Discussion on: Challenge - Print Spiral

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Replies for: Six cases, four of them with explicit return values (at the edges), two of them recursive: import Text.Printf type Size = Int type Row = Int ty...

Instead of functorial lifting into the list (resulting in O( n² )) I use the loop function iterateUntilM to iterate the slightly modified printElement function until the spiral is printed: O(1)

import Text.Printf (printf)
import Control.Monad.Loops (iterateUntilM)

type Size = Int
type Row = Int
type Column = Int
type Counter = Int

element :: Size -> Row -> Column -> Int
element n row col
  | even n && row == 0   = sqr - 1 - col
  | even n && col == n-1 = sqr - n - row
  | even n               = element (n-1) (row-1) col
  | odd n && row == n-1  = sqr - n + col
  | odd n && col == 0    = sqr - n - (n-1) + row
  | otherwise            = element (n-1) row (col-1)
  where sqr = n^2

printElement :: Size -> Counter -> IO Counter
printElement n counter =
  printf ("%*v" ++ end) len (element n row col) >> return (counter+1)
  where len = (length . show) (n^2-1)
        end = if col == n-1 then "\n" else " "
        (row,col) = quotRem counter n

printSpiral n = iterateUntilM (>=n^2) (printElement n) 0 >> return ()

main = printSpiral 10

In the printElement function, elements are indexed by a counter from 0 to n² - 1, their position in the spiral can be derived from the index by division modulo n: (row,col) = quotRem counter n.