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  1. Practice
  2. Functional Programming
  3. Recursion
  4. Pascal's Triangle
  5. Discussions

Pascal's Triangle

  • strawstack + 1 comment

    UPDATE: The program now prints a nested array. Still trying to get Haskell to format basic output? (Output of below is [[1],[1,1],[1,2,1],[1,3,3,1]])

    -- print tri value at coords
pas m n
    | n == 0 = 1 
    | m == n = 1
    | otherwise = pas (m-1) (n-1) + pas (m-1) n
      
-- print row of triangle 
row m = map (pas m) [0..m]   
    
-- print triangle until row
tri m = map (row) [0..(m-1)] 
    
main = do   
   m <- readLn :: IO Int
   print (tri m) -- print tri to row m
    • strawstack + 2 comments

      UPDATE: Finally, figured out how to print a [[Char]] in the correct format with Haskell...

      -- PRINT with FORMAT    
two :: [Int] -> [String]
two m = map (show) m

three :: [[Int]] -> [[String]]
three m = map (two) m

main = do   
   m <- readLn :: IO Int
   --send (tri m) -- print tri to row m
   mapM_ (putStrLn . intercalate " " $) (three (tri m))

      Where (tri m) produces [["1"], ["1", "1"], ["1", "2", "1"], ["1", "3", "3", "1"]]

      • ende76 + 0 comments

        Thanks for this chain of posts. I was struggling through the same problems with how to properly output.
        Much longer than to solve the actual problem!

      • tury345 + 1 comment

        Achieved something similar here:

        putStr (unlines (map (\line -> unwords (map (\(x:[]) -> show x) line)) rows))

        where rows is [[1], [1, 1], [1, 2, 1], [1, 3, 3, 1]]

        • egoek + 0 comments

          I concocted this:

          mapM_ putStrLn (map (\x -> unwords (map show x)) result)

          where result is [[1], [1, 1], [1, 2, 1] ...] I'm sure I could get rid of the parenthesis. Any suggestions welcome.