Re: [Haskell-cafe] Re: Haskell and C++ program
On 17 jan 2009, at 22:22, Derek Elkins wrote: On Thu, 2009-01-15 at 13:40 +0100, Apfelmus, Heinrich wrote: Eugene Kirpichov wrote: Well, your program is not equivalent to the C++ version, since it doesn't bail on incorrect input. Oops. That's because my assertion show . read = id is wrong. We only have read . show = id show . read = id (in the less defined than sense) No, you only have read . show = id which often doesn't hold in practice. show . read /= id You do not even have that; the read may remove surplus parentheses which will not be reinserted by the show. Doaitse Assuming the first identity holds, you do of course have show . read . show = show and this probably holds even in most cases where read . show = id does not hold. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Haskell and C++ program
On Mon, 2009-01-19 at 22:12 -0500, S. Doaitse Swierstra wrote: On 17 jan 2009, at 22:22, Derek Elkins wrote: On Thu, 2009-01-15 at 13:40 +0100, Apfelmus, Heinrich wrote: Eugene Kirpichov wrote: Well, your program is not equivalent to the C++ version, since it doesn't bail on incorrect input. Oops. That's because my assertion show . read = id is wrong. We only have read . show = id show . read = id (in the less defined than sense) No, you only have read . show = id which often doesn't hold in practice. show . read /= id You do not even have that; the read may remove surplus parentheses which will not be reinserted by the show. Doaitse My notation is show . read is not less than or equal to id. That covers that case. The particular example I was thinking of was actually simply whitespace. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Haskell and C++ program
On Thu, 2009-01-15 at 13:40 +0100, Apfelmus, Heinrich wrote: Eugene Kirpichov wrote: Well, your program is not equivalent to the C++ version, since it doesn't bail on incorrect input. Oops. That's because my assertion show . read = id is wrong. We only have read . show = id show . read = id (in the less defined than sense) No, you only have read . show = id which often doesn't hold in practice. show . read /= id Assuming the first identity holds, you do of course have show . read . show = show and this probably holds even in most cases where read . show = id does not hold. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Haskell and C++ program
Jonathan Cast wrote:
reverseDouble =
unlines
. intro
. map show
. reverse
. map (read :: String - Double)
. takeWhile (/= end)
. words
where
intro l =
(read ++ show (length l) ++ elements) :
elements in reversed order :
l
This can be simplified to
... . map show . reverse . map read . ...
= { map f . reverse = reverse . map f }
... . reverse . map show . map read . ...
= { map f . map g = map (f . g) }
... . reverse . map (show . read) . ...
= { show . read = id }
... . reverse . map id . ...
= { map id = id }
... . reverse . ...
In other words,
reverseDouble =
unlines. intro . reverse . takeWhile (/= end) . words
where
intro xs =
(read ++ show (length xs) ++ elements) :
elements in reversed order :
xs
And the doubles disappeared completely. :)
Regards,
H. Apfelmus
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Re: [Haskell-cafe] Re: Haskell and C++ program
Well, your program is not equivalent to the C++ version, since it
doesn't bail on incorrect input.
2009/1/15 Apfelmus, Heinrich apfel...@quantentunnel.de:
Jonathan Cast wrote:
reverseDouble =
unlines
. intro
. map show
. reverse
. map (read :: String - Double)
. takeWhile (/= end)
. words
where
intro l =
(read ++ show (length l) ++ elements) :
elements in reversed order :
l
This can be simplified to
... . map show . reverse . map read . ...
= { map f . reverse = reverse . map f }
... . reverse . map show . map read . ...
= { map f . map g = map (f . g) }
... . reverse . map (show . read) . ...
= { show . read = id }
... . reverse . map id . ...
= { map id = id }
... . reverse . ...
In other words,
reverseDouble =
unlines. intro . reverse . takeWhile (/= end) . words
where
intro xs =
(read ++ show (length xs) ++ elements) :
elements in reversed order :
xs
And the doubles disappeared completely. :)
Regards,
H. Apfelmus
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[Haskell-cafe] Re: Haskell and C++ program
Eugene Kirpichov wrote: Well, your program is not equivalent to the C++ version, since it doesn't bail on incorrect input. Oops. That's because my assertion show . read = id is wrong. We only have read . show = id show . read = id (in the less defined than sense) Regards, H. Apfelmus ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re[2]: [Haskell-cafe] Re: Haskell and C++ program
Hello Eugene,
Thursday, January 15, 2009, 3:27:59 PM, you wrote:
but at least length.map show is eq to length :)
Well, your program is not equivalent to the C++ version, since it
doesn't bail on incorrect input.
2009/1/15 Apfelmus, Heinrich apfel...@quantentunnel.de:
Jonathan Cast wrote:
reverseDouble =
unlines
. intro
. map show
. reverse
. map (read :: String - Double)
. takeWhile (/= end)
. words
where
intro l =
(read ++ show (length l) ++ elements) :
elements in reversed order :
l
This can be simplified to
... . map show . reverse . map read . ...
= { map f . reverse = reverse . map f }
... . reverse . map show . map read . ...
= { map f . map g = map (f . g) }
... . reverse . map (show . read) . ...
= { show . read = id }
... . reverse . map id . ...
= { map id = id }
... . reverse . ...
In other words,
reverseDouble =
unlines. intro . reverse . takeWhile (/= end) . words
where
intro xs =
(read ++ show (length xs) ++ elements) :
elements in reversed order :
xs
And the doubles disappeared completely. :)
Regards,
H. Apfelmus
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--
Best regards,
Bulatmailto:bulat.zigans...@gmail.com
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