How does Set actually work internal when checking for values?

问题

This is a very general computer science based question, but one that doesn't seem intuitive based on the literature about how they work. It's a language agnostic question, but relates to how a Set data type works internally.

I have used them many times, and it is recommended to use them to store unique values and quickly access them. Supposedly in Big-O notation its time and complexity is O(1) every time the Set is accessed. How is that possibly if the Set may contain thousands of items? Even if the items are unique.

In order to find an item in the Set, it still has to scan each and every unique item, which in Big-O is O(n) in time and complexity. Is there anything I'm missing here?

Thanks in advance for your help! Most thorough answer gets the up-vote!

回答1:

A Set is an example of a more general kind of objects collectively known as HashedCollections. These use some sort of HashTable to actually store and retrieve their elements.

Given any element, these tables compute an integer value for it named its hash. There are several well-known techniques to define the mapping between elements and their hash values. Some are intrinsic, in the sense that the hash does not depend on the attributes of the element, which might change, and hence the hash remains the same along the life of the element. Others are extrinsic, in the sense that they may depend on the attributes. However, in the latter case, it is supposed that the particular elements will not be modified while they are referenced from a HashedCollection (otherwise the HashedCollection must be rehashed).

The procedure for storing an element works as follows:

1. The hash is computed for the element.
2. An index to the table is computed as the remainder of the hash modulo the length of the table.
3. If the slot at the index so calculated is already taken, some policy is applied to resolve the collision.

Step 1 is supposed to be really fast (e.g., the hash has not cryptographic strength).

Step 2 assumes (in most of the cases) that the length of the table is a prime number (powers of 2 are also used)

Step 3 can be resolved in basically two different ways:

• The table is sequentially scanned j times until the slot at index + j is free, or
• The element is added to the collection of elements colliding at the given index (bucket)

In addition, if there aren't enough empty slots (which increases the probability of collision), the table is enlarged and rehashed (because the modulo changed).

With sufficient free slots and a fairly random distribution of the indexing mechanism, the probability of finding the desired slot in O(1) is very high. Of course, if too many elements collide, the average complexity is no longer O(1), but this is supposedly mitigated by the growing policy (+ rehash).

The retrieval is similar. To check whether an element belongs in the collection, its hash and modulo are computed and the element is compared with the contents of the target slot. If the comparison fails, the search proceeds linearly in the bucket.

The removal of elements is somewhat more difficult when there is no bucket and instead the indexes are increased, but you get the idea.

If you really want to see all of this at work, go ahead and debug the basic operations of HashedCollections in any Smalltalk dialect. Lots of fun guaranteed.

来源：https://stackoverflow.com/questions/55750234/how-does-set-actually-work-internal-when-checking-for-values