# Constraint on Universal set of hash functions

I was reading hashing from CLRS. In it author says:

Let $\mathscr{H}$ be a finite collection of hash functions that map a given universe $U$ of keys into the range ${0,1,...,m-1}$. Such a collection is said to be universal if for each pair of distinct keys $k,l\in U$, the number of hash functions $h\in\mathscr{H}$ for which $h(k)= h(l)$ is at most $|\mathscr{H}|/m$.

So basically in universal set of hash functions, the number of hash functions a finite collection of hash functions is said to be universal if number of hash functions $h$ for which $h(k)=h(l)$ is at most $\frac{\text{number of hash functions}}{\text{size of hash table}}=\frac{|\mathscr{H}|}{m}$

However next the author says:

In other words, with a hash function randomly chosen from $\mathscr{H}$, the chance of a collision between distinct keys $k$ and $l$ is no more than the chance $1/m$ of a collision if $h(k)$ and $h(l)$ were randomly and independently chosen from the set $\{0,1,...,m-1\}$.

I didnt get the last part "$h(k)$ and $h(l)$ were randomly and independently chosen from the set $\{0,1,...,m-1\}$". How $|\mathscr{H}|=$[$h(k)$ and $h(l)$ were randomly and independently chosen from the set $\{0,1,...,m-1\}$]

• What didn't you get? It says if (such-and-such). I can't tell what your question is. Please edit the question to be clearer about yoru question: focus exactly on what you interpret the writing to mean and what your specific question is. The last two sentences of the question need elaboration. What guesses have you formed about what the text might mean?
– D.W.
May 26, 2016 at 23:43
• In first statement $|\mathscr{H}|/m$ involves $|\mathscr{H}|$ which is number of hash functions in universal set. In second statement, it says "collision if $h(k)$ and $h(l)$ were randomly and independently chosen from the set ${0,1,...,m−1}$." I am not able to understand how these two turn out to be the same.
– Maha
May 28, 2016 at 10:36

Let's plug real numbers into the passage:

$|\mathscr{H}|$ = 10;

$m$ (size of hash table) = 100

In the first statement:

$\frac{|\mathscr{H}|}{m}$ = $\frac{10}{100}$ = 0.1 = the maximum number of hash functions that will cause a collision. (Assuming the count of functions in $\mathscr{H}$ can be represented as decimals for sake of argument.)

The important thing to note here is that the chance of selecting a collision causing hash function is: $\frac{0.1}{|\mathscr{H}|}$ = $\frac{0.1}{10}$ = 1%.

In the second statement:

The argument being made is that the chance of collision "is no more than..":

$\frac{1}{m}$ = $\frac{1}{100}$ = 1% = the chance of generating a hash value $h(key)$ from the set {0, 1..., 99}. AKA the chance of selecting a specific value from 100 values.

Conclusion:

So, from our example, a collection of 10 hash functions that map a given universe of keys into the range {0, 1..., 100 - 1} is said to be a universal collection if: by randomly choosing a hash function, the chance of collision between distinct keys is no more than 1%.

In the second paragraph:

"In other words, with a hash function randomly chosen from $$\mathscr{H}$$, the chance of a collision between distinct keys $$k$$ and $$l$$ is no more than the chance $$1/m$$ of a collision if $$h(k)$$ and $$h(l)$$ were randomly and independently chosen from the set $$\{0,1,...,m-1\}$$."

Here, "the chance of a collision between distinct keys", is simply specified to be $$\le 1/m$$.

Further the author says that, this $$1/m$$ is also the chance (probability) as below:

If two values (here $$h(k), h(l)$$) are randomly/independently picked from 0 to $$m-1$$, what is the chance of they being same. This is actually:

(probability that both equal certain value $$i$$) * (all possible values of $$i$$)

= $$((1/m)*(1/m)) * m$$

= $$1/m$$

So the author is giving a meaning to the value $$1/m$$.