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

Pascal's Triangle

  • strawstack + 1 comment

    I'm trying to explain to Haskell that a triangle is not a stright line. Can anyone help? (The output of the following code is 1 1 1 1 2 1 1 3 3 1 (without any new lines))

    -- FORMAT    
mat v = putStr ( (show v) ++ " " ) -- print value with space     

-- 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 = mapM_ (mat . (pas m) $) [0..m]   
    
-- print triangle until row
tri m = mapM_ (row) [0..(m-1)] 
    
main = do   
   m <- readLn :: IO Int
   tri m -- print tri to row m
    • 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 + 1 comment

            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.

            • twostepted + 0 comments

              Nice. Vscode extension 'Haskutil' recommended shorten slightly to this

                n <- readLn
  mapM_ (putStrLn . unwords . map show) $ tri n