# Exercise 11.3.5

$\star$ Define a family $\mathscr{H}$ of hash functions from a finite set $U$ to a finite set $B$ to be

$\epsilon$-universalif for all pairs of distinct elements $k$ and $l$ in $U$,$$ \Pr\{ h(k) = h(l) \} \le \epsilon $$

where the probability is over the choice of the hash function $h$ drawn at random from the family $\mathscr{H}$. Show that an $\epsilon$-universal family of hash functions must have

$$ \epsilon \ge \frac{1}{|B|} - \frac{1}{|U|} $$

I've found this a bit tricky, especially with (1) trying to wrap my brain around the notation, and (2) sticking to the formalism. Let's try to build some intuition here, and then use it to fill potential holes in a more formal proof.

To save some typing, let $b = |B|$ and $u = |U|$.

The assertion tells use that $Pr \{ h(k) = h(l) \} \ge \frac{1}{b} - \frac{1}{u} = (u - b) / bu$. If $b \ge u$, this means that we can design the function so that there are no collisions. If $b$ is lower, there are bound to be some collisions, however. For example, if $b = 5$ and $u = 8$, at most 4 elements can be stored without any collisions. The other 4 will collide to the same hash value. Alternatively, 6 elements can be mapped into three pairwise collisions, and the other two can be without any collisions.

Let's sketch both, assuming without any loss of generality, that $B = \{1, 2, 3, 4, 5\}$ and $U = \{ 1, 2, \ldots, 8 \} $. $h_1$ will spread the collisions evenly, where $h_2$ will put them in a single bucket.

$v \in U$ | $h_1(v)$ | $h_2(v)$ |
---|---|---|

1 | 1 |
1 |

2 | 1 |
2 |

3 | 2 |
3 |

4 | 2 |
4 |

5 | 3 |
5 |

6 | 3 |
5 |

7 | 4 | 5 |

8 | 5 | 5 |

Colliding elements are in **bold**.

Which one is better? Note that it depends on whether we're talking about non-colliding elements, or no-colliding pairs. $h_2$ has fewer colliding elements (4), but more colliding pairs ($\{5, 6\}, \{5, 7\}, \{5, 8\}, \{6, 7\}, \{6, 8\}, \{7, 8\}$). On the other hand, $h_1$ has 6 colliding elements, but only 4 colliding pairs. Note that if $k$ elements are mapped to the same value in $B$, there are $\binom{k}{2}$ pairs that collide.

Let's consider how to minimize the number of colliding pairs.

Again, let's assume that $U = \{1, 2, \ldots, |U| \}$ and $B = \{1, 2, \ldots, |B| \}$. This is OK, because however we pick the original $U$ and $B$, there is a trivial bijection to the first few natural numbers.

Let $c_i$ be the number of pairs for elements that map to $i \in B$. The total number of collisions, $C$, is then:

$$ C = \sum_{i=1}^{b}{\binom{c_i}{2}} = \frac{1}{2} \sum_{i=1}^{k}{c_i (c_i - 1)} $$

Let's do a bit of hand-waving. What's better in order to minimize $C$ - (1) to distribute the collisions evenly across the buckets, or (2) to let $b - 1$ elements be without collisions, and put the remaining $u - b + 1$ elements go in the same bucket? In the first approach ($C_1$), each bucket would have at most $\binom{\lceil u/b \rceil}{2}$ pairs that collide. In the other approach, we would have $C_2 = \binom{u - b + 1}{2}$ total collisions for a single bucket.

For $h_1(x)$, we have:

$$ C_1 = \sum_{i=1}^{b} \binom{\lceil u/b \rceil}{2} = b \frac{ \lceil u/b \rceil ( \lceil u/b - 1 \rceil ) } { 2 } \le b \frac{ (u/b) (u/b - 1) } { 2 } = \frac { u (u - b) } { 2b } $$

For $h_2(x)$ we have:

$$ C_2 = \binom{u - b + 1}{2} = \frac{ (u - b + 1)(u - b) }{2} = \frac{u^2 + u + b^2 - 2bu - b }{2} $$

It appears that $C_1$ is smaller and always minimizes the number of collisions. Plug in some concrete values to convince yourself (e.g. with $u = 1000$ and $b = 970$, $C_1$ is $15$, but $C_2$ is $456$; with a bigger in $u$ and $b$, we get a much bigger difference between the number of collisions).

Let's try this another way – lower bound for number of colliding pairs:

$$ C = \sum_{i=1}^{b}{c_i (c_i - 1) } = \sum_{i=1}^{b}{c_i^2} - \sum_{i=1}^{b}{c_i} = \sum_{i=1}^{b}{c_i^2} - |U| $$

The final bit, $|U| = \sum{c_i}$, is because each element from $U$ maps to a particular bucket $i$ and is counted once and only once.

We also notice that (sort of):

$$ c_i \ge \frac{u}{b} $$

And:

$$ \sum_{i=1}^{b}{c_i} \ge \sum_{i=1}^{b}{\frac{u^2}{b^2}} = \frac{u^2}{b} $$

Thus, finally:

$$ \epsilon \ge \Pr\{h(k) = h(l)\} = \frac{ \sum_{i=1}^{b}{ c_i (c_i - 1) } } { u(u - 1) } \ge \frac { \sum_{i=1}^{b}{c_i^2} - \sum_{i=1}^{b}{c_i} } {u^2} \ge \frac { \frac{u^2}{b} - u } {u^2} = \frac {u^2 - ub }{u^2b} = \frac{u - b}{ub} = \frac{1}{b} - \frac{1}{u} $$