So, thanks to Henning Thielemann I was able to make a code a little
more functional.
I did find ByteString module that really speed things up.
I got 0.04 seconds with the following snippet:
-- code start
import qualified Data.ByteString.Char8 as BS
import Data.Char (toLower)
main :: IO ()
main = interact' $ unlines' . solveAll . takeWhile ((/= '*') . head') . lines'
solveAll :: [String'] -> [String']
solveAll = map $ toStr . solve
toStr :: Bool -> String'
toStr True = makeString' "Y"
toStr False = makeString' "N"
solve :: String' -> Bool
solve = isTautogram . words'
isTautogram :: [String'] -> Bool
isTautogram (x:xs) = all ((== firstChar) . normalizeHead) xs
where firstChar = normalizeHead x
normalizeHead :: String' -> Char
normalizeHead = toLower . head'
-- optimizations
type String' = BS.ByteString
interact' = BS.interact
unlines' = BS.unlines
lines' = BS.lines
head' = BS.head
words' = BS.words
makeString' = BS.pack
-- code end
Thanks all,
Thiago.
2011/8/11 Henning Thielemann <schlepp...@henning-thielemann.de>:
> On 09.08.2011 01:43, Thiago Negri wrote:
>>
>> Hello all,
>>
>> I'm relatively new to Haskell and trying to solve some online judge's
>> problems in it.
>> One of the problems is to say if a given sentence is a tautogram or not.
>> A tautogram is just a sentence with all the words starting with the same
>> letter.
>>
>> My first try (solution is ok) was to do it as haskeller as possible,
>> trying to overcome my imperative mind.
>> But it did bad at performance (0.30 secs of runtime, 4.6 mb of memory):
>>
>> -- code start
>> import Data.Char (toLower)
>>
>> main = getContents>>= mapM_ (putStrLn . toStr . isTautogram . words)
>> . takeWhile (/= "*") . lines
>
> That's still imperative! :-)
>
> How about 'interact' and using 'unlines' instead of 'putStrLn' ?
>
>
>> toStr :: Bool -> [Char]
>
> You may want to write String instead of [Char] for clarity.
>
>> toStr True = "Y"
>> toStr False = "N"
>>
>> isTautogram :: [[Char]] -> Bool
>> isTautogram (x:[]) = True
>
> I assume this case is not necessary, since all [] == True anyway.
>
>> isTautogram (x:xs) = all ((== firstChar) . toLower . head) xs
>> where firstChar = toLower . head $ x
>
> It is maybe more elegant, not to compare all words with the first one, but
> to compare adjacent words in the list:
>
> all (zipWith (...) xs (drop 1 xs))
>
>
>> Note that the only thing that changed between the two tries was the
>> main-loop.
>> The second version runs faster (got 0.11 secs) and with less memory (3.6
>> mb)
>>
>> Can someone explain to me what is really going on?
>> Maybe pointing out how I can achieve these optimizations using
>> profiling information...
>
> Interesting observation. I do not see a problem quickly.
>
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