I'm experimenting with guard recursion, but I'm having a difficult time with Haskell syntax.
I'm trying to use recursion in Haskell to reduce an integer 'n' number of times until it reaches my desired base case, while keeping track of 'n', then return the reduced integer and 'n' as a tuple for my piece of information.
My base case would be when the integer is within 1 to 13.
For example,
Prelude> reduce 42
(3, 2)
Prelude> reduce 3
(3, 0)
To explain the logic,
Prelude> reduce 42
( (((42 - 13) - 13) - 13), 1+1+1+1 ) --> assume the accumulator = 0, hence 0,1,2.
(3, 3)
Prelude> reduce 3
(3, 0) --> already in base case!
Currently, I have
reduce (x, 0)
| x `elem` [1 .. 13] = (x, acc)
| otherwise = reduce (x -13, acc + 1)
where
acc = 0
But of course, my IDE is yelling that I'm wrong. I have no clue how to implement initialize tuples into this recursion so that it's like reduce :: a -> (a, b)
EDIT: I think I'm getting closer...but of course, I'm trying to implement it as a -> (a,b).
You need to initialise it with a helper function:
reduce::Int->(Int,Int)
reduce x = reduce' (x,0)
reduce'::(Int,Int)->(Int,Int)
reduce' (x,n)
| elem x [1..13] = (x,n)
| otherwise = reduce' (x-13,n+1)