Worst and average case of open addressing

Question

I read chapter 11 of CLRS and there are three theorems provided regarding the analysis of open addressing:

11.6: Given an open-address hash table with load factor α=n/m<1 the expected number of probes in an unsuccessful search is at most 1/1-α assuming uniform hashing.

11.7: Inserting an element into an open-address hash table with load factor α requires at most 1/1-α probes on average, assuming uniform hashing.

11.8: Given an open-address hash table with load factor α<1, the expected number of probes in a successful search is at most (1/α)ln(1/1-α) assuming uniform hashing and assuming that each key in the table is equally likely to be searched for.

Is it correct to say that:

Search worst case is O(n) where n is the hash table size.

Succesful search averege case is O(1/1-α).

Unsuccesful search averege case is O((1/α)ln(1/1-α)).

Insert averege case is O(1/1-α).

At last, what is insert worst case?

Answer 1

Yes, the worst-case cost of a search is O(n), but you would need to be pretty unlucky for that to happen! That would mean that you happened to hit most of the elements during your probe sequence, which is very unlikely to occur. The worst-case cost of an insert is similarly O(n) and happens if, as above, you hit a huge fraction of the existing items on the probe sequence.

Your other bounds are all correct except for unsuccessful searches. Theorem 11.6 says that the expected cost of an unsuccessful search is O(1 / (1 - α)).