Universal hash function is defined as follow:

$f_{(k_0,k_1)}(x) = (k_0\cdot x + k_1) \mod p$, where $p$ is prime

In Wikipedia, it is mentioned that the above function is pairwise independent. Pairwise independence is defined here

I wonder what are the random variables in $f$ and what pairwise independence means here.

  • $\begingroup$ Could you give the link for the definition? $\endgroup$ – kelalaka Oct 6 '19 at 19:06
  • $\begingroup$ @kelalaka the link is in the text, but here is in plain en.wikipedia.org/wiki/Pairwise_independence $\endgroup$ – Reza Oct 6 '19 at 19:11
  • $\begingroup$ This is in Universal_hashing . Carter and Wegman; $h_{a,b}(x) = ((a \cdot x +b) \bmod p) \bmod m $ where $a , b$ are randomly chosen integers modulo $p$ with $a \neq 0$. $\endgroup$ – kelalaka Oct 6 '19 at 19:18
  • $\begingroup$ please include the definition in the question itself for improved readability. after all you are asking others to put in an effort to answer your question. $\endgroup$ – kodlu Oct 6 '19 at 21:26

The random variables are $k_0$ and $k_1$, typically taken to be uniformly distributed in $\mathbb Z/p\mathbb Z$ in this context.

(Sometimes we take $k_0$ to be uniform in $(\mathbb Z/p\mathbb Z)^\times$ instead, i.e. exclude $k_0 = 0$, but as long as $p \gg 2^{100}$ this is not important.)

  • $\begingroup$ Thanks! Why do not you consider the $x$ as a random variable? $\endgroup$ – Reza Oct 6 '19 at 19:49
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    $\begingroup$ The defining property of a universal hash family is a bound on the collision probability $\Pr[H(x) = H(y)]$ for all $x \ne y$. That is, there's a for all quantifier on $x$ and $y$, but a probability quantifier on the hash function $H$ (or, equivalently, over the internal parameters $k_0$ and $k_1$). $\endgroup$ – Squeamish Ossifrage Oct 6 '19 at 19:51
  • $\begingroup$ So, why $f_{(k_0,k_1)}(x) = (k_0\cdot x + k_1) \mod p$ is a pairwise independent but $f_{(k_0,k_1)}(x) = (k_0\cdot x \cdot k_1)$ is not? $\endgroup$ – Reza Oct 6 '19 at 20:36
  • $\begingroup$ @Reza $f_{(k_0,k_1)}(x) = (k_0 \cdot x \cdot k_1) \bmod p$ is equivalent to $g_{k_0\cdot k_1}$ where $g_k(x) = k\cdot x \bmod p$. Suppose you know what $g_k(x)$ is; can you deduce, then, what $g_k(y)$ is? If you can, is it possible for $g$ to be pairwise independent? (Remember what the definition of pairwise independence is!) $\endgroup$ – Squeamish Ossifrage Oct 6 '19 at 21:24

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