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.