# Problem 11.3

## Quadratic probing

Suppose that we are given a key $k$ to search for in a hash table with positions $0, 1, \ldots, m - 1$, and suppose that we have a hash function $h$ mapping the key space into the set $\{0, 1, \ldots, m - 1\}$. The search scheme is as follows:

- Compute the value of $j = h(k)$, and set $i = 0$.
- Probe in position $j$ for the desired key $k$. If you find it, or if this position is empty, terminate the search.
- Set $i = i + 1$. If $i$ now equals $m$, the table is full, so terminate the search. Otherwise, set $j = (i + j) \bmod m$, and return to step 2.
Assume that $m$ is a power of $2$.

a. Show that this scheme is an instance of the general "quadratic probing" scheme by exhibiting the appropriate constants $c_1$ and $c_2$ for equation (11.5).

b. Prove that this algorithm examines every table position in the worst case.

### a. Is it quadratic?

Observing what this process yields, and let's calculate a few values for $f(k, i)$ produced by this method.

$$ \begin{aligned} & f(k, 0) = h(k) + 0 & = C + 0 \\ & f(k, 1) = f(0) + 1 = h(k) + 1 &= C + 1 \\ & f(k, 2) = f(1) + 2 = f(0) + 1 + 2 = h(k) + 3 &= C + 3 \\ & f(k, 3) = f(2) + 3 = f(1) + 2 + 3 = f(0) + 1 + 2 + 3 = h(k) + 6 &= C + 6 \\ \end{aligned} $$

Or generally:

$$ f(k, i) = f(k, i - 1) + i $$

Which is a recurring relation that we can use induction to prove that

$$ f(k, i) = h(k) + \sum_{j=0}^{j} j = h(k) + \frac{n(n+1)}{2} $$

(because the sum is just an arithmetic progression)

This fits equation (11.5) with $c_1 = c_2 = 1/2$, because:

$$ f(k, i) = h(k) + \frac{1}{2}i + \frac{1}{2}i^2 $$

### b. Does it examine every position?

Had to consult the Instructor Manual yet again.

Let's assume it doesn't. This means that there are two separate values $a$ and $b$ such that $0 \le a < b < m$ for which $f(k, a) = f(k, b)$ modulo $m$, that is:

$$ h(k) + \frac{a(a + 1)}{2} = h(k) \frac{b(b + 1)}{2} \mod m \\ \Downarrow \\ \frac{a(a + 1)}{2} = \frac{b(b + 1)}{2} \mod m \\ \Downarrow \\ \frac{a(a + 1)}{2} - \frac{b(b + 1)}{2} = 0 \mod m \\ \Downarrow \\ \frac{a^2 + a - b^2 - b}{2} = 0 \mod m \\ \Downarrow \\ \frac{a^2 + a + ab - b^2 - b -ab}{2} = 0 \mod m \\ \Downarrow \\ \frac{(a - b)(a + b + 1)}{2} = 0 \mod m $$

The final part means that there is an integer $r$ so that:

$$ \frac{(a - b)(a + b + 1)}{2} = rm \\ \Downarrow \\ (a - b)(a + b + 1) = 2rm \\ \Downarrow \\ (a - b)(a + b + 1) = r \cdot 2^{p+1} $$

The last step is because we know that $m$ is a power of $2$, that is there is an integer $p$ such that $m = 2^p$.

Now, since $a$ and $b$ are both integers, this means that $a - b$ and $a + b + 1$ are integers as well, and more importantly, one of them is even, and another one is odd. That is, at least one of them is not dividable by $2$. This means the other is dividable by $2^{p+1}$. But it can't be $(a - b)$ because $a - b < m < 2^{p+1}$. It can't be $(a + b + 1)$, because $a + b + 1 \le (m - 1) + (m - 2) + 1 = 2m - 2 < 2^{p+1}$. We've reacted a contradiction, which means that $f(k, a) \ne f(k, b)$, for $0 \le a < b < m$, which means that the algorithm will exhaust all slots before it reaches $m$.