Intersection complexity

Question

In Python you can get the intersection of two sets doing:

>>> s1 = {1, 2, 3, 4, 5, 6, 7, 8, 9}
>>> s2 = {0, 3, 5, 6, 10}
>>> s1 & s2
set([3, 5, 6])
>>> s1.intersection(s2)
set([3, 5, 6])

Anybody knows the complexity of this intersection (&) algorithm?

EDIT: In addition, does anyone know what is the data structure behind a Python set?

Answer 1

The answer appears to be a search engine query away. You can also use this direct link to the Time Complexity page at python.org. Quick summary:

Average:     O(min(len(s), len(t))
Worst case:  O(len(s) * len(t))

EDIT: As Raymond points out below, the "worst case" scenario isn't likely to occur. I included it originally to be thorough, and I'm leaving it to provide context for the discussion below, but I think Raymond's right.

Answer 2

The intersection algorithm always runs at O(min(len(s1), len(s2))).

In pure Python, it looks like this:

    def intersection(self, other):
        if len(self) <= len(other):
            little, big = self, other
        else:
            little, big = other, self
        result = set()
        for elem in little:
            if elem in big:
                result.add(elem)
        return result

[Answer to the question in the additional edit] The data structure behind sets is a hash table.

Answer 3

Set intersection of two sets of sizes m,n can be achieved with O(max{m,n} * log(min{m,n})) in the following way: Assume m << n

1. Represent the two sets as list/array(something sortable)
2. Sort the **smaller** list/array (cost: m*logm)
3. Do until all elements in the bigger list has been checked:
    3.1 Sort the next **m** items on the bigger list(cost: m*logm)
    3.2 With a single pass compare the smaller list and the m items you just sorted and take the ones that appear in both of them(cost: m)
4. Return the new set

The loop in step 3 will run for n/m iterations and each iteration will take O(m*logm), so you will have time complexity of O(nlogm) for m << n.

I think that's the best lower bound that exists