# Confusion with analysis of hashing with chaining

I was attending a class on analysis of hash tables implemented using chaining, and the professor said that:

In a hash table in which collisions are resolved by chaining, an search (successful or unsuccessful) takes average-case time θ(1 + α), under the assumption of simple uniform hashing.

and

The worst-case time for searching is θ(n) plus the time to compute the hash function.

These are quoted from out textbook, ITA. Here α is the load factor, which is equal to n/m where n is the total number of elements to be inserted to the hash table and m is the size of the hash table (which is a constant for each implementation).

My confusion arises from the thinking that

θ(1 + α) = θ(α) = θ(n/m) = θ(n)

So I claimed that θ(1 + α) = θ(n), ie the average case and worst case time complexities are the same and the professor disagrees with me.

She asked me to find out by myself why I am wrong. I did some brainstorming and the only deduction I can make is that either I am wrong in assuming that m is a constant or asymptotic notations can't be used to compare between worst-case and average-case running time in this case. Please help me find what is wrong here. Also I don't understand why θ(1 + α) is not written simply as θ(α) and also why the professor insists on the value of α being less than, equal to, or greater than 1

The reason that we are using $\Theta(1+\alpha)$ rather than $\Theta(\alpha)$ is that $\alpha$ could be very small. Imagine for example that the hidden constant in both $\Theta$s is one, and that $\alpha=0.000001$. There is a large difference between $1.000001$ and $0.000001$.
• We don't think of $m$ as constant, since it's not really constant. Different instances of the data structure can have widely different $m$. It's as if you would say that the space is $O(1)$ rather than $\Theta(m)$. Mar 2, 2016 at 9:42
• Whenever you run an algorithm on a particular input, all the parameters are fixed. That doesn't mean that when we compute the asymptotic performance of an algorithm we assume that they are constant. In other words, $m$ and $n$ are both values of the same "type". You can call them parameters, but that's not a particularly good term. Mar 2, 2016 at 12:02