I want to understand Haskell better by trying to stretch possibilities of Functor, Applicative, and Monad as much as possible and study how they behave. So in Learn You a Haskell there's an exercise to generate all possible moves for a knight from a given position on chessboard:
moveKnight :: KnightPos -> [KnightPos] moveKnight (c,r) = filter onBoard [(c+2,r-1),(c+2,r+1),(c-2,r-1),(c-2,r+1) ,(c+1,r-2),(c+1,r+2),(c-1,r-2),(c-1,r+2) ] where onBoard (c,r) = c `elem` [1..8] && r `elem` [1..8]
Now I thought to myself, wait a second, I can actually generate that list
[(c+2,r-1),(c+2,r+1),(c-2,r-1),(c-2,r+1),(c+1,r-2),(c+1,r+2),(c-1,r-2),(c-1,r+2)] instead of hardcoding it. So this is how I did it with list comprehensions:
invert x | x == 1 = 2 | x == 2 = 1 | otherwise = 0 moves = [(c `f` a, r `g` (invert a)) | c <- , r <- , f <- [(+), (-)], g <- [(+), (-)], a <- [1, 2]] -- generates: [(7,4),(8,3),(7,0),(8,1),(5,4),(4,3),(5,0),(4,1)]
So the above works correctly, but the code looks ugly. Because every list comprehension is equivalent to a lift, the above code could be rewritten with
liftA5 :: Applicative g => (a -> b -> c -> d -> e -> f) -> g a -> g b -> g c -> g d -> g e -> g f liftA5 f a b c d e = f <$> a <*> b <*> c <*> d <*> e liftA5 (\c r f g a -> (c `f` a, r `g` (invert a)))   [(+), (-)] [(+), (-)] [1,2]
My questions are:
- Are there any way to generate the above list in a more elegant way?
- How to generalize it to arbitrary possible permutations? (So that my code could parametrized and not fixed to just numbers 1 and 2, or just + and -, etc)
- What intuitions, insights and lessons can I get from this exercise?
I know that this is a massively contrived situation, but they serve me as an exercise to understand the language better.