# Haskell Quiz/PP Pascal/Solution Burton

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```{--
*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 = putStrLn \$ unlines \$ 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```