CODEWITHSUNDEEP
haskell
Alex Smith
Alex Smith

Reputation: 51

Checking if two lists are equal in Haskell

I am trying to create function setEqual.. This is what I have so far:

setEqual :: Eq a => [a] -> [a] -> Bool
setEqual []    _                    = True
setEqual _     []                   = False
setEqual a@(x:a') (y:b) | x == y    = setEqual a' b
                        | otherwise = setEqual a b   

setEqualTest = [ not ("ABCD" `setEqual` "ABC"),
                 not ("ABC" `setEqual` "ABCD"),
                 [0,2,1,3] `setEqual` [0,1,2,3] ]

When I run the test I get [True, False, False]. Am I doing this correctly?

Upvotes: 1

Views: 482

Answers (1)

thor
thor

Reputation: 22480

I think you just need to get two things right for writing your recursive function (to check set equality here): the base case and the recursive case.

You base case is incorrect. You don't know that an empty set [] is equal to an arbitrary set _ unless it's also empty. So it should be:

setEqual []    []                    = True

plus that empty set is not equal to anything else:

setEqual []    _                    = False
setEqual _    []                    = False

(Note that the setEqual [] [] ... line takes precedence as it is written before the others.

Your recursive case has a typo, and the last line should be:

| otherwise = setEqual a' b

Upvotes: 1

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