
You also don't need mutual recursion for this explicit recursion since
I imagine it would use up more stack space.
-- Doing the one dimensional case.
f011 :: [Int] -> Int
f011 (t:u:v:xs) = f011helper t u v xs
f011helper :: Int -> Int -> Int -> [Int] -> Int
f011helper t u v (w:ws)
| ws == [] = t*u*v*w
| otherwise = max (t*u*v*w) (f011helper u v w ws)
-- Note: f011helper does not call f011.
On Mon, Aug 29, 2011 at 9:40 AM, Oscar Picasso
Got it.
f :: [Int] -> Int f (t:u:v:xs) = helper t u v xs
helper :: Int -> Int -> Int -> [Int] -> Int helper t u v (w:ws) | ws == [] = t*u*v*w | otherwise = max (t*u*v*w) (f (u:v:w:ws))
I tend to overlook mutual recursion in my toolbox.
Good going! :)
Thanks for the enlightenment.
On Sun, Aug 28, 2011 at 4:54 PM, KC
wrote: Try something like the following:
-- Project Euler 11
-- In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
-- <snip>
-- The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
-- What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?
import Data.List
-- Doing the one dimensional case. f011 :: [Int] -> Int f011 (t:u:v:xs) = f011helper t u v xs
f011helper :: Int -> Int -> Int -> [Int] -> Int f011helper t u v (w:ws) | ws == [] = t*u*v*w | otherwise = yada nada mada
-- What are yada nada mada?
-- The 20x20 grid case will become: f0112D :: [[Int]] -> Int -- where [[Int]] is a list of lists of rows, columns, major diagonals, & minor diagonals.
-- -- Regards, KC