Difference between revisions of "Haskell Quiz/Posix Pangrams/Solution Burton"
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Revision as of 08:44, 27 October 2006
This is a naive solution - it works by taking the sortest subsequence of the original list which is a pangram then trying to replace words in it with shorter, unused ones. The pangrams it produces aren't very short.
module Main
where
import IO
import Char
import List
{--
*Main> main
pangram: admin bc ex fg lp sh tr zcat yacc write who vi uux unlink uniq join
wordcount: 16, total chars: 52, repeated chars: 41
--}
makePangram :: [String] -> String -> [String] -> Int -> IO String
makePangram [] pg rejects pass | isPangram pg = return pg
| pass == 0 = makePangram rejects pg [] 1
| otherwise = return ("ran out of words! " ++ pg)
makePangram (w:ws) pg rejects pass | isPangram pg = return pg
| hasNew w (nub pg) = handleDupes
| otherwise = makePangram ws pg rejects pass
where handleDupes | not (hasDupes (filter isAlpha w) (nub pg)) || pass == 1 = makePangram ws (pg ++ " " ++ w) rejects pass
| otherwise = makePangram ws pg (w:rejects) pass
hasNew :: String -> String -> Bool
hasNew x [] = True
hasNew [] _ = False
hasNew (x:xs) w | not $ isAlpha x = hasNew xs w
| not $ elem x w = True
| otherwise = hasNew xs w
hasDupes :: String -> String -> Bool
hasDupes _ [] = False
hasDupes [] _ = False
hasDupes (x:xs) w = if elem x w then True else hasDupes xs w
numDupes :: String -> Int
numDupes s = numDupes' s 0
where numDupes' [] c = c
numDupes' (x:xs) c = if elem x xs then numDupes' xs (c+1) else numDupes' xs c
isPangram :: String -> Bool
isPangram = (==26) . length . nub . filter isAlpha
stats :: String -> String
stats s = "wordcount: " ++ (show (length (words s))) ++ ", total chars: " ++ (show (length (filter isAlpha s))) ++ ", repeated chars: " ++ (show (numDupes s))
main = do hdl <- openFile "posix-utils.txt" ReadMode
cs <- hGetContents hdl
pg <- (makePangram (lines cs) "" [] 0)
putStrLn ("pangram: " ++ pg)
putStrLn (stats pg)