y r | otherwise = Branch l y (insertT x r) Hope that helps. Bob

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Re: Gitit - Encoding
by Arnaud Bailly 31 Dec '08

31 Dec '08

Hi Jeremy and John, Thank you very much Jeremy for your very detailed answer ! I checked in Gitit source and found various occurences of usage of Byte8 strings, which John as confirmed. I switched template.html to advertising encoding of iso-8859-1 and things are working ok. I setup LANG (and LC_ALL) to fr_FR.UTF-8. When I set meta tag with utf-8 encoding, things get messy. John: I checked your url and, oddly enough, characteres appears OK when I manually change page display encoding to latin-1. I came to the possibly obvious conclusion that: -gitit communicated with HApps hence the outside world in latin-1 - talk to git in utf-8 Is this the right picture ? If so, then I have part of the problem solved. Using latin-1 for input/display is annoying only for the € (euro) symbol. Regards, Arnaud BTW, my accents problem solved, I would like to work with RSS feeds so that: 1. a user could put a feed inside a page 2. gitit could produce feeds for changes/adds I have done things like that before in Java but would need some directions to incorporate it in gitit. -- Arnaud Bailly, PhD OQube - Software Engineering web> http://www.oqube.com

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