Functional Sets, Part 6: Union and Remove
With folds done, our sets are shaping up. Folds unlock some more interesting things for us. Namely: unions!
Union—Two Sets Become One
The union of two sets is the set of all the items in both. So if we have a set with 1 and 2, and a set with 2 and 3, the union of the two sets is 1, 2, and 3. In psuedocode, it looks like this:
union [1, 2] [2, 3] == [1, 2, 3]
So how would we go about that?
How about we
insert all the items from the first list into the second?
insert will deduplicate for us, we won’t have to rewrite any logic!
We can combine
fold to get our implementation.
union : Set comparable -> Set comparable -> Set comparable union = foldl insert
Seriously, that’s the whole thing.
I’m just as surprised as you!
It was quite a shock when I replaced the version I wrote before
foldl with that tiny thing and the tests all passed.
So what’s going on here?
Why is this so simple?
Remember how fold takes an accumulator value? So far we’ve only used base values like an empty list or the number 0. But there’s nothing that prevents us from something more complex!
We’ll use the first set passed in as the accumulator value.
foldl walks over all the items of the second set and insert them into the first.
The result is the union of the two sets!
By the way: yes, I used point-free style again here. If you want to do without:
union : Set comparable -> Set comparable -> Set comparable union a b = foldl insert a b
But if you do it in point-free style, you don’t have to worry about what to call the two sets. Bonus!
Now that we can get the union of two sets, we can remove items. This seems counterintuitive at first. Why would we need to join two sets to remove an item? Isn’t that the opposite of what we want to do? Let’s see!
We want this call to look roughly like:
remove 1 [1, 2] == 
If the set is empty or doesn’t contain the value, we shouldn’t do anything. Otherwise, we should return a new set without the value in it.
remove : comparable -> Set comparable -> Set comparable remove item set = case set of Empty -> set Tree _ head left right -> if item < head then tree head (remove item left) right |> balance else if item > head then tree head left (remove item right) |> balance else union left right
This looks like member! When we have an empty subtree, we return it unchanged. When we find a tree where the head is not equal to the item we’re looking for, we try and remove from the relevant subtree. And finally, if the head we’re looking at is the one we’re looking for, we have to construct a new tree without it.
In this case, we can’t return an
Empty, because the subtrees might have values in them.
We also can’t return a
Tree, because we don’t have a head.
So we use our newly-minted
union function to combine the left and right subtrees into a new
The function calls then bubble up like normal, and we’re done.
Update: You also have to rebalance the parent trees after removal, or the set becomes unbalanced.
Try this yourself by removing the
balance calls above and creating a set, then removing all the items in the left side.
The tree will become more and more unbalanced the more items you remove from it.
An easy replication:
List.range 1 10 |> fromList |> remove 1 |> remove 2 |> remove 3.
Thanks to Ilias Van Peer for the catch!
Union’d and Remove’d!
So this week, we’ve seen:
- Exactly how terse a fold can make a function. We specify what function we want to combine values with and our work is done. (This is the second week in a row we’ve seen this. It’s important!)
- How to remove an item from a tree: make a new tree without that item.
Next week we’re going to make another useful function:
We’ll use it to do some more set operations!