Exercise 11.2.5

Suppose that we are storing a set of $n$ keys into a hash table of size $m$. Show that if the keys are drawn from a universe $U$ with $| U | > nm$ then $U$ has a subset of size $n$ consisting of keys that all hash to the same slot, so that the worst-case searching time for hashing with chaining is $\Theta(n)$.

Obvious statement is obvious. Oh well.

We are hashing elements into $m$ distinct buckets. Let $h$ be the hash function, that is for $u \in U$ we have $0 \le h(u) < m$.

Furthermore, let $C(k)$ be the number of elements in $U$ that hash to $k$.

We need to demonstrate that there is some $j: 0 \le j < m$ for which there are are at least $n$ elements $x_i \in U$ (where for $i \in \{1, 2, \dots, n\} )$ such that $h(x_i) = j$, that is, there exists a $k$ such that $C(k) \ge n$.

Let's assume this is incorrect, that is, every $j$ has at most $n - 1$ elements hashing into it, or in other words $C(x) \le n - 1$ for every $x$. We know that every element of $U$ needs to hash to one of $m$ values, that is:

$$\sum_{i = 0}^{i<m}{ C(i) } = | U | \gt nm \\ \Downarrow \\ C(0) + \sum_{i = 1}^{i<m}{ C(i) } \gt nm \\ \Downarrow \\ C(0) \gt nm - \sum_{i = 1}^{i<m}{ C(i) } \ge nm - \sum_{i=1}^{i<m}{(n - 1)} = nm - (m - 1)(n - 1)\\$$