I've got a few exercises to prepare for the exam in Haskell/Prolog.
One Haskell task is to rewrite the function below:
original :: [Integer] -> Integer
original [] = 0
original (x:xs) | x < 20 = 5 * x - 3 + original xs
| otherwise = original xs
But the condition is that I am only allowed to remove the two "undifined" in the scheme below:
alternative :: [Integer] -> Integer
alternative = foldr undefined undefined
My problem is that I dont know how this could match the normal foldr structure with 3 parameters (function, "start value" or how is it called?,list)?
Maybe an equivalent example would be helpfull, not the full soultion please!
Futhermore I am not allowed to use "let" or "where".
Thank you for any help!
Sooo... I just followed the idea from @hugo to just first complete the task on the "normal" way, which works but is not allowed by our university correction tool:
alternative :: [Integer] -> Integer
alternative list = foldr (\ x y -> if x < 20 then 5*x -3 + y else y) 0 list
AND after try end error i got the solution:
alternative :: [Integer] -> Integer
alternative = foldr (\ x y -> if x < 20 then 5*x -3 + y else y) 0
A list like [1,4,2,5] is syntactical sugar for (:) 1 ((:) 4 ((:) 2 ((:) 5 []))). foldr f z basically replaces the (:) data constructor with f, and the empty list data constructor [] with z. So foldr f z will result in f 1 (f 4 (f 2 (f 5 z))).
Since you write original [] = 0, this thus means that for z, we can use 0. For f we can use if x < 20 then (+) (5*x-3) else id, since in case x < 20, we add 5*x-3 to the value, and otherwise, we do nothing with the recursively computed value.
We can thus make an alternative implementation that looks like:
alternative :: (Foldable f, Num a, Ord a) => f a -> a
alternative = foldr f 0
where f x ys | x < 20 = 5*x - 3 + ys
| otherwise = ysor without the where clause with an inline lambda expression:
alternative :: (Foldable f, Num a, Ord a) => f a -> a
alternative = foldr (\x -> if x < 20 then (+) (5*x-3) else id) 0