March 2009 - Haskell-Cafe

Re: [Haskell-cafe] Rational and % operator remix
by michael rice 30 Mar '09

30 Mar '09

cf2 :: Rational -> [Int] cf2 a = let ai = floor a <-- Doesn't this make ai an Int? -Michael in if a == (toRational ai) then [ai] else ai : cf2 (1 / (a - ai)) --- On Sun, 3/29/09, Brandon S. Allbery KF8NH <allbery(a)ece.cmu.edu> wrote: From: Brandon S. Allbery KF8NH <allbery(a)ece.cmu.edu> Subject: Re: [Haskell-cafe] Rational and % operator remix To: "michael rice" <nowgate(a)yahoo.com> Cc: "Brandon S. Allbery KF8NH" <allbery(a)ece.cmu.edu>, "Lennart Augustsson" <lennart(a)augustsson.net>, haskell-cafe(a)haskell.org Date: Sunday, March 29, 2009, 4:27 PM On 2009 Mar 29, at 16:19, michael rice wrote:but I'm still at least one error short of a clean run Main> :load cf.hs ERROR "cf.hs":7 - Type error in application *** Expression : ai : cf2 (1 / (a - ai)) *** Term : ai *** Type : Ratio Integer *** Does not match : Int Where did I go wrong or what did I leave out? cf2 returns [Int], but you're using it as the tail of a list starting with ai, which is a Rational. Need to decide which you actually want. At a guess: > cf2 a = let ai = floor a> rai = toRational ai> in> if a == rai> then [ai]> else rai : cf2 (1 / (a - ai)) -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery(a)kf8nh.comsystem administrator [openafs,heimdal,too many hats] allbery(a)ece.cmu.eduelectrical and computer engineering, carnegie mellon university KF8NH

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Is Text.XHtml.Table usable?
by Thomas Hartman 29 Mar '09

29 Mar '09

I was playing with Text.XHtml.Table but couldn't use it to output tables. ( cell . toHtml $ " a " ) `beside` (cell . toHtml $ " b " ) <tr > a b </tr > already seems wrong -- should be two cells, right? And the result doesn't get embedded in a table tag? Is there something I'm missing? Working code samples would be great.

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Re: [Haskell-cafe] Rational and % operator remix
by michael rice 29 Mar '09

29 Mar '09

Works like a charm. Here's a final copy for those interested. Michael ===== import Data.Ratio cf2 :: Rational -> [Int] cf2 a = let ai = floor a air = toRational ai in ai : if a == air then [] else cf2 (1 / (a - air)) --- On Sun, 3/29/09, Miguel Mitrofanov <miguelimo38(a)yandex.ru> wrote: From: Miguel Mitrofanov <miguelimo38(a)yandex.ru> Subject: Re: [Haskell-cafe] Rational and % operator remix To: "michael rice" <nowgate(a)yahoo.com> Cc: "Lennart Augustsson" <lennart(a)augustsson.net>, haskell-cafe(a)haskell.org Date: Sunday, March 29, 2009, 4:25 PM cf2 a = let ai = floor a air = toRational ai in ai : if a == air then [] else cf2 (1 / (a - air)) On 30 Mar 2009, at 00:19, michael rice wrote: > I'm a Lisper, kind of feeling my way around here in Haskell, so please bear with me. > > I did the things you suggested, but I think the last one may have gotten garbled. Anyway, this is what I ended up with > > cf2 :: Rational -> [Int] > cf2 a = let ai = floor a > in > if a == (toRational ai) > then [ai] > else ai : cf2 (1 / (a - ai)) > > but I'm still at least one error short of a clean run > > Main> :load cf.hs > ERROR "cf.hs":7 - Type error in application > *** Expression : ai : cf2 (1 / (a - ai)) > *** Term : ai > *** Type : Ratio Integer > *** Does not match : Int > > > Where did I go wrong or what did I leave out? > > Michael > > > --- On Sun, 3/29/09, Lennart Augustsson <lennart(a)augustsson.net> wrote: > > From: Lennart Augustsson <lennart(a)augustsson.net> > Subject: Re: [Haskell-cafe] Rational and % operator remix > To: "michael rice" <nowgate(a)yahoo.com> > Cc: haskell-cafe(a)haskell.org > Date: Sunday, March 29, 2009, 2:29 PM > > You can use floor in a Rational directly, no need to take it apart and divide. > > There is no need to write (toRational 1), just write 1. > > Don't write (subtract ai a), write (ai - i). > > You also have a type error; the ai should no be a Rational, so you > need to move to toRational call to the comparison. > > -- Lennart > > 2009/3/29 michael rice <nowgate(a)yahoo.com>: > > Hi, > > > > Thanks again for the help last night. > > > > The second function cf2 is an attempt to reverse the process of the first > > function, i.e., given a rational number it returns a list of integers, > > possibly infinite, but you shouldn't get into trouble if you use 98%67 as > > input (output should be [1,2,6,5]). The interpreter is complaining about the > > '=' following the 'in' keyword. Is there a better way to state this? > > > > Michael > > > > import Data.Ratio > > cf :: [Int] -> Rational > > cf (x:[]) = toRational x > > cf (x:xs) = toRational x + 1 / cf xs > > > > cf2 :: Rational -> [Int] > > cf2 a = let ai = toRational (floor ((numerator a) / (denominator a))) > > in > > if a = ai > > then [a] > > else ai : cf2 ((toRational 1) / (subtract ai a)) > > > > > > > > _______________________________________________ > > 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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Re: [Haskell-cafe] Rational and % operator remix
by michael rice 29 Mar '09

