This is a question that’s been lingering in my mind for some time …
Suppose I have a list of items and an equivalence relation on them, and comparing two items takes constant time.
I want to return a partition of the items, e.g. a list of linked lists, each containing all equivalent items.
One way of doing this is to extend the equivalence to an ordering on the items and order them (with a sorting algorithm); then all equivalent items will be adjacent.
But can it be done more efficiently than with sorting? Is the time complexity of this problem lower than that of sorting? If not, why not?
You seem to be asking two different questions at one go here.
1) If allowing only equality checks, does it make partition easier than if we had some ordering? The answer is, no. You require Omega(n^2) comparisons to determine the partitioning in the worst case (all different for instance).
2) If allowing ordering, is partitioning easier than sorting? The answer again is no. This is because of the Element Distinctness Problem. Which says that in order to even determine if all objects are distinct, you require Omega(nlogn) comparisons. Since sorting can be done in O(nlogn) time (and also have Omega(nlogn) lower bounds) and solves the partition problem, asymptotically they are equally hard.
If you pick an arbitrary hash function, equal objects need not have the same hash, in which case you haven’t done any useful work by putting them in a hashtable.
Even if you do come up with such a hash (equal objects guaranteed to have the same hash), the time complexity is expected O(n) for good hashes, and worst case is Omega(n^2).
Whether to use hashing or sorting completely depends on other constraints not available in the question.
The other answers also seem to be forgetting that your question is (mainly) about comparing partitioning and sorting!