# Exercise 11.2.6

Suppose we have stored $n$ keys in a hash table of size $m$, with collisions resolved by chaining, and that we know the length of each chain, including the length $L$ of the longest chain. Describe a procedure that selects a key uniformly at random from among the keys in the hash table and returns it in expected time $\O(L \cdot (1 + 1 / \alpha))$.

Think about the hash table as a matrix with $m$ rows and $L$ columns. The first element of chain with hash $k$ is put in the first column of the $k$-th row, the second is put in the second column of the $k$-th row and so on. Essentially, each row contains a chain followed by some empty elements.

The procedure is:

1. Keep picking a random cell in that table (that is, pick a number s = rand(m * L), row s / L and column s % L) until we get a non-empty one.
2. Walk the linked list for the chosen row until we get to the chosen column.

Note that when step 1 picks a non-empty element, it does so uniformly. That is, each element has equal chance of getting picked.

We need to calculate the time for the first step (say $A$) and add it to the time of the second step (say $B$).

The probability to pick an element in step 1 is:

$$\Pr \{ \text{not empty} \} = \frac{n}{mL} = \frac{\alpha}{L}$$

The expected number of trials to pick a non-empty element is modelled by the geometric distribution (Bernoulli trials, (C.32)), and has $\E[X] = 1 / p$. That is, on average we expect the following number of trials:

$$A = \frac{L}{\alpha} = \O( L / \alpha )$$

Once we have picked up the ith row and the jth column, we need to walk the linked list on row i and advance j elements. Worst case, that takes $L$ steps, because the longest chain has that many elements. That is:

$$B = \O(L)$$

And thence, the expected time is:

$$A + B = \O(L / \alpha) + \O(L) = \O( L \cdot ( 1 + 1 / \alpha ) )$$