29 Mar '09

I'm a Lisper, kind of feeling my way around here in Haskell, so please bear with me. I did the things you suggested, but I think the last one may have gotten garbled. Anyway, this is what I ended up with cf2 :: Rational -> [Int] cf2 a = let ai = floor a in if a == (toRational ai) then [ai] else ai : cf2 (1 / (a - ai)) but I'm still at least one error short of a clean run Main> :load cf.hs ERROR "cf.hs":7 - Type error in application *** Expression : ai : cf2 (1 / (a - ai)) *** Term : ai *** Type : Ratio Integer *** Does not match : Int Where did I go wrong or what did I leave out? Michael --- On Sun, 3/29/09, Lennart Augustsson <lennart(a)augustsson.net> wrote: From: Lennart Augustsson <lennart(a)augustsson.net> Subject: Re: [Haskell-cafe] Rational and % operator remix To: "michael rice" <nowgate(a)yahoo.com> Cc: haskell-cafe(a)haskell.org Date: Sunday, March 29, 2009, 2:29 PM You can use floor in a Rational directly, no need to take it apart and divide. There is no need to write (toRational 1), just write 1. Don't write (subtract ai a), write (ai - i). You also have a type error; the ai should no be a Rational, so you need to move to toRational call to the comparison. -- Lennart 2009/3/29 michael rice <nowgate(a)yahoo.com>: > Hi, > > Thanks again for the help last night. > > The second function cf2 is an attempt to reverse the process of the first > function, i.e., given a rational number it returns a list of integers, > possibly infinite, but you shouldn't get into trouble if you use 98%67 as > input (output should be [1,2,6,5]). The interpreter is complaining about the > '=' following the 'in' keyword. Is there a better way to state this? > > Michael > > import Data.Ratio > cf :: [Int] -> Rational > cf (x:[]) = toRational x > cf (x:xs) = toRational x + 1 / cf xs > > cf2 :: Rational -> [Int] > cf2 a = let ai = toRational (floor ((numerator a) / (denominator a))) > in > if a = ai > then [a] > else ai : cf2 ((toRational 1) / (subtract ai a)) > > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > >

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Re: [Haskell-cafe] Rational and % operator remix
by michael rice 29 Mar '09

29 Mar '09

Correct, if it can be stated as a Rational then it terminates. I was messing around last night with the first function on a sequence that approximates sqrt(2) cf (take 25 (1: [2,2..])) forgetting that while the sequence is infinite, I only grabbed the first 25 elements. Your replacement code works fine except for a missing import Data.Ratio It's quite removed from what I was trying so I'm going to have to read ahead a bit to understand it. Thanks, Michael --- On Sun, 3/29/09, Daniel Fischer <daniel.is.fischer(a)web.de> wrote: From: Daniel Fischer <daniel.is.fischer(a)web.de> Subject: Re: [Haskell-cafe] Rational and % operator remix To: haskell-cafe(a)haskell.org Cc: "michael rice" <nowgate(a)yahoo.com> Date: Sunday, March 29, 2009, 2:35 PM -----Inline Attachment Follows----- Am Sonntag 29 März 2009 19:40:19 schrieb michael rice: > Hi, > > Thanks again for the help last night. > > The second function cf2 is an attempt to reverse the process of the first > function, i.e., given a rational number it returns a list of integers, > possibly infinite, Not for rational numbers. > but you shouldn't get into trouble if you use 98%67 as > input (output should be [1,2,6,5]). The interpreter is complaining about > the '=' following the 'in' keyword. That should be '=='. > Is there a better way to state this? > > Michael > > import Data.Ratio > cf :: [Int] -> Rational > cf (x:[]) = toRational x > cf (x:xs) = toRational x + 1 / cf xs > > cf2 :: Rational -> [Int] > cf2 a = let ai = toRational (floor ((numerator a) / (denominator a))) > in > if a = ai > then [a] > else ai : cf2 ((toRational 1) / (subtract ai a)) import Data.List (unfoldr) cf3 :: Rational -> [Integer] -- Int may overflow cf3 0 = [0] cf3 x = a0:unfoldr f r where a0 = floor x r = x - fromInteger a0 f 0 = Nothing f y = Just (properFraction $ recip y)

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Template Haskell messes up scoping?
by Stephan Friedrichs 29 Mar '09

29 Mar '09

Hi, when I tried to reorganize some code to use the data-accessor and data-accessor-template packages, i stumbled across a strange effect: When using template haskell some things are out of scope that really shouldn't be. Let me give an example: ============== T.hs ============== {-# LANGUAGE TemplateHaskell #-} data Foo = Foo { bar :: Bar } $( return [] ) data Bar = Bar ================================== > ghci T.hs GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Main ( T.hs, interpreted ) T.hs:4:13: Not in scope: type constructor or class `Bar' Failed, modules loaded: none. No matter what you do in $( ... ), the code doesn't compile. Even in the simple case above, it doesn't work. What's happening here? Bug or feature? And do you know a workaround? //Stephan -- Früher hieß es ja: Ich denke, also bin ich. Heute weiß man: Es geht auch so. - Dieter Nuhr

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Rational and % operator remix
by michael rice 29 Mar '09

29 Mar '09

Hi, Thanks again for the help last night. The second function cf2 is an attempt to reverse the process of the first function, i.e., given a rational number it returns a list of integers, possibly infinite, but you shouldn't get into trouble if you use 98%67 as input (output should be [1,2,6,5]). The interpreter is complaining about the '=' following the 'in' keyword. Is there a better way to state this? Michael import Data.Ratio cf :: [Int] -> Rational cf (x:[]) = toRational x cf (x:xs) = toRational x + 1 / cf xs cf2 :: Rational -> [Int] cf2 a = let ai = toRational (floor ((numerator a) / (denominator a))) in if a = ai then [a] else ai : cf2 ((toRational 1) / (subtract ai a))

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Re: [Haskell-cafe] Re: Exception handling in numeric computations
by Jonathan Cast 29 Mar '09

29 Mar '09

On Sat, 2009-03-28 at 12:51 +0300, Gregory Petrosyan wrote: > On Sat, Mar 28, 2009 at 10:53 AM, Ketil Malde <ketil(a)malde.org> wrote: > > So the difference between an exception or an error type is mainly what > > you intend to do about it. There's no point in wrapping divisions in > > Maybe unless you actually are able to do something useful to recover > > from a zero denominator. > > That is exactly the point I was trying to make. > > When I write a code, I can't say in advance, in what way it will be used. > So, for dealing with errors, I have to choose one way or another, mostly > without that knowledge. When I'm using e.g. C++, it's easy: > something like mantra "when in doubt, throw an exception" :-) > combined with RAII, works good (but not ideal, of course). > > So, I'll ask again: when I program in Haskell, what mechanism should I use? If you don't know, use a (true) exception. That is, Left or Exception or throwIO. Only use error or throw when you *know* the condition is un-recoverable. jcc

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Re: [Haskell-cafe] type Rational and the % operator
by michael rice 29 Mar '09

29 Mar '09

This seems to work: import Data.Ratio cf :: [Integer] -> Rational cf (x:[]) = toRational x cf (x:xs) = toRational x + (1 / cf xs) Hugs> :load cf.hs Main> cf [1,2,6,5] 98 % 67 Main> cf [2,3] 7 % 3 Main> Thanks, guys! Michael --- On Sun, 3/29/09, Luke Palmer <lrpalmer(a)gmail.com> wrote: From: Luke Palmer <lrpalmer(a)gmail.com> Subject: Re: [Haskell-cafe] type Rational and the % operator To: "michael rice" <nowgate(a)yahoo.com> Cc: "Ryan Ingram" <ryani.spam(a)gmail.com>, haskell-cafe(a)haskell.org Date: Sunday, March 29, 2009, 12:23 AM 2009/3/28 michael rice <nowgate(a)yahoo.com> Still the same problem. The book says the answer is supposed to be an exact 98/67. Michael import Data.Ratio cf :: [Integer] -> Rational cf (x:xs) = (x % 1) + (1 % (cf xs)) cf (x:[]) = x % 1 cf [] = 0 % 1 Use (/) for division: cf :: [Integer] -> Rationalcf (x:xs) = toRational x + 1 / cf xscf [x] = toRational x cf [] = 0 Luke

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Re: [Haskell-cafe] type Rational and the % operator
by michael rice 29 Mar '09

29 Mar '09

Still the same problem. The book says the answer is supposed to be an exact 98/67. Michael import Data.Ratio cf :: [Integer] -> Rational cf (x:xs) = (x % 1) + (1 % (cf xs)) cf (x:[]) = x % 1 cf [] = 0 % 1 Hugs> :load cf.hs ERROR "cf.hs":3 - Type error in application *** Expression : x % 1 + 1 % cf xs *** Term : x % 1 *** Type : Ratio Integer *** Does not match : Ratio (Ratio Integer) Data.Ratio> --- On Sat, 3/28/09, Ryan Ingram <ryani.spam(a)gmail.com> wrote: From: Ryan Ingram <ryani.spam(a)gmail.com> Subject: Re: [Haskell-cafe] type Rational and the % operator To: "michael rice" <nowgate(a)yahoo.com> Cc: "Brandon S. Allbery KF8NH" <allbery(a)ece.cmu.edu>, haskell-cafe(a)haskell.org Date: Saturday, March 28, 2009, 11:17 PM Just use "/" for division. % is for construction of rationals from the "underlying" numeric type. For example, instead of "toRational x" you can write "x % 1". -- ryan 2009/3/28 michael rice <nowgate(a)yahoo.com>: > I may be missing something here, but this is what I intended. > > An expression of the form > > 1 > a1 + ------ > 1 > a2 + ------ > 1 > a3 + -- > a4 + ... > > Where the ai's are positive integers is called > a continued fraction. > > Function cf should take [1,2,6,5] to produce > > 1 > 1 + ----- > 1 > 2 + ----- > 1 > 6 + -- > 5 > > Michael > > > --- On Sat, 3/28/09, Brandon S. Allbery KF8NH <allbery(a)ece.cmu.edu> wrote: > > From: Brandon S. Allbery KF8NH <allbery(a)ece.cmu.edu> > Subject: Re: [Haskell-cafe] type Rational and the % operator > To: "michael rice" <nowgate(a)yahoo.com> > Cc: "Brandon S. Allbery KF8NH" <allbery(a)ece.cmu.edu>, "Duane Johnson" > <duane.johnson(a)gmail.com>, haskell-cafe(a)haskell.org > Date: Saturday, March 28, 2009, 10:39 PM > > On 2009 Mar 28, at 22:36, michael rice wrote: > > import Data.Ratio > cf :: [Integer] -> Rational > cf (x:xs) = (toRational x) + (1 % (cf xs)) > cf (x:[]) = toRational x > cf [] = toRational 0 > > Data.Ratio> :load cf.hs > ERROR "cf.hs":3 - Type error in application > *** Expression : toRational x + 1 % cf xs > *** Term : toRational x > *** Type : Ratio Integer > *** Does not match : Ratio (Ratio Integer) > > Your function cf produces a Rational (Ratio Int); you're using it in the > denominator of another Ratio, which makes that Ratio's type Ratio (Ratio > Int). This is almost certainly not what you intended, but I couldn't say > what you actually want. > -- > brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery(a)kf8nh.com > system administrator [openafs,heimdal,too many hats] allbery(a)ece.cmu.edu > electrical and computer engineering, carnegie mellon university KF8NH > > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > >

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