Haskell - implement powerset function of a set

Question

I've just started to learn Haskell and had difficult times of solving this problem: powerSet :: Set a -> Set (Set a)

Here's my attempt:

powerSet :: Tree a -> Tree (Tree a)
powerSet Empty = Empty
powerSet tree = fromList' [fromList' p | p <- powerSet' (toList tree)]

powerSet' :: [a] -> [[a]]
powerSet' [] = [[]]
powerSet' (x:xs) = [x:ps | ps <- powerSet' xs] ++ powerSet' xs

I should expect the output in such order: powerset {1,2,3} => { {}, {1}, {2}, {1,2,3}, {1,3}, {2}, {2,3}, {3} }

But instead I got this: powerset {1,2,3} => { {1,2,3}, {1,2}, {1,3}, {1}, {2,3}, {2}, {3}, {} }

Is there any way I can change it?

A set is an unordered collection. Likely, you should disregard the order of the elements as they are printed. — chi
– chi, Commented Jan 14, 2020 at 22:41
Your implementation furthermore deals with a Tree, instead of a Set. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Jan 14, 2020 at 22:44
@WillemVanOnsem For what it's worth, if you change the first {2} to {1,2} in the expected output, then the subsets appear in lexicographical order; so perhaps the exercise is intended to encourage a search tree implementation. — Daniel Wagner
– Daniel Wagner, Commented Jan 14, 2020 at 23:05
@DanielWagner: ah yes. But then the fromList should thus insert the items in a ordered way. Then the "responsibility" is more for the tree generation part than for the powerSet' function. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Jan 15, 2020 at 8:45

willeM_ Van Onsem · Accepted Answer · 2020-01-15 15:44:15Z

0

Wondering if there's any way I can change it?

Well the order is { {1,2,3}, {1,2}, {1,3}, {1}, {1,3}, {2,3}, {2}, {3}, {} } because that is how you implemented this. Indeed for an empty list, you return as powerSet' [], a singleton list with an empty list. For a non-empty list, you recurse on the tail of the list, and first you yield that list where you each time prepend x to that list, and then you yield the result of the recursive call without prepending.

Hence for a list [2], the recursive case is [[]], so we first prepend it with 2 yielding [2], and then we do not prepend this, yielding [].

If we thus calculate the powerset of [1,2], the recursive call will result in [[2], []]. We first emit the list where we each time prepend 1, so [1,2] and [1], and then we emit the list without prepending, making it [2] and []. Or putting this together [[1,2], [1], [2], []].

You can yield the results in a different order, for example with:

powerSet' :: [a] -> [[a]]
powerSet' l = [] : go l
    where go [] = []
          go (x:xs) = [x] : map (x:) (go xs) ++ go xs

But as @DanielWagner says in his comment:

so perhaps the exercise is intended to encourage a search tree implementation.

You likely should not alter the order of the items that are yielded by the powerSet', but make sure your Tree is a (binary) search tree, and thus implement the fromList function in such way that it insert the items in an ordered manner in your Tree.

answered Jan 15, 2020 at 15:44

willeM_ Van Onsem

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1 Comment

likeliterally Over a year ago

yes I did use a BST and this explained my question very clearly. Many thanks!

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