Question

I am currently learning Haskell. I am trying to write a function which given two list of n distinct elements returns true if one is a permutation of the other. I am doing this as an exercise.

I first wrote it as :

isPermut :: (Eq a)=>[a]->[a]->Bool
isPermut u v = foldl (\acc x -> acc && (elem x u)) True v

This seems to work. I now try to rewrite it without lambda expression. So I try :

isPermut :: (Eq a)=>[a]->[a]->Bool
isPermut u v = foldl (&& (flip $ elem u)) True v

This gives me a compile error :

Couldn't match expected type `b0 -> Bool' with actual type `Bool'
Expected type: Bool -> b0 -> Bool
  Actual type: Bool -> Bool
In the first argument of `foldl', namely `(&& (flip $ elem u))'
In the expression: foldl (&& (flip $ elem u)) True v

What does this error mean? What is the correct way to write the function without lambda?
Thanks.

Answer 1

Folds have to take a function of two arguments. You gave a function of one argument since (&& whatever) is the same as \ x -> x && whatever.

To not use a lambda here, you need some way to compose &&, a two-argument function, with (`elem` u), a one-argument function. (Note that (`elem` u) is the same as (flip elem) u or \ x -> elem x u.)