CODEWITHSUNDEEP
listhaskellshuffle
Markeazy
Markeazy

Reputation: 76

Shuffling a List Based On an Int List

Given the code:

data Error a = Fail|Ok a
               deriving (Eq, Ord, Show)

split :: Int -> [a] -> (Error ([a],[a]))
split 0 list = Ok ([], list)
split n list
    | n < 0 = Fail
    | n > length (list) = Fail
    | otherwise = Ok (take n list, drop n list)

interleave ::  [a] ->  [a] -> [a]
interleave list [] = list
interleave [] list = list
interleave (x:xs) (y:ys) = x : y : interleave xs ys


shuffle :: [Int] -> [a] -> Error [a]

How I write the function shuffle which will take a list of Ints, and split another list based on those ints. Examples of the int list would be intList = [20,23,24,13] where shuffle will split a list after the 20th element, interleave, split after the 23rd element, interleave, and so on.

Upvotes: 0

Views: 137

Answers (2)

Markeazy
Markeazy

Reputation: 76

I figured it out: 

shuffle [] xs = Ok (xs)
shuffle (x:xs) list = case (split x list) of
                           Fail        -> Fail
                           Ok (l1, l2) -> shuffle xs (interleave l1 l2)

Upvotes: 0

ThreeFx
ThreeFx

Reputation: 7360

Okay, what you want is basically the following:

Given a list xs and indices [a1, a2, a3, ..], split xs at a1, interleave, split it at a2 and interleave, and so on...

Now that leaves us with two functions:

step :: Int -> [a] -> [a]
step index xs = ??? -- something with split and interleave

shuffle :: [a]
shuffle [] xs = xs
shuffle (i:indices) xs = let newList = step i xs
                         in ??? -- something with recursion

Try to write the rest of these functions on yourself.

step can easily be expressed as let (x1, x2) = split index xs in interleave x1 x2. Basically, the rest of shuffle can be written as shuffle indices newList.

Upvotes: 1

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