# Haskell Quiz/PP Pascal/Solution Burton

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< Haskell Quiz | PP Pascal(Difference between revisions)

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## Latest revision as of 11:36, 13 December 2009

{-- *Main> ppPascal 10 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 --} --Pascal's triangle from http://www.haskell.org/haskellwiki/Blow_your_mind pascal :: [[Integer]] pascal = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1] ppPascal :: Int -> IO () ppPascal n = mapM_ (putStrLn . ppRow) tr where tr = take n pascal mRow = last tr pad = cols $ maximum $ mRow cols n = if n < 10 then 1 else 1 + cols (n `div` 10) ppRow r@(x:xs) = padRow ++ (showFirst x) ++ (concatMap showRest xs) ++ padRow where showFirst e = (show e) ++ (padR e) showRest e = padL ++ (show e) ++ padR e padRow = replicate ((length mRow - length r)*pad) ' ' padL = replicate pad ' ' padR d = replicate (pad-(cols d)) ' '

A bit shorter than the above, though certainly not prettier. (reverse._.reverse) makes it easier to calculate the leading padding for each line. -- AlsonKemp

pascal :: [[Integer]] pascal = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1] ppPascal :: Int -> IO () ppPascal n = mapM_ putStrLn $ reverse $ ppRow 0 (reverse tr) where ppRow _ [] = [] ppRow lpads (x:xs) = (replicate lpads ' ' ++ concatMap (padNum pad) x) : (ppRow (lpads+pad) xs) tr = take n pascal padNum pads n = take (2*pads) (show n ++ (repeat ' ')) pad = 1 + (length $ show $ maximum $ last tr) `div` 